MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Does a universal central extension exist for the fundamental group of a Compact Riemann Surface of genus1? Please give a detailed explanation.I am unable to justify the statements in Atiyah-Bott Phil Trans Roy Soc 1982 p559 (apparently it exists and some quotient operations are performed)

share|cite|improve this question
It would help if you could quote the text of the Atiyah--Bott article that confuses you. – HJRW Feb 13 '14 at 11:54

Well, the OP is right: Atiyah and Bott indeed use the term "universal central extension" for the fundamental group of a compact surface $\Sigma_g$ of genus $g\geq 1$ -- which is incorrect since that group is not perfect. What they mean is that there is in this case a canonical central extension by $\mathbb{Z}$: indeed these extensions are parameterized by $H^2(\pi _1(\Sigma _g),\mathbb{Z})$; since $\Sigma _g$ is a $K(\Pi ,1)$, this is canonically isomorphic to $H^2(\Sigma _g, \mathbb{Z})=\mathbb{Z}$, and they consider the extension associated to the generator of this group. They describe explicitly this extension $$0\rightarrow \mathbb{Z}\rightarrow \tilde{\pi }_g\rightarrow \pi _1(\Sigma _g)\rightarrow 1 $$as follows: $\tilde{\pi }_g $ is the largest quotient of the free group $F_{2g}$, with generators $a_1,\ldots ,a_g,b_1,\ldots ,b_g$, such that the element $[a_1,b_1]\ldots [a_g,b_g]$ is central.

share|cite|improve this answer

A compact Riemann surface of genus one (with a chosen basepoint) has fundamental group $\mathbb{Z}^2$. A group has a universal central extension if and only if it is perfect, and $\mathbb{Z}^2$ is not perfect.

share|cite|improve this answer
Now I see that you were asking about genus strictly greater than one, but the "greater than" sign disappeared from the text of the question. At any rate, the abelianization of $\pi_1(\Sigma_g)$ is free abelian on $2g$ generators, so the group is not perfect. – S. Carnahan Jul 21 at 23:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.