Let $G$ be an affine group scheme of finite type over an algebraically closed field $k$. Suppose that $V$ is a finite dimensional representation of $G$.

For every $k$-algebra $A$ we have the base change representation $V_A = V\otimes_kA$ of the group $G_A = \mathrm{Spec}\, A\times_{\mathrm{Spec}\,k}G$.

Definition 1. $V$ is absolutely irreducible if for every $k$-algebra $A$ there are no nontrivial subspaces $W\nsubseteq V$ such that $W_A = A\otimes_kW$ is $G_A$-stable in $V_A$.

Definition 2. $V$ is absolutely irreducible if for every field extension $L$ of $k$ the representation $V_L$ of the group $G_L$ is irreducible.

Two definitions above capture the same notion. Indeed, if $W_A$ is $G_A$-stable for some $k$-algebra $A$, then pick a nonzero morphism $A\rightarrow L$ into a field $L$ over $k$. Then $W_L$ is $G_L$-stable. So if there are no stable subspaces in the second sense, there are also no stable subspaces in the first. The other implication is trivial.

Fact. Suppose that $G$ is reduced. If $V$ is irreducible, then $V$ is absolutely irreducible.

Sketch of proof. Since $G$ is reduced and $k$ is algebraically closed, we deduce that $V$ is an irreducible representation of the abstract group $G(k)$. Suppose that the dimension of $V$ is $n$ and consider a morphism $\rho:G(k)\rightarrow \mathbb{M}_n(k)$ inducing action of $G(k)$ on $V$. Extend $\rho$ to a morphism of $k$-algebras $\tilde{\rho}:k[G(k)]\rightarrow \mathbb{M}_n(k)$. By Jacobson's density theorem we derive that $\tilde{\rho}$ is surjective. This property is stable under base change. This implies that $V_L$ is irreducible over $G_L(L)$ and hence $V_L$ is irreducible over $G_L$.

Now my question:

Suppose that $G$ is nonreduced. If $V$ is irreducible, then is $V$ absolutely irreducible?

  • $\begingroup$ Your definition does not make much sense to me: if $I$ is a proper ideal of $A$, then $V\otimes_kI$ is non-trivial $G_A$-stable submodule of $V\otimes_kA$. The standard definition assumes that $V\otimes_kA$ is irreducible for all field extensions $A$ of $k$; but the fact that every irreducible representation is absolutely irreducible is standard, since $k$ is algebraically closed $\endgroup$ – Angelo Jan 7 at 18:15
  • $\begingroup$ @Angelo The definition I wrote was incorrect and was not the one I had in mind. Thank you for your comment. Can you please provide reference for irreducible $\Rightarrow$ absolutely irreducible for every group scheme of finite type over $k$? $\endgroup$ – Slup Jan 8 at 6:57
  • $\begingroup$ @JimHumphreys It seems that this answers my question. Can you provide reference or maybe write an answer? $\endgroup$ – Slup Jan 8 at 6:58

Suppose $G$ is an affine group scheme over an algebraically closed field $k$, and $V$ is a finite dimensional representation of $G$. Let $K$ be an extension of $k$, and assume that $V_K$ is reducible as a representation of $G_K$. Then I claim that $V$ is reducible as a representation of $G$.

Let $r$ be an integer with $0 < r < \dim V$, such that $V_K$ has a $G_K$-invariant $r$-dimensional sub-vector space. Let $\mathbb{G}(r,V)$ the Grassmannian of $r$-dimensional vector spaces of $V$, with its induced action of $G$. Since formation of scheme-theoretic fixed point loci commutes extension of base field, the fix point locus $\mathbb{G}(r,V)^G$ has a $K$-valued point. This means that is non-empty; since it is of finite type, and $k$ is algebraically closed, this means that $\mathbb{G}(r,V)^G$ has a $k$-valued point, so $V$ is reducible.

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  • $\begingroup$ Thank you. One more question. Can the proof be generalized to arbitrary group functors on the category of $k$-algebras? $\endgroup$ – Slup Jan 9 at 14:27
  • $\begingroup$ LSpice: thanks, I corrected it. $\endgroup$ – Angelo Jan 10 at 4:45
  • $\begingroup$ Slup: the problem with actions of arbitrary group functors is that the fixed points functor might be not representable. $\endgroup$ – Angelo Jan 11 at 14:36

The classical viewpoint is captured well in the 1962 book Representation Theory of Finite Groups and Associative Algebras by Curtis and Reiner (Wiley), Corollary 29.15. This of course doesn't directly answer the question about non-reduced representations of affine group schemes. But I suspect this generalizes. See for example II, $\S2$ of the 1970 book by Demazure and Gabriel, Groupes Algebriques (Masson, Paris). This book is written in the language of affine group schemes.

(By the way, my comment on the question here was being edited when I was interrupted by a phone call and didn't finish the editing.)

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  • $\begingroup$ @LSpice: Yes, I've now deleted the comment. $\endgroup$ – Jim Humphreys Jan 12 at 2:01

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