This must be an easy question but I don't have a good argument for it and have not found a counterexample: Let $G$ be a connected semisimple algebraic group over $\mathbb{Q}$ such that the center of $G$, $Z(G)$ is trivial, i.e. $Z(G)=\{1\}$ is it true that the center of $G_{\mathbb{R}}$, denoted $Z(G_{\mathbb{R}})$ is also trivial?
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4$\begingroup$ 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, $\endgroup$– EmertonCommented Dec 22, 2012 at 16:48
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3$\begingroup$ If $G$ is not assumed connected, some argument is maybe needed to justify that $G_\mathbf{R}$ is Zariski dense. $\endgroup$– YCorCommented Dec 22, 2012 at 17:00
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2$\begingroup$ 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 $Z(G)$. $\endgroup$– user29720Commented Dec 22, 2012 at 17:19
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3$\begingroup$ 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 $g \in G(k_s)$! $\endgroup$– user29720Commented Dec 22, 2012 at 17:23
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3$\begingroup$ Dear Jack, My answer addressed the centre of $Z(G(\mathbb R))$, so addressed a different question. Cheers, $\endgroup$– EmertonCommented Dec 22, 2012 at 23:44
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2 Answers
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The short answer is that the center Z(G) of a semisimple algebraic group is a well-defined (finite) algebraic subgroup which commutes with extension of the base field, so if it's trivial over $\mathbb{Q}$, then its trivial over every field. But presumably, that is not what you meant to ask. Perhaps you mean: does $Z(\mathbb{Q})$ trivial imply $Z(\mathbb{R})$ trivial? The answer is no. You have a finite group $Z(\bar{\mathbb{Q}})$ with an action of the absolute Galois group of $\mathbb{Q}$ and you are asking: if only 1 is fixed by the full Galois group is only 1 fixed by complex conjugation? Obviously, not necessarily. It's easy to makes lots of different Galois modules as centers of semisimple groups over $\mathbb{Q}$. Or perhaps you mean: if $Z(G)$ is trivial (as an algebraic subgroup), is the center of $G(\mathbb{R})$ trivial? The answer is yes, because $G(\mathbb{R})$ is dense in $G$ for the Zariski topology.
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3$\begingroup$ Dear anon: Jack's comments (please look at them) clarify what he meant; it deviates from what you address, and you should probably give some connected semisimple $G$ with $Z(\mathbf{R})\ne 1$ but $Z(\mathbf{Q})=1$ (e.g., the Weil restriction $G={\rm{R}}_{K/\mathbf{Q}}({\rm{SL}}_{2n+1})$ for $K$ not totally real with $\mu_{2n+1}(K)=1$). Much deeper is that $G(k)$ has trivial center when $G$ is an adjoint connected semisimple group over an arbitrary field $k$ since for finite $k$ (OK even if $(k^{\times})^2 = 1$!) the proof uses Borel-Tits structure theory (Google "Kneser-Tits Conjecture"). $\endgroup$ Commented Dec 22, 2012 at 19:45
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1$\begingroup$ Actually, having read Jack's comments, I still haven't a clue what he asking. $\endgroup$– anonCommented Dec 22, 2012 at 22:17
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$\begingroup$ @anon: he says that by $G_\mathbf{R}$ he means $G$ viewed as an $\mathbf{R}$-group. So it seems that the first 3 lines of your answer are what he actually did ask. $\endgroup$– YCorCommented Dec 22, 2012 at 23:14
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$\begingroup$ Dear anon, Thank you very much for your answer. As Yves said my question was exactly the compatibility of $Z(G)$ with base field extension and so again as he said the first 3 lines of your answer are exactly what I wanted. Anyhow, the rest of your answer is interesting in it's own right, so I again thank all of you for your detailed answers. Jack $\endgroup$– JackCommented Dec 22, 2012 at 23:36
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Since the question and its answer have both gone through a number of edits, and are accompanied by numerous comments, it may be helpful (or not) to sort the question out further.
1) The interface between semisimple Lie groups (real or complex) and semisimple algebraic groups over $\mathbb{C}$ is somewhat tricky and has been approached historically from different directions. (Borel for instance got heavily involved in correlating these theories.) There is no single reference which takes all viewpoints into account, but it may be helpful to look at Chapter 5 of Onishchik-Vinberg Lie Groups and Algebraic Groups. See for example Section 1 and the first few problems there, with hints and cross-references given later. How you think about the question here is partly determined by your upbringing.
2) It's worth noting that "defined over $\mathbb{Q}$" in this context is not really needed, since a connected semisimple algebraic group in characteristic 0 always arises (by Chevalley's work) from a $\mathbb{Z}$-form and is automatically defined over $\mathbb{Q}$. Aside from this, the essential problem is to relate the complex group $G$ to its subgroup of real points. Here is where the interaction with Lie groups comes in, adding a layer of complication in terminology and notation.
3) To amplify Emerton's comment in matrix terms, think of each center as the kernel of an adjoint representation. Say $G$ is a complex algebraic matrix group, so its Lie algebra (in the sense of either algebraic groups or Lie groups) contains the Lie algebra of the Lie group $G(\mathbb{R})$. Under the adjoint (conjugation) action any real matrix in the center of $G(\mathbb{R})$ then centralizes the complex Lie algebra of $G$ and thus lies in the (trivial) center of $G$.
4) This kind of relationship between real and complex groups goes back to E. Cartan and Chevalley, but the interaction with algebraic group language owes most to Borel. In any case, the theory involved was in place over half a century ago. It doesn't require the explicit classificaiton of semsiimple groups or Lie algebras, nor does it require any scheme language. (In the characteristic 0 Borel-Chevalley theory, it's usually safe to identify an algebraic group with its points over $\mathbb{C}$.)
answered Dec 23, 2012 at 16:50
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2$\begingroup$ I'm confused by (3). In a Lie group context, I would not say that the (non-connected) group $PGL_2(\mathbf{R})$ is a real form of $PGL_2(\mathbf{C})$ (it does not fulfill the categorical definition of complexification using the forgetful functor from complex Lie groups to real Lie group). What did you have in mind? $\endgroup$– YCorCommented Dec 23, 2012 at 18:17
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1$\begingroup$ Yes, I oversimplified the connectedness issue here. The ideas sketched in (3) seem OK when
$G(\mathbb{R})$is connected in the euclidean topology, but I'm not quite sure what happens otherwise. It seems there is no effect on the center under the conditions placed on$G$, but I'll have to do more thinking about this interplay of algebraic groups and Lie groups (including the Iwasawa decomposition, where the center lies in the maximal compact subgroup). $\endgroup$ Commented Dec 23, 2012 at 20:16 -
$\begingroup$ As I have understood from the comments and discussions above, my question about center has nothing to do with connectedness. The center $Z(G)$, commutes with the extension of the base field for arbitrary $G$ (or at least $G$ semisimple) without the assumptions on connectedness (as indicated by Yves, anon and others). $\endgroup$– JackCommented Dec 23, 2012 at 20:52
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1$\begingroup$ @Yves: I substituted simpler language in (3), not wanting to add more confusion to the discussion. $\endgroup$ Commented Dec 23, 2012 at 21:42