5

What is the fastest way to compute $$H_*((S^n\times S^m)/\mathbb{Z}_2;\mathbb{Z}),$$ where $\mathbb{Z}_2$ acts on both factors by the antipode map? Is there a better way than using the Serre spectral sequence?

Sophie
  • 51
  • Using the Serre spectral sequence should be quick : there is a fiber bundle with fiber $S^m$ and base $\Bbb RP^n$. In particular, the spectral sequence degenerates at $E^2$ and you get an exact sequence. You can probably identify explicitely the map since you get the Gysin exact sequence. – Nicolas Hemelsoet Jul 13 '18 at 16:50
  • @Nicolas This requires some care identifying the $E^2$ page; because the base is not simply connected and the antipodal map sometimes acts nontrivially on the homology of the fibers, one needs to use twisted coefficients in writing down the $E^2$ page. –  Jul 13 '18 at 22:19
  • @MikeMiller : definitely, but the local systems will be representations of $\Bbb Z/2 \Bbb Z$ so if i'm not mistaken $H^0$ is trivial and I think $H^n$ corresponds to the non-trivial representation but indeed one should think about it. – Nicolas Hemelsoet Jul 13 '18 at 22:25
  • @NicolasHemelsoet Beyond that I agree your comment works nicely; I only wanted to make a word of warning to OP, who may have only seen Serre from references not mentioning local coefficients. I have been bitten by this in my past :) –  Jul 13 '18 at 22:31
  • @MikeMiller : me too in fact, so this is definitely something worth mentioning :-) – Nicolas Hemelsoet Jul 13 '18 at 22:34
  • 3
    God bless those who understand spectral sequences so I don’t have to. – Randall Jul 13 '18 at 22:59

0 Answers0