A Beautiful Sine Formula Author(S): Pantelis A

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A Beautiful Sine Formula Author(s): Pantelis A. Damianou Source: The American Mathematical Monthly, Vol. 121, No. 2 (February), pp. 120-135 Published by: Mathematical Association of America Stable URL: http://www.jstor.org/stable/10.4169/amer.math.monthly.121.02.120 . Accessed: 29/04/2015 16:59

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This content downloaded from 205.175.118.33 on Wed, 29 Apr 2015 16:59:12 PM All use subject to JSTOR Terms and Conditions A Beautiful Sine Formula Pantelis A. Damianou

Abstract. Let g be a complex simple Lie algebra of rank `, h the Coxeter number, m1, m2,..., m` the exponents of g, and C the Cartan matrix. Then

` Y mi π 22` sin2 = det C. 2h i=1

1. INTRODUCTION. Here, we attempt to explain in simple terms the meaning of the symbols appearing in the formula

` Y mi π 22` sin2 = det C. 2h i=1

In particular, we address the Cartan matrix, the rank, the exponents, and the Coxeter number of a complex simple Lie algebra. For more details, see [8] and [9]. In this article, we consider only ﬁnite-dimensional Lie algebras. Cartan matrices appear in the classiﬁcation of simple Lie algebras over the complex numbers. A Cartan matrix is associated to each such Lie algebra. It is an ` × ` square matrix, where ` is the rank of the Lie algebra. The Cartan matrix encodes all the properties of the simple Lie algebra it represents. Let g be a complex simple Lie algebra, h a Cartan subalgebra, and 5 = {α1, . . . α`} a basis of simple roots for the root system 1 of h in g. The elements of the Cartan matrix C are given by αi , α j ci j := 2 , (1) α j , α j where the inner product is induced by the Killing form. The ` × `-matrix C is invert- ible and called the Cartan matrix of g. The detailed machinery for constructing the Cartan matrix from the root system can be found, e.g., in [8, p. 55] or [10, p. 111]. In the following example, we give full details for the case of sl(4, C), which is of type A3.

4 Example 1. Let E be the hyperplane of R for which the coordinates sum to√ 0 (i.e., vectors orthogonal to (1, 1, 1, 1)). Let 1 be the set of vectors in E of length 2 with integer coordinates. There are 12 such vectors in all. We use the standard inner product 4 in R and the standard orthonormal basis {1, 2, 3, 4}. Then, it is easy to see that 1 = {i − j | i 6= j}. The vectors

α1 = 1 − 2,

α2 = 2 − 3, and

α3 = 3 − 4

http://dx.doi.org/10.4169/amer.math.monthly.121.02.120 MSC: Primary 17B20, Secondary 20F55; 33C45

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This content downloaded from 205.175.118.33 on Wed, 29 Apr 2015 16:59:12 PM All use subject to JSTOR Terms and Conditions form a basis of the root system in the sense that each vector in 1 is a linear combination of these three vectors with integer coefﬁcients, either all nonnegative or all nonposi- tive. For example, 1 − 3 = α1 + α2, 2 − 4 = α2 + α3, and 1 − 4 = α1 + α2 + α3. + Therefore, 5 = {α1, α2, α3}, and the set of positive roots 1 is given by

+ 1 = {α1, α2, α3, α1 + α2, α2 + α3, α1 + α2 + α3}.

Deﬁne the matrix C using (1). It is clear that cii = 2 and

(αi , αi+1) ci,i+1 = 2 = −1 for i = 1, 2. (αi+1, αi+1)

Similar calculations lead to the following form of the Cartan matrix

 2 −1 0  C = −1 2 −1 . 0 −1 2

The complex simple Lie algebras are classiﬁed as:

Al , Bl , Cl , Dl , E6, E7, E8, F4, G2.

