# An algebraic group has how many representations?

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)?

• Does "up to isomorphism" mean the isomorphism type of the sub-representation, or up to conjugation by a $\rho$-equivariant automorphism of $V$? May 15 at 14:48
• Up to isomorphism as representations of $G$. May 15 at 14:53
• Side note: a reductive group has only countably many representations up to isomorphism. And so does the 1-dimensional unipotent group.
– YCor
May 15 at 14:56
• @PseudoNeo, re, Jantzen - Representations of algebraic groups is, I think, a standard reference. May 15 at 19:23
• @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.
– YCor
May 15 at 22:46

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.