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A version of the Borsuk--Ulam theorem states that a continuous antipodal map from the M-sphere into euclidean N-space has a zero provided that M is at least N. Clearly the general case follows from the case when M = N. But is the case when M >> N any easier to prove than the equidimensional case?

edited Jan 5 2011 at 14:58

Theo Buehler
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asked Jan 5 2011 at 14:20

Tony Carbery
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Are you asking if there is a direct and easier proof when $M >> N$? – Somnath Basu Jan 5 2011 at 14:38

(Hi Tony!) Please explain: why do you expect the proof to be easier when M>>N? – Piero D'Ancona Jan 5 2011 at 14:44

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It's a shame about that answer, as the question was a very nice one. I suppose there's still just about room for the optimist to wonder whether there is a short proof of the Borsuk--Ulam theorem that starts by proving it in very high dimensions. – gowers Jan 5 2011 at 17:53

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Hi Tony! Could you specify what you mean by "easier", ie are you looking for a proof that doesn't appeal to the fact that an antipode-preserving map $f\colon S^n\to S^n$ is of odd degree, therefore homotopically non-trivial? – Mark Grant Jan 5 2011 at 22:39

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Tim, there is a relatively short argument via dimensional reduction which is a hybrid of one of Shchepin's together with a low-dimensional argument reminiscent of the Stokes' theorem proof of the Brouwer fixed point theorem. See maths.ed.ac.uk/~carbery/analysis/links.html and click on Notes on the Borsuk--Ulam theorem. – Tony Carbery Jan 6 2011 at 15:05

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BS · Accepted Answer · 2011-01-05 15:22:28Z UTC

I don't think so, since any antipodal (non-existent) map $S^n\to S^{n-1}$ would easily be "suspended" to an antipodal map $S^{n+1}\to S^n$. Iterating and composing would then yield antipodal maps $S^m\to S^{n-1}$ with arbitrarily large $m$. The $n=2$ case is somewhat easier though.

Borsuk--Ulam question

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