Traditionally, Al , Bl , Cl , Dl are called the classical Lie algebras, while E6, E7, E8, F4, G2 are called the exceptional Lie algebras. Moreover, for any Cartan matrix there exists just one complex simple Lie algebra up to isomorphism that gives rise to it. The classiﬁcation is due to Killing and Cartan. According to A. J. Coleman [4], the classiﬁcation paper of Killing is the greatest mathematical writing of all time (after Euclid’s Elements and Newton’s Principia). In Table1, we list the determinants for the Cartan matrices of complex simple Lie algebras.

Table 1. Determinants of Cartan matrices

Lie algebra An Bn Cn Dn E6 E7 E8 F4 G2 det n + 1 2 2 4 3 2 1 1 1

Simple Lie algebras over C are classiﬁed by using the associated Dynkin diagram, a graph whose vertices correspond to the elements of 5. Each pair of vertices αi , α j are connected by

2 4(αi , α j ) mi j = (αi , αi )(α j , α j ) edges, where

mi j ∈ {0, 1, 2, 3}.

To a given Dynkin diagram 0 with n nodes, we associate the Coxeter adjacency matrix, which is the n × n matrix A = 2I − C, where C is the Cartan matrix. The

February 2014] A BEAUTIFUL SINE FORMULA 121

This content downloaded from 205.175.118.33 on Wed, 29 Apr 2015 16:59:12 PM All use subject to JSTOR Terms and Conditions characteristic polynomial an(x) of the adjacency matrix is simply related to the characteristic polynomial of the Cartan matrix pn(x). In fact,

an(x) = pn(x + 2). (2)

It follows that, if λ is an eigenvalue of the adjacency matrix, then 2 + λ is an eigenvalue of the corresponding Cartan matrix. Let us recall the deﬁnition of exponents for a simple complex Lie group G; see [3], [5], [12]. We form the de Rham cohomology groups H i (G, C) and the corresponding Poincare´ polynomial of G:

` X i pG (t) = bi t , i=1

i where bi = dim H (G, C) are the Betti numbers of G. The De Rham groups encode topological information about G. Following work of Cartan, Ponrjagin, and Brauer, Hopf proved that the cohomology algebra is isomorphic to that of a ﬁnite product of ` spheres of odd dimension, where ` is the rank of G. This result implies that

` Y 2mi +1 pG (t) = 1 + t . i=1

The positive integers {m1, m2,..., m`} are called the exponents of G. They are also the exponents of the Lie algebra g of G. We can also extract the exponents from the root space decomposition of g following methods developed by R. Bott, A. Shapiro, R. Steinberg, A. J. Coleman, and B. Kostant. We describe a method that was observed empirically by A. Shapiro and R. Steinberg, and proved in a uniform way by B. Kostant. If α ∈ 1+, then we can express α uniquely as a sum of simple roots, = P` α i=1 ni αi , where ni are nonnegative integers. The height of α is deﬁned to be the number

` X ht (α) = ni . i=1

The procedure follows.

+ Proposition 1. Let b j be the number of α ∈ 1 such that ht (α) = j. Then b j − b j+1 is the number of times j appears as an exponent of g. Furthermore,

` = b1 ≥ b2 ≥ · · · ≥ bh−1, where h is the Coxeter number. Moreover, the partition of r = |1+| as deﬁned by the above sequence is conjugate to the partition

h − 1 = m` ≥ m`−1 ≥ · · · ≥ m1 = 1.

The Coxeter number is determined by the relation h` = 2r.

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This content downloaded from 205.175.118.33 on Wed, 29 Apr 2015 16:59:12 PM All use subject to JSTOR Terms and Conditions Example 2. Consider the previous example of A3. Recall that 5 = {α1, α2, α3} and the set of positive roots 1+ is given by

+ 1 = {α1, α2, α3, α1 + α2, α2 + α3, α1 + α2 + α3}.

