## Weyl algebras; definition & automorphisms (2)

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

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

and

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

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 $i. \ \Box$