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\title{Commutation
relations and Vandermonde determinants}
\author {Yuri Bahturin
\\Department of Mathematics\\Memorial University, St John's,
Canada,\\ and \\Moscow state University, Moscow, Russia
\\{\tt yuri@math.mun.ca}\\ and
\\AmitaiRegev\\Department of Theoretical Mathematics\\Weizmann Institute, Rehovot
76100, Israel\\
{\tt amitai.regev@weizmann.ac.il}\\ and
\\Doron Zeilberger\\Department of Mathematics\\Rutgers
University, New Brunswick 08903, NJ\\{\tt
zeilberg@math.rutgers.edu}} \maketitle

\medskip
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 $n^2\times n^2$ 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
$n\times n$ Vandermonde matrices. In particular this allows us to
evaluate the determinant of $M^{M_n(F)}$.

\newpage

\section{Regular algebras}
Let $A$ be an algebra over the field $F$.

\begin{defn}~\cite[Definition 2.3]{ReSe}
The decompositions
\begin{eqnarray}\label{eq.directsum1}A=A_1\oplus\cdots\oplus
A_r\end{eqnarray} of the algebra $A$ as a direct sum of subspaces
is called {\it regular} if it satisfies the following two
conditions.

\medskip
1) Given $n$ indices $1\le i_1,\ldots,i_n\le r$, there exist
elements $x_{i_j}\in A_{i_j}$, $1\le j\le n$ such that
$x_{i_1}\cdots x_{i_n}\ne 0$.

\medskip
2) Given $1\le i,j\le r$, there exist $0\ne \theta_{i,j}^{(A)}\in
F\setminus\{0\}$ such that for any $x_i\in A_i$ and $y_j\in A_j$

\begin{eqnarray}\label{eq.directsum2}
x_iy_j=\theta_{i,j}^{(A)} y_jx_i.
\end{eqnarray}

 Then
\begin{eqnarray}\label{eq.directsum01}
M^{A}:=(\theta_{i,j}^{(A)})_{1\le i,j\le r}
\end{eqnarray}
 is the  matrix of the commutation
relations of the regular decomposition~\eqref{eq.directsum1}.
\end{defn}

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 $2\times 2$ are P.I. equivalent. The importance of
regular decompositions stems from the following theorem.

\begin{thm}\label{directsum12}~\cite[Theorem 3.1]{ReSe}

Let $A, B$ be algebras with regular decompositions
$A=\oplus_{i=1}^rA_i$ and $B=\oplus_{i=1}^rB_i$ and with
corresponding $r\times r$ matrices $M^A$ and $M^B$ of the
commutation relations. Assume $M^B=P^{-1}M^AP$ where $P$ is an
$r\times r$ permutation matrix. Then $A$ and $B$ satisfy the same
multi-linear identities. Hence  $A$ and $B$ are P.I.~equivalent
when $char(F)=0$.
\end{thm}

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




\begin{example}\label{directsum3}

Let $A=M_n(F)$, the $n\times n$ matrices over $F$. Note first that
with respect to the decomposition $M_n(F)=\oplus_{i,j=1}^n
span_F\{e_{i,j}\}$, $M_n(F)$ is {\it not} regular.  Here $e_{i,j}$
are the matrix units. However, $M_n(F)$ {\it is} regular -- with
respect to the following decomposition. Let $\xi$ be a primitive
$n$-th root of 1 and assume that $\xi\in F$. Let
$X_a=diag(\xi^{n-1},\xi^{n-2},\ldots,\xi,1)$ and
$X_b=e_{n,1}+\sum_{i=1}^{n-1}e_{i,i+1}$, then we obtain the
following decomposition:
\begin{eqnarray}\label{eq.directsum5}
M_n(F)=\bigoplus_{i,j=0}^{n-1}A_{i,j}\quad\mbox{where}\quad
 A_{i,j}=span_F\{X_a^iX_b^j\}.
\end{eqnarray}
Indeed, since $1,\xi,\xi^2,\ldots,\xi^{n-1}$ are distinct, by a
standard Vandermonde argument, the matrices $\{X_a^iX_b^j\mid 0\le
i\le n-1\}$ are linearly independent for any fixed $j$. It follows
that $\{X_a^iX_b^j\mid 0\le i,j\le n-1\}$ are linearly
independent, which implies the
decomposition~\eqref{eq.directsum5}. We verify that $M_n(F)$ is
regular with respect to that decomposition: The relation
$X_aX_b=\xi X_bX_a$ implies that
\begin{eqnarray}\label{eq.directsum6}
(X_a^iX_b^j)(X_a^kX_b^\ell)=\xi^{jk}X_a^{i+k}X_b^{j+\ell},
\end{eqnarray}
which then implies that
\begin{eqnarray}\label{eq.directsum7}
(X_a^iX_b^j)(X_a^kX_b^\ell)=\xi^{jk-i\ell}(X_a^kX_b^\ell)(X_a^iX_b^j).
\end{eqnarray}
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 ~\eqref{eq.directsum5}. The commutation relations
given by~\eqref{eq.directsum7} yield the $n^2\times n^2$ matrix
\begin{eqnarray}\label{directsum8}
M^{M_n(F)}=(\xi^{jk-i\ell})_{0\le i,j,k,\ell\le n-1}.
\end{eqnarray}
This matrix has row indices $(i,j)$ and column indices $(k,\ell)$,
with $0\le i,j,k,\ell\le n-1$.


