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I am looking for a compact $2k+1$-dimensional manifold $M$ ($k\ge1$) with boundary $\partial M$, such that

a) $H_k(\partial M;\mathbb{Q}) \neq 0$,

b) $\iota_*\colon H_k(\partial M;\mathbb{Q})\rightarrow H_k( M;\mathbb{Q})$ is injective. (Where $\iota\colon \partial M\hookrightarrow M$ is the inclusion).

I'm mainly interested in seeing whether a manifold like that even exists, so one particular example would be good enough. So far I have considered:

  • $\partial M$ a closed surface of genus $g$, embedded into $\mathbb{R}^3$ in the usual way, such that it bounds a compact $3$-manifold $M$. For $g=0$, condition a) is not satisfied and for $g>0$ condition b) fails because $\dim H_1(M;\mathbb{Q})=g$ and $\dim H_1(\partial M;\mathbb{Q})=2g$.

  • $M=X\times Y$, where $X$ is closed and even (resp. odd) dimensional and $Y$ has a boundary $\partial Y$ and is odd (resp. even) dimensional. Looking at the Künneth theorem it seems like injectivity of $H_i(\partial Y;\mathbb{Q})\rightarrow H_i( Y;\mathbb{Q})$ is required in all degrees $i$, so the problem becomes even harder (?). In order to find a low dimensional example I've considered $Y$ to be a bounded planar domain, but then again again b) fails for dimension reasons.

Questions:

  • How can I approach this problem and find an example for such an $M$?

  • Are there any weird ways a surface can bound a $3$-manifold different from the above mentioned?

  • Is there a table of manifolds with boundary and their homology that I can consider?
Jan Bohr
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    If I recall correctly, you can use the long exact sequence of the pair and Poincare duality to show that half of the homology has to get killed by inclusion. – Cheerful Parsnip May 15 '18 at 19:56

2 Answers2

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For three dimensional manifolds there are no examples. Here I assume $M$ is orientable and $\partial M$ is connected. Then looking at the long exact sequence of the pair, we get $$0\to H_2(M)\to H_2(M,\partial M)\to H_1(\partial M)\to H_1(M)\to H_1(M,\partial M)\to 0$$ Let $a$ be the rank of $H_2(M)$ and $b$ the rank of $H_2(M,\partial M)$. Then by Poincare-Lefschetz duality, $b$ is the rank of $H_1(M)$ and $a$ is the rank of $H_1(M,\partial M)$. Let $c$ be the rank of $H_1(\partial M)$. Then since the sequence is exact, the Euler characteristic is $0$, meaning $c=2b-2a$. Now if $H_1(\partial M)\to H_1(M)$ were injective, then $H_1(M)\cong H_1(M,\partial M)$ by the above sequence, implying $b=a$. But then $c=0$. I think you can generalize this to higher dimensions but I'm running short on time.

