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Is it true that the universal cover of SL2(ℝ) has no non-trivial central extensions... as an abstract group?
(that's certainly true as a Lie group)

Motivation:
I have a projective action of SL2(ℝ) on some Hilbert space H and I'd like to know that it induces an honest action of its universal cover. But it's a hassle to show that the action is continuous. So I'm wondering if there is an alternative argument that uses solely group theory.

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Would "K-theory" and "Steinberg group" be appropriate tags, rather than "Lie groups"? –  darij grinberg Apr 30 '11 at 14:58
    
I hate to be nothing more than Watson, but search gives the following part of an article linkinghub.elsevier.com/retrieve/pii/S0096300311000907 "[an example of something] was given in [12] for a reductive Lie group which is a quotient Lie group of a one-dimensional central extension of the universal covering group of $SL_2(R)$." Unless my parsing is wrong, it seems as though there is a 1-dim central extension. –  Junkie Apr 30 '11 at 15:03
    
Tanks. I Fixed the tags. –  André Henriques Apr 30 '11 at 15:04
    
Can't you prove that your action is measurable? That's enough to guarantee continuity... –  Alain Valette Apr 30 '11 at 15:15
    
Yes. I can even show that the action is continuous. But my question is about whether or not there exists an alternative argument, that uses no knowledge at all about the action. –  André Henriques Apr 30 '11 at 15:21
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7 Answers

up vote 15 down vote accepted

The answer should be negative, because the $K_2$ of the reals is humongous. That is, there are nontrivial central extensions. (Please do not ask a question and then explain its negation.) Algebraic $K$ theory detects transcendentals. There is a Chern class map from $K_2(\Bbb R)$ towards $\Omega^2_{\Bbb R}$, where the Kähler differentials are taken over the integers. It maps the Steinberg symbol {t,u} to $dlog\ t\wedge dlog\ u$, with $dlog\ t=dt/t$. It hits much more than a cyclic group. The map from the Schur multiplier of $SL_2(\Bbb R)$ to the stable $K$ group is surjective in this case, by Steinberg. So the universal cover as a Lie group realizes only a very small part of the Schur multiplier.

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I also agree this is rather convincing. –  Junkie May 1 '11 at 20:06
    
This argument looks very convincing. I'll toggle my acceptation of Richard Borcherd's answer, and I'll wait until things are clarified. –  André Henriques May 1 '11 at 20:09
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Just to confirm that my answer was indeed wrong and this one is correct. –  Richard Borcherds May 2 '11 at 0:04
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As van der Kallen mentions, the answer is: definitely no. Here's argument which is less precise than his, but with a checkable basic reference: thm 11.10 from Milnor's book "Introduction to Algebraic K-theory": if a field F is uncountable then $K_2(F)$ is uncountable as well. Moreover $K_2(F)$ is generated by symbols which already define cocycles on $\text{SL}_2$. So the Schur multiplier of the discrete group $\text{SL}_2(\mathbf{R})$ is uncountable, hence certainly not cyclic.

In the same spirit, the Schur multiplier of the simply connected Lie group, $\text{SL}_2(\mathbf{C})$, viewed as a discrete group, is uncountable.

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@ Junkie: This once more illustrates how unreliable Wikipedia can be... –  Alain Valette May 1 '11 at 20:51
    
Or maybe, how much better at self-corrections MO can be, given the multiple wrong answers to start... :) It seems that the whole difficulty was sorting out definitions, and uses of implied terminology vis-a-vis abstract and topological groups. –  Junkie May 1 '11 at 21:14
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EDIT: My beliefs about all of this are subject to change, since it's been decades since I was involved in this literature. I apologize for adding to the clutter here, but the question itself (though different in header and body of the post) is worth pursuing in the absence of a clearcut reference in the literature, which is what I'd really like to see. The early work on algebraic K-theory is beautiful but has faded from sight to some extent as the subject got specialized. Speaking of that pioneering work, my colleague Eduardo Cattani at UMass has just forwarded an email from Dan Quillen's wife informing friends of his death at age 70. The work itself should certainly be kept alive and accessible to future mathematicians.

EARLIER: I believe the answer to the basic question here is yes, but if so the result is probably embedded in a much broader discussion going back to work of Steinberg from 1962 on. This has close connections with the work of Moore, Matsumoto, and others on the congruence subgroup problem. It's clearly essential to make a careful distinction between the theory of topological covering groups of Lie groups (going back to E. Cartan) and the theory of universal central extensions for abstract groups (which owes much to Schur and then Steinberg).

