I'm trying to see why the Chevalley groups over $\mathbb C$ have no nontrivial character? I know that a compact connected semisimple Lie group has no nontrivial character but is the compactness necessary?
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3$\begingroup$ A Chevalley group over $\mathbb{C}$, like $\mathrm{SL}_n(\mathbb{C})$, is always perfect, i.e., equals its derived subgroup. So this forces no non-trivial characters because any commutator is in the kernel of a character, no? $\endgroup$– Jay TaylorCommented Jul 3, 2019 at 21:47
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In the original sense, Chevalley groups are generated by copies of the additive group of the field and are in fact simple as abstract groups if the field is not too small. (This was the motive for constructing them in Chevalley's 1955 paper.) But it's nontrivial to make the connection with compact semisimple Lie groups (always connected by assumption), which are determined by root data. The Chevalley seminar of 1956-58 classified the semisimple algebraic ggroups over an arbitrary algebraically closed field in a similar way, allowing one to match these with the (connected) semisimple Lie groujps over $\mathbb{C}$.
In particular, it's not neessary to consider just the compact groups here, if you are willing to cite the Chevalley seminar. Similarly, you could follow Steinberg's more general version of "Chevalley group" in his Yale lectures of 1967-68, starting with an arbitrary rational/algebraic irreducible representation of a similar algebraic group (which for Chevalley was just of adjoint type) Here there can be a finite center.
Note that typeset, corrected versions of the Chevalley seminar and the Steinberg lectures have been published respectively by Springer-Verlag and AMS.
answered Jul 3, 2019 at 11:34
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$\begingroup$ Ok let me see if I got it, by Steinberg's lectures(page 126) there is a maximal compact subgroup K of a complex Chevalley group G, and K is Zariski-dense in G. And so the rational irreducible representations of G remain distinct and irreducible on restriction to K. K is a compact semisimple Lie group and so have no nontrivial rational characters and so does G(because different representations have different characters) am I correct? $\endgroup$– AmiCommented Jul 3, 2019 at 17:01
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$\begingroup$ I'm actually not sure why would K be semisimple... $\endgroup$– AmiCommented Jul 3, 2019 at 23:04
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$\begingroup$ An arbitrary maximal compact subgroup of a Lie group might well be reductive (almost-direct product of a semisimpe group with a compact torus) rather than semisimple. This is a drawback to using maximal compact subgrouips here. $\endgroup$ Commented Jul 4, 2019 at 13:21