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...)

  • $\begingroup$ If you edit in that way based on answers, then answers stop being related to the question! $\endgroup$ – Mariano Suárez-Álvarez Jul 28 '10 at 21:59
  • $\begingroup$ 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... $\endgroup$ – 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?

  • $\begingroup$ 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? $\endgroup$ – user717 Jul 28 '10 at 15:19
  • $\begingroup$ The Weyl algebra is a PBW deformation of the symmetric algebra. $\endgroup$ – Mariano Suárez-Álvarez 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.

  • $\begingroup$ 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 $A_\kappa$ $\endgroup$ – user717 Jul 29 '10 at 19:54
  • $\begingroup$ 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). $\endgroup$ – user717 Jul 29 '10 at 20:00
  • $\begingroup$ So, I think my first comment is just exactly what you explained in general (thanks for this). $\endgroup$ – user717 Jul 29 '10 at 20:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.