Volumes of Orthogonal Groups and Unitary Groups Are Very Useful in Physics and Mathematics [3, 4]

Home , Orthogonal group, Up to

arXiv:1509.00537v5 [math-ph] 3 Nov 2017 ∗ -al oyln13cm [email protected] [email protected]; E-mail: o ue fOtooa rusadUiayGroups Unitary and Groups Orthogonal of Volumes fteqatmifrainapiain ftevlmso or of volumes the co of to applications and information pedagogical, quantum is the article of the of purpose The results. ba metric a of volume The well. group. mixe as unitary computed of of set volume the the of using volume by The Specifically, well. unit as and presented groups orthogonal of W volumes group. the derives Lie easily compact respe can a computed, for formula are volume groups Mcdonald’s present unitary and groups matr orthogonal about knowledge of background key detailed presenting by h arxitga a ayapiain ndvrefields. diverse in applications many has integral matrix The hr r onwrslsi hsatce u nydtie an detailed only but article, this in results new no are There nttt fMteais aghuDaz nvriy Han University, Dianzi Hangzhou Mathematics, of Institute i Zhang Lin Abstract 1 ∗ lc noepaemn,i o all, not if many, place one in llect r rus plctosaealso are Applications groups. ary hgnladuiaygroups. unitary and thogonal lmnaypof fexisting of proofs elementary d unu ttsi computed is states quantum d zo 108 RChina PR 310018, gzhou xitga.Te h volumes the Then integral. ix tvl.A nfiain we unification, a As ctively. t hsvlm oml,one formula, volume this ith hsrve ril begins article review This li ntr ru salso is group unitary in ll Contents

1 Introduction 3

2 Volumes of orthogonal groups 5 2.1 Preliminary ...... 5 2.2 Thecomputationofvolumes ...... 17

3 Volumes of unitary groups 28 3.1 Preliminary ...... 28 3.2 Thecomputationofvolumes ...... 44

4 The volume of a compact Lie group 55

5 Applications 73 5.1 Hilbert-Schmidt volume of the set of mixed quantum states...... 73 5.2 Areaoftheboundaryofthesetofmixedstates ...... 77 5.3 Volumeofametricballinunitarygroup ...... 79

6 Appendix I: Volumes of a sphere and a ball 85

7 Appendix II: Some useful facts 89 7.1 Matrices with simple eigenvalues form open dense sets of fullmeasure ...... 89 7.2 Resultsrelatedtoorthogonalgroups ...... 93 7.3 Resultsrelatedtounitarygroups ...... 96

8 Appendix III: Some matrix factorizations 107

9 Appendix IV: Selberg’s integral 112

2 1 Introduction

Volumes of orthogonal groups and unitary groups are very useful in physics and mathematics [3, 4]. In 1949, Ponting and Potter had already calculated the volume of orthogonal and unitary group [27]. A complete treatment for group manifolds is presented by Marinov [17], who ex- tracted the volumes of groups by studying curved path integrals [18]. There is a general closed formula for any compact Lie group in terms of the root lattice [?]. Clearly the methods used previously are not followed easily. Zyczkowski˙ in his paper [38] gives a re-derivation on the volume of unitary group, but it is still not accessible. The main goal of this paper is to compute the volumes of orthogonal groups and unitary groups in a systematic and elementary way. The obtained formulae is more applicable. As an application, by re-deriving the volumes of orthogonal groups and unitary groups, the authors in [38] computes the volume of the convex (n2 1)-dimensional set D (Cn) of the density − matrices of size n with respect to the Hilbert-Schmidt measure. Recently, the authors in [30] give the integral representation of the exact volume of a metric ball in unitary group; and present diverse applications of volume estimates of metric balls in manifolds in information and coding theory. Before proceeding, we recall some notions in group theory. Assume that a group acts on G the underlying vector space via g x for all g and x . Let x be any nonzero vector in X | i ∈ G ∈X | i . The subset x := g x : g is called the -orbit of x . Denote X G| i { | i ∈ G} G | i∈X xG := g : g x = x . { ∈ G | i | i} We have the following fact:

x /xG . G| i ∼ G If x = e , where e is a unit element of , then the action of on x is called free. In this case, G { } G G | i x / e x . G| i ∼ G { }⇐⇒G ∼G| i Now we review a fast track of the volume of a unitary group. Let us consider the unitary groups = (n + 1), n = 1,2,... To establish their structure, we look at spaces in which the group acts G U transitively, and identify the isotropy subgroup. The unitary group (n + 1) acts naturally in U the complex vector space = Cn+1 through the vector or "deﬁning" representation; the image X of any nonzero vector x = ψ Cn+1 is contained in the maximal sphere S2n+1 of radius | i | i ∈ ψ since ψ = Uψ for U (n + 1), and in fact it is easy to see that it sweeps the whole k k k k k k ∈ U sphere when U runs though (n + 1), i.e. the group (n + 1) acts transitively in this sphere. U U The isotropy group of the vector n + 1 is easily seen to be the unitary group with an entry less, | i that is x = (n) 1. Indeed, the following map is onto: G U ⊕ ϕ : (n + 1) (n + 1) n + 1 = S2n+1. U −→ U | i

3 If we identity (n) with (n) 1, as a subgroup of (n + 1), then U U ⊕ U (n) n + 1 = n + 1 , U | i | i implying that ker ϕ = (n), and thus U (n + 1)/ker ϕ (n + 1) n + 1 = S2n+1 = ( (n + 1)/ker ϕ) n + 1 . U ∼ U | i U | i Therefore we have the equivalence relation

(n + 1)/ (n)= S2n+1. (1.1) U U This indicates that

vol ( (n + 1)) = vol S2n+1 vol ( (n)) , n = 1, 2, . . . (1.2) U · U That is,

1 3 2n 1 vol ( (n)) = vol S vol S vol S − . (1.3) U × ×···× We can see this in [8]. The volume of the sphere of unit radius embedded in Rn(n > 1), is calculated from the Gaussian integral (see also Appendix I for the details):

+∞ t2 √π = e− dt. ∞ Z− Now +∞ n n t2 v 2 √π = e− dt = e−k k dv ∞ Rn Z− Z +∞ +∞ r2 r2 = e− dσdr = σn 1(r)e− dr, Sn 1 − Z0 Z − (r) Z0 Sn 1 where σn 1(r)= Sn 1 dσ is the volume of sphere − (r) of radius r. Since − − (r) R n 1 σn 1(r)= σn 1(1) r − , − − × it follows that +∞ n n 1 r2 √π = σn 1(1) r − e− dr, − × Z0 implying that

n 2π 2 vol Sn 1 := . (1.4) − Γ n 2

4 Finally, we get the volume formula of a unitary group:

n k n n(n+1) 2π 2 π 2 vol ( (n)) := ∏ = . (1.5) U Γ(k) 1!2! (n 1)! k=1 · · · −

2 Volumes of orthogonal groups

2.1 Preliminary

The following standard notations will be used [19]. Scalars will be denoted by lower-case letters, vectors and matrices by capital letters. As far as possible variable matrices will be denoted by X, Y,... and constant matrices by A, B,.... n Let A = [aij] be a n n matrix, then Tr (A) = ∑j=1 ajj is the trace of A, and det(A) is the T× determinant of A, and over a vector or a matrix will denote its transpose. Let X = [xij] be a m n matrix of independent real entries x ’s. We denote the matrix of differentials by dX, i.e. × ij dx dx dx 11 12 · · · 1n  dx dx dx  21 22 · · · 2n dX := [dxij]= . . . . .  . . .. .       dxm1 dxm2 dxmn   · · ·    Then [dX] stands for the product of the m n differential elements × m n [dX] := ∏ ∏ dxij (2.1) i=1 j=1 and when X is a real square symmetric matrix, that is, m = n, X = XT, then [dX] is the product of the n(n + 1)/2 differential elements, that is,

n n [dX] := ∏ ∏ dxij = ∏ dxij. (2.2) j=1 i=j i>j

Throughout this paper, stated otherwise, we will make use of conventions that the signs will be ignored in the product [dX] of differentials of independent entries. Our notation will be the following: Let X = [x ] be a m n matrix of independent real entries. Then ij × [dX]= m n dx (2.3) ∧i=1 ∧j=1 ij

5 when [dX] appears with integrals or Jacobians of transformations;

m n [dX]= ∏i=1 ∏j=1 dxij (2.4) when [dX] appears with integrals involving density functions where the functions are nonnegative and the absolute value of the Jacobian is automatically taken. As Edelman said [7] before, many researchers in Linear Algebra have little known the fact that the familiar matrix factorizations, which can be viewed as changes of variables, have simple Ja- cobians. These Jacobians are used extensively in applications of random matrices in multivariate statistics and physics. It is assumed that the reader is familiar with the calculation of Jacobians when a vector of scalar variables is transformed to a vector of scalar variables. The result is stated here for the sake of completeness. Let the vector of scalar variables X be transformed to Y, where

x1 y1 . . X =  .  and Y =  .  ,  x   y   n   n      by a one-to-one transformation. Let the matrix of partial derivatives be denoted by ∂Y ∂y = i . ∂X ∂x j The the determinant of the matrix ∂yi is known as the Jacobian of the transformation X going ∂xj to Y or Y as a function of X it is writtenh i as ∂y J(Y : X)= det i or [dY]= J(Y : X)[dX], J = 0 ∂x 6 j and 1 J(Y : X)= or 1 = J(Y : X)J(X : Y). J(X : Y) Note that when transforming X to Y the variables can be taken in any order because a permutation brings only a change of sign in the determinant and the magnitude remains the same, that is, J remains the same where J denotes the absolute value of J. When evaluating integrals | | | | involving functions of matrix arguments one often needs only the absolute value of J. Hence in all the statements of this notes the notation [dY] = J[dX] means that the relation is written ignoring the sign.

Proposition 2.1. Let X, Y Rn be of independent real variables and A Rn n be a nonsingular matrix ∈ ∈ × of constants. If Y = AX, then

[dY]= det(A)[dX]. (2.5)

6 Proof. The result follows from the deﬁnition itself. Note that when Y = AX, A = [aij ] one has

y = a x + + a x , i = 1,..., n i i1 1 · · · in n where xj’s and yj’s denote the components of the vectors X and Y, respectively. Thus the partial derivative of yi with respect to xj is aij, and then the determinant of the Jacobian matrix is det(A).

In order to see the results in the more complicated cases we need the concept of a tensor product.

Deﬁnition 2.2 (Tensor product). Let A = [a ] Rp q and B = [b ] Rm n. Then the tensor ij ∈ × ij ∈ × product, denoted by , is a pm qn matrix in Rpm qn, formed as follows: ⊗ × × a B a B a B 11 12 · · · 1q  a B a B a B  21 22 · · · 2q B A = . . . . (2.6) ⊗  . . .. .       ap1B ap2B apqB   · · ·    and

b A b A b A 11 12 · · · 1n  b A b A b A  21 22 · · · 2n A B = . . . . . (2.7) ⊗  . . .. .       bm1 A bm2 A bmn A   · · ·    Deﬁnition 2.3 (Vector-matrix correspondence). Let X = [x ] Rm n matrix. Let the j-th column ij ∈ × of X be denoted by Xj. That is, X = [X1,..., Xn], where

x1j . Xj =  .  .  x   mj    mn Consider an mn-dimensional vector in R , formed by appending X1,..., Xn and forming a long string. This vector will be denoted by vec(X). That is,

X1 . vec(X)=  .  . (2.8)  X   n    From the above deﬁnition, we see that the vec mapping is a one-to-one and onto correspondence from Rm n to Rn Rm. We also see that vec(AXB)= (A BT) vec(X) if the product AXB × ⊗ ⊗ exists.

7 Proposition 2.4. Let X, Y Rm n be of independent real variables and A Rm m and B Rn n ∈ × ∈ × ∈ × nonsingular matrices of constants. If Y = AXB, then

[dY]= det(A)n det(B)m[dX]. (2.9)

Proof. Since Y = AXB, it follows that vec(Y)=(A BT) vec(X). Then by using Proposition 2.1, ⊗ we have ∂Y ∂(vec(Y)) J(Y : X) = det := det ∂X ∂(vec(X)) = det(A BT)= det(A 1 ) det(1 BT) ⊗ ⊗ n m ⊗ = det(A)n det(B)m, implying that [dY]= J(Y : X)[dX]= det(A)n det(B)m[dX].

This completes the proof.

Remark 2.5. Another approach to the proof that [dZ] = det(A)n[dX], where Z = AX, is described as follows: we partition Z and X, respectively, as: Z = [Z1,..., Zn], X = [X1,..., Xn].

Now Z = AX can be rewritten as Zj = AXj for all j. So

∂Z ∂(Z1,..., Zn)  A  = = . , (2.10) ∂X ∂(X1,..., Xn)  ..       A      implying that A A   n [dZ]= det . [dX]= det(A) [dX].  ..       A      Proposition 2.6. Let X, A, B Rn n be lower triangular matrices where A = [a ] and B = [b ] are ∈ × ij ij constant matrices with ajj > 0, bjj > 0, j = 1,..., n and X is a matrix of independent real variables. Then

Y = X + XT = [dY]= 2n[dX], (2.11) ⇒ n j Y = AX = [dY]= ∏ ajj [dX], (2.12) ⇒ j=1 ! n n j+1 Y = XB = [dY]= ∏ bjj− [dX]. (2.13) ⇒ j=1 !

8 Thus

n j n j+1 Y = AXB = [dY]= ∏ ajjbjj− [dX]. (2.14) ⇒ j=1 !

Proof. Y = X + XT implies that

x 0 0 x x x 2x x x 11 · · · 11 21 · · · n1 11 21 · · · n1  x x 0   0 x x   x 2x x  21 22 · · · 22 · · · 2n 21 22 · · · n2 . . . . + . . . . = . . . . .  . . .. .   . . .. .   . . .. .               xn1 xn2 xnn   0 0 xnn   xn1 xn2 2xnn   · · ·   · · ·   · · ·        When taking the partial derivatives the n diagonal elements give 2 each and others unities and hence [dY] = 2n[dX]. If Y = AX, then the matrices of the conﬁgurations of the partial derivatives, by taking the elements in the orders (y11, y21 ..., yn1); (y22,..., yn2);...; ynn and

(x11, x21 ..., xn1); (x22,..., xn2);...; xnn are the following: ∂(y , y ..., y ) ∂(y ,..., y ) 11 21 n1 = A, 22 n2 = A[1ˆ 1ˆ],..., ∂(x11, x21 ..., xn1) ∂(x22,..., xn2) | ∂ynn [ [ = A[1ˆ n 1 1ˆ n 1]= ann, ∂xnn · · · − | · · · − where A[iˆ iˆ jˆ jˆ ] means that the obtained submatrix from deleting both the i ,..., i -th 1 · · · µ| 1 · · · ν 1 µ rows and the j1,..., jν-th columns of A. Thus

∂(y11,y21,...,yn1) 0 0 ∂(x11,x21,...,xn1) · · · ∂(y ,...,y ) ∂Y  0 22 n2 0  ∂(x22,...,xn2) · · · = . . . . ∂X  . . .. .   . . .     0 0 ∂ynn   · · · ∂xnn    A 0 0 · · ·  0 A[1ˆ 1ˆ] 0  | · · · = . . . . .  . . .. .     [ [   0 0 A[1ˆ n 1 1ˆ n 1]   · · · · · · − | · · · −    We can also take another approach to this proof. In fact, we partition X, Y by columns, respectively, Y = [Y1,..., Yn] and X = [X1,..., Xn]. Then Y = AX is equivalent to Yj = AXj, j = 1,..., n. Since Y, X, A are lower triangular, it follows that

y11 x11 y22 x22  y   x  21 21 . ˆ ˆ . ˆ [ ˆ [ . = A . ,  .  = A[1 1]  .  ,..., ynn = A[1 n 1 1 n 1]xnn = ann xnn.  .   .  | · · · − | · · · −          yn2   xn2   yn1   xn1                  9 Now

n [ [ n [dY] = ∏[dYj]= det(A) det(A[1ˆ 1ˆ]) det(A[1ˆ n 1 1ˆ n 1]) ∏[dXj] j=1 | · · · · · · − | · · · − j=1 n j = ∏ ajj [dX]. j=1 !

Next if Y = XB, that is,

x 0 0 b 0 0 11 · · · 11 · · ·  x x 0   b b 0  21 22 · · · 21 22 · · · Y = XB = ......  . . .. .   . . .. .           xn1 xn2 xnn   bn1 bn2 bnn   · · ·   · · ·      x b 0 0 11 11 · · ·  x b + x b x b 0  21 11 22 21 22 22 · · · = . . . .  . . .. .     n n   ∑ xnjbj1 ∑ xnjbj2 xnnbnn   j=1 j=1 · · ·    The matrices of the conﬁgurations of the partial derivatives, by taking the elements in the orders y11; (y21, y22);...; (yn1,..., ynn) and x11; (x21, x22);...; (xn1,..., xnn) are the following:

T ∂y11 ∂(y21, y22) b11 b21 b11 0 = b11, = = , ∂x11 ∂(x21, x22) " 0 b22 # " b21 b22 # T b11 b21 b31 b11 0 0 ∂(y31, y32, y33) =  0 b22 b32  =  b21 b22 0  ,..., ∂(x31, x32, x33)  0 0 b33   b31 b32 b33          T b 0 0 11 · · · ( )  b21 b22 0  ∂ yn1, yn2,..., ynn · · · = . . . . . ∂(xn1, xn2,..., xnn)  . . .. .       bn1 bn2 bnn   · · ·    Thus

∂y11 0 0 ∂x11 · · · ∂(y21,y22) ∂Y  0 0  ∂(x21,x22) · · · = . . . . ∂X  . . .. .     ∂(yn ,yn2,...,ynn)   0 0 1   · · · ∂(xn1,xn2,...,xnn)    Denote by B[i ... i j ... j ] the sub-matrix formed by the i ,..., i -th rows and j ,..., j -th 1 µ| 1 ν 1 µ 1 ν

10 columns of B. Hence T B[1 1] 0 0 | · · ·  0 B[12 12] 0  ∂Y | · · · = . . . . . ∂X  . . .. .       0 0 B[1 n 1 n]   · · · · · · | · · ·    The whole conﬁguration is a upper triangular matrix with b11 appearing n times and b22 appearing n 1 times and so on in the diagonal. Also we give another approach to derive the Jacobian − for Y = XB. Indeed, we partition Y, X by rows, respectively,

Y1 X1

 Y2   X2  Y = . , X = . ,  .   .           Yn   Xn          where Yj, Xj are row-vectors. So Y = XB is equivalent to Yj = Xj B, j = 1,..., n. Moreover

y = x B[1 1]= x a , [y , y ] = [x , x ]B[12 12],..., 11 11 | 11 11 21 22 21 22 | [y ,..., y ] = [y ,..., y ]B[1... n 1... n]. n1 nn n1 nn | Therefore n n [dY] = ∏[dYj]= det(B[1 1]) det(B[12 12]) det(B[1 n 1 n]) ∏[dXj] j=1 | | · · · · · · | · · · j=1 n n+j 1 = ∏ bjj − [dX]. j=1 ! We are done.

Proposition 2.7. Let X be a lower triangular matrix of independent real variables and A = [aij] and

B = [bij] be lower triangular matrices of constants with ajj > 0, bij > 0, j = 1,..., n. Then

n T T n j Y = AX + X A = [dY]= 2 ∏ ajj [dX], (2.15) ⇒ j=1 ! n T T n n j+1 Y = XB + B X = [dY]= 2 ∏ bjj− [dX]. (2.16) ⇒ j=1 !

Proof. Let Z = AX. Then Y = Z + ZT. Thus [dY] = 2n[dZ]. Since Z = AX, it follows from Proposition 2.6 that n j [dZ]= ∏ ajj [dX], j=1 ! implying the result. The proof of the second identity goes similarly.

11 Proposition 2.8. Let X andY be n n symmetric matrices of independent real variables and A Rn n × ∈ × nonsingular matrix of constants. If Y = AXAT, then

[dY]= det(A)n+1[dX]. (2.17)

Proof. Since both X and Y are symmetric matrices and A is nonsingular we can split A and AT as products of elementary matrices and write in the form

Y = E E XETET · · · 2 1 1 2 · · ·

T where Ej, j = 1, 2, . . . are elementary matrices. Write Y = AXA as a sequence of transformations of the type

Y = E XET, Y = E Y ET,..., = 1 1 1 2 2 1 2 ⇒ [dY1]= J(Y1 : X)[dX], [dY2]= J(Y2 : Y1)[dY1],...

Now successive substitutions give the ﬁnal result as long as the Jacobians of the type J(Yk :

Yk 1) are computed. Note that the elementary matrices are formed by multiplying any row (or − column) of an identity matrix with a scalar, adding a row (column) to another row (column) and combinations of these operations. Hence we need to consider only these two basic elementary matrices. Let us consider a 3 3 case and compute the Jacobians. Let E be the elementary × 1 matrix obtained by multiplying the ﬁrst row by α and E2 by adding the ﬁrst row to the second row of an identity matrix. That is,

α 0 0 1 0 0

E1 =  0 1 0  , E2 =  1 1 0   0 0 1   0 0 1          and

2 α x11 αx12 αx13 T E1XE1 =  αx21 x22 x23  ,

 αx31 x32 x33      u11 u11 + u12 u13 T E2Y1E2 =  u11 + u21 u11 + u21 + u12 + u22 u13 + u23  ,

 u31 u31 + u32 u33      T T where Y1 = E1XE1 and Y2 = E2Y1E2 and the elements of Y1 are denoted by uij’s for convenience.

12 The matrix of partial derivatives in the transformation Y1 written as a function of X is then

α2 00000  0 α 0 0 00 

∂Y1  0 01000  =   . ∂X    0 0 0 α 0 0     0 00010       0 00001      This is obtained by taking the xij’s in the order x11; (x21, x22); (x31, x32, x33) and the uij’s also in the same order. Thus the Jacobian is given by

4 3+1 3+1 J(Y1 : X)= α = α = det(E1) .

Or by deﬁnition it follows directly that

2 4 [dY1]= d(α x11)d(αx12)d(αx13)dx22dx23dx33 = α [dX].

For a n n matrix it will be αn+1. Let the elements of Y be denoted by v ’s. Then again taking × 2 ij the variables in the order as in the case of Y1 written as a function of X the matrix of partial derivatives in this transformation is the following:

100000  110000 

∂Y2  121000  =   . ∂Y   1  000100     000110       000001      3+1 3+1 The determinant of this matrix is 1 = 1 = det(E2) . In general such a transformation gives the Jacobian, in absolute value, as 1 = 1n+1. Thus the Jacobian is given by

J(Y : X)= det( E E )n+1 = det(A)n+1. · · · 2 1 We are done.

Example 2.9. Let X Rn n be a real symmetric positive deﬁnite matrix having a matrix-variate ∈ × gamma distribution with parameters (α, B = BT > 0). We show that

α 1 α n+1 Tr(BX) n 1 det(B)− = [dX] det(X) − 2 e− , Re(α) > − . (2.18) Γ (α) > n ZX 0 2

13 Indeed, since B is symmetric positive deﬁnite there exists a nonsingular matrix C such that B = CCT. Note that

Tr (BX) = Tr (CTXC) .

Let

U = CTXC = [dU]= det(C)n+1[dX] ⇒ 1 from Proposition 2.8 and det(X) = det(B)− det(U). The integral on the right reduces to the following:

α n+1 Tr(BX) α α n+1 Tr(U) [dX] det(X) − 2 e− = det(B)− [dU] det(U) − 2 e− . > > ZX 0 ZU 0 But

α n+1 Tr(U) [dU] det(U) − 2 e− = Γn(α) > ZU 0 > n 1 for Re(α) −2 . The result is obtained. Proposition 2.10. Let X, Y Rn n skew symmetric matrices of independent real variables and A ∈ × ∈ n n T R × nonsingular matrix of constants. If Y = AXA , then

n 1 [dY]= det(A) − [dX]. (2.19)

Note that when X is skew symmetric the diagonal elements are zeros and hence there are n(n 1) only 2− independent variables in X. Proposition 2.11. Let X, A, B Rn n be lower triangular matrices where A = [a ] and B = [b ] ∈ × ij ij are nonsingular constant matrices with positive diagonal elements, respectively, and X is a matrix of independent real variables. Then

n T T n j Y = A X + X A = [dY]= 2 ∏ ajj [dX], (2.20) ⇒ j=1 ! n T T n n j+1 Y = XB + BX = [dY]= 2 ∏ bjj− [dX]. (2.21) ⇒ j=1 !

T T T 1 1 Proof. Consider Y = A X + X A. Premultiply by (A )− and postmultiply by A− to get the following:

T T T 1 1 T 1 T 1 Y = A X + X A = (A )− YA− =(A )− X + XA− . ⇒ Let

T n 1 1 n (n j+1) Z = XA− + XA− = [dZ]= 2 ∏ a−jj − [dX] ⇒ ! j=1

14 by Proposition 2.7 and

T 1 1 (n+1) Z = A− YA− = [dZ]= det(A)− [dY] ⇒ by Proposition 2.8. Now writing [dY] in terms of [dX] one has

n n n (n j+1) n n+1 n j [dY]= ∏ a−jj − 2 ∏ ajj [dX]= 2 ∏ ajj [dX] j=1 ! j=1 ! j=1 !

n since det(A) = ∏j=1 ajj because A is lower triangular. Thus the ﬁrst result follows. The second is proved as follows. Clearly,

T T T 1 T 1 1 1 Y = BX + XB = B− Y(B )− = B− X + B− X := Z. ⇒ (n+1) n n j Thus [dZ] = det(B)− [dY] and [dZ] = 2 ∏j=1 a−jj [dX]. Therefore expressing [dY] in terms of [dX] gives the second result.

Proposition 2.12. Let X Rn n be a symmetric positive deﬁnite matrix of independent real variables ∈ × and T = [tij] a real lower-triangular matrix with tjj > 0, j = 1,..., n, and tij, i > j independent. Then

n T n j X = T T = [dX]= 2 ∏ tjj [dT], (2.22) ⇒ j=1 ! n T n n j+1 X = TT = [dX]= 2 ∏ tjj− [dT]. (2.23) ⇒ j=1 !

Proof. By considering the matrix of differentials one has

X = TTT = dX = dT TT + T dTT. ⇒ · · Now treat this as a linear transformation in the differentials, that is, dX and dT as variables and T a constant. This completes the proof.

n n n 1 Example 2.13. Let X R symmetric positive deﬁnite matrix and Re(α) > − . Show that ∈ × 2

α n+1 Tr(X) Γn(α) := [dX] det(X) − 2 e− > ZX 0 n(n 1) 1 n 1 = π 4− Γ(α)Γ α Γ α − . (2.24) − 2 · · · − 2

15 Let T be a real lower triangular matrix with positive diagonal elements. Then the unique representation (see Theorem 8.5)

n T n n j+1 X = TT = [dX]= 2 ∏ tjj− [dT]. ⇒ j=1 !

Note that

Tr (X) = Tr (TTT) = t2 +(t2 + t2 )+ + t2 + + t2 , 11 21 22 · · · n1 · · · nn n T 2 det(X) = det(TT )= ∏ tjj. j=1

When X > 0, we have TTT > 0, but t > 0, j = 1,..., n which means that ∞ < t < ∞, i > jj − ij < < ∞ n(n 1) j, 0 tjj , j = 1,..., n. The integral splits into n integrals on tjj’s and 2− integrals on tij’s, i > j. That is,

n ∞ α n+1 ∞ 2 n j+1 t2 t2 Γ 2 − jj ij n(α)= ∏ 2 tjj tjj− e− dtjj ∏ e− dtij . 0 ! × > ∞ ! j=1 Z i j Z− But ∞ j t2 j 1 2 α 2 jj 2 (tjj) − e− dtjj = Γ α − , 0 − 2 Z > j 1 for Re(α) −2 , j = 1,..., n and

∞ 2 tjj e− dtjj = √π. ∞ Z− Multiplying them together the result follows. Note that

j 1 n 1 Re(α) > − , j = 1,..., n = Re(α) > − . 2 ⇒ 2 The next result is extremely useful and uses the fact (see Theorem 8.5) that any positive deﬁnite n n matrix X has a unique decomposition as X = TT T, where T is an upper-triangular × n n matrix with positive diagonal elements. × Proposition 2.14. If X is an n n positive deﬁnite matrix and X = TTT, where T is upper-triangular × with positive diagonal elements, then

n n n j+1 [dX]= 2 ∏ tjj− [dT]. j=1

Proof. Since X = TTT, now express each of the elements of X on and above the diagonal in terms of each of the elements of T and take differentials. Remember that we are going to take the exterior product of these differentials and that products of repeated differentials are zero;

16 hence there is no need to keep track of differentials in the elements of T which have previously occurred. We get

2 x11 = t11, dx11 = 2t11dt11,

x12 = t11t12, dx12 = t11dt12 + . · · · . x = t t , dx = t dt + 1n 11 1n 1n 11 1n · · · x = t2 + t2 , dx = 2t dt + 22 12 22 22 22 22 · · · . .

x2n = t12t1n + t22t2n, dx2n = t22dt2n + . · · · . x = t2 + + t2 , dx = 2t dt + nn 1n · · · nn nn nn nn · · · Hence taking exterior products gives

n n n n n n 1 n n j+1 [dX]= dxij = 2 t11t22− tnn dtij = 2 ∏ tjj− [dT], 6 · · · 6 = i^j i^j j 1 as desired.

2.2 The computation of volumes

Deﬁnition 2.15 (Stiefel manifold). Let A be a n m(n > m) matrix with real entries such that T × A A = 1m, that is, the m columns of A are orthonormal vectors. The set of all such matrices A is known as the Stiefel manifold, denoted by (m, n). That is, for all n m matrices A, O × (m, n)= A Rn m : AT A = 1 , (2.25) O { ∈ × m} where Rn m denotes the set of all n m real matrices. × × T The equation A A = 1m imposes m(m + 1)/2 conditions on the elements of A. Thus the number of independent entries in A is mn m(m + 1)/2. − If m = n, then A is an orthogonal matrix. The set of such orthogonal matrices form a group. This group is known as the orthogonal group of m m matrices. × Deﬁnition 2.16 (Orthogonal group). Let B be a n n matrix with real elements such that BT B = × 1 . The set of all B is called an orthogonal group, denoted by (n). That is, n O (n)= B Rn n : BT B = 1 . (2.26) O { ∈ × n}

17 Clearly (n, n) = (n). Note that BT B = 1 imposes n(n + 1)/2 conditions and hence the O O n number of independent entries in B is only n2 n(n + 1)/2 = n(n 1)/2. − − Deﬁnition 2.17 (A symmetric or a skew symmetric matrix). Let A Rn n. If A = AT, then A is ∈ × said to be symmetric and if AT = A, then it is skew symmetric. − Proposition 2.18. Let V (n) with independent entries and the diagonal entries or the entries in the ∈ O T ﬁrst row of V all positive. Denote dG = V dV where V = [v1,..., vn]. Then

n 1 n − [dG] = ∏ ∏ vi,dvj (2.27) i=1 j=i+1h i n(n 1)/2 (n 1) = 2 − det(1n + X)− − [dX], (2.28)

1 where X is a skew symmetric matrix such that the ﬁrst row entries of (1n + X)− , except the ﬁrst entry, are negative.

Proof. Let the columns of V be denoted by v1,..., vn. Since the columns are orthonormal, we have v , v = δ . Then h i ji ij v ,dv + dv , v = 0, h i ji h i ji implying v ,dv = 0 since v ,dv is a real scalar. We also have h j ji h j ji v ,dv = v ,dv for i = j. h i ji −h j ii 6 Then VTdV is a skew symmetric matrix. That is,

T v1 T T  v2  dG = V dV = . [dv1,dv2, ,dvn]  .  · · ·    T   v   n    v ,dv v ,dv v ,dv h 1 1i h 1 2i ··· h 1 ni  v ,dv v ,dv v ,dv  h 2 1i h 2 2i ··· h 2 ni = . . . . .  . . .. .       vn,dv1 vn,dv2 vn,dvn   h i h i ··· h i    This indicates that

0 v ,dv v ,dv h 1 2i ··· h 1 ni  v ,dv 0 v ,dv  −h 1 2i ··· h 2 ni dG = . . . .  . . .. .       v1,dvn v2,dvn 0   −h i −h i ···    18 Then there are only n(n 1)/2 independent entries in G. Then [dG] is the wedge product of the − entries upper the leading diagonal in the matrix VTdV:

n 1 n [dG]= − v ,dv ∧i=1 ∧j=i+1 h i ji

This establishes (2.27). For establishing (2.28), take a skew symmetric matrix X, then V = 2(1n + X) 1 1 is orthonormal such that VVT = 1 . Further the matrix of differentials in V is given − − n n by dV = 2(1 + X) 1 dX (1 + X) 1, i.e. − n − · · n − 1 dV = (1 + V) dX (1 + V). −2 n · · n Thus

T 1 T dG = V dV = (1 + V ) dX (1 + V) −2 n · · n and the wedge product is obtained

1 T n 1 n 1 n + V − 1 − [dG]= det [dX]= det √2(1n + X)− [dX]. √2 Therefore the desired identity (2.28) is proved.

Proposition 2.19. LetX bean n symmetric matrix of independent real entries and with distinct and × nonzero eigenvalues λ > > λ and let D = diag(λ ,..., λ ). LetV (n) be a unique such that 1 · · · n 1 n ∈ O X = VDVT. Then

n 1 n [dX]= ∏ − ∏ λ λ [dD][dG], (2.29) i=1 j=i+1 i − j where dG = VTdV.