Therefore, r = 6 and the Coxeter number is determined by h` = 2r, i.e., 3h = 12, or h = 4. The number of roots of height 1 is 3; therefore, b1 = 3. Similarly, b2 = 2 and b3 = 1. Therefore, the exponents are m3 = 3, m2 = 2 and m1 = 1.

The exponents of a complex simple Lie algebra are given in Table2.

Table 2. Exponents and Coxeter number for root systems

Root system Exponents Coxeter number

An 1, 2, 3,..., n n + 1

Bn 1, 3, 5,..., 2n − 1 2n Cn 1, 3, 5,..., 2n − 1 2n Dn 1, 3, 5,..., 2n − 3, n − 1 2n − 2 E6 1, 4, 5, 7, 8, 11 12 E7 1, 5, 7, 9, 11, 13, 17 18 E8 1, 7, 11, 13, 17, 19, 23, 29 30 F4 1, 5, 7, 11 12 G2 1, 5 6

Note in the table, the well-known duality of the set of exponents

mi + m`+1−i = h, (3) where h is the Coxeter number. An alternative way to view the exponents is by considering the corresponding Cox- eter group. A Coxeter graph is a simple graph 0 with n vertices and edge weights mi j ∈ {3, 4,..., ∞}. We deﬁne mii = 1 and mi j = 2 if node i is not connected with node j. By convention, if mi j = 3, then the edge is often not labeled. If 0 is a Coxeter graph with n vertices, then we deﬁne a bilinear form B on Rn by choosing a basis e1, e2,..., en and setting π B(ei , e j ) = −2 cos . mi j

If mi j = ∞, then we define B(ei , e j ) = −2. We also define, for i = 1, 2,..., n, the reflections

σi (e j ) = e j − B(ei , e j )ei .

Let S = {σi | i = 1,..., n}. The Coxeter group W(0) is the group generated by the reﬂections in S. The group W has the presentation

2 mi j W = hσ1, σ2, . . . , σn | σi = 1, (σi σ j ) = 1i. It is well known that W(0) is ﬁnite if and only if B is positive deﬁnite. A Coxeter element (or transformation) is a product of the form

σα(1)σα(2) . . . σα(n) α ∈ Sn.

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This content downloaded from 205.175.118.33 on Wed, 29 Apr 2015 16:59:12 PM All use subject to JSTOR Terms and Conditions If the Coxeter graph is a tree, then the Coxeter elements are in a single conjugacy class in W.A Coxeter polynomial for the Coxeter system (W, S) is the characteristic polynomial of the matrix representation of a Coxeter element. For Coxeter systems whose graphs are trees, the Coxeter polynomial is uniquely determined.

Example 3. Consider a Coxeter system with graph A3. The bilinear form is deﬁned by the Cartan matrix

 2 −1 0  C = −1 2 −1 . 0 −1 2

Figure 1.

The reﬂection σ1 is determined by the action

σ1(e1) = −e1, σ1(e2) = e1 + e2, σ1(e3) = e3.

It has the matrix representation

−1 1 0 σ1 =  0 1 0 . 0 0 1

Similarly,

1 0 0 1 0 0  σ2 = 1 −1 1 , and σ3 = 0 1 0  . 0 0 1 0 1 −1

A Coxeter element is deﬁned by

0 0 −1 R = σ1σ2σ3 = 1 0 −1 . 0 1 −1

The Coxeter polynomial is the characteristic polynomial of R, which is

x 3 + x 2 + x + 1.

Note that R has order 4, i.e., R4 = I . The order of the Coxeter element is called the Coxeter number. The roots of the Coxeter polynomial are −1, −i, i. We can write i = i 1, −1 = i 2, −i = i 3. The integers 1, 2, 3 are the exponents of W. In general, the order of the Coxeter (or Weyl or Killing) group is

(m1 + 1)(m2 + 1) ··· (ml + 1), where mi are the exponents. In the present example, W = S4. To prove the sine formula, we use the following two facts.

