In the general case, quiver cycles are of the form of orbit closures of $GL\cdot V_{\vec{r}}$, where $GL= \prod_{i=0}^n GL_{r_i}$ is the possible changes of basis on all of the vector spaces on each of the vertices and $V_{\vec{r}}$ is any representation of the quiver with fixed dimension vector $\vec{r}$. In the equioriented An case, these are well understood by Zelevinsky and Lakshmibai-Magyar by showing them isomorphic to open sets in Schubert varieties. Bobinski and Zwara claim to reduce the non-equioriented case to the equioriented case, but I don't see how they are doing that.

In the introduction to ``Normality of Orbit Closures for Dynkin Quivers" (manuscripta math. 2001), Bobinski and Zwara say that they will generalize the result that equioriented An quivers have the same singularities as Schubert varieties to non-equioriented An quivers. They claim that they will do this by reducing the non-equioriented case to the equioriented case. So far, so good. But then, they say that this result follows from the proposition that they will prove, which I don't see has to do with the theorem at all.

The proposition is about a Dynkin quiver, Q, of type Ap+q+1 with p arrows in one direction and q arrows in the other and Q' an equioriented Dynkin quiver of type Ap+2q+1, their respective path algebras B=kQ and A=kQ', and respective Auslander-Reiten quivers &GammaB and &GammaA over the category of finite dimensional left modules over A and B. The proposition says ``Let A=kQ' and B=kQ be the path algebras of quivers Q' and Q, respectively, where Q and Q' are Dynkin quivers of type A. Assume there exists a full embedding of translation quivers $F: \Gamma_B \to \Gamma_A$. Then there exists a hom-controlled exact functor $\mathcal{F}: \text{mod }B \to \text{mod }A$."

Can anyone tell me how (or if) their results translate into a result that tells me a recipe for constructing a Kazhdan-Lustzig variety from my non-equioriented quiver? (By K-L variety, I mean a Schubert variety intersect an opposite Bruhat cell.) Alternately, is there a way to see which particular sub-variety of the representation variety of equioriented Ap+2q+1 I get out of this theorem and how that is (maybe a GIT quotient away from) a Kazhdan-Lustzig variety?



  • $\begingroup$ His name is Bobinski, without the extra n :P $\endgroup$ – Mariano Suárez-Álvarez Jun 9 '10 at 13:29

The relevance of hom-controlled functors comes from Zwara's paper "Smooth morphisms of module schemes" (Theorem 1.2). The definition there is that two schemes with basepoints $(X,x)$ and $(Y,y)$ have identical singularities if there is a smooth morphism $f \colon X \to Y$ such that $f(x) = f(y)$.

Let $F$ be a hom-controlled functor. He shows that when we're dealing with module varieties, and $X$ is an orbit closure $\overline{O}_M$ with basepoint $x$ some closed point of $O_N$ (so $x$ represents the isomorphism class of a module $N$), then $(\overline{O}\_M, x)$ has identical singularities as $(\overline{O}\_{FM}, y)$ where $y$ is a closed point of $O_{FN}$.

So if one starts with non-equioriented ${\rm A}_n$ and picks an orbit closure $\overline{O}_M$ together with a closed point in it, then one knows that there is a smooth morphism to some orbit closure in a bigger ${\rm A}_m$. The orbit closure is just the image of $M$ under the hom-controlled functor constructed in Bobinski and Zwara's paper (though this construction is long and I don't remember the details). Then one can use Lakshmibai–Magyar to get a smooth morphism from this orbit closure to some Schubert variety.

So it's enough to understand how to construct $F$, which I remember being explicit but requiring quite a few steps, if we just want the varieties together with the singularities, but constructing the smooth morphism itself would take a lot more digging to construct explicitly.

  • $\begingroup$ "...then one knows that there is a smooth morphism to some orbit closure in a bigger $A_m$." Is it really to, and not from? $\endgroup$ – Allen Knutson Jun 9 '10 at 17:54
  • $\begingroup$ Looking back at the proof, it might actually be neither. In the proof of Theorem 1.2, Zwara factors $F$ into several hom-controlled functors, and it seems that each of them produces maps going in opposite directions. $\endgroup$ – Steven Sam Jun 9 '10 at 18:54

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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