7
$\begingroup$

I am looking at Problem 7 in Milnor's Topology from the Differentiable Viewpoint, namely :

Show that any smooth map $S^n \to S^n$ of odd degree must carry some pair of antipodal points into a pair of antipodal points.

Here is my attempt : consider the contraposition and take $f : S^n \to S^n$, smooth, such that for all $x \in S^n$, $$ f(x) \neq -f(-x) $$ (this is saying that no pair of antipodal points is carried to another pair of antipodal points by $f$). We will try to show that $deg(f)$ is even.

Another way to interpret this is to say that for all $x \in S^n$,

$$ ||f(x)-f(-x)||<2 $$ as the equality only holds if $f(x)$ and $f(-x)$ are antipodal. Thus we can build a smooth homotopy $F$ between $f(x)$ and $f(-x)$ (cf Problem 3), given for example by $$ F(x,t)=\frac{tf(x)+(1-t)f(-x)}{||tf(x)+(1-t)f(-x)||} $$ Now if $f(x)$ and $f(-x)$ are smoothly homotopic, then they must have the same degree. Rewriting $f(-x)$ as $f(a(x))$, where $a : S^n \to S^n$ is the antipodal map, we have that $$ deg(f)=deg(f\circ a)=deg(f)deg(a)=(-1)^{n+1}deg(f) $$ If $n$ is even, then $deg(f)=0$. However is $n$ is odd we cannot conclude with this method. Is there something obvious that I am missing, or is this the wrong approach ? Any small tip would be appreciated.

I have also proven that (cf Problem 6) if $g : S^n \to S^n$ is smooth and such that $deg(g) \neq (-1)^{n+1}$, then $g$ must have a fixed point. In our case we must have $deg(f)\neq(-1)^{n+1}$, so $f$ has a fixed point.

$\endgroup$
  • $\begingroup$ As @R. Alexandre suggests, you really want to be concentrating on the map $g(x) = \dfrac{f(x)+f(-x)}{\|f(x)+f(-x)\|}$, and you can easily show that $\deg_2 g = 0$. $\endgroup$ – Ted Shifrin Apr 9 '17 at 16:35
  • 1
    $\begingroup$ Yes, I read the answer without even trying to figure out what $F(x,1/2)$ would look like (numbers ? Ew !). It's much clearer now. Thank you for your comment. $\endgroup$ – Bass Apr 9 '17 at 16:50
4
$\begingroup$

The Brouwer theorem (p. 51) says that two functions $g,h:S^n\to S^n$ have same degree iff they are (smoothly) homotopic. So $f(x)$ and $f(-x)$ can't be smoothly homotopic if $n$ is odd.

So you "really" need a new idea to finish your proof for the case $n$ odd.

(Sorry I don't have any tip to give you right now, I need to think a bit. And since I don't have enough reputation I can't just post this as a comment...)

edit :

Ok I think I have a nice clue for you. You should compare degrees of $F(x,t)$ for $t=0$ and $t=1/2$ ;)

$\endgroup$
  • $\begingroup$ Thank you for your remark. However I don't really see how the clue relates to the problem... I'm not even sure I can say anything about $F(x,1/2)$ ! $\endgroup$ – Bass Apr 9 '17 at 14:37
  • 1
    $\begingroup$ Try to count the number of preimages. They come by pairs. So the degree is even for t=1/2 $\endgroup$ – R. Alexandre Apr 9 '17 at 14:53
  • $\begingroup$ This is great. Thanks ! $\endgroup$ – Bass Apr 9 '17 at 16:47

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.