Replacing $\xi$ by the variable $x$ we obtain the $n^2\times n^2$
matrix
\begin{eqnarray}\label{eq.directsum9}
B(x)=(x^{jk-i\ell})_{0\le i,j,k,\ell\le n-1}.
\end{eqnarray}
\end{example}

\section{Tensor factorization of $M^{M_n(F)}$}
We show that $B(x)$ of~\eqref{eq.directsum9} is very close to the
tensor product of two generic $n\times n$ 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
$M^{M_n(F)}=B(\xi)$, it follows that $M^{M_n(F)}$ admits a similar
tensor decomposition. In fact, by~\eqref{mat1} below,
$B(x)=D(x,x^{-1})$, and $D(x,y)$ is obtained from the tensor
product $V(x)\otimes V(y)$
%of the two generic Vandermonde matrices $V(x)$ and $V(y)$
by a natural row permutation. It follows that $M^{M_n(F)}$ is
obtained by the same row permutation from $V(\xi)\otimes
V(\xi^{-1})$. This is Theorem~\ref{more1} below.

\medskip
Consider the tensor product $A\otimes B$ of two $n\times n$
matrices $A$ and $B$.
%If $B=(b_{r,s})$ then (by definition)
%$A\otimes B=(A\cdot b_{r,s})$.
By definition $A\otimes B=(a_{j_1,j_2}\cdot b_{i_1,i_2})$. This is
an $n^2\times n^2$ matrix. Let its row indices be $(i_1,j_1)$ and
column indices $(i_2,j_2)$, both ordered lexicographically, where
$1\le,i_1,j_1,i_2,j_2\le n$. Then $A\otimes B$ is an array of
$n\times n$ blocks, each block being an $n\times n$ matrix. Each
such a block is determined by fixing some $i_1$ and $i_2$ and
letting $1\le j_1,j_2\le n$, which is then the $(i_1,i_2)$ block
$A\cdot b_{i_1,i_2}$ in $A\otimes B$. Thus, if
$A=(a_{i_1,j_1})_{1\le,i_1,j_1\le n}$ and $B=(b_{i_2,j_2})_{1\le
i_2,j_2\le n}$, then $$A\otimes B=(a_{j_1,j_2}\cdot
b_{i_1,i_2})_{1\le,i_1,j_1,i_2,j_2\le n}.$$


\medskip

It is well known that when both $A$ and $B$ are $n\times n$
matrices,
$$\det (A\otimes
B)=(\det(A)\cdot\det(B))^n$$



The key observation here is the following lemma.

