Posts Tagged ‘linear automorphism’

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.

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