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

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    $\begingroup$ 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

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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.

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

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