Why is $W(V)\simeq D(k[X_1,\dots,X_n])$? 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.
 A: 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.