\begin{lem}\label{work1}
Let $V(x)=(x_i^{j-1})_ {1\le i,j\le n}$ denote a generic $n\times
n$ Vandermonde matrix. Thus $$V(x)\otimes
V(y)=(x_{j_1}^{j_2-1}\cdot y_{i_1}^{i_2-1})_{1\le
i_1,j_1,i_2,j_2\le n}.$$ Let
\begin{eqnarray}\label{mat1}
D(x,y)=(x_{i_1}^{j_2-1}\cdot y_{j_1}^{i_2-1})_{1\le
i_1,j_1,i_2,j_2\le n}.
\end{eqnarray}
Then the matrix $D(x,y)$ is obtained from $V(x)\otimes V(y)$ by a
row permutation. That row permutation is a product of $n\choose 2$
disjoint transpositions (hence is an involution).
\end{lem}
\begin{proof}
Recall that $V(x)\otimes V(y)$ has row indices $(i_1,j_1)$. Apply
on $V(x)\otimes V(y)$ the row permutation $\sigma\in S_{n^2}$
given by $(i_1,j_1)\longrightarrow (j_1,i_1)$. Thus
$x_{j_1}^{j_2-1}\cdot y_{i_1}^{i_2-1}\longrightarrow
x_{i_1}^{j_2-1}\cdot y_{j_1}^{i_2-1}$ by $\sigma$, so $V(x)\otimes
V(y)$ is changed by $\sigma$ into  $D(x,y)$. Since $\sigma$ has
$n$ fixed points $(r,r)$, it is the product of $(n^2-n)/2$
disjoint transpositions.
\end{proof}
The above implies
\begin{thm}\label{more1}
Let $\sigma\in S_{n^2}$ denote the permutation given by $\sigma
((i,j))=(j,i)$. It induces the corresponding row permutation on
$n^2\times n^2$ matrices, which we also denote by $\sigma$. Let
$B(x)$ and $D(x,y)$ be given by~\eqref{eq.directsum9}
and~\eqref{mat1}. Then
$$D(x,y)=\sigma(V(x)\otimes V(y)),$$ hence $$B(x)=D(x,x^{-1})=\sigma(V(x)\otimes
V(x^{-1})),$$ and therefore $$B(\xi)=\sigma(V(\xi)\otimes
V(\xi^{-1})).$$
\end{thm}
\subsection{The determinants}
We prove here
\begin{prop}\label{work010}
The matrix $B(\xi)=M^{M_n(F)}$ of~\eqref{directsum8} has the
following determinant:
$$det(M^{M_n(F)})=det((\xi^{jk-i\ell})_{0\le i,j,k,\ell\le
n-1})=\pm n^{(n^2)}.$$
\end{prop}

\medskip
The proof is given below. Note first that by Theorem~\ref{more1}
and Lemma~\ref{work1}
\begin{eqnarray}\label{det1}
\det (D(x,y))=(-1)^{n\choose 2}\cdot\det(V(x)\otimes V(y))
=(-1)^{n\choose 2}(\det (V(x))\cdot\det(V(y)))^n.
\end{eqnarray}

\begin{cor}
Since $V(x), V(y)$ are Vandermonde matrices, it follows
from~\eqref{det1} that
\begin{eqnarray}\label{det2}
\det (D(x,y))=(-1)^{n\choose 2}\left(\prod_{1\le i<j\le
n}(x_i-x_j)\cdot\prod_{1\le i<j\le n}(y_i-y_j) \right)^n.
\end{eqnarray}
\end{cor}

\begin{thm}\label{directsum10}
The matrix $B(x)$ of~\eqref{eq.directsum9} has the following
determinant:
\begin{eqnarray}\label{eq.directsum12}
\det(B(x))=(-1)^{{n\choose 2}(n+1)}\cdot \left(\prod_{0\le i<j\le
n-1}(x^i-x^j)\right)^{2n}\cdot\left(\frac{1}{\prod_{0\le i<j\le
n-1}x^{i+j}}\right)^n.
\end{eqnarray}
\end{thm}
\begin{proof}
Substitute $x_i\longrightarrow x^{i-1}$ and $y_i\longrightarrow
1/x^{i-1}$, then
$$x_{i_1}^{j_2-1}\cdot y_{j_1}^{i_2-1}\longrightarrow
x^{(i_1-1)(j_2-1)-(j_1-1)(i_2-1)}.$$ Denote
$i_1-1=j,~j_1-1=i,~i_2-1=k$ and $j_2-1=\ell$, then
$$x^{(i_1-1)(j_2-1)-(j_1-1)(i_2-1)}=x^{jk-i\ell},$$ $0\le
i,j,k,\ell$, hence $D(x,y)\longrightarrow B(x)$,
see~\eqref{eq.directsum9}, and $\det(D(x,y))\longrightarrow \det
(B(x))$.


\medskip
By these same substitutions,
$$\prod_{1\le i<j\le n}(x_i-x_j)\longrightarrow \prod_{0\le i<j\le
n-1}(x^i-x^j).$$