One of the most readable sources, with references, would be the 1967-68 mimeographed Yale lecture notes by Steinberg, Lectures on Chevalley Groups (section 7 on central extensions). Here the abstract theory of universal central extensions is laid out, with special reference to generators and relations for Chevalley groups over arbitrary fields (in principle including $\mathbb{R}$).

Steinberg's 1962 Brussels conference paper, which gave rise to "Steinberg symbols", is collected with his other papers in an AMS volume. Besides these sources, the IHES paper by Moore can be downloaded via www.numdam.org. But in the study of the congruence subgroup problem, the field $\mathbb{R}$ is only one of the local fields involved. And your group is just the rank one case of a more general theory; here the Lie type $A_1$ is part of the symplectic family which requires special treatment.

ADDED: As Richard points out in his answer, the basic argument in the real case goes back to Steinberg's 1962 Brussels paper. The treatment in section 7 of his Yale lecture notes leads to his Theorem 13 (for complex Lie groups) and the following concise remark on adapting the argument to the real case; this starts at the bottom of page 90. Anyway, the gist of the matter in the rank one case seems to be the conclusion that the algebraic universal central extension has an infinite cyclic kernel just as in the topological situation (and with the same generator).

By the way, a scanned PDF version of the Yale lecture notes was available at one time on the old Wisconsin homepage of Arun Ram (now at Melbourne) but seems to have disappeared. Probably it's still living somewhere in cyberspace.

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I actually ran across one of Moore's paper in my searches, though I'm not sure it's one you say. archive.numdam.org/article/PMIHES_1968__35__5_0.pdf –  Junkie Apr 30 '11 at 19:17
    
Yes, that's the relevant paper by Moore. –  Jim Humphreys Apr 30 '11 at 20:44
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The pdf version of Steinberg's Yale notes is still available on Arun Ram's Wisconsin page: math.wisc.edu/~ram/YaleNotes.pdf –  Peter McNamara May 1 '11 at 1:45
    
:( ${}{}{}{}{}$ –  Mariano Suárez-Alvarez May 1 '11 at 23:00
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(This is incorrect: see below)

A proof of this can be extracted from Steinberg's paper "Générateurs, relations et revêtements de groupes algébriques". If Ive understood it correctly, he shows that a Cartan subgroup of the universal central extension of $SL_2(K)$ (for a field $K$ with at least 4 elements) is generated by elements $h(t)$ for $t$ a nonzero element of the field, that in particular satisfy $h(tu^2)=h(t)h(u^2)$, and the center of the universal central extension is the kernel of the map from the Cartan subgroup to $K^*/\pm 1$ taking $h(t)$ to $t$. In the case when $K$ is the reals, this kernel is generated by $h(-1)$ as every element is a square or a square times $-1$. So the center of the universal central extension is generated by $h(-1)$ and in particular is cyclic. The center is known be be at least $Z$, so the center of the universal central extension is exactly $Z$, and the universal central extension is therefore the same as the universal cover.

Added later: this answer is wrong. I overlooked that the proof of $h(tu^2)=h(t)h(u^2)$ in Steinberg's paper requires that $t$ and $u$ generate a cyclic subgroup of $K^*$, which holds in the finite field case he was considering but not over the reals.

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This is not the correct reading of that fundamental paper. What Steinberg introduces here are `symplectic Steinberg symbols'. There is no rule $h(t^2u)=h(t)h(u^2)$. Instead there is a rule {$t^2u,v$}={$t^2 ,v$}{u,v}. There is a map towards the ordinary bilinear Steinberg symbols. In any case the difference between the abstract universal central extension and the universal cover is huge. –  Wilberd van der Kallen May 1 '11 at 14:50
    
You can find the rule $h(tu^2)=h(t)h(u^2)$ on page 121 of Steinberg's paper, as the special case of 7.3 (e) when r=s and d=2. –  Richard Borcherds May 1 '11 at 16:35
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This was my field when I was younger. Will have to check where the confusion comes from. –  Wilberd van der Kallen May 1 '11 at 17:07
    
@Wilberd Looking at the remarks after Theorem 4.2 in users.ictp.it/~pub_off/lectures/lns023/Rehmann/Rehmann.pdf (from Matsumoto), it seems that $SL_2$ should be considered "symplectic" for some purposes. –  Junkie May 1 '11 at 19:35
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Steinberg 7.3(e) says: $h_r(t)h_s(u)h_r(t)^{-1}=h_s(t^du)h_s(t^d)^{-1}$ where $d=c(r,s)=2(r,s)/(s,s)$. When $r=s$, then $d=2$, and this becomes $h(t)h(u)h(t)^{-1}=h(t^2u)h(t^2)^{-1}$, and even switching $t$ and $u$, I am not able to see how this gives your claim? –  Junkie May 1 '11 at 21:41
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I have uploaded a copy of Steinberg's Yale lecture notes on Chevalley groups to Google Docs.