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This content downloaded from 205.175.118.33 on Wed, 29 Apr 2015 16:59:12 PM All use subject to JSTOR Terms and Conditions Lemma 1. The eigenvalues of C occur in pairs {ξ, 4 − ξ}.

See [2, p. 345] for a general proof. A proof of this lemma is given in section8 using the duality of the exponents and the following lemma.

Lemma 2. The roots of an(x) are m π 2 cos i , h where mi are the exponents of g and h is the Coxeter number of g.

In this article, we demonstrate this lemma in an elementary fashion using only basic properties of Chebyshev polynomials. Nevertheless, for the interested reader we give some references on how to prove this lemma without a case by case veriﬁcation; there is an empirical procedure due to Coxeter, which can be used to ﬁnd the roots of the Coxeter polynomial. If ζ is a primitive h root of unity (where h is the Coxeter number), then the roots of the Coxeter polynomial are ζ m , where m runs over the exponents of the corresponding root system [3], [6]. This observation allows the calculation of the Coxeter polynomial for each root system. It also explains the duality (3) since non-real eigenvalues of R appear in conjugate pairs. Coxeter also observed that

h` = 2r, (4) where r is the number of positive roots. Using (4) as the only empirical fact, Coleman proved in [3] the procedure of Coxeter. The proof of (4) is in the classic 1959 paper of Kostant [12]. Knowing the roots of the Coxeter polynomial, it is straightforward to determine the spectrum of the Cartan matrix C; see, e.g., [2, Theorem 2]. This in turn determines the roots of an(x) via the relation (2). Alternatively, we can compute the Coxeter polynomial as follows. Deﬁne the associated polynomial

1 Q (x) = x na x + . n n x

Qn(x) turns out to be an even, reciprocal polynomial of the form

2 Qn(x) = fn(x ).

The polynomial fn is the Coxeter polynomial of the underlying graph. For more details, see [7]. The procedure in [7] allows us to connect the roots of the Coxeter polyno- = m j π mial with the roots of an(x). Let θ j h . Then the roots of the Coxeter polynomial m π 2iθ j = j are e , while the roots of an(x) are 2 cos θ j 2 cos h .

2. CHEBYSHEV POLYNOMIALS. To compute pn(x) explicitly, we use the following result (see [6] and [11]).

Proposition 2. Let C be the n × n Cartan matrix of a simple Lie algebra over C. Let pn(x) be its characteristic polynomial and deﬁne qn(x) = det(2x I + A). Then x x p (x) = q − 1 and a (x) = q . n n 2 n n 2

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This content downloaded from 205.175.118.33 on Wed, 29 Apr 2015 16:59:12 PM All use subject to JSTOR Terms and Conditions The polynomial qn is related to Chebyshev polynomials as follows:

An : qn = Un,

Bn, Cn : qn = 2Tn, and

Dn : qn = 4xTn−1 where Tn and Un are the Chebyshev polynomials of ﬁrst and second kind, respectively.

Proof. We give an outline of the proof. Note that

x x − 2 q − 1 = det 2 I + A n 2 2 n

= det (x In − 2In + A)

= det (x In − 2In + 2In − C)

= det (x In − C) = pn(x).

Furthermore, the matrix A for classical Lie algebras has the form

0 1  1 0 1   . . .   ......  A =   ,  . . .   ......     1 0 1  1 D where D is

0 1 1 0 2 0 1 (0), , , 1 0 0 1 0 2 0   1 0 0 for the cases An, Bn, Cn, Dn, respectively. The proof is by induction on n. Suppose that the result is proved for n and n − 1. To get qn+1(x), expand the determinant of 2x In+1 + A by the ﬁrst two rows to obtain

qn+1(x) = 2xqn(x) − qn−1.

