## Derviations of Weyl algebras are inner

Posted: September 23, 2011 in Noncommutative Ring Theory Notes, Weyl Algebras
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Let $k$ be a field. We proved here that every derivation of a finite dimensional central simple $k$-algebra is inner. In this post I will give an example of an infinite dimensional central simple $k$-algebra all of whose derivations are inner. As usual, we will denote by $A_n(k)$ the $n$-th Weyl algebra over $k.$ Recall that $A_n(k)$ is the $k$-algebra generated by $x_1, \ldots , x_n, y_1, \ldots , y_n$ with the relations $x_ix_j-x_jx_i=y_iy_j-y_jy_i=0, \ y_ix_j-x_jy_i= \delta_{ij},$ for all $i,j.$ When $n = 1,$ we just write $x,y$ instead of $x_1,y_1.$ If $\text{char}(k)=0,$ then $A_n(k)$ is an infinite dimensional central simple $k$-algebra and we can formally differentiate and integrate an element of $A_n(k)$ with respect to $x_i$ or $y_i$ exactly the way we do in calculus.  Let me clarify “integration” in $A_1(k).$ For every $u \in A_1(k)$ we denote by $u_x$ and $u_y$ the derivations of $u$ with respect to $x$ and $y$ respectively. Let $f, g, h \in A_1(k)$ be such that $g_x=h_x=f.$ Then $[y,g-h]=0$ and so $g-h$ lies in the centralizer of $y$ which is $k[y].$ So $g-h \in k[y].$ For example, if $f = y + (2x+1)y^2,$ then $g_x=f$ if and only if $g= xy + (x^2+x)y^2 + h(y)$ for some $h(y) \in k[y].$ We will write $\int f \ dx = xy+(x^2+x)y^2.$

Theorem. If $\text{char}(k)=0,$ then every derivation of $A_n(k)$ is inner.

Proof. I will prove the theorem for $n=1,$ the idea of the proof for the general case is similar. Suppose that $\delta$ is a derivation of $A_1(k).$ Since $\delta$ is $k$-linear and the $k$-vector space $A_1(k)$ is generated by the set $\{x^iy^j: \ i,j \geq 0 \},$ an easy induction over $i+j$ shows that $\delta$ is inner if and only if there exists some $g \in A_1(k)$ such that $\delta(x)=gx-xg$ and $\delta(y)=gy-yg.$ But $gx-xg=g_y$ and $gy-yg=-g_x.$ Thus $\delta$ is inner if and only if there exists some $g \in A_1(k)$ which satisfies the following conditions

$g_y=\delta(x), \ \ g_x = -\delta(y). \ \ \ \ \ \ \ (1)$

Also, taking $\delta$ of both sides of the relation $yx=xy+1$ will give us

$\delta(x)_x = - \delta(y)_y. \ \ \ \ \ \ \ \ (2)$

From $(1)$ we have $\delta(x) = - \int \delta(y)_y \ dx + h(y)$ for some $h(y) \in k[y].$ It is now easy to see that

$g = - \int \delta(y) \ dx + \int h(y) \ dy$

will satisfy both conditions in $(1). \ \Box$