A Fixed Point Theorem for Volume Preserving Linear Transformations

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International Journal of Pure and Applied Mathematics Volume 106 No. 3 2016, 957-964 ISSN: 1311-8080 (printed version); ISSN: 1314-3395 (on-line version) url: http://www.ijpam.eu AP doi: 10.12732/ijpam.v106i3.19 ijpam.eu

A FIXED POINT THEOREM FOR VOLUME PRESERVING LINEAR TRANSFORMATIONS

Martin Moskowitz Ph.D. Program in Mathematics City University of New York Graduate Center 365 Fifth Ave, New York, NY 10016-4309, USA

Abstract: In this note we derive consequences of the fact that if g ∈ SL(n), where n ≥ 2, and σi(g) are the coeﬃcients of its characteristic polynomial, then g has 1 as an eigenvalue if n−1 i−1 − σi g n and only if Pi=1 ( 1) ( ) = 0 or 2 according to the parity of . These are: Corollary 2. Let D be a real or complex n × n matrix of tr(D) = 0. If D is singular, then one of the following equations holds (according to the parity of n).

n−1 n−1 i−1 i−1 X(−1) σi(Exp D) = 0 X(−1) σi(Exp D) = 2. i=1 i=1 Conversely, if D has no eigenvalues in 2πiZ and the appropriate one of these equations holds, D must be singular.

Corollary 3. Let A ∈ Mn(R) = M which acts on M by adA(X) = [A,X]. If A has only non-negative eigenvalues and one of the following equations holds (according to the parity of n) 2 2 n −1 n −1 i−1 i−1 X (−1) σi(AdExp(A)) = 0 X (−1) σi(AdExp(A)) = 2, i=1 i=1 then [A,X] = 0 for some X ∈ M, not a linear combination of A and I.

Corollary 4. Let A ∈ M and have all non-negative eigenvalues. Then the action of Exp(A) on M by conjugation has a ﬁxed point which is not a linear combination of A and I.

AMS Subject Classiﬁcation: 15A04, 15A16, 15A24, 15A27, 17B40, 20G20, 22E15 Key Words: ﬁxed point, eigenvalue, determinant, trace, exponential map, singular matrix, adjoint action, Lie algebra, centralizer

Received: December 17, 2015 c 2016 Academic Publications, Ltd. Published: March 4, 2016 url: www.acadpubl.eu 958 M. Moskowitz

Let K be a ﬁeld, g GL(n, K), where n 2 and ∈ ≥ − − χ (x) = xn σ (g)xn 1 + σ (g)xn 2 ... + ( 1)n det(g) g − 1 2 − − be its characteristic polynomial. Since g is invertible, it not only acts on Kn, but also on Kn (0). Here we shall be interested in when g has a ﬁxed point under \ this action, or equivalently when g has 1 as an eigenvalue. We consider the − map GL(n, K) Kn 1 given by g (σ (g), . . . , σ − (g)). For this to occur → 7→ 1 n 1 the image of g under this map must lie in a certain hyperplane and conversely. We consider the following equation.

n−1 − ( 1)i 1σ (g) = 1 + ( 1)n det(g). (1) X − i − i=1 When g SL(n, K) equation 1 becomes ∈

n−1 − ( 1)i 1σ (g) = 1 + ( 1)n. (2) X − i − i=1 Theorem 1. Let g GL(n, K). Equation 1 holds, if and only if g has a ∈ ﬁxed point in this action. (Here the parity of n doesn’t matter.) Hence when g SL(n, K), g has a ﬁxed point if and only if equation 2 holds. That is, ∈n−1 i−1 ( 1) σi(g) = 0 or 2 according to the parity of n. Pi=1 − Proof. If g GL(n, K) and the equation holds, then evidently χ (1) = 0 ∈ g so 1 is an eigenvalue and conversely.

When g SL(n, K) this result seems remarkably similar to the fact that ∈ the Euler characteristic, χ(Sn), of the n sphere is 1+( 1)n expressing the − fact that the Euler characteristic based on chains is the same as that based on homology. We illustrate it with some examples.

