## Character group of Frobenius kernels

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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 2011 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 2011 at 20:17

## 2 Answers

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.

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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 2011 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 2011 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 2011 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 2011 at 15:32

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@George: In his 1960 papers Curtis studied restricted Lie algebra representations and started to work on representations of finite Chevalley groups over the prime field. But Steinberg's 1963 paper covered all groups of Lie type more systematically, including the twisted tensor product theorem. Curtis did work in the BN-pair setting in the early to mid-1960s (inspiring Carter-Lusztig), but algebraic group methods went much farther. The rank one case was studied concretely and much earlier by Brauer, with steps toward higher ranks by his students; this inspired both Curtis and Steinberg. – Jim Humphreys Mar 8 2011 at 16:45
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Frobenius Kernel cannot be ever trivial, only for g = -1. But at such g, the complement H copletely loses its meaning as a complement and thus the kernel too. Tomas Perna (Pecik)

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