Character group of Frobenius kernels

2011-03-08T02:18:14Z

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?

Answer by Jim Humphreys for Character group of Frobenius kernels

2011-03-08T11:57:08Z

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.

Answer by tomas perna for Character group of Frobenius kernels

2012-09-10T11:19:15Z

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)

Character group of Frobenius kernels - MathOverflow

Character group of Frobenius kernels

Answer by Jim Humphreys for Character group of Frobenius kernels

Answer by tomas perna for Character group of Frobenius kernels