Why do non-equioriented A_n quivers have singularities identical to the singularities of Schubert varieties?

2010-06-09T13:27:08Z

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 A_n 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 A_n quivers have the same singularities as Schubert varieties to non-equioriented A_n 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 A_p+q+1 with p arrows in one direction and q arrows in the other and Q' an equioriented Dynkin quiver of type A_p+2q+1, their respective path algebras B=kQ and A=kQ', and respective Auslander-Reiten quivers &Gamma_B and &Gamma_A 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 A_p+2q+1 I get out of this theorem and how that is (maybe a GIT quotient away from) a Kazhdan-Lustzig variety?

Thanks,

Anna

Answer by Steven Sam for Why do non-equioriented A_n quivers have singularities identical to the singularities of Schubert varieties?

2010-06-09T17:32:46Z

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.

Why do non-equioriented An quivers have singularities identical to the singularities of Schubert varieties? - MathOverflow

Why do non-equioriented An quivers have singularities identical to the singularities of Schubert varieties?

Answer by Steven Sam for Why do non-equioriented An quivers have singularities identical to the singularities of Schubert varieties?

Why do non-equioriented A_n quivers have singularities identical to the singularities of Schubert varieties? - MathOverflow

Why do non-equioriented A_n quivers have singularities identical to the singularities of Schubert varieties?

Answer by Steven Sam for Why do non-equioriented A_n quivers have singularities identical to the singularities of Schubert varieties?