Let g SL(2,K), where K = R or C acting as usual on K2. Then g has • ∈ a ﬁxed point if and only if tr(g) = 2. This means g has 1 as an eigenvalue with multiplicity 2 and so is conjugate to a unitriangular matrix and conversely.

Let G = SO(3, R). By Euler’s theorem every g G has a rotation axis • ∈ and eigenvalue 1. Hence tr(g) = σ (g) for all g G. 2 ∈ A FIXED POINT THEOREM FOR VOLUME... 959

Let A = (a ) be a stochastic matrix of order n (all a 0 and a = • i,j i,j ≥ j i,j 1 for all i). Then the theorem of Perron-Frobenius (see [2] pg. 17-18)P tells us, among other things, that A has 1 as an eigenvalue. Hence equation 1 holds (whether A is invertible, or not).

One can construct a large number of matrices in SL(3, Z) all of which • have 1 as an eigenvalue. For g SL(3, Z), χ (x) has its coeﬃcients in Z ∈ g and, 1 is an eigenvalue if and only if tr(g) = σ2(g). Since, as we shall see tr(g) below, σ2(g) = 0, this means = 1. Let 6 σ2(g) 0 0 1 g =  1 m p   0 1 n 

where n, m and p are integers. Then χ (x) = x3 (n+m)x2+(mn p)x 1. g − − − Thus det(g) = 1, tr(g) = n + m and σ (g) = mn p. We consider 2 − Pythagorean triples of integers, r2 = s2 + t2, and their multiples by ν = 1, 2,..., ν2r2 = ν2s2 + ν2t2. Let r and ν be chosen so that tr(g) = 2νr 2 2 and σ2(g) = ν s p. We choose p so that tr(g)/σ2(g) = 1. Since tr(g) 2νr − 2 2 = 2 2− , this will be 1 exactly when p = ν s 2νr. This determines σ2(g) ν s p − the integer p. Hence 2νr = σ (A) = mn p = mn (ν2s2 2νr). Hence 2 − − − mn = ν2s2 = ν2r2 ν2t2 = (νr + νt)(νr νt). Taking m = νr + νt − − and n = νr νt (or vice versa), g has entries m = νr + νt n = − νr νt p = ν2s2 2νr (and tr(A) = m + n = 2νr). Thus for each − − multiple of a Pythagorean triple we have a solution and there are inﬁnitely many Pythagorean triples as well.

Our theorem has the following corollary:

Corollary 2. Let D be a real or complex n n matrix of tr(D) = 0. If D × is singular, then one of the following equations holds (depending on whether n is odd or even).

n−1 n−1 − − ( 1)i 1σ (Exp D) = 0 ( 1)i 1σ (Exp D) = 2. X − i X − i i=1 i=1

Conversely, if D has no eigenvalues in 2πiZ and the appropriate one of these equations holds, D must be singular. 960 M. Moskowitz

In particular, suppose g is a real or complex Lie algebra and D is a derivation of trace 0. Let α = Exp(D) be the corresponding automorphism. If one of the equations above holds for α and α has no non-trivial eigenvalues which are roots of unity, then the derivation D is singular.1

Proof. Proof. Because tr(D) = 0, it follows that det(Exp D) = etr(D) = 1 so we can apply the theorem for SL(n, K). If D is singular, then D(v) = 0 for some v = 0. Hence Exp D(v) = v. Since Exp D has v as a ﬁxed point and 6 determinant 1, one or the other of these two equations holds. Conversely, if equation 2 holds and therefore by Theorem 1 Exp D(v) = v for some v = 0, Exp D−I 6 then D (D(v)) = 0. By our assumption on the eigenvalues of D, it follows Exp D−I that D is invertible and hence D(v) = 0.

