2
$\begingroup$

Finite dimensional, irreducible representations of simply connected, complex semisimple algebraic groups can be classified by their highest weight. I was wondering if there is an analogous classification of representations of reductive groups.

This can be made precise as follows: let $\mathfrak g = \operatorname{Lie}(G)$ with $G$ a complex simply connected algebraic group. Let $\mathfrak t$ be a Cartan subalgebra of $\mathfrak g$ with set of simple roots $\Delta$ and simple coroots $\Delta^{\vee} \subset \mathfrak t^{\ast}$. We define a partial order on $\mathfrak t^{\ast}$ by saying that $\lambda \geq \mu$ if $\lambda - \mu$ is a nonnegative linear combination of simple coroots.

We say that $\mu \in \mathfrak t^{\ast}$ is dominant if $\langle \mu,\alpha \rangle \geq 0$ for all $\alpha \in \Delta$, and we say that $\mu$ is integral if $\langle \mu, \alpha \rangle$ is an integer for all $\alpha \in \Delta$.

Then there is a bijection between isomorphism classes of irreducible representations of $G$, ones of $\mathfrak g$, and dominant integral weights. The explicit correspondence is we begin with a representation $(\pi, V)$ of $G$, and let $(d\pi, V)$ be the associated tangent space map on $\mathfrak g$. Now $V$ has an associated weight space decomposition $V = \bigoplus\limits_{\mu \in \mathfrak t^{\ast}} V_{\mu}$, where $$V_{\mu} = \{ v \in V : d\pi(X)v = \mu(X)v \textrm{ for all } X \in \mathfrak t\}$$ There is a unique maximal dominant, integral element $\mu \in \mathfrak t^{\ast}$ for which $V_{\mu}$ is nonzero, and this $\mu$ determines $\pi$ and $d\pi$ up to isomorphism.

If we assume that $G$ is connected and semisimple, but not necessarily simply connected, then the classification is the same, but $\pi \mapsto d \pi$ is no longer a bijection.

Question: Is there a classification along similar lines to irreducible, finite dimensional representations of connected, complex reductive groups?

When $G$ is reductive, but not semisimple, the notion of highest weight still makes sense, since we can write the Cartan subalgebra as the direct sum of that of the derived group of $G$ and the Lie algebra of the center.

$\endgroup$
  • $\begingroup$ Sure. A connected reductive group is generated by a central torus and a semisimple group. On any irreducible representation space, the torus must act by a character, so the restriction to the semisimple group stays irreducible. $\endgroup$ – Zero Oct 10 at 5:36
  • $\begingroup$ The comolex seting adds nothing here. For sources, see the 1967-68 Steinberg lectures at Yale Lectures on Chevalley groups (in the somewhat edited and tpeset version now puboished by the AMS), or seeJantezn's book Representations of Algebraic Groups (now also published by the AMS), for a more comprehensive treatment. $\endgroup$ – Jim Humphreys Oct 15 at 0:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.