5

Edit: I already received a good answer to my second question. I'd be interested in a hint about the first one, as well. Thanks in advance!

I'm interested in compact Riemann surfaces and their homology. In this question, Kundor proposes a nice drawing of the connected sum of tori, saying that it is clearer than the traditional drawing of regular $4g$-gon whose sides are to be identified.
At first I was convinced that this was a smarter way to draw the $4g$-gon (just a continuous deformation), but then I realized that this is not the case (it works for genus 2, though).

So I was wondering if there exists a way to make the transition from the $4g$-gon to the "connected sum of squares", that makes the construction of the genus $g$ surface a lot easier to visualize.

Also, I had in mind a theorem, saying that the homology groups of the connected sum of two spaces are the direct sum of the homology groups of the single spaces (well, apart from $H_0$ and $H_n$).

Only, I wasn't able to find a reference: I was convinced I saw the result in Nakahara, Geometry, Topology and Physics, but I couldn't find it anymore. After some googling, I found these two sources (link1, link2), but nothing conclusive on any book I consulted.

Does this theorem have a name? Do you know a book where it is stated/proved?

Community
  • 1
Andrea Orta
  • 1,122

1 Answers1

10

Computing the homology of a connected sum is a matter of applying Mayer--Vietoris.

We have (let's say closed, connected, orientable) manifolds $M_1$ and $M_2$. In each of them we choose an $n$-ball $B_i$, and a slightly smaller $n$-ball $B_i'$ contained in $B_i$ (here $i = 1,2$).

For a moment, let's omit the subscripts and just let $M$ be a closed $n$-manifold. Let $B$ be an $n$-ball in $M$, let $B'$ be a slightly smaller ball contained in $B$, so that $B \setminus B'$ is a spherical shell, while $M = M\setminus B' \cup B.$ Then by Mayer--Vietotris, one computes that $H_n(M\setminus B') = 0$, while $H_i(M \setminus B') \cong H_i(M)$ for $i < n$.

Now reintroduce the subscripts: the connected sum $M_1 \# M_2$ of $M_1$ and $M_2$ is obtained by gluing $M_1 \setminus B'_1$ and $M_2 \setminus B_2'$ along their (homeomorphic) spherical shells $B_1 \setminus B_1'$ and $B_2 \setminus B_2'$.

Applying Mayer--Vietoris again, we find that $H_n(M_1 \# M_2) = H_0(M_1 \# M_2) = \mathbb Z$, and that $H_i(M_1 \# M_2) = H_i(M_1) \oplus H_i(M_2)$ if $0 < i < n$.

Matt E
  • 123,735
  • Thanks, this is interesting. Have you got any advice on my first question, please? – Andrea Orta Jul 27 '13 at 13:31
  • @Matt E: Is the orientability assumption needed for your argument in the connected sum? in other words, if you were to assume that $M_1$ and $M_2$ are non-orientable, can you still glue them easily along the boundary and apply Mayer-Vietoris? that seems to be not the case, but i don't see where your argument fails. – adrido Jan 31 '14 at 12:27
  • @adrido: Dear adrido, Implicit in my argument is that the boundary map $H_n(M) \to H_{n-1}(B\setminus B') = H_{n-1}(S^{n-1})$ is an isomorphism. This uses the the description of the fundamental class in the top cohomology of a closed connected orientable manifold. If $M$ were not orientable, then this map certainly wouldn't be an isomorphism, and what we would find instead is that $H_{n-1}(M\setminus B')$ is different from $H_{n-1}(M)$. E.g. think about removing a disk from the real projective plane to get a Mobius band.
    Regards,     – Matt E Jan 31 '14 at 19:49
  • Dear @Matt: this makes sense. Thank you! but I do not know how to compute $H_{n-1}(B\backslash B')$ in the case of non-orientable manifolds. My main problem is that I don't know how to picture the intersection. Can you please tell me how this is done? – adrido Feb 03 '14 at 01:33
  • @adrido: Dear adrido, $B$ is still a ball, and $B'$ is still a smaller ball, so $B\setminus B'$ is still a spherical shell (i.e. an $n-1$-sphere times an interval). The ambient manifold $M$ doesn't play any role in this computation. Regards, – Matt E Feb 03 '14 at 02:50
  • @Matt: Dear Matt, thank you for the clarification. how about the map $H_n(M)\to H_n(S^{n-1})$? it is not an iso. as you mentioned, but how do I compute it? – adrido Feb 03 '14 at 03:02
  • @adrido: Dear adrido, For a non-orientable manifold, the top homology vanishes, so it is just the zero map from $0$ to $H_{n-1}(S^{n-1}) = \mathbb Z$. Regards, – Matt E Feb 03 '14 at 04:06
  • @adrido: Dear adrido, For an orientable manifold, the basic theory of the fundamental class shows that $H_n(M)$ maps isomorphically to $H_{n-1}(S^{n-1})$ (indeed, this is part of how you show that $H_n(M) = \mathbb Z$). Thus the Mayer--Vietoris sequence shows that deleting the ball just kills $H_n$, but doesn't affect other homology. However, in the non-orientable case, deleting the ball adds $1$ to the rank of $H_{n-1}$. Again, I'd encourage you to work out carefully the case of removing a disk from $\mathbb R P^2$. Regards, – Matt E Feb 03 '14 at 04:11
  • @Matt: Dear Matt, now I see it! When I remove a disk from $\mathbb{R}P^2$, I send $(\mathbb{R}P^2\backslash B)\cap B'$ to the boundary circle of the Mobius band. Then Mayer-Vietoris gives $H_1(\mathbb{R}P^2\backslash B)=\mathbb{Z}$ (which we already knew). But this helps me see the exactness of $$0\to \mathbb{Z}\stackrel{(2,2)}{\to} \mathbb{Z}\oplus \mathbb{Z}\to H_1(\mathbb{R}P^2#\mathbb{R}P^2)\to 0.$$ This gives $$H_1(\mathbb{R}P^2#\mathbb{R}P^2)=\mathbb{Z}\oplus \mathbb{Z}_2.$$ Thank you for your patience! best – adrido Feb 03 '14 at 05:28
  • @adrido: Dear adrido, You're welcome. Best wishes, – Matt E Feb 03 '14 at 05:40
  • 1
    @MattE dear MattE, thanks for this precise answer, could you please give me a reference for the result that $H_n(M)$ maps isomorphically to $H_{n-1}(S^{n-1})$? because I proved the result of $H_n(M) \cong Z$ using compactly supported section if the orientation bundle... And can you explain to me, how do you apply such result to the boundary map (which is purely algebraic)? thanks in advance! – Luigi M Apr 12 '15 at 08:38