It is well known that there is a parallel between Galois theory and covering theory. So I wonder whether there is a deep similarity between the following two facts:

Artin-Schreier theorem. The only non-trivial finite field extensions $U \supset V$ with $U$ an algebraically closed field are $U=V(\sqrt {-1}),$ so among non-trivial finite groups only $\mathbb Z/2\mathbb Z$ may "Galois-act" on an algebraically closed field.

Corollary of Lefschetz fixed point theorem. The only non-trivial finite coverings $U \to V$ with $U$ an even-dimensional sphere are given by antipodal maps, so among non-trivial finite groups only $\mathbb Z/2\mathbb Z$ may act freely on (these particular) spaces that are the universal coverings of themselves.

Unfortunately, the statement from covering theory is very particular. Maybe, there is a simple case of the first statement where the similarity may be seen even better?

  • $\begingroup$ P. S. This question was unsuccesully asked by me on MSE:878944 two years ago. $\endgroup$ – evgeny Nov 9 '16 at 8:21
  • $\begingroup$ Probably a very naive comment but the example of the Riemann sphere viewed as a compactification of the complex plane could suggest that an even dimensional sphere corresponds to the compactification of a vector space over an algebraically closed field. $\endgroup$ – Sylvain JULIEN Nov 9 '16 at 11:33
  • $\begingroup$ Haven't we unintentionally lost the word "finite" from the statement of Artin-Schreier? $\endgroup$ – Oliver Nash Nov 9 '16 at 12:28
  • $\begingroup$ @OliverNash, thank you, I corrected! $\endgroup$ – evgeny Nov 9 '16 at 12:33
  • 1
    $\begingroup$ The second fact can be proved a different way, as a special case of the fact that in an $n$ to $1$ covering $U\to V$ of (for example) finite cell complexes the Euler characteristics satisfy $\chi(U)=n\chi(V)$. Is there some sense in which an algebraically closed field, like an even-dimensional sphere, has Euler characteristic $2$? $\endgroup$ – Tom Goodwillie Nov 9 '16 at 20:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.