# Orbit closures in smooth irreducible components

$$\DeclareMathOperator\mod{mod}\DeclareMathOperator\GL{GL}$$ 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.

• There is always a closed orbit and that is non-singular. Did you miss a condition? Dec 22, 2021 at 8:48
• @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. Dec 22, 2021 at 16:33

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.