What is the low degree cohomology of the mapping class group of a non-orientable surface? More specifically, what is the universal central extension of the mapping class group of a non-orientable surface?

(I've done some googling, but so far have only found presentations of the MCG.)

(I would be happy to have the answer in just the negative Euler characteristic case.)


1 Answer 1


Let me write $\mathcal{N}_g$ for the mapping class group of the connect sum of $g$ projective planes. Nathalie Wahl proved that these groups enjoy homological stability, and in

O. Randal-Williams, The homology of the stable non-orientable mapping class group, Algebraic & Geometric Topology 8 (2008) 1811-1832.

I calculated the stable (co)homology. Using the best currently available general homological stability ranges, one has $$H_1(\mathcal{N}_g ; \mathbb{Z}) = \mathbb{Z}/2 \quad \text{for} \quad g \geq 7$$ and $$H_2(\mathcal{N}_g ; \mathbb{Z}) = (\mathbb{Z}/2)^2 \quad \text{for} \quad g \geq 10.$$

In fact, by

M. Korkmaz, First homology group of mapping class groups of nonorientable surfaces, Math. Proc. Cambridge Philos. Soc. 123 (1998), 487-499.

the abelianisations of all $\mathcal{N}_g$ are known: starting at $g=1$ they are $$0, (\mathbb{Z}/2)^2, (\mathbb{Z}/2)^2, (\mathbb{Z}/2)^3, (\mathbb{Z}/2)^2, (\mathbb{Z}/2)^2, \mathbb{Z}/2, \mathbb{Z}/2, \ldots.$$

As the abelianisation does not vanish for $g > 1$ these groups are not perfect, so they do not have a universal central extension.

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    $\begingroup$ Great -- thanks. $H^2 = (Z/2)^2$ was the answer I was hoping to hear. For orientable surfaces, one can form the signature cocycle/entension using 3-manifolds bounded by the surface and the Wall non-additivity formula. This leads to an extension by $Z$ (oriented 4d bordism). In the unoriented case, the corresponding bordism group is $Z/2 \times Z/2$, and one is lead to guess that a similar construction would yield an extension by $Z/2 \times Z/2$. Has this been worked out by someone? $\endgroup$ Sep 29, 2016 at 17:44
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    $\begingroup$ @Kevin: where has this been worked out in the oriented case? I am just curious. $\endgroup$ Sep 29, 2016 at 21:11
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    $\begingroup$ @KevinWalker: I doubt it. $\endgroup$ Sep 30, 2016 at 10:34
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    $\begingroup$ @JE: in my 1991 TQFT notes available on my web page. See also arxiv.org/pdf/0912.4706v3.pdf $\endgroup$ Sep 30, 2016 at 15:40

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