Commutation relations and Vandermonde determinants

Yuri Bahturin Department of Mathematics Memorial University, St John’s, Canada, and Moscow state University, Moscow, Russia

                       yuri@math.mun.ca

and AmitaiRegev Department of Theoretical Mathematics Weizmann Institute, Rehovot 76100, Israel

                  amitai.regev@weizmann.ac.il

and Doron Zeilberger Department of Mathematics Rutgers University, New Brunswick 08903, NJ

                   zeilberg@math.rutgers.edu

September 28, 2008

Abstract: We consider a certain decomposition of the matrix algebra M_{n}(F), where F is a field. The commutation relations of that decomposition yield an n^{2} × n^{2} matrix M^{M_{n}(F)}, which determines the multilinear polynomial identities of M_{n}(F). Thus if char(F) = 0, the matrix M^{M_{n}(F)} determines the polynomial identities of M_{n}(F). We show that M^{M_{n}(F)} is very close to the tensor product of two n × n Vandermonde matrices. In particular this allows us to evaluate the determinant of M^{M_{n}(F)}.

1 Regular algebras

Let A be an algebra over the field F.

Definition 1.1. [1, Definition 2.3] The decompositions

A = A_{1} ⊕ ⋯ ⊕ A_{r}  (1)

of the algebra A as a direct sum of subspaces is called regular if it satisfies the following two conditions.

1) Given n indices 1 ≤ i_{1}, …, i_{n} ≤ r, there exist elements x_{i_{j}} ∈ A_{i_{j}}, 1 ≤ j ≤ n such that x_{i_{1}} ⋯ x_{i_{n}} ≠ 0.

2) Given 1 ≤ i, j ≤ r, there exist 0 ≠ θ_{i,j}^{(A)} ∈ F \ {0} such that for any x_{i} ∈ A_{i} and y_{j} ∈ A_{j}

x_{i}y_{j} = θ_{i,j}^{(A)}y_{j}x_{i}.  (2)

Then

M^{A} := (θ_{i,j}^{(A)})_{1≤i,j≤r}  (3)

is the matrix of the commutation relations of the regular decomposition (1).

Recall that algebras satisfying polynomial identities are called P.I. algebras. Two P.I. algebras are P.I. equivalent if they satisfy the same polynomial identities. For example, it can be shown that in characteristic zero, the quaternions algebra and the algebra of 2 × 2 are P.I. equivalent. The importance of regular decompositions stems from the following theorem.

Theorem 1.2. [1, Theorem 3.1] Let A, B be algebras with regular decompositions A = ⊕_{i=1}^{r}A_{i} and B = ⊕_{i=1}^{r}B_{i} and with corresponding r × r matrices M^{A} and M^{B} of the commutation relations. Assume M^{B} = P^{−1}M^{A}P where P is an r × r permutation matrix. Then A and B satisfy the same multilinear identities. Hence A and B are P.I. equivalent when char(F) = 0.

The main object studied in this paper is given in the following example.

Example 1.3. Let A = M_{n}(F), the n × n matrices over F. Note first that with respect to the decomposition M_{n}(F) = ⊕_{i,j=1}^{n}span_{F} {e_{i,j}}, M_{n}(F) is not regular. Here e_{i,j} are the matrix units. However, M_{n}(F) is regular – with respect to the following decomposition. Let ξ be a primitive n-th root of 1 and assume that ξ ∈ F. Let X_{a} = diag(ξ^{n−1}, ξ^{n−2}, …, ξ, 1) and X_{b} = e_{n,1} + ∑_{i=1}^{n−1} e_{i,i+1}, then we obtain the following decomposition:

M_{n}(F) = ⨁_{i,j=0}^{n−1} A_{i,j} where A_{i,j} = span_{F} {X_{a}^{i}X_{b}^{j}}.  (4)