Here is the link

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There is a more complete version available at Steinberg's home page math.ucla.edu/~rst –  Richard Borcherds May 1 '11 at 4:46
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I would have prefered to write this as a comment, but I'm lacking rep. Does this help?: The universal cover for $SL_2(R)$ is also the universal central extension and hence centrally closed (i.e. it coincides with its universal central extension) as an abstract group.

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Can you provide a reference? Thanks. –  Alain Valette Apr 30 '11 at 15:49
    
If what you say is correct, then that answers my question. But I would like to see an argument (or a reference) for the fact that the universal cover for SL2(R) is also its universal central extension (as an abstract group). –  André Henriques Apr 30 '11 at 16:02
    
Not happy with Wikipedia, with no other source given? :) "The Schur multiplier of PSL(2,R) is Z, and the universal central extension is the same as the universal covering group." en.wikipedia.org/wiki/SL2%28R%29 –  Junkie Apr 30 '11 at 16:10
    
I guess they are not happy with that. I will look for an appropriate reference. –  Benjamin Westrich Apr 30 '11 at 16:14
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I suspect that, in all those references, the words "universal central extension" mean "universal central extension in the category of Lie groups", and not "universal central extension in the category of groups". –  André Henriques Apr 30 '11 at 17:33
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Non-answer I: just a reference for the difference between topological and abstract groups: I think the first part of Chapter I of the above Moore reference addresses the question, but I'm not sure it exactly answers it w/o more work. He notes at the start of chapter 1 that he is talking about abstract groups there, and "For topological groups [the abstract fundamental group] need not coincide with the usual [topological] fundamental group although it does in the most important cases (e.g. semi-simple Lie groups)", but I see no specific proof of this latter fact.

An abstract group $G$ is "simply connected" if for every central extension $$1\rightarrow A\rightarrow E\rightarrow G\rightarrow 1$$ there is a unique (splitting) homomorphism $\phi: G\rightarrow E$ that composes with the quotient map to yield ${\rm id}_G$. He shows this is equivalent to $H^1(G,T)=H^2(G,T)=0$ where $T$ is the circle group, and notes in passing that $H^1(G,T)=0$ is equivalent to $[G,G]=G$.

A cover $E$ of $G$ (more properly defined on the morphism) is when $E=[E,E]$ and the kernel of the map from $E$ to $G$ is central in $E$. He then proves that any group with $G=[G,G]$ has a (unique) simply connected covering group $F$, and that this is characterized by the inflation property that $H^2(G,T)\rightarrow H^2(F,T)$ is the zero map. He then shows that that $F$ is universal in the sense expected. All of this is at the abstract group level from my reading.

Notably the "universal central extension" at the group level is the "universal covering group" at this level. In section 2 of Chapter I, he then compares to the topological case, and with Theorem 2.3, noting the ("somewhat amazing") fact that: "if $$1\rightarrow A\rightarrow E\rightarrow G\rightarrow 1$$ is a topological extension of $G$ by $A$ which splits as extension of abstract groups, then it splits as extension of topological groups", assuming $G=[G,G]$ here (and not just that the derived subgroup is merely dense). In fact, he shows that a splitting map $\phi: G\rightarrow E$ must be continuous.

However, as he assumes that the extension is topological, it doesn't seem to show (from what I can tell) that the abstract universal covering group is indeed the topological universal covering group as would be desired.

Answer II: Another reference, more specific to $SL_n$ is Chapter 1 of http://users.ictp.it/~pub_off/lectures/lns023/Rehmann/Rehmann.pdf

Therein (first page) the central extension $$1\rightarrow K_2(n,k)\rightarrow St_n(k)\rightarrow SL_n(k)\rightarrow 1$$ gives the Steinberg group as the universal central extension, and $K_2(n,k)$ as the (algebraic) fundamental group. There is a epimorphism surjecting $K_2(n,k)\rightarrow K_2(k)$, from the inductive limit $St(k)\rightarrow SL(k)$, and for $n>2$ this is isomorphic. Milnor notes that $K_2(R)$ is the direct sum of $Z/2$ and an uniquely divisible group, so in particular larger than the $Z$ from topological coverings. See Example 1.6 of "Algebraic K-theory and quadratic forms" http://www.springerlink.com/content/t025u1152j330325/ Note: I think he speaks of the Milnor $K$-group in general, but that it is not a bother at $K_2$.

In Theorem 1.2 of Rehmann's paper (and Theorem 4.2 for general Chevalley), the generators of the kernel are given, from Matsumoto's work.

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