5
$\begingroup$

The cohomology ring of $\mathrm{SL}_2(\mathbb C)$ as a topological group is straightforward (it's generated by a Chern class), but what is known in the discrete case? I'm particularly interested in $H^2$ with coefficients in $\mathbb C^\times = \mathbb C \setminus \{0\}$, especially if there are explicit formulas for the classes.

$\endgroup$
4
  • 2
    $\begingroup$ As a discrete group $H_2(\mathrm{SL}_2(\mathbf{C}))$ has continuum cardinal. Probably this is still true modulo torsion. If so, this $H^2$ is huge (maybe even of cardinal power of the continuum). A full description possibly exists in terms of Kähler differentials. Related: mathoverflow.net/a/63599/14094 $\endgroup$
    – YCor
    Mar 24, 2021 at 16:00
  • 2
    $\begingroup$ I think with Matsumoto's theorem, you can identify $H_2(SL_2(\mathbb{C}),\mathbb{Z})$ with $K^M_2(\mathbb{C})$, and then you could use universal coefficient formulas to get statements about second cohomology with $\mathbb{C}^\times$-coefficients. Not sure if this is specific enough, but the torsion should be controllable, the uniquely divisible part less so. $\endgroup$ Mar 24, 2021 at 19:11
  • 1
    $\begingroup$ As a follow-up on the link to Milnor K-theory, elements of $H^2({\rm SL}_2(\mathbb{C}),\mathbb{C}^\times)$ should be given by $\mathbb{C}^\times$-valued Steinberg coycles. Those are maps $\mathbb{C}^\times\times\mathbb{C}^\times\to \mathbb{C}^\times:(u,v)\mapsto \{u,v\}$ which are bimultiplicative and satisfy $\{a,1-a\}=1$. That point of view may help with computations, the K-book has more information in this direction in the $K_2$-section. $\endgroup$ Mar 25, 2021 at 10:46
  • $\begingroup$ Thanks! I know very little about K-theory, so your suggested point of view is helpful. $\endgroup$ Mar 25, 2021 at 14:10

1 Answer 1

8
$\begingroup$

There is a paper by Milnor called On the homology of Lie groups made discrete which contains many results on this kind of problem, as well as references to related literature. However, I don't think that it directly answers your question.

$\endgroup$
3
  • 1
    $\begingroup$ Could you please add some information on the state of art of the Milner conjecture (which I presume is made in this paper) to your answer?! $\endgroup$
    – user51223
    Mar 25, 2021 at 13:04
  • 1
    $\begingroup$ I don't know the current status. I am not an expert in this area, I merely reported something that I remembered reading a long time ago. $\endgroup$ Mar 25, 2021 at 15:02
  • $\begingroup$ MathSciNet will give you a list of all the papers that cite the one that I mentioned, and that will presumably include all papers that make progress on the conjecture. $\endgroup$ Mar 25, 2021 at 15:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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