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Let $G$ be a reductive, linear algebraic group (variety) over an algebraically closed field $\Bbbk$ of characteristic zero. If $G$ is connected, I know from Humphrey's book that for any Borel subgroup $B\subseteq G,$ there is some opposite Borel subgroup $B^-$ such that $T=B\cap B^-$ is a Torus, and denoting by $U$ and $U^-$ the unipotent radicals of $B$ and $B^-$ respectively, there is an open immersion \begin{align*} \beta: U\times T\times U^- &\longrightarrow G \\ (u,t,v) &\longmapsto utv \end{align*} I am now in a situation where my group is not connected, and I am wondering if the above still holds (to some degree). Searching MO, I found this post which suggests that the above holds for "split reductive" groups. I am not familliar with the term "split reductive", and searching for a definition has yielded very little so far, but wikipedia seems to suggest that this is some issue of separability which I would not have to worry about over an algebraically closed field of characteristic zero.

So, my question is: If $G$ is not connected, can something similar or even the same be said? A reference would be the icing on the cake, I would love to read more about this. Thanks a lot in advance.

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    $\begingroup$ I think that "split reductive" means that $G$ contains a "split maximal torus", in other words a torus $T$ such that the base change of $T$ to an algebraic closure $k^{alg}$ is a maximal torus of the base change of $G$ to $k^{alg}$. Thus if your field is alg. closed, then $G$ is split reductive. $\endgroup$ – Matthieu Romagny Oct 31 '13 at 19:25
  • $\begingroup$ @MatthieuRomagny: I do like the sound of that. Seriously though, why can't I find a single book with the definition in it? $\endgroup$ – Jesko Hüttenhain Oct 31 '13 at 20:10
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    $\begingroup$ @Matthieu: A split (maximal or not) torus is one which is isomorphic to a power of the multiplicative group (over k). $\endgroup$ – Peter McNamara Oct 31 '13 at 21:34
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    $\begingroup$ @MatthieuRomagny: There is nothing in "Basic Theory of Affine Group Schemes", but he states the Bruhat decomposition for reductive groups in his script "Reductive Groups" (Theorem 3.41). However, he does not give a proof, he references the book by Springer (Linear Algebraig Groups) instead - I happen to have looked at that one, and it also contains the connectedness assumption, so basically Milne's reference states less than he claims =/. This brings me back to square one, I still don't have a reference. $\endgroup$ – Jesko Hüttenhain Nov 1 '13 at 20:26
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    $\begingroup$ @Jesko: you're absolutely right. What I wanted to say is that Milne's notes are a good reference for the basic theory of group schemes over an arbitrary base field. I am afraid that I don't know any reference for the Bruhat decomposition in this setting. But you may be able to do it by yourself? $\endgroup$ – Matthieu Romagny Nov 2 '13 at 9:39
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Maybe I can clarify some of the issues here. First, the distinction between "reductive" and "semisimple" is minor, since the difference between these lies in the maximal torus and doesn't really affect the Bruhat decomposition or the big cell $U^- T U$. But most books and survey articles do assume that $G$ is connected, mainly to avoid trivialities which complicate the notation. So it's hard to find a formulation of the Bruhat decomposition or the big open cell without the connecteness assumption.

Fortunately the issue is minor, since a disconnected reductive (or semisimple) group is a finite union of translates of the identity component $G^\circ$ which are geometrically the same. The most common example of a disconnected semisimple group is the full orthogonal group, in which the special orthogonal group has index 2. Here you can choose a diagonalizable matrix $g$ of determinant $-1$ lying outside the identity component as a coset representative for $G/G^\circ$ and get two disjoint copies of the big cell whose union is still open in the Zariski (or euclidean) topology. A similar thing happens in general for disconnected groups.

Concerning the notion of a "split" maximal torus, this is covered in all textbooks when fields of definition are discussed. As in other algebraic theories, you may need to extend an arbitrary field in order to diagonalize a semisimple matrix or such. (Typically this is always possible for an algebraic structure by taking a finite separable extension, which is where "separable" probably comes in here.) Over an algebraically closed field there is no problem. For a torus, as mentioned in comments, it just means being isomorphic over the given field to a direct product of copies of the multiplicative group. The contrast is with non-split situations that occur for compact Lie groups, etc.

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  • $\begingroup$ Surely you don't mean to write that $G$ equals $U^- T U$? $\endgroup$ – LSpice Aug 10 '15 at 22:33
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    $\begingroup$ @ L Spice: No, that was just too hurriedly written. Will edit. Thanks. $\endgroup$ – Jim Humphreys Aug 11 '15 at 13:12

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