This is the recurrence relation satisﬁed by the Chebyshev polynomials Tn and Un. Sections5 and6 cover in detail the cases An and Bn, respectively. In section7, we outline the case of Dn. Finally, note that

x + 2 x a (x) = p (x + 2) = q − 1 = q . n n n 2 n 2

Orthogonal polynomials are associated with three-term recurrences, which in turn correspond to tridiagonal matrices. A sequence of orthogonal polynomials Pn(x) is

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This content downloaded from 205.175.118.33 on Wed, 29 Apr 2015 16:59:12 PM All use subject to JSTOR Terms and Conditions characterized by a three-term recurrence of the form

x Pn(x) = αn Pn+1(x) + βn Pn(x) + γn Pn−1(x).

We associate to each such recurrence, a tridiagonal matrix:

β0 α0  γ1 β1 α1   . . .   ......  J + =   . n 1  . . .   ......     γn−1 βn−1 αn−1 γn βn

Each zero of the polynomial Pn(x) is an eigenvalue of Jn, and the characteristic polynomial of Jn is precisely Pn(x). In our case, the Cartan matrices are either tridiagonal or close to tridiagonal. The corresponding family of orthogonal polynomials are the Chebyshev polynomials. It is remarkable that the Chebyshev polynomials give the universal solution of any second-order recurrence relation with constant coefficients; see [1, Theorem 3.1, p. 617]. Moreover, it is clear from what follows that Chebyshev polynomials were designed in order to fit nicely the theory of complex simple Lie algebras. The precise definition and some basic properties of Un and Tn will be given in sections5 and6, respectively. In the case of exceptional Lie algebras, we can directly compute (preferably using a symbolic manipulation package) the characteristic polynomial an(x) for each exceptional type:

2 G2 : a2(x) = x − 3, 4 2 F4 : a4(x) = x − 4x + 1, 6 4 2 E6 : a6(x) = x − 5x + 5x − 1, 7 5 3 E7 : a7(x) = x − 6x + 9x − 3x, and 8 6 4 2 E8 : a8(x) = x − 7x + 14x − 8x + 1.

Lemma2 is then easily veriﬁed.

3. THE An SINE FORMULA. Toeplitz matrices have constant entries on each diagonal parallel to the main diagonal. Tridiagonal Toeplitz matrices are commonly the result of discretizing differential equations. The eigenvalues of the Toeplitz matrix

b a  c b a   . . .   ......    (5)  . . .   ......     c b a c b

February 2014] A BEAUTIFUL SINE FORMULA 127

This content downloaded from 205.175.118.33 on Wed, 29 Apr 2015 16:59:12 PM All use subject to JSTOR Terms and Conditions are given by

r c jπ λ = b + 2a cos for j = 1, 2 ..., n. (6) j a n + 1

See, e.g., [14, p. 59]. The Cartan matrix of type An is a tridiagonal Toeplitz matrix of the form

 2 −1  − −  1 2 1   . . .   ......  C =   . (7) An  . . .   ......     −1 2 −1 −1 2

Taking a = c = −1, b = 2 in (6) immediately yields the following.

Proposition 3. The eigenvalues of C An are given by

jπ jπ λ = 2 − 2 cos = 4 sin2 for j = 1, 2,..., n. j n + 1 2(n + 1)

Let dn be the determinant of C An . We can compute dn by using expansion on the ﬁrst row and induction. The result is dn = 2dn−1 − dn−2, d1 = 2, d2 = 3. This is a simple linear recurrence with solution dn = n + 1. We conclude that

n Y jπ 4 sin2 = n + 1. 2(n + 1) j=1

Equivalently,

n Y jπ 22n sin2 = n + 1 (the A sine formula). (8) 2(n + 1) n j=1

We refer to this relation as the An sine formula. This formula will be generalized for each complex simple Lie algebra.

4. PRODUCT OF DISTANCES FROM 1 TO THE OTHER ROOTS OF UNITY. The following problem is well known. Consider a regular polygon inscribed in the unit circle with one vertex at the point (1, 0). Find the product of the distances from the point (1, 0) to the other n − 1 vertices. The answer is n.

