# Find $(M,\partial M)$ with injective $H_k(\partial M)\rightarrow H_k(M)$.

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?
• 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. May 15, 2018 at 19:56

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.

• 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. May 16, 2018 at 9:26
• Why is there a zero instead $H_2(\partial M)$ of on the left of your exact sequence? May 16, 2018 at 9:38
• @ 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. May 16, 2018 at 9:58
• 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 May 16, 2018 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? May 16, 2018 at 15:33

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 $a_k-2d=0$.

• Glad to point you in the right direction. May 16, 2018 at 18:09