Timeline for Center of the algebraic group $G_{\mathbb{R}}$ for a centerless $G$

Current License: CC BY-SA 4.0

20 events

when toggle format	what		by	license	comment
S Mar 6, 2023 at 19:20	history	suggested	Kan't	CC BY-SA 4.0	Typesetting title
Mar 6, 2023 at 18:14	review	Suggested edits
S Mar 6, 2023 at 19:20
Dec 23, 2012 at 16:50	answer	added	Jim Humphreys		timeline score: 4
Dec 22, 2012 at 23:44	comment	added	Emerton		Dear Jack, My answer addressed the centre of $Z(G(\mathbb R))$, so addressed a different question. Cheers,
Dec 22, 2012 at 23:31	vote	accept	Jack
Dec 22, 2012 at 18:06	answer	added	anon		timeline score: 4
Dec 22, 2012 at 17:23	comment	added	user29720		Dear Jack: I think what you are seeking is to know that there is a good notion of "scheme-theoretic center" (as Yves indicates): if $G$ is a smooth $k$-group then the functor that assigns to any $k$-algebra $A$ the set of $g \in G(A)$ whose conjugation action on $G_A$ is trivial is represented by a closed $k$-subgroup scheme $Z(G)$ of $G$. Once this is known, the formation of $Z(G)$ trivially commutes with ground field extension. Build $Z(G)$ using Galois descent and my previous comment: descend the schematic intersection inside $G_{k_s}$ of schematic centralizers of all $g \in G(k_s)$!
Dec 22, 2012 at 17:23	comment	added	Jack		Dear Kreck, thank you for your beautiful answer.
Dec 22, 2012 at 17:20	comment	added	Jack		Dear Yves, thanks a lot for your illuminating answer. In fact this was exactly what I had in my mind-that the triviality of center should be independent of the base field - but i thought maybe this is very naive and is not a proof, that's why I said in the beginning that I don't have a good argument. Again thank you very much.
Dec 22, 2012 at 17:19	comment	added	user29720		Dear Jack: Your question is about the definition of $Z(G)$ and its compatibility with ground field extension. If $X$ is geometrically reduced lft over a field $k$ then $X(k_s)$ is relatively schematically dense in $X_{k_s}$: for a $k_s$-scheme $S$, any $S$-map from $X_S$ to a separated $S$-scheme is uniquely determined by its restriction to $X(k_s)$. This is an instructive exercise. It implies that if $G$ is a smooth $k$-group and $A$ is a $k$-algebra then any $g \in G(A)$ centralizing $G(k_s)$ inside $G(A_{k_s})$ centralizes the scheme $G_A$! This underlies good behavior of $Z(G)$.
Dec 22, 2012 at 17:11	comment	added	YCor		Actually Emerton's argument used implicitly a result of Rosenlicht of density. But what you ask is easier. The center is defined in a way that is independent of the base field. When you ask if a certain variety is trivial, you don't have to refer to a base field. In particular the answer for your question is true for an arbitrary algebraic group over any field of characteristic zero (in characteristic $p$ it holds if you refer to the schematic center, otherwise there may be a reduceness issue).
Dec 22, 2012 at 17:04	comment	added	Jack		Dear emmerton, Thanks for your comment, but by $G_{\mathbb{R}}$, I mean the base change of the group $G$ (which is a group over $Q$) to $\mathbb{R}$. Is this argument also true in this situation?
Dec 22, 2012 at 17:03	history	edited	Jack	CC BY-SA 3.0	added 10 characters in body
Dec 22, 2012 at 17:02	comment	added	Jack		dear Yves, $G$ is connected. I will add it in the question.
Dec 22, 2012 at 17:00	comment	added	YCor		If $G$ is not assumed connected, some argument is maybe needed to justify that $G_\mathbf{R}$ is Zariski dense.
Dec 22, 2012 at 16:48	comment	added	Emerton		Dear Jack, The $\mathbb R$ points will be Zariski dense in $G$, and so any real-valued element of the centre will be in the centre of $G$. Thus if $G$ has trivial centre, the real points of the group will also have trivial centre. Regards,
Dec 22, 2012 at 16:47	history	edited	Jack		edited tags
Dec 22, 2012 at 16:29	history	edited	Jack	CC BY-SA 3.0	added 26 characters in body
Dec 22, 2012 at 16:22	history	edited	Jack	CC BY-SA 3.0	added 10 characters in body; edited title
Dec 22, 2012 at 16:17	history	asked	Jack	CC BY-SA 3.0

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