\bigskip

Similarly,
$$\prod_{1\le i<j\le n}(y_i-y_j)\longrightarrow \prod_{0\le i<j\le
n-1}\frac{x^j-x^i}{x^{i+j}}=(-1)^{n \choose 2}\cdot\prod_{0\le
i<j\le n-1}\frac{x^i-x^j}{x^{i+j}}$$ Together with~\eqref{det2} it
implies that $$\det(B(x))=(-1)^{{n\choose 2}}\left(\prod_{0\le
i<j\le n-1}(x^i-x^j)\cdot(-1)^{{n\choose 2}}\cdot\prod_{0\le
i<j\le n-1}\frac{(x^i-x^j)}{x^{i+j}}\right)^n=~~~~~~~~~~~~~~~~
$$
$$~~~~~~~~~~~
(-1)^{{n\choose 2}(n+1)}\cdot \left(\prod_{0\le i<j\le
n-1}(x^i-x^j)\right)^{2n}\cdot\left(\frac{1}{\prod_{0\le i<j\le
n-1}x^{i+j}}\right)^n.
$$
\end{proof}


Note that $$\sum_{0\le i<j\le n-1}i+j=\frac{(n-1)^2n}{2},$$ hence
$$\left(\frac{1}{\prod_{0\le i<j\le
n-1}x^{i+j}}\right)^n=\left(\frac{1}{x}\right)^{\frac{(n-1)^2n^2}{2}}.$$
Thus~\eqref{eq.directsum12} can be rewritten as

\begin{eqnarray}\label{eq.directsum1212}
\det(B(x))=(-1)^{{n\choose 2}(n+1)}\cdot \left(\prod_{0\le i<j\le
n-1}(x^i-x^j)\right)^{2n}\cdot
\left(\frac{1}{x}\right)^{\frac{(n-1)^2n^2}{2}}.
\end{eqnarray}

{\bf The proof of Proposition~\ref{work010}}
\begin{proof}
 Let $\xi$ be a
primitive $n$-th root of unit. Substituting $x\longrightarrow \xi$
in $B(x)$ we obtain the matrix $B(\xi)$ of~\eqref{directsum8}.
Since $\xi^n=1$, the factor
$$\left(\frac{1}{\xi}\right)^{\frac{(n-1)^2n^2}{2}}$$ equals 1 and
can be discarded. Thus by~\eqref{eq.directsum1212} we need to show
that
\begin{eqnarray}\label{kodem1}
\left(\prod_{0\le i<j\le n-1}(\xi^i-\xi^j)\right)^{2n}= \pm
n^{(n^2)}. \end{eqnarray} Taking $n$-th root of both sides, we
need to prove that for some integer $r$,
$$\left(\prod_{0\le i<j\le n-1}(\xi^i-\xi^j)\right)^{2}= \prod_{0\le i<j\le
n-1}(\xi^i-\xi^j)^2=\pm \xi^rn^n. $$

Rewrite the left side (up to a $\pm$ sign) as
\begin{eqnarray}\label{bdika1}
\prod_{0\le i\neq j\le n-1}(\xi^i-\xi^j).
\end{eqnarray}
Given $0\le i\neq j\le n-1$, write the factor $\xi^i-\xi^j$ as
$\xi^i-\xi^j=\xi^i(1-\xi^{j-i})$ if $i<j$, and as
$\xi^i-\xi^j=\xi^i(1-\xi^{n-(i-j)})$ if $i>j$. Thus, for each
$1\le a\le n-1$ the factor $1-\xi^a$ is obtained in~\eqref{bdika1}
from $1-\xi^a,~\xi-\xi^{a+1},~\ldots ,~\xi^{n-a-1}-\xi^{n-1}$
($n-a$ factors) and from
$\xi^{n-a}-1,~\xi^{n-a+1}-\xi,~\ldots,~\xi^{n-1}-\xi^{a-1}$ ($a$
factors). After pulling out the appropriate power of $\xi$, it
follows that for each $1\le a\le n-1$, the factor $1-\xi^a$
appears in~\eqref{bdika1} exactly $n$ times. Also, for that $1\le
a\le n-1$ we pull out the factors $\xi^i,~0\le i\le n-1$, namely
we pull out
$$\xi^{0+1+2+\cdots +n-1}=\xi^{\frac{n(n-1)}{2}}.$$
Since there are $n-1$ $a$'s, in total we pull out
$$\xi^{\frac{n(n-1)^2}{2}}.$$


It follows that
\begin{eqnarray}\label{bdika2}\prod_{0\le i<j\le
n-1}(\xi^i-\xi^j)^2=\pm\xi^{\frac{n(n-1)^2}{2}   }\left(
\prod_{i=1}^{n-1}(1-\xi^i)\right)^n.\end{eqnarray}

