# $\mathbb S^2$ or $\mathbb RP^2$ on boundary of a 3-manifold

Let $M$ be a 3-manifold with boundary $\partial M$. Suppose $\partial M$ contains a sphere or a projective plane, which is contractable in $M$. Show that $M$ is also contractable.

The above statement is shown in the proof of Sphere Theorem, so it should not be used to show the above argument.

## 2 Answers

As Stallings writes in his book Group theory and three-dimensional manifolds, where this proof of the Sphere Theorem originates, "It is easy to see that, if $$\partial M$$ contains a $$2$$-sphere which is contractible in $$M$$, then $$M$$ is itself contractible." I gave a talk about his proof recently, and this was the easiest way I could see it:

(Let $$M$$ denote a compact connected $$3$$-manifold in what follows.)

Claim. If there is a nullhomotopic $$S^2\subset \partial M$$, then $$\pi_1(M)=1$$.

Proof. If not, take copies $$M_1,M_2$$ of $$M$$ and join them along the $$S^2$$ boundary to form $$M'$$. By van Kampen, $$\pi_1(M')=\pi_1(M_1)*\pi_1(M_2)$$, which is infinite since we are assuming $$\pi_1(M)\neq 1$$, so the universal cover $$\widetilde{M'}$$ is non-compact. Lift $$S^2$$ to a sphere $$\Sigma\subset \widetilde{M'}$$, which separates $$\widetilde{M'}$$ since $$S^2$$ separates $$M'$$. Call the pieces $$N_1,N_2$$, which are both noncompact by construction. By excision with $$\Sigma$$ being thought of as the boundary of a $$3$$-ball, $$H_3(N_i,\Sigma)=H_3(N_i)=0$$, so the LES gives that $$H_2(\Sigma)\to H_2(N_i)$$ is injective. Mayer Vietoris gives $$0\to H_2(\Sigma)\to H_2(N_1)\oplus H_2(N_2)\to H_2(\widetilde{M'})$$ so the kernel of the map to $$H_2(\widetilde{M'})$$ is generated by $$([\Sigma],[\Sigma])$$, hence $$([\Sigma],0)$$ is not in the kernel, and so $$H_2(\Sigma)\to H_2(\widetilde{M'})$$ has non-trivial image. But $$S^2$$ being nullhomotopic in $$M\subset M'$$ implies $$\Sigma$$ is nullhomotopic in $$\widetilde{M'}$$ by lifting the nullhomotopy, implying $$[\Sigma]$$ is zero in $$H_2(\widetilde{M'})$$! Therefore, $$M$$ must have been simply connected.

Claim. If $$M$$ is simply connected, then $$M$$ is orientable.

Proof. If $$M$$ is non-orientable, then $$H^1(M;\mathbb{Z}/2\mathbb{Z})$$ is non-trivial since there is an oriented connected double-cover.

Claim. If $$M$$ is simply connected with a nullhomologous boundary component $$\Sigma\subset \partial M$$, then $$H_2(M)=0$$.

Proof. Let $$[M]\in C_3(M)$$ denote the fundamental class, and recall that $$\partial[M]$$ is a sum of fundamental classes of each boundary component. If $$\Sigma$$ is a boundary of some $$3$$-chain $$A$$, then $$A$$ is non-trivial in $$H_3(M;\partial M)=\mathbb{Z}$$, so $$A$$ and $$[M]$$ must be homologous, hence there is only a single boundary component. By Poincare duality, $$H_2(M)=H^1(M,\Sigma)$$, and the LES gives $$\widetilde{H}^0(\Sigma)\to H^1(M,\Sigma)\to \widetilde{H}^1(M)$$, and since $$\Sigma$$ is connected and $$M$$ is simply connected, we get $$H_2(M)=0$$.

Claim. If $$M$$ has a nullhomotopic $$S^2\subset \partial M$$, then $$M$$ is contractible.

