I am attempting to show that the connected sum $M_1 \# M_2$ of two closed, non-orientable, connected $n$-manifolds has $H_{n-1} \cong H_{n-1}(M_1)\oplus H_{n-1}(M_2)$, IF one swaps a $\mathbb{Z}_2$ summand for a $\mathbb{Z}$. This is the last part of exercise 3.3.6 of Hatcher:

enter image description here

His hint is to use Euler characteristics. But, I do not see how they enter into this problem. I know that $M_1 \# M_2$ is non-orientable, thus its $H_{n-1}$ has torsion part $\mathbb{Z}_2$. I also know that the Euler characteristic of an odd-dimensional, non-orientable manifold is zero. But how can I use the Euler characteristic in this case to say something about the homology?


I am going to assume you already know $H_k(M_1\sharp M_2;R) \cong H_k(M_1;R)\oplus H_k(M_2;R)$ for any ring $R$ and any $0 < k < n-1$. I'm also going to assume all the cohomology groups are finitely generated.

Since Euler characteristic is independent of the field coefficients, let's take $R = \mathbb{Z}/2\mathbb{Z}$ and see what happens. I'm going to let $c_i$ denote the $i$-th Betti number with $\mathbb{Z}/2\mathbb{Z}$ coefficients, reserving $b_i$ for the usual rational Betti numbers. Note that $M_1\sharp M_2$ has Euler characteristic $\chi(M_1)+\chi(M_2) - (1+(-1)^n)$. Thus, $$\sum_{k=0}^n (-1)^k c_k(M_1\sharp M_2) = \left(\sum_{k=0}^n (-1)^k c_k(M_1) + (-1)^kc_k(M_2)\right) - (1+(-1)^n).$$

However, $c_0(M_1\sharp M_2) = c_0(M_1) + c_0(M_2) - 1$, while $c_k(M_1\sharp M_2) = c_k(M_1) + c_k(M_2)$ for $0 < k < n-1$. In addition, as you've already noted, $c_n(M_1\sharp M_2) = c_n(M_1)+c_n(M_2)-1$. Substituting this into the above, we deduce that $c_{n-1}(M_1\sharp M_2) = c_{n-1}(M_1) + c_{n-1}(M_2)$.

Now, think about universal coefficients theorem. Writing $H_{n-1}(M_1\sharp M_2)$ as a free part summed with the torsion $\mathbb{Z}/2\mathbb{Z}$, we find that $c_{n-1}(M_1\sharp M_2) = b_{n-1}(M_1\sharp M_2) + 1$, while $c_{n-1}(M_i) = b_{n-1}(M_i) + 1$.

So, we have $b_{n-1}(M_1\sharp M_2) + 1 = b_{n-1}(M_1) + 1 + b_{n-1}(M_2) + 1$, so $$b_{n-1}(M_1\sharp M_2) = b_{n-1}(M_1)+b_{n-1}(M_2) + 1.$$ In other words, $H_{n-1}(M_1\sharp M_2)$ has an extra $\mathbb{Z}$-summand, and, by the result you already mentioned, one less $\mathbb{Z}/2\mathbb{Z}$ summand.

  • $\begingroup$ One question. How are we using the UCT where you mention? $\endgroup$ Oct 5 at 1:30
  • $\begingroup$ @DescartesBeforetheHorse: Good question! That is definitely a mistake on my part: the formula $c_{n-1} = b_{n-1} + 1$ is only valid if we know that $H_{n-2}$ has no $2$ -torsion. However, the contribution from $Tor(H_{n-2}(M_1\sharp M_2), \mathbb{Z}/2\mathbb{Z})$ cancels with the contribution from $Tor(H_{n-2}(M_1),\mathbb{Z}/2\mathbb{Z})\oplus Tor(H_{n-2}(M_2), \mathbb{Z}/2\mathbb{Z})$ because $Tor$ distributes over direct sums in the first slot and we already know $H_{n-2}(M_1\sharp M_2)\cong H_{n-2}(M_1)\oplus H_{n-2}(M_2)$. $\endgroup$ Oct 5 at 1:40

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.