Indeed, since 1, ξ, ξ^{2}, …, ξ^{n−1} are distinct, by a standard Vandermonde argument, the matrices {X_{a}^{i}X_{b}^{j} | 0 ≤ i ≤ n − 1} are linearly independent for any fixed j. It follows that {X_{a}^{i}X_{b}^{j} | 0 ≤ i, j ≤ n−1} are linearly independent, which implies the decomposition (4). We verify that M_{n}(F) is regular with respect to that decomposition: The relation X_{a}X_{b} = ξX_{b}X_{a} implies that

(X_{a}^{i}X_{b}^{j})(X_{a}^{k}X_{b}^{ℓ}) = ξ^{jk}X_{a}^{i+k}X_{b}^{j+ℓ},  (5)

which then implies that

(X_{a}^{i}X_{b}^{j})(X_{a}^{k}X_{b}^{ℓ}) = ξ^{jk−iℓ}(X_{a}^{k}X_{b}^{ℓ})(X_{a}^{i}X_{b}^{j}).  (6)

Since any product of the matrices X_{a}^{i} and X_{b}^{j} is non zero, it follows that M_{n}(F) is regular – with respect to the decomposition (4). The commutation relations given by (6) yield the n^{2} × n^{2} matrix

M^{M_{n}(F)} = (ξ^{jk−iℓ})_{0≤i,j,k,ℓ≤n−1}.  (7)

This matrix has row indices (i, j) and column indices (k, ℓ), with 0 ≤ i, j, k, ℓ ≤ n − 1. Replacing ξ by the variable x we obtain the n^{2} × n^{2} matrix

B(x) = (x^{jk−iℓ})_{0≤i,j,k,ℓ≤n−1}.  (8)

2 Tensor factorization of M^{M_{n}(F)}

We show that B(x) of (8) is very close to the tensor product of two generic n×n Vandermonde matrices V (x) and V (y), and is obtained from such a tensor product by a natural row permutation and by a natural substitution. Since M^{M_{n}(F)} = B(ξ), it follows that M^{M_{n}(F)} admits a similar tensor decomposition. In fact, by (9) below, B(x) = D(x, x^{−1}), and D(x, y) is obtained from the tensor product V (x)⊗V (y) by a natural row permutation. It follows that M^{M_{n}(F)} is obtained by the same row permutation from V (ξ) ⊗ V (ξ^{−1}). This is Theorem 2.2 below.

Consider the tensor product A ⊗ B of two n × n matrices A and B. By definition A ⊗ B = (a_{j_{1},j_{2}} · b_{i_{1},i_{2}}). This is an n^{2} × n^{2} matrix. Let its row indices be (i_{1}, j_{1}) and column indices (i_{2}, j_{2}), both ordered lexicographically, where 1 ≤, i_{1}, j_{1}, i_{2}, j_{2} ≤ n. Then A ⊗ B is an array of n × n blocks, each block being an n × n matrix. Each such a block is determined by fixing some i_{1} and i_{2} and letting 1 ≤ j_{1}, j_{2} ≤ n, which is then the (i_{1}, i_{2}) block A · b_{i_{1},i_{2}} in A ⊗ B. Thus, if A = (a_{i_{1},j_{1}})_{1≤,i_{1},j_{1}≤n} and B = (b_{i_{2},j_{2}})_{1≤i_{2},j_{2}≤n}, then

A ⊗ B = (a_{j_{1},j_{2}} · b_{i_{1},i_{2}})_{1≤,i_{1},j_{1},i_{2},j_{2}≤n}.

It is well known that when both A and B are n × n matrices,

det(A ⊗ B) = (det(A) · det(B))^{n}

The key observation here is the following lemma.

Lemma 2.1. Let V (x) = (x_{i}^{j−1})_{1≤i,j≤n} denote a generic n × n Vandermonde matrix. Thus

V (x) ⊗ V (y) = (x_{j_{1}}^{j_{2}−1} · y_{i_{1}}^{i_{2}−1})_{1≤i_{1},j_{1},i_{2},j_{2}≤n}.

