MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ and $H$ be affine algebraic groups over a scheme $S$ of characteristic 0 and let $\textbf{Hom}_{S,gp}(G,H)$ be the functor $T \mapsto \text{Hom}\_{T,gp}(G,H)$

Theorem (SGA 3, expose XXIV, 7.3.1(a)): Suppose G is reductive. Then $\textbf{Hom}_{S,gp}(G,H)$ is representable by a scheme.

Can this fail if $G$ is not reductive? I worked out a few example with $G = \mathbb{G}_a$, but they were representable.

share|cite|improve this question
up vote 15 down vote accepted

Hom(Ga, Gm) is not representable.

Let R be a ring of characteristic zero. I claim that Hom(Ga, Gm)(Spec R) is {Nilpotent elements of R}. Intuitively, all homs are of the form x -> e^{nx} with n nilpotent.

More precisely, the schemes underlying Ga and Gm are Spec R[x] and Spec R[y, y^{-1}] respectively. Any hom of schemes is of the form y -> \sum f_i x^i for some f_i in R. The condition that this be a hom of groups says that \sum f_k (x_1+x_2)^k = (\sum f_i x_1^i) (\sum f_j x_2^j). Expanding this, f_{i+j}/(i+j)! = f_i/i! f_j/j!. So every hom is of the form f_i = n^i/i!, and n must by nilpotent so that the sum will be finite.

Now, let's see that this isn't representable. For any positive integer k, let R_k = C[t]/t^k. The map x -> e^{tx} is in Hom(Ga, Gm)(Spec R_k) for every k. However, if R is the inverse limit of the R_k, there is no corresponding map in Hom(Ga, Gm)(Spec R). So the functor is not representable.

share|cite|improve this answer

There is a reasonable salvage, at least if the base scheme is a field: Hom(G,H) is a direct limit of representable subfunctors. See Lemma A.8.13 in the book "pseudo-reductive groups" (where it is used to prove that the scheme-theoretic fixed locus for a linearly reductive group acting on a connected reductive group is always reductive (possibly disconnected) provided the base is a field.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.