Proof. We have that $$M$$ is simply connected with $$H_2(M)=0$$. By Hurewicz, $$\pi_2(M)=0$$. (Note that this is all we need for the Sphere Theorem, since $$\pi_2(M)\neq 0$$ is a hypothesis.) Since $$M$$ has boundary, $$H_3(M)=0$$. Again by Hurewicz, $$\pi_n(M)=0$$ for $$n\geq 3$$. Since $$M$$ has a CW structure, the map $$*\to M$$ is a homotopy equivalence by Whitehead's theorem, so $$M$$ is contractible.

Claim. If $$M$$ has an $$\mathbb{R}\mathrm{P}^2\subset\partial M$$ that as a map $$S^2\to M$$ is nullhomotopic, then $$\pi_2(M)=0$$.

Proof. Let $$P$$ denote this $$\mathbb{R}\mathrm{P}^2$$ in the boundary. If it were the case that $$\pi_1(P)\to \pi_1(M)$$ has trivial image, then the Loop Theorem gives an embedded disk $$D$$ in $$M$$ with $$\partial D$$ being nontrial in $$\pi_1(P)$$. The normal bundle of $$D$$ is trivial since $$H^1(D;\mathbb{Z}/2\mathbb{Z})=0$$, so $$D$$ orients the normal bundle for $$\partial D$$ in $$P$$, which is a contradiction. So, the oriented double cover of $$M$$ has $$P$$ lift to an $$S^2$$ in the boundary. By hypothesis, $$P$$ is nullhomologous in $$M$$, so this $$S^2$$ would be nullhomologous in the oriented double cover, and the preceding claims imply that the oriented double cover is contractible, hence $$\pi_2(M)=0$$.

Hence, if $$\pi_2(M)\neq 0$$:

• If there is an $$S^2$$ in the boundary, it is not nullhomotopic.

• If there is a projective plane in the boundary, its map $$S^2\to M$$ is not nullhomotopic.

That is, such boundary components satisfy the conclusion of the Sphere Theorem, when pushed into $$M$$ slightly so they become properly embedded $$2$$-sided surfaces.

First I'll prove that given the condition that a $S^2$ boundary contracts inside $M$ implies $M$ is simply connected.

If $M$ has a sphere in the boundary, then glue two copies of $M$ together with respect to boundary sphere. If $M$ is not simply connected then $\pi_1(M_1\cup M_2)$ is a free group. Then consider two non-trivial loops $a_1\in M_1$ and $a_2\in M_2$. Let $\gamma$ be $a_1*a_2$. Since $S^2$ is contractible in $M$, that implies we can do some homotopy such that $\gamma$ doesnot intersect $S^2$. Which implies $\gamma$ either lies in $M_1$ or $M_2$. And this contracdicts the free property of $\pi_1(M_1\cup M_2)$. Thus $M$ is simply connected.

If $M$ is simply connected with $S^2$ boundary, then Hurewicz theorem (since all its homology groups are zero by Poincare Dulality) implies that it is contractible.

BTW, $M$ cannot have an non-orientable boundary componenet such as $\mathbb RP^2$. Then $M$ will not be contractible, since any simply connected manifold is orientable. And boundary of any orientable manifold is orientable.

• very nice reasoning, but it doesn't work in case $M$ has another boundary components but one $S^2$ (in this case $M$ can be simply connected but not contractible) – Andrey Ryabichev Nov 8 '17 at 19:51
• How is it that $S^2$ being nullhomologous means there is a homotopy of $\gamma$ such that $\gamma$ does not intersect $S^2$? – Kyle Miller Feb 24 at 3:50
• @KyleMiller nullhomotopic is stronger than nullhomologous – Anubhav Mukherjee Feb 24 at 5:49
• @AnubhavMukherjee Sorry, I meant to type "nullhomotopic," or rather "contractible." I don't see how you can homotope $\gamma$ to avoid the $S^2$. Is there some result that the nullhomotopy of $S^2$ can be through isotopies? – Kyle Miller Feb 24 at 8:15
• @KyleMiller S^2 is homotopic to a point. And we can assume that S^2 and the loop $\gamma$ intersect transversally but then so is the loop and the point. So we can assume they are infact disjoint. – Anubhav Mukherjee Feb 24 at 16:06