2πi Proof. Consider the unit circle |z| = 1 in the complex plane. Let ω = e n . Then, the other vertices of the polygon are the roots of unity ω, ω2, . . . , ωn−1. These numbers, together with 1, are roots of the polynomial zn − 1. Therefore,

zn − 1 = (z − 1)(z − ω) z − ω2 ··· z − ωn−1.

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This content downloaded from 205.175.118.33 on Wed, 29 Apr 2015 16:59:12 PM All use subject to JSTOR Terms and Conditions Divide both sides by z − 1 to obtain

zn−1 + zn−2 + · · · + z + 1 = (z − ω) z − ω2 ··· z − ωn−1 .

Substitute z = 1 and take absolute values to obtain the result.

= 2π k = ikθ k Let θ n . Then, ω e and the distance from 1 to ω is

√ kθ πk |1 − ωk | = |1 − (cos kθ + i sin kθ) | = 2 − 2 cos kθ = 2 sin = 2 sin . 2 n

As a result, we obtain the formula

n−1 Y kπ n sin = . (9) n 2n−1 k=1

This formula was standard in textbooks in the 19th century. For example, it is derived in S. L. Loney’s book Plane Trigonometry, one of the books that Ramanujan used to learn mathematics. We are now in a position to give a different proof of the An sine formula (8). Con- sider the formula (9) with n replaced by 2(n + 1). It becomes

2n+1 Y kπ n + 1 n + 1 sin = = . 2(n + 1) 22n 4n k=1

We split the product on the left-hand side in the following way:

n 2n+1 Y kπ Y kπ sin sin . 2n + 2 2n + 2 k=1 k=n+2

Note also that

2n+1 2n+1 Y kπ Y kπ sin = sin π − 2n + 2 2n + 2 k=n+2 k=n+2

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This content downloaded from 205.175.118.33 on Wed, 29 Apr 2015 16:59:12 PM All use subject to JSTOR Terms and Conditions 2n+1 Y 2n + 2 − k = sin π 2n + 2 k=n+2

2n−1 Y 2n − j = sin π. 2n + 2 j=n

This last product is the same as

n Y kπ sin , 2n + 2 k=1 in reversed order. Consequently,

n Y kπ n + 1 sin2 = , 2(n + 1) 4n k=1 which yields formula (8) exactly.

5. CARTAN MATRIX OF TYPE An. The Chebyshev polynomials form an inﬁnite sequence of orthogonal polynomials. The Chebyshev polynomial of the second kind of degree n is usually denoted by Un. We list some properties of Chebyshev polynomials following [13], [15]. A fancy way to deﬁne the nth Chebyshev polynomial of the second kind is

2x 1   1 2x 1   . . .   ......  U (x) = det   , (10) n  . . .   ......     1 2x 1  1 2x where n is the size of the matrix. By expanding the determinant with respect to the ﬁrst row, we get the recurrence

U0(x) = 1, U1(x) = 2x, Un+1(x) = 2xUn(x) − Un−1(x). (11)

It is easy then to compute recursively the ﬁrst few polynomials:

U0(x) = 1,

U1(x) = 2x, 2 U2(x) = 4x − 1, 3 U3(x) = 8x − 4x, 4 2 U4(x) = 16x − 12x + 1, 5 3 U5(x) = 32x − 32x + 6x, and 6 4 2 U6(x) = 64x − 80x + 24x − 1.

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This content downloaded from 205.175.118.33 on Wed, 29 Apr 2015 16:59:12 PM All use subject to JSTOR Terms and Conditions Letting x = cos θ, we obtain

sin((n + 1)θ) U (x) = . n sin θ

Knowledge of the roots of Un implies that

n Y jπ U (x) = 2n x − cos . (12) n n + 1 j=1

Setting x = 1 in this equation, we obtain again the An sine formula (8). We note that the matrix 2x I + A, where A is the adjacency matrix of the An graph, coincides with the matrix in (10). Consequently, we conclude that

qn(x) = Un(x).

