Algebra endomorphisms and centers of Weyl algebras

LEMMA 1.7   Let $k$ be a field of characteristic $p\neq 0$ . For any $k$ -algebra endomorphism $\phi$ of $A_n(k)$ ,

1. $\Phi_{c}\circ\phi$ is a surjective homomorphism for any $c\in k^{2n}$ .
2. $\phi(Z_n(k))\subset Z_n(k)$ .

We may assume that $k$ is an algebraically closed field.

(1) The composition $\phi_c=\Phi_{c} \circ \phi$ is a representation of $A_n$ . By Lemma 1.4 we see that any irreducible sub representation of $\phi_c$ is equivalent to $\Phi_{\bar{c}}$ for some $\bar{c}\in k^{2n}$ . By a dimensional argument, we conclude that $\phi_c$ itself is equivalent to $\Phi_{\bar{c}}$ . (In other words, there exists $g_c\in \operatorname{GL}_{n}(k)$ such that

$$\phi_c(x)=g_c \Phi_{\bar{c}}(x)g_c^{-1} \qquad (\forall x\in A_n)$$

holds.) Thus $\phi_{c}$ is surjective as required.

(2) Let $z \in Z_n$ , $x\in A_n$ . For any $c\in k^{2n}$ , we have (using the same notation as above)

$$[\phi_c(z),\phi_c(x)]= \phi_c([z,x])=0$$

Since we know by (1) that $\phi_c$ is surjective, we see that $\phi_c(z)$ belongs to the center of $M_{p^n}(k)$ . In particular for any $y\in A_n$ , we have

$$\Phi_c([\phi(z),y])=[\Phi_c(\phi(z)),\Phi_c(y)]=[\phi_c(z),\Phi_c(y)] =0.$$

Thus

$$[\phi(z),y]\in \bigcap_c \operatorname{Ker}(\Phi_c)=0.$$

So $\phi(z)\in Z(A_n)=Z_n$ as required.

COROLLARY 1.8  

Let $\phi:A_n\to A_n$ be a $k$ -algebra endomorphism of $A_n$ . Then by restriction we obtain a homomorphism

$$\phi_{Z_n}: Z_n \to Z_n.$$

Furthermore, if the base field $k$ is perfect, then $\phi$ may be uniquely extended to its $p$ -th root.

$$\phi_{S_n}: S_n \to S_n.$$

In precise, Let us write down $\phi_{Z_n}$ like

$$\phi_{Z_n}(\gamma_j^p)=\sum_I f_{j,I} (\gamma^p)^I \quad(j=1,2,\dots, 2 n).$$

Then $\phi_{S_n}$ is given by the following formula.

$$\phi_{S_n}(T_j)=\sum_I (f_{j,I})^{1/p} T^I \quad(j=1,2,\dots, 2 n).$$

Here comes a geometric interpretation of endomorphisms of Weyl algebras.

COROLLARY 1.9   Let $k$ be a perfect field of characteristic $p\neq 0$ . Let $\phi:A_n\to A_n$ be a $k$ -algebra endomorphism of $A_n$ . Then we have a matrix valued function $G\in \operatorname{GL}_{p^n}(S_n)$ and a morphism $f: \operatorname{Spec}(S_n)\to \operatorname{Spec}(S_n)$ which enables the following diagram commute.

$$\begin{CD}A_n @>\phi » A_n @V\Phi VV @V\Phi VV M_{p^n}(S_n) @>\bar\phi» M_{p^n}(S_n) \end{CD}$$ (*)

Where $\bar{\phi}$ is defined as

$$\bar{\phi}(x)=G f^*(x) G^{-1}.$$ (**)

We may write down the commutative diagram above as the following equation.

$$\Phi(\phi(a))=G f^*(\Phi(a)) G^{-1}$$

PROOF.. Let us define

$$\phi_{Z_n}: Z_n\to Z_n$$

and

$$\phi_{S_n}: S_n\to S_n$$

as in the previous Corollary. We put $f=\operatorname{Spec}(\phi_{S_n})$ .

We have an well-defined $Z_n$ -algebra homomorphism

$$\phi_{S_n}\otimes \phi: S_n \otimes_{Z_n} A_n \to S_n \otimes_{Z_n} A_n.$$

By using the isomorphism in , we obtain a $Z_n$ -homomorphism

$$\bar{\phi}= \hat\Phi (\phi_{S_n}\otimes \phi) {\hat \Phi}^{-1}: M_{p^n}(S_n) \to M_{p^n} (S_n)$$

which is compatible with $\phi$ in the sense that it satisfies the commutative diagram (*) of the statement.

It remains to prove that the map $\bar{\phi}$ is represented as (**). By pull-back, we obtain an $S_n$ -algebra homomorphism

$$\rho: M_{p^n}(S_n)\cong S_n\otimes_{\phi_{S_n},S_n} M_{p^n}(S_n) \ni y \otimes x \mapsto y \bar{\phi}(x) \in M_{p^n}(S_n).$$

Where the first isomorphism in the above line is the inverse of the following $S_n$ -algebra homomorphism.

$$S_n\otimes_{\phi_{S_n},S_n} M_{p^n}(S_n) \ni y\otimes x \mapsto y f^*(x) \in M_{p^n}(S_n).$$

by an argument similar to that in (I,Lemma 7.9), we see that there exists $G\in GL_{p^n}(S_n)$ such that

$$\rho(x)= G x G^{-1} \qquad (\forall y \in M_{p^n} (S_n))$$

holds. (See appendix for the detail.)

$\qedsymbol$