How canonical is the triangular decomposition of a rational Cherednik algebra?

Introduction: Let $V$ be a finite-dimensional $\mathbb{C}$-vector space, let $G \leq \mathrm{GL}(V)$ be a finite subgroup and let $\kappa:V \times V \rightarrow \mathbb{C}G$ be an alternating bilinear map. Let $I$ be the ideal in $\mathrm{T}(V) \sharp \mathbb{C}G$ generated by the elements $vw-wv-\kappa(v,w)$ with $v,w \in V$. ($\sharp$ denotes the smash product). If we let elements of $V$ have degree $1$ and elements of $\mathbb{C}G$ have degree 0, we get a grading on $\mathrm{T}(V) \sharp \mathbb{C}G$ and so a filtration on $A := \mathrm{T}(V) \sharp \mathbb{C} G/I$. As $\mathrm{gr}(A)$ is commutative in degree 1, the quotient morphism $\mathrm{T}(V) \sharp \mathbb{C} G$ induces a surjective graded algebra morphism $\xi: \mathrm{S}(V) \sharp \mathbb{C} G \rightarrow \mathrm{gr}(A)$.

Now, $A$ is called a rational Cherednik algebra (or perhaps in this more general setting it's called a graded Hecke algebra or Drinfeld-Hecke algebra) if $\xi$ is an isomorphism.

If I did not make a mistake, the condition that $\xi$ is an isomorphism is equivalent to the following: for any linear section $s$ of the quotient map $\mathrm{T}(V) \rightarrow \mathrm{S}(V)$, the vector space morphism $\theta_s:\mathrm{S}(V) \otimes \mathbb{C} G \rightarrow A$, $x \otimes g \mapsto q(s(x) \otimes g)$, is an isomorphism, where $q:\mathrm{T}(V) \otimes \mathbb{C} G \rightarrow A$ is the quotient map.

Hence, a rational Cherednik algebra $A$ is $\mathrm{S}(V) \otimes \mathbb{C}G$ as a vector space, but only up to the choice of a linear section. Now, is there a canonical way to identify $A$ with $\mathrm{S}(V) \otimes \mathbb{C}G$?

This might be a little pedantic, but nobody ever mentions a choice of a section and I fear that I miss an important point here.

In an earlier version of my question I also asked if there is also a canonical way to view $S(V)$ as a subalgebra of $A$ but I see that this can't be true in general.

An additional (not totally unrelated) question: If the PBW morphism $\xi$ is an isomorphism, the algebra $A$ is already pretty nice because several properties of $\mathrm{S}(V) \sharp \mathbb{C}G$ (noetherian, prime, finite homolgical dimension, Cohen-Macaulay) are transported to $A$. But $\xi$ would in the same way as above already exist if $\kappa$ would be a map to $V \cdot \mathbb{C} G \subseteq \mathrm{T}(V) \sharp \mathbb{C}G$, i.e. if it would involve degree 1 elements. Why aren't those algebras interesting? (Perhaps because they don't have a triangular decomposition or they aren't deformations of $\mathrm{S}(V) \sharp \mathbb{C}G$?) I'm just looking for a natural reason to look at the $A_\kappa$ (I remember very well my question and the answers about reasons for studying rational Cherednik algebras but let's try it this way...)

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If you edit in that way based on answers, then answers stop being related to the question! – Mariano Suárez-Alvarez Jul 28 '10 at 21:59
At this point I wasn't referring to your answer but to the nonsense I was writing (the isomorphism). As I didn't want to leave something wrong there, I removed this. But you're right; I will rewrite this... – user717 Jul 29 '10 at 10:55

Pick $\dim V=2$, $G$ trivial, and $\kappa$ a symplectic form. Then $A$ is the first Weyl algebra, which does not contain any commutative algebra isomorphic to $S(V)$. The answer to your further problem is thus no. A reference for this result is [Dixmier, Jacques. Sur les algèbres de Weyl. (French) Bull. Soc. Math. France 96 1968 209--242. MR0242897 (39 #4224)], where Dixmier shows that whenever $x$ is a non-scalar element in $A_1$ the centralizer $C(x)$ of $x$ in $A_1$ is a module of finite type over $k[x]\subseteq A_1$.

