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A matrix representation of Weyl algebra

From now on in this section we fix a prime number $ p$ and we assume $ k$ is a field of characteristic $ p$ . We will simply write $ A_n$ instead of $ A_n(k)$ .

Let us define operators(matrices) $ \{\mu_i\}_{i=1}^{2 n}$ acting on $ p^n$ -dimensional vector space

$\displaystyle V=k[x_1,x_2,\dots,x_n]/(x_1^p,x_2^p,\dots x_n^p)
$

by

\begin{equation*}
\left .
\begin{aligned}
&\mu_i=\text{multiplication by $x_i$}\...
...rtial/\partial x_i.
\end{aligned}\right \}
\qquad i=1,2,\dots,n.
\end{equation*}

Let

$\displaystyle S_n=k[T_1,T_2,\dots,T_{2n-1},T_{2n}]
$

be a polynomial ring of $ 2 n$ -variables over $ k$ .

Then we have a faithful representation

$\displaystyle \Phi: A_n \to M_{p^n}(S_n)
$

of the Weyl algebra $ A_n$ by putting

$\displaystyle \Phi(\gamma_i)=T_i\cdot 1_{p^n}+\mu_i.
$

Furthermore, for any $ c=(c_1,c_2,\dots,c_{2n}) \in k^{2n}$ , we have by specialization the following representation of $ A_n$ .

$\displaystyle \Phi_c: A_n \to M_{p^n}(k)
$

by putting

$\displaystyle \Phi_c(\gamma_i)=c_i\cdot 1_{p^n}+\mu_i.
$

We recall also that

LEMMA 1.3 (Part I, Lemma 7.3)   The center $ Z_n$ of $ A_n$ is isomorphic to a polynomial algebra with $ 2 n$ indeterminates. Namely,

$\displaystyle Z_n=k[\gamma_1^p,\gamma_2^p,\dots, \gamma_{2n}^p] .
$

LEMMA 1.4 (Part I, Corollary 7.8)   Let $ k$ be an algebraically closed field of characteristic $ p\neq 0$ . Then every finite dimensional irreducible representation $ \alpha:A_n(k)\to \operatorname{End}_k(V)$ of $ A_n(k)$ is equivalent to a representation $ \Phi_c$ for some $ c\in k^{2n}$ .

LEMMA 1.5   Let $ k$ be a field of characteristic $ p$ . If $ k$ contains infinite number of elements, then:

$\displaystyle \bigcap_{c\in k^{2 n} } \operatorname{Ker}(\Phi_c)=0.
$

Here is another thing we need to know.

LEMMA 1.6   $ \Phi$ gives a $ k$ -algebra isomorphism

$\displaystyle \hat{\Phi}:S_n\otimes_{Z_n} A_n \cong M_{p^n} (S_n).
$

PROOF.. One may easily verify that we have a well-defined homomorphism given by

$\displaystyle \hat{\Phi}:
S_n\otimes_{Z_n} A_n \ni f \otimes a \mapsto f \Phi(a)\ni M_{p^n} (S_n).
$

For any $ j\in \{1,2,3,\dots,2 n\}$ , we have

$\displaystyle \hat{\Phi} (1\otimes \gamma_j -T_j\otimes 1)=\mu_j.
$

So the image $ \operatorname{Image}(\hat{\Phi})$ contains $ \mu_j$ . Since we know that $ \{\mu_j\}_{j=1}^{2 n}$ generates $ M_{p^n}(k)$ as $ k$ -algebra (Part I, Corollary 7.10) , we conclude that the map $ \hat{\Phi}$ is surjective.

Now, both $ S_n\otimes_{Z_n}A_n$ and $ M_{p^n}(S_n)$ is free $ S_n$ -modules of rank $ p^{2 n}$ . So the map $ \hat{\Phi}$ is generically injective. (That means, if we take the quotient field $ Q(S_n)$ of $ S_n$ and consider

$\displaystyle 1_{Q(S_n)}\otimes\hat{\Phi}:
Q(S_n)\otimes_{S_n}(S_n\otimes_{Z_n} A_n )\to M_{p^n} (Q(S_n)).
$

then by an elementary theorem in linear algebra, we see that it is an isomorphism.)

Since $ S_n$ (a polynomial algebra over $ k$ ) is an integral domain, $ S_n\otimes_{Z_n}A_n$ is $ Z_n$ -torsion free. (That means,

$\displaystyle S_n \otimes_{Z_n} A_n\to Q(S_n)\otimes_{S_n}(S_n\otimes_{Z_n} A_n )
$

is injective.)

So we see that $ \hat{\Phi}$ is injective.

$ \qedsymbol$


next up previous
Next: Algebra endomorphisms and centers Up: A review Previous: perfect field
2008-03-15