For simplicity, I will be talking only about connected groups over an algebraically closed field of characteristic zero.

The basic theorem of affine algebraic groups is that they all admit faithful, finite-dimensional representations. The *fundamental* theorem for semisimple groups is that these representations are all completely reducible, but unfortunately there is no reason that any irreducible summand of a faithful representation should be faithful, only that the kernels of all these representations intersect trivially.

My question is whether such a representation does, in fact, exist.

(Answered: iff the center is cyclic.)

This does not hold of general reductive groups for the following reason: if $T$ is any torus of rank $r > 1$, then its irreducible representations are all characters $\chi \colon T \cong \mathbb{G}_m^r \to \mathbb{G}_m$, which therefore have nontrivial kernels. More generally, any reductive group $G$ has connected center a torus of some rank $r$, so by Schur's lemma this center acts by a character $\chi$ in any irreducible representation of $G$ and if $r > 1$, therefore does not act faithfully.

The exceptional case $r = 1$ *does* have an example, namely $\operatorname{GL}_n$, whose standard representation is faithful and irreducible and whose center has rank 1. A more general version of this question might be, then:

Does any reductive group whose center has rank at most 1 have a faithful irreducible representation?

(Answered: when not semisimple, iff the center is connected.)

Another special case is that if $G$ is simple and of adjoint type, then its adjoint representation is irreducible and faithful by definition (or, depending on your definition, because the center is trivial). A constructive version of this question for any $G$ (semisimple or reductive of central rank 1) is then:

Can we give a construction of a faithful, irreducible representation of $G$ from its adjoint representation?

(Not yet answered!)

This is deliberately a little vague since I don't want to restrict the possible form of such a construction, only that it not start out with "Throw away the adjoint representation and take another one such that..."

Finally, suppose the answer is "no".

What is the obstruction to such a representation existing?

(Answered: for $Z$ the center, it is the existence of a generator for $X^*(Z)$.)

`\em`

command that I could set to whichever kind of emphasis I like. I think the best way to offset a question is to put it on its own line with a greater than sign (and a space)`>`

in front. This should make it indented and on a gray box. $\endgroup$ – Theo Johnson-Freyd Jun 29 '10 at 8:12