Weyl algebras; definition & automorphisms (2)

Posted: January 25, 2011 in Noncommutative Ring Theory Notes, Weyl Algebras
Tags: ,

A non-linear automorphism of A_n(k). Let k be a field. For any u, v \in A_n(k) we let [u,v]=uv-vu. So the realtions that define A_n(k) become [x_i,x_j]=[y_i,y_j]=0, \ [y_i,x_j]=\delta_{ij}, for all i,j.

Lemma 1. Let f,g \in k[x_1, \cdots , x_n] and 1 \leq r,s \leq n. Then

1) [fy_r,g] = f \frac{\partial{g}}{\partial{x_r}}.

2) [fy_r,gy_s] = f \frac{\partial{g}}{\partial{x_r}}y_s - g \frac{\partial{f}}{\partial{x_s}}y_r.

Proof. An easy induction shows that y_r x_r^{\ell} = x_r^{\ell}y_r + \ell x_r^{\ell -1} for all \ell. Applying this, we will get that if h = x_1^{\alpha_1} \cdots x_n^{\alpha_n}, then y_r h = \frac{\partial{h}}{\partial{x_r}} + hy_r. So, since every element of k[x_1, \cdots , x_n] is a finite linear combination of monomials in the form h, we will get

y_r g = \frac{\partial{g}}{\partial{x_r}} + gy_r, \ \ \ \ \ \ \ \ \ \ \ \ (*)

for all g \in k[x_1, \cdots , x_n]. Both parts of the lemma are straightforwad results of (*). \Box

Notation. Let n \geq 2 and fix an integer 1 \leq m < n. For every 1 \leq i \leq m choose f_i \in k[x_{m+1}, \cdots , x_n] and put f_{m+1} = \cdots = f_n = 0.

Lemma 2. For any 1 \leq r,s,t \leq n we have \frac{\partial{f_r}}{\partial{x_s}} \cdot \frac{\partial{f_t}}{\partial{x_r}} = 0.

Proof. If r > m, then f_r = 0 and we are done. If r \leq m, then x_r will not occur in f_t and so \frac{\partial{f_t}}{\partial{x_r}} = 0. \ \Box

Now define the maps \varphi : A_n(k) \longrightarrow A_n(k) and \psi : A_n(k) \longrightarrow A_n(k) on the generators by

\varphi (x_i) = x_i + f_i, \ \varphi(y_i)= y_i - \sum_{r=1}^n \frac{\partial{f_r}}{\partial{x_i}}y_r


\psi (x_i)=x_i-f_i, \ \psi(y_i)=y_i + \sum_{r=1}^n \frac{\partial{f_r}}{\partial{x_i}}y_r,

for all 1 \leq i \leq n and extend the definition homomorphically to the entire A_n(k) to get k-algebra homomorphisms of A_n(k). Of course, we need to show that these maps are well-defined i.e. the images of x_i,y_i under \varphi and \psi satisfy the same relations that x_i, y_i do. Before that, we prove an easy lemma.

Lemma 3. \varphi(f) = \psi(f)=f for all f \in k[x_{m+1}, \cdots , x_n].

Proof. Let f = \sum c_{\alpha} x_{m+1}^{\alpha_{m+1}} \cdots x_n ^{\alpha_n}, where c_{\alpha} \in k and \alpha_i \geq 0. Then

\varphi(f) = \sum c_{\alpha} (x_{m+1} + f_{m+1})^{\alpha_{m+1}} \cdots (x_n + f_n)^{\alpha_n}.

But by our choice f_{m+1} = \cdots = f_n = 0 and thus \varphi(f)=f. A similar argument shows that \psi(f)=f. \ \Box

Lemma 4. The maps \varphi and \psi are well-defined.

Proof. I will only prove the lemma for \varphi because the proof for \psi is identical. Since f_i \in k[x_1, \cdots , x_n], we have \varphi(x_i) \in k[x_1, \cdots , x_n], for all i, and thus \varphi(x_i) and \varphi(x_j) commute. The relations [\varphi(y_i), \varphi(x_j)] = \delta_{ij} follow from the first part of Lemma 1 and Lemma 2. The relations [\varphi(y_i), \varphi(y_j)]=0 follow from the second part of Lemma 1 and Lemma 2. \Box.

Theorem. The k-algebra homomorphisms \varphi and \psi are automorphisms.

Proof. We only need to show that \varphi and \psi are the inverse of each other. Lemma 3 gives us \varphi \psi(x_i) = \psi \varphi(x_i)=x_i and Lemma 2 with Lemma 3 will give us \varphi \psi(y_i)=\psi \varphi (y_i)=y_i, for all i. \ \Box


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