Now let A M (R) = and have it act on by ad (X) = [A, X]. ∈ n M M A When X is in the linear span of I and A, adA(X) is automatically 0. Here we shall be interested in non-trivial such X.2 Taking the matrix units E as a basis for we see i,j M n n tr(ad ) = ad (E ) = ad ( E ) = ad (I) = 0. A X A i,i A X i,i A i=1 i=1 Let = X : tr(AX) = 0 . X { ∈ M } Then is a subspace of invariant under ad . This is because tr(A[A, X]) = X M A tr(A2X) tr(AXA) and since tr(YZ) = tr(ZY ) for all Y and Z, this is zero − (whether X , or not). ∈ X Assume tr(A2) = 0. Then 6 = , A ⊕ X M where is the line through A. = (0) because if tr(AλA) = 0, then A A ∩ X λ tr(A2) = 0 and so λ = 0. The sum of these subspaces is since if X , M ∈ M there is a real λ with tr((X λA)A) = 0. For this we take − tr(XA) λ = . tr(A2)

1Since α has no eigenvalues which are roots of unity, a theorem of N. Jacobson tells us g is actually nilpotent. See [2], Theorem 5.4.2. 2It is a pleasure for the author to acknowledge a number of useful conversations on this material with Laura Geatti. A FIXED POINT THEOREM FOR VOLUME... 961

Choosing a basis for and, by adding A, extending this to a basis of X M we see

ad X 0 A |  0  ad = . (3) A  0     0 0 0 0 

Hence Spec(ad ) = Spec(ad X ) (0) so since tr(ad ) = 0, tr(ad X ) is A A | ∪ A A | also 0. Moreover, exponentiating equation 3 we get,

Exp(ad ) X 0 A |  0  Exp(ad ) = . (4) A  0     0 0 0 1  Now since A has all real eigenvalues, it can be put in triangular form over R. Hence tr(A2) is the sum of the squares of its eigenvalues. This means tr(A2) > 0 unless all eigenvalues are zero, i.e. A is nilpotent.

We recall that Exp adA = AdExp(A) and ﬁrst assume A is not nilpotent. In [1] Section 3.3 an equivariant isomorphism φ is established,

φ : Rn Rn, M → ⊗ with ad = φ−1(A I I At)φ. Because tr(A) = tr(At), it follows that A ⊗ − ⊗ tr(ad ) = tr(A)n n tr(At) = 0. Since A has only real eigenvalues, the eigen- A − values of ad acting on being diﬀerences of the eigenvalues of A, are also A M real. Because is invariant under ad , the eigenvalues of its restrictions to X A X being a subset of these eigenvalues are also real and so ad X has no purely A | imaginary eigenvalues at all. Since adA has trace zero, it follows from equation 3, tr(ad X ) = 0. A | Moreover, from the Jordan form of A one sees easily that in each Jordan block Exp(A) is a product of a scalar matrix and a unipotent matrix. Hence on each of these subspaces AdExp(A) acts (by conjugation) as a unipotent matrix. But the eigenvalues of this action by conjugation are quotients of the eigenvalues of this unipotent matrix. Hence all the eigenvalues on each block are all 1 and so AdExp(A) = Exp(adA) is itself unipotent. Equation 4 then tells us Ad X = Exp(ad ) X is unipotent as well. By the previous corollary Exp(A) | A | there is a non-zero X with [A, X] = 0. On the other hand, if A were ∈ X nilpotent, since the eigenvalues of adA are diﬀerences of those of A, adA is a nilpotent operator on and so adj = 0 for some integer j 1. If j 2, M A ≥ ≥ 962 M. Moskowitz

− ad (X) = 0 for all X adj 1. Thus [A, X] = 0 for all these X. If j = 1, then A ∈ A [A, X] = 0 for all X . ∈ M This proves that if A has only real eigenvalues and one of the following ∈ M equations holds (according to whether n is odd or even).

n2−1 n2−1 − − ( 1)i 1σ (Ad ) = 0 ( 1)i 1σ (Ad ) = 2, X − i Exp(A) X − i Exp(A) i=1 i=1 then [A, X] = 0 for some X . ∈ M \ A We now sharpen this result to obtain non trivial X Ker(ad ). For this ∈ A we shall have to assume the eigenvalues of A are not merely real, but actually non-negative. Let = lin.sp.R A, I . If A and I are linearly dependent, the statement of B { } Corollary 3 below follows from the above so we now assume A and I are linearly independent. Consider I + λ∗A = Y∗. We choose λ∗ so that Y∗ . Since ∈ X ∩ B this is in for any λ∗, we need B 2 0 = tr((I + λ∗A)A) = tr(A) + λ∗ tr(A ).

