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 $G^0$.

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    $\begingroup$ 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

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


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