TITLE: Are representations of a linearly reductive group discretely parameterized?
QUESTION [12 upvotes]: Suppose $G$ is a linearly reductive group over a field (say $\mathbb C$). Does somebody know of a proof that any flat family of finite-dimensional representations of $G$ must be locally constant?
In other words, I want to prove that if $A$ is a commutative $\mathbb C$-algebra (without idempotents) and $\rho:G\to GL_n(A)$ is an algebraic group homomorphism (roughly, a family of representations parameterized by $Spec(A)$), then after conjugating by some element of $GL_n(A)$, the image of $\rho$ is actually contained in $GL_n(\mathbb C)\subset GL_n(A)$.

Remark 1: A finite-dimensional representation of $G$ is completely determined by the dimensions of its highest weight spaces. For a long time I thought this "discrete" parameterization somehow proved the result, but it doesn't. For example, nilpotent matrices are "discretely" parameterized by Jordan type, but it's obviously possible to have a flat family of nilpotent matrices in which Jordan type jumps, like $\begin{pmatrix}0&t\\ 0& 0\end{pmatrix}$ over the affine line with coordinate $t$.
Remark 2: Another reason I thought this was clear is that deformations of a representation $V$ are controlled by cohomology groups $H^{>0}(G,V\otimes V^*)$, which all vanish when $G$ is linearly reductive. This implies that any formal family—any flat family over an Artin ring—of representations has to be constant. However, this isn't enough to say that any family has to be locally constant. For example, deformations of $G$, as a group, are controlled by $H^{>0}(G,Ad)$, which vanish when $G$ is linearly reductive. So any formal family of linearly reductive groups has to be constant, but it is possible to have a flat family of linearly reductive groups which is not locally constant. Specifically, the affine line with a doubled origin is a flat group over the affine line. The fiber over the origin is $\mathbb Z/2$, but all the other fibers are trivial.

REPLY [11 votes]: I don't think this is literally true. For example, suppose that $G$ is a finite cyclic group of order 2 generated by $s$, suppose that $L$ is an non-trivial 2-torsion invertible $A$-module over a Dedekind ring $A$. Then $L^{\oplus 2}$ is isomorphic to $A^{\oplus 2}$; you can let $G$ act on $L^{\oplus 2}$ with the rule $s(a,b) = (a, -b)$. This gives a family of representations that does not come from $\mathbb C$. You can give a similar example in which $G$ is a torus.
On the other hand, what you want is true Zariski-locally; would that be enough for your needs?
[Edit] Let me add a proof that the statement is Zariski-locally true. The key point is the following: given a locally free sheaf on a scheme $X$ with an action of $G$, the subsheaf of invariants is also locally free. This is easy: if $X = \mathop{\rm Spec} A$ and $E$ corresponds to a projectve $A$-module $M$, then $M^G$ is a direct summand of $M = M^G \oplus M_G$, because $G$ is linearly reductive (and $M$ is a locally finite representation of $G$, this is standard); and $M_G$ is also an $A$-submodule, so $M^G$ is a projective $A$-module. Now, let $E$ be a locally free sheaf with an action of $G$. Let $p$ be a rational point of $X$, and $V$ be the representation of $G$ appearing as the fiber $E(p)$ of $E$ at $p$. Consider the locally free sheaf $Hom_{\mathcal O_X}(V \otimes \mathcal O_X, E)$; the sheaf of invariants, that is, the sheaf of $G$-equivariant homomorphisms $V \otimes \mathcal O_X \to E$, is locally free. At the point $p$ we have a section of its fiber, given by the tautological isomorphism $V \simeq E(p)$; since the sheaf is locally free, this extends to a section in a neighborhood. By further restricting the neighborhood, this sectioin will be an isomorphism, and this concludes the proof.