Let $G$ be a connected, linear algebraic group over $\mathbb{C}$. Let $\rho:G\to \mathrm{GL}(V)$ be a rational representation. Does $\rho$ have at most countably many subrepresentations (up to isomorphism)?
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2$\begingroup$ Does "up to isomorphism" mean the isomorphism type of the sub-representation, or up to conjugation by a $\rho$-equivariant automorphism of $V$? $\endgroup$– Kenta SuzukiCommented May 15 at 14:48
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$\begingroup$ Up to isomorphism as representations of $G$. $\endgroup$– Doug LiuCommented May 15 at 14:53
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3$\begingroup$ Side note: a reductive group has only countably many representations up to isomorphism. And so does the 1-dimensional unipotent group. $\endgroup$– YCorCommented May 15 at 14:56
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1$\begingroup$ @PseudoNeo, re, Jantzen - Representations of algebraic groups is, I think, a standard reference. $\endgroup$– LSpiceCommented May 15 at 19:23
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1$\begingroup$ @KentaSuzuki I should have said "constructible", not "Zariski-closed"; it is enough to have this finite-continuum dichotomy. Let $E$ be the (Zariski-closed) subset of the Grassmanian of $V$ consisting of invariant subspaces. The set of triples $(W_1,W_2,f,g)$ where $W_1,W_2\in E$ and $G$-invariant, and $f,g$ are inverse equivariant homomorphisms $W_1\to W_2$, $W_2\to W_1$, is Zariski-closed in $E^2\times\mathrm{Mat}(V)^2$. So its projection to $E^2$ is constructible. This is exactly the equivalence relation. $\endgroup$– YCorCommented May 15 at 22:46
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1 Answer
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No. Let $G$ be the 2-dimensional unipotent abelian group, and its 3-dimensional representation given by $$(t,s)\mapsto \begin{pmatrix}1 & t & s\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix}.$$
Let $G_u$ be the subgroup of $G$ of those $(t,ut)$, $t\in\mathbf{C}$. Then the fixed point set $V_u$ of $G_u$ consists of triples $(x,-uz,z)$, $x,z\in\mathbf{C}$. Then $V_u$ is a subrepresentation of $V$ whose kernel is exactly $G_u$. In particular, the $V_u$, for $u\in\mathbf{C}$, are pairwise non-isomorphic representations.