19
$\begingroup$

The group $S_{24}$ of permutations of $24$ things has fourth integral cohomology $\mathrm H^4(S_{24};\mathbb Z) \cong \mathbb Z/2 \oplus \mathbb Z/2 \oplus \mathbb Z/12$. According to Sikiric and Ellis the largest Mathieu group $M_{24}$ has $\mathrm H^4(M_{24};\mathbb Z) \cong \mathbb Z/12$. The Mathieu group is defined in terms of a permutation representation on $24$ things (namely, the coordinate vectors in the extended binary Golay code), and so there is a restriction map $\mathrm H^4(S_{24};\mathbb Z) \to \mathrm H^4(M_{24};\mathbb Z)$.

The number $12$ being somewhat magical, I expect that this map is a surjection. Is it? Is the answer to this question known? My impression of the literature is that $M_{24}$ is just beyond where current technology can fully work out its (2-local, or even $\mathbb F_2$) cohomology. Compare this older MO question.

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ Does M_{24} contain a copy of a medium-sized symmetric group, like S_6? Maybe it is less intimidating to compute the composite map H^4(S_{24}) --> H^4(M_{24}) --> H^4(S_6), than either map separately. $\endgroup$
    – David Treumann
    Commented Sep 15, 2016 at 2:41
  • $\begingroup$ According to Wikipedia, here are nine conjugacy classes of maximal subgroups, including the "octad subgroup" (stabilizer of an octad in the $(5,8,24)$ Steiner system, which is isomorphic with $2^4:A_8$) and the "sextet group" $2^6 : (3.S_6)$. $\endgroup$
    – Noam D. Elkies
    Commented Sep 15, 2016 at 2:46
  • $\begingroup$ @DavidTreumann Good idea. But if so, I would expect that the map H^4(S_24) --> H^4(S_6) to be multiplication by 4. It depends, of course, on how the S_6 sits inside the S_24. $\endgroup$
    – Theo Johnson-Freyd
    Commented Sep 15, 2016 at 2:51
  • 4
    $\begingroup$ Out of curiosity: why do you need this? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Sep 15, 2016 at 7:24
  • 2
    $\begingroup$ @MarianoSuárez-Álvarez David Treumann and I decided to answer to my older question. We have basically got it to the point where the present question is the only (or one of the few) missing ingredient(s). (Compare this question.) I want these answers in order to understand (and prove!) a conjectural behavior in certain conformal field theories --- that behavior is predicted based on expected connections to TMF. I will let David speak to why he is interested. $\endgroup$
    – Theo Johnson-Freyd
    Commented Sep 15, 2016 at 12:54

1 Answer 1

Reset to default
11
$\begingroup$

The answer to my question is No. The generator of the $\mathbb Z/12$ part of $H^4(S_{24})$ is $p_1$ of the permutation representation. That representation restricts to $M_{24}$ to a Spin representation, i.e. one with $w_1=w_2=0$. For any such representation, $p_1$ is automatically even.

Thus the map $H^4(S_{24}) \to H^4(M_{24})$ has image within the $\mathbb Z/6 \subset \mathbb Z/12$.

Improve this answer
$\endgroup$
1

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .