Character group of Frobenius kernels - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T06:27:54Z http://mathoverflow.net/feeds/question/57761 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57761/character-group-of-frobenius-kernels Character group of Frobenius kernels Christopher Drupieski 2011-03-08T02:18:14Z 2012-09-10T11:19:15Z <p>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 $k$.)</p> <p>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 $G(A) = SL_n(A)$.</p> <p>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 $(a_{ij}^p)$.</p> <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.</p> <blockquote> <p>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?</p> </blockquote> http://mathoverflow.net/questions/57761/character-group-of-frobenius-kernels/57815#57815 Answer by Jim Humphreys for Character group of Frobenius kernels Jim Humphreys 2011-03-08T11:57:08Z 2011-03-08T12:19:01Z <p>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 <em>Representations of Algebraic Groups</em> (Chapter II.3). Given a connected, simply connected semisimple group <code>$G$</code>, the irreducible representations of its Frobenius kernel <code>$G_r$</code> are parametrized naturally by <code>$p^r$</code> of the highest weights for <code>$G$</code> relative to a fixed maximal torus. Only the zero weight corresponds to a 1-dimensional representation (i.e., character of <code>$G_r$</code>) because <code>$G$</code> is semisimple. </p> http://mathoverflow.net/questions/57761/character-group-of-frobenius-kernels/106805#106805 Answer by tomas perna for Character group of Frobenius kernels tomas perna 2012-09-10T11:19:15Z 2012-09-10T11:19:15Z <p>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)</p>