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This is my first question.

Take a simple, connected, compact, simply connected Lie group $ G$ (dim $G\geq 3$).
The cohomology of $G$ with integer coefficients is $H^{1,2}(G,\mathbb{Z})\cong 0$, $H^{3}(G,\mathbb{Z})\cong\mathbb{Z}$.
I'm looking for a generator of $H^{3}(G,\mathbb{Z})$.
It should be something of the form $\alpha K(\theta,[\theta,\theta])$, where $\theta$ denotes the left Maurer-Cartan-Form, $K$ is the Killing-Form and $\alpha$ is a scaling factor.

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  • $\begingroup$ Set $\sigma(x,y,z):=cK([x,y],z)$ such that $cK(h_\alpha,h_\alpha)\in 2\mathbb{Z}$ for each coroot $h_\alpha$ of G. Then $[\sigma]\in H^3(G;\mathbb{Z})$. This is proved in Loop Groups by Segal-Pressley (page $49$). $\endgroup$
    – Somnath Basu
    Apr 6, 2011 at 3:49
  • $\begingroup$ If I am not mistaken, any subgroup of $G$ isomorphic to $SU_2$ induces an isomorphism $H_3(SU_2;\mathbb Z)\cong H_3(G;\mathbb Z)$. Now topologically $SU_2$ is just a 3-sphere, so each such subgroup can be viewed as an integral 3-cycle, and this 3-cycle will thus represent a generator of $H_3(G;\mathbb Z)$. $\endgroup$
    – მამუკა ჯიბლაძე
    Sep 5, 2014 at 18:36
  • $\begingroup$ I think the answer in Pressley, Segal might be off by a factor of $48 \pi^2$. See my question on Stack Exchange for the discussion: math.stackexchange.com/questions/3603441/… $\endgroup$
    – Blazej
    Apr 2, 2020 at 7:48

1 Answer 1

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The third cohomology group of the Lie algebra (with coefficients in the trivial module) is generated by the cocyle $\phi(x,y,z)=K(x,[y,z])$. The usual proof that the de Rham cohomology of the group is the same as the Lie algebra cohomology of the corresponding Lie algebra gives you a more or less explicit $3$-form generating $H^3_{\mathrm{dR}}(G,\mathbb R)$. It gives you a generator of integral cohomology up to scaling. I don't know what the factor should be, though...

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    $\begingroup$ Is there an explicit formula for the scaling factor? $\endgroup$
    – user14120
    Apr 5, 2011 at 19:20
  • $\begingroup$ @mikethspike Something like $1/2\pi$ or similar (perhaps an $i$ in there as well). $\endgroup$
    – David Roberts ♦
    Apr 5, 2011 at 23:40
  • $\begingroup$ The scalar is a cute invariant of the group. Hopefully there is some more precise description than that :) $\endgroup$
    – Mariano Suárez-Álvarez
    Apr 6, 2011 at 1:04
  • $\begingroup$ I thought there was a 1/12th in there, but I'm blanking on where I saw that. $\endgroup$
    – Allen Knutson
    Apr 6, 2011 at 2:31

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