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Let $G$ be a group. In An introduction to homological algebra, Chapter 6.9 Weibel defines a universal central extension as a central extension $$0 \to A \to X \to G \to 1,$$ which is initial with respect to all central extensions. In other words, if $0 \to B \to Y \to G \to 1$ is any other central extension, then there is a unique homomorphism $X \to Y$ which is compatible with the two maps to $G$.

The first basic lemma (6.9.2) Weibel proves is that if there exists a universal central extension, then both $X$ and $G$ are perfect (i.e. $[G,G] = G$).

Now both Wikipedia and this notes (Exercise 10) claim that the extension $$0 \to \mathbb Z \to \mathcal B_3 \to \operatorname{PSl}_2(\mathbb Z) \to 1$$ is universal. (For the definition of the extension, see my other question.)

But that apparently contradicts Weibel's lemma, because the group $\mathcal B_3$ is not perfect, it's abelization is $\mathbb Z$ (by exercise 7 of the above notes). So I guess there is a different notion of universal central extension? This section in Wikipedia seems to also indicate that, but I don't quite understand what they mean, and it seems to mostly care about finite groups? Does that make a difference?

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  • $\begingroup$ For those of us who (like me) don't know much about your question, it would be helpful it you could state what your symbols stand for. Like: What kind of algebraic objects are your A, X, G ? Groups? Abelian groups? Something else? Etc. $\endgroup$
    – Dan Asimov
    Sep 23 at 0:58
  • $\begingroup$ @DanAsimov Take a look at Wikipedia for the definition of a central extension. My question was aimed at people who already know that, which is why I didn't include it. $\endgroup$
    – red_trumpet
    Sep 24 at 10:53

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This question had been discussed at this MO-post. The "usual" definition of a universal central extension $X$ of a group $G$ assumes that $G$ is perfect, and that it is the unique (up to isomorphism) group $X$ that is a Schur covering group of $G$. In other words, the universal central extension of a perfect group is also perfect - see here.

However, the terminology has been also used as follows: In a universal central extension $1 \to M \to X \to G \to 1$ of $G$, the image of $M$ in $X$ is required to be in the commutator subgroup $[X,X]$ of $X$. In our case, the intersection of the central cyclic subgroup $\langle x^2 \rangle = \langle y^3 \rangle$ of $B_3$ with $[B_3,B_3]$ is trivial.

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  • $\begingroup$ Just to verify I understand this correctly: Any central extension $1 \to M \to X \to G \to 1$ with $M \subset [X,X]$ is called universal? So the extension $0 \to c \mathbb Z \to B_3 \to \operatorname{PSl}_2(\mathbb Z) \to 1$ is not universal in this sense, but still called universal for other reasons? $\endgroup$
    – red_trumpet
    May 24 at 12:40
  • $\begingroup$ Yes, some people say that there is no requirement in general for a universal central extension to be perfect - only that its central subgroup $M$ is in its commutator subgroup. $\endgroup$
    – Dietrich Burde
    May 24 at 12:45
  • $\begingroup$ Do you have a reference for the non-perfect case? I checked the books by Aschbacher and Rotman, but I think they only cover the perfect case. $\endgroup$
    – red_trumpet
    May 25 at 11:19
  • $\begingroup$ For a reference see for example here. It is now mostly called a "Schur cover". But in topology it is called "universal central extension". We have a reference already, i.e., the one with Exercise $10$. $\endgroup$
    – Dietrich Burde
    May 25 at 11:56
  • $\begingroup$ Thanks, that is way clearer than the Wikipedia article. The notes by Jenny Wilson do not contain a definition for "universal central extension", or did I miss anything? $\endgroup$
    – red_trumpet
    May 25 at 14:13

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