Proof. Take the differentials in X = VDVT to get

dX = dV D VT + V dD VT + V D dVT, · · · · · · implying that VT dX V = VTdV D + dD + D dVTV. · · · · Let dY = VT dX V for ﬁxed V. Since dG = VTdV, it follows that [dX] = [dY] and · · dY = dG D + dD + D dGT · · = dD + dG D D dG · − · = dD + [dG, D]

19 where we used the fact that dGT = dG and the concept of commutator, deﬁned by [M, N] := − MN NM. Clearly the commutator of M and N is skew symmetric. Now − dy = dλ and dy =(λ λ )dg , i < j = 1,..., n. jj j ij j − i ij Then

n n n [dY] = ∏ dyjj ∏ dyij = ∏ dλj ∏ λi λj dgij j=1 ! i

= ∏ λi λj [dD][dG], i

n where [dD]= ∏j=1 dλj and [dG]= ∏i

Clearly X = VDVT, where D = diag(λ ,..., λ ) (λ > > λ ), VVT = 1 , is not a one- 1 n 1 · · · n n n to-one transformation from X to (D, V) since X determines 2 matrices [ v1,..., vn], where T ± ± v1,..., vn are the columns of V, such that X = VDV . This transformation can be shown to be unique if one entry from each row and column are of a speciﬁed sign, for example, the diagonal entries are positive. Once this is done we are integrating with respect to dG, where dG = VTdV over the full orthogonal group (n), the O result must be divided by 2n to get the result for a unique transformation X = VDVT.

Remark 2.20. Now we can try to compute the following integral based on Eq. (2.29):

n n [dX] = δ ∑ λj 1 ∏ λi λj ∏ dλj [dG] > > > > ZX 0:Tr(X)=1 Zλ1 λn 0 j=1 − ! i

n(n 1) − n+1 n π 4 Γ j vol (D (Rn)) := [dX]= 2 ∏ Γ . X>0:Tr(X)=1 Γ n(n+1) Γ 1 2 Z 4 2 j=1 ( ) See below for the notation C 1,1 and vol ( (n)). n O Proposition 2.21. Let X be a p n(p 6 n) matrix of rank p and let X = TUT, where T is a p p × 1 × lower triangular matrix with distinct nonzero diagonal entries and U is a unique n p semi-orthogonal 1 × matrix, UTU = 1 , all are of independent real entries. Let U be an n (n p) semi-orthogonal matrix 1 1 p 2 × −

20 T 1 T such that U1 augmented with U2 is a full orthogonal matrix. That is, U = [U1 U2], U U = n, U2 U2 = T 1n p, U U2 = 0. Let uj be the j-th column of U and duj its differential. Then − 1

p n j − [dX]= ∏j=1 tjj [dT][dU1], (2.30) where [dU ]= p n u ,du . 1 ∧j=1 ∧i=j+1 h i ji Proof. Note that

T T T 1 T U1 U2 U2 U1 U2 p 0 U U = T [U1 U2]= T T = . " U # " U U1 U U2 # " 0 1n p # 2 2 2 − T Take the differentials in X = TU1 to get

dX = dT UT + T dUT. · 1 · 1 Then

dX U = dT UTU + T dUTU · · 1 · 1 = dT UT[U , U ]+ T dUT[U , U ] · 1 1 2 · 1 1 2 = [dT + T dUT U , T dUT U ] · 1 · 1 · 1 · 2 T 1 T since U1 U1 = p, U1 U2 = 0. Make the substitutions

dW = dX U,dY = dUT U ,dS = dUT U ,dH = T dS. · 1 · 1 1 · 2 · Now we have dW = [dT + T dY, dH]. · Thus

dt11 0 0 t11 0 0 0 dy dy · · · · · · 12 · · · 1p . .  dt21 dt22 .   t21 t22 .   dy12 0 dy2p  dT + T dY = · · · + · · · − · · · ...... ·  . . .   . . .   . . .. .   . . . 0   . . . 0           dt dt dt   t t t   dy1p dy2p 0   p1 p2 pp   p1 p2 pp   − − · · ·   · · ·   · · ·    Let us consider, for example, the case where p = 2,3 in order for computing the wedge product of dT + T dY. Now for p = 2, we have ·

dt11 0 t11 0 0 dy12 dT + T dY = + · " dt dt22 # " t t22 #" dy 0 # 21 21 − 12 dt t dy = 11 11 12 " dt t22dy dt22 + t dy # 21 − 12 21 12 21 Thus the wedge product of dT + T dY is: · [dT + T dY] = dt (t dy ) (dt t dy ) (dt + t dy ) · 11 ∧ 11 12 ∧ 21 − 22 12 ∧ 22 21 12 = t dt dy dt dt = t [dT][dY] 11 11 ∧ 12 ∧ 21 ∧ 22 11 2 2 j = ∏ tjj − [dT][dY]. j=1 !

For p = 3, we have

dt11 0 0 t11 0 0 0 dy12 dy13 dT + T dY =  dt dt 0  +  t t 0   dy 0 dy  · 21 22 21 22 − 12 23  dt31 dt32 dt33   t31 t32 t33   dy13 dy23 0       − −        dt11 t11dy12 t11dy13 =  dt t dy dt + t dy t dy + t dy  , 21 − 22 12 22 21 12 21 13 22 23  dt31 t32dy12 t33dy13 dt32 + t31dy12 t33dy23 dt33 + t31dy13 + t32dy23   − − −    implying the wedge product of dT + T dY is: · 3 2 3 j t11t22[dT][dY]= ∏ tjj − [dT][dY]. j=1 !

For the general p, by straight multiplication, and remembering that the variables are only

dy12,...,dy1p,dy23,...,dy2p,...,dyp 1p. −

Thus the wedge product of dT + T dY gives · p p j ∏ tjj − [dY][dT], j=1 !

ignoring the sign, and p 1 p [dY]= − u ,du . ∧j=1 ∧i=j+1 h i ji Now consider dH = T dS. Since dS is a p (n p) matrix, we have · × − p n p n p [dH]= det(T) − [dS]= ∏ tjj − [dS]. j=1 !

The wedge product in dS is the following:

[dS]= p n u ,du . ∧j=1 ∧i=p+j h i ji

22 Hence from the above equations,

[dX] = [dW]= [dT + TdY] [dH] ∧ ∧ p p p j n p = ∏ tjj − [dY][dT] ∏ tjj − [dS] j=1 ! j=1 !

p n j = ∏ tjj − [dY][dT][dS]. j=1 !

Now p 1 p p n [dY][dS]= − u ,du u ,du . ∧j=1 ∧i=j+1 h i ji ∧j=1 ∧i=p+jh i ji Substituting back one has

p n j p − n [dX]= ∏ tjj [dT] j=1 i=j+1 ui,duj j=1 ! ∧ ∧ h i

which establishes the result.

If the triangular matrix T is restricted to the one with positive diagonal entries, that is, tjj > 0, j = 1,..., p, then while integrating over T using Proposition 2.21, the result must be multiplied by 2p. Without the factor 2p, the t ’s must be integrated over ∞ < t < ∞, j = 1,..., p. If the jj − jj expression to be integrated contains both T and U, then restrict tjj > 0, j = 1,..., p and integrate U over the full Stiefel manifold. If the rows of U are u ,..., u , then u ,..., u give 2p choices. 1 p ± 1 ± p p 2p Similarly tjj > 0, tjj < 0 give 2 choices. But there are not 2 choices in X = TU. There are only 2p choices. Hence either integrate out the t ’s over ∞ < t < ∞ and a unique U or over jj − jj 0 < tjj < ∞ and the U over the full Stiefel manifold. For uniqueness of matrix factorization, see Theorem 8.6.

Proposition 2.22. Let X be an n p(n > p) matrix of rank p of independent real entries and let 1 × X = U T where T is a real p p upper triangular matrix with distinct nonzero diagonal entries and 1 1 1 1 × T 1 U1 is a unique real n p semi-orthogonal matrix, that is, U1 U1 = p. Let U = [U1 U2] such that T T × T U U = 1n, U U2 = 1n p, U U2 = 0. Let uj be the j-th column of U and duj its differential. Then 2 − 1

p n j − [dX1]= ∏j=1 tjj [dT1][dU1], (2.31) where p n [dU1]= ∏ ∏ ui,duj . j=1 i=j+1h i

Proposition 2.23. If X1, T1 and U1 are as deﬁned in Proposition 2.22, then the surface area of the full Stiefel manifold (p, n) or the total integral of the wedge product p n u ,du over (p, n) is O ∧j=1 ∧i=j+1 h i ji O given by

23 p pn p n 2 π 2 j=1 i=j+1 ui,duj = n , (2.32) (p,n) ∧ ∧ h i Γp ZO 2 where p(p 1) 1 p 1 Γ (α)= π 4− Γ(α)Γ α α − p − 2 · · · − 2 > p 1 for Re(α) −2 .

Proof. Note that since X is n p, the sum of squares of the np variables in X is given by 1 × 1 n p T 2 Tr (X1 X1) = ∑ ∑ xij. i=1 j=1

Then

∞ ∞ n p T ∑n ∑p 2 Tr(X X1) i=1 j=1 xij [dX1]e− 1 = e− ∏ ∏ dxij X ∞ · · · ∞ Z 1 Z− Z− i=1 j=1 p ∞ n 2 xij = ∏ ∏ e− dxij ∞ i=1 j=1 Z− np = π 2 by direct evaluation of the exponential integrals. Make the transformation as in Proposition 2.22:

X = U T = XT X = TT T , 1 1 1 ⇒ 1 1 1 1 where T is a real p p upper triangular matrix with distinct nonzero diagonal entries and U is 1 × 1 a unique real n p semi-orthogonal matrix—UTU = 1 , implying × 1 1 p T T 2 Tr (X1 X1) = Tr (T1 T1) = ∑ tij. i6j

Note that dX1 is available from Proposition 2.22. Now

p T 2 Tr X X n j ∑ 6 t p ( 1 1) − i j ij n [dX1]e− = ∏ tjj e− [dT1] j=1 i=j+1 ui,duj . X T (p,n) ∧ ∧ h i Z 1 Z 1 j=1 ! ZO

But for 0 < t < ∞, ∞ < t < ∞(i < j) and U unrestricted, jj − ij 1 p 2 n n j ∑i6j tij p ∏ t − e− [dT ]= 2− Γp jj 1 2 ZT1 j=1

24 observing that for j = 1,..., p, the p integrals

∞ 2 n j 1 n j tjj 1 tjj − e− dtjj = 2− Γ − , n > j 1, 0 2 − 2 − Z and each of the p(p 1)/2 integrals − ∞ 2 tij e− dtij = √π, i < j. ∞ Z− Thus the result that follows.

Theorem 2.24. Let X be a full-ranked and n n matrix of independent real entries and let X = UT, × where T is a real n n upper triangular matrix with distinct nonzero diagonal entries and U is a unique × real orthogonal matrix. Let uj be the j-th column of U and duj its differential. Then the volume content of the full orthogonal group (n) is given by O

2 n n n n(n+1) T 2 π 2 2 π 4 vol( (n)) = [U dU]= n = . (2.33) O (n) ∧ Γn n Γ k ZO 2 ∏k=1 2

Proposition 2.25. Let X be a p n(p 6 n) matrix of rank p and let X = TUT, where T is a p p lower × 1 × triangular matrix with distinct positive diagonal entries t > 0, j = 1,..., p and U is a unique n p jj 1 × T 1 T T semi-orthogonal matrix, U1 U1 = p, all are of independent real entries. Let A = XX = TT . Then

p n p+1 p n [dX]= 2− det(A) 2 − 2 [dA] u ,du . ∧j=1 ∧i=j+1h i ji Proof. Since A = TTT, it follows that

p p p+1 j [dA]= 2 ∏ tjj − [dT], j=1 ! i.e. p p p 1+j [dT]= 2− ∏ t−jj − [dA]. j=1 ! But p n j [dX]= ∏ tjj − [dT][dU1], j=1 !

where p n [dU1]= ∏ ∏ ui,duj . j=1 i=j+1h i 2 p 2 Note that det(A)= det(T) = ∏j=1 tjj. The desired conclusion is obtained.

25 Proposition 2.26. Let X be a m n matrix of rank m(m 6 n),T = [t ] a m m lower triangular × jk × matrix with t > 0, j = 1,..., mandLan m matrix satisfying LT L = 1 , where the matrices are of jj × m independent real entries. Then show that, if X = TLT, then

m n j ˆ [dX]= ∏j=1 tjj− [dT][dL] (2.34) where m n dLˆ = ∏ ∏ li,dlj , j=1 i=j+1h i l is the j-th column of Lˆ = [LL ] (n); dl the differential of the i-th column of L. j 1 ∈ O i Proposition 2.27 (Polar decomposition). Let X be a m n(m 6 n) matrix, S a m m symmetric × × positive deﬁnite matrix and L a n m matrix with LT L = 1 , all are of independent real entries. Then × m show that, if X = √SLT, then

m 1 n m 1 [dX]= (det(S)) −2 − [dS][dLˆ ] (2.35) 2 where dLˆ is deﬁned in Proposition 2.26.

T T T T Proof. Now if X = √SL and L L = 1m, then XX = S. By Proposition 2.26, we have X = TL , where T = [t ] is a m m lower triangular matrix with t > 0, j = 1,..., m and L a n m matrix jk × jj × satisfying LT L = 1 . Denote Lˆ = [LL ] (n). Hence m 1 ∈ O m n j ˆ [dX]= ∏ tjj− [dT][dL]. j=1 !

It also holds that S = TTT implies

m m m+1 j [dS]= 2 ∏ tjj − [dT]. j=1 !

Both expressions indicate that

m m n m 1 ˆ [dX]= 2− ∏ tjj− − [dS][dL]. j=1 !

T m 2 Since det(S)= det(TT )= ∏j=1 tjj, it follows that

m 1 n m 1 [dX]= (det(S)) −2 − [dS][dLˆ ]. 2 We are done.

Proposition 2.28. With the same notations as in Proposition 2.27, it holds that

26 m mn 2 π 2 [dLˆ ]= (2.36) T 1 Γ n ZL L= m m 2 and for m = n

2 n n ˆ 2 π 2 [dV]= n . (2.37) (n) Γn ZO 2 Deﬁne the normalized orthogonal measures as n n Γn Γn dµ(V) := 2 [dVˆ ]= 2 v ,dv (2.38) n2 n2 ∏ i j 2nπ 2 ! 2nπ 2 ! i>j h i or n Γn dµ(V) := 2 [dG], (2.39) n2 2nπ 2 ! where dG = VTdV for V = [v ,..., v ] (n). It holds that [dVˆ ] = (VTdV) is invariant under 1 n ∈ O ∧ simultaneous translations V UVW, where U, W (n). That is, dµ(V) is an invariant measure → ∈ O under both left and right translations, i.e. Haar measure over (n). O Proof. We know that T Tr(XX ) mn [dX]e− = π 2 . ZX From Proposition 2.27, via the transformation X = √SLT, we see that m T n m 1 Tr(XX ) 1 − − Tr(S) [dX]e− = det(S) 2 e− [dS] [dLˆ ]. X 2 S>0 × LT L=1 Z Z Z m We also see from the deﬁnition of Γp(α) that

n n m+1 Tr(S) Γm = det(S) 2 − 2 e− [dS]. 2 S>0 Z Then m mn 2 π 2 [dLˆ ]= . T 1 Γ n ZL L= m m 2 For m = n, the result follows easily. For ﬁxed U, W (n), we have ∈ O (UVW)Td(UVW)=(WTVTUT) (U dV W) = WT dG W, · · · · implying that [(UVW)Td(UVW)] = [dG]. ∧ That is, dµ(V) = dµ(UVW) for all U, V, W (n), dµ(V) is an invariant measure under both ∈ O left and right translations over (n). U

27 3 Volumes of unitary groups

3.1 Preliminary

In Section 2, we dealt with matrices where the entries are either real constants or real variables. Here we consider the matrices whose entries are complex quantities. When the matrices are real, we will use the same notations as in Section 2. In the complex case, the matrix variable X will be denoted by X to indicate that the entries in X are complex variables so that the entries of theorems in Section 3 will not be confused with those in Section 2. The complex conjugate of a e matrix A will be denoted by A and the conjugate transpose by A∗. The determinant of A will be denoted by det(A). The absolute value of a scalar a will also be denoted by a . The wedge e e e | | e product of differentials in X will be denoted by [dX] and the matrix of differentials by dX. e It is assumed that the reader is familiar with the basic properties of real and complex matrices. e e e Some properties of complex matrices will be listed here for convenience. A matrix X with complex elements can always be written as X = X + √ 1X where 1 − 2 X1 = Re(X) and X2 = Im(X) are real matrices. Let us examine the wedge product of the e 2 e 2 differentials in X. In general, there are n real variables in X1 and another n real variables and e e the wedge product of the differentials will be denoted by the following: e

Notation 1:

[dX] := [dX1][dX2] or [dX] := d Re(X) d Im(X) (3.1) h ih i e e e e where [dX1] is the wedge product in dX1 and [dX2] is the wedge product in dX2. In this notation an empty product is interpreted as unity. That is, when the matrix X is real then X2 is null and

[dX] := [dX1]. If X is a hermitian matrix, then X1 is symmetric and X2 is skew symmetric, and in e this case e e (1) (2) [dX ]= > dx and [dX ]= > dx (3.2) 1 ∧j k jk 2 ∧j k jk

( ) ( ) where X = x 1 and X = x 2 . If Y is a scalar function of X = X + √ 1X then Y can be 1 jk 2 jk 1 − 2 written as Y h= Y1i+ √ 1Y2 whereh i Y1 and Y2 are real. Thus if Y = F(X) it is a transformation − e e e of (X1, X2) to (Y1, Y2) or where (Y1, Y2) is written as a function of (X1, X2) then we will use the e e e following notation for the Jacobian in the complex case.

Notation 2: (Jacobians in the complex case). J(Y1, Y2 : X1, X2): Jacobian of the transformation where Y and Y are written as functions of X and X or where Y = Y + √ 1Y is a 1 2 1 2 1 − 2 function of X = X1 + √ 1X2. − e e 28 Lemma 3.1. Consider a matrix A Cn n and (2n) (2n) matrices B and C where ∈ × × e A1 A2 A1 A2 A = A1 + √ 1A2, B = , C = − − " A2 A # " A2 A # − 1 1 where A , A Rne n. Then for det(A ) = 0 1 2 ∈ × 1 6 1 1 det(A) = det(B) 2 = det(C) 2 . (3.3) | | | |

Proof. Let det(A)= a + √ 1 b where a and b are real scalars. Then the absolute value is available − e as (a + √ 1b)(a √ 1b). If det(A + √ 1A ) = a + √ 1b, then det(A √ 1A ) = a − − − 1 − 2 − 1 − − 2 − √ q1b. Hence − (a + √ 1b)(a √ 1b)= det(A + √ 1A ) det(A √ 1A ) − − − 1 − 2 1 − − 2 A1 + √ 1A2 0 = det − . " 0 A1 √ 1A2 #! − − Adding the last n columns to the ﬁrst n columns and then adding the last n rows to the ﬁrst n rows we have

A1 + √ 1A2 0 2A1 A1 √ 1A2 det − = det − − . " 0 A1 √ 1A2 #! " A1 √ 1A2 A1 √ 1A2 #! − − − − − − Using similar steps we have

2A1 A1 √ 1A2 2A1 A1 √ 1A2 det − − = det − − 1 1 " A √ 1A2 A √ 1A2 #! " √ 1A2 A √ 1A2 #! 1 − − 1 − − − − 2 1 − 2 − 2A1 √ 1A2 = det − − 1 " √ 1A2 A1 #! − − 2 1 = det(A1) det(A1 + A2 A1− A2)

A1 A2 A1 A2 = det = det − . " A2 A #! " A2 A #! − 1 1 by evaluating as the determinant of partitioned matrices. Thus the absolute value of det(A) is given by e 1 1 1 det(A) = det(A ) det(A + A A− A )= det(B) 2 = det(C) 2 . 1 1 2 1 2 | | | | q This establishes the result. e

Remark 3.2. Now we denote A1 = Re(A) and A2 = Im(A). Clearly both Re(A) and Im(A) are real matrices. Each complex matrix A := Re(A)+ √ 1Im(A) can be represented faithfully as a e − e e e block-matrix e e e Re(A) Im(A) A − . (3.4) −→ " Im(A) Re(A) # e e e e 29 e Thus

Re(A)T Im(A)T A∗ T T . (3.5) −→ " Im(A) Re(A) # − e e e Then Y = AX can be rewritten as, via block-matrixe technique,e

e e e Re(Y) Im(Y) Re(A) Im(A) Re(X) Im(X) − = − − . (3.6) " Im(Y) Re(Y) # " Im(A) Re(A) #" Im(X) Re(X) # e e e e e e From the above, wee see thate the mentionede representatione is ane injectivee ring homomorphism which is continuous. sometimes we use the following representation:

Re(Y) Re(A) Im(A) Re(X) = − . (3.7) " Im(Y) # " Im(A) Re(A) #" Im(X) # e e e e Lemma 3.1 can be reexpressede as e e e

Re(A) Im(A) 2 det − = det(A) = det(AA∗) . (3.8) " Im(A) Re(A) #!

e e e e e Proposition 3.3. Let X, Y Cen be of n independente complex variables each, A Cn n a nonsingular ∈ ∈ × matrix of constants. If Y = AX, then e e e

e e e [dY]= det(AA∗) [dX]. (3.9)

If Y∗ = X∗ A∗, then e e e e

n e e e [dY∗]=( 1) det(AA∗) [dX]. (3.10) −

Proof. Let X = X + √ 1X , wheree X Rn, me=e 1,2. e Let Y = Y + √ 1Y , where Y 1 − 2 m ∈ 1 − 2 m ∈ n R , m = 1, 2 are real. Y = AX implies that Ym = AXm, m = 1, 2 if A = A is real. This e e transformation is such that the 2n real variables in (Y1, Y2) are written as functions of the 2n real e e e e variables in (X1, X2). Let

T T X1 = [x11,..., xn1], X2 = [x12,..., xn2], T T Y1 = [y11,..., yn1], Y2 = [y12,..., yn2].

Then the Jacobian is the determinant of the following matrix of partial derivatives: ∂(Y , Y ) ∂(y ,..., y , y ,..., y ) 1 2 = 11 n1 12 n2 . ∂(X1, X2) ∂(x11,..., xn1, x12,..., xn2) Note that ∂(Y , Y ) ∂(y ,..., y , y ,..., y ) 1 2 = 11 n1 12 n2 . ∂(X1, X2) ∂(x11,..., xn1, x12,..., xn2)

30 Note that ∂Y ∂(y ,..., y ) 1 = 11 n1 = A, ∂X1 ∂(x11,..., xn1) ∂Y ∂Y 1 = 0 = 2 , ∂X2 ∂X1 ∂Y 2 = A. ∂X2 Thus the Jacobian is

A 0 J = det = det(A)2. " 0 A #!

If A is complex, then let A = A + √ 1A where A and A are real. Then 1 − 2 1 2

e Y e= Y1 + √ 1Y2 =(A1 + √ 1A2)(X1 + √ 1X2) − − − = (A X A X )+ √ 1(A X + A X ) e 1 1 − 2 2 − 1 2 2 1 implies that

Y = A X A X , Y = A X + A X . 1 1 1 − 2 2 2 1 2 2 1 Then

∂Y1 ∂Y1 = A1, = A2, ∂X1 ∂X2 − ∂Y2 ∂Y2 = A2, = A1. ∂X1 ∂X2 Thus the Jacobian is

A1 A2 2 J = det − = det(A) = det(AA∗) , " A2 A1 #!

e e e which establishes the result. The second result follows by noting that [d Y ] = [dY ]( 1)n[dY ]= ∗ 1 − 2 ( 1)n[dY]. − e e e Propositione 3.4. Let X, Y Cm n of mn independent complex variables each. Let A Cm m and ∈ × ∈ × n n B C × nonsingular matrices of constants. If Y = AXB, then ∈ e e e n m e [dY]= det(AA∗)e dete e(BeB∗) [dX]. (3.11)

Proof. Let Y = Y1 + √ 1Y2 eand X = eXe1 + √ 1Xe2.e Indeed, e let Y = [Y1,..., Yn] and X = − − [X1,..., Xn], then Yj = AXj, j = 1,..., n when Y = AX. Thus [dYj] = det(AA∗) [dXj] for each e e e e e e j, therefore, ignoring the signs,

e e e e e e e e e e e e n n [dY]= ∏[dYj]= det(AA∗) [dX]. j=1

e e e e e 31 Denoting A = A + √ 1A , the determinant is 1 − 2 n e A1 A2 J = det − . " A2 A1 #!

Hence the Jacobian in this case, denoting B = B + √ 1B , is given by 1 − 2 m e B1 B2 J = det . " B2 B #! − 1 For establishing our result, write Y = AZ where Z = XB. That is,

n n m [dY]= det(AA∗e) [deZe]= det(eAA∗e) e det(BB∗) [dX].

This completes thee proof. e e e e e ee e n Remark 3.5. Another approach to the fact that [dY] = det(AA∗) [dZ], where Y = AZ, is described as follows: e e e e e ee Re(Y) = Re(A)Re(Z) Im (A)Im(Z) − Im(Y) = Im(A)Re(Z)+ Re(A)Im(Z)  e e e e e leading to  e e e e e ∂ Re(Y), Im(Y) Re(A)(n) Im(A)(n) = − , (n) (n) ∂ Re(Z), Im(Z) " Im(A) Re(A) # e e e e where e e e e Re(A) Im(A) (n) . (n) . Re(A) :=  ..  , Im(A) :=  ..  . e e  Re(A)   Im(A)  e   e       Then the Jacobian of this transformation cane be computed as e

Re(A)(n) Im(A)(n) J Re(Y), Im(Y) : Re(Z), Im(Z) = det − (3.12) " Im(A)(n) Re(A)(n) #! e e e e e e 2 = det Re(A)(n) + √ 1Im(A)(n) . (3.13) e − e

That is, e e

2 2n n (n) J Re(Y), Im(Y) : Re(Z), Im(Z) = det A = det(A) = det(AA∗) .

e e e e e e e e

32 Proposition 3.6. Let X, A, B Cn n be lower triangular matrices where X is matrix of n(n+1) indepen- ∈ × 2 dent complex variables, A, B are nonsingular matrices of constants. Then e e e e Y = X + XT e=e [dY]= 22n[dX], (3.14) ⇒ = [dY]= 2n[dX] if the x ’s are real; (3.15) e e e ⇒ e e jj n 2j Y = AX = [dYe]= ∏eajj e[dX], (3.16) ⇒ j=1 !

e e e e n e e 2j 1 = [dY]= ∏ ajj − [dX] if the ajj’s and xjj’s are real; (3.17) ⇒ j=1 !

e n e 2(n j+1) e e e − Y = XB = [dY]= ∏ bjj [dX], (3.18) ⇒ ! j=1

e e e e n e 2(n j)+1 e − = [dY]= ∏ bjj [dX] if the bjj’s and xjj’s are real; (3.19) ⇒ ! j=1

e e e e e Proof. Results (3.14) and (3.15) are trivial. Indeed, note that

2xjj, if j = k yjk = . x , if j > k  jke e By the deﬁnition, ignoring the sign, we havee

n [dY]= > dy = dy > dy , ∧j k jk ∧j=1 jj ∧j k jk (1) (2) (1) (2) (m) (m) where dy := dy dy for ye = y +e√ 1y . Soe for j =e1,..., n, we get y = 2x , m = jk jk jk jk jk − jk jj jj 1,2. Hence the result. If xjj’s are real, the result follows easily by deﬁnition. Let e e Y = Y + √e 1Y , X = X + √ 1X , A = A + √ 1A , B = B + √ 1B , 1 − 2 1 − 2 1 − 2 1 − 2 ( ) ( ) ( ) ( ) Y = [y m ], X = [x m ], A = [a m ], B = [b m ], m = 1, 2. em jk m e jk m jk e m jk e where Ym, Xm, Am, Bm, m = 1, 2 are all real. When Y = AX we have Y = A X A X and Y = A X + A X . The matrix of partial 1 1 1 − 2 2 2 1 2 2 1 derivative of Y with respect to X , that is ∂Y1 , can be seen to be a lower triangular matrix with 1 1 ∂X1 (1) e e e ajj repeated j times, j = 1,..., n, on the diagonal. Let this matrix be denoted by G1. That is,

A1  A [1ˆ 1ˆ]  ∂Y1 ∂Y2 1 | = := G1 = . . ∂X1 ∂X2  ..     [ [   A1[1ˆ n 1 1ˆ n 1]   · · · − | · · · −   

33 (2) Let G2 be a matrix of the same structure with ajj ’s on the diagonal. Similarly,

A2  A [1ˆ 1ˆ]  ∂Y1 ∂Y2 2 | = := G2 = . . − ∂X2 ∂X1  ..     [ [   A2[1ˆ n 1 1ˆ n 1]   · · · − | · · · −    Then the Jacobian matrix is given by

∂(Y , Y ) G1 G2 1 2 = − . ∂(X1, X2) " G2 G1 #

Let G = G + √ 1G . Then 1 − 2 e A  A[1ˆ 1ˆ]  | G = e . ,  ..   e   [ [  e  A[1ˆ n 1 1ˆ n 1]   · · · − | · · · −    where e

A = A + √ 1A , A[1ˆ 1ˆ]= A [1ˆ 1ˆ]+ √ 1A [1ˆ 1ˆ],..., 1 − 2 | 1 | − 2 | [ [ [ [ [ [ A[1ˆ n 1 1ˆ n 1] = A [1ˆ n 1 1ˆ n 1]+ √ 1A [1ˆ n 1 1ˆ n 1]. · · · − | · · · − e 1 · · · − | e· · · − − 2 · · · − | · · · − Thuse

∂(Y , Y ) G1 G2 det 1 2 = det − . ∂(X , X ) 1 2 " G2 G1 #!

2 From Lemma 3.1, the determinant is available as det(G) where G = G + √ 1G . Since G is 1 − 2 triangular the absolute value of the determinant is given by

e e e 2 2 2 2 det(G) = det(A) det(A[1ˆ 1ˆ]) det(A[1ˆ n[1 1ˆ n[1]) | · · · · · · − | · · · − n 2 j e = ∏ ( aejj ) . e e j=1

This establishes (3.16). Another approach is presented also: Let Y = [Y1,..., Yn], where Yj, j =

1,..., n, is the j-th column of the matrix Y. Similarly for X = [X1,..., Xn]. Now Y = AX implies e e e e that e e e e e e e

Yj = AXj, j = 1,..., n.

e e e 34 That is,

y11 x11 0 0 0 0

 y21   x21   y22   x22   0   0  e. = A e. , . = A . ,..., . = A . .  .   .   .   .   .   .               e   e   e   e       yn1  e  xn1   yn2  e  xn2   ynn  e  xnn                          Since Y, X, Aeare all lowere triangular,e it follows thate e e

e e e y11 x11 y22 x22  y   x  21 21 . ˆ ˆ . e. = A e. ,  .  = A[1 1]  .  ,...,  .   .  e | e      e   e   yn2   xn2   yn1  e  xn1    e              [ [ ynn = A[1...ˆ n 1 1...e n 1]xnn, e e e − | − where A[iˆjˆ iˆjˆ] stands fore a matrixe obtained fromb deletinge the i, j-th rows and columns of A, | respectively. We can now draw the conclusion that e [ [ [ [ [dY ]= det(A[1...ˆ j 1] 1...ˆ j 1])A[1...ˆ j 1] 1...ˆ j 1])∗ [dX ], j − | − − | − j

that indicates thate e n n [ [ [ [ [dY] = ∏[dYj]= ∏ det(A[1...ˆ j 1] 1...ˆ j 1])A[1...ˆ j 1] 1...ˆ j 1])∗ [dXj] − | − − | − j=1 j=1 e e e = det(AA∗) det (A[1ˆ 1ˆ]A[1ˆ 1ˆ]∗) a a∗ [dX] | | · · · | nn nn | 2 2 2 = a11a22 a nn a22a33 ann ann [dX] | e·e · · | × e| e· · · | ×···×e e| | e n 2j = e∏e ajj e [dX]. e e e e e j=1 !

e e If the xjj’s and ajj’s are real then note that the xjk’s for j > k contribute ajj twice that is, corre- (1) (2) (1) sponding to xjk and xjk , whereas the ajj’s appear only once corresponding to the xjj ’s since the (2) e e e xjj ’s are zeros. This establishes (3.17). If Y = XB and if a matrix H1 is deﬁned corresponding to ( ) e G then note that the b 1 ’s appear n j + 1 times on the diagonal for j = 1,..., n. Results (3.18) 1 jj − e e e and (3.19) are established by using similar steps as in the case of (3.16) and (3.17).

