boundary of a simply connected, compact manifold with boundary

Question

For a simply connected, compact n-manifold with boundary (n > 1), is its boundary connected?

It’s obviously false when n = 1, but how to prove or disprove the statement when n > 1? I’m especially interested in n = 2. Thanks.

Answer 1

$\mathbb{S}^{n-1}\times[0,1]$ is a counterexample for $n>2$.

Answer 2

Let $X$ be a simply connected and compact $2$-manifold, hence a surface. According to "Riemann Surfaces" by Simon Donaldson (See here, Chapter 7, page 92), we have the following inequality for surfaces: $$b_0(\partial X)\leq 2-\chi(X),$$ where $b_0(\partial X)$ (zeroth Betti number) denotes the number of path-connected components of $\partial X$. We need to compute those for $X$ to compute its Euler characteristic:

• Since $X$ is simply connected and therefore per definition path-connected, we have $b_0(X)=1$.
• Since $X$ is simply connected, we have $H_1^\text{sing}(X,\mathbb{Z})\cong\pi_1(X)^\mathrm{ab}\cong 1$ and therefore $b_1(X)=0$.
• Since the Betti number are the ranks of their respective homology groups, we have $b_2(X)\geq 0$.
• Since $X$ is a $2$-manifold, we have $b_n(X)=0$ for $n>2$.

We therefore have: $$\chi(X)=b_0(X)-b_1(X)+b_2(X)\geq 1$$ resulting in: $$b_0(\partial X)\leq 1,$$ which proves that $\partial X$ is path-connected and therefore connected.

Answer 3

$\require{AMScd}$ I believe that for $n=2$ there are no such manifolds $M$, at least if you require $M$ to be orientable. Assume such an $M$ exists and assume that we can decompose the boundary as $\partial M = U \sqcup V$, where now $U,V$ are connected $1$-manifolds. Then, using Poincaré-Lefschetz (which requires the orientability), we have the following commutative diagram, where the rows are long exact sequences of the pair $(M,\partial M)$: \begin{CD} \cdots @>>> H^1(M) @>>> H^1(\partial M) @>>> H^2(M,\partial M) @>>> \cdots\\ & @V\cong VV @V\cong VV @V\cong VV \\ \cdots @>>> H_1(M,\partial M) @>>> H_0(\partial M) @>>> H_0(M) @>>> \cdots\\ \end{CD} Now we observe that:

• $H^1(M) \cong 0$, since $M$ is simply connected.
• $H_0(M) \cong \mathbb{Z}$, since $M$ is in particular connected.
• $H_0(\partial M) \cong \mathbb{Z}^2$, by definition of $\partial M$.

It follows that $H_1(M, \partial M) \cong 0$ and hence by exactness the homomorphism $$ H_0(\partial M) \cong \mathbb{Z}^2 \to \mathbb{Z} \cong H_0(M) $$ is injective. But this is a contradiction, because there are no injective homomorphisms $\mathbb{Z}^2 \to \mathbb{Z}$.

I hope this was helpful!

Edit: As user ecrin points out, the orientability of $M$ follows from it being simply connected. Then my proof should work in general.

Answer 4

Here's a slightly more geometrically flavored argument for the case $n=2$. Let $M$ be a simply connected, compact $2$-manifold with boundary. Its boundary $\partial M$ is a compact $1$-manifold (without boundary), hence it has finitely many components, each homeomorphic to $S^1$ by the classification of $1$-manifolds. If $S^1\cong N\subseteq\partial M$ is such a component, we can form the manifold $M\cup_ND^2$, which is a compact $2$-manifold with one less boundary component. Furthermore, it is the union of embedded copies of the simply connected $M$ and $D^2$ intersecting along an embedded copy of the path-connected $S^1$, so it is simply connected (to be rigorous, one has to use collar neighborhoods here). Repeating this step finitely many times, we obtain a simply connected, compact $2$-manifold $M^{\prime}$ without boundary, so $M^{\prime}\cong S^2$ by the classification of surfaces. Now, working backwards, we see that $M$ is $S^2$ with a finite number of open disks having disjoing closures removed, i.e. it is has the homotopy type of a finitely punctured sphere (the number of punctures being the same thing as the number of boundary components of $\partial M$). However, this homotopy type is simply connected if and only if the number of punctures is $0$ or $1$, so $\partial M$ is connected. More precisely, this argument shows that $M\cong S^2$ or $M\cong D^2$.