Let

D(x, y) = (x_{i_{1}}^{j_{2}−1} · y_{j_{1}}^{i_{2}−1})_{1≤i_{1},j_{1},i_{2},j_{2}≤n}.  (9)

Then the matrix D(x, y) is obtained from V (x) ⊗ V (y) by a row permutation. That row permutation is a product of \binom{n}{2} disjoint transpositions (hence is an involution).

Proof. Recall that V (x) ⊗ V (y) has row indices (i_{1}, j_{1}). Apply on V (x) ⊗ V (y) the row permutation σ ∈ S_{n^{2}} given by (i_{1}, j_{1}) ⟶ (j_{1}, i_{1}). Thus x_{j_{1}}^{j_{2}−1} · y_{i_{1}}^{i_{2}−1} ⟶ x_{i_{1}}^{j_{2}−1} · y_{j_{1}}^{i_{2}−1} by σ, so V (x) ⊗ V (y) is changed by σ into D(x, y). Since σ has n fixed points (r, r), it is the product of (n^{2} − n)/2 disjoint transpositions.

The above implies

Theorem 2.2. Let σ ∈ S_{n^{2}} denote the permutation given by σ((i, j)) = (j, i). It induces the corresponding row permutation on n^{2} × n^{2} matrices, which we also denote by σ. Let B(x) and D(x, y) be given by (8) and (9). Then

D(x, y) = σ(V (x) ⊗ V (y)),

hence

B(x) = D(x, x^{−1}) = σ(V (x) ⊗ V (x^{−1})),

and therefore

B(ξ) = σ(V (ξ) ⊗ V (ξ^{−1})).

2.1 The determinants

We prove here

Proposition 2.3. The matrix B(ξ) = M^{M_{n}(F)} of (7) has the following determinant:

det(M^{M_{n}(F)}) = det((ξ^{jk−iℓ})_{0≤i,j,k,ℓ≤n−1}) = ±n^{(n^{2})}.

The proof is given below. Note first that by Theorem 2.2 and Lemma 2.1

det(D(x, y)) = (−1)^{\binom{n}{2}} · det(V (x) ⊗ V (y)) = (−1)^{\binom{n}{2}}(det(V (x)) · det(V (y)))^{n}.  (10)

Corollary 2.4. Since V (x), V (y) are Vandermonde matrices, it follows from (10) that

det(D(x, y)) = (−1)^{\binom{n}{2}} ( ∏_{1≤i<j≤n}(x_{i} − x_{j}) · ∏_{1≤i<j≤n}(y_{i} − y_{j}))^{n} .  (11)

Theorem 2.5. The matrix B(x) of (8) has the following determinant:

det(B(x)) = (−1)^{\binom{n}{2}(n+1)} · ( ∏_{0≤i<j≤n−1}(x^{i} − x^{j}))^{2n} · (\frac{1}{∏_{0≤i<j≤n−1} x^{i+j}} )^{n} .  (12)

Proof. Substitute x_{i} ⟶ x^{i−1} and y_{i} ⟶ 1/x^{i−1}, then

x_{i_{1}}^{j_{2}−1} · y_{j_{1}}^{i_{2}−1} ⟶ x^{(i_{1}−1)(j_{2}−1)−(j_{1}−1)(i_{2}−1)}.

Denote i_{1} − 1 = j, j_{1} − 1 = i, i_{2} − 1 = k and j_{2} − 1 = ℓ, then

x^{(i_{1}−1)(j_{2}−1)−(j_{1}−1)(i_{2}−1)} = x^{jk−iℓ},

0 ≤ i, j, k, ℓ, hence D(x, y) ⟶ B(x), see (8), and det(D(x, y)) ⟶ det(B(x)). By these same substitutions,

∏_{1≤i<j≤n}(x_{i} − x_{j}) ⟶ ∏_{0≤i<j≤n−1}(x^{i} − x^{j}).

