2
$\begingroup$

$\DeclareMathOperator\mod{mod}\DeclareMathOperator\GL{GL}$ Consider a basic connected finite dimensional algebra $A$ over an algebraically closed field $k$, with $n$ distinct isomorphism classes of simple (left) $A$-modules. For a fixed integer $d$, let $\mod(A,d)$ denote the variety of all left $A$-modules of $k$-dimension $d$. View $\mod(A,d)$ as a variety under the action of the general linear group $\GL(d)$, via conjugation. The connected components of $\mod(A,d)$ are well-known: They are given by the module varieties $\mod(A,\underline{e})$, for a dimension vector $\underline{e} \in \mathbb{Z}_{\geq 0}^{n}$. Namely, $\underline{e}={(e_i)}_{i=1}^{n}$ with $e_1+\dotsb+e_n=d$.

Thinking about such module varieties, I was wondering if the following question has a known answer:

Let $\mathcal{Z}$ be an irreducible component in a representation variety $\mod(A,\underline{e})$. In $\mathcal{Z}$, one can talk about some orbits in $\mathcal{Z}$ which are trivially closed and smooth (such as the orbits of semisimple modules, or those of similar nature). I can make the preceding sentence more precise if needed, but experts should be able to identify such "trivial" cases. It is often easier if $A$ is viewed as a quotient of a path algebra, of the form $kQ/I$, where $Q$ is a finite connected quiver. Thus, $\mod(A,\underline{e})$ can be seen as the representation variety.

That said, I am wondering if $\mathcal{Z}$ can be smooth such that all "Non-trivial" orbit closures in $\mathcal{Z}$ are singular. If that can happen, I would like to see an example of this phenomenon, and furthermore, know if there is any non-trivial set of conditions which guarantees that every smooth irreducible component $\mathcal{Z}$ contains at least one smooth orbit closure.

$\endgroup$
2
  • 1
    $\begingroup$ There is always a closed orbit and that is non-singular. Did you miss a condition? $\endgroup$ Dec 22, 2021 at 8:48
  • $\begingroup$ @FriedrichKnop Thank you for bringing that to my attention. Yes, I forgot to exclude this type of closed orbits from my question (such as orbits of semisimple modules, which are closed and smooth). I made the necessary changes. $\endgroup$
    – Kaveh
    Dec 22, 2021 at 16:33

1 Answer 1

1
$\begingroup$

Let $V$ be an irreducible representation of a reductive group $H$ such that every orbit has codimension $\ge2$ (e.g., when $\dim V\ge\dim H+2$). Put $G:=\mathbb C^*H$ with $\mathbb C^*$ acting by scalar multiplication. Then every orbit closure $X\subset V$ contains $0\in V$. The tangent space $T:=T_0X$ is a submodule of $V$. Hence $T=0$ or $T=V$ by irreducibility. In the second case, we have $\dim X\le\dim V-2<\dim V=\dim T$. Hence $X$ is not smooth in $0$. The first case leads to $X=\{0\}$ which is a closed orbit.

Minimal examples include $H=\text{quaternion group}$ acting on $V=\mathbb C^2$ or $H=SL(3)$ and $V=\text{adjoint representation}=\text{$3\times3$-matrices of trace $0$}$.

In the case above, the only way to get a smooth orbit closure is when $V$ itself is one. So I guess, a smooth non-trivial orbit closure exists if and only if there is a Luna slice having a summand with an open orbit.

$\endgroup$
1
  • $\begingroup$ Thanks a lot for the clear argument and explicit examples. This was very helpful! $\endgroup$
    – Kaveh
    Dec 30, 2021 at 7:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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