6
$\begingroup$

Let $k$ be an algebraically closed field of characteristic zero and let $G$ be a non-connected linear algebraic group with reductive connected component $G^0$ over $k$. What is known about the semi-simple elements in $G$ ? Are they dense in every connected component of $G$ ? Do they contain an open dense subset of every connected component of $G$ ?

If $G$ is instead connected, then the answer is yes, as the regular semi-simple elements form an open dense subset. Also I am aware of a result of Guralnick and Malle [Lemma 6.9 in "Simple groups admit Beauville structures" J. Lond. Math. Soc. (2) 85 (2012), no. 3, 694-721] which gives the answer YES in a special case. Note that in characteristic zero all unipotent elements lie in $G^0$.

$\endgroup$
  • 1
    $\begingroup$ Probably the answer to all your questions is yes in characteristic 0, but I will have to check some of the details including older literature when I have time. (Of course, prime characteristic causes some big problems here, though the disconnected case has been examined pretty well, starting with Spaltenstein's 1982 monograph, Springer Lecture Notes 946.) $\endgroup$ – Jim Humphreys Sep 11 '17 at 13:54
10
$\begingroup$

The $G^0$-action on a coset is the same as a so-called twisted action which is pretty well understood. See, e.g., Mohrdieck, S.: Conjugacy classes of non-connected semisimple algebraic groups, Transformation Groups, 8, (2003) 377-395 (MSN).

More precisely, let $C=G^0a$ be a connected component of $G$. Then conjugation by $a$ induces an automorphism $\tau$ on $G^0$. Identifying $C$ with $G^0$ via $g\mapsto ga$ converts conjugation on $C$ to twisted conjugation on $G^0$ $$ u(ga)u^{-1}=(ug\tau(u)^{-1})a $$ It is possible to choose $a$ in such a way that $\tau$ preserves a Borel $B$, a maximal torus $T$ and a pinning of $G^0$, i.e., $\tau$ is induced by an automorphism of the Dynkin diagram of $G^0$. Let $T_0:=(T^\tau)^0$ be the connected component of the $\tau$-fixed points in $T$. Then it is not difficult to see that the map $G^0\times T_0\to G^0:(g,t)\mapsto gt\tau(g)^{-1}$ is dominant. Thus the conjugacy classes of $T_0a$ contain an open subset of $C$ and they are all semisimple.

Edit: Urs Hartl pointed out to me that the proof of Prop. 3.8 in loc.cit may contain a gap (it is unclear that $t$ exists such that $t^{\mathrm ord\,\tau}$ is regular semisimple). Therefore, I am adding a direct argument for the claim that $G^0\times T_0\to G^0$ is dominant. This is done in two stages: Let $C:=(G^\tau)^0\subseteq G^0$. First one shows that $$ G^0\times C\to G^0:(g,c)\mapsto gc\tau(g)^{-1} $$ is dominant. For this it suffices to show that the map on tangent spaces in $(1,1)$ is surjective. Because of $\mathrm{Lie}\,C=\ker(1-\tau)$ that follows from $$ (1-\tau)\mathfrak g\oplus \ker(1-\tau)=\mathfrak g $$ (observe that $\tau$ is of finite order). So all elements in an open subset of $G^0$ are twisted conjugate to an element of $C$. For the second step observe that twisted conjugation on $C$ is ordinary conjugation and that $T_0$ is a maximal torus of $C$. So all elements in an open subset of $C$ are (twisted) conjugate to an element of $T_0$. q.e.d

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.