Derviations of Weyl algebras are inner

Posted: September 23, 2011 in Noncommutative Ring Theory Notes, Weyl Algebras
Tags: , , ,

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

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