# Character group of Frobenius kernels

Let $G$ be a semisimple algebraic group over an algebraically closed field $k$ of characteristic $p$ (e.g., $G=SL_n(k)$). Then $G$ is equal to its derived subgroup $[G,G]$. Consequently, the character group $X(G)$ of all algebraic group homomorphisms $G \rightarrow \mathbb{G}_m$ is trivial, because any character $\chi \in X(G)$ will vanish on the derived subgroup $[G,G]$. (Here $\mathbb{G}_m$ is the multiplicative group of units in $k$.)

Now I want to think of $G$ as an algebraic group scheme. Thus, $G$ is a representable functor from the category of commutative $k$-algebras to the category of groups. Given a commutative $k$-algebra $A$, $G(A) = \textrm{Hom}_{k-alg}(k[G],A)$, where $k[G]$ is the (usual) coordinate ring of $G$. For the example $G=SL_n$, we can be more explicit and say $G(A) = SL_n(A)$.

Since the characteristic of $k$ is positive, the group $G$ comes equipped with its Frobenius morphism $F: G \rightarrow G$. This is induced by a certain map of $k$-algebras $k[G] \rightarrow k[G]$, which, roughly speaking, is just the $p$-th power map $f \mapsto f^p$. In our example $G(A) = SL_n(A)$, the image of a matrix $(a_{ij}) \in SL_n(A)$ under $F$ is the matrix $(a_{ij}^p)$.

We can consider the scheme-theoretic kernel $G_1$ of $F$, and, more generally, the kernel $G_r$ of the $r$-th iterate $F^r$. These are the Frobenius kernels of $G$. They are normal subgroup schemes of $G$. They are not interesting algebraic groups in the classical sense (e.g., if $A=k$, then $(a_{ij}^p)=1$ only if $(a_{ij})=1$ and the kernel is trivial), but they are interesting as algebraic group schemes.

Let $G_r$ be the $r$-th Frobenius kernel of $G$. What is the structure of the character group $X(G_r)$ of algebraic group homomorphisms $G_r \rightarrow \mathbb{G}_m$? If $G$ is semisimple and simply-connected, is $X(G_r)$ trivial?

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I suspect that the Frobenius kernels are all equal to their derived subgroups, and that the natural first attempt to prove this should be to mimic the proof that G is equal to its derived subgroup. –  Peter McNamara Mar 8 '11 at 5:29
An addendum to my previous comment, via the in-exile Bcnrd, the notion of derived group is somewhat problematic for non-smooth group schemes (but quesiton has affirmative answer for G 1-connected by reduction to SL_2 subgroups). –  Peter McNamara Mar 8 '11 at 20:17

It's probably most natural to consider this as a question about the (rational) representations of Frobenius kernels, in the spirit of Jantzen's book Representations of Algebraic Groups (Chapter II.3). Given a connected, simply connected semisimple group $G$, the irreducible representations of its Frobenius kernel $G_r$ are parametrized naturally by $p^r$ of the highest weights for $G$ relative to a fixed maximal torus. Only the zero weight corresponds to a 1-dimensional representation (i.e., character of $G_r$) because $G$ is semisimple.
Nice answer. I feel silly for not seeing that approach myself. By the way, I assume you mean that the irreducible representations of $G_r$ are parametrized naturally by the $p^r$-restricted highest weights for $G$, not by $p^r$ of the highest weights for $G$. –  Christopher Drupieski Mar 8 '11 at 14:38
Incidentally, this argument would also work for the finite group of Lie type $G(\mathbb{F}_{p^r})$ (the fixed points in $G$ under $F^r$). –  Christopher Drupieski Mar 8 '11 at 14:48
Yes, I'm using the relevant restricted weights here, though technically the "weights" for the Frobenius kernel are taken mod $p^r$ in the weight lattice (character group of maximal torus). It's more straightforward here for the finite groups, since you only have to restrict the $p^r$ irreducible representations of $G$, whereas passing to Frobenius kernels imitates in an enriched way passage to the Lie algebra in characteristic 0 theory. –  Jim Humphreys Mar 8 '11 at 15:13
Historical question: it was Curtis who first described the simples for the Frobenius kernel when $r=1$ (or rather, he described the simple restricted Lie(G) representations). I guess the "analogous" description of simples for $G(\mathbf{F}_p)$ was done earlier? Maybe by Steinberg? I've somehow never known the sequence of events... –  George McNinch Mar 8 '11 at 15:32