So it suffices to prove that
$$\prod_{i=1}^{n-1} (1-\xi^i)=n.$$
To prove this, note that, since $\xi$ is primitive,
$\xi,\xi^2,\ldots, \xi^{n-1}$ are the roots of the monic
polynomial $(x^n-1)/(x-1)$, hence
$$\prod_{i=1}^{n-1} (x-\xi^i)=(x^n-1)/(x-1)=\sum_{i=0}^{n-1}x^i.$$
Now substitute $x\longrightarrow 1$.
\end{proof}

\begin{remark}\label{pm1}
Let $\xi_1$ and $\xi_2$ be two primitive $n$-th roots of 1, then
\begin{eqnarray}\label{bdika02}\prod_{0\le i<j\le
n-1}(\xi_1^i-\xi_1^j)=\pm\prod_{0\le i<j\le
n-1}(\xi_2^i-\xi_2^j).\end{eqnarray}
\end{remark}
\begin{proof}
There is an integer $r>0$ such that $\xi_2=\xi_1^r$, and since
both are primitive, $(r,n)=1$.
 Thus
 $$
\prod_{0\le i<j\le n-1}(\xi_2^i-\xi_2^j)=\prod_{0\le i<j\le
n-1}(\xi_1^{ri}-\xi_1^{rj}).
$$
Since $(r,n)=1$, the product on the right contains all the factors
$\xi_1^k-\xi_1^\ell, ~0\le k<\ell\le n-1$, each factor with a
$\pm$ sign.
\end{proof}



\section{A generalization}

More generally, given $0\le a,b\in\mathbb{Z}$, denote
\begin{eqnarray}\label{eq.directsum13}
D(x,a,b)=(x^{ajk+bi\ell})_{0\le i,j,k,\ell\le n-1}.
\end{eqnarray}
\begin{prop}\label{directsum10}
The above matrix $D(x,a,b)$ has the following determinant:
\begin{eqnarray}\label{eq.directsum121}
det(D(x,a,b)) = \left(\prod_{0\le i<j\le
n-1}(x^{ai}-x^{aj})\right)^{n} \cdot\left(\prod_{0\le i<j\le
n-1}(x^{bi}-x^{bj})\right)^{n}
\end{eqnarray}
\end{prop}
\begin{proof}
Start with~\eqref{det2} and apply the substitution
$x_i\longrightarrow x^{a(i-1)}$ and $y_i\longrightarrow
 x^{b(i-1)}$.
\end{proof}
\begin{prop}\label{ab1}
Let $\xi$ be a primitive $n$-th root of unity and let $a,b$ be
integers. If either $(a,n)>1$ or $(b,n)>1$ then $\det
(D(\xi,a,b))=0$. If both $(a,n)=(b,n)=1$ then
$$
\det (D(\xi,a,b))=\pm n^{(n^2)}
$$


\end{prop}

\begin{proof}
By~\eqref{eq.directsum121}
\begin{eqnarray}\label{eq.directsum0121}
det(D(\xi,a,b)) = \left(\prod_{0\le i<j\le
n-1}(\xi^{ai}-\xi^{aj})\right)^{n} \cdot\left(\prod_{0\le i<j\le
n-1}(\xi^{bi}-\xi^{bj})\right)^{n}
\end{eqnarray}
Assume $(a,n)=d>1$, then  $(\xi^a)^{\frac{n}{d}}=1$, and since $0<
\frac{n}{d}\le n-1$, the product
$$
\prod_{0\le i<j\le n-1}(\xi^{ai}-\xi^{aj})
$$
contains the factor $1-(\xi^a)^{\frac{n}{d}}=0$ hence is zero.
Similarly if $(b,n)>1$.

\medskip
If both $(a,n)=(b,n)=1$ then $\xi_1=\xi^a$ and $\xi_2=\xi^b$ are
both primitive. By~\eqref{bdika02}, the two factors
of~\eqref{eq.directsum0121} differ by a $\pm$ sign, so
by~\eqref{eq.directsum0121} and by~\eqref{kodem1}
%\begin{eqnarray}\label{square1}
$$det(D(\xi,a,b)) = \pm\left(\prod_{0\le i<j\le
n-1}(\xi^{i}-\xi^{j})\right)^{2n}=\pm n^{(n^2)}.$$

\end{proof}

\section{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 $n^2 \times n^2$ matrix, and one should get
expressions featuring Schur polynomials evaluated at roots of
unity.

\begin{thebibliography}{99}

\bibitem{ReSe}
  Regev, A. and Seeman, T., $\mathbb{Z}_2$-graded tensor products
  of p.i.~algebras,
   {\it J. Algebra } {\bf 291} (2005), 274--296.


\end{thebibliography}













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