Proposition 3.7. Let X Cn n be hermitian matrix of independent complex entries and A Cn n be a ∈ × ∈ × nonsingular matrix of constants. If Y = AXA∗, then e e e e e e n [dY]= det(AA∗) [dX]. (3.20)

e e e e 35 Proof. Since A is nonsingular it can be written as a product of elementary matrices. Let E1,..., Ek be elementary matrices such that e e e

A = EkEk 1 E1 = A∗ = E1∗E2∗ Ek∗. − · · · ⇒ · · ·

For example, let E1 be suche thate e the j-the row ofe an identitye e e matrix is multiplied by a scalar c = a + √ 1b where a, b R. Then E1XE∗ means that the j-th row of X is multiplied by − e ∈ 1 a + √ 1b and the j-th column of X is multiplied by a √ 1b. Let e − e e e − − e

U1 = E1XE1∗e, U2 = E2U1E2∗, ..., Uk = EkUk 1Ek∗. − Then the Jacobian of Yewrittene e ase a functione e Xe ise given by e e e e

e J(Y : X)= J(Y : Ue k 1) J(U1 : X). − · · ·

Let us evaluate [dU1] in termse of e[dX] bye directe computation.e e Since X is hermitian its diagonal elements are real and the elements above the leading diagonal are the complex conjugates of e e e those below the leading diagonal, and U is also of the same structure as X. Let U = U + √ 1V 1 1 − and X = Z + √ 1W where U = [ujk], V = [vjk], Z = [zjk ], W = [wjk] are all real and the diagonal − e e e elements of V and W are zeros. Take the ujj’s and zjj’s separately. The matrix of partial derivatives e 2 2 of u11,..., unn with respect to z11,..., znn is a diagonal matrix with the j-th element a + b and all other elements unities. That is,

1 .  ..  ∂(diag(U)) ∂(u11,..., unn) = =  a2 + b2 = c 2  := C, ∂(diag(Z)) ∂(z11,..., znn)  | |   .   ..     e   1      where diag(X) means the diagonal matrix, obtained by keeping the diagonal entries of X and ignoring the off-diagonal entries. n(n 1) n(n 1) The remaining variables produce a − − matrix of the following type 2 × 2 ∂(U , V ) A B 0 0 = 0 0 ∂(Z0, W0) " B0 A0 # − where U0, V0, Z0, W0 mean that the diagonal elements are deleted, A0 is a diagonal matrix with n 1 of the diagonal elements equal to a and the remaining unities and B is a diagonal matrix − 0 such that corresponding to every a in A there is a b or b with j 1 of them equal to b and 0 − − −

36 n j of them equal to b. Thus the Jacobian of this transformation is: − C 0 0 ∂(diag(U), U0, V0) J(U1 : X) = = det  0 A0 B0  ∂(diag(Z), Z0, W0)  0 B0 A0  e e  −    A0 B0 = det(C) det " B0 A0 #! − 2 From Lemma 3.1, the determinant is det(A + √ 1B ) . That is, 0 − 0

∂( U , V ) A0 B0 det 0 0 = det ∂(Z0, W0) " B0 A0 #! − = det((A + √ 1B )(A + √ 1B )∗) 0 − 0 0 − 0 2 2 n 1 =( a + b ) − .

Thus

n 2 2 n [dU1]=(a + b ) [dX]= det(E1E1∗) [dX].

Note that interchanges of rowse and columnse can produce e e only a changee in the sign in the determinant, the addition of a row (column) to another row (column) does not change the determinant n and elementary matrices of the type E1 will produce det(E1E1∗) in the Jacobian. Thus by computing J(U : X), J(U : U ) etc we have 1 2 1 e e e n e e e e [dY]= det(AA∗) [dX].

As a speciﬁc example, the conﬁguratione ofe thee partial e derivatives for n = 3 with j = 2 is the following: If

z 1 0 0 11 ∗ ∗ X =  z + √ 1w z  and E1 =  0 c 0  , 21 − 21 22 ∗  z31 + √ 1w31 z32 + √ 1w32 z33   0 0 1  e  − −  e      e  then

z11 2∗ ∗ U1 =  c(z + √ 1w ) c z  21 − 21 | | 22 ∗  z31 + √ 1w31 c(z32 + √ 1w32) z33  e  − −   e e  u 11 e ∗ ∗ =  u + √ 1v u  , 21 − 21 22 ∗  u31 + √ 1v31 u32 + √ 1v32 u33   − −    37 thus

u = z , u = c 2 z , u = z , 11 11 22 | | 22 33 33 u21 = az21 bw21, u31 = z31, u32 = az32 + bw32, − e v = aw + bz , v = w , v = aw bz . 21 21 21 31 31 32 32 − 32 Now

u11 1 0 0000 000 z11  u   0 c 2 000 0 0 00   z  22 | | 22  u33   0 0 1000 000   z33           e     u21   0 0 0 a 0 0 b 0 0   z21     −     u  =  0 0 0010 000   z  .  31     31         u32   0 0 000 a 0 0 b   z32               v21   0 0 0 b 0 0 a 0 0   w21         v   0 0 0000 010   w   31     31         v32   0 0 000 b 0 0 a   w32     −          Now

1 a b 2 − C =  c  , A0 =  1  , B0 =  0  . | |  1   a   b         e      We are done.

Remark 3.8. If X is skew hermitian then the diagonal elements are purely imaginary, that is, the real parts are zeros. It is easy to note that the structure of the Jacobian matrix for a transformation e of the type Y = AXA , where X = X, remains the same as that in the hermitian case of ∗ ∗ − Proposition 3.7. The roles of (ujj, zjj)’s and (vjj, wjj)’s are interchanged. Thus the next theorem e e e e e e will be stated without proof.

Proposition 3.9. Let X Cn n skew hermitian matrix of independent complex entries. Let A Cn n ∈ × ∈ × be a nonsingular matrix of constants. If Y = AXA∗, then e e e e e e n [dY]= det(AA∗) [dX]. (3.21)

Some simple nonlinear transformationse wille e be considerede here. These are transformations which become linear transformations in the differentials so that the Jacobian of the original transformation becomes the Jacobian of the linear transformation where the matrices of differentials are treated as the new variables and everything else as constants.

38 Proposition 3.10. Let X Cn n be hermitian positive deﬁnite matrix of independent complex variables. ∈ × n n n n Let T C × be lower triangular and Q C × be upper triangular matrices of independent complex ∈ e ∈ variables with real and positive diagonal elements. Then e e n n 2(n j)+1 X = TT∗ = [dX]= 2 ∏ tjj − [dT], (3.22) ⇒ j=1 ! e ee e n e n 2(j 1)+1 X = QQ∗ = [dX]= 2 ∏ tjj − [dQ]. (3.23) ⇒ j=1 ! e e e e e Proof. When the diagonal elements of the triangular matrices are real and positive there exist unique representations X = TT∗ and X = QQ∗. Let X = X1 + √ 1X2 and T = T1 + √ 1T2, (m) (m)− − where X = [xjk], T = [tjk], tjk = 0, j < k, Xm = [x ], Tm = [t ], m = 1,2. Note that X1 is e ee e e e jk e jk e symmetric and X2 is skew symmetric. The diagonal elements of X2 and T2 are zeros. Hence when e e e e e considering the Jacobian we should take xjj, j = 1,..., p and xjk, j > k separately.

T T X = TT∗ = X1 + √e 1X2 =(T1 + √ 1eT2)(T √ 1T ) ⇒ − − 1 − − 2 T T X1 = T1T + T2T e ee = 1 2 , ⇒ X = T TT T TT  2 2 1 − 1 2 (1) (2)  with tjj = tjj, tjj = 0, j = 1,..., n. Note that

(1) (1) 2 (1) 2 (2) 2 (2) 2 xjj = tj1 + + tjj + tj1 + + tj,j 1 · · · · · · − implies that

(1) ∂xjj ( ) = 2t 1 = 2t , j = 1,..., n. (1) jj jj ∂tjj So

( ) ( ) 2t11 ∂(x 1 ,..., x 1 ) 11 nn = .. := Z. (1) (1)  .  ∂(t ,..., tnn ) 11  2t   nn    (1) > Now consider the xjk ’s for j k. It is easy to note that

∂(X , X ) UV 10 20 = , ∂(T10, T20) " WY #

(1) where a zero indicates that the xjj ’s are removed and the derivatives are taken with respect to the ( ) ( ) t 1 ’s and t 2 ’s for j > k. U and Y are lower triangular matrices with t repeated n j times along jk jk jj −

39 (2) (1) the diagonal and V is of the same form as U but with tjj = 0 along the diagonal and the tjk ’s (2) (1) (1) (1) (1) (1) (1) replaced by the tjk ’s. For example, take the xjk ’s in the order x21 , x31 ,..., xn1 , x32 ,..., xn,n 1 − (1) n(n 1) n(n 1) and t ’s also in the same order. Then we get the − − matrix jk 2 × 2

t 0 0 11 · · ·  t 0  ∂X10 ∗ 11 · · · U = = . . . ∂T10  . . .. 0       tn 1,n 1   ∗ ∗ ··· − −    where the ’s indicate the presence of elements some of which may be zeros. Since U and V are ∗ lower triangular with the diagonal elements of V being zeros, one can make W null by adding suitable combinations of the rows of (U, V). This will not alter the lower triangular nature or the diagonal elements of Y. Then the determinant is given by

n UV 2(n j) det = det(U) det(Y)= ∏ tjj − . " WY #! j=1

Multiply with the 2tjj’s for j = 1,..., n to establish the result. As a speciﬁc example, we consider the case where n = 3. Let X C3 3. Denote X = X + √ 1X . Thus ∈ × 1 − 2 (1) (1) (1) (2) (2) ex11 x21 x31 e 0 x21 x31 (1) (1) (1) (2) − − (2) X1 =  x21 x22 x32  , X2 =  x21 0 x32  . (1) (1) (1) (2) (2) −  x x x   x x 0   31 32 33   31 32      Similarly, let T = T + √ 1T . We also have: 1 − 2 (1) e t11 0 0 0 0 0 (1) (1) (2) T1 =  t21 t22 0  , T2 =  t21 0 0  . (1) (1) (1) (2) (2)  t t t   t t 0   31 32 33   31 32      T T Now X1 = T1T1 + T2T2 can be expanded as follows:

(1) (1) 2 (1) (1) (1) (1) (1) (1) x11 = t11 , x21 = t21 t11 , x31 = t31 t11 ;

(1) (1)2 (1) 2 (2) 2 (1) (1) (1) (1) (1) (2) (2) x22 = t21 + t22 + t21 , x32 = t31 t21 + t32 t22 + t31 t21 ;

(1) (1)2 (1)2 (1)2 (2) 2 (2) 2 x33 = t31 + t32 + t33 + t31 + t32 . Then X = T TT T TT can be expanded as follows: 2 2 1 − 1 2 (2) (2) (1) (2) (2) (1) (2) (2) (1) (2) (1) x21 = t21 t11 , x31 = t31 t11 , x32 = t31 t21 + t32 t22 .

40 From the above, we see that

(1) (1) (1) dx11 2t11 00000000 dt11 (1) (1) (1) (2) (1)  dx22   0 2t22 0 2t21 0 0 2t21 0 0   dt22  (1) (1) (1) (1) (2) (2) (1)  dx   0 0 2t 0 2t 2t 0 2t 2t   dt   33   33 31 32 31 32   33   (1)   (1) (1)   (1)   dx21   t21 0 0 t11 0 0 0 0 0   dt21   ( )   ( ) ( )   ( )   dx 1  =  t 1 0 0 0 t 1 0 0 0 0   dt 1  .  31   31 11   31   (1)   (1) (1) (1) (1) (2) (2)   (1)   dx   0 t 0 t t t t t 0   dt   32   32 31 21 22 31 21   32   (2)   (2) (1)   (2)   dx21   t21 0 0 0 0 0 t11 0 0   dt21   ( )   ( ) ( )   ( )   dx 2   t 2 000000 t 1 0   dt 2   31   31 11   31   (2)   (2) (2) (1) (1)   (2)   dx   0 t 0 t 0 0 0 t t   dt   32   32 31 21 22   32        In what follows, we compute the Jacobian of this transformation:

(1) 2t11 00000000 (1) (1) (2)  0 2t22 0 2t21 0 0 2t21 0 0  (1) (1) (1) (2) (2)  0 0 2t 0 2t 2t 0 2t 2t   33 31 32 31 32   (1) (1)   t21 0 0 t11 0 0 0 0 0   ( ) ( )  J(X : T) = det  t 1 0 0 0 t 1 0 0 0 0   31 11   (1) (1) (1) (1) (2) (2)   0 t 0 t t t t t 0  e e  32 31 21 22 31 21   (2) (1)   t21 0 0 0 0 0 t11 0 0   ( ) ( )   t 2 000000 t 1 0   31 11   (2) (2) (1) (1)   0 t 0 t 0 0 0 t t   32 31 21 22    100000000 (1) (1) (2)  0 2t22 0 2t21 0 0 2t21 0 0  (1) (1) (1) (2) (2)  0 0 2t 0 2t 2t 0 2t 2t   33 31 32 31 32   (1) (1)   t21 0 0 t11 0 0 0 0 0  (1)  ( ) ( )  = 2t det  t 1 0 0 0 t 1 0 0 0 0  , 11  31 11   (1) (1) (1) (1) (2) (2)   0 t 0 t t t t t 0   32 31 21 22 31 21   (2) (1)   t21 0 0 0 0 0 t11 0 0   ( ) ( )   t 2 000000 t 1 0   31 11   (2) (2) (1) (1)   0 t 0 t 0 0 0 t t   32 31 21 22    by adding the corresponding multiples of the ﬁrst row to the second row through the last one,

41 respectively, we get

100000000 (1) (1) (2)  0 2t22 0 2t21 0 0 2t21 0 0  (1) (1) (1) (2) (2)  0 0 2t 0 2t 2t 0 2t 2t   33 31 32 31 32   (1)   0 0 0 t11 0 0 0 0 0  (1)  ( )  J(X : T)= 2t det  00 0 0 t 1 0 0 0 0  , 11  11   (1) (1) (1) (1) (2) (2)   0 t 0 t t t t t 0  e e  32 31 21 22 31 21   (1)   00 0 0 0 0 t11 0 0   ( )   00 0 0 0 0 0 t 1 0   11   (2) (2) (1) (1)   0 t 0 t 0 0 0 t t   32 31 21 22    iteratively, finally we get the final result. We also take a simple approach (i.e. by definition) with a tedious computation as follows:

(1) (1) (1) (1) (1) (1) (1) dx11 = 2t11dt11, dx21 = t21 dt11 + t11dt21 , dx31 = t31 dt11 + t11dt31 ; (1) (1) (1) (2) (2) dx22 = 2t22dt22 + 2t21 dt21 + 2t21 dt21 , (1) (1) (1) (1) (1) (1) (1) (2) (2) (2) (2) dx32 = t31 dt21 + t21 dt31 + t32 dt22 + t22dt32 + t31 dt21 + t21 dt31 ; (1) (1) (1) (1) (1) (2) (2) (2) (2) dx33 = 2t31 dt31 + 2t32 dt32 + 2t33dt33 + 2t31 dt31 + 2t32 dt32 , and

(2) (2) (2) (2) (2) (2) dx21 = t21 dt11 + t11dt21 , dx31 = t31 dt11 + t11dt31 , (2) (2) (1) (1) (2) (2) (2) dx32 = t31 dt21 + t21 dt31 + t22dt32 + t32 dt22.

Hence we can compute the Jacobian by deﬁnition as follows:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) [dX] = dx 1 dx 1 dx 1 dx 2 dx 2 dx 1 dx 2 dx 1 dx 1 11 ∧ 21 ∧ 31 ∧ 21 ∧ 31 ∧ 22 ∧ 32 ∧ 32 ∧ 33 3 3 2(n j)+1 e = 2 ∏ tjj − [dT]. j=1 ! e The proof in the case of X = QQ∗ is similar but in this case it can be seen that the triangular matrices corresponding to U and Y will have tjj repeated j 1 times along the diagonal for e e e − j = 1,..., n.

Example 3.11. If X = X Cn n is positive deﬁnite , and Re(α) > n 1, ∗ ∈ × − α n − Tr(X) e eΓn(α) := [dX] det X e− > ZX 0 n(n 1) e e = π 2− Γe(α )Γ(α e1) Γ(α n + 1). (3.24) e − · · · −

42 Indeed, let T = [tjk], tjk = 0, j < k be a lower triangular matrix with real and positive diagonal elements tjj > 0, j = 1,..., n such that X = TT∗. Then from Proposition 3.10 e e e n en ee 2(n j)+1 [dX]= 2 ∏ tjj − [dT] j=1 ! e e Note that n 2 2 2 2 Tr X = Tr TT∗ = ∑ tjj + t21 + + tn1 + + tn,n 1 − j=1 · · · · · · e ee e e e and

α n n − n 2α 2j+1 det(X) [dX]= 2 ∏ tjj − [dT]. ! j=1 e e e n(n 1) > The integral over X splits into n integrals over the tjj’s and 2− integrals over the tjk’s, j k. (1) (2) (1) (2) Note that 0 < tjj < ∞, ∞ < t < ∞, ∞ < t < ∞, where tjk = t + √ 1t . But e − jk − jk jk − jk e ∞ 2α 2j+1 t2 jj Γ e > 2 tjj − e− dtjj = (α j + 1), Re(α) j 1, Z0 − − for j = 1,..., n, so Re(α) > n 1 and − ( ) 2 ( ) 2 2 ∞ ∞ t 1 + t 2 tjk − jk jk (1) (2) e−| | dtjk = e dtjk dtjk = π. tjk ∞ ∞ Z e Z− Z− The desired result ise obtained.e

Deﬁnition 3.12 (Γn(α): complex matrix-variate gamma). It is deﬁned as stated in Example 3.11. We will write with a tilde over Γ to distinguish it from the matrix-variate gamma in the real case. e Example 3.13. Show that

α α n − Tr(BX) det(B) det(X) e− ( )= f X e e Γn(α) e e e for B = B > 0, X = X > 0,Re(α) > n 1 and f (X) = 0 elsewhere, is a density function for X ∗ ∗ − e where B is a constant matrix, with Γn(α) as given in Deﬁnition 3.12. Indeed, evidently f (X) > 0 e e e e e e for all X and for all X and it remains to show that the total integral is unity. Since B is hermitian e e e positive deﬁnite there exists a nonsingular C such that B = C∗C. Then e e e

Tr BX =e Tr CXC∗e . e e Hence from Proposition 3.10 e e e e e

n n Y = CXC∗ = [dY]= det CC∗ [dX]= det(B) [dX], ⇒

e e e e e e43e e e e and

1 1 1 X = C− YC∗− = det(X) = det(CC∗) − det(Y) ⇒ | |

Then e e e e α n − Tr(Y) det(Y) e− f (X)[dX]= [dY] = 1. > > e X 0 Y 0 Γ n(α) Z Z e But from Example 3.11e , the righte e side ise unitye for Re(α) > n 1. This density f (X) is known as e − the complex matrix-variate density with the parameters α and B. e e 3.2 The computation of volumes

Deﬁnition 3.14 (Semiuniatry and unitary matrices). A p n matrix U is said to be semiunitary × if UU∗ = 1p for p < n or U∗U = 1n for p > n. When n = p and UU∗ = 1n, then U is called a e unitary matrix. The set of all n n unitary matrices is denoted by (n). That is, e e e e× Ue e e (n) := U Cn n : UU = 1 , (3.25) U ∈ × ∗ n n o where Cn n denotes the set of all n n ecomplex matrices.e e × × Deﬁnition 3.15 (A hermitian or a skew hermitian matrix). Let A Cn n. If A = A , then A is ∈ × ∗ said to be hermitian and if A∗ = A, then it is skew hermitian. − e e e e When dealing with unitarye matricese a basic property to be noted is the following:

UU∗ = 1 = U∗dU = dU∗U. ⇒ − ∗ But U∗dU = dU∗U, whiche meanse that U∗edU eis a skewe hermitiane matrix. The wedge product of U dU, namely, U dU enters into the picture when evaluating the Jacobians involving ∗e e e∧ e ∗ e e unitary transformations. Hence this will be denoted by dG for convenience. Starting from e e e e U∗U = 1n one has dU U∗. · e √ 1θj Assume that U = [uij] (n) where uij C. Let ujj = ujj e − by Euler’s formula, where e e e e ∈ U ∈ θj [ π, π]. Then ∈ − e e e e e √ 1θ2 √ 1θn √ 1θ u11 u12e− − u1ne− − e − 1 0 0 | √| · · · √ √ · · ·  u e 1θ1 u u e 1θn   0 e 1θ2 0  21 − − | 22 | · · · 2n − − − · · · U = e. e . . e ...... (3.26)  . . .. .   . . .. .   e e e     √ 1θ √ 1θ2   √ 1θn  e  un1e− − 1 un2e− − unn   0 0 e −   · · · | |   · · ·      This indicates that any U (n) can be factorized into a product of a unitary matrix with e e ∈ U e diagonal entries being nonnegative and a diagonal unitary matrix. It is easily seen that such e 44 factorization of a given unitary matrix is unique. In fact, we have a correspondence which is one-to-one:

(n) ( (n)/ (1) n) (1) n. U ∼ U U × × U × Notation. When U is a n n unitary matrix of independent complex entries, U its conjugate × ∗ transpose and dU the matrix of differentials then the wedge product in dG := dU U∗ will be e e · denoted by [dG]. That is, ignoring the sign, e e e e e [dG] := dU U = U dU . ∧ · ∗ ∧ · ∗ If the diagonal entries or the entriese in onee rowe of thise unitaery matrix U are assumed to be real, then the skew hermitian matrix dU U∗ will be denoted by dG1 and its wedge product by · e [dG ] := dU U∗ . e e1 ∧ · e Indeed, [dG] here means the wedge product over (n), but however [dG ] means the wedge e e Ue 1 n product over (n)/ (1)× . Therefore, for any measurable function f over (n), [dG] = [dG1][dD], e U U Ue f (U)[dG]= f (VD)[dG1][dD], e e e (n) (n) (1) n ZU ZU1 ZU × where U = VD for V (n),e dG =e U dU and dG =eVe dVe. Furthermore,e let V = [v ] for ∈ U1 ∗ 1 ∗ ij + vij C and vjj = vjj R . Then it holds still that V (n), thus vjj is not an independent ∈ e e e e∈ e e e e ∈e U e e 2 2 e variable, for example, v11 = 1 v21 vn1 . From this, we see that e e − | | −···− | | e q [dG1]= ∏e d Re(V∗dVe)ij d Im(V∗dV)ij < i j and e e e e e n [dG]= ∏ Im(U∗dU)jj ∏ Re(U∗dU)ijIm(U∗dU)ij. j=1 ! × i

n e2(n j)+1 [dX]= ∏j=1 tjj − [dT][dG] (3.27) where dG = dU U ; and e e e · ∗ (ii) for all thee diagonale e entries in U being real,

n e 2(n j) [dX]= ∏ tjj − [dT][dG1] (3.28) j=1 ! ·

e e e e where dG = dU U . 1 · ∗

e e e 45 Proof. Taking differentials in X = TU one has

e edeX = dT U + T dU. · ·

Postmultiplying by U∗ and observinge that Ue Ue∗ = e1n wee have

e dX U∗ = deTe+ T dU U∗. (3.29) · · · (i). Let the diagonal elements in eT bee real ande e positivee e and all other elements in T and U be complex. Let e e e

dV = dX U∗ = [dV] = [dX] (3.30) · ⇒ ignoring the sign, since U is unitary.e Lete deG = dU eU ande its wedge product be [dG]. Then · ∗ e dV =edT + eT deG e (3.31) · where dG is skew hermitian. Write e e e e

( ) ( ) e V = [v ], v = v 1 + √ 1v 2 , jk jk jk − jk ( ) ( ) ( ) ( ) T = [t ], t = t 1 + √ 1t 2 , j > k, t 1 = t > 0, t 2 = 0, e ejk ejk jk − jk jj jj jj ( ) ( ) ( ) ( ) ( ) ( ) dG = [dg ], dg = dg 1 + √ 1dg 2 , dg 1 = dg 1 , dg 2 = dg 2 . e e jk e jk jk − jk jk − kj jk kj From Eq. (3.31e ), e e

dtjk + tj1dg1k + + tjjdgjk , j > k dv = · · · jk  tj dg + + tjjdg , j < k.  e1 1ke ·e · · jk e e e (m)  e(m) e (m) (me) e (m) (m) > The new variables are dvjk = vjk ,dtjk = tjk ,dgjk = gjk , j k, m = 1, 2. The matrices of partial derivatives are easily seen to be the following: b b b (1) (m) ∂vjj ∂vjk = 1, , j > k = 1, m = 1, 2, (m)  ∂tjj   ∂t  b bjk  (1)   (2) ∂vb ∂v kj > = b kj > = ( ) , j k A, ( ) , j k B  ∂g 1  ∂g 2  bkj bkj     where A and B are triangular matrices with t repeated n j times, b jj b − ∂v(2) jj = ( ) ( ) diag t11,..., tnn .  ∂g 2  bjj   b 46 By using the above identity matrices one can wipe out other submatrices in the same rows and columns and using the triangular blocks one can wipe out other blocks below it when evaluating the determinant of the Jacobian matrix and ﬁnally the determinant in absolute value reduces to the form n 2(n j)+1 det(A) det(B)t11 tnn = ∏ tjj − . · · · j=1

Hence the result. As a speciﬁc example, we consider the case where n = 3. We expand the expression: dV = dT + T dG. That is, · e e edV e = dT + T d Re(G) T d Im(G) , 1 1 1 − 2 dV2 = dT2 + T2d Re(Ge) + T1d Im(Ge) . Furthermore, e e

(1) (1) (1) dv11 dv12 dv13 (1) (1) (1)  dv21 dv22 dv23  (1) (1) (1)  dv dv dv   31 32 33   (1)  (1) (1) (1) dt11 0 0 t11 0 0 0 dg12 dg13 (1) (1) (1) (1) (1) (1) =  dt21 dt22 0  +  t21 t22 0   dg12 0 dg23  (1) (1) (1) (1) (1) (1) − (1) (1)  dt dt dt   t t t   dg dg 0   31 32 33   31 32 33   − 13 − 23    (2) (2) (2)   0 0 0 dg11 dg12 dg13 (2) (2) (2) (2)  t21 0 0   dg12 dg22 dg23  − (2) (2) (2) (2) (2)  t t 0   dg dg dg   31 32   13 23 33      and

(2) (2) (2) dv11 dv12 dv13 (2) (2) (2)  dv21 dv22 dv23  (2) (2) (2)  dv dv dv   31 32 33    (1) (1) 0 0 0 0 0 0 0 dg12 dg13 (2) (2) (1) (1) =  dt21 0 0  +  t21 0 0   dg12 0 dg23  (2) (2) (2) (2) − (1) (1)  dt dt 0   t t 0   dg dg 0   31 32   31 32   − 13 − 23   (1)   (2) (2)  (2)  t11 0 0 dg11 dg12 dg13 (1) (1) (2) (2) (2) +  t21 t22 0   dg12 dg22 dg23  . (1) (1) (1) (2) (2) (2)  t t t   dg dg dg   31 32 33   13 23 33     

47 Thus

(1) (1) (1) (1) (1) dv11 = dt11, dv12 = t11dg12 , dv13 = t11dg13 , ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) dv 1 = dt 1 t dg 1 t 2 dg 2 , dv 1 = dt + t 1 dg 1 t 2 dg 2 , 21 21 − 22 12 − 21 11 22 22 21 12 − 21 12 ( ) ( ) ( ) ( ) ( ) ( ) dv 1 = t 1 dg 1 + t dg 1 t 2 dg 2 , 23 21 13 22 23 − 21 13 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) dv 1 = dt 1 t 1 dg 1 t dg 1 t 2 dg 2 t 2 dg 2 , 31 31 − 32 12 − 33 13 − 31 11 − 32 12 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) dv 1 = dt 1 + t 1 dg 1 t dg 1 t 2 dg 2 t 2 dg 2 , 32 32 31 12 − 33 23 − 31 12 − 32 22 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) dv 1 = dt + t 1 dg 1 + t 1 dg 1 t 2 dg 2 t 2 dg 2 33 33 31 13 32 23 − 31 13 − 32 23 and

(2) (2) (2) (2) (2) (2) dv11 = t11dg11 , dv12 = t11dg12 , dv13 = t11dg13 , ( ) ( ) ( ) ( ) ( ) ( ) ( ) dv 2 = dt 2 + t 1 dg 2 + t dg 2 t 2 dg 1 , 21 21 21 11 22 12 − 21 12 (2) (2) (1) (1) (2) (2) dv22 = t21 dg12 + t21 dg12 + t22dg22 , (2) (2) (1) (1) (2) (2) dv23 = t21 dg13 + t21 dg13 + t22dg23 , ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) dv 2 = dt 2 t 2 dg 1 + t 1 dg 2 + t 1 dg 2 + t dg 2 , 31 31 − 32 12 31 11 32 12 33 13 (2) (2) (2) (1) (1) (2) (1) (2) (2) dv32 = dt32 + t31 dg12 + t31 dg12 + t32 dg22 + t33dg23 , (2) (2) (1) (2) (1) (1) (2) (1) (2) (2) dv33 = t31 dg13 + t32 dg23 + t31 dg13 + t32 dg23 + t33dg33 .

( ) ( ) According to the deﬁnition, we now have dv = dv 1 dv 2 , then jk jk ∧ jk

[dV] = dv dv dv edv dv22 dv23 dv dv32 dv33 11 ∧ 12 ∧ 13 ∧ 21 ∧ ∧ ∧ 31 ∧ ∧ 3 2(3 j)+1 − e = e∏ tjj e e[dT][dGe ]. e e e e e j=1 ! e e (ii). Let the diagonal elements of U be real and all other elements in U and T complex. Starting from Eq. (3.31), observing that dG is dG1 in this case with the wedge product [dG1], and taking e (1) (1) (2) e (2) e the variables dvjk’s in the order dv , j > k,dv , j < k,dv , j > k,dv , j < k and the other e jk e jk jk jk e (1) > (1) > (2) > (2) > variables in the order dtjk , j k,dgjk , j k,dtjk , j k,dgjk , j k, we have the following e conﬁguration in the Jacobian matrix:

1 ∗ ∗ ∗  0 A 0 A  − 1 2  0 1   ∗ ∗     0 A2 0 A1   − −    where the matrices marked by can be made null by operating with the ﬁrst and third column ∗ submatrices when taking the determinant. Thus they can be taken as null matrices, and A1 and

48 ( ) ( ) A are triangular matrices with respectively t 1 and t 2 repeated n j times in the diagonal. The 2 jj jj − Jacobian matrix can be reduced to the form

AB 1 0 0 0 , A = , B = . " BA # " 0 A # " 0 A2 # − − 1 Then the determinant is given by

AB = det((A + √ 1B)(A + √ 1B)∗) BA − − −

= det(( A + √ 1A )( A + √ 1A )∗) − 1 − 2 − 1 − 2 p 2(n j) = ∏ tjj − j=1

e ( ) ( ) since A + √ 1A is triangular with the diagonal elements t 1 + √ 1t 2 repeated n j − 1 − 2 − jj − jj − ( ) 2 ( ) 2 2 times, giving t 1 + t 2 = t repeated n j times in the ﬁnal determinant and hence jj jj jj − the result. e By using Proposition 3.16 one can obtain expressions for the integral over U of [dG] and

[dG1]. These will be stated as corollaries here and the proofs will be given after stating both the e e corollaries. e Theorem 3.17. Let dG = dU U , where U (n). Then · ∗ ∈ U e e e e

n n2 n n(n+1) 2 π 2 π 2 vol ( (n)) = [dG]= = . (3.32) U (n) Γ (n) 1!2! (n 1)! ZU n · · · − e e

Proof. Let X be a n n matrix of independent complex variables. Let ×

2 2 Tr(XX∗) ∑j,k xjk n e B = [dX]e− = [dX]e− = π X X | | Z e e Z e since e e e e

( ) 2 ( ) 2 2 +∞ +∞ x 1 + x 2 xjk − jk jk (1) (2) e−| | dxjk = e dxjk dxjk = π. xjk ∞ ∞ Z e Z− Z−

Consider the transformatione e used in Proposition 3.16 with tjj’s real and positive. Then

Tr XX∗ = Tr TT∗ e e ee 49 and let

n 2 Tr(XX∗) 2(n j)+1 ∑j>k tjk B = [dX]e− = ∏ tjj − e− [dT][dG]. X T U | | Z e e Z Z j=1 ! e But e e e e e e ∞ 2(n j)+1 t2 1 jj Γ > tjj − e− dtjj = (n j + 1) for n j + 1 0 Z0 2 − − and for j > k

2 tjk e−| | dtjk = π. tjk Z e Then the integral over T gives e e

n(n 1) n n − n e 2− π 2 ∏ Γ(n j + 1)= 2− Γn(n). j=1 − Hence e 2 2nπn [dG]= . (n) Γ (n) ZU n e e Theorem 3.18. Let U (n) with the diagonal elements real. Let dG = dU U . Let the full unitary 1 ∈ U 1 1 · 1∗ n group of such n n matrices U1 be denoted by 1(n)= (n)/ (1)× . Then × e U U U e e e n(n 1) n(n 1) − e n π − π 2 vol ( 1(n)) = vol (n)/ (1)× = [dG1]= = . (3.33) U U U 1(n) Γ (n) 1!2! (n 1)! ZU n · · · − Proof. Now consider the transformation used in Propositione 3.16 with all the elements in T com- e plex and the diagonal elements of U real. Then e n 2 2(n j) ∑ t e − j>k jk B = ∏ tjj e− | | [dT][dG1]. T U1 Z Z j=1 ! e Note that e e e e

2 n(n 1) tjk − ∏ e−| | dtjk = π 2 . > tjk j k Z e ! Let t = t = t + √ 1t . Put t = r cose θ and te = r sin θ. Let the integral over t be denoted by jj 1 − 2 1 2 jj aj. Then e e e 2 +∞ +∞ 2(n j) 2 2 t 2 2 n j (t1+t2) aj = t − e− dt = (t1 + t2) − e− dt1dt2 t | | ∞ ∞ Z ∞ ∞ e Z− Z− + + 2 2 2 2 n j (t1+t2) = 4e e (t1 +et2) − e− dt1dt2 Z0 Z0 π ∞ 2 2 n j r2 = 4 (r ) − e− r drdθ Zθ=0 Zr=0 · · · = πΓ(n j + 1) for n j + 1 > 0. − −

50 Then

n n2 B = π Γn(n) [dG1]= π ZU1 implies that e e e

πn(n 1) [dG ]= − , 1 Γ ZU1 n(n) which establishes the result. e e e Example 3.19. Evaluate the integral

α 2 Γ Tr(XX∗) n n(α + n) ∆(α)= [dX] det(XX∗) e− = π (3.34) X · Γn(n) Z e e e e e e e e for Re(α) > 1, where X Cn n matrix of independent complex variables. Indeed, put X = TU, − ∈ × where U (n), T is lower triangular with real distinct and positive diagonal elements. Then ∈ U e e e e α n e e 2α det(XX∗) = ∏ tjj , j=1 n e e 2 2 2 Tr XX∗ = Tr TT∗ = ∑ tjk = ∑ tjj + ∑ tjk , j>k j=1 j>k e e n ee e e 2(n j)+1 [dX] = ∏ tjj − [dT][dG], j=1 ! 2 e 2nπn e e [dG] = . Γ ZU n(n) Thus e e e

n 2 2 2α+2(n j)+1 t ∑ > t ∆(α) = [dT][dG] ∏ t − e− jj e− j k jk T,U | | Z j=1 ! e ∞ ∞ e ne e e 2 + 2 2α+2(n j)+1 tjj tjk = ∏ t − e− dtjj ∏ e−| | dtjk [dG] 0 · > ∞ · U j=1 Z ! j k Z− e ! Z 2 n n(n 1) n n e e e n − 2 π = 2− ∏ Γ(α + n j + 1) π 2 j=1 − ! · · Γn(n) for Re(α + n) > n 1 or Re(α) > 1. That is, − − e

2 Γ (α + n) ∆(α)= πn n for Re(α) > 1. Γn(pn) − e e 51 Proposition 3.20. Let X Cn n be a hermitian matrix of independent complex entries with real distinct ∈ × eigenvalues λ1 > λ2 > > λn. Let U (n) with real diagonal entries and let X = UDU∗, where e· · · ∈ U D = diag(λ1,..., λn). Then e e e e 2 [dX]= ∏ > λ λ [dD][dG ], (3.35) j k k − j · 1 where dG = U dU. e e 1 ∗ ·

Proof. Takee thee differentialse in X = UDU∗ to get

dX = dUe De Ue∗ + U dD U∗ + U D dU∗. · · · · · ·

Premultiply by U∗, postmultiplye e by Ueand observee thate dGe 1 is skewe hermitian. Then one has

e dW =edG D + dD D deG 1 · − · 1 where dW = U dXU with [dWe ] = [edX]. Using the samee steps as in the proof of Proposi- ∗ · tion 7.10, we have e e e e e e 2 [dW]= ∏ λk λj [dD][dG1]. j>k − !

e e Hence the result follows.

