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Jan 4, 2019 at 5:35 vote accept Shivani Sengupta
Jul 9, 2018 at 22:14 comment added K.K. @AchimKrause Indeed :) I think my argument actually shows that if $X$ is a sufficiently nice compact space all of whose Betti numbers are $< p-1$ and with non-zero Euler characteristic, then $\mathbb{Z}/p\mathbb{Z}$ cannot act freely.
Jul 9, 2018 at 19:03 comment added Achim Krause I think if you take into account the cup product structure, you can directly see that the action on cohomology is trivial.
Jul 9, 2018 at 18:54 history answered K.K. CC BY-SA 4.0