2
$\begingroup$

First, I know that any compact connected surface is either $\mathbb S^2$, $\mathbb P^2 \# \cdots \# \mathbb P^2$ ($n$ copies), or $\mathbb T^2 \# \cdots \# \mathbb T^2$ ($n$ copies).

Also, suppose I know the following result:

-Given $X$ is a connected space with a contractible universal covering space, $f : Y \to X$ is null homotopic if and only if the induced homomorphism $f_* : \pi_1(Y, y) \to \pi_1(X, f(y))$ is trivial for each $y \in Y$.

For the first case, I know that since the fundamental group of $\mathbb S^2$ is trivial, it must be that any continuous $f$ from $\mathbb S^2$ to $\mathbb S^1$ should be trivial.

For the second case, when $n$ is just one, since the fundamental group of the projective plane is just a group of order $2$.

However, I am not sure how to proceed with other cases, and show that there exists/does not exist non-null homotopic maps.

Also, what should I do to prove the fact I am assuming?

$\endgroup$
2
+50
$\begingroup$

(This should be a comment but I dont have enough reputation to make a comment)

The non-orientable surface $\Sigma$ has a contractible covering space hence it is $K(\pi_{1}(\Sigma),1)$ in order to give a map to $S^1$ which is $K(\mathbb{Z},1)$ you can give a map at the level of groups. As you pointed out the fundemental group of $\mathbb{RP}^2$#$\mathbb{RP}^2$ is the infinite dihedral group, hence you get a map to $\mathbb{Z}$.

Even in the case of orientable surfaces you can produce a lot of map by repeating the arguement above, as the universial cover of orientable surfuce of genus $\geq$ 2 is disc (or equivalently the upper half plane). Hence the question boils down to a purely group theoretical question. In particular if you have elements of infinite order in the fundemental group you have a map. Which might have several interesting geometric intepretations.

$\endgroup$
4
  • $\begingroup$ Why is having a non-trivial homomorphism between fundamental groups sufficient? I think it would be sufficient only if such homomorphism guarantees that there is a continuous map that induces it. $\endgroup$ Dec 6 '20 at 22:20
  • $\begingroup$ Because if you have two spaces which are $K(G,1)$ it is enough to give a map at the level of groups. If you have a group homomophism between two groups say H to G then you can make a principle G bundle over any space a principle H bundle via the homomorphism. But principle G bundles for discrete groups are classified by $K(G,1)$ hence it lift to the level of spaces. Hope this helps. $\endgroup$ Dec 7 '20 at 0:28
  • $\begingroup$ To prove the fact that the covering space, of compact surfaces of higher genus is contractible you may take a look at John Lee's book called Geometry of Surfaces. The basic idea is to get a tesselation of the disc or upper half plane with the $4g$-gon identification of the surface. $\endgroup$ Dec 7 '20 at 0:54
  • $\begingroup$ I am studying using John Lee’s introduction to topological manifold (this is the problem 11-21) and not sure if I ever saw words like $K(G, 1)$ or bundles yet. Is there a way to get the result you are mentioning with what is in the book I mentioned? $\endgroup$ Dec 7 '20 at 1:21
0
$\begingroup$

You can use the Van Kampen theorem to determine the fundamental group of a connected sum, this should answer your question for the projective planes.

For the torus, what do you think, if $n=1$, of the projection $\mathbb T^2 \cong S^1\times S^1\to S^1$ ? What does it tell you for $n\geq 2$ ?

To prove the result you are assuming, you should use the lifting criterion: suppose $\tilde X\to X$ is a covering space, with say both $\tilde X$ and $X$ connected; then what is a criterion for when a map $Y\to X$ can be lifted to a map $Y\to \tilde X$ ?

What does it tell you when $\tilde X$ is contractible ?

$\endgroup$
8
  • $\begingroup$ Could you elaborate bit more? I think I see thee case for $\mathbb P^2$, but not sure for $\mathbb P^2 \# ... \# \mathbb P^2$ in general. Also, for the torus case, do I look at $\mathbb T^2 \# ... \# \mathbb T^2 \to \mathbb T^2 \to \mathbb S^1$? $\endgroup$ Dec 5 '20 at 22:57
  • $\begingroup$ For the torus case yes, that is a possible example. For the projective plane, do you know how to compute the fundamental group of this connected sum ? Start with two projective planes $\endgroup$ Dec 5 '20 at 23:02
  • $\begingroup$ Shouldn't the fundamental group be $\mathbb Z / 2 * ... * \mathbb Z / 2$? $\endgroup$ Dec 5 '20 at 23:17
  • $\begingroup$ What should be the map from a connected sum of T^2 to T^2? Also, I am not sure how to show any continuous map from P^2 # ... # P^2 should be null-homotopic from the result I am using; for example, P^2 # P^2 has a fundamental group isomorphic to an infinite dihedral group, which has non-trivial homomorphism to Z. $\endgroup$ Dec 6 '20 at 5:11
  • $\begingroup$ So you figured out yourself that the answer was not the free product of Z/2's. In fact, for a connected sum of more than 1 projective planes there will be non nullhomotopic maps to $S^1$, using the fact that those come from maps on $\pi_1$. For tori, note that you can kill a copy of T^2 in the connected sum, and this leave you with a T^2 where a small disk has been shrunk to a point, but this is homotopy equivalent to T^2 $\endgroup$ Dec 6 '20 at 9:48

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.