Example 3.21. Let D = diag(λ1,..., λn) where the λj’s are real distinct and positive or let λ1 > > λ > 0. Show that · · · n (i)

2 2 Γn(n) n 2 Tr(D) [dD] ∏ λk λj e− = = ∏ Γ(j) ; > > > n(n 1) Zλ1 λn 0 j>k − ! π − j=1 ! ··· e

(ii)

n 2 α n Tr(D) Γn(α)Γn(n) [dD] ∏ λk λj ∏ λj − e− = . > > > n(n 1) Zλ1 λn 0 j>k − ! · j=1 ! · π − ··· e e

In fact, let Y be a n n hermitian positive deﬁnite matrix, U a unitary matrix with real diagonal × elements such that e e

U∗YU = D = diag(λ1,..., λn).

From the matrix-variate gammae integrale e

α n − Tr(Y) Γn(α)= [dY] det(Y) e− for Re(α) > n 1. Y=Y >0 · − Z ∗ e e e e e e 52 Hence

Tr(Y) Γn(n)= [dY]e− . Y=Y >0 Z ∗ e Then Y = UDU∗ implies that e e e e

e e e 2 [dY]= ∏ λk λj [dD][dG1], dG1 = U∗ dU. j>k − ! ·

Note that e e e e e

Tr Y = Tr UDU∗ = Tr (D) , α n α n n − e− e e det(Y) = ∏ λj . j=1 !

Then e

2 Tr(D) Γn(n)= [dD] ∏ λk λj e− [dG1]. > > > Zλ1 λn 0 j>k − ! × ZU ··· e Clearly e e

πn(n 1) [dG ]= − . 1 Γ ZU n(n) Substituting this, results (i) and (ii) follow.e e e Remark 3.22. We can try to compute the following integral in (3.35):

n n 2 [dX] = δ 1 ∑ λj ∏(λi λj) ∏ dλj [dG1] > > > > X:Tr(X)=1 λ1 λ2 λn 0 − < − × 1(n) Z Z ··· j=1 ! i j j=1 ZU which ise equivalente e to the following e

n n 1 2 [dX] = δ 1 ∑ λj ∏(λi λj) ∏ dλj [dG1] X:Tr(X)=1 n! − < − × 1(n) Z Z j=1 ! i j j=1 ZU n 1 n(n 1) e e e ∏ − Γ(n j)Γ(n j + 1) 2− e 1 j=0 − − π = 2 n n! Γ(n ) × ∏j=1 Γ(j) 1 n(n 1) Γ(1)Γ(2) Γ(n + 1) = π 2− · · · , n! Γ(n2) where we used the integral formula:

n 1 n n ∏ − Γ(n j)Γ(n j + 1) 2 j=0 δ 1 ∑ λj ∏(λi λj) ∏ dλj = − − . Γ(n2) Z − j=1 ! i

53 n(n 1) Γ Γ Γ n − (1) (2) (n) vol (D (C )) = π 2 · · · . (3.36) Γ(n2)

We make some remarks here: to obtain the volume of the set of mixed states acting on Cn, 1 1 one has to integrate the volume element [dX]. By deﬁnition, the ﬁrst integral gives HS , where n! Cn ( ) CHS = C 1,2 and n n e n ∞ ∞ n n n 1 α 1 β = δ 1 ∑ λj ∏ λk− ∏(λi λj) ∏ dλj (α,β) 0 · · · 0 − − Cn zZ }| Z { j=1 ! k i λ > > λ (a generic 1 2 · · · n density matrix is not degenerate), which corresponds to a choice of a certain Weyl chamber of the eigenvalue simplex ∆n 1. In other words, different permutations of the vector of n generically − 1 different permutations (Weyl chambers) equals to n!. This is why the factor n! appears in the right hand side in the above identity. In summary, the transformation

X (D, U) 7→ such that X = UDU , is one-to-one if ande only if De = diag(λ ,..., λ ), where λ > > λ ∗ 1 n 1 · · · n n and U (n)/ (1)× . ∈ Ue Ue e Remarke 3.23. In this remark, we will discuss the connection between two integrals [14]: for α, β > 0,

n ∞ ∞ n n n (1) α 1 β n (α, β) = δ 1 ∑ λj ∏ λk− ∏ (λi λj) ∏ dλj, I zZ0 ·}| · · Z0 { − j=1 ! k=1 16i

( ) We introduce an auxiliary variable t in the expression of 1 (α, β), and deﬁne I(t) as In n ∞ ∞ n n n α 1 β I(t)= δ t ∑ λj ∏ λk− ∏ (λi λj) ∏ dλj. (3.37) Zz0 ·}| · · Z0 { − j=1 ! k=1 16i

54 ( ) Then I(1)= 1 (α, β). Taking the Laplace transform, denoted by L , of I(t), we obtain In n ∞ ∞ n n n L st α 1 β (I) := δ t ∑ λj e− dt ∏ λk− ∏ (λi λj) ∏ dλj (3.38) zZ0 ·}| · · Z0 { "Z − j=1 ! # k=1 16i

αn β(n) (w) I(s) := L (I)= s− − 2 (α, β). (3.41) ·I Therefore e αn+β(n) 1 t 2 I(t)= L 1(L (I)) = L 1(I)= − (w)(α, β). (3.42) − − Γ n αn + β(2) ·I Letting t = 1 gives the conclusion: e

( ) 1 1 (α, β)= (w)(α, β). (3.43) n Γ n I αn + β(2) ·I ( ) In order to calculate the integral 1 (α, β), it sufﬁces to calculate the integral (w)(α, β), which is In I derived in Corollary 9.4 via Selberg’s integral (See Appendix):

β β n Γ Γ 1 + j 2 α +(j 1) 2 (w)(α, β)= ∏ − . (3.44) I Γ β j=1 1 + 2 Matrix integrals, especially over unitary groups, are very important. We recently present some results of this aspect [32]. Generally speaking, the computation of matrix integrals is very difficult from the first principle. Thus frequently we need to perform variable substitution in computing integrals. The first step in substitution is to compute Jacobians of this transformation, this is what we present. We also apply the matrix integrals over unitary groups to a problem [33, 34] in quantum information theory.

4 The volume of a compact Lie group

The content of this section is mainly from [22]. Some missing details are provided. Note that Macdonald’s result presents unifying treatment for computing the volume of a compact Lie group. Before proceeding, it is necessary to recall the notion of root system and its related properties.

55 Definition 4.1. A root system (E, R) is a finite-dimensional real vector space E with an inner product , , together with a finite collection R of nonzero vectors in E satisfying the following h· ·i properties:

(i) span R = E. { } (ii) If α R, then L R = α , where L = rα : r R . ∈ α ∩ {± } α { ∈ } (iii) If α, β R, then s (β) R, where s is the linear transformation of E deﬁned by ∈ α ∈ α x, α s (x)= β 2h i α, x E. α − α, α ∈ h i

β,α h i Z (iv) For all α, β R, 2 α,α . ∈ h i ∈ The dimension of E is called the rank of the root system and the elements of R are called roots.

Deﬁnition 4.2. If (E, R) is a root system, the Weyl group W of R is the subgroup of the orthogonal group of E generated by the reﬂections s (α R). That is, for arbitrary positive integer k N α ∈ ∈ and any non-negative integer n ,..., n , sn1 snk W, i.e., 1 k α1 · · · αk ∈

n1 nk W = s s : sα R for j = 1,..., k N . α1 · · · αk j ∈ ∈ n o Deﬁnition 4.3. If (E, R) is a root system, then for each root α R, the co-root α is the vector ∈ ∨ given by α α∨ := 2 . α, α h i The set of all co-roots is denoted R∨ and is called the dual root system to R.

β,α Z It is easily seen that β, α∨ = 2 hβ,βi for α, β R. Besides, we see that h i h i ∈ ∈ 4 α∨, α∨ = . α, α h i In fact, if R is a rooty system, then R∨ is also a root system and the Weyl group for R∨ is the same as the Weyl group for R. Furthermore, R∨∨ = R. Thus

α α = 2 ∨ . α , α h ∨ ∨i Deﬁnition 4.4. If (E, R) is a root system, a subset ∆ of R is called a base if the following conditions hold:

(i) ∆ is a basis for E as a vector space.

56 (ii) Each root α R can be expressed as a linear combination of elements of ∆ with integer ∈ coefﬁcients and in such a way that the coefﬁcients are either all non-negative or all non- positive.

The roots for which the coefﬁcients are non-negative are called positive roots and the others are called negative roots (relative to the base ∆). The set of positive roots relative to a ﬁxed base ∆ is denoted by R+ and the set of negative roots is denote by R . Thus R = R+ R . The elements − ⊔ − of ∆ are called the positive simple roots.

Note that if ∆ is a base for R, then the set of all co-roots α (α ∆) is a base for the dual root ∨ ∈ system R . We also see that if ∆ is a base, then W is generated by the reflections s with α ∆. ∨ α ∈ Definition 4.5. An element µ of E is an integral element if for all α R, µ, α Z. If ∆ is a base ∈ h ∨i ∈ for R, an element µ of E is dominant (relative to ∆) if µ, α > 0 for all α ∆ and strictly dominant h i ∈ if µ, α > 0 for all α ∆. h i ∈ Definition 4.6. Let ∆ = α ,..., α be a base. Then the fundamental weights (relative to ∆) are { 1 n} the elements µ1,..., µn with the property that

µi, α∨j = δij, i, j = 1,..., n. D E That is, µ ,..., µ can be viewed as the dual base of ∆ := α ,..., α . { 1 n} ∨ { 1∨ n∨} Deﬁnition 4.7. Let ∆ = α ,..., α be a base for R and R+ the associated set of positive roots. { 1 n} We then let ρ denote half the sum of the positive roots and let σ denote half the sum of the positive co-roots: 1 1 ρ = ∑ α, σ = ∑ α∨. 2 α R+ 2 α R+ ∈ ∈ They are often called the Weyl vectors.

One important result for Weyl vector is that ρ are strictly dominant integral element; indeed,

ρ, α∨ = 1, α∨ ∆∨. (4.1) ∀ ∈

Similarly,

σ, α = 1, α ∆. (4.2) h i ∀ ∈ It is well-known that the Weyl vector can be expressed as the sum of fundamental weights:

n ρ = ∑ µj. (4.3) j=1

57 Deﬁnition 4.8. Let ∆ = α ,..., α be a base for R. For a positive root α R+, the height of α is { 1 n} ∈ deﬁned by

n ht(α)= ∑ kj Z+, j=1 ∈

n where α = ∑j=1 kjαj for non-negative integers k1,..., kn. That is,

n n ht ∑ kjαj = ∑ kj Z+. j=1 ! j=1 ∈

Apparently, ht(α)= 1 if and only if α ∆. ∈ With these notations, we can show that ρ, α = ht(α ) for any α R+. Note that the base h ∨i ∨ ∈ ∆ = α ,..., α for the root system R is given, then ∆ = α ,..., α is known for the dual { 1 n} ∨ { 1∨ n∨} root system R . Now for any α R+, α is also positive co-root and it is can be written as ∨ ∈ ∨ n N α∨ = ∑ kjα∨j , kj , j = 1,..., n. j=1 ∈

By using (4), we see that

n n n n n ρ, α∨ = ∑ µi, ∑ kjα∨j = ∑ kj µi, α∨j = ∑ kjδij = ∑ kj = ht(α∨). *i=1 j=1 + i,j=1 i,j=1 j=1 D E Similarly, σ, α = ht(α) for α R+. h i ∈ We also have to recall the following polynomial function P : E R given by → P(X)= ∏ α, X . α R+ h i ∈ Note that P has an important property:

P(w X)= det(w)P(X), w W, X E. · ∀ ∈ ∀ ∈ Indeed, for any w W, we have ∈ 1 P(w X)= ∏ α, w X = ∏ w− α, X . · α R+ h · i α R+ · ∈ ∈ D E 1 Suppose ﬁrst that w = w− = sβ, where β is a positive simple root. Since sβ permutes the positive roots different from β, whereas s β = β. Thus β · −

P(sβ X)= ∏ α, X β, X = ∏ α, X . · α R+ β h i × h− i − α R+ h i ∈ \{ } ∈

58 This because the determinant of a reﬂection is 1. Note W is generated by all positive simple − roots α. Thus for w W, it can be written as the product of some positive simple roots: w = ∈ s s . Then α1 · · · αk

P(w X) = P(sα1 sαk X)= P(sα1 sαk 1 X) · · · · · − · · · − · = · · · = ( 1)k P(X)= det(s s )P(X) − α1 · · · αk = det(w)P(X).

This means that P is alternating, or skew-symmetric. Next we can start our focus, i.e., computing the volume of a compact Lie group. Let G be a compact Lie group and let g be its Lie algebra, thought of as the tangent space TeG to G at the identity element e. Choose a Lebesgue measure µL on the vector space g. By means of a chart of G at e, we can construct from µL a translation-invariant measure on a neighborhood of e in

G, and then we can extend this by translation in G to a Haar measure µHaar on G. The purpose of this section is to establish a formula (Eq. (4.4) below), obtained by Macdonald in 1980, for

µHaar(G), the volume of G relative to the measure µHaar, as a function of µL. There are two ingredients in the formula. Firstly, from a suitably chosen Chevalley basis (i.e. root vectors) of the complexiﬁcation of g, we can construct an "integer lattice" gZ, which is a lattice in g and a Lie algebra over Z. By abuse of notation, let µL(g/gZ) denote the volume

(with respect to µL) of a fundamental parallelepiped for gZ in g. Secondly, it is well-known that the manifold G has the same cohomology, apart from torsion, as a product of odd-dimensional spheres, say of dimensions r1,..., rn. We shall prove the following result:

Theorem 4.9 ([22]). It holds that

n rj µHaar(G)= µL(g/gZ) ∏ vol (S ) (4.4) j=1

rj rj rj+1 where vol (S ) is the superﬁcial measure of the unit sphere S in R , that is (since rj = 2mj + 1 is mj+1 odd) vol (Srj ) = vol S2mj+1 = 2π . By abuse of notation, we record the above fact as Γ(mj+1)

n r vol(G)= vol(g/gZ) ∏ vol (S j ) . j=1

Proof. Let T be a maximal torus in G and let t g be its Lie algebra. Let d = dim(G), n = dim(T). ⊂ The Lebesgue measure µL on g determines a Lebesgue measure (also denoted by µL) and hence 1 a Haar measure (also denoted by µHaar) on T . Let tZ be the lattice in t such that the kernel of

1Note: an inner product on g does determine Lebesgue measures on both g and t.

59 exp : t T is kerexp = 2πtZ, then clearly → n µHaar(T)= µL(t/2πtZ)=(2π) µL(t/tZ), (4.5) n vol(T)= vol(t/2πtZ)=(2π) vol(t/tZ). (4.6)

Let R be the set of roots of G relative to T, and let W be the Weyl group. The roots are real linear forms on t, integer-valued on the lattice tZ. Fix a system of positive roots, and let P = ∏α>0 α, a homogeneous polynomial function on t of degree N = 1 (d n), i.e., the number of all positive 2 − roots. Let ξ, η be a positive deﬁnite inner product on g which is invariant under the adjoint action h i of G and such that the cube generated by an orthonormal basis of g has unit volume relative to µ . Let ξ = ξ, ξ . L k k h i If φ is a suitablep G-invariant function on g, we have [6]

vol(G/T) 2 φ(ξ)dµL(ξ)= φ(τ)P(τ) dµL(τ) (4.7) g W t Z | | Z be the counterpart for g of Weyl’s integration formula. By taking φ(ξ) = exp 1 ξ, ξ , we − 2 h i obtain

d 1 1 vol(G) 1 2 (2π) = exp ξ, ξ dµL(ξ)= exp τ, τ P(τ) dµL(τ) (4.8) g −2 h i W vol(T) t −2 h i q Z | | Z To calculate this integral, we proceed as follows. Let x, y = ∑ x y be the usual inner product h i j j j on Rn. This extends to a scalar product on the algebra S(Rn) of polynomial functions on Rn, such that xα, xβ = α!δ for any two multi-indices α =(α ,..., α ), β =(β ,..., β ) Nn. Here αβ 1 n 1 n ∈ ∏n ∏n α!:= j= 1 αj! and δαβ := j=1 δαj βj . Indeed, such scalar product is equivalently deﬁned [12] by

p, q := p(∂)q = , h i∂ |x 0 where p, q R[x ,..., x ] are homogeneous polynomials of the same order and p(∂) means that ∈ 1 n x is replaced by ∂ in p(x) p(x ,..., x ). For instance, a generic polynomial in R[x ,..., x ] j j ≡ 1 n 1 n can be written as n α α αj p(x)= ∑ cαx , x := ∏ xj . α j=1

α ∂ j βj Thus we know that αj xj = αj!δαj βj , and ∂xj xj=0

n αj n n αj n α β ∂ βj ∂ βj x , x = ∏ ∏ x = ∏ x = ∏ αj!δα β = α!δαβ.  αj  j αj j j j j=1 ∂xj j=1 ! j=1 ∂xj j=1 D E (x1,...,xn)=0  

60 Let γ be the (Gaussian) measure on Rn deﬁned by

1 1 dγ(x)= exp x, x [dx] n −2 h i (2π) where [dx] is Lebesgue measure. For ap function f on Rn, let f = f γ, i.e., ∗ ∗

f ∗(x)= f (x y)dγ(y) Rn Z − whenever the integral is deﬁned. Then for all f , g S(Rn) we have ∈ ∆ f ∗ = e 2 f (4.9) where ∆ is the Laplace operator, and

f ∗(ix)g∗(ix)dγ(x)= f , g . (4.10) Rn ∂ Z h i α ξα To prove Eq. (4.9), we may assume by linearity that f is a monomial x , i.e. the coefﬁcient of α! x,ξ in eh i. Indeed, n ∞ (∑n x ξ )m x,ξ j=1 j j eh i = exp ∑ xjξj = ∑ , j=1 ! m=0 m! where m n m α α ∑ xjξj = ∑ x ξ j=1 ! α: α =m α | | m m! α α1 αn α α1 αn n for ( α ) = α ! α ! , x = x1 xn , and ξ = ξ1 ξn ; α = ∑j=1 αj and α! = α1! αn!. Thus 1 ··· n · · · · · · | | · · · ∞ α x,ξ ξ α eh i = ∑ ∑ x . m=0 α: α =m α! | | x,ξ α This means that eh i is a linear combination of monomials x . Hence it is enough to verify x,ξ Eq. (4.9) when f (x)= eh i; indeed,

f ∗(x) = f (x y)dγ(y) Rn Z − x y,ξ 1 1 = eh − i exp y, y [dy] Rn n −2 h i Z (2π) x,ξ y,ξ 1 1 = eh i eh−p i exp y, y [dy] Rn n −2 h i Z (2π) implying p

n x,ξ 1 1 2 f ∗(x) = eh i ∏ exp (yj + 2ξjyj) dyj √ Rn −2 j=1 2π Z n 1 2 1 1 x,ξ 2 ξj 2 = eh i ∏ e exp (yj + ξj) d(yj + ξj) √ Rn −2 j=1 2π Z

61 then a simple calculation shows that f (x)= e x,ξ exp 1 ξ, ξ . Then ∗ h i 2 h i n ∂2 n n ∂2 n ∆ x,ξ eh i = ∑ 2 exp ∑ ξj xj = ∑ 2 exp ∑ ξj xj j=1 ∂xj ! j=1 ! j=1 ∂xj j=1 !! n 2 x,ξ x,ξ = ∑ ξj eh i = ξ, ξ eh i, j=1 h i

∆ which means that f ∗(x)= e 2 f (x) since

∞ m ∆ ∆ x,ξ 1 ∆ x,ξ e 2 f (x) = e 2 eh i = ∑ eh i m! 2 m=0 ∞ 1 ∞ 1 ∆m x,ξ m x,ξ = ∑ m eh i = ∑ m ξ, ξ eh i m=0 m!2 m=0 m!2 h i 1 ξ,ξ x,ξ = e 2 h ieh i = f ∗(x).

x,ξ x,η Likewise, it is enough to verify Eq. (4.10) when f (x) is replaced by eh i and g(x) by eh i (and f , g by e ξ,η ). Indeed, for this case, h i∂ h i ∞ α ∞ β x,ξ x,η ξ α η β f , g ∂ = eh i, eh i = ∑ ∑ x , ∑ ∑ x ∂ h i *i=0 α: α =i α! j=0 β: β =j β! + D E | | | | ∂ ∞ ∞ ξα ηβ = ∑ ∑ ∑ ∑ xα, xβ ∂ i=0 α: α =i j=0 β: β =j α! β! | | | | D E ∞ α α ξ η ξ,η = ∑ ∑ = eh i j=0 α: α =j α! | | and

1 ξ,ξ ix,ξ 1 η,η ix,η f ∗(ix)g∗(ix)dγ(x) = e 2 h ieh ie 2 h ieh idγ(x) Rn Rn Z Z 1 ξ,ξ 1 η,η i x,ξ η = e 2 h ie 2 h i e h − idγ(x) Rn Z 1 ξ,ξ 1 η,η 1 ξ η,ξ η ξ,η = e 2 h ie 2 h ie− 2 h − − i = eh i where we used the fact that

∞ 2 1 at2 ity 1 y e− e− dt = exp . √2π ∞ √2a −4a Z− Suppose now that f is a harmonic homogeneous polynomial, i.e. ∆ f = 0. Then ∆k f = 0 for k > 1 and ∞ ∆ ∆k = 2 = = f ∗ e f ∑ k f f k=0 2 k!

62 by Eq. (4.9), and therefore from Eq. (4.10) we obtain

f , f = f ∗(ix) f ∗(ix)dγ(x)= f (ix) f (ix)dγ(x) ∂ Rn Rn h i Z Z = f (x)2dγ(x). Rn Z Now we have already proven that if f is a harmonic homogeneous polynomial, then we have

2 1 2 1 x,x f , f = f (x) dγ(x)= f (x) e− 2 h i[dx]. (4.11) ∂ Rn n Rn h i Z (2π) Z The polynomial P is skew-symmetric with respectp to the Weyl group W, and the inner product

(on t or t∗) is W-invariant. It follows that ∆P = 0, and hence from Eq. (4.11) that

1 2 n exp τ, τ P(τ) dµL(τ)= (2π) P, P ∂ , t −2 h i h i Z q where

N N P, P ∂ = ∏ α, ∏ α = ∏ αj, ∏ αj h i *α>0 α>0 + *j=1 j=1 +

def = α α , α 1 α 1 . ∑ 1 N σ− (1) σ− (N) σ S ⊗···⊗ ⊗···⊗ ∈ N D E Now substituting this integral value in Eq. (4.8), we obtain

1 vol(G) (2π)d = (2π)n P, P . W vol(T) h i∂ q | | q That is,

vol(G) d n W N W =(2π) −2 | | =(2π) | | . (4.12) vol(T) P, P P, P h i∂ h i∂ d n Note here that N = − . Now the number P, P has already been calculated by Steinberg [9]: 2 h i n N P, P ∂ = W P(ρ)= 2− W ∏ mj! ∏ α, α h i | | | | j=1 α>0 h i

1 where ρ := 2 ∑α>0 α is half the sum of the positive roots, and the mj are the exponents of G. Indeed, from Weyl’s denominator formula:

wρ 1 α 1 α ∑ sign(w)e = ∏ e 2 e− 2 , w W α>0 − ∈ we see that for any chosen X such that α, X = 0 for all α > 0, we have that h i 6 wρ,tX ρ,tX α,tX ∑ sign(w)eh i = e−h i ∏ eh i 1 , t R. w W α>0 − ∀ ∈ ∈ 63 That is,

t wρ,X t ρ,X t α,X ∑ sign(w)e h i = e− h i ∏ e h i 1 w W α>0 − ∈ t α,X N t ρ,X e h i 1 = t e− h i ∏ − . α>0 t

Now using the Taylor expansion of ex:

∞ xk ex = ∑ , k=0 k! we see that et α,X 1 ∞ tk 1 α, X k h i − = α, X + ∑ − h i = α, X + o(t). t h i k=2 k! h i And thus

t α,X N t ρ,X e h i 1 N N t e− h i ∏ − = t (1 + o(t))(∏ α, X + o(t)) = t (P(X)+ o(t)). α>0 t α>0 h i

Taking the N-th derivative with respect to t on both sides, we get that

N t wρ,X ∑ sign(w) wρ, X e h i = N!P(X)+ o(t). w W h i ∈ Now taking the limit for t 0, we get that → ∑ sign(w) wρ, X N = N!P(X) ∑ sign(w)(wρ)N = N!P. w W h i ⇐⇒ w W ∈ ∈ Thus

1 N 1 N P, P ∂ = ∑ sign(w)(wρ) , ∏ α = ∑ sign(w) (wρ) , ∏ α h i N! *w W α>0 + N! w W * α>0 + ∈ ∈ 1 1 = ∑ sign(w)2 ρN, ∏ α = W N! ∏ ρ, α = W P(ρ) N! w W * α>0 + N! | | α>0 h i | | ∈ where we used the fact that the polynomial P is skew-symmetric with respect to the Weyl group W. We can also present another approach to the identity P, P = W P(ρ). Denote q (X) = h i∂ | | ρ wρ,X ∑w W sign(w)eh i Note that ∈

wρ,X wρ,X P(∂)qρ(0) = ∑ sign(w) P(∂)eh i = ∑ sign(w) ∏ ∂αeh i X=0 w W w W α>0 ∈ ∈ X=0

d wρ,tα = ∑ sign(w) ∏ t=0eh i = ∑ sign(w) ∏ wρ, α , w W α>0 dt w W α>0 h i ∈ ∈

64 i.e.,

P(∂)qρ(0) = ∑ sign(w) ∏ α, wρ = ∑ sign(w)P(wρ) w W α>0 h i w W ∈ ∈ = ∑ sign(w)2P(ρ)= W P(ρ). w W | | ∈ In addition,

1 1 2 α,X 2 α,X P(∂) ∏ e h i e− h i = P(∂)P(0)= P, P ∂ . > − h i α 0 X=0

Indeed,

∞ 2n+1 1 1 α 2 α 2 α ∏ e e− = ∏ α + ∑ 2n = ∏ α + terms of degree at least N + 1. > − > 2 (2n + 1)! ! > α 0 α 0 n=1 α 0 Note that P(∂) is the linear partial operator of order N, hence

1 1 2 α,X 2 α,X P(∂) ∏ e h i e− h i = P(∂)P(0)+ 0 = P, P ∂ . > − h i α 0 X=0

Thus

P, P = W P(ρ). h i∂ | | Now for any root α, 2 ρ, α h i = ρ, α∨ = ht(α∨), α, α h i the height of α , and it is known that if d 6 d 6 6 d are the degrees of the basic invariants, ∨ 1 2 · · · n where dj = mj + 1 [11], the number of roots of height k minus the number of roots of height k + 1 is just the number of mj’s equal to k (or the number of dj’s equal to k + 1) [15, 28, 21, 11]. That is,

α∨ > 0:ht(α∨)= k α∨ > 0:ht(α∨)= k + 1 = j [n] : m = k . − ∈ j Indeed, in order to show this result, we partition the number N of all positive roots via the height of all positive co-roots: h 1 − α∨ α > 0 = α∨ > 0:ht(α∨)= j , | j=1 G where h is the Coxeter number [11] for Weyl group W. Denote by n = α > 0:ht(α )= k . k |{ ∨ ∨ }| Thus we see that h 1 − N = ∑ nk, n1 > n2 > > nh 1. − k=1 · · · ℓ(w) ℓ It is shown that the Poincaré polynomial W(t) := ∑w W t , where (w) is the length of w W, ∈ ∈ can be factorized by two ways [21]:

1 t1+ht(α) n 1 tmj+1 W(t)= = . (4.13) ∏ − ht(α) ∏ − > 1 t 1 t α 0 − j=1 −

65 This implies that 1 + ht(α) n W = lim W(t)= ∏ = ∏(mj + 1). | | t 1 ht(α) → α>0 j=1 Note the above identity holds also for dual root system since W corresponds to both the original root system R and its dual root system R∨, it follows that n 1 + ht(α∨) ∏ = ∏(mj + 1). α>0 ht(α∨) j=1

Now partition [n] by the number of times k as exponents of W:

h 1 − [n]= , := j [n] : m = k . Ik Ik ∈ j k=1 G Then n2 nh 1 1 + ht(α ) 3 h − ∨ n1 n1 n2 n2 n3 nh 1 ∏ = 2 = 2 − 3 − h − ht(α ) 2 · · · h 1 · · · α>0 ∨ − and n 1 2 h 1 ∏(mj + 1)= 2|I |3|I | h|I − |. j=1 · · · Therefore we have

nk nk+1 = k , k = 1,..., h 2; nh 1 = h 1 . − |I | − − |I − |

In what follows, we calculate the value of ∏ > ht(α ). In fact, we have ht(α ) [1, h 1] N α 0 ∨ ∨ ∈ − ∩ and

n1 n2 nh 1 ∏ ht(α∨)= 1 2 (h 1) − . (4.14) α>0 · · · − Clearly,

n1 = h 1 + + 2 + 1 , |I − | · · · |I | |I | n2 = h 1 + + 2 , |I − | · · · |I | . .

nh 1 = h 1 . − |I − | Furthermore,

1 2 h 1 ∏ ht(α∨) = (1!)|I |(2!)|I | [(h 1)!]|I − | α>0 · · · −

= ∏ mj! ∏ mj! ∏ mj! × ×···× j 1 j 2 j h 1 ∈I ∈I ∈I − n = ∏ mj! = ∏ mj! = ∏ mj!. j 1 h 1 j [n] j=1 ∈I ⊔···⊔I − ∈

66 Thus 2 ρ, α n n ∏ h i = ∏ ρ, α∨ = ∏(dj 1)! = ∏ mj!. α>0 α, α α>0 j=1 − j=1 h i Therefore,

n 2 ρ, α α, α N P(ρ)= ∏ α, ρ = ∏ h i ∏ h i = 2− ∏ mj! ∏ α, α . h i α, α 2 h i α>0 α>0 h i α>0 j=1 α>0 That is,

n N P(ρ)= 2− ∏ mj! ∏ α, α . (4.15) j=1 α>0 h i

That is,

n N P, P = 2− ∏ dj! ∏ α, α . (4.16) h i j=1 α>0 h i

Hence we have vol(G) (2π)N 2π = = ∏ (4.17) vol(T) P(ρ) α, ρ α>0 h i a formula due to Harish-Chandra, and also, using Eq. (4.6),

+ (2π)Nvol(T) (2π)N nvol(t/tZ) vol(G) = = n P(ρ) 2 N ∏ m ! ∏ > α, α − j=1 j α 0 h i 4 n 2πmj+1 = vol(t/tZ) ∏ ∏ α, α m ! α>0 h i j=1 j n r = vol(t/tZ) ∏ α∨, α∨ ∏ vol (S j ) α>0 j=1

∑n 4 where rj = 2mj + 1 and j=1 mj = N and α,α = α∨, α∨ for α∨ is the coroot corresponding to α. h i h i Let gC, tC be the complexiﬁcations of g, t. For X = ξ + iη gC(ξ, η g) Let X¯ = ξ iη. There ∈ ∈ − exists a Chevalley basis of gC relative to tC in which the root vectors Xα satisfy X α = X¯ α. Let − ξα = Xα + X α, ηα = i(Xα X α), then the "integer lattice" gZ is spanned by the ξα and ηα for − − − α > 0, together with a basis of tZ; the 2N vectors ξα, ηα are orthogonal to each other and to t, and ξ = η = α . It follows that k α k k α k k ∨ k

vol(g/gZ)= vol(t/tZ) ∏ α∨, α∨ ; α>0

67 substituting this, we derive the desired formula:

n r vol(G)= vol(g/gZ) ∏ vol (S j ) . j=1

We are done.

