FeaturedChallenge Advanced Math Problem of the Week 9/14/2017

1. Sep 14, 2017

PF PotW Robot

Here is this week's advanced math problem of the week. We have several members who will check solutions, but we also welcome the community in general to step in. We also encourage finding different methods to the solution. If one has been found, see if there is another way. Occasionally there will be prizes for extraordinary or clever methods.

Prove that if $n > 0$, an even map between $n$-spheres has even homological degree.

(PotW thanks to our friends at MHB)

2. Sep 18, 2017

PF PotW Robot

PF T-Shirt to the member that solves this!

3. Sep 18, 2017

Infrared

Let $f:S^n\to S^n$ be even, i.e. $f(x)=f(-x)$ for all $x\in S^n$. This condition means that $f$ factors through the quotient map $p:S^n\to\mathbb{RP}^n$ (explicitly, we can write $f=g\circ p$ for a continuous map $g:\mathbb{RP}^n\to S^n$).

If $n$ is odd, $\mathbb{RP}^n$ is orientable and the quotient map $p$ has homological degree $2$ (since it is a local homeomorphism and is 2:1). Degrees multiply with composition, so $\deg(f)$ is even.

If $n$ is even (and positive), then $\mathbb{RP}^n$ isn't orientable, so $H_n(\mathbb{RP}^n)\cong\mathbb{Z}/2$. We can still factor the map $f_*: H_n(S^n)\to H_n(S^n)$ as $H_n(S^n)\xrightarrow{p_*} H_n(\mathbb{RP}^n)\xrightarrow{g_*} H_n(S^n)$. The only homomorphism $\mathbb{Z}/2\to\mathbb{Z}$ is the trivial one, so $g_*=0$ and $\deg(f)=0$ is even in this case too.

All coefficients are in $\mathbb{Z}$.

Last edited: Sep 18, 2017
4. Sep 18, 2017

lavinia

Note that if $f$ is smooth then the inverse image of any regular value has an even number of points in it since for each point in the inverse image its antipode is also in the inverse image. This means that the homology degree of the map is even.

Last edited: Sep 18, 2017
5. Sep 18, 2017

lavinia

The top integer homology of any even dimensional real projective space - in fact of any compact non-orientable manifold without boundary- is zero - not Z/2. You could use Z/2 if you used integer cohomology instead. But zero works just as well in your argument
.

Last edited: Sep 18, 2017
6. Sep 18, 2017

Infrared

Thanks, of course you're right.

7. Sep 18, 2017

lavinia

@Infrared An equivalent way to state your argument for the even dimensional case is to observe that the antipodal map on an even dimensional sphere has degree negative one So if $f= f \circ A$ then its degree must be its own negative.

Last edited: Sep 18, 2017