= x Hence, an(x) Un 2 . It is well known that the roots of Un are given by kπ x = cos for k = 1, 2,..., n, k n + 1 as we already observed in (12). We can write them in the form xk = cos kθ, where = π θ n+1 . The roots of an(x) are then

kπ λ = 2 cos = 2 cos kθ for k = 1, 2,..., n, k n + 1 i.e., the roots of an(x) are m π 2 cos i h where mi are the exponents of An and h is the Coxeter number.

6. CARTAN MATRIX OF TYPE Bn AND Cn. The Chebyshev polynomials of the ﬁrst kind are denoted by Tn(x). They are deﬁned by the recurrence:

T0(x) = 1, T1(x) = x, Tn+1(x) = 2xTn(x) − Tn−1(x).

Using this recursion, we may compute the ﬁrst few polynomials as follows:

T0(x) = 1,

T1(x) = x, 2 T2(x) = 2x − 1, 3 T3(x) = 4x − 3x, 4 2 T4(x) = 8x − 8x + 1, 5 3 T5(x) = 16x − 20x + 5x, and 6 4 2 T6(x) = 32x − 48x + 18x − 1.

February 2014] A BEAUTIFUL SINE FORMULA 131

This content downloaded from 205.175.118.33 on Wed, 29 Apr 2015 16:59:12 PM All use subject to JSTOR Terms and Conditions It is well known that cos(nθ) can be expressed as a polynomial in cos(θ). For example,

cos(0θ) = 1, cos(1θ) = cos θ, cos(2θ) = 2(cos θ)2 − 1, and cos(3θ) = 4(cos θ)3 − 3(cos θ).

More generally, we have that

cos(nθ) = Tn(cos θ).

The roots of Tn(x) are well known: n − Y (2 j − 1)π T (x) = 2n 1 x − cos . (13) n 2n j=1

Setting x = 1 in this equation quickly leads to the formula

n Y (2 j − 1)π 22n sin2 = 2 (the B sine formula) . (14) 4n n j=1

We refer to this relation as the Bn sine formula. The Cartan matrix is a tridiagonal matrix of the form

 2 −1  − −  1 2 1   . . .   ......  CB =   . (15) n  − −   1 2 1   −1 2 −2 −1 2

Since the Cartan matrix of type Cn is the transpose of this matrix, we consider only the Cartan matrix of type Bn. Using expansion on the ﬁrst row, it is easy to prove that det(CBn ) = 2. Likewise, by expanding the determinant of the matrix 2x I + A with respect to the ﬁrst row, we obtain the recurrence

2 q1(x) = 2x, q2(x) = 4x − 2, and qn+1(x) = 2xqn(x) − qn−1(x).

Deﬁne q0(x) = 2. The recurrence implies that qn(x) = 2Tn(x), where Tn is the nth Chebyshev polynomial of the ﬁrst kind. The roots of Tn(x) are given by

(2k − 1)π x = cos for k = 1, 2,..., n. k 2n That is,

xk = cos(2k − 1)θ for k = 1, 2,..., n,

132 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121

This content downloaded from 205.175.118.33 on Wed, 29 Apr 2015 16:59:12 PM All use subject to JSTOR Terms and Conditions = π where θ 2n . We conclude that the roots of an(x) are (2k − 1)π λ = 2 cos = 2 cos (2k − 1)θ for k = 1, 2,..., n. k 2n We have veriﬁed that Lemma2 holds in this case.