Remark 4.10. Let (W, S) be an irreducible finite Coxeter system of rank n with S = s ,..., s { α1 αn } its set of simple reflections. The element c := s s W is called Coxeter transformation. α1 · · · αn ∈ Since all Coxeter transformations are conjugate, they have the same order, characteristic polynomial and eigenvalues. The order h of Coxeter elements is called the Coxeter number of W. For a fixed Coxeter element c W, if its eigenvalues are of the form: ∈

exp (2πim1/h) , ,exp (2πimn/h) with 0 < m 6 6 m < h, then the integers m ,..., m are called the exponents of W. We have 1 · · · n 1 n already known that for any irreducible root system R of rank n,

(i) mj + mn j+1 = h for 1 6 j 6 n; − (ii) m = 1, m = h 1; 1 n − (iii) the height of the highest root in R+ is given by m = h 1. n − Let k := α R+ ht(α)= j . The height distribution of R+ is deﬁned as a multiset of positive j |{ ∈ | }| integers:

k1,..., kh 1 . { − } + Apparently, (k1,..., kh 1) N = R . It is well-known that the exponents of the Weyl group − ⊢ | | W are given by the dual partition of the height distribution of R+ is given by a multiset of non- negative integers [15, 28, 21, 1]:

n k1 k1 k2 kh 2 kh 1 kh 1 (0) − , (1) − ,..., (h 2) − − − , (h 1) − , − − n o where (a)b means the integer a appears exactly b times. If k k > 0, then m appears exactly j − j+1 j k k times. Otherwise, m does not appear if k = k . j − j+1 j j j+1 Note that in the Macdonald’s method to the volume of a compact Lie group, we see that n n n d n d = ∑j=1 rj = 2 ∑j=1 mj + n. It follows that ∑j=1 mj = − = N. Besides, since mj + mn j+1 = h, 2 − it follows that n 1 N = ∑ mj = nh. j=1 2

68 F F C Remark 4.11. Denote all polynomials in x = (x1,..., xd) with coefﬁcients in , where = or R F , by [x1,..., xd]. A polynomial is called homogeneous if all the monomials appearing in this polynomial have the same total degree. Denote the space of homogeneous polynomials of degree n in d variables by (Fd). That is, Pn

d α d n(F )= P P(x)= ∑ cαx , where x F . P ( | α: α =n ∈ ) | |

d n+d 1 d Note that dim (R )= ( − ). For any two homogeneous polynomials P, Q (F ), a scalar Pn n ∈Pn product may be deﬁned as follows:

P, Q := ∑ p¯αqαα!, h i α: α =n | | α α α β where P(x)= ∑α: α =n pα x and Q(x)= ∑α: α =n qα x . From this deﬁnition, we see that x , x = | | | | α!δ . In particular, if P, Q (Rd), the scalar product can be rewritten as [12, 13] αβ ∈Pn P, Q =(P(∂)Q)(0). h i α β Indeed, since (∂ x )x=0 = α!δαβ and

α β α β P(∂)Q = ∑ pα∂ ∑ qβx = ∑ pαqβ ∂ x , α: α =n ! β: β =n ! α,β: α = β =n | | | | | | | | it follows that

(P(∂)Q)(0)= ∑ pαqβ α!δαβ = ∑ pαqαα!, α,β: α = β =n α: α =n | | | | | | which means that P, Q =(P(∂)Q)(0). h i As an example, let P(x) = α , x α , x and Q(x) = β , x β , x be real polynomials in h 1 ih 2 i h 1 ih 2 i (R3), where α = (a1, a2, a3) and β = (b1, b2, b3) for j = 1, 2. In what follows, we calculate the P2 j j j j j j j j scalar product P, Q . Apparently, h i 3 3 i j i i 2 i j j i P(∂) = ∑ a1a2∂i∂j = ∑ a1a2∂i + ∑ a1a2 + a1a2 ∂i∂j, < i,j=1 i=1 i j 3 3 i j i i 2 i j j i Q(x) = ∑ b1b2xixj = ∑ b1b2xi + ∑ b1b2 + b1b2 xi xj. < i,j=1 i=1 i j It follows that

3 i i i i i j j i i j j i P(∂)Q(x) = 2 ∑ a1a2b1b2 + ∑ a1a2 + a1a2 b1b2 + b1b2 < i=1 i j

69 Now we see that

α , β α , β + α , β α , β h 1 1ih 2 2i h 1 2ih 2 1i 3 3 3 3 i i j j i i j j = ∑ a1b1 ∑ a2b2 + ∑ a1b2 ∑ a2b1 i=1 ! j=1 ! i=1 ! j=1 ! 3 3 i i i i i i j j i i i i i i j j = ∑ a1b1a2b2 + ∑ a1b1a2b2 + ∑ a1b2a2b1 + ∑ a1b2a2b1 "i=1 i=j # "i=1 i=j # 6 6 3 i i i i i i j j i i j j = 2 ∑ a1a2b1b2 + ∑ a1b1a2b2 + ∑ a1b2a2b1. i=1 i=j i=j 6 6 That is,

3 i i i i i i j j i i j j α1, β1 α2, β2 + α1, β2 α2, β1 = 2 ∑ a1a2b1b2 + ∑ a1b1a2b2 + ∑ a1b2a2b1 h ih i h ih i i=1 i=j i=j 6 6 3 i i i i i i j j j j i i i i j j j j i i = 2 ∑ a1a2b1b2 + ∑ a1b1a2b2 + a1b1a2b2 + ∑ a1b2a2b1 + a1b2a2b1 < < i=1 i j i j 3 i i i i i i j j j j i i i i j j j j i i = 2 ∑ a1a2b1b2 + ∑ a1b1a2b2 + a1b1a2b2 + a1b2a2b1 + a1b2a2b1 < i=1 i j 3 i i i i i j j i i j j i = 2 ∑ a1a2b1b2 + ∑ a1a2 + a1a2 b1b2 + b1b2 . < i=1 i j Therefore, we can conclude that

(P(∂)Q)(0) = α , β α , β + α , β α , β h 1 1ih 2 2i h 1 2ih 2 1i = α α , β 1 β 1 . ∑ 1 2 σ− (1) σ− (2) σ S ⊗ ⊗ ∈ 2 D E n k We can generalize this result to the case where P(x) = ∏j=1 αj, x and Q(x) = ∏j=1 βj, x , in k k short, P = ∏j=1 αj and Q = ∏j=1 βj. We have the following:

P, Q =(P(∂)Q)(0)= α α , β 1 β 1 . ∑ 1 k σ− (1) σ− (k) h i σ S ⊗···⊗ ⊗··· ∈ k D E Remark 4.12. Consider the real vector space V generated by the roots. , denotes the inner h· ·i product on V which is invariant under the operation of the Weyl group W. The given inner product on V extends to one on the symmetric algebra of V by the formula:

k k k α , β = α , β 1 = α , β 1 α , β 1 (4.18) ∏ j ∏ j ∑ ∏ j σ− (j) ∑ 1 σ− (1) k σ− (k) *j=1 j=1 + σ S j=1 σ S · · · ∈ k D E ∈ k D E D E

= α α , β 1 β 1 (4.19) ∑ 1 k σ− (1) σ− (k) σ S ⊗···⊗ ⊗··· ∈ k D E

70 where α , β V. In view of this, we see that j j ∈

∏ α, ρN = ∑ ∏ α, ρ = N! ∏ α, ρ . *α>0 + σ S α>0 h i α>0 h i ∈ N In the following, we show that [23]

n ∏ α, ∏ α∨ = ∏ dj!, (4.20) *α>0 α>0 + j=1

2α 1 2α σ− (j) where α∨ = α,α such that α, α∨ = 2. Indeed, by replacing each ασ∨ 1(j) by in the h i − α 1 ,α 1 h i σ− (j) σ− (j) deﬁnition, we see that D E

def ∏ α, ∏ α∨ = ∑ α1 αN, α∨ 1 α∨ 1 σ− (1) σ− (N) *α>0 α>0 + σ S ⊗··· ⊗··· ∈ N D E 2N = ∑ α1 αN, ασ 1(1) ασ 1(N) ⊗··· − ⊗··· − N σ S ∏ α 1 , α 1 ∈ N D E j=1 σ− (j) σ− (j) 2N D E = ∑ α1 αN, α 1 α 1 . ∏ σ− (1) σ− (N) α>0 α, α σ S ⊗··· ⊗··· h i ∈ N D E Therefore 2N n ∏ α, ∏ α∨ = P, P = ∏ dj!. ∏ > α, α h i *α>0 α>0 + α 0 h i j=1 Note that in deriving the volume formula of a compact Lie group, we used the following fact which is necessarily recorded here:

Theorem 4.13 ([6]). The mapping that assigns to φ in C0(g), the space of continuous functions with compact support on g, the function on G/T t × (gT, X) φ(Ad (X)) det(ad ) (4.21) 7−→ g X g/t extends to a topological isomorphism: L1(g) L1(G/T t)W, where now sT W acts on (gT, X) by → × ∈ sending it to (gs 1T,Ad (X)). Moreover, if φ L1(g), then: − s ∈ 1 φ(X)dX = φ(Adg(X))d(gT) det(adX)g/t dX. (4.22) g W t Z Z ZG/T | | It is easily seen that if φ is G-invariant, i.e., φ(Ad (X)) = φ(X) for all g G, then g ∈ vol(G) φ(Ad (X))d(gT)= φ(X)vol(G/T)= φ(X) . g (T) ZG/T vol Thus 1 vol(G) φ(X)dX = φ(X) det(adX)g/t dX, g W (T) t Z vol Z | |

71 2 where Jacobian det(adX)g/t = P(X) for P = ∏α>0 α. More simpler proof about Macdonald’s volume formula can be found in [10]. With Macdon- ald’s volume formula, one can derive all volume formulas for orthogonal groups and unitary groups. Although these methods is a digression to the subject of the present paper, I also collect these material together for reference. Once again, we will use the Macdonald’s volume formula for a compact Lie group to derive the volume formula for the special unitary group G := SU(n). The special unitary group SU(n) is of rank n 1. Its root system consists of 2(n) roots spanning a (n 1)-dimensional Euclidean − 2 − space. Here, we use n redundant coordinates instead of n 1 to emphasize the symmetries of − the root system (the n coordinates have to add up to zero). In other words, we are embedding this n 1 dimensional vector space in an n-dimensional one. The maximal torus T of G is − equivalently identiﬁed with SU(1) n, and its Lie algebra is t = Rn 1 of rank n 1. The lattice × ∼ − − Zn 1 tZ ∼= − . Thus vol(t/tZ)= 1. The set of all positive roots of g is given by

+ R = α t∗ : α (X)= x x for any X = diag(x ,..., x ), i < j . (4.23) ij ∈ ij i − j 1 n Thus for any α R+, we have α, α = 2. Note that N = R+ = (n). Then the Coxeter number ∈ h i | | 2 2N h = n 1 = n. The exponents of the special unitary group SU(n) are: 1,..., n 1 . Note that − { − } n 1 4 − r vol (G)= vol(t/tZ) ∏ ∏ vol (S j ) , HS α, α α>0 h i j=1 where rj = 2j + 1. It follows that

n 1 (n) − 2j+1 volHS(G)= 2 2 ∏ vol S . j=1 Since U(n)= U(1) SU(n), it follows from vol (U(1)) = vol S1 that × HS n 1 (n) 2j 1 volHS(U(n)) = vol S volHS(SU(n)) = 2 2 ∏ vol S − . j=1 That is, n(n+1) (2π) 2 volHS(U(n)) = n . ∏j=1 Γ(j) Besides, we can also employ Harish-Chandra’s volume formula for ﬂag manifold, we get that

N N Tn 2π (2π) (2π) volHS(U(n)/ ) = ∏ = = n α, ρ P(ρ) 2 N ∏ m ! ∏ > α, α α>0 h i − j=1 j α 0 h i n(n 1) N − (2π) (2π) 2 = = , 2 N0!1! (n 1)!2N ∏n Γ(j) − · · · − j=1 72 implying that

n(n 1) n(n 1) − − Tn (2π) 2 n (2π) 2 volHS(U(n)) = vol( ) n =(2π) n . ∏j=1 Γ(j) ∏j=1 Γ(j) Again, we also see that n(n+1) (2π) 2 volHS(U(n)) = n . ∏j=1 Γ(j) Remark 4.14. We can rewrite the above results as two partitions of N, the number of all positive roots, (n1,..., nh 1) N and (mn,..., m1) N, where nj denotes the number of j-th row; mi the − ⊢ ⊢ number of n j + 1-th column or the n j + 1-th row of the conjugate partition of the original − − partition (m ,..., m ) N. n 1 ⊢

5 Applications

The present section is directly written based on [30, 37, 38]. The results were already obtained. We just here add some interpretation, from my angle, about them since the details concerning computation therein are almost ignored.

5.1 Hilbert-Schmidt volume of the set of mixed quantum states

Any unitary matrix may be considered as an element of the Hilbert-Schmidt space of operators with the scalar product U, V = Tr U V . This suggests the following deﬁnition of an h iHS ∗ invariant metric of the unitary group (n): denote dG := U∗dU, then e e U e e ds2 := dG,dG = Tre dGe 2 e, (5.1) HS − D E implying e e e

n 2 n 2 n 2 2 ds = ∑ dGij = ∑ dGjj + 2 ∑ dGij . (5.2) < i,j=1 j=1 i j e e e Since dG = dG, it follows that ∗ − n 2 n 2 n 2 2 e des = ∑ dGjj + 2 ∑ d(Re(Gij)) + 2 ∑ d(Im(Gij)) . (5.3) < < j=1 i j i j e e e This indicates that the Hilbert-Schmidt volume element is given by

n(n 1) n n(n 1) − − dν = 2 2 ∏ d Im(Gjj) ∏ d(Re(Gij))d(Im(Gij)) = 2 2 [dG], (5.4) × < j=1 i j e e e e 73 that is,

n(n 1) − volHS ( (n)) := dν = 2 2 [dG] (5.5) U (n) (n) ZU ZU n n(n+1) n(n 1) 2 π 2 e = 2 2− (5.6) × 1!2! (n 1)! n(n+1) · · · − (2π) 2 = . (5.7) 1!2! (n 1)! · · · − Finally we have obtained the Hilbert-Schmidt volume of unitary group:

n(n+1) n(n 1) (2π) 2 vol ( (n)) = 2 2− vol ( (n)) = . (5.8) HS U U 1!2! (n 1)! · · · −

We compute the volume of the convex (n2 1)-dimensional set D (Cn) of density matrices of − size n with respect to the Hilbert-Schmidt measure. The set of mixed quantum states D (Cn) consists of Hermitian, positive matrices of size n, normalized by the trace condition

n n n D (C ) = ρ : C C ρ∗ = ρ, ρ > 0, Tr (ρ) = 1 . (5.9) { → | } It is a compact convex set of dimensionalitye (en2 e1)e. Any densitye matrix may be diagonalized − by a unitary rotation, ρ = UΛU , where U (n) and Λ = diag(λ ,..., λ ) for λ R+. ∗ ∈ U 1 n j ∈ n Since Tr (ρ) = 1, it follows that ∑j=1 λj = 1, so the spectra space is isomorphic with a (n 1)- e e e − e R+ n dimensional probability simplex ∆n 1 := p : ∑j=1 pj = 1 . e − ∈ Let B be a diagonal unitary matrix.n Since ρ = UBΛB∗Uo∗, in the generic case of a non- degenerate spectrum (i.e. with distinct non-negative eigenvalues), the unitary matrix U is deter- e e e e e e mined up to n arbitrary phases entering B. On the other hand, the matrix Λ is deﬁned up to a e permutation of its entries. The form of the set of all such permutations depends on the character e of the degeneracy of the spectrum of ρ.

Representation ρ = UBΛB∗U∗ makes the description of some topological properties of the 2 n e (n 1)-dimensional space D (C ) easier. Identifying points in ∆n 1 which have the same com- − e e e e e − ponents (but ordered in a different way), we obtain an asymmetric simplex ∆n 1. Equivalently, − one can divide ∆n 1 into n! identical simplexes and take any one of them. The asymmetric − e simplex ∆n 1 can be decomposed in the following natural way: −

e ∆n 1 = δd ,...,d , (5.10) − 1 k d + +d =n 1 ···[ k e 74 where k = 1,..., n denotes the number of different coordinates of a given point of ∆n 1, d1 − the number of occurrences of the largest coordinate, d2 the number of occurrences of the sec- e ond largest etc. Observe that δd1,...,dk is homeomorphic with the set Gk, where G1 is a single point, G2 is a half-closed interval, G3 an open triangle with one edge but without corners and, generally, G is an (k 1)-dimensional simplex with one (k 2)-dimensional hyperface without k − − boundary (the latter is homeomorphic with an (k 2)-dimensional open simplex). There are n − ordered eigenvalues: λ > λ > > λ , and n 1 independent relation operators "larger(>) or 1 2 · · · n − n 1 n 1 equal(=)", which makes altogether 2 − different possibilities. Thus, ∆n 1 consists of 2 − parts, − n 1 out of which (m− 1) parts are homeomorphic with Gm, when m ranges from 1 to n. − e Cn Let us denote the part of the space D ( ) related to the spectrum in δd1,...,dk (k different eigenvalues; the largest eigenvalue has d1 multiplicity, the second largest d2 etc) by Dd1,...,dk . A mixed state ρ with this kind of the spectrum remains invariant under arbitrary unitary rotations performed in each of the dj-dimensional subspaces of degeneracy. Therefore the unitary matrix e B has a block diagonal structure with k blocks of size equal to d1,..., dk and e D [ (n)/( (d1) (d ))] G , (5.11) d1,...,dk ∼ U U ×···×U k × k where d + + d = n and d > 0 for j = 1,..., k. Thus D (Cn) has the structure 1 · · · k j D (Cn) D [ (n)/( (d ) (d ))] G , (5.12) ∼ d1,...,dk ∼ U U 1 ×···×U k × k d + +d =n d + +d =n 1 ···[ k 1 ···[ k where the sum ranges over all partitions (d , , d ) n of n. The group of rotation matrices B 1 · · · k ⊢ equivalent to (d1) (dk) is called the stability group of (n). U ×···×U U e Note also that the part of D1,...,1 represents a generic, non-degenerate spectrum. In this case all elements of the spectrum of ρ are different and the stability group is equivalent to an n-torus

n D (n)/( (1)× G . (5.13) e1,...,1 ∼ U U × n The above representation of generic states enables us to define a product measure in the space D (Cn) of mixed quantum states. To this end, one can take the uniform (Haar) measure on (n) U and a certain measure on the simplex ∆n 1. − The other 2n 1 1 parts of D (Cn) represent various kinds of degeneracy and have measure − − zero. The number of non-homeomorphic parts is equal to the number P(n) of different representations of the number n as the sum of positive natural numbers. Thus P(n) gives the number of different topological structures present in the space D (Cn). To specify uniquely the unitary matrix of eigenvectors U, it is thus sufficient to select a point on the coset space (n) ne Fl := (n)/ (1)× , C U U called the complex flag manifold. The volume of this complex flag manifold is:

75 n(n 1) − (n) volHS ( (n)) (2π) 2 volHS FlC = U n = . (5.14) volHS ( (1)) 1!2! (n 1)! U · · · −

The generic density matrix is thus determined by (n 1) parameters determining eigenvalues − and (n2 n) parameters related to eigenvectors, which sum up to the dimensionality (n2 1) − − of D (Cn). Although for degenerate spectra the dimension of the flag manifold decreases, these cases of measure zero do not influence the estimation of the volume of the entire set of density matrices. In this subsection, we shall use the Hilbert-Schmidt metric. The infinitesimal distance takes a particularly simple form

ds2 = dρ 2 = dρ,dρ (5.15) HS k kHS h iHS valid for any dimension n. Making use ofe the diagonale forme ρ = UΛU†, we may write

dρ = U dΛ + [UdU, Λ] U∗ e e e (5.16) Thus the inﬁnitesimal distance cane bee rewrittene as e e n n 2 2 2 2 dsHS = ∑ dλj + 2 ∑(λi λj) i U∗dU j (5.17) < − j=1 i j D E n n 2 2 2 e e = ∑ dλj + 2 ∑(λi λj) i dG j , (5.18) < − j=1 i j D E e n n where dG = U∗dU. Apparently, ∑j=1 dλj = 0 since ∑j=1 λj = 1. Thus

n 1 n 2 e e e 2 − 2 dsHS = ∑ dλimijdλj + 2 ∑(λi λj) dGij . (5.19) < − i,j=1 i j

e The corresponding volume element gains a factor det(M), where M = [mij] is the metric in the 2 (n n)-dimensional simplex ∆n 1 of eigenvalues.p Note that − − 1 1 1 · · ·  1 1 1  1 · · · M = n + . . . . .  . . .. .       1 1 1   · · ·    Therefore the Hilbert-Schmidt volume element is given by

n 1 − 2 dVHS = √n ∏ dλj ∏(λi λj) ∏ 2d Re(Gij) d Im(Gij) . (5.20) < − < j=1 i j i j

e e 76 Then

n(n 1) n 1 − 2 − dVHS = √n2 2 ∏(λi λj) ∏ dλj [dG1] Z Z i

That is, respect to Hilbert-Schmidt measure, the volume of the set of mixed quantum states is

n(n 1) n − n volHS (D (C )) = √n2 2 vol (D (C )) , i.e.

n(n 1) Γ Γ Γ n − (1) (2) (n) vol (D (C )) = √n(2π) 2 · · · . (5.21) HS Γ(n2)

We see from the above discussion that the obtained formula of volume depends the used measure. If we used the Hilbert-Schmidt measure, then we get the Hilbert-Schmidt volume of the set of quantum states [38]; if we used the Bures measure, then we get the Bures volume of the set of quantum states [29]. A special important problem is to compute the volume of the set of all separable quantum states, along this line, some investigation on this topic had already been made [35, 36]. There are some interesting topics for computing volumes of the set of some kinds of states, for instance, Milz also considered the volumes of conditioned bipartite state spaces [24], Link gave the geometry of Gaussian quantum states [16] as well. We can also propose some problems like this. Consider the following set of all states being of the form: 1 1 1 1 , 1 := ρ D C2 C2 : Tr (ρ )= 1 = Tr (ρ ) . C 2 2 2 2 12 ∈ ⊗ 1 12 2 2 2 12 Paratharathy characterized the extremale points of this sete[26]. He obtainede that all the extremal points of this convex set is maximal entangled states. That is, 0 ψ + 1 ψ | i| 0i | i| 1i. √2 It remains open to compute the volume of this convex set 1 1 , 1 1 . C 2 2 2 2 5.2 Area of the boundary of the set of mixed states

The boundary of the set of mixed states is far from being trivial. Formally it may be written as a solution of the equation det(ρ)= 0

77e which contains all matrices of a lower rank. The boundary ∂D (Cn) contains orbits of different dimensionality generated by spectra of different rank and degeneracy. Fortunately all of them are of measure zero besides the generic orbits created by unitary rotations of diagonal matrices with all eigenvalues different and one of them equal to zero;

Λ = 0, λ < < λ . { 2 · · · n}

Such spectra form the (n 2)-dimensional simplex ∆n 2, which contains (n 1)! the Weyl − − − chambers—this is the number of possible permutations of elements of Λ which all belong to the same unitary orbits. Hence the hyper-area of the boundary may be computed in a way analogous to (5.21):

n n 2 2 [dX]= δ ∑ λj 1 ∏ λi λj ∏(λ dλj) [dG1] < < < j Zrank(X)=n 1 Z0 λ2 λn j=2 − ! 2=i

n(n 1) n(n 1) Γ Γ (n 1) − (n 1) d − (1) (n + 1) vol − = √n 12 2 vol − D C = √n 1(2π) 2 · · · . HS − − Γ(n)Γ(n2 1) −

In an analogous way, we may ﬁnd the volume of edges, formed by the unitary orbits of the vector of eigenvalues with two zeros. More generally, states of rank N n are unitarily similar − to diagonal matrices with n eigenvalues vanishing,

Λ = λ = = λ = 0, λ < < λ . { 1 · · · m m+1 · · · n} These edges of order m are n2 m2 1 dimensional, since the dimension of the set of such spectra − − is n m 1, while the orbits have the structure of (n)/ ( (m) (1)n m) and dimensionality − − U U × U − n2 m2 (n m). We obtain the volume of the hyperedges − − − ( ) vol Fl n (n m) √n m 1 HS C vol − = − HS (n m)! (1+2m,2) (m) Cn m volHS FlC − − 78 5.3 Volume of a metric ball in unitary group

Consider a metric ball around the identity 1 in the n-dimensional unitary group (n) with n U Euclidean distance ǫ,

Bǫ := U (n) : U 1n 6 ǫ , (5.22) ∈ U − 2 n o where is the p-norm for p =e 2. We consider e the invariant Haar-measure µ, a unform k∗kp √ 1θ distribution deﬁned over (n). Denote the eigenvalues of U by e − j . The joint density of the U angles θj is given by e 1 2 √ 1θi √ 1θj p(θ1,..., θn)= n ∏ e − e − , (5.23) (2π) n! 6 < 6 − 1 i j n

where θ [ π, π], j = 1,..., n. In what follows, we check the correctness of the integral formula: j ∈ − p(θ)dθ = 1. Z 2 √ 1θ √ 1θ √ 1θ Indeed, set J(θ)= ∏ < e i e − j and ζ = e − j , so 16i j6n − − j

2 1 1 J(θ) = ∏ ζi ζ j = ∏(ζi ζj)(ζi− ζ−j ) (5.24) i

(n 1) 2 = (sign τ)(ζ1 ζn)− − ∏(ζi ζj) , (5.25) · · · i

n(n 1) Note that sign τ =( 1) 2− . We see that the integral is the constant term in − (n 1) 2 Cn(sign τ)(ζ1 ζn)− − ∏(ζi ζj) . (5.26) · · · i

Thus our task is to identify the constant term in this Laurent polynomial. To work on the last factor, we recognize

V(ζ)= ∏(ζi ζj) (5.27) i

σ(1) 1 σ(n) 1 V(ζ)= ∑ (sign σ)ζ1 − ζn − . (5.28) σ S · · · ∈ n

79 Hence

2 2 σ(1)+π(1) 2 σ(n)+π(n) 2 ∏(ζi ζj) = V(ζ) = ∑ (sign σ)(sign π)ζ1 − ζn − . (5.29) i

(n 1) 2 J(θ)=(sign τ)(ζ ζ )− − V(ζ) . (5.30) 1 · · · n We see this constant term is equal to

n 1 2π 2π J(θ)dθ ( π)n 2 Zz0 ·}| · · Z0 { n 2π 2π 1 σ(1)+π(1) n 1 σ(n)+π(n) n 1 − − − − =(sign τ) n ∑ (sign σ)(sign π)ζ1 ζn dθ (2π) z0 ·}| · · 0 { σ,π S · · · ! Z Z ∈ n 2π 2π 1 σ(1)+π(1) n 1 1 σ(n)+π(n) n 1 =(sign τ) ∑ (sign σ)(sign π) ζ1 − − dθ1 ζn − − dθn 2π 0 ×···× 2π 0 σ,π Sn Z Z ∈ =(sign τ) ∑ (sign σ)(sign π)=(sign τ) ∑ (sign σ)(sign π). (σ,π) S S : j,σ(j)+π(j)=n+1 (σ,π) S S :π=τσ ∈ n× n ∀ ∈ n× n Note that the sum is over all (σ, π) S S such that σ(j)+ π(j) = n + 1 for each j 1,..., n . ∈ n × n ∈ { } In other words, we get π(j)= n + 1 σ(j) = τ(σ(j)) for all j 1,..., n , i.e. π = τσ. Thus the − ∈ { } sum is equal to (sign τ) ∑ (sign σ)(sign τσ)= n!, σ S ∈ n which gives rise to

n 1 2π 2π 1 J(θ)dθ = . ( π)n n 2 Zz0 ·}| · · Z0 { !

Now the condition on the distance measure U 1n 6 ǫ is equivalent to − 2

n 2 √ 1θj e 2 ∑ e − 1 6 ǫ . − j=1

Using Euler’s formula e√ 1θ = cos θ + √ 1sin θ and the fact that (cos θ 1)2 + sin2 θ = 4sin2 θ , − − − 2 we get n 2 2 θj ǫ U 1n 6 ǫ ∑ sin 6 . (5.31) − 2 ⇐⇒ 2 4 j=1 e Thus the (normalized) volume of the metric ball Bǫ equals the following:

vol(Bǫ) := µ(Bǫ)= dµ(U). (5.32) ZBǫ e 80 By spectral decomposition of unitary matrix, we have

θ 0 0 1 · · ·  0 θ 0  √ 1Θ Θ 2 · · · U = VDV∗, D = e − , = . . . , θj [ π, π](j = 1,..., n).  . . .. 0  ∈ −     e e e e e  0 0 θn   · · ·    Hence

V∗ dU V = dD + V∗dV, D , · · h i implying that e e e e e e e V∗ U∗dU V = D∗ dD + V∗dV, D . · · h i Let dG = U dU,dG = V dV and dX = V dG V. Thus ∗ 1 ∗ e e e e ∗ ·e · e e e e e e e e e e e [deX] =e [dGe] (5.33) because of V (n)/ (1) n. We also havee e ∈ U U ×

e dX = D∗ dD + dG , D = D∗ V∗ dU V, · 1 · · · h i it follows that e e e e e e e e e

[dX] = [dU]. (5.34)

Apparently, e e

n Bǫ = VDV∗ (n) : D (n), V (n)/ (1)× , D 1n 6 ǫ . (5.35) ∈ U ∈ U ∈ U U − 2 n o

But e e e e e e

2 √ 1θi √ 1θj [dU]= ∏ e − e − [dD][dG1], (5.36) 6 < 6 − 1 i j n e e e therefore

2 √ 1θi √ 1θj [dG]= ∏ e − e − [dD][dG1], (5.37) 6 < 6 − 1 i j n

e e e together with the facts that the region in which (θ ,..., θ ) lies is symmetric and V (n)/ (1) n, 1 n ∈ U U × implying e 1 2 √ 1θi √ 1θj [dG] = θ [ π,π](j=1,...,n); ∏ e − e − [dD] [dG1]. j∈ − Bǫ n! · · · < − × (n) Z Z Z n 2 6 ǫ2 16i j6n ZU ∑j=1 sin (θj/2) 4 e e e

81 That is, 1 2 √ 1θi √ 1θj dµ(U) = θ [ π,π](j=1,...,n); ∏ e − e − [dD], n j∈ − Bǫ (2π) n! · · · < − Z Z Z n 2 6 ǫ2 16i j6n ∑j=1 sin (θj/2) 4 e e where [dG] n dµ(U)= and [dG]=(2π) [dG1]. [dG] (n) 1(n) (n) ZU ZU U e From this, we get e e e R e n

vol (Bǫ) = θ [ π,π](j=1,...,n); p(θ1,..., θn) ∏ dθj, (5.38) · · · j∈ − Z Z n 2 6 ǫ2 j=1 ∑j=1 sin (θj/2) 4

2 where ǫ [0, 2√n]. For the maximal distance ǫ = 2√n, the restriction ∑n sin2(θ /2) 6 ǫ ∈ j=1 j 4 becomes irrelevant and vol B2√n = 1. We start by rewriting then-dimensional integral (5.38), with the help of a Dirac delta function, as 2 ǫ π π n n 4 2 θj vol (Bǫ) = δ t ∑ sin p(θ1,..., θn) ∏ dθjdt. (5.39) 0 π · · · π − 2 Z Z− Z− j=1 ! j=1 We know that ∞ δ(t a) f (t)dt = f (a). ∞ − Z− 1 If f (t) t:α6t6β , then ≡ { } β ∞ 1 1 δ(t a)dt = δ(t a) t:α6t6β dt = t:α6t6β . α − ∞ − { } { } Z Z− By using the Fourier representation of Dirac Delta function ∞ 1 √ 1(t a)s δ(t a)= e − − ds, − 2π ∞ Z− we get

ǫ2 ∞ ǫ2 4 1 4 √ 1(t a)s δ(t a)dt = e − − dt ds 0 − 2π ∞ 0 Z Z− Z ! 2 √ 1 ǫ s 1 ∞ √ 1 1 e − 4 = − − ds. 2π ∞ se√ 1as Z− − θ Let a = ∑n sin2 j = n ∑n 1 cos θ . Indeed, j=1 2 2 − j=1 2 j n n n n 2 n 2 n 2 ∑ sin (θj/2) = + ∑(1 sin (θj/2)) = + ∑ cos (θj/2) 2 − j=1 − 2 j=1 − − 2 j=1 n n 1 n 1 = + ∑ 1 + cos θj = ∑ cos θj. − 2 j=1 2 j=1 2 82 Therefore

2 2 √ ǫ ǫ 1 4 s n ∞ √ 1 1 e − cos θj 4 2 θj 1 √ 1s ∑n δ t sin dt = − − e − j=1 2 ds ∑ √ n 0 − 2 2π ∞ 1 2 s Z j=1 ! Z− se − 2 √ 1 ǫ s ∞ √ 1 1 e − 4 n cos θ 1 √ 1s j = − − e 2 ds √ n ∏ − 2π ∞ 1 2 s Z− se − j=1 ! Inserting this formula into (5.39) and performing the integration over t ﬁrst, we have

2 √ 1 ǫ s ∞ √ 1 1 e − 4 n cos θ 1 √ 1s j vol(B ) = − − p(θ ,..., θ ) e 2 dθ ǫ √ n 1 n ∏ − j 2π ∞ 1 2 s [ π,π]n Z− se − Z − j=1 ! 2 √ 1 ǫ s 1 ∞ √ 1 1 e − 4 = − − D (s)ds, √ n n 2π ∞ 1 2 s Z− se − where

n cos θ √ 1s j Dn(s)= p(θ1,..., θn) ∏ e − 2 dθj. [ π,π]n Z − j=1 Since

2 √ 1θi √ 1θj √ 1(i 1)θk √ 1(i 1)θ ∏ e − e − = det e − − det e − − k 6 < 6 − 1 i j n is a product of two Vandermonde determinants where i, k = 1,..., n. The following fact will be used.

