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I was reading about the Weyl algebra, but don't get a certain isomorphism.

For a little background, let $V$ be a vector space of dimension $2n$, with a bilienar form $\omega$, nondegenerate, and let $W(V)$ be the corresponding Weyl algebra. So recall that $W(V)=T(V)/I$ where $I$ is the ideal generated by $x\otimes y-y\otimes x-\omega(x,y)$ for $x,y\in V$ and $T(V)$ denotes the tensor algebra.

If $k$ is a field of characteristic $0$, then $W(V)\simeq D(k[X_1,\dots,X_n])$, but I'm having trouble realizing this. Does anyone have a clear proof, or reference to a proof to bring this isomorphism to light? Many thanks.

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1 Answer 1

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If $V^{2n}$ is a symplectic vector space (and we assume characteristic 0), then we can always find a basis $\{x_1, p_1, \cdots, x_n, p_n\}$ of $V$ so that $$ \omega(p_i, x_j) = \delta_{ij} $$ and $\omega(x_i, x_j) = 0 = \omega(p_i, p_j)$. Then the Weyl algebra is the algebra with generators $x_i, p_j$ with the relations $x_i x_j = x_j x_i$, $p_i p_j = p_j p_i$, and $p_i x_j -x_j p_i = \delta_{ij}$.

Now consider the algebra of polynomial differential operators. This has generators $x_i, \partial_j$, satisfying $x_i x_j = x_j x_i$, $\partial_i \partial_j = \partial_j \partial_i$, and $\partial_i x_j - x_j \partial_i = \delta_{ij}$.

So in this basis for $V$, we see that both algebras have the same generators and relations, so the isomorphism is given by $p_j \mapsto \partial_j$.

The fact that such a basis of $V$ exists is so basic that I'm not sure it even has a name! (For symplectic manifolds this is the Darboux theorem, but that already presupposes the result about vector spaces.) Any book or lectures notes on symplectic geometry should prove it. It's easy enough that you should be able to prove it yourself (hint: use nondegeneracy of $\omega$ and induction on $n$).

Update:

Here is a proof that we can always choose a symplectic basis $\{x_1, p_1, \cdots, x_n, p_n\}$. Pick some vector $x_1 \in V$. Since $\omega$ is nondegenerate, we can find some $p_1 \in V$ so that $\omega(p_1, x_1) \neq 0$. Then by rescaling $p_1$ if necessary, we can assume $\omega(p_1, x_1) = 1$. If $\dim V = 2$ we're done, and if not assume by induction that the result is true for symplectic vector spaces of strictly smaller dimension. Now consider the subspaces $W$ and $W^\perp$ defined by \begin{align} W &= \mathrm{span}\{x_1, p_1\} \\\ W^\perp &= \{v \in V \ | \ \omega(v, w) = 0 \ \forall\ w \in W \} \end{align} A quick calculation shows that $W \cap W^\perp = 0$, and that the symplectic form restriced to $W^\perp$ is nondegenerate. Hence by induction there is a basis $\{x_2, p_2, \cdots, x_n, p_n\}$ of $W^\perp$ satisfying $\omega(p_i, x_j) = \delta_{ij}$. Since $V = W \oplus W^\perp$, we're done.

An good reference is the lecture notes by Cannas da Silva.

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Thanks Jonathan. I'm not familiar with symplectic vector spaces really, so I've been having a hard time proving this claim these past days. Do you mind including it if you have the time, or possibly giving a reference where I could look it up? (I'm not familiar with texts on symplectic geometry either.) –  Buble Apr 12 '12 at 23:46
    
Updated to include a sketch of the proof. You can find the details in the lecture notes by Cannas da Silva. –  Jonathan Apr 13 '12 at 22:43
    
This is a very slick proof, more so than the one I'm familiar with. Out of curiosity, how do you know that the algebra of polynomial differential operators has generators $x_i,\partial_i$ satisfying those claimed relations? –  Kally Apr 14 '12 at 21:20
    
@Kally what is your working definition of polynomial differential operators, if not the algebra generated by $x_i, \partial_j$ subject to the above relations? There are several equivalent definitions, so the answer depends on which you take as fundamental. –  Jonathan Apr 25 '12 at 19:24

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