7. CARTAN MATRIX OF TYPE Dn. The Cartan matrix of type Dn is a matrix of the form  2 −1  − −  1 2 1   . . .   ......  CD =   . (16) n  − − −   1 2 1 1  −1 2 0  −1 0 2

Note that the matrix is no longer tridiagonal. Using expansion on the first row and induction, we find that det(CDn ) = 4. By expanding the determinant of the matrix 2x I + A with respect to the first row, we deduce the recurrence

2 3 q2(x) = 4x , q3(x) = 8x − 4x, and qn+1(x) = 2xqn(x) − qn−1.

Deﬁne q1(x) = 4x. It is clear that qn(x) = 4xTn−1(x), where Tn is the nth Chebyshev polynomial of the ﬁrst kind. We now proceed to verify Lemma2 in the case of Dn. We have that x a (x) = 2xT − . n n 1 2

The equation an(x0) = 0 is therefore equivalent to 2x0Tn−1(x0) = 0. As a result, either = = (2k−1)π = − x0 0 or x0 2 cos 2(n−1) for k 1, 2,..., n 1.

8. THE SINE FORMULA. We have seen that the roots of an(x) are m π 2 cos i , h

= mi π where mi are the exponents of g and h is the Coxeter number of g. Let θi h . Then the roots of an(x) are λi = 2 cos θi , for i = 1, 2, . . . , `. Recall the duality property of the exponents (3):

mi + m`+1−i = h.

It follows that π π m + m + − = π. i h ` 1 i h This leads to

θi + θ`+1−i = π.

February 2014] A BEAUTIFUL SINE FORMULA 133

This content downloaded from 205.175.118.33 on Wed, 29 Apr 2015 16:59:12 PM All use subject to JSTOR Terms and Conditions Using this formula, we can infer a relationship satisﬁed by the roots of pn(x), which are θ ξ = 4 cos2 i . i 2 Namely,

2 θi 2 θ`+1−i ξ + ξ + − = 4 cos + 4 cos i ` 1 i 2 2 θ π − θ = 4 cos2 i + cos2 i 2 2 θ θ = 4 cos2 i + sin2 i 2 2 = 4.

In other words, we have proved Lemma1: The eigenvalues of the Cartan matrix occur in pairs ξ, 4 − ξ.

Theorem 1. Let g be a complex simple Lie algebra of rank `, h the Coxeter number, m1, m2,..., m` the exponents of g, and C the Cartan matrix. Then

` Y mi π 22` sin2 = detC. 2h i=1

Proof. The roots of an(x) are m π 2 cos i , h where mi are the exponents of g and h is the Coxeter number. Hence,

` Y mi π a (x) = x − 2 cos . n h i=1 Setting x = 2, we obtain

` Y mi π a (2) = 2 − 2 cos n h i=1

` Y mi π = 2` 1 − cos h i=1

` Y mi π = 2` 2 sin2 2h i=1

` Y mi π = 22` sin2 . 2h i=1

134 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121

This content downloaded from 205.175.118.33 on Wed, 29 Apr 2015 16:59:12 PM All use subject to JSTOR Terms and Conditions To establish the formula, we calculate an(2). We have

pn(x) = (x − ξ1)(x − ξ2) ··· (x − ξ`)

= (x − (4 − ξ1))(x − (4 − ξ2)) ··· (x − (4 − ξ`)).

This implies that pn(4) = ξ1ξ2 ··· ξ` = det C. Since an(x) = pn(x + 2), we conclude that

an(2) = pn(4) = det C.

ACKNOWLEDGMENTS. The author wishes to thank the reviewers of this article for their helpful com- ments, which led to a considerable improvement of the presentation.

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PANTELIS A. DAMIANOU has a B.S. degree from the University of New Hampshire, a masters degree from UCLA, and a Ph.D. in mathematics from the University of Arizona. His research interests include Lie theory, Poisson geometry, integrable systems, and number theory. He is currently a professor of mathematics at the University of Cyprus. Department of Mathematics and Statistics, University of Cyprus, P.O. Box 20537, 1678, Nicosia, Cyprus [email protected]

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