Proposition 5.1 (Andréief’s identity). For two n n matrices M(x) and N(x), deﬁned by the following: × M (x ) M (x ) M (x ) N (x ) N (x ) N (x ) 1 1 1 2 · · · 1 n 1 1 1 2 · · · 1 n  M (x ) M (x ) M (x )   N (x ) N (x ) N (x )  2 1 2 2 · · · 2 n 2 1 2 2 · · · 2 n M(x)= . . . . , N(x)= . . . .  . . .. .   . . .. .           Mn(x1) Mn(x2) Mn(xn)   Nn(x1) Nn(x2) Nn(xn)   · · ·   · · ·      and a function w( ) such that the integral · b Mi(x)Nj(x)w(x)dx Za exists, then the following multiple integral can be evaluated as

n b det(M(x)) det(N(x)) ∏ w(xj)dxj = det Mi(t)Nj(t)w(t)dt , (5.40) · · · ∆ a Z Z a,b j=1 Z where ∆ := x =(x ,..., x ) Rn : b > x > x > > x > a . a,b { 1 n ∈ 1 2 · · · n }

83 By invoking this identity, we know that

n cos θ 1 √ √ √ j 1(i 1)θk 1(i 1)θk 1s 2 Dn(s) = n det e − − det e− − − ∏ e − dθj (2π) n! · · · [ π,π]n Z Z − j=1 π 1 √ √ cos θ 1(i j)θ 1s 2 = n det e − − e − dθ . (2π) π Z− In what follows, we need the Bessel function of the ﬁrst kind. The deﬁnition is

π ∞ j 2j+n 1 √ 1(nθ x sin θ) ( 1) x Jn(x)= e − − dθ = ∑ − . 2π π Γ(j + n + 1)j! 2 Z− j=0

We can list some properties of this Bessel function:

π √ 1(nθ+x sin θ) π √ 1( nθ x sin θ) (i) π e − dθ = π e − − − dθ − − R 1 π √ 1(nθ Rx sin θ) 1 π √ 1( nθ+x sin θ) (ii) Jn(x)= 2π π e − − dθ = 2π π e − − dθ − − 1R π √ 1( nθ x sin θ) R1 π √ 1(nθ+x sin θ) (iii) J n(x)= 2π π e − − − dθ = 2π π e − dθ − − − We claim that R R

π √ 1(nθ+x cos θ) √ 1 nπ e − dθ = 2πe − 2 Jn(x). (5.41) π Z−

Indeed, since

π π √ 1(nθ+x cos θ) √ 1 nπ √ 1(nθ+x sin θ) √ 1 nπ e − dθ = e− − 2 e − dθ = e− − 2 2πJ n(x) π π · − Z− Z− n and J n(x)=( 1) Jn(x), the desired claim is obtained. − − Thus

π π √ 1(i j)θ √ 1s cos θ √ 1[(i j)θ+ s cos θ] e − − e − 2 dθ = e − − 2 dθ Z π Z π − − i j √ 1 − π s = 2πe − 2 Ji j . − 2 We now have

84 i j 1 √ 1 − π s 1 s Dn(s)= det 2πe − 2 Ji j = det Ji j . (2π)n − 2 (2π)n − 2

By the deﬁnition of J (x), we have J ( x)=( 1)n J (x), and furthermore, n n − − n D ( s)= D (s). n − n Therefore

2 √ 1 ǫ s 1 ∞ √ 1 1 e − 4 vol(B ) = − − D (s)ds ǫ √ n n 2π ∞ 1 2 s Z− se − 2 ∞ sin ns + sin ǫ n s 1 2 4 − 2 s = det Ji j ds, π 0 s − 2 Z where we used Euler’s formula. In fact, we have

2 ǫ2 n √ ǫ n √ 1 s √ 1 s 1 4 2 s 1 e − 4 e− − 2 e − − − = − √ 1 n s se − 2 s n ǫ2 n n ǫ2 n cos 2 s cos 4 2 s sin 2 s sin 4 2 s = − − + √ 1− − − . s − s Finally, we get the volume of a metric ball in unitary group (n): U

2 ∞ sin ns + sin ǫ n s 1 2 4 − 2 s vol(Bǫ)= det Ji j ds. (5.42) π 0 s − 2 Z

6 Appendix I: Volumes of a sphere and a ball

The following result is very important in deriving the volumes of a sphere and a ball in Fn where F = R, C.

Proposition 6.1. It holds that

n π n 1 δ(t ψ ψ )[dψ]= t − (t > 0). (6.1) Cn Γ(n) Z −h | i

85 In particular, πn δ(1 ψ ψ )[dψ]= . (6.2) Cn Γ(n) Z −h | i Proof. Recall that n [dψ] = [d(Re(ψ))][d(Im(ψ))] = ∏ dxjdyj j=1 for ψ = ∑n ψ j and ψ = x + iy R + iR(j = 1,..., n). We now have | i j=1 j| i j j j ∈ n n 2 2 δ(1 ψ ψ )[dψ]= δ 1 ∑(xj + yj ) ∏ dxjdyj. −h | i − j=1 ! j=1

Let n n Γ(n) 2 2 F(t)= δ t ∑(x + y ) ∏ dxjdyj. πn j j Z − j=1 ! j=1 Then the Laplace transform of F(t) is given by:

∞ n n Γ(n) 2 2 st L (F)(s) = δ t ∑(x + y ) e− dt ∏ dxjdyj πn j j Z "Z0 − j=1 ! # j=1 ∞ n ∞ n Γ(n) 2 2 sxj syj n = ∏ e− dxj ∏ e− dyj = s− Γ(n) πN ∞ ∞ "Z− j=1 #"Z− j=1 #

n 1 implying via the inverse Laplace transform that F(t)= t − . This indicates that n π n 1 δ(t ψ ψ )[dψ] = t − . Γ(n) Z −h | i In particular, for t = 1, we get the desired result.

In the following, we will give a method via Dirac delta function to calculate the volume of n-dimensional ball in Rn. Moreover, based on this formula, we then give the surface of (n 1)- − dimensional sphere in Rn. Denote the ball of radius R in Rn as

n B Rn 2 2 n(R) := (x1,..., xn) : ∑ xj 6 R ( ∈ j=1 ) and the sphere of radius R in Rn as

n Sn 1 Rn 2 2 − (R) := (x1,..., xn) : ∑ xj = R . ( ∈ j=1 )

Now the volume of the ball of radius R is given by

R2 n n B 2 vol( n(R)) = dt δ t ∑ xj ∏ dxj. (6.3) Rn Z0 Z − j=1 ! j=1

86 Denote n n 2 F(t)= δ t ∑ xj ∏ dxj. Z − j=1 ! j=1 Then its Laplace transform is given by

n n L (F)(s)= π 2 s− 2 .

This implies that n π 2 n 1 F(t)= t 2 . Γ n − 2 Thus 2 n n n R π 2 R π 2 vol(B (R)) = F(t)dt = = Rn. (6.4) n Γ n n Γ n Z0 2 2 2 + 1 By differentiating the above formula with respect to the radius R, we get that B n n n 1 dvol( n(R)) 2 π 2 n 1 2π 2 n 1 vol S − (R) = = 2R F(R )= nR − = R − . (6.5) dR · Γ n + 1 Γ n 2 2 From this, we see that

n n Sn 1 2 2 vol − (R) = 2R δ R ∑ xj ∏ dxj, (6.6) · − ! Z j=1 j=1 where

n n n π 2 δ R2 x2 dx = Rn 2. (6.7) ∑ j ∏ j Γ n − Z − j=1 ! j=1 2 In particular, for R = 1, we see that

n n Sn 1 2 vol − (1) = 2 δ 1 ∑ xj ∏ dxj = 2 δ (1 u u ) [du], (6.8) − ! Rn −h | i Z j=1 j=1 Z where

n n n 2 π 2 δ 1 ∑ xj ∏ dxj = δ (1 u u ) [du]= . (6.9) Rn Γ n Z − j=1 ! j=1 Z −h | i 2 Cn R2n We also see that the volume of the ball of radius R in ∼= is given by R2 n π 2n vol(Bn(R, C)) = dt δ (t ψ ψ ) [dψ]= R . (6.10) Cn Γ (n + ) Z0 Z −h | i 1 The surface of the sphere of radius R in Cn is given by

n 2n 1 2 2π 2n 1 vol S − (R, C) = 2R δ R ψ ψ [dψ]= R − , (6.11) Cn Γ (n) · Z −h | i 87 where n 2 π 2n 2 δ R ψ ψ [dψ]= R − . (6.12) Cn Γ (n) Z −h | i Recall a property of Dirac delta function: 1 δ a2 x2 = (δ(a x)+ δ(a + x)). (6.13) − 2 a − | | From this, we see that 1 δ R2 ψ ψ = δ R2 ψ 2 = (δ(R ψ )+ δ(R + ψ )). (6.14) −h | i − k k 2R − k k k k Clearly δ(R + ψ )= 0 since R + ψ > 0. Therefore we obtain k k k k

2R δ R2 ψ ψ = δ(R ψ ). (6.15) · −h | i − k k

In summary, we have the following result.

Proposition 6.2. The surface of sphere of radius R in Cn can be represented by

n 2n 1 2π 2n 1 vol S − (R, C) = δ (R ψ ) [dψ]= R − . (6.16) Cn − k k Γ (n) Z

We remark here that

∞ n 1 f (x)[dx]= dr [du]r − f (r u)δ(1 u ). (6.17) Rn Rn Z Z0 Z · − k k

In particular, if f is independent of u, i.e., f (ru)= f (r), then we have that

∞ n ∞ n 1 2π 2 n 1 f (x)[dx]= r − f (r)dr δ(1 u )[du]= r − f (r)dr. (6.18) Rn Rn Γ n Z Z0 × Z − k k 2 Z0 Indeed, For a Gaussian random vector ω = [ω ,..., ω ]T Rn with i.i.d. standard normal 1 n ∈ random variable ω N(0, 1), we have j ∼ n 1 2 1 =(2π)− 2 exp ω [dω]. (6.19) Rn −2 k k Z 88 By using polar coordinate of ω, we see ω = ω u = r u, where u = 1. Thus k k · · k k n 1 [dω]= dr δ(r ω )[dω]= r − dr δ(1 u )[du], (6.20) × − k k × − k k where

δ(1 u )[du]= 2δ(1 u u )[du]. − k k −h | i Indeed, the truth is checked as follows,

n 1 2 (2π)− 2 exp ω [dω] Rn Z −2 k k ∞ n 1 2 n 1 =(2π)− 2 exp r r − dr δ(1 u )[du] 0 −2 Rn − k k Z Z n n 1 n =(2π)− 2 2 2 − Γ δ(1 u )[du] Rn 2 Z − k k n n n Γ Γ 2 2 2 2π = n δ(1 u )[du]= n n = 1. 2 Rn 2 Γ 2π Z − k k 2π 2 7 Appendix II: Some useful facts

7.1 Matrices with simple eigenvalues form open dense sets of full measure

The present subsection is written based on the Book by Deift and Gioev [5]. Let H (C) be the set of all n n Hermitian matrices with simple spectrum, i.e. the multi- n × plicity of each eigenvalue are just one. Next we show that Hn(C) is an open and dense set of 2 full measure (i.e. the Lebesgue measure of the complement is vanished) in H (Cn) Rn , the 2 ≃ set of all n n Hermitian matrices, that is, Rn H (C) is a set of zero-measure in the sense of × \ n Lebesgue measure.

Assume that M is an arbitrary Hermitian matrix in Hn(C). with simple spectrum µ1 < < µ , then by standard perturbation theory, all matrices in a neighborhood of M have simple · · · n e n spectrum. Moreover if H H (C ) with eigenvalues hj : j = 1,..., n , then by spectral theorem, ∈ { } e H = UΛU∗ for some unitary U and Λ = diag(h1,..., hn). Now we can always ﬁnd ǫj arbitrarily e small for all j so that hj + ǫj are distinct for all j. Thus e e e e

Hǫ := Udiag(h1 + ǫ1,..., hn + ǫn)U∗ is a Hermitian matrix with simplee spectrum,e arbitrarily close toe H. The above two facts show that Hn(C) is open and dense. In order to show that Hn(C) is of full measure, consider the e discriminant: 2 ∆(λ) := ∏ (λj λi) . 16i

89 By the fundamental theorem of symmetric functions, ∆ is a polynomial function of the ele- 2 mentary symmetric functions of the λj’s, and hence a polynomial function of the n entries

H11,..., Hnn,Re(Hij), Im(Hij)(i < j) of H. Now if Hn(C) were not of full measure, then ∆ would 2 2 vanish on a set of positive measure in Rn . It follows that ∆ 0 because ∆ is polynomial in Rn . e e e e e ≡ Let H = diag(1,2,..., n) is a Hermitian matrix with distinct spectrum and ∆(H) = 0. This 6 n gives a contradiction and so Hn(C) is of full measure in H (C ). Similarly, the set S (R) of all n n real symmetric matrices with simple spectrum is an open n × and dense set of full measure in the set S (Rn) of all real symmetric matrices. From the above discussion, we conclude that the set of all density matrices with distinct positive eigenvalues is an open and dense set of full measure in the set of all density matrices. We begin by considering real symmetric matrices of size n, which is the simplest case. Because

Sn(R) is of full measure, it is sufficient for the purpose of integration to restrict our attention to matrices M Sn(R). Let µ1 < < µn denote the eigenvalues of M. By spectral theorem, T ∈ · · · M = UΛU , where Λ = diag(µ1,..., µn). Clearly the columns of U are defined only up to multiplication by 1, and so the map M (Λ, U): ± 7→ n Sn(R) M (Λ, U) R (n) ∋ 7→ ∈ ↑ × O is not well-defined, where

n n R = (µ1,..., µn) R : µ1 < < µn , (n)= n n orthogonal group. ↑ { ∈ · · · } O × We consider instead the map

n φ1 : Sn(R) M (Λ, U) R ( (n)/K1) (7.1) ∋ 7→ ∈ ↑ × O where K is the closed subgroup of (n) containingb 2n elements of the form diag( 1,..., 1), 1 O ± ± (n)/K is the homogeneous manifold obtained by factoring (n) by K , and U = UK is the O 1 O 1 1 coset containing U. The map φ1 is now clearly well-deﬁned. b The differentiable structure on (n)/K is described in the following general result about O 1 homogeneous manifolds:

Proposition 7.1. Let K be a closed subgroup of a Lie group G and let G/K be the set gK : g G of { ∈ } left cosets module K. Let π : G G/K denote the natural projection π(g) = gK. Then G/K has a → unique manifold structure such that π is C∞ and there exist local smooth sections of G/K in G, i.e., if gK G/K, there is a neighborhood W of gK and a C∞ map τ : W G such that π τ = id. ∈ → ◦ In other words, for each gK, it is possible to choose a g gK G such that the map ′ ∈ ⊂ gK g τ(gK)) is locally deﬁned and smooth. 7→ ′ ≡

90 For example, if G = R×, the multiplication group of nonzero real numbers, and if K is the subgroup 1 , then G/K = x > 0 and π(a) = a = a . If gK = a for some a > 0, {± } ∼ { } | | ∼ {± } {± } then τ( a ) = a. Also, if G = (2) and K consists of four elements of the form {± } O 1

1 0 u11 u12 ± , g = (2), " 0 1 # " u21 u22 # ∈ O ± then u11 u12 u11 u12 u11 u12 u11 u12 gK1 = , − , − , − − . (" u21 u22 # " u21 u22 # " u21 u22 # " u21 u22 #) − − − − 2 2 2 2 Since u11 + u21 = 1 = u12 + u22, each column of g contains at least one nonzero number: For example, suppose u and u are nonzero; then there exists a unique g gK such that the 21 22 ′ ∈ elements in the second row of g′ are positive. The same is true for all g′′K close to gK. Then g K g is the desired (local map) τ. We can generalize the above construction on (n)/K . ′′ 7→ ′′ O 1 We will prove the following result:

n Proposition 7.2. φ1 is a diffeomorphism from Sn(R) onto R ( (n)/K1). ↑ × O Proof. Note that (here and below we always speak of the real dimensions)

n N(N 1) N(N + 1) dim R ( (n)/K1) = N + − = (7.2) ↑ × O 2 2 as it should. n n Deﬁne the map φˆ1 : R ( (n)/K1) Sn(R) as follows: If (Λ, U) lies in R ( (n)/K1) ↑ × O → ↑ × O → Sn(R), then b T φˆ1(Λ, U)= UΛU (7.3) where U is any matrix in the coset U. If U isb another such matrix, then U = Uh for some h K ′ ′ ∈ 1 and so b T T T T U′Λ(U′) = UhΛh U = UΛU .

Hence φˆ1 is well-deﬁned. We will show that

ˆ n φ1 φ1 = idR ( (n)/K1) (7.4) ◦ ↑ × O and

φˆ φ = idS R . (7.5) 1 ◦ 1 n( ) Indeed,

φ φˆ Λ, U = φ (UΛUT) , U U (7.6) 1 1 1 ∈ b = (Λ, UK1) = Λ, U b. (7.7) 91 b Conversely, if M = UΛUT S (R), then ∈ n

T φˆ1(φ1(M)) = φˆ1 Λ, U = UK1 = UΛU = M. (7.8) This proves Eq. (7.4) and Eq. (7.5). In orderb to prove that φ1 is a diffeomorphism, it sufﬁces to show that φ1 and φˆ1 are smooth. The smoothness of φ follows from perturbation theory: Fix M = U Λ UT S (R), where 1 0 0 0 0 ∈ n Λ = diag(µ ,..., µ ) for µ < < µ , and U (n). Then for M near M , M 0 01 0n 01 · · · 0n 0 ∈ O 0 ∈ S (R), the eigenvalues of M, µ (M) < < µ (M), are smooth functions of M. Moreover, the n 1 · · · n associated eigenvectors uj(M),

Muj(M)= λj(M)uj(M), 1 6 j 6 n, can be chosen orthogonal

ui(M), uj(M) = δij, (1 6 i, j 6 n), and smooth in M. Indeed, for any j with 1 6 j 6 n, let Pj be the orthogonal projection

1 1 Pj(M)= ds (7.9) 2πi Γ s M I j − where Γ is a small circle of radius ǫ around µ , µ µ > ǫ for i = j. Then, where u (M ) is j 0j 0i − 0j 6 j 0 the j-th column of M0,

Pj(M)uj(M0) uj(M)= , 1 6 j 6 n, uj(M0), Pj(M)uj(M0) q is the desired eigenvector of M. Set U(M) = [u (M),..., u (M)] (n). Then M (Λ(M), U(M)) is smooth and hence 1 n ∈ O 7→ φ (M)=(Λ(M), U(M)) π(U(M)) 1 ≡ is smooth, as claimed. n Finally, we show that φˆ1 is smooth. Fix Λ0, U0 R ( (n)/K1) and let τ be the lifting ∈ ↑ × O map from some neighborhood W of U0 to (n). Now for all U W, τ(U) U by Proposition 7.1. O b ∈ ∈ Hence b b b b T φˆ1 Λ, U = τ(U)Λτ(U) (7.10) b b b n from which it is clear that φˆ1 is smooth near Λ0, U0 , and hence everywhere on R ( (n)/K1). ↑ × O This completes the proof. b

92 Now consider Hn(C). The calculations are similar to Sn(R). Deﬁne the map

n φ2 : Hn(C) M Λ, U R ( (n)/K2). (7.11) ∋ 7→ ∈ ↑ × U T T iθ1 iθn R Here K2 is the closed subgroup of (n) givenb by = diag e ,..., e : θj , U ×···× ∈ Λ = diag(λ ,..., λ ), λ < < λ as before, and 1 n 1 · · · n

M = UΛU∗, U = π(U), where π : (n) (n)/K2 is the natural projectionb U π(U) = UK2 of the unitary U → U 7→ group (n) onto the homogeneous manifold (n)/K . Let τ : (n)/K (n) denote the U U 2 U 2 → U (locally deﬁned) lifting map as above. As before, φ2 is a well-deﬁned map that is one-to- one from H (C) to (n) (n)/K . Note that because dim( (n)) = n2, dim( (n)/K ) = n U → U 2 U U 2 dim( (n)) dim(K )= n2 n, as it should. U − 2 − As before, the proof of the following result is similar to that of Proposition 7.2.

n Proposition 7.3. φ2 is a diffeomorphism from Hn(C) onto R ( (n)/K2). ↑ × U

7.2 Results related to orthogonal groups

Proposition 7.4. Let Y, X, T Rn n, where Y and T are nonsingular, X is skew symmetric and T is ∈ × lower triangular of independent real entries y ’s, x ’s and t ’s respectively. Let t > 0, j = 1,..., n ij ij ij jj − 1, ∞ < t < ∞; ∞ < t < ∞, j > k; ∞ < x < ∞, j < k or ∞ < t < ∞, j > k and − nn − jk − jk − jk 1 the ﬁrst row entries, except the ﬁrst one, of (1n + X)− are negative. Then the unique representation Y = T 2(1 + X) 1 1 = T(1 X)(1 + X) 1 implies that n − − n n − n − n n(n 1)/2 n j (n 1) [dY]= 2 − ∏ tjj − (det(1n + X))− − [dT][dX]. (7.12) j=1 !

Proof. Take the differentials to get

1 1 1 dY = dT 2(1 + X)− 1 + T 2(1 + X)− dX (1 + X)− · n − n · − n · · n 1 1 = dT 2(1 + X)− 1 + T (1 + Z) dX (1 + Z) · n − n · −2 n · · n 1 = dT Z T(1 + Z) dX (1 + Z). · − 2 n · · n where Z = 2(1 + X) 1 1 . Then we have n − − n 1 T 1 1 T T− dY Z = T− dT (1 + Z) dX (1 + Z ). · · · − 2 n · · n The Jacobian of the transformation of T and X going to Y is equal to the Jacobian of dT,dX going to dY. Now treat dT,dX,dY as variables and everything else as constants. Let

1 T 1 T dU = T− dY Z , dV = T− dT, dW =(1 + Z) dX (1 + Z ). · · · n · · n

93 Thus

n T n n [dU]= det(T)− det(Z ) [dY]= det(T)− [dY] n n n = [dY]= det(T) [dU]= ∏ tjj [dU]. ⇒ j=1 !

Note that det(ZT)= 1 since Z is orthogonal. Since X is skew symmetric, one has ± n 1 n(n 1) (n 1) [dW]= det(1n + Z) − [dX]= 2 − det(1n + X)− − [dX].

One also has n j [dV]= ∏ tjj − [dT]. j=1 !

Now we see that 1 1 dU = dV dW = U = V W. − 2 ⇒ − 2 Let U = [uij], V = [vij], W = [wij]. Then since T is lower triangular tij = 0, i < j and thus V is lower triangular, and since X is skew symmetric xjj = 0 for all j and xij= x for i = j and thus W − ji 6 is skew symmetric. Thus we have 1 u = v , u = w , i < j ii ii ij −2 ij 1 u = v + w , i > j. ij ij 2 ij

Take the u-variables in the order uii, i = 1,..., n; uij, i < j; uij, i > j and vii, i = 1,..., n; vij, i < j; vij, i > j. Then the Jacobian matrix is of the following form:

1n 0 0 1 1  0 (n) 0  − 2 2 1 1 1  0 (n) (n)   2 2 2    n(n 1)/2 1 − and the determinant, in absolute value, is 2 . That is, n(n 1)/2 1 − [dU]= [dV][dW]. 2 Now substitute for [dU], [dW] and [dV], respectively to obtain the result.

Proposition 7.5. Let Y, X, D Rn n be of independent real entries, where Y is symmetric with distinct ∈ × 1 and nonzero eigenvalues, X is skew symmetric with the entries of the ﬁrst row of (1n + X)− , except the ﬁrst entry, negative and D = diag(λ ,..., λ ), with λ > > λ . Then, excluding the sign, 1 n 1 · · · n Y = 2(1 + X) 1 1 D 2(1 + X) 1 1 implies that n − − n n − − n n(n 1)/2 (n 1) [dY]= 2 − (det(1n + X))− − ∏ λi λj [dX][dD]. (7.13) i

94 Proof. Let Z = 2(1 + X) 1 1 . Take the differentials and reduce to get n − − n T 1 T 1 T Z dY Z = (1 + Z ) dX (1 + Z) D + dD + D (1 + Z ) dX (1 + Z). · · −2 n · · n · 2 · n · · n Put

dU = ZT dY Z, dW = dD, dV =(1 + ZT)dX(1 + Z). · · n n Thus 1 1 dU = dV D + D dV + dW. −2 · 2 · But since dY is symmetric and Z is orthogonal, [dU] = det(Z)n+1[dY] = [dY], excluding the sign. Clearly [dW] = [dD]. Since X is skew symmetric we have

n 1 n(n 1) (n 1) dV = det(1n + Z) − [dX]= 2 − det(1n + X)− − [dX].

We see that

duii = dwii, 1 du = (λ λ )dv , i < j. ij 2 i − j ij

Take the u-variables in the order uii, i = 1,..., n; uij, i < j and the w and v-variables in the order wii, i = 1,..., n; vij, i < j. Then the matrix of partial derivatives is of the following form:

1 0 , " 0 M # where M is a diagonal matrix with the diagonal elements 1 (λ λ ), i < j. There are n(n 1)/2 2 i − j − n(n 1)/2 elements. Hence the determinant of the above matrix, in absolute value, is 2 ∏ < λ λ . − − i j i − j That is,

n(n 1)/2 [dU]= 2− − ∏ λi λj [dV][dD]. i

Hence

n(n 1)/2 (n 1) [dY]= 2 − det(1n + X)− − ∏ λi λj [dX][dD]. i

Remark 7.6. When integrating over the skew symmetric matrix X using the transformation in Proposition 7.5, under the unique choice for Z = 2(1 + X) 1 1 , observe that n − − n n2 π 2 2n(n 1)/2 det(1 + X) (n 1)[dX]= . (7.14) − n − − Γ n ZX n 2 Note that the λj’s are to be integrated out over ∞ > λ1 > >λn and X over a unique choice · · · of Z.

95 7.3 Results related to unitary groups

When X is skew hermitian, that is, X = X, it is not difﬁcult to show that 1 X are both ∗ − ± 1 nonsingular and Z = 2(1 + X)− 1 is unitary, that is, ZZ∗ = 1. This property will be made use e − e e e of in the ﬁrst result that will be discussed here. Also note that e e e e 1 1 1 2(1 + X)− 1 =(1 + X)− (1 X)=(1 X)(1 + X)− . − − − When X is skew hermitiane and of functionallye e independente complexe variables, then there n(n 1) are p + 2 − = n2 real variables in X. Let T be a lower triangular matrix of functionally e2 independent complex variables with the diagonal elements being real. Then there are n2 real e e variables in T also. Thus combined, there are 2n2 real variables in T and X. It can be shown that

1 e Y = T 2(1 + X)− 1 e e − can produce a one-to-one transformatione e when theetjj’s are real and positive.

Y = TZ,

1 where ZZ∗ = , T = [tjk], tjj = tjj > 0, j = 1,...,e ene. Then

n e e e e e 2 2 YY∗ = TT∗ = t11 = ∑ y1k . ⇒ k=1 | | e e ee e Note that when t11 is real and positive it is uniquely determined in terms of Y. Now consider the 2 ﬁrst row elements of TT∗ that is t , t11t21,..., t11tn1. Hence t21,..., tn1, that is, the ﬁrst column 11 e of T is uniquely determined in terms of Y. Now consider the second row of TT∗ and so on. ee e e 1 e e Thus T is uniquely determined in terms of Y. But Z = T− Y and hence Z, thereby X is uniquely e e ee determined in terms of Y with no additional restrictions imposed on the elements of Z. e e e e e e e In this chapter the Jacobians will also be written ignoring the sign as in the previous chapters. e e Proposition 7.7. Let Y, X and T = [t ] be n n matrices of functionally independent complex variables jk × where Y is nonsingular, X is skew hermitian and T is lower triangular with real and positive diagonal e e e e elements. Ignoring the sign, if e e e 1 1 Y = T 2(X + 1)− 1 = T(1 X)(1 + X)− , − − then e e e e e e

n n n2 2(n j)+1 1 1 − [dY]= 2 ∏ tjj − det(( + X)( X)) [dX][dT]. (7.15) · ! · − · j=1

e e e e e

96 Proof. Taking differentials in Y = T 2(1 + X) 1 1 , one has − − 1 1 1 dY = T 2(1 +eX)−e dX (1e+ X)− + dT 2(1 + X)− 1 . − · · · − Let e e e e e e e

1 1 1 Z = 2(1 + X)− 1 = (1 + X)− = (1 + Z) − ⇒ 2 and observe that ZZ∗ =e 1. Then e e e

1 1 1 e eT− dY Z∗ = (1 + Z) dX (1 + Z∗)+ T− dT. (7.16) · · −2 · · · Let dW =(1 + Z) deX (1e+ Ze ). Then e e e e e · · ∗ n n 2n2 − e[dW]= edet((1e + Z)(1e+ Z∗)) [dX]= 2 det((1 + X)(1 X)) [dX]. (7.17) · · − ·

Let dU =eT 1 dT, then e e e e e e − · n e e e (2j 1) [dU]= ∏ t−jj − [dT]. (7.18) j=1 ! · e e Let dV = T 1 dY Z , then − · · ∗ n − e e e e [dV]= det(TT∗) [dY]. (7.19) ·

Eq. (7.16) reduces to e ee e 1 dV = dW + dU. (7.20) −2 Note that dW is skew hermitian. Denotee e e

( ) ( ) e V = [v ] = [v 1 ]+ √ 1[v 2 ], jk jk − jk ( ) ( ) W = [w ] = [w 1 ]+ √ 1[w 2 ], e e jk jk − jk ( ) ( ) U = [u ] = [u 1 ]+ √ 1[u 2 ], e ejk jk − jk (m) (m) (m) where v , w , u , m = 1, 2e are all real.e Then from Eq. (7.20), we see that V = 1 W + U. jk jk jk − 2 Thus e e e ( ) 1 ( ) ( ) v m = w m + u m , j > k, m = 1, 2, jk 2 kj jk ( ) 1 ( ) v m = w m , j < k, m = 1, 2, jk −2 jk (1) (1) vjj = ujj , ( ) 1 ( ) v 2 = w 2 . jj −2 jj

97 The matrices of partial derivatives are the following:

(1) (2) (1) (2) < (1) (2) > ∂ vjj , vjj ; vjk , vjk (j k); vjk , vjk (j k) (1) (2) (1) (2) < (1) (2) > ∂ ujj , wjj ; wjk , wjk (j k); ujk , ujk (j k) 1n 0 0 000  0 1 1 0 0 00  − 2 n 1 1  0 0 (n) 0 0 0  =  − 2 2   1 1  0 0 0 (n) 0 0  − 2 2   1 1 1   0 0 (n) 0 (n) 0   2 2 2   1 1 1   0 0 0 (n) 0 (n)   2 2 2    where

(1) ∂vjj A = , j = 1,..., n = 1 , 11  (1)  n ∂ujj  (2)  ∂vjj 1 A = , j = 1,..., n = 1 22  (2)  −2 n ∂wjj   and

( ) ∂v 1 jk < 11 A33 = , j k = (n),  (1)  −2 2 ∂wjk  (2)  ∂v jk < 11 A44 = , j k = (n)  (2)  −2 2 ∂wjk  (1)  ∂v jk > 1 A55 = , j k = (n),  (1)  2 ∂ujk  (2)  ∂v jk > 1 A66 = , j k = (n).  (2)  2 ∂ujk   ∂v(1) ∂v(2) jk > 11 jk > 11 A53 = : j k = (n), A64 = : j k = (n).  (1)  2 2  (2)  2 2 ∂wkj ∂wkj     The determinants of A11,..., A66 contribute towards the Jacobian and the product of the deter- n(n 1) n n2 minants, in absolute value, is 2− − − = 2− . Without going through the above procedure 2 1 one may note from (7.16) that since dX has n real variables, multiplication by 2 produces the

98 n2 factor 2− in the Jacobian. From (7.17),(7.18), (7.19) and (7.20) we have

n n 2n 2n n2 [dY] = ∏ tjj [dV]= ∏ tjj 2− [dW][dU] j=1 ! · j=1 ! · ·

e n e e ne n 2n 2n2 n2 1 1 − (2j 1) = ∏ tjj 2 − det(( + X)( X)) ∏ t−jj − [dX][dT] ! · · − · ! · j=1 j=1 n 2 e e n e e n 2(n j)+1 1 1 − = 2 ∏ tjj − det(( + X)( X)) [dX][dT]. ! · − · j=1 e e e e This completes the proof.

