# Are the semi-simple elements in a non-connected reductive algebraic group dense?

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 $G^0$.

• 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.) – Jim Humphreys Sep 11 '17 at 13:54

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