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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

$\displaystyle \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)

$\displaystyle [\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

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

Thus

$\displaystyle [\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

$\displaystyle \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.

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

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

$\displaystyle \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.

$\displaystyle \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.

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

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

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

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

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

PROOF.. Let us define

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

and

$\displaystyle \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

$\displaystyle \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

$\displaystyle \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

$\displaystyle \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.

$\displaystyle 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

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

holds. (See appendix for the detail.)

$ \qedsymbol$


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Next: appendix Up: A review Previous: A matrix representation of
2007-12-11