Newsgroups: sci.math.research From: hirsch@mathcs.emory.edu (Michael Hirsch) Subject: Re: Borsuk's theorem Date: Fri, 21 Apr 1995 21:01:03 GMT >>>>> In article <1995Apr21.162724.22986@leeds.ac.uk>, pmt6jrp@gps.leeds.ac.uk (J R Partington) writes: jrp> I will be eternally grateful if someone could point me to an jrp> *elementary* proof of Borsuk's theorem: jrp> Given any continuous map f: S^n --> R^n there is a point x in S^n jrp> such that f(x)=f(-x). There is a not too bad differential topology proof. Presumably it was discovered before by others, but I don't have a reference. A generalization of this argument can be found in "A geometric approach to a class of equilibrium existence theorems" in J. of Math. Econ, 1990, pp. 95-106, by Hirsch, Magill, and Mas-Colell. It is easy to see that this theorem is equivalent to: Any continuous map f:S^n --> R^n which commutes with multiplication by -1 must have a zero. Proof: Let f be such a function. Without too much work you can show that f induces a section of the n-plane bundle over projective n-space, P^n, which is the n-fold direct product of the tautological line bundle over P^n. (P^n is the space of lines in R^{n+1}. The Tautological line bundle has the each line as the fibre over itself.) f has a zero iff the induced bundle has a zero. Now we can construct a section of the bundle which has exactly one zero. Choose n linearly independent vectors v_1, ..., v_n in R^{n+1}. Let L be a point (i.e. a line) in P^n. Let s_i(L) = the projection of v_i to L. Each s_i is then a section of the tautological line bundle, so s = (s_1, ..., s_n) is a section of the big bundle. s is zero only when L is perpendicular to all the v_i. Only one line is perpendicular to all, so s has a unique zero. Finally, it is not hard to show that s(P^n) is transverse to P^n (the zero section of the bundle). Thus any section homotopic to s must have algebraic crossing number 1 (mod 2) with the zero section, hence must have a zero. The punch line is that any two sections are homotopic, thus the original function f must have had a zero. I don't know if this is elementary enough, but it doesn't use much algebraic machinery. It needs transversality, a little bundle theory, and invariance of mod 2 intersection under homotopy. Hope that helps, -- Michael Hirsch Work: (404) 727-7940 Emory University, Atlanta, GA 30322 FAX: (404) 727-5611 Internet: hirsch@mathcs.emory.edu BITNET: hirsch@emory.bitnet UUCP: {rutgers,gatech}!emory!hirsch http://www.mathcs.emory.edu/~hirsch/ Public key for encrypted mail available upon request (or finger hirsch@cssun.mathcs.emory.edu). ============================================================================== Newsgroups: sci.math.research From: "Mark D. Meyerson" Subject: sci.math.research/Re: Borsuk's theorem Date: Wed, 26 Apr 1995 18:18:58 GMT >>>>> In article <1995Apr21.162724.22986@leeds.ac.uk>, pmt6jrp@gps.leeds.ac.uk (J R Partington) writes: jrp> I will be eternally grateful if someone could point me to an jrp> *elementary* proof of Borsuk's theorem: jrp> Given any continuous map f: S^n --> R^n there is a point x in S^n jrp> such that f(x)=f(-x). For an elementary (in the sense of little advanced theory) and constructive proof, see my article with Alden Wright: "A New and Constructive Proof of the Borsuk-Ulam Theorem", Mark D. Meyerson and Alden H. Wright, Proceedings of the American Math Soc, vol 73, No 1, Jan 1979, pp. 134-136. It has some references for other proofs too. --------------------------------------------------------------------------- Mark D. Meyerson, Math Dept, US Naval Academy, Annapolis, MD 21402-5002 USA Phone: (410) 293-6710 Fax: (410) 293-4883 Email: mdm@usna.navy.mil "Mathematics possesses not only truth, but supreme beauty ..." --- Bertrand Russell ---------------------------------------------------------------------------