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.

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    $\begingroup$ Keep in mind that "linearly reductive" = "reductive" in characteristic 0, so algebraic tori are examples. What is true in that case? Also, in your Remark 1 (relevant only to reductive groups with a nontrivial derived group), "dimensions of the highest weight spaces" needs to omit "highest". $\endgroup$ Feb 20, 2011 at 22:25
  • $\begingroup$ while we're at it, what about projective representations? $\endgroup$ Feb 21, 2011 at 20:44

2 Answers 2


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.

  • $\begingroup$ Awesome! This is exactly the sort of answer I was hoping for. The explanation of why my attempt to re-express the problem over $Spec(A)$ failed is a bonus. $\endgroup$ Feb 21, 2011 at 18:40

Having a discrete set of parameters is not enough, as you explain, but having a discrete set of parameters which are continuous rather than semicontinuous in families certainly is. In the case of a reductive group [in characteristic zero] the entire category of representations sheafifies over the spectrum of the center of the enveloping algebra (this is true for any category over the spectrum of its center). In this case we know exactly what this spectrum is (dual Cartan mod Weyl group), and moreover we know the allowable parameters in there form a discrete subset (integral points), and moreover the fiber of the category of group representations over any such parameter is just Vect.

[Edit: this means that the moduli stack of irreducible objects is a disjoint union of $BG_m$'s, coming from the Schur lemma automorphisms.. of course one can twist a family of irreducibles by a line bundle on the base, so the assertion can literally only ever be true up to line bundles. Also I was ignoring connected components of the group, which contribute some additional finite parameters to the spec of the center of the enveloping algebra.]

Edit: to put it slightly more concisely, there are idempotents in the center of the enveloping algebra, when specialized to any G-representation [edit: to be precise, when completed at the augmentation associated to any irrep], which give projections onto integral eigenspaces of the Casimirs. So given a family of irreducibles over a ring you can pick out connected components on which the group acts by a given highest weight representation [edit: up to Schur twist].

[Edit: of course there's a much simpler way to say all this: the values of the Casimirs give locally constant functions on the base of any family of irreducibles, which allow you to pick up the isotypic components on various connected components]

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    $\begingroup$ The center of the enveloping algebra of a reductive group is a mixed polynomial Laurent polynomial algebra so does not contain any non-trivial idempotents (or consider PBW for the whole algebra which thus has no non-trivial idempotents). However, a suitable completion will still act on finite dimensional representations and contain enough idempotents. $\endgroup$ Feb 21, 2011 at 5:18

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