Cheerful Parsnip
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  • Thanks, I was not familiar with Poincare-Lefschetz-duality! It's easy to generalize this to higher dimensions, see my answer. I'm still wondering if it is possible to exactly determine the rank of the kernel of $H_k(\partial M)\rightarrow H_k(M)$ in higher dimensions. You've suggested that it's half of the rank of $H_k(\partial M)$, but I don't see how this would follow from the duality results. – Jan Bohr May 16 '18 at 09:26
  • Why is there a zero instead $H_2(\partial M)$ of on the left of your exact sequence? – Georges Elencwajg May 16 '18 at 09:38
  • @ Georges: I've been wondering the same thing and I think that the sequence above is wrong. However, since we're just interested in the ranks, it should still work out as described in my answer. – Jan Bohr May 16 '18 at 09:58
  • I still think we can maybe construct a 5 manifold...lets start with $\mathbb {CP^2}$ #$ \bar{\mathbb CP^2}$... it bounds a 5 manifold..if we somehow surgery out the 3 cells of that 5 manifold, then we can get our desired manifold...I am trying to figuring out the way I can do this surgery...you can think in this direction as well – Anubhav Mukherjee May 16 '18 at 14:25
  • @ Anubhav: Would this $5$-manifold then be non-orientable? Or do you claim that the argument suggested by Cheerful Parsnip does not generalize to higher dimensions? – Jan Bohr May 16 '18 at 15:33
  • @Jan: $X:=\mathbb{C}P^2\sharp \overline{C}P^2$ bounds an orientable manifold, (in fact, $X$ is a linear $S^2$ bundle over $S^2$, then just fill in the fibers). So it seems as though your argument shows that the surgery approach won't work. (I don't know anything about surgery theory, so I cannot say why it wouldn't work as Anubhav suggests) – Jason DeVito - on hiatus May 16 '18 at 16:24
  • It is easy to construct such a manifold...take $\mathbb CP^2 \times I$ and dig a hole from one side to another – Anubhav Mukherjee May 16 '18 at 17:50
  • @GeorgesElencwajg: I omitted the $H_3(M,\partial M)$ and $H_2(\partial M)$ terms since they are both rank $1$ and the connecting homomorphism is an isomorphism. – Cheerful Parsnip May 16 '18 at 18:11
  • Anyway I understand. Lefschitz dualityu will give me an obstruction for this surgery – Anubhav Mukherjee May 16 '18 at 19:34
  • @Cheerful Parsnip: what you write in the comment may be true but it is not obvious and it would be helpful to justify these assertions. – Georges Elencwajg May 16 '18 at 21:32
  • @GeorgesElencwajg: These are fairly standard observations (think fundamental class), but as the OP realized, you don't really need them for the argument to work. You just have to include the $0$-dimensional homology in the sequence as well. – Cheerful Parsnip May 16 '18 at 21:41
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Cheerful Parsnips argument generalizes to higher dimensions: Again assume that $M$ is orientable and $\partial M$ is connected. Denote $a_i=\mathrm{rank}(H_i(\partial M))$, $b_i=\mathrm{rank}(H_i( M))$, $c_i=\mathrm{rank}(H_i(M,\partial M))$, then by Poincare duality for $\partial M$ we have $a_i=a_{n-1-i}$ and by Poincare-Lefschetz duality for $(M,\partial M)$ we have $b_{n-i}=c_{i}$.

Assuming that $H_k(\partial M)\rightarrow H_k(M)$ is injective, the long exact sequence for the pair $(M,\partial M)$ breaks up into two pieces and yields $$ 0=\sum_{i=0}^k (-1)^ i a_i - \sum_{i=0}^k (-1)^ i b_i + \sum_{i=0}^k (-1)^ i c_i $$ and $$ 0=\sum_{i=k+1}^{n-1} (-1)^ i a_i - \sum_{i=k+1}^n (-1)^ i b_i + \sum_{i=k+1}^n (-1)^ i c_i. $$ Change indices in the latter equation and use the duality results to obtain $$ 0=\sum_{i=0}^{k-1} (-1)^ i a_i + \sum_{i=0}^k (-1)^ i c_i - \sum_{i=0}^k (-1)^ i b_i, $$ where the sign changes are due to the fact that $n$ is odd and $n-1$ is even. Subtract this from the first equation, then $$ a_k=0 $$ follows.

EDIT: In the comments it was claimed that $d:=\mathrm{rank}\left(\ker(H_k(\partial M)\rightarrow H_k(M)\right)=\frac{1}{2}\mathrm{rank}H_k(\partial M)$, this also follows by taking the long exact sequence for $(M,\partial M)$ and spitting it into:

$$\dots \rightarrow H_{k+1}(M,\partial M) \rightarrow\ker(H_k(\partial M)\rightarrow H_k(M)) \rightarrow 0 $$ and $$ 0\rightarrow \ker(H_k(\partial M)\rightarrow H_k(M)) \hookrightarrow H_k(\partial M)\rightarrow H_k(M) \rightarrow \dots $$ and then doing essentially the same computations as above to obtain $a_k-2d=0$.

Jan Bohr
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