Similarly,

∏_{1≤i<j≤n}(y_{i} − y_{j}) ⟶ ∏_{0≤i<j≤n−1} \frac{x^{j} − x^{i}}{x^{i+j}} = (−1)^{\binom{n}{2}} · ∏_{0≤i<j≤n−1} \frac{x^{i} − x^{j}}{x^{i+j}}

Together with (11) it implies that

det(B(x)) = (−1)^{\binom{n}{2}} ( ∏_{0≤i<j≤n−1}(x^{i} − x^{j}) · (−1)^{\binom{n}{2}} · ∏_{0≤i<j≤n−1} \frac{(x^{i} − x^{j})}{x^{i+j}})^{n} =

(−1)^{\binom{n}{2}(n+1)} · ( ∏_{0≤i<j≤n−1}(x^{i} − x^{j}))^{2n} · (\frac{1}{∏_{0≤i<j≤n−1} x^{i+j}} )^{n} .

Note that

∑_{0≤i<j≤n−1} i + j = \frac{(n − 1)^{2}n}{2},

hence

(\frac{1}{∏_{0≤i<j≤n−1} x^{i+j}} )^{n} = (\frac{1}{x})^{\frac{(n−1)^{2}n^{2}}{2}} .

Thus (12) can be rewritten as

det(B(x)) = (−1)^{\binom{n}{2}(n+1)} · ( ∏_{0≤i<j≤n−1}(x^{i} − x^{j}))^{2n} · (\frac{1}{x})^{\frac{(n−1)^{2}n^{2}}{2}} .  (13)

The proof of Proposition 2.3

Proof. Let ξ be a primitive n-th root of unit. Substituting x ⟶ ξ in B(x) we obtain the matrix B(ξ) of (7). Since ξ^{n} = 1, the factor

(\frac{1}{ξ})^{\frac{(n−1)^{2}n^{2}}{2}}

equals 1 and can be discarded. Thus by (13) we need to show that

( ∏_{0≤i<j≤n−1}(ξ^{i} − ξ^{j}))^{2n} = ±n^{(n^{2})}.  (14)

Taking n-th root of both sides, we need to prove that for some integer r,

( ∏_{0≤i<j≤n−1}(ξ^{i} − ξ^{j}))^{2} = ∏_{0≤i<j≤n−1}(ξ^{i} − ξ^{j})^{2} = ±ξ^{r}n^{n}.

Rewrite the left side (up to a ± sign) as

∏_{0≤i≠j≤n−1}(ξ^{i} − ξ^{j}).  (15)

Given 0 ≤ i ≠ j ≤ n − 1, write the factor ξ^{i} − ξ^{j} as ξ^{i} − ξ^{j} = ξ^{i}(1 − ξ^{j−i}) if i < j, and as ξ^{i} − ξ^{j} = ξ^{i}(1 − ξ^{n−(i−j)}) if i > j. Thus, for each 1 ≤ a ≤ n − 1 the factor 1 − ξ^{a} is obtained in (15) from 1 − ξ^{a}, ξ − ξ^{a+1}, …, ξ^{n−a−1} − ξ^{n−1} (n − a factors) and from ξ^{n−a} − 1, ξ^{n−a+1} − ξ, …, ξ^{n−1} − ξ^{a−1} (a factors). After pulling out the appropriate power of ξ, it follows that for each 1 ≤ a ≤ n − 1, the factor 1 − ξ^{a} appears in (15) exactly n times. Also, for that 1 ≤ a ≤ n − 1 we pull out the factors ξ^{i}, 0 ≤ i ≤ n − 1, namely we pull out

ξ^{0+1+2+···+n−1} = ξ^{\frac{n(n−1)}{2}} .

Since there are n − 1 a’s, in total we pull out

ξ^{\frac{n(n−1)^{2}}{2}} .

It follows that

∏_{0≤i<j≤n−1}(ξ^{i} − ξ^{j})^{2} = ±ξ^{\frac{n(n−1)^{2}}{2}} (∏_{i=1}^{n−1}(1 − ξ^{i}))^{n} .  (16)

So it suffices to prove that

∏_{i=1}^{n−1}(1 − ξ^{i}) = n.

