= Solution
The <dimension of a topological space by irreducible chains> is the supremum of the integers $d$ for which there is a chain $Z_0\subsetneq Z_1\subsetneq\cdots\subsetneq Z_d$ of nonempty irreducible closed subsets of the space. Closedness here is relative to the given locally closed space. For varieties this is their <Krull dimension>.
An <algebraic group> is a group whose underlying space is an <algebraic variety> and whose multiplication $G\times G\to G$ and inversion $G\to G$ are morphisms. An <algebraic group action> on a variety $V$ is a morphism $G\times V\to V$ satisfying the identity and associativity axioms of a <group action>.
For the homomorphism $\varphi$, the kernel is $\varphi^{-1}(1_H)$, so \b[the kernel is closed]. To prove closedness of the image, use the <Chevalley constructibility theorem>: the image of a morphism of varieties is constructible. Thus $S=\varphi(G)$ is a <constructible subset of a variety> and an abstract subgroup. Its closure $C$ is also a subgroup: translation by elements of $S$ preserves $C$, and continuity then extends multiplication and inversion to $C$.
A dense constructible subset contains a dense open subset $U$ of its closure. For $c\in C$, both $U$ and $cU$ are dense open subsets of $C$, so their intersection is nonempty. If $u=cv$ with $u,v\in U\subseteq S$, then $c=uv^{-1}\in S$. Hence $S=C$, proving that \b[the image is closed]. This is the principle that a <constructible subgroup is closed>.
Every nonempty fiber of $G\to S$ is a translate of $\ker\varphi$ and has that same dimension. The <fiber dimension theorem> therefore gives
$$
\boxed{\dim G=\dim\ker\varphi+\dim\operatorname{im}\varphi.}
$$
This <dimension formula for an algebraic group homomorphism> is a dimension statement, so it does not require separability of $\varphi$.
For a <dimension vector of a quiver representation> $\mathbf n$, set
$$
\operatorname{Rep}_Q(\mathbf n)=\prod_{\rho:i\to j}\operatorname{Hom}_k(k^{n_i},k^{n_j}),\qquad\operatorname{GL}(\mathbf n)=\prod_i\operatorname{GL}_{n_i}(k).
$$
The <base change action on quiver representations> is
$$
\boxed{(g\cdot f)_\rho=g_{t(\rho)}f_\rho g_{s(\rho)}^{-1}.}
$$
The entries are regular functions on the product of the <general linear groups> and the <quiver representation space>, because inverse entries are cofactors divided by the invertible determinant. Thus this is an <algebraic group action>.
For nonzero $\mathbf n$, let $\Delta k^\times$ be the common scalar subgroup $(\lambda I_{n_i})_i$ and define $\operatorname{PGL}(\mathbf n)=\operatorname{GL}(\mathbf n)/\Delta k^\times$. Common scalars act trivially, so the formula descends to an <algebraic group action> of this <projective base change group of a quiver>. This is a quotient by one common scalar, not a product of the individual projective groups. If every $n_i=0$, the representation space is a point and both actions are taken to be trivial.
Back to article page
