## Weyl algebras; definition & automorphisms (1)

Posted: January 24, 2011 in Noncommutative Ring Theory Notes, Weyl Algebras
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Let $R$ be a ring and let $n \geq 0$ be an integer. The $n$-th Weyl algebra over $R$ is defined as follows. First we define $A_0(R)=R.$ For $n \geq 1,$ we define $A_n(R)$ to be the ring of polynomials in $2n$ variables $x_i, y_i, \ 1 \leq i \leq n,$ with coefficients in $R$ and subject to the relations

$x_ix_j=x_jx_i, \ y_iy_j=y_jy_i, \ y_ix_j = x_jy_i + \delta_{ij},$

for all $i,j,$ where $\delta_{ij}$ is the Kronecker delta. We will assume that every element of $R$ commutes with all $2n$ variables $x_i$ and $y_i.$ So, for example, $A_1(R)$ is the ring generated by $x_1,y_1$ with coefficients in $R$ and subject to the relation $y_1x_1=x_1y_1+1.$ An element of $A_1(R)$ is in the form $\sum r_{ij}x_1^iy_1^j, \ r_{ij} \in R.$. It is not hard to prove that the set of monomials in the form

$x_1^{\alpha_1} \ldots x_n^{\alpha_n}y_1^{\beta_1} \ldots y_n^{\beta_n}$

is an $R$-basis for $A_n(R).$ Also note that $A_n(R)=A_1(A_{n-1}(R)).$ If $R$ is a domain, then $A_n(R)$ is a domain too. It is straightforward to show that if $k$ is a field of characteristic zero, then $A_n(k)$ is a simple noetherian domian.

Linear automorphisms of $A_n(k).$ Now suppose that $k$ is field. Define the map $\varphi : A_1(k) \longrightarrow A_1(k)$ on the generators by $\varphi(x_1)=ax_1+by_1, \ \varphi(y_1)=cx_1+dy_1, \ a,b,c,d \in k.$ We would like to see under what condition(s) $\varphi$ becomes a $k$-algebra homomorphism. Well, if $\varphi$ is a homomorphism, then since $y_1x_1=x_1y_1+1,$ we must have

$\varphi(y_1)\varphi(x_1)=\varphi(x_1)\varphi(y_1)+1.$

Simplifying the above will give us $(ad-bc)y_1x_1=(ad-bc)x_1y_1 + 1$ and since $y_1x_1=x_1y_1+1,$ we get $ad-bc=1.$  We can now reverse the process to show that if $ad-bc=1,$ then $\varphi$ is a homomorphism. So $\varphi$ is a homomorphism if and only if $ad-bc=1.$ But then the map $\psi : A_1(k) \longrightarrow A_1(k)$ defined by

$\psi(x_1)=dx_1 - by_1, \ \psi(y_1)=-cx_1+ay_1$

will also be a homomorphism and $\psi = \varphi^{-1}.$ Thus $\varphi$ is an automorphism of $A_1(k)$ if and only if $ad-bc=1.$ In terms of matrices, the matrix $S=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ defines a linear automorphism of $A_1(k)$ if and only if $\det S=1.$

We can extend the above result to $A_n(k), \ n\geq 1.$ Let $S \in M_{2n}(k),$ a $2n \times 2n$ matrix with entries in $k.$ Let ${\bf{x}}=[x_1, \ldots , x_n, y_1, \ldots , y_n]^T$ and define the map $\varphi: A_n(k) \longrightarrow A_n(k)$ by ${\bf{x}} \mapsto S {\bf{x}}.$ Clearly $\varphi$ is a $k$-algebra homomorphism if and only if $\varphi(x_i), \varphi(y_i)$ satisfy the same relations that $x_i,y_i$ do, i.e.

$\varphi(x_i)\varphi(x_j)=\varphi(x_j) \varphi(x_i), \ \varphi(y_i) \varphi(y_j)=\varphi(y_j) \varphi(y_i),$  $\ \varphi(y_i) \varphi(x_j)=\varphi(x_j) \varphi(y_i) + \delta_{ij}, \ \ \ \ \ \ \ \ \ (1)$

for all $1 \leq i,j \leq n.$ Let $I_n \in M_n(k)$ be the identity matrix and let ${\bf{0}} \in M_n(k)$ be the zero matrix. Let $J=\begin{pmatrix} {\bf{0}} & I_n \\ -I_n & {\bf{0}} \end{pmatrix}.$ Then, in terms of matrices, $(1)$ becomes

$SJS^T=J. \ \ \ \ \ \ \ \ \ \ (2)$

Clearly if $S$ satisfies $(2),$ then $S$ is invertible and thus $\varphi$ will be an automorphism. So $(2)$ is in fact the necessary and sufficient condition for $\varphi$ to be an automorphism of $A_n(k).$

A $2n \times 2n$ matrix which satisfies $(2)$ is called symplectic. See that if $S$ is a $2 \times 2$ matrix, then $S$ is symplectic if and only if $\det S=1.$