I know that for any surfaces, $K, L$ forming $K$#$L$ (where # denotes the connected sum) is done by deleting a disk from each and gluing together the boundary. I also know a few facts about the homology groups of surfaces:

1) $H_0(K)=H_0(L) \simeq \mathbb Z,$ since each is connected, the connected sum is connected.

2) $H_1(K$#$L) = H_1(K) \bigoplus H_1(L)$

3)$H_2(K$#$L)$ is either isomorphic to $\mathbb Z$ or to 0, depending on the orientability of $K$ and $L$.

I came to this conclusion (which my professor has confirmed) by considering the case of connected sum of tori, and the connected sum of projective planes, and using the fact that all surfaces are homeomorphic to one of these, or the sphere.

This seemed pretty okay when dealing with connected sums of tori and projective planes. The problem arises when dealing with classifications of surfaces of the form $T$#$P$, where $T$ is a torus and $P$ is a projective plane. To illustrate, we note that $H_1 (T)= \mathbb Z \bigoplus \mathbb Z,$ and that $H_1(P) = \mathbb Z/ 2 \mathbb Z$

So by the above, forming the connect sum would give $H_1(T$#$P)=\mathbb Z \bigoplus \mathbb Z\bigoplus (\mathbb Z /2\mathbb Z)$.

However, it a well known fact that $T$#$P$=$P$#$P$#$P$. Using statement (2) above on the right of this equation gives $H_1$$(P$#$P$#$P)$=$(\mathbb Z/ 2 \mathbb Z) \bigoplus (\mathbb Z/ 2 \mathbb Z) \bigoplus (\mathbb Z/ 2 \mathbb Z)$

But the problem is this. We computed the homology of both sides of the above equation, and together they imply that:

$\mathbb Z \bigoplus \mathbb Z\bigoplus (\mathbb Z /2\mathbb Z) = (\mathbb Z/ 2 \mathbb Z) \bigoplus (\mathbb Z/ 2 \mathbb Z) \bigoplus (\mathbb Z/ 2 \mathbb Z)$.

But this is false, since the group on the right side have exactly 8 elements, and the group on the left has an infinite number of elements.

So the question is, what went wrong in this analysis?

  • $\begingroup$ What you are discovering is that some of the patterns that you perceive to hold in simpler situations (orientable surfaces) fall apart when they are generalized to other situations (non-orientable surfaces). The bottom line is to use the Mayer-Vietoris theorem: that's a general pattern which holds universally. $\endgroup$ – Lee Mosher Jan 18 '15 at 16:28

2) is not true: $X\#Y \cup D \cong X\vee Y$ (the first union is along the boundary of the disc D along which we perform the connected sum) and from the Mayer-Vietoris sequence ($n$ is the dimension of the manifold): $$0\rightarrow H_n(X\#Y)\oplus H_n(D) \rightarrow H_n(X\vee Y) \rightarrow H_{n-1}(S^{n-1}) \rightarrow H_{n-1}(X\#Y) \oplus H_{n-1}(D) \rightarrow H_{n-1}(X\vee Y)\rightarrow 0$$ And you can see how the (n-1)th homology group changes. For instance, if both manifolds are not orientable, then 2) cannot hold.

  • $\begingroup$ According to this, topospaces.subwiki.org/wiki/Homology_of_connected_sum , it appears that (2) is true if both are compact and at least one is orientable. This is the case for the torus and projective plane. $\endgroup$ – Alfred Yerger Jan 18 '15 at 1:23
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    $\begingroup$ But not the case of the three projective planes. $\endgroup$ – user17786 Jan 18 '15 at 8:37
  • $\begingroup$ AH you're right. That's what I forgot. $\endgroup$ – Alfred Yerger Jan 18 '15 at 17:42

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