That is, λ∗ = tr(A)/ tr(A2) and, as above, this can be done as long as A − is not nilpotent. Then Y∗ and is non-zero since A and I are lin- ∈ X ∩ B early independent. Call ∗ the 1 dimensional subspace of generated by Y∗. Y X Evidently, adA(Y∗) = 0 and in a similar manner to the above we will further decompose into a direct sum of ad -invariant spaces = ∗ ∗, where X A X Y ⊕ X ∗ = X : tr(X) = 0 . This is so since tr(X(I + λ∗A)) = 0 if and only if X { ∈ X } tr(X) + λ∗ tr(XA) = 0 and since X , tr(XA) = 0 automatically. ∈ X Now these spaces are disjoint because if I + λ∗A had trace zero, then n + λ∗ tr(A) = 0 so λ∗ = n/ tr(A) and hence − n = tr(A)2/ tr(A2), (5)

the denominator being positive since A is not nilpotent. Taking ai to be the non-negative eigenvalues of A and applying the Cauchy-Schwarz inequality to the vectors (a1, . . . , an) and (1,..., 1), we see

a √n( a2)1/2. X i ≤ X i i i

But from equation 5 it follows that a √n( a2)1/2. Hence these quan- i i ≥ i i tities are equal. By the case of equalityP of Cauchy-SchwarzP it follows that all A FIXED POINT THEOREM FOR VOLUME... 963

the ai are equal so A is a scalar and hence A and I are linearly dependent, a contradiction. Thus we get a direct sum decomposition of adA-invariant subspaces,

= ∗. M B ⊕ X

Since ad X∗ must have trace zero and Exp(ad ) has all its eigenvalues A | A equal to 1 on , the same is true when this operator is restricted to ∗. Ap- X X plying Corollary 2 (assuming A is not nilpotent) we see Corollary 3. Suppose A has only non-negative eigenvalues and one ∈ M of the following equations holds (according to whether n is odd or even)

n2−1 n2−1 − − ( 1)i 1σ (Ad ) = 0 ( 1)i 1σ (Ad ) = 2. X − i Exp(A) X − i Exp(A) i=1 i=1

Then [A, X] = 0 for some X . ∈ M \ B On the other hand if A is nilpotent then, as above, adA is nilpotent and remains so when restricted to any invariant subspace. The argument then proceeds just as before.

Denote by t(n, R) the real triangular matrices of order n and T (n, R) those with positive diagonal entries. We ﬁrst observe

Exp : t(n, R) T (n, R) → is a global diffeomorphism because T (n, R)is a connected and simply connected real solvable Lie group with all real Ad roots. Now let A M (n, R) and ∈ n assume it has all real eigenvalues. Then, as above, A can be put in triangular form over R as can Exp(A), which has all its eigenvalues positive. Since AdExp(A) = Exp(adA) and Exp is global diffeomorphism, the action of conjugation by Exp(A) on is equivariantly equivalent to the action of ad on . M A M Now assume further that the eigenvalues of A are actually non-negative. Then by Corollary 3 above there is an X so that [A, X] = 0. Therefore X ∈ M \ B also commutes with Exp(A). This yields, Corollary 4. Let A and and have all non-negative eigenvalues. ∈ M Then the action of Exp(A) on by conjugation has a non-trivial fixed point. M 964 M. Moskowitz

References

[1] H. Abbaspour, M. Moskowitz, Basic Lie Theory, World Scientiﬁc, Singapore (2007), doi: 10.1142/6462. [2] I. Farmakis, M. Moskowitz, Fixed Point Theorems and their Applications, World Scien- tiﬁc, Singapore (2013), doi: 10.1142/8748.