To prove this, note that, since ξ is primitive, ξ, ξ^{2}, …, ξ^{n−1} are the roots of the monic polynomial (x^{n} − 1)/(x − 1), hence

∏_{i=1}^{n−1}(x − ξ^{i}) = (x^{n} − 1)/(x − 1) = ∑_{i=0}^{n−1} x^{i}.

Now substitute x ⟶ 1.

Remark 2.6. Let ξ_{1} and ξ_{2} be two primitive n-th roots of 1, then

∏_{0≤i<j≤n−1}(ξ_{1}^{i} − ξ_{1}^{j}) = ± ∏_{0≤i<j≤n−1}(ξ_{2}^{i} − ξ_{2}^{j}).  (17)

Proof. There is an integer r > 0 such that ξ_{2} = ξ_{1}^{r}, and since both are primitive, (r, n) = 1. Thus

∏_{0≤i<j≤n−1}(ξ_{2}^{i} − ξ_{2}^{j}) = ∏_{0≤i<j≤n−1}(ξ_{1}^{ri} − ξ_{1}^{rj}).

Since (r, n) = 1, the product on the right contains all the factors ξ_{1}^{k} − ξ_{1}^{ℓ}, 0 ≤ k < ℓ ≤ n − 1, each factor with a ± sign.

3 A generalization

More generally, given 0 ≤ a, b ∈ ℤ, denote

D(x, a, b) = (x^{ajk+biℓ})_{0≤i,j,k,ℓ≤n−1}.  (18)

Proposition 3.1. The above matrix D(x, a, b) has the following determinant:

det(D(x, a, b)) = ( ∏_{0≤i<j≤n−1}(x^{ai} − x^{aj}))^{n} · ( ∏_{0≤i<j≤n−1}(x^{bi} − x^{bj}))^{n}  (19)

Proof. Start with (11) and apply the substitution x_{i} ⟶ x^{a(i−1)} and y_{i} ⟶ x^{b(i−1)}.

Proposition 3.2. Let ξ be a primitive n-th root of unity and let a, b be integers. If either (a, n) > 1 or (b, n) > 1 then det(D(ξ, a, b)) = 0. If both (a, n) = (b, n) = 1 then

det(D(ξ, a, b)) = ±n^{(n^{2})}

Proof. By (19)

det(D(ξ, a, b)) = ( ∏_{0≤i<j≤n−1}(ξ^{ai} − ξ^{aj}))^{n} · ( ∏_{0≤i<j≤n−1}(ξ^{bi} − ξ^{bj}))^{n}  (20)

Assume (a, n) = d > 1, then (ξ^{a})^{\frac{n}{d}} = 1, and since 0 < \frac{n}{d} ≤ n − 1, the product

∏_{0≤i<j≤n−1}(ξ^{ai} − ξ^{aj})

contains the factor 1 − (ξ^{a})^{\frac{n}{d}} = 0 hence is zero. Similarly if (b, n) > 1.

If both (a, n) = (b, n) = 1 then ξ_{1} = ξ^{a} and ξ_{2} = ξ^{b} are both primitive. By (17), the two factors of (20) differ by a ± sign, so by (20) and by (14)

det(D(ξ, a, b)) = ± ( ∏_{0≤i<j≤n−1}(ξ^{i} − ξ^{j}))^{2n} = ±n^{(n^{2})}.

4 Minors

Since the minors of a generic Vandermonde matrix are Schur polynomials multiplied by the product of differences of the relevant variables (times some monomial), analogous statements can be proved for any minor of the considered n^{2} ×n^{2} matrix, and one should get expressions featuring Schur polynomials evaluated at roots of unity.

References

[1] Regev, A. and Seeman, T., ℤ_{2}-graded tensor products of p.i. algebras, J. Algebra 291 (2005), 274–296.