3
$\begingroup$

For a smooth map $f:M\to N$ from an orientable closed surface $M$ to a non-orientable closed surface $N$, we define its parity (also called modulo 2 degree, and denoted $\deg_2(f)$) as the parity of the number of preimages of any regular value of $f$.

By a geometric construction, I am able to convince myself that in fact $f$ is even, but I suspect that there exists an argument of algebraic topology to show this more easily.

Avoiding the construction, I can say the following:

  • If there exists an odd map $f:M\to N$, then there exists an odd map $gf:M\to\mathbb P^2(\mathbb R)$, since for any non-orientable $N$ we can construct an odd map $g:N\to\mathbb P^2(\mathbb R)$ by collapsing to a point the complement of a tubular neighbourhood of an orientation reversing loop. So we can assume that $N=\mathbb P^2$, which has odd Euler characteristic. This is interesting because $M$ has even characteristic.

  • If $f$ is a local homeomorphism, then it's a covering map, so it factors via the orientation covering of $N$, which is even. So in this case it's easy to show that $f$ is even.

Of course, it would also be good to know what happens in greater dimensions. I think that a similar construction shows again that the map must be even.

And although it really isn't necessary for the construction, I would like to know what happens when you collapse to a point the border of a compact manifold-with-border. When can I say that I get a topological manifold? In this case I collapsed a tubular neighbourhood of an orientation-reversing curve. In even dimensions I obtain the projective space, but in odd dimensions I have no idea. I can't obtain the projective space because I can't obtain an orientable manifold.

$\endgroup$
1
  • 1
    $\begingroup$ I don't have time to write a full answer, but look at the induced map $f^\ast :H^\ast(\mathbb{R}P^2,\mathbb{Z}_2) = \mathbb{Z}_2[x]/x^3 \rightarrow H^\ast(M_g) = \mathbb{Z}_2[y_1,...y_g, z_1,..., z_g]/I$ for appropriate ideal $I$. Note that $I$ contains the squares of the $y_i$ and $z_j$. Then $f^\ast(x) = \sum a_i y_i + \sum b_j z_j$, so $f^\ast(x^2) = f^\ast(x)^2 = \sum a_i^2 y_i^2 + \sum b_j^2 z_j^2 + 2\sum a_i b_j y_i z_j = 0 (mod\text{ }I)$ since everything is mod $2$. $\endgroup$ – Jason DeVito Sep 26 '14 at 12:56
0
$\begingroup$

I think that you can prove it in the following way. Every non-orientable manifold $Y$ has an orientable double cover $\tilde{Y}$. It should be possible to prove that if you have a map from an orientable manifold $X$ to $Y$, then this lifts to the orientable double cover, and so the degree of the map $f : X \to Y$ must be even.

Recall that a map $f : X \to Y$ lifts to $\tilde{Y}$ if and only if we have that $$ f_*\pi_1X \subset p_*\pi_1\tilde{Y} $$ where $p : \tilde{Y} \to Y$ is the covering map. This subgroup consists of exactly those loops in $Y$ which do not reverse orientation.

So it would suffice to show that if you have a path in an orientable space $X$, then its image in $Y$ cannot reverse orientation. However, this would be true as long as you can represent the path $\gamma$ by one which is a local isomorphism (i.e. the map restricted to a neighbourhood of $\gamma$ is a covering map), which should be true.

$\endgroup$
3
  • 1
    $\begingroup$ It is not that easy: I can show examples of maps from orientable manifolds to non-orientable manifolds that don't lift. But of course, they are even (in fact, non-surjective). Example: Project the torus to $S^1$, and send $S^1$ to an orientation-reversing loop. $\endgroup$ – Marcos Cossarini Sep 26 '14 at 12:41
  • $\begingroup$ Also, one cannot in general modify a loop $\gamma$ in the domain of $f$ so that it avoids the singularities of $f$. For example, we can perturb the identity $1_{S^1}$ to obtain a singular map $g$. Then, the map $f=1_{S^1}\times g$ has a wrinkle that can't be avoided by curves homotopic to $\{p\}\times 1_{S^1}$. $\endgroup$ – Marcos Cossarini Sep 26 '14 at 12:50
  • $\begingroup$ I think that I was a little too focused on the case of algebraic varieties over $\mathbb{C}$, where the singularities are always in high enough codimension that it works. So I suppose this isn't really what you're looking for... $\endgroup$ – Simon Rose Sep 26 '14 at 13:01

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.