Example 7.8. Let X Cn n skew hermitian matrix of independent complex variables. Then ∈ × show that e

2 n πn [dX] det((1 + X)(1 X)) − = . X − 2n(n 1)Γ (n) Z − n

e e e e e Indeed, let Y = [y ] be a n n matrix of independent complex variables. Consider the jk × integral e e n ∞ + 2 2 Tr(YY∗) yjk n [dY]e− = ∏ e− dyjk = π . (7.21) ∞ | | Y j,k=1 Z e e Z− e e e e Now consider a transformation Y = T 2(1 + X) 1 1 , where T = [t ] is lower triangular with − − jk tjj’s real and positive and no restrictions on X other than that it is skew hermitian. Then since e e e e e Tr YY∗ = Tr TT∗ , from Proposition 7.7 and Eq. (7.21), we have e n e e 2 ee 2 ( )+ n n Tr(TT∗) n 2 n j 1 1 1 − π = [dX][dT]e− 2 ∏ tjj − det(( + X)( X)) T,X · · ! · − Z ee j=1 e e n e e e e 2 ( )+ n Tr(TT∗) n 2 n j 1 1 1 − = [dT]e− 2 ∏ tjj − [dX] det(( + X)( X)) T · · ! × X − Z ee j=1 Z e e e e Note that e e

n Tr(TT∗) 2 2 e− = exp ∑ tjj ∑ tjk . − j=1 − j>k ! ee e But

+∞ 2 +∞ 2(n j)+1 t2 1 tjk jj Γ e− dtjk = π and tjj − e− dtjj = (n j + 1). ∞ | | 0 2 − Z− e Z e 99 Hence

n ( )+ Tr(TT∗) 2 n j 1 nΓ [dT]e− ∏ tjj − = 2− n(n). T Z ee j=1 ! e e Substituting this the resulte follows.

Remark 7.9. When a skew hermitian matrix X is used to parameterize a unitary matrix such as Z in Proposition 7.7, can we evaluate the Jacobian by direct integration? This will be examined e here. Let e n − In := [dX] det((1 + X)(1 X)) . X − Z

Partition 1 + X as follows: e e e e

1 + x11 X12 e 1 + X = , 1 " X12∗ + X1 # − e e e where X12 represents the first row of 1 + X eexcludinge the first element 1 + x11, and 1 + X1 is obtained from 1 + X by deleting the first row and the first column. Note that e e e e 1 1 1 1 dete ( + X)= det( + X1) 1 + x11 + X12( + X1)− X12∗ . Similarly, e e e e e e

1 x11 X12 1 X = − − − 1 " X12∗ X1 # e −e e and e e

1 det(1 X)= det(1 X ) 1 x + X (1 + X )− X∗ . − − 1 − 11 12 1 12 1 1 1 For ﬁxed ( + X1) let U12 :e= X12( + X1)e− , thene e e e 1 e e [dU e]= dete((1 + X )(1 X )) − [dX ] 12 1 − 1 · 12 and observing that X = eX we have e e e 1∗ − 1 1 e eX (1 + X )− X∗ = U (1 X )U∗ . 12 1 12 12 − 1 12 Let Q be a unitary matrix suche that e e e e e

e Q∗X1Q = diag(√ 1λ1,..., √ 1λn 1) − − − where λ1,..., λn 1 are real.e Lete e −

V12 = U12Q = [v1,..., vn 1]. −

e e e 100e e Then

1 2 2 U12( X1)U12∗ =(1 √ 1λ1) v1 + +(1 √ 1λn 1) vn 1 − − − | | · · · − − − | − | and e e e e e

1 1 + x + X (1 + X )− X∗ = a √ 1b 11 12 1 12 − − where e e e e

2 2 a = 1 + v1 + + vn 1 | | · · · | − | (2) 2 2 b = x + λ1 v1 + + λn 1 vn 1 − 11e | | e· · · − | − | (2) (2) observing that x is purely imaginary, that is,ex = √ 1x , wheree x is real. Thus 11 11 − 11 11 n e det((1 + Xe)(1 X)) − −

yields the factor e e

n 2 2 n [(a √ 1b)(a + √ 1b)]− =(a + b )− . − − − So

n − In = det(1 + X1)(a √ 1b) det(1 X1)(a + √ 1b) [dX1][dX12]dx11 ZX1 ZX12 Zx11 − − − − n − 2 2 n = e e e det (1 +eX1)(1 X1) (a + be )− [dX1][dX12]dx11.e e e X1 X12 x11 − Z Z Z

Since e e e e e e e e 1 [dV ] = [dU ] and [dU ]= det (1 + X )(1 X ) − [dX ], 12 12 12 1 − 1 12 it follows that e e e e e e [dX ]= det (1 + X )(1 X ) [dV ]. 12 1 − 1 12 Based on this, we have e e e e (n 1) − − 2 2 n In = det (1 + X1)(1 X1) (a + b )− [dX1][dV12]dx11. X1 V12 x11 − Z Z Z

Then e e e e e e e e

I I 2 2 n (2) n = n 1 ( ) (a + b )− [dV12]dx11 − V x 2 Z 12 Z 11 2 n e e b − ( ) = I [dV ] a 2n 1 + dx 2 . n 1 12 − (2) 2 11 − V x a Z 12 Z 11 ! e e

101 ( ) ( ) ( ) Consider the integral over x 2 , ∞ < x 2 < +∞. Change x 2 to b and then to c = b/a. Then 11 − 11 11 2 n 2 n +∞ b − ( ) b − 1 + dx 2 = 1 + db = a (1 + c2) pdc (2) 2 11 2 − x a b a c= ∞ Z 11 Z Z − ∞ Γ 1 Γ 1 2 n 2 n 2 = 2a (1 + c )− dc = a − := k, Γ(n) Z0 by evaluating using a type-2 beta integral after transforming u = c2. Hence

(2n 1) In = kIn 1 [dV12]a− − − ZV12 +∞ +∞ (2n 1) 2 2 − − = kIn 1 e e 1 + v1 + + vn 1 dv1 dvn 1. − ∞ · · · ∞ | | · · · | − | · · · − Z− Z− ( ) ( ) ( ) For evaluating the integral use the polar coordinates.e Letev = v 1 + √ e1v 2 , wheree v 1 and j j − j j (2) vj are real. Let e (1) vj = rj cos θj, 0 6 rj < ∞, 0 6 θj 6 2π.  (2) vj = rj sin θj,

Then denoting the multiple integral by n 1, we have  I − +∞ +∞ (2n 1) 2 2 − − n 1 = 1 + v1 + + vn 1 dv1 dvn 1 I − ∞ · · · ∞ | | · · · | − | · · · − Z− Z− ∞ ∞ + + (2n 1) n 1 2 2 − − = (2π) − e r1 rn 1e 1 + r1 + + ren 1 e dr1 drn 1. − − Zr1=0 · · · Zrn 1=0 · · · · · · − · · · − Evaluating this by a Dirichlet integral we have

n 1 Γ(n) n 1 = π − for n > 2. I − Γ(2n 1) − Hence for n > 2, Γ Γ 1 n 1 (n) n 2 In = In 1π − √π − . − Γ(2n 1)Γ(n) − By using the duplication formula for gamma functions

2n 2 1 Γ(2n 1)= √π2 − Γ n Γ(n). − − 2 Hence n I I π n = n 1 2n 2 . − 2 − Γ(n) Repeating this process we have

n n 1 n2 π π − π π In = = . 22n 2Γ(n) 22(n 1) 2Γ(p 1) · · · 22 2Γ(1) 2n(n 1)Γ (n) − − − − − − n This is what we obtained in Example 7.8. e

102 Next we consider a representation of a hermitian positive definite matrix Y in terms of a skew hermitian matrix X and a diagonal matrix D such that e 1 1 ∗ e Y = 2(1 + X)− 1 D 2(1 + X)− 1 − − where D = diag(λ1,..., λep) with theeλj’s real distinct ande positive, and the first row elements of (1 + X) 1 real and of specified signs, which amounts to require 2(1 + X) 1 1 (n)/ (1) n. − − − ∈ U U × In this case it can be shown that the transformation is unique. Note that e e

Y = ZDZ∗ = λ Z Z∗ + + λ Z Z∗ 1 1 1 · · · p n n it indicates that e e e e e e e

(Y λ 1)Z = 0, j = 1,..., n − j j where Z ,..., Z are the columnse of Z suche that Z , Z = δ . Since λ ,..., λ are the eigen- 1 n h j ki jk 1 n values of Y, which are assumed to be real distinct and positive, D is uniquely determined in e e e e e terms of Y. Note that Zj is an eigenvector corresponding to λj such that Zj, Zj = 1, j = 1,..., n. e h i Hence Zj is uniquely determined in terms of Y except for a multiple of 1, √ 1. If any par- e e ±e e± − ticular element of Zj is assumed to be real and positive, for example the first element, then Zj is e e uniquely determined. Thus if the first row elements of Z are real and of specified signs, which is e 1 e equivalent to saying that the first row elements of (1 + X)− are real and of specified signs, then e the transformation is unique. e Proposition 7.10. Let Y and X ben n matrices of functionally independent complex variables such that × 1 Y is hermitian positive definite, X is skew hermitian and the first row elements of (1 + X)− are real and e e of specified signs. Let D = diag(λ1,..., λn), where the λj’s are real distinct and positive. Ignoring the e e e sign, if

1 1 ∗ Y = 2(X + 1)− 1 D 2(X + 1)− 1 , − − then e e e

n n(n 1) 2 1 1 − [dY]= 2 − ∏ λk λj det(( + X)( X)) [dX][dD]. · > − ! · − · j k

e e e e

Proof. Let Z = 2(1 + X) 1 1, X = X. Taking the differentials in Y = ZDZ we have − − ∗ − ∗

e e dY = deZ D Ze∗ + Z dD Z∗ + Z D dZ∗.e e e (7.22) · · · · · ·

e e e e 103 e e e But

1 1 dZ = 2(1 + X)− dX (1 + X)− − · · 1 = (1 + Z) dX (1 + Z) e −2 e · ·e e and e e e

1 1 dZ∗ = 2(1 X)− dX (1 X)− − · · − 1 = (1 + Z∗) dX (1 + Z∗). e 2 e · e· e From (7.22), one has e e e 1 1 Z∗ dY Z = (1 + Z∗) dX (1 + Z) D + dD + D (1 + Z∗) dX (1 + Z) · · −2 · · · 2 · · · observinge thate Ze∗Z = 1. Let e e e e e e

deUe = Z∗ dY Z = [dU] = [dY] since Z∗Z = 1, (7.23) · · ⇒ 1 1 dV = (1 + Z∗) dX (1 + Z)=(1 + X∗)− 4dX (1 + X)− = e e e ·e · e e e· e · ⇒ 2 n [dV] = 4n det((1 + X)(1 X)) − [dX] (7.24) e · e e −e · e e e if there are n2 freee real variables in X.e But in oure case theree are only n2 n real variables in X − when X is uniquely chosen and hence e e n n2 n − e [dV] = 4 − det((1 + X)(1 X)) [dX], (7.25) − ·

and e e e e 1 1 dU = dV D + D dV + dD. (7.26) −2 · 2 · From (7.26) and using the fact thate dV is skewe hermitian ande dU is hermitian we have

( ) 1 ( ) du = dλ , du m e= (λ λ )dv m , j > ke, m = 1, 2. jj j jk ±2 k − j jk Thus the determinant of the Jacobian matrix, in absolute value, is

2 1 n(n 1) 2 ∏ λk λj = 2− − ∏ λk λj . j>k 2 − ! j>k −

That is,

n(n 1) 2 [dU]= 2− − ∏ λk λj [dV][dD]. · j>k − ·

e e Substituting for [dU] and [dV] from (7.23) and (7.24) the result follows.

e e 104 Example 7.11. For X an n n skew hermitian matrix with the ﬁrst row elements of (1 + X) 1 × − real and of speciﬁed signs show that e e n Γ (n) [dX] det((1 + X)(1 X)) − = n , X − ∆ Z e where e e e e

n(n 1) 2 Tr(D) ∆ := 2 − [dD] ∏ λi λj e− , > > > Zλ1 λn 0 "16i > λ > 0. Consider a n n hermitian positive deﬁnite 1 n 1 · · · n × matrix Y of functionally independent complex variables. Let

n n Tr(Y) − Tr(Y) e B = [dY]e− = [dY] det(Y) e− > > ZY=Y∗ 0 ZY 0 n(n 1) e e Γ 2− Γ Γ Γ = ne(ne)= πe (n) (n e 1) e (1) e n(n 1) − · · · = π 2− (n 1)!(n 2)! 1! e − − · · · evaluating the integral by using a complex matrix-variate gamma integral. Put

1 Y = ZDZ∗, Z = 2(1 + X)− 1 − as in Proposition 7.10. Then e e e e e

n n(n 1) 2 1 1 − [dY]= 2 − ∏ λk λj det(( + X)( X)) [dX][dD] · j>k − ! · − ·

e e e e and

n B = [dX] det((1 + X)(1 X)) − X − Z

e e e n(n 1) e 2 Tr(D) [dD]2 − ∏ λk λj e− . > > > × Zλ1 λn 0 · j>k − ! ···

Hence the result. From (i) in Example 3.21, we see that

n(n 1) 2 − 2 ∆ = Γ (n) , π n e which implies that, when X is taken over all skew hermitian under the restriction that the ﬁrst 1 row elements of (1 + X)− are real and of speciﬁed signs, e e

105 n π n(n 1) 1 [dX] det((1 + X)(1 X)) − = − . X − 2 Γn(n) Z

e e e e e Remark 7.12. In fact, we can derive the volume formula (3.36) from Proposition 7.10. The reasoning is as follows:

n n(n 1) 2 [dY] = 2 − [dD]δ ∑ λj 1 ∏ λk λj > > > > ZY 0:Tr(Y)=1 Zλ1 λn 0 j=1 − ! j>k − ! ··· e n e e [dX] det((1 + X)(1 X)) − × X − Z 2n(n 1) Γ(1) Γ(n)Γ(1) Γ (n + 1) π n(n 1) 1 = e− e · · · e ·e · · − Γ 2 n! × (n ) × 2 Γn(n) n(n 1) Γ(1) Γ(n) = π 2− · · · . Γ(n2) e

Example 7.13. Let X Cn n be a hermitian matrix of independent complex variables. Show that ∈ × n(n 1) n2 Tr(XX∗) − e [dX]e− = 2− 2 π 2 . X Z e e Indeed, X∗ = X implies that e e

n e e 2 2 Tr XX∗ = ∑ xjj + 2 ∑ xij . j=1 i

n ∞ 2 2 Tr(XX∗) xjj 2 xij [dX]e− = ∏ e− dxjj ∏ e− dxij ∞ | | X j=1 ! × i

e e n n(n 1) 2 e − xij = π 2 2− 2 ∏ e−| | dxij × i

n n 1 2 n(n 1) n − 2 2− 2 ∏ λk λj exp ∑ λj dλ1 dλn = 2− π ∏ j!. ∞> > > > ∞ Z λ1 λn j>k − ! − j=1 ! · · · j=1 ··· −

Indeed, in Example 7.13 letting X = UDU , where D = diag(λ ,..., λ ) and U (n), gives ∗ 1 n ∈ U1 rise to e e e e 2 [dX]= ∏(λi λj) [dD][dG1], λ1 > > λn, i

n(n 1) n2 − Tr(XX∗) 2− 2 π 2 = [dX]e− X Z e e n e e 2 2 = ∏(λi λj) exp ∑ λj [dD] [dG1]. ∞>λ1> >λn> ∞ < − − × 1(n) Z ··· − i j ! j=1 ! ZU e That is,

n n n 1 n(n 1) n − 2 2 2− 2 ∏(λi λj) exp ∑ λj ∏ dλj = 2− π ∏ j!. ∞>λ1> >λn> ∞ < − − Z ··· − i j ! j=1 ! j=1 j=1 Therefore

n n n n(n 1) n 2 2 2− 2 ∏(λi λj) exp ∑ λj ∏ dλj = 2− π ∏ j!. Z i

8 Appendix III: Some matrix factorizations

The materials in this section are collected from Muirhead’s book [25]. It is the necessary underlying basis for computing some Jacobians . Firstly, we recall the Gram-Schmidt orthogonalization process which enables us to construct m m an orthonormal basis of R given any other basis X1,..., Xm of R . We deﬁne

Y1 = X1,

 Y1,X2 Y2 = X2 hY ,Y i Y1,  − h 1 1i  Y2,X3 Y1,X3 Y3 = X3 h i Y2 h i Y1,  − Y2,Y2 − Y1,Y1  h i h i ············  m 1 Yj,Xm Ym = Xm ∑ − h i Yj,  j=1 Yj,Yj  − h i  1  Rm and put Zj = Y ,Y Yj, where j = 1,..., m. Then Z1,..., Zm form an orthonormal basis for . h j ji Next matrix factorization utilizes this process.

Proposition 8.1. If A is a real m m matrix with real characteristic roots, then there exists an orthogonal × matrix H such that HT AH is an upper-triangular matrix whose diagonal elements are the characteristic roots of A.

Proof. Let λ1,..., λm be the characteristic roots of A := A1 and let X1 be a characteristic vector of A corresponding to λ1. This is real since the characteristic roots are real. Let X2,..., Xm be m any other vectors such that X1, X2,..., Xm for a basis for R . Using the Gram-Schmidt orthogonalization process, construct from X1, X2,..., Xm an orthonormal basis given as the columns of

107 the orthogonal matrix H1, where the first column h1 is proportional to X1, so that h1 is also a characteristic vector of A corresponding to λ1. Then the first column of AH1 is Ah1 = λ1h1, and T T T 1 hence the first column of H1 A1H1 is λ1 H1 h1. Since this is the first column of λ1H1 H1 = λ1 m, it T is (λ1,0,...,0) . Hence

T λ1 B1 H1 A1H1 = , " 0 A2 # where A is (m 1) (m 1). Since 2 − × −

det(A1 λ1m)=(λ1 λ) det(A2 λ1m 1) − − − −

T and A1 and H1 A1H1 have the same characteristic roots, the characteristic roots of A2 are λ2,..., λm. Now, using a construction similar to that above, ﬁnd an orthogonal (m 1) (m 1) matrix − × − H2 whose ﬁrst column is a characteristic vector of A2 corresponding to λ2. Then

T λ2 B2 H2 A2H2 = , " 0 A3 # where A is (m 2) (m 2) with characteristic roots λ ,..., λ . 3 − × − 3 m Repeating this procedure an additional m 3 times we now deﬁne the orthogonal matrix −

H = H1(1 H2)(12 H3) (1m 2 Hm 1) ⊕ ⊕ · · · − ⊕ −

T and note that H AH is upper-triangular with diagonal elements equal to λ1,..., λm.

Proposition 8.2. If A is an m m non-negative deﬁnite matrix of rank r then: × (i) There exists an m r matrix B of rank r such that A = BBT. × (ii) There exists an m m nonsingular matrix C such that × 1 r 0 T A = C C . " 0 0 #

Proof. As for (i), let D1 = diag(λ1,..., λr) where λ1,..., λr are the nonzero characteristic roots of A, and let H be an m m orthogonal matrix such that HT AH = diag(λ ,..., λ ,0,...,0). × 1 r Partition H as H = [H , H ],where H is m r and H is m (m r); then 1 2 1 × 2 × −

D1 0 T T A = H H = H1D1H1 . " 0 0 #

Putting √D1 = diag(√λ1,..., √λr), we then have

T T A = H1 D1 D1H1 = BB , p p 108 where B = H √D is m r of rank r. As for (ii), let C bee an m m nonsingular matrix whose 1 1 × × ﬁrst r columns are the columns of the matrix B in (i). Then

1 r 0 T A = C C . " 0 0 #

The following result is used often in the text.

Proposition 8.3 (Vinograd, 1950). Suppose that A and B are real matrices, where A is k m and B is × k n, with m 6 n. Then AAT = BBT if and only if there exists an m n matrix H with HHT = 1 × × m such that AH = B.

Proof. First suppose there exists an m n matrix H with HHT = 1 such that AH = B. Then × m BBT = AHHT AT = AAT. Now suppose that AAT = BBT. Let C be a k k nonsingular matrix such that × 1 T T r 0 T AA = BB = C C , " 0 0 #

T 1 1 where rank(AA )= r. Now put D = C− A, E = C− B and partition these as

D1 E1 D = , E = , " D2 # " E2 # where D is r m, D is (k r) m, E is r n, and E is (k r) n. Then 1 × 2 − × 1 × 2 − × T T 1 T E1E1 E1E2 1 T 1,T r 0 EE = T T = C− BB C− = " E2E1 E2E2 # " 0 0 # and T T 1 T D1D1 D1D2 1 T 1,T r 0 DD = T T = C− AA C− = " D2D1 D2D2 # " 0 0 # T T 1 which imply that E1E1 = D1D1 = r and D2 = 0, E2 = 0, so that

D1 E1 D = , E = . " 0 # " 0 #

Now let E be an (n r) n matrix such that 2 − ×

e E1 E = " E2 # e e 109 is an n n orthogonal matrix, and choose an (n r) m matrix D and an (n r) (n m) × − × 2 − × − matrix D3 such that e D1 0 e D = " D2 D3 # is an n n orthogonal matrix. Then e × e e

E1 1r 0 D1 0 1r 0 E = = E, [D, 0]= = D, " 0 # " 0 0 # " 0 0 # " 0 0 # e e and hence E = [D, 0]DT E = [D, 0]Q, where Q = DTE is n n orthogonal. Partitioning Q as × e e eH e Q = , " P # where H is m n and P is (n m) n, we then have HHT = 1 and × − × m 1 1 C− B = E = DH = C− AH so that B = AH, completing the proof.

Proposition 8.4. Let A be an n m real matrix of rank m(6 n). Then: × (i) A can be written as A = H B, where H is n m with HT H = 1 and B is m m positive 1 1 × 1 1 m × deﬁnite.

(ii) A can be written as 1m A = H B, " 0 # where H is n n orthogonal and B is m m positive definite. × × Proof. As for (i), let B := √AT A be the positive definite square root of the positive definite matrix AT A, so that AT A = B2 = BT B.

Now by using Theorem 8.3, A can be written as A = H B, where H is n m with HT H = 1 . 1 1 × 1 1 m As for (ii), let H be the matrix in (i) such that A = H B and choose an n (n m) matrix H so 1 1 × − 2 that H = [H , H ] is n n orthogonal. Then 1 2 ×

1m A = H1B = H B. " 0 #

We are done.

We now turn to decompositions of positive deﬁnite matrices in terms of triangular matrices.

110 Theorem 8.5. If A is an m m positive deﬁnite matrix, then there exists a unique m m upper-triangular × × matrix T with positive diagonal elements such that A = TT T.

Proof. An induction proof can easily be constructed. The stated result holds trivially for m = 1. Suppose the result holds for positive deﬁnite matrices of size m 1. Partition the m m matrix − × A as A11 a12 A = T , " a12 a22 # where A is (m 1) (m 1). By the induction hypothesis there exists a unique (m 1) 11 − × − − × (m 1) upper-triangular matrix T with positive diagonal elements such that A = TT T . − 11 11 11 11 Now suppose that

T T T A11 a12 T 0 T11 x T T11 T x A = = 11 = 11 11 , T T T T 2 " a12 a22 # " x y #" 0 y # " x T11 x x + y #

1 T 1 where x is (m 1) 1 and y R . For this to hold we must have x = (T )− a , and then − × ∈ 11 12 2 T T 1 T 1 T 1 y = a x x = a a T− (T )− a = a a A− a . 22 − 22 − 12 11 11 12 22 − 12 11 12 Note that this is positive, and the unique y > 0 satisfying this is

T 1 y = a a A− a . 22 − 12 11 12 q This completes the proof.

Theorem 8.6. If A is an n m real matrix of rank m(6 n), then A can be uniquely written as A = H T, × 1 where H is n m with HT H = 1 and T is m m upper-triangular wit positive diagonal elements. 1 × 1 1 m × Proof. Since AT A is m m positive deﬁnite it follows form Theorem 8.5 that there exists a unique × m m upper-triangular matrix with positive diagonal elements such that AT A = TTT. By The- × orem 8.3, there exists an n m matrix H with HT H = 1 such that A = H T. Note that H is × 1 1 1 m 1 1 unique because T is unique and rank(T)= m.

Theorem 8.7. If A is an m m positive definite matrix and B is an m m symmetric matrix, there exists × × an m m nonsingular matrix L such that A = LLT and B = LDLT, where D = diag(d ,..., d ), with × 1 m 1 d1,..., dm being the characteristic roots of A− B. If B is positive definite and d1,..., dm are all distinct, L is unique up to sign changes in the first row of L.

Proof. Let √A be the positive deﬁnite square root of A. There exists an m m orthogonal matrix × H such that 1/2 1/2 T A− BA = HDH ,

111 1/2 T T where D = diag(d1,..., dm). Putting L = A H, we now have A = LL and B = LDL . Note 1 that d1,..., dm are the characteristic roots of A− B.

Now suppose that B is positive deﬁnite and the dj are all distinct. Assume that as well as A = LLT and B = LDLT, we also have A = MMT and B = MDMT, where M is m m × nonsingular. Then

T 1 1 1 T 1,T 1 1,T 1 T T, 1 M− L M− L = M− LL M− = M− AM− = M− MM M − = 1m 1 so that the matrix Q = M− L is orthogonal and QD = DQ. If Q =(qij) we then have qijdi = qijdj so that q = 0 for i = j. Since Q is orthogonal it must then have the form Q = diag( 1,..., 1), ij 6 ± ± and L = MQ. e Theorem 8.8e (SVD). If A is an m n real matrix (m 6 n), there exist an m m orthogonal matrix H × × and an n n orthogonal matrix Q such that ×

T HAQ = [Σm, 0],

Σ > 2 2 T where m = diag(d1,..., dm) for dj 0, j = 1,..., m and d1,..., dm are the characteristic roots of AA .

Proof. Let H be an orthogonal m m matrix such that AAT = HT D2H, where D2 = diag(d2,..., d2 ), × 1 m 2 > T with dj 0 for j = 1,..., m because AA is non-negative deﬁnite. Let D = diag(d1,..., dm) with d > 0 for j = 1,..., m; then AAT = (HT D)(HT D)T, and by Theorem 8.3, there exists an m n j × matrix Q with Q QT = 1 such that A = HT DQ . Choose an (n m) n matrix Q so that the 1 1 1 m 1 − × 2 n n matrix × Q1 Q = " Q2 #

T T T is orthogonal; we now have A = H DQ1 = H [Σm, 0]Q so that HAQ = [Σm, 0], and the proof is complete.

Theorem 8.9. If Z SO(m), i.e., Z is an orthogonal matrix with determinant one, then there exists an ∈ m m skew-symmetric X such that × Z = eX.

9 Appendix IV: Selberg’s integral

This section is rewritten based on Mehta’s book [20]. The well-known Selberg’s integral is calculated, and some variants and consequences are obtained as well.

112 Theorem 9.1 (Selberg’s integral). For any positive integer N, let [dx]= dx dx , 1 · · · N > ∏16i

N Φ Φ α 1 β 1 ∆ 2γ (x) (x1,..., xN)= ∏ xj − (1 xj) − (x) . (9.2) ≡ j=1 − ! | |

Then

1 1 N 1 − Γ(α + γj)Γ(β + γj)Γ(γ + 1 + γj) SN(α, β, γ) Φ(x)[dx]= ∏ , (9.3) ≡ 0 · · · 0 Γ(α + β + γ(N + j 1))Γ(1 + γ) Z Z j=0 − and for 1 6 K 6 N,

1 1 K K α + γ(N j) 1 1 ∏ xj Φ(x)[dx]= ∏ − Φ(x)[dx], (9.4) 0 · · · 0 α + β + γ(2N j 1) 0 · · · 0 Z Z j=1 ! j=1 − − Z Z valid for integer N and complex α, β, γ with

1 Re(α) Re(β) Re(α) > 0, Re(β) > 0, Re(γ) > min , , . (9.5) − N N 1 N 1 − − Aomoto’s proof. For brevity, let us write Φ [0,1]N f (x1,..., xN) (x1,..., xN)dx1 dxN f (x ,..., x ) := · · · . (9.6) h 1 N i Φ(x ,..., x )dx dx R [0,1]N 1 N 1 · · · N Firstly, we note that R

d a (x1x2 xKΦ) (9.7) dx1 · · · a N a a 1 x1x2 xK x1x2 xK =(a + α 1)x − x2 xKΦ (β 1) · · · Φ + 2γ ∑ · · · Φ. (9.8) − 1 · · · − − 1 x x x − 1 j=2 1 − j Indeed, d d a Φ a 1 Φ a Φ (x1x2 xK ) = ax1− x2 xK + x1x2 xK (x1,..., xN), (9.9) dx1 · · · · · · · · · dx1 where

d d N Φ ∆ 2γ α 1 β 1 = (x) ∏ xj − (1 xj) − . (9.10) dx1 dx1 | | j=1 − !

Since

a x1=1 [x x2 xKΦ(x1,..., xN)] = x2 xKΦ(1, x2,..., xN) 0 = 0, (9.11) 1 · · · x1=0 · · · −

113 it follows from integrating between 0 and 1 in (9.7), we get

a N a a 1 x1x2 xK x1x2 xK 0 =(a + α 1) x − x2 xK (β 1) · · · + 2γ ∑ · · · . (9.12) − h 1 · · · i− − 1 x x x − 1 j=2 1 − j Now for a = 1 and a = 2, we get

N x1x2 xK x1x2 xK 0 = α x2 xK (β 1) · · · + 2γ ∑ · · · , (9.13) h · · · i− − 1 x x x − 1 j=2 1 − j 2 N 2 x1 x2 xK x1 x2 xK 0 = (α + 1) x1x2 xK (β 1) · · · + 2γ ∑ · · · . (9.14) h · · · i− − 1 x x x − 1 j=2 1 − j Clearly

x x x x2x x x x x x2x x 1 2 · · · K 1 2 · · · K = 1 2 · · · K 1 2 · · · K x x − x x x x − x x 1 − j 1 − j 1 − j 1 − j = x x x ; (9.15) h 1 2 · · · Ki interchanging x1 and xj and observing the symmetry,

x x x xj x2 xK 0, if 2 6 j 6 K, 1 2 · · · K = · · · = (9.16) x1 xj − x1 xj  1 < x2 xK , if K j 6 N, − −  2 h · · · i and 

2 2 1 x x x xj x2 xK x1x2 xK , if 2 6 j 6 K, 1 2 · · · K = · · · = 2 h · · · i (9.17) x1 xj − x1 xj  * + x1x2 xK , if K < j 6 N. − − h · · · i Performing the difference: (9.13) (9.14) and using the above, we get: − (α + 1 + γ(K 1)+ 2γ(N K)+ β 1) x x x =(α + γ(N K)) x x , (9.18) − − − h 1 2 · · · Ki − h 2 · · · Ki or repeating the process

(α + γ(N K)) x1x2 xK = − x1 xK 1 = (9.19) h · · · i α + β + γ(2N K 1) h · · · − i · · · − − K α + γ(N j) = ∏ − . (9.20) α + β + γ(2N j 1) j=1 − − For K = N, (9.4) can be written as

K SN(α + 1, β, γ) α + γ(N j) = x1x2 xN = ∏ − (9.21) S (α, β, γ) h · · · i α + β + γ(2N j 1) N j=1 − −

114 or for a positive integer α,

S (α, β, γ) S (α, β, γ) S (2, β, γ) N = N N (9.22) S (1, β, γ) S (α 1, β, γ) · · · S (1, β, γ) N N − N K α 1 + γ(N j) K 1 + γ(N j) = ∏ − − ∏ − (9.23) α 1 + β + γ(2N j 1) · · · 1 + β + γ(2N j 1) j=1 − − − j=1 − − K (α 1 + γ(N j)) (1 + γ(N j)) = ∏ − − · · · − . (9.24) (α 1 + β + γ(2N j 1)) (1 + β + γ(2N j 1)) j=1 − − − · · · − − Note that Γ(m + ζ)=(ζ) Γ(ζ), where (ζ) := ζ(ζ + 1) (ζ + m 1). By using this identity, m m · · · − we get Γ(α + γ(N j)) (α 1 + γ(N j)) (1 + γ(N j)) = − − − · · · − Γ(1 + γ(N j)) − and Γ(α + β + γ(2N j 1)) (α 1 + β + γ(2N j 1)) (1 + β + γ(2N j 1)) = − − . − − − · · · − − Γ(1 + β + γ(2N j 1)) − − It follows that

N Γ(α + γ(N j))Γ(1 + β + γ(2N j 1)) sN(α, β, γ)= SN(1, β, γ) ∏ − − − Γ(α + β + γ(2N j 1))Γ(1 + γ(N j)) j=1 − − − N Γ(1 + β + γ(2N j 1)) N Γ(α + γ(N j))Γ(β + γ(N j)) = SN(1, β, γ) ∏ − − ∏ − − . Γ(β + γ(N j))Γ(1 + γ(N j)) Γ(α + β + γ(2N j 1)) j=1 − − ! j=1 − − As S (α, β, γ)= S (β, α, γ) by the fact that ∆(1 x)= ∆(x), that is, S (α, β, γ) is a symmetric N N − ± N function of α and β, at the same time, the following factor is also symmetric in α and β:

N Γ(α + γ(N j))Γ(β + γ(N j)) ∏ − − . Γ(α + β + γ(2N j 1)) j=1 − − It follows that the factor

N Γ(1 + β + γ(2N j 1)) SN(1, β, γ) ∏ − − Γ(β + γ(N j))Γ(1 + γ(N j)) j=1 − − should be a symmetric function of α and β. But, however, this factor is independent of α, therefore it should be also independent of β by the symmetry. Denote this factor by c(γ, N), we get

N Γ(α + γ(N j))Γ(β + γ(N j)) SN(α, β, γ) = c(γ, N) ∏ − − Γ(α + β + γ(2N j 1)) j=1 − − N Γ(α + γ(j 1))Γ(β + γ(j 1)) = c(γ, N) ∏ − − (9.25) Γ(α + β + γ(N + j 2)) j=1 − where c(γ, N) is independent of α and β.

