7
$\begingroup$

In Milnor's book Topology from the Differentiable Viewpoint there's the following problem:

Problem $6$ (Brouwer). Show that any map $S^n\to S^n$ with degree different from $(-1)^{n+1}$ must have a fixed point.

My solution: Assume that the map $f:S^n\to S^n$ has no fixed points. Let $a:S^n\to S^n$ denote the antipodal map $a(x)=-x$. Then the map $a\circ f$ is homotopic to the identity as follows:

Since $x$ is never mapped to $-x$ (by assumption), there exist a unique shortest great circle arc from $a(f(x))$ to $x$, simply take the straight line homotopy flowing along such arches to get a homotopy $a\circ f\simeq\operatorname{id}$.

Now we have: $$1=\deg(\operatorname{id})=\deg(a\circ f)=\deg(a)\deg(f)=(-1)^{n+1}\deg(f)$$ and thus $\deg(f)=(-1)^{n+1}$.

My questions:

  1. Is this solution correct?
  2. How to show that the homotopy is continuous? It seems intuitively true to me, but I'm not sure about how to proceed to show it. Maybe saying that it is the flow of some good tangent field?
  3. Are there other (elegant/short/interesting) proofs for this fact?
$\endgroup$
  • $\begingroup$ For 2, following your intuition, you can try using the geodesic flow on $TS^n$, but that requires showing that the assignment of the tangent vector field corresponding to the great circle arc from $a(f(x))$ to $x$ is a continuous vector field. $\endgroup$ – Willie Wong May 23 '14 at 10:59
  • $\begingroup$ @WillieWong Yes, that's what I was thinking about. Does it work for the continuous case, or only for the smooth one? $\endgroup$ – Daniel Robert-Nicoud May 23 '14 at 11:29
  • $\begingroup$ The geodesic flow can be thought of as a parametrised ODE, and as long as the metric is sufficiently smooth you get continuous dependence on the initial data. Another way to think about it is that the exponential map from $TS^n \to S^n$ is a smooth map. Thus the image of continuous sections is still continuous. $\endgroup$ – Willie Wong May 23 '14 at 15:46
8
$\begingroup$

I think, what you gave is (a sketch of) the simplest proof, unless you know about the Lefshetz fixed point theorem. Using the latter, then the argument goes like this: $H_k(S^n)$ is nonzero only in dimensions $0$ and $n$ and $H_0(S^n)\cong H_n(S^n)\cong Z$. The action of $f$ on $H_0$ is trivial and the action on $H_n$ is by multiplication by $d=\deg(f)$. The Lefschetz number of $f$ then equals $$ \Lambda_f= (-1)^0 + (-1)^n (d)= 1+ d(-1)^n. $$ This number is nonzero unless $$ d= (-1)^{n+1} $$ as required. If $\Lambda_f\ne 0$ then $f$ has a fixed point (this is the Lefschetz fixed point theorem).

As for your solution, do not bother with geodesic flow, that's unnecessarily complicated; you can use linear algebra instead. For $t\in [0,1]$ and $x, y\in S^n$ non-antipodal, define $$ z_t= \frac{tx+(1-t)y}{|tx+(1-t)y|}\in S^n. $$ This is a point on the shortest arc of the great circle connecting $x$ and $y$. The rest you can figure out yourself.

$\endgroup$
  • $\begingroup$ Thanks! I heard of the Lefschetz fixed point theorem, but as the book by Milnor didn't treat it, I wanted to find an elementary proof. Also thanks for pointing out the necessary homotopy (the book doesn't treat flows either, so doing it without is much much, better :) ). $\endgroup$ – Daniel Robert-Nicoud May 23 '14 at 22:48
3
$\begingroup$

Best solution: Notice that $\|f(x)-(a \circ f) (x)\| < 2$ for every $x.$ But then a problem above this one in the book, asks to prove that such maps must be homotopic (smoothly so if maps themselves are smooth.) Then your computation goes thru with nothing on conscience!

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.