# Universal Central Extension of pi(X), X a compact Riemann surface of genus>1

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)

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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.
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.
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