115 To determine c(γ, N), put α = β = 1;

N 2 2γ Γ(1 + γ(j 1)) SN(1, 1, γ)= ∆(x) dx = c(γ, N) ∏ − . (9.26) [0,1]N | | Γ(2 + γ(N + j 2)) Z j=1 −

Let y be the largest of the x1,..., xN and replace the other xj by xj = ytj, where 0 6 tj 6 1. Without loss of generality, we assume that y = x . Then x = yt for j = 1,..., N 1. Then N j j − N 1 2γ γN(N 1) 2γ − 2γ ∆(x1,..., xN) = y − ∆(t1,..., tN 1) ∏ (1 tj) , (9.27) − | | · | | · j=1 − and the Jacobian of change of variables is

t1 t2 tN 1 1 · · · − y 0 0 0 · · · ∂(x1,..., xN) N 1 det = 0 y 0 0 = y − . (9.28) ∂(y, t ,..., t ) · · · 1 N 1 . . . . . − ......

0 0 y 0 · · ·

Now we have

2γ SN(1, 1, γ)= N! ∆(x) dx 06x 6 6x 61 | | Z 1 ··· N 1 N 1 γN(N 1) N 1 2γ − 2γ = N! y − y − dy ∆(t1,..., tN 1) ∏ (1 tj) dt1 ...dtN 1 0 · 06t16 6tN 161 | − | − − Z Z ··· − j=1 1 N 1 γN(N 1) N 1 2γ − 2γ = N y − y − dy ∆(t1,..., tN 1) ∏ (1 tj) dt1 ...dtN 1 0 [0,1]N 1 | − | − − Z Z − j=1 ! 1 = SN 1(1, 2γ + 1, γ), γ(N 1)+ 1 − − that is 1 SN(1, 1, γ) = SN 1(1, 2γ + 1, γ) (9.29) γ(N 1)+ 1 − − N 1 c(γ, N 1) − Γ(1 + γ(j 1))Γ(2γ + 1 + γ(j 1)) = − ∏ − − (9.30) γ(N 1)+ 1 · Γ(1 + 2γ + 1 + γ(N 1 + j 2)) − j=1 − − N 1 c(γ, N 1) − Γ(1 + γ(j 1))Γ(1 + γ + γj) = − ∏ − . (9.31) γ(N 1)+ 1 · Γ(2 + γ(N + j 1)) − j=1 − Thus

N 2 N 1 Γ(1 + γ(j 1)) c(γ, N 1) − Γ(1 + γ(j 1))Γ(1 + γ + γj) c(γ, N) ∏ − = SN(1, 1, γ)= − ∏ − . Γ(2 + γ(N + j 2)) γ(N 1)+ 1 · Γ(2 + γ(N + j 1)) j=1 − − j=1 −

116 This implies that c(γ, N) Γ(1 + γN) = , (9.32) c(γ, N 1) Γ(1 + γ) − or c(γ, N) c(γ, 2) Γ(1 + γN) Γ(1 + γ2) c(γ, N)= c(γ, 1) = c(γ, 1) . (9.33) · c(γ, N 1) · · · c(γ, 1) · Γ(1 + γ) · · · Γ(1 + γ) − Finally we get

N Γ(1 + γj) N Γ(1 + γj) c(γ, N)= c(γ, 1) = , (9.34) ∏ Γ ∏ Γ · j=2 (1 + γ) j=2 (1 + γ) where the fact that c(γ, 1)= 1 is trivial.

Remark 9.2. Note that the conclusion is derived here for integers α, β and complex γ. A slight change of reasoning due to Askey gives (9.25) directly for complex α, β and γ as follows. (9.88) and the symmetry identity that SN(α, β, γ)= SN(β, α, γ) give the ratio of SN(α, β, γ) and SN(α, β + m, γ) for any integer m, S (α, β + m, γ) S (α, β + m, γ) S (α, β + 1, γ) N = N N (9.35) S (α, β, γ) S (α, β + m 1, γ) · · · S (α, β, γ) N N − N N (β + m 1)+ γ(N j) N β + γ(N j) = ∏ − − ∏ − (9.36) α +(β + m 1)+ γ(2N j 1) · · · α + β + γ(2N j 1) j=1 − − − j=1 − − N (β + γ(N j)) = ∏ − m , (9.37) (α + β + γ(2N j 1)) j=1 − − m where we have used the notation Γ(a + m) (a) = ; m > 0; m Γ(m) i.e. (a) = 1 and (a) := a(a + 1) (a + m 1) for m > 1. Now 0 m · · · − N 2γ α 1 β+m 1 SN(α, β + m, γ) = ∆(x) ∏ x − (1 xj) − dxj (9.38) N j Z[0,1] | | j=1 − N x β+m 1 αN γN(N 1) ∆ 2γ α 1 j − = m− − − (x) ∏ xj − 1 dxj. (9.39) [0,m]N | | − m Z j=1 Thus N (α + β + γ(2N j 1))m SN(α, β, γ)= SN(α, β + m, γ) ∏ − − (9.40) · (β + γ(N j)) j=1 − m N β+m 1 N xj − (α + β + γ(2N j 1)) αN γN(N 1) ∆ 2γ α 1 m = m− − − (x) ∏ xj − 1 dxj ∏ − − (9.41) [0,m]N | | − m (β + γ(N j)) Z j=1 j=1 − m N N β+m 1 (α + β + γ(2N j 1)) xj − m α γ(N 1) ∆ 2γ α 1 = ∏ − − m− − − (x) ∏ xj − 1 (9.42)dxj. (β + γ(N j)) [0,m]N | | − m j=1 − m Z j=1

117 Denote a = α + γ(N j), b = β + γ(N j) and c = α + β + γ(2N j 1), we have j − j − j − − Γ(c+m) (c) Γ(b) Γ(c + m) Γ(b) Γ(m)mc m mb c = mb c = . − − Γ(b+m) (b)m Γ(c) Γ(b + m) Γ(c) Γ(m)mb

By using the fact that Γ(m + c) lim = 1 ( c R), m ∞ Γ( ) c → m m ∀ ∈ it follows that

(c)m b c Γ(b) lim m − = , (9.43) m ∞ ( ) Γ( ) → b m c therefore N N Γ (cj)m b c (bj) lim ∏ m j− j = ∏ . (9.44) m ∞ ( ) Γ( ) → j=1 bj m j=1 cj

Taking m ∞ in (9.40) gives rise to the following: → N Γ ∞ ∞ N (bj) 2γ α 1 SN(α, β, γ)= ∏ ∆(x) ∏ x − exp xj dxj. (9.45) Γ(c ) j j=1 j ! Z0 · · · Z0 | | j=1 − Furthermore,

∞ ∞ N ∆ 2γ ∏N α 1 Γ(aj)Γ(bj) 0 0 (x) j=1 xj − exp xj dxj SN(α, β, γ)= ∏ · · · | | − . (9.46) Γ N Γ j=1 (cj) ! R R ∏j=1 (aj)

By the symmetry of α and β, it follows that the factor

∞ ∞ ∆ 2γ ∏N α 1 0 0 (x) j=1 xj − exp xj dxj · · · | | − (9.47) N Γ R R ∏j=1 (aj) is a symmetric function of α and β. Since it is independent of β, it is also independent of α by the symmetry of α and β. Thus

∞ ∞ 2γ N ∞ ∞ 2γ N β 1 ∆ ∏ α 1 ∆ ∏ − 0 0 (x) j=1 xj − exp xj dxj 0 0 (x) j=1 xj exp xj dxj · · · | | − = · · · | | − ,(9.48) N Γ N Γ R R ∏j=1 (aj) R R ∏j=1 (bj) which is denoted by c(γ, N). This indicates that

∞ ∞ N N ∆ 2γ α 1 Γ (x) ∏ xj − exp xj dxj = c(γ, N) ∏ (aj) (9.49) Z0 · · · Z0 | | j=1 − j=1 N Γ(1 + γj) N N Γ(α + γ(j 1))Γ(1 + γj) = Γ(a )= . (9.50) ∏ Γ ∏ j ∏ Γ − j=2 (1 + γ) j=1 j=1 (1 + γ)

118 Remark 9.3. Now we show that Γ(n + α) lim = 1 ( α R+ 0 ). n ∞ Γ( ) α → n n ∀ ∈ ∪ { } Indeed, in order to prove this fact, we need a limit representation of the gamma function given 1 n ∞ (1+ k ) by Carl Friedrich Gauss via Euler’s representation of n! = ∏k=1 n : 1+ k n!nz Γ(z)= lim . n ∞ z(z + 1) (z + n) → · · ·

Γ(n + α) Γ(α)α(α + 1) (α + n 1) lim = lim · · · − (9.51) n ∞ Γ(n)nα n ∞ (n 1)!nα → → − α(α + 1) (α + n) n = Γ(α) lim · · · (9.52) n ∞ α + → n!n n α α(α + 1) (α + n) n = Γ(α) lim · · · lim (9.53) n ∞ α n ∞ + → n!n → n α 1 = Γ(α) 1 = 1. (9.54) · Γ(α) · Another short and elementary proof of this fact can be derived from a result related to inequalities for Gamma function ratios [31]:

a 1 Γ(x + a) a x(x + a) − 6 6 x ( a [0, 1]). (9.55) Γ(x) ∀ ∈ Indeed, Γ(x + α) xα (9.56) Γ(x) ∼ as x ∞ with α ﬁxed.(This was the objective of Wendel’s article.) To show this, ﬁrst suppose → that α [0, 1]. Then Eq. (9.55) gives ∈ α α 1 Γ(x + α) 1 + − 6 6 1, x Γ(x)xα Γ(x+α) α α 1 > leasing to limx ∞ Γ α = 1 since limx ∞ 1 + − = limx ∞ 1 = 1. For α 1, the statement → (x)x → x → now follows from the fact that Γ(x + α) Γ(x + α 1) =(x + α 1) − . Γ(x) − Γ(x) We are done.

Corollary 9.4 (Laguerre’s integral). By letting xj = yj/L and β = L + 1 in Selberg’s integral and taking the limit L ∞, we obtain → ∞ ∞ N N 2γ α 1 Γ(α + γ(j 1))Γ(1 + γj) ∆(x) ∏ x − exp xj dxj = ∏ − . (9.57) j Γ(1 + γ) Z0 · · · Z0 | | j=1 − j=1 119 2 Corollary 9.5 (Hermite’s integral). By letting xj = yj/L and α = β = λL + 1 in Selberg’s integral and taking the limit L ∞, we obtain → +∞ +∞ N N 2γ 2 N/2 N(γ(N 1)+1)/2 Γ(1 + γj) ∆(x) ∏ exp λxj dxj =(2π) (2λ)− − ∏ . (9.58) ∞ · · · ∞ | | − Γ(1 + γ) Z− Z− j=1 j=1 Remark 9.6. For an integer γ, the last equation can also be written as a ﬁnite algebraic identity. Firstly, we note that +∞ i d n +∞ exp a2x2 2iaxλ xndx = exp a2x2 2iaxλ dx (9.59) ∞ − − 2a dλ ∞ − − Z− Z− i d n +∞ = exp( λ2) exp (ax + iλ)2 dx (9.60) 2a dλ − ∞ − Z− √π i d n = exp( λ2), (9.61) a 2a dλ − letting λ = 0 in the above reasoning, we get +∞ √π i d n √π i d n exp a2x2 xndx = exp( λ2) = exp( x2) , (9.62) ∞ − a 2a dλ − a 2a dx − Z− λ=0 x=0 we replace a by √a in the last equation, we get

+∞ π i d n exp ax2 xndx = exp( x2) , (9.63) ∞ − a 2√a dx − Z− r x=0 thus (9.58) therefore takes the form:

γN(N 1) 2γ N N i − ∂ ∂ 2 γN(N 1)/2 (jγ)! ∏ exp ∑ xj =(2a)− − ∏ .(9.64) 2√a 16p

N 2 ℓ Replacing the exponential by its power series expansion, one notes that the term ( ∑ x ) − j=1 j gives zero on differentiation if ℓ< γN(N 1)/2, and leaves a homogeneous polynomial of order − ℓ γN(N 1)/2 in the variables x ,..., x , if ℓ> γN(N 1)/2. On setting x = 0, j = 1,..., N, − − 1 N − j one sees that therefore that there is only one term, corresponding to ℓ = γN(N 1)/2, which − gives a non-zero contribution. So ℓ 2γ N N ∂ ∂ ℓ (jγ)! ∏ ∑ x2 = 2 ℓ! ∏ , (9.65) ∂x − ∂x j γ! 16p

120 Corollary 9.7. It holds that

2γ N dθ 2π 2π 2γ N dθ ∆(eiθ1 ,..., eiθN ) ∏ k = ∏ eiθi eiθj ∏ k (9.67) [0,2π]N 2π 0 · · · 0 6 < 6 − 2π Z k=1 Z Z 1 i j N k=1 (Nγ)! = , (9.68) (γ!)N where γ is non-negative integer.

Corollary 9.8. It holds that when there is no overlap in the two sets of factors, the result is

1 1 K1 K1+K2 (K1, K2) = ∏ xi ∏ (1 xj)Φ(x)dx (9.69) B 0 · · · 0 − Z Z i=1 j=K1+1

K1 K2 ∏i=1(α + γ(N i)) ∏j=1(β + γ(N j)) = N(α, β, γ) − − , (9.70) I ∏K1+K2 (α + β + γ(2N k 1)) k=1 − − where K1, K2 > 0, K1 + K2 6 N, and when there is overlap

K K +K K 1 1 1 1 2− 3 (K1, K2, K3) = ∏ xi ∏ (1 xj)Φ(x)dx (9.71) C 0 · · · 0 − Z Z i=1 j=K1+1 K3 − K3 α + β + γ(N k 1) = (K1, K2) ∏ − − , (9.72) B α + β + 1 + γ(2N k 1) k=1 − − where K , K , K > 0, K + K K 6 N. 1 2 3 1 2 − 3 Remark 9.9. Still another integral of interest is the average value of the product of traces of the matrix in the circular ensembles. For example,

2p 1 (γ!)N 2π 2π N 2γ N iθk iθi iθj SN(p, γ) := N ∑ e ∏ e e ∏ dθk (9.73) (2π) (Nγ)! 0 · · · 0 6 < 6 − Z Z k=1 1 i j N k=1

is known for γ = 1 that SN(p, 1) gives the number of permutations of (1,..., k) in which the length of the longest increasing subsequence is less than or equal to N. One has in particular,

SN(p, 1)= p!, 0 6 p 6 N.

It is desirable to know the integrals SN(p, γ) for a general γ.

The following short proof of Selberg’s formula is from [2].

Anderson’s proof of Selberg’s Integral. Anderson’s proof depends on Dirichlet’s generalization of the beta integral given in the following: For Re(αj) > 0,

n ∏ Γ(αj) α0 1 α1 1 αn 1 j=1 p0 − p1 − pn − dp0 dpn 1 = , (9.74) · · · V · · · · · · − n Z Z Γ ∑j=1 αj 121 n where V is the set pj > 0, ∑j=0 pj = 1. The formula is used after a change of variables. To see this, ﬁrst consider Selberg’s integral, which may be written as

1 x x n 2 α 1 β 1 γ Sn = n!An(α, β, γ) := n! φ(0) − φ(1) − ∆φ dx1 dxn, (9.75) Z0 Z0 · · · Z0 | | | | · · · where 0 < x < x < < x < 1, 1 2 · · · n n n n 1 n φ(t)= ∏(t xj)= t φn 1t − + +( 1) φ0 (9.76) − j=1 − − · · · − and ∆φ is the discriminant of φ, so that

2 n ∆φ = ∏ φ′(xj) = ∏ (xi xj) . 6 < 6 − j=1 1 i j n

We now change the variables from x 1,..., xn to φ0,..., φn 1, which are the elementary symmetric − functions of the xi’s. In fact, we have:

1 α 1 β 1 γ 2 An(α, β, γ)= φ(0) − φ(1) − ∆φ − dφ0dφ1 dφn 1, (9.77) Z | | | | · · · − where the integration is over all points (φ0, φ1,..., φn 1) in which the φj are elementary sym- − metric functions of x ,..., x with 0 < x < < x . Indeed, it is sufﬁcient to prove that the 1 n 1 · · · n Jacobian ∂φ det i = ∆ . ∂x φ j q Observe that two columns of the Jacobian are equal when x = x . Thus ∏ < (x x ) is a factor i j i j i − j of the determinant. Moreover, the Jacobian and ∏ < (x x ) are homogeneous and of the same i j i − j degree. This proves the above Jacobian. We make a similar change of variables in Eq. (9.74). To accomplish this, set

n ϕ(t)= ∏(t ζj) (0 6 ζ0 < ζ1 < < ζn < 1) j=0 − · · · and let

n = ∏(t xj) : ζj 1 < xj < ζj; j = 1,..., n . (9.78) − D (j=1 − )

n n 1 n Next, we show that: For all φ(t)= t φn 1t − + +( 1) φ0 , the following map − − · · · − ∈ D

φ(ζ0) φ(ζn) n+1 (φ0, φ1,..., φn 1) ,..., (p0, p1,..., pn) R − 7→ ϕ (ζ ) ϕ (ζ ) ≡ ∈ ′ 0 ′ n > n where ϕ′(t) denotes the derivative of ϕ(t), is a bijection and pj 0 with ∑j=0 pj = 1.

122 Observe that

φ(ζj) (ζj x1)(ζj x2) (ζj xn) pj = = − − · · · − > 0 (9.79) ϕ′(ζj) (ζj ζ0) (ζj ζj 1)(ζj ζj+1) (ζj ζn) − · · · − − − · · · − ϕ(t) since the numerator and denominator have exactly n j negative factors. Now let ϕj(t) = t ζ . − − j By Lagrange’s interpolation formula

n n ϕj(t) φ(t)= ∑ pj ϕj(t) ∑ φ(ζj). (9.80) j=0 ≡ j=0 ϕ′(ζj)

One can directly verify this by checking that both sides of the equation are polynomials of degree n n and are equal at n + 1 points t = ζj, j = 0,..., n. Equate the coefﬁcients of t on both sides to n n > get ∑j=0 pj = 1. Now for a given point (p0, p1,..., pn) with ∑j=1 pj = 1 and pj 0, j = 1,..., n, deﬁne φ(t) by Eq. (9.80). The expressions

φ(ζj)= pj ϕj(ζj)= pj(ζj ζ0) (ζj ζj 1)(ζj ζj+1) (ζj ζn) − · · · − − − · · · − and

φ(ζ )= p ϕ (ζ )= p (ζ ζ ) (ζ ζ )(ζ ζ ) (ζ ζ ) j+1 j+1 j+1 j+1 j+1 j+1 − 0 · · · j+1 − j j+1 − j+2 · · · j+1 − n show that φ(ζj) and φ(ζj+1) have different signs and φ vanishes at some point xj+1 between ζj and ζ . Thus φ . This proves the bijection. j+1 ∈ D We can now restate Dirichlet’s formula Eq. (9.74) as:

α 1 n ∏n ϕ (ζ ) j− 2 Γ(α ) αj 1 j=0 ′ j j ∏ φ(ζj) − dφ0 dφn 1 = . (9.81) φ(t) · · · − n Z j=0 Γ ∑ αj ∈D j= 1 Indeed, αj 1 αj 1 α 1 φ(ζj) − φ(ζj) − j− pj = = α 1 , ϕ′(ζj) ϕ (ζ ) j− ′ j hence

α0 1 α1 1 αn 1 p0 − p1 − pn − dp0 dpn 1 (9.82) Z · · · ZV · · · · · · − n 1 α 1 j− = α 1 ∏ φ(ζj) dp0 dpn 1. (9.83) n j φ(t) · · · − ∏j=0 ϕ′(ζj) − Z j=0 ∈D

We need to verify that the Jacobian

n 1 ∂(p0,..., pn 1) 2 det − = ∏ ϕ′(ζj) − , ∂(φ ,..., φ ) 0 n 1 j=0 −

123 1 1 n 2 n 2 that is, dp0 dpn 1 = ∏j=0 ϕ′(ζj) − dφ0 dφn 1 or dφ0 dφn 1 = ∏j=0 ϕ′(ζj) dp0 dpn 1. · · · − · · · − · · · − · · · − Since 1 n n 1 n pj = ζj φn 1ζj − + +( 1) φ0 , ϕ′(ζj) − − · · · − the Jacobian is

j 1 j n det ζi det ζ ∏ ϕ (ζ ) 2 ∂(p0,..., pn 1) n 1 i n j=0 ′ j det − = − = = . n 1 n n ∂(φ0,..., φn 1) ∏ − ϕ (ζj) ∏j=0 ϕ′(ζj) ∏j=0 ϕ′(ζj) − j=0 ′

The numerator is a Vandermonde determinant and therefore the result follows:

α0 1 α1 1 αn 1 p − p − pn − dp0 dpn 1 0 1 − Z · · · ZV · · · · · · n n 1 1 αj 1 2 = 1 ∏ φ(ζj) − ∏ ϕ′(ζj) dp0 dpn 1 n αj − ∏ ϕ (ζ ) − 2 Zφ(t) j=0 j=0 · · · ! j=0 ′ j ∈D n 1 αj 1 = 1 ∏ φ(ζj) − dφ0 dφn 1. n αj − ∏ ϕ (ζ ) − 2 Zφ(t) j=0 · · · j=0 ′ j ∈D

The ﬁnal step is to obtain the ( 2n 1)-dimensional integral. Let φ(t) and Φ(t) be two polynomials − such that

n 1 n − φ(t)= ∏(t xi) and Φ(t)= ∏(t yj), (9.84) i=1 − j=1 −

0 < y1 < x1 < y2 < < xn 1 < yn < 1. (9.85) · · · − The resultant of φ and Φ, denoted R(φ, Φ), is given by

n n 1 − R(φ, Φ) = ∏ (xi yj) = ∏ φ(yj) = ∏ Φ(xi) . (9.86) | | − i [n 1];j [n] j=1 i=1 ∈ − ∈

The absolute value of the discriminant of φ can be written as R( φ, φ ) . That is, | ′ | n 1 2 − ∆φ = ∆(x) = ∏ φ′(xj). | | j=1

The (2n 1)-dimensional integral is − α 1 β 1 γ 1 Φ(0) − Φ(1) − R(φ, Φ) − dφ0 dφn 2dΦ0 dΦn 1 Φ − Z(φ, ) | | | | | | · · · · · · − γ 1 n − α 1 β 1 = Φ(0) − Φ(1) − ∏ φ(yj) dφ0 dφn 2dΦ0 dΦn 1. (9.87) (φ,Φ) | | | | · · · − · · · − Z j=1

124 Here the integration is over all φ and Φ deﬁned by Eq. (9.84). Then we show that Selberg’s integral An (α, β, γ) satisﬁes the recurrence relation: Γ(α)Γ(β)Γ(γn) An (α, β, γ) = An 1 (α + γ, β + γ, γ) . (9.88) Γ(α + β + γ(n 1)) − − In fact, integrate the (2n 1)-dimensional integral Eq. (9.87) with respect to dφ0 dφn 2 and − · · · − use Φ(t) instead of ϕ(t) in Eq. (9.81) to get

γ 1 n − α 1 β 1 Φ(0) − Φ(1) − ∏ φ(yj) dφ0 dφn 2dΦ0 dΦn 1 (φ,Φ) | | | | · · · − · · · − Z j=1 n α 1 β 1 γ 1 = Φ(0) − Φ(1) − ∏ φ (yj) − dφ0 dφn 2 dΦ0 dΦn 1 Φ − − Z | | | | Zφ j=1 · · · ! · · ·

γ 1 ∏n Φ − 2 Γ α 1 β 1 j=1 ′(yj) (γ) = Φ(0) − Φ(1) − dΦ0 dΦn 1 Φ | | | |  Γ(γn )  · · · − Z  γ 1  n n − 2 n Γ(γ) α 1 β 1 Γ(γ) = Φ(0) − Φ(1) − ∏ Φ′(yj) dΦ0 dΦn 1 = An(α, β, γ). Γ(γn) Φ | | | | · · · − Γ(γn) Z j=1

It remains to compute Eq. (9.87) in another way, set φ(t)= t ∏n (t x ), and j=1 − j α = α, α = γ(j = 1,..., n 1), α = β; x = 0, x = 1 0 j − en 0 n so that Eq. (9.87) is equal to

γ 1 n 1 − α 1 β 1 − Φ(0) − Φ(1) − ∏ Φ(xj) dΦ0 dΦn 1dφ0 dφn 2 (φ,Φ) | | | | · · · − · · · − Z j=1 n αj 1 = ∏ Φ(xj) − dΦ 0 dΦn 1dφ0 dφn 2. (9.89) Φ − − Z(φ, ) j=0 · · · · · ·

Now integrate Eq. (9.89) with respect to dΦ0 dΦn 1 and use φ instead of ϕ(t) in Eq. (9.81) to · · · − obtain 1 e γ 2 Γ Γ Γ n 1 n 1 − 1 1 (α) (β) (γ) − − α 2 β 2 ∏ φ′(xj) φ′(0) − φ′(1) − dφ0 dφn 2. Γ(α + β + γ(n 1)) φ · · · − − Z j=1

e e e Since n 1 − φ′(0) = ∏ xj , j=1

n 1 e − φ′(1) = ∏(1 xj) , j=1 −

n n 1 n 1 e − − ∏ φ′(xj) = ∏ xj ∏ 1 xj ∆φ , j=1 j=1 j=1 −

125 the last integral can be written as

n 1 n 1 Γ(α)Γ(β)Γ(γ) − − α+γ 1 + γ 1 β γ 1 ∆ − 2 ∏ xj − (1 xj) − φ dφ0 dφn 2 Γ(α + β + γ(n 1)) φ − · · · − − Z j=1 ! n 1 Γ(α)Γ(β)Γ(γ) − = An 1(α, β, γ). Γ(α + β + γ(n 1)) − − Equate the two different evaluations of the (2n 1)-dimensional integral to obtain the result. − Finally, Selberg’s formula is obtained by iterating Eq. (9.88) (n 1) times. −

Acknowledgement

Partial material of the present work is completed during a research visit to Chern Institute of Mathematics, at Nankai University. The author would like, in particular, to thank Seunghun Hong for his remarks about the approach toward the volume of unitary group via Macdonald’s method for the volume of a compact Lie group. Both Zhen-Peng Xu and Zhaoqi Wu are acknowl- edged for valuable comments for the earlier version of the paper. LZ is grateful to the ﬁnancial support from National Natural Science Foundation of China (No.11301124).

References

[1] T. Abe, M. Barakat, M. Cuntz, T. Hoge, and H. Terao, The freeness of ideal subarrangements of Weyl arrangements, J. Eur. Math. Soc. 18, 1339-1348 (2016).

[2] G.E. Andrews, R. Askey, and R. Roy, Special Functions, Cambridge University Press (1999).

[3] L.J. Boya, The geometry of compact Lie groups, Rep. Math. Phys. 30(2), 149-162 (1991).

[4] L.J. Boya, E.C.G. Sudarshan, T. Tilma, Volumes of compact manifolds, Rep. Math. Phys. 52(3), 401-422 (2003).

[5] P. Deift, D. Gioev, Random Matrix Theory: Invariant Ensembles and Universality, AMS (2009).

[6] J.J. Duistermaat, J.A.C Kolk, Lie Groups, Springer-Verlage New York, Inc. (2000).

[7] A. Edelman, Eigenvalues and condition numbers of random matrices, Ph.D. Thesis (1989).

[8] K. Fujii, Introduction to Grassmann manifolds and quantum computation, J. Appl. Math. 8, 371-405 (2002).

[9] G. Harder, A Gauss-Bonnet formula for discrete arithmetically deﬁned groups, Ann. Ec. Norm. Sup. 4, 409-445 (1971).

126 [10] Y. Hashimoto, On Macdonald’s formula for the volume of a compact Lie group, Comment. Math. Helv. 72, 660-662 (1997).

[11] J.E. Humphreys, Reﬂection groups and Coxeter groups, Cambridge University Press (1990).

[12] K. Iwasaki, Basic invariants of ﬁnite feﬂection groups, J. Algebra 59 538-547 (1997).

[13] A.W. Knapp, Lie groups beyound an introduction, 2nd, Birkhäuser (2002).

[14] S. Kumar and A. Pandey, Entanglement in random pures states: spectral density and average von Neumann entropy, J. Phys. A : Math. Theor. 44, 445301 (2011).

[15] B. Kostant, The principal three-dimensional subgroup and the Betti numbers of a complex simple Lie group, Amer. J. Math. 81, 973-1032 (1959).

[16] V. Link and W.T. Strunz, Geometry of Gaussian quantum states, J. Phys. A : Math. Theor. 48, 275301 (2015).

[17] M.S. Marinov, Invariant voulmes of compact groups, J. Phys. A : Math. Gen. 13, 3357-3366 (1980); Correction to ‘Invariant voulmes of compact groups’, J. Phys. A : Math. Gen. 14, 543-544 (1981).

[18] M.S. Marinov and M.V. Terentyev, Dynamics of the group manifold and path integrals, Fort. D. Phys. 27, 511-545 (1979).

[19] A.M. Mathai, Jacobians of Matrix Transformations and Functions of Matrix Arguments, World Scientiﬁc (1997).

[20] M.L. Mehta, Random Matrices, 3nd Edition, Elsevier Academic Press (2004).

[21] I.G. Macdonald, The Poincaré series of a Coxeter group, Math. Ann. 199, 161-174 (1972).

[22] I.G. Macdonald, The volume of a compact Lie group, Inven. Math. 56: 93-95 (1980).

[23] I.G. Macdonald, Some conjectures for root systems, SIAM J. Math. Anal. 13(6), (1982).

[24] S. Milz and W.T. Strunz, Volumes of conditioned bipartite state spaces, J. Phys. A : Math. Theor. 48, 035306 (2015).

[25] R.J. Muirhead, Aspects of Multivariate Statistics Theory, John Wiley & Sons, New York (2005).

[26] K.R. Parthasarathy, Extremal quantum states in coupled systems, Ann. I. H. Poincaré–PR 41, 257-268 (2005).

127 [27] F.W. Ponting, H.S.A. Potter, The volume of orthognal and unitary space, Q. J. Math os-20(1): 146-154 (1949).

[28] R. Steinberg, Finite reﬂection groups, Trans. Amer. Math. Soc. 91(3), 493-504 (1959).

[29] H.J. Sommers and K. Zyczkowski,˙ Bures volume of the set of mixed quantum states, J. Phys. A : Math. Gen. 36, 10083-10100 (2003).

[30] L. Wei, R.A. Pitaval, J. Corander, and O. Tirkkonen, From random matrix theory to coding theory: volume of a metric ball in unitary group, arXiv: 1506.07259

[31] J.G. Wendel, Note on the gamma function, Amer. Math. Monthly 55, 563-564 (1948).

[32] L. Zhang, Matrix integrals over unitary groups: An application of Schur-Weyl duality, arXiv: 1408.3782

[33] L. Zhang, A.K. Pati, and J. Wu, Interference visibility, entanglement, and quantum correlation, Phys. Rev. A 92, 022316 (2015).

[34] L. Zhang, L. Chen, and K. Bu, Fidelity between a bipartite state and another one undergoing local unitary dynamics, Quant. Inf. Processing 14, 4715 (2015).

[35] K. Zyczkowski,˙ P. Horodecki, A. Sanpera, and M. Lewenstein, Volume of the set of separable states, Phys. Rev. A 58, 883 (1998).

[36] K. Zyczkowski,˙ Volume of the set of separable states. II, Phys. Rev. A 60, 3496 (1999).

[37] K. Zyczkowski,˙ W. Slomczynski, The Monge metric on the sphere and geometry of quantum states, J. Phys. A : Math. Gen. 34, 6689 (2001).

[38] K. Zyczkowski,˙ H.-J. Sommers, Hilbert-Schmidt volume of the set of mixed quantum states, J. Phys. A : Math. Gen. 36, 10115 (2003).

128