The algebras you mention in your related problem are indeed worthy of study! They are deformations of $S(V)\\#\mathbb CG$, and some of them are even PBW deformations. You can write down the explicit condition for this to happen by unrolling the so called Jacobi condition; you'll find this discussed, for example, in Roland Berger and Victor Ginzburg's paper on non homogeneous PBW deformations of $N$-Koszul algebras (you case is quadratic, so simpler, but they discuss it it)

As for your pedantic point, I don't know. I guess you want someting like the symmetrization map that exists for enveloping algebras?

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I missed a point in my first comment: Is $\xi$ in the case of the Weyl algebra still an isomorphism? If so, then is the restriction to the particular $\kappa$ just something to make life simpler when considering deformations of $S(V) \sharp \mathbb{C} G$? As for the pedantic point: Perhaps I could have just summarized this in the question "what precisely is the vector space isomorphism $\mathrm{S}(V) \otimes \mathbb{C}G \rightarrow A$ the PBW-property induces"? Is it unique? – user717 Jul 28 '10 at 15:19
The Weyl algebra is a PBW deformation of the symmetric algebra. – Mariano Suárez-Alvarez Jul 28 '10 at 15:24

Perhaps part of the confusion is that you didn't actually discuss triangular decompositions at all (that certainly confused me at first). A triangular decomposition for a Cherednik algebra comes from a choice of extra data: two transverse $\kappa$-Lagrangian subspaces $M_1,M_2$. These exist as long as $\kappa$ is a symplectic form (I've never heard of considering the case where it might be degenerate). Ideally these would be $G$-invariant, but that's not necessary if all you want is PBW (for other things, it will mess you up).

For a subspace $W\subset V$ on which $\kappa$ is trivial, there is a map of algebras $S(W)\to A$, since the corresponding elements commute. Thus, our Lagrangian subspaces give us algebra maps $S(M_1),S(M_2)\to A$, and every element of $A$ can be uniquely written as $abc$ with $a\in S(M_1), b\in \mathbb{C} G$ and $c\in S(M_2)$. That is, there is a natural vector space isomorphism $A\cong S(M_1)\otimes \mathbb{C}G \otimes S(M_2)$.

Now, this isn't absolutely canonical (I had to choose the subspaces, and I had to choose what order to put them in), but I think it's about as good as you're likely to get.

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Well, I started too general and this generality was source of my confusion. Above I mentioned (if it's not wrong!) that for any section I have a vector space isomorphism $S(V) \otimes \mathbb{C}G \rightarrow A_\kappa$ (equivalent to PBW for $A_\kappa$). Now, in the case of a rational Cherednik algebra I have as vector space $V \oplus V^*$ and the above gives a vector space isomorphism $S(V) \otimes \mathbb{C}G \otimes S(V^*) \rightarrow A_\kappa$ (the triangular decomposition!?). AND, this also gives an algebra embedding of $S(V)$ and $S(V^*)$ into $A_\kappa$ because $[V,V^*]=0$ in $A_\kappa$ – user717 Jul 29 '10 at 19:54
My problem was that I was searching for an embedding of $S(V)$ into a general graded Hecke algebra and this cannot work. In case of rational Cherednik algebras I'm only embedding 'half' of this algebra and I didn't realize this. Anyways, the problem remains if I have to choose a section as above to get my vector space decomposition or if this can be done canonically (it's still very pedantic but I'm still not sure if I miss a point here). – user717 Jul 29 '10 at 20:00
So, I think my first comment is just exactly what you explained in general (thanks for this). – user717 Jul 29 '10 at 20:02