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I have been reading chapter 8 of Ralf Schiffler's 'Quiver Representations'. There he has defined, for a finite quiver $Q=(Q_0,Q_1)$ without oriented cycles and for a (fixed) dimension vector $\textbf{d}:= (d_i)_{i \in Q_0} \in \mathbb{Z}_{\geq 0}^n$, the space $E_\textbf{d}$ of representations $M$ of $Q$ having dimension vector equal to $\textbf{d}$, and the group $G_\textbf{d} := \prod_{i \in Q_0} GL_{d_i} (k)$, which acts on $E_\textbf{d}$ by conjugation: that is, for $g:=(g_i)_{i \in Q_0} \in G_\textbf{d}$ and $M := (M_i, \phi_\alpha)_{i \in Q_0, \alpha \in Q_1} \in E_\textbf{d}$, $$g \cdot M := (M_i, \hspace{1mm} g_{t(\alpha)} \hspace{0.5mm} \phi_\alpha \hspace{0.5mm} g_{s(\alpha)}^{-1})_{i \in Q_0, \alpha \in Q_1} \in E_\textbf{d}$$ where the source and target of arrow $\alpha \in Q_1$ are given by $s(\alpha)$ and $t(\alpha)$ respectively.

For a representation $M \in E_\textbf{d}$, we denote the orbit of $M$ under the above action by $\mathcal{O}_M$. Then $G_\textbf{d}$, $Aut(M)$ and $\mathcal{O}_M$ have the structure of an algebraic variety, and as such their dimension (as algebraic varieties) is defined.

Now in Proposition 8.10, in order to prove that the codimension of $\mathcal{O}_M$ is equal to the dimension of $Ext^1(M,M)$, he says that the "group of automorphisms of $M$ is an open subgroup of the group of endomorphisms thus $dim \hspace{0.51mm} Aut(M) = dim \hspace{0.51mm} End(M)$". The proof also uses Proposition 8.4 which is a relation between the vector space dimensions of $End(M)$ and $Ext^1(M,M)$.

Unfortunately all I know about algebraic varieties and topological groups is the definitions of the dimension of an algebraic variety $V$ (as one less than the length of the longest chain of distinct non-empty irreducible subvarieties of $V$) and of irreducible varieties, hence it is not obvious to me why the following are true:

$1.$ How can we topologize $End(M)$ to make it a topological group such that $Aut(M)$ turns out to be an open subset of $End(M)$?

$2.$ Does a general vector space become an algebraic variety? (If so, how?) If not, why do $End(M)$ and $Ext^1(M,M)$ in particular have structures of algebraic varieties? (For $End(M)$, I can still see that we could make it an algebraic variety by noting that the diagram commutativity conditions for morphisms between quiver representations gives some polynomial relations between the $\phi_{\alpha}$'s, and I wanted to confirm if this is right. And what about $Ext^1(M,M)$?) And why is the dimension of these vector spaces equal to their dimension as algebraic varieties (because that is what is being subtly used when Proposition 8.4 is employed to establish 8.10)?

$3.$ What does $Aut(M)$ being open in $End(M)$ have to do with the equality of the (algebraic variety) dimensions of $Aut(M)$ and $End(M)$ or with the vector space dimension of $End(M)$?

I would really appreciate some help in this regard.

asrxiiviii
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1 Answers1

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  1. The general idea here is that automorphisms are just endomorphisms of nonzero determinant, so any topology so that the determinant polynomial is continuous and $\{0\}\subset k$ is closed will do. In practice, the Zariski topology on $k^n$ suffices: this is the topology where the closed sets are $\{x\in k^n\mid p(x)=0\}$ for $p$ a polynomial $k[x_1,\cdots,x_n]$, and if you're working over a field that comes with a natural topology (like $\Bbb R$, $\Bbb C$, or $\Bbb Q_p$) those will all work as well, because they're finer (that is, more open sets) than the Zariski topology.

  2. This depends on your definition of an algebraic variety. If you take the "classical" definition from the linked post, it's obvious: set the topology to be the Zariski topology, then boom, variety. For other notions there, you'll have to do a little more, but it's not usually too bad.

  3. Open subsets of irreducible varieties in the Zariski topology have the same dimension as the whole space - see here for example. As the dimension of a vector space as a vector space and a vector space as a variety are equal (see here), we have the requested result.

KReiser
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