2
$\begingroup$

If $G$ is a connected reductive group over a perfect field $k$ (The definition given in Milne's "Algebraic Groups": $G$ is a connected group variety containing no non-trivial connected unipotent normal subgroup variety, even over the algebraic closure of $k$), why is the radical of $G$ (that is, the maximal connected solvable normal subgroup) equal to the connected component of the center? This is stated on page 7 of Shahidi's "Eisenstein series and automorphic L-functions." On page 357 of Milne's Algebraic Groups it says that the center of $G$ is of multiplicative type, and its largest subtorus is equal to the radical of $G$.

Borel's book (Linear Algebraic Groups) has a proof if $k$ is algebraically closed -- one can characterize the center of $G$ as the intersection of the Borels. But I don't know if that generalizes in some way over an arbitary field.

$\endgroup$
  • 1
    $\begingroup$ What do you mean by "radical"? I know the unipotent radical (which will be trivial for a reductive group) and the solvable radical (which can be nontrivial, but is typically not equal to the connected component of the center). $\endgroup$ – Jason Starr Jun 5 '17 at 8:51
  • $\begingroup$ I just added some clarifying comments. $\endgroup$ – Grad student Jun 5 '17 at 9:33
  • 2
    $\begingroup$ In fact $k$ needs not be perfect. This is the proof from Milne's AGS course. The radical $RG_{\bar k}$ of $G_{\bar k}$ is a connected smooth solvable group with trivial unipotent radical (since $G$ is reductive), thus it is a torus. Since $(RG)_{\bar k}$ is a smooth connected subgroup of $RG_{\bar k}$, it is a subtorus. Since $G$ is connected, its action by conjugation on $RG$ is trivial (by rigidity). Hence $RG \subseteq (ZG)_{\mathrm{red}}^\circ$. The other inclusion is clear (since $(ZG)_{\mathrm{red}}^\circ$ is a smooth connected commutative group) . $\endgroup$ – Arkandias Jun 5 '17 at 9:35
  • $\begingroup$ @Arkandias Ah, it's Theorem 5.1(b) on page 261 in those notes. Thanks! $\endgroup$ – Grad student Jun 5 '17 at 9:56
  • $\begingroup$ A comment on your last sentence: Relative to an arbitrary field of definition $k$ (as treated by Borel and Tits), the immediate problem is that $G$ need not possess any Borel subgroups defined over $k$. Indeed, in the extreme case where $G$ is $k$-anisotropic, little can be said about $G$ relative to $k$ from the viewpoint of the Borel-Tits structure theory. (A smaller comment is that "the connected component of the center" doesn't make sense; I'd prefer to say "the identity component of the center", meaning that component which contains the identity element.) $\endgroup$ – Jim Humphreys Jun 5 '17 at 17:44

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.