# Torsion in a manifold embedded in the 3 sphere

This is an old qual question.

Let $K$ be a simplicial complex homeomorphic to the $3$-sphere. Let $L$ be a subcomplex homeomorphic to a manifold with nonempty boundary. Show that $H_1(L,\mathbb Z)$ has no torsion.

I am totally lost as to how to go about this question. Here is what I have tried/know:

First, the torsion in $H_1(L,\mathbb Z)$ is the same as the torsion in $H^2(L,\mathbb Z)$.

Second, since $L$ is a manifold, we have poincare duality $H_1(L,\mathbb Z) \cong H^2(L,\partial L)$ and $H^2(L,\mathbb Z) \cong H_1(L,\partial L)$.

Third, we have a long exact sequence for the pair $(L,\partial L)$.

However, I am not sure how to put this information together or how to use that it has a CW structure/substructre...

• Can you use Alexander duality? Commented Aug 6, 2018 at 18:19
• @SteveD I would guess that is beyond 'qual material', but it does provide a very nice and quick proof.
– user98602
Commented Aug 6, 2018 at 18:26
• Alexander duality should probably be okay and I did consider using it but I didn't quite see how @SteveD Commented Aug 6, 2018 at 18:37

I would want to run Mayer-Vietoris on $L$ and $L' = \overline{S^3 \setminus L}$, which have $L \cap L' = \partial L$, some oriented surface. I prefer taking reduced homology.

Then we see $0 \to H_1(\partial L) \to H_1(L) \oplus H_1(L') \to H_1(S^3) \to H_2(\partial L)$ is exact. In particular, since $H_1(S^3) = 0$, $H_1(\partial L) \to H_1(L) \oplus H_1(L')$ is an isomorphism, and since $H_1(L)$ is a summand of a free abelian group, it must itself be free abelian.

To incorporate Steve D's comment, $L$ is a nice (compact, locally contractible) subset of $S^3$, so Alexander duality dictates that $H_k(L) \cong H^{3 - k - 1}(S^3 \setminus L)$. Because $S^3 \setminus L$ is homotopy equivalent to $L'$, this says $H_1(L) \cong H^1(L')$. (Note that this says more than the Mayer-Vietoris argument above!) Now the universal coefficient sequence dictates that $H^1(L') \cong \text{Hom}(H_1 L', \Bbb Z)$ - there is no Ext term because $H_0$ is always free. In particular, this term $\text{Hom}(H_1 L', \Bbb Z)$ is torsion free; we know it is a finitely generated free abelian group because $L'$ is compact.

• Ah very nice, the only real piece of information we need here seems to be that it is contained in $S^3$ and that $\partial L$ is a oriented surface. Commented Aug 6, 2018 at 18:40
• @ArithmeticGeometer And that it is well-enough embedded so that the closures of the two sides are still manifolds (you could imagine something nasty like the Alexander horned ball). I will edit to incorporate the Alexander duality comment.
– user98602
Commented Aug 6, 2018 at 18:41
• Everything in sight here uses $\Bbb Z$ coefficients.
– user98602
Commented Aug 6, 2018 at 18:44
• Notice that this says simplicial, which is even stronger than cellular: the picture is that there is some codimension-1 hypersurface in $S^3$, made of linear simplices, that cuts it into two pieces. But the inclusion of a sub-CW-complex into the whole thing is a well-behaved embedding, like you say.
– user98602
Commented Aug 6, 2018 at 18:50
• @Arithmetic I had imagined that you were only talking about codimemsion 0 submanifolds! My mistake. Your fix sounds like a good idea:in general, pick a small open neighborhood of L and take the closure. (You could do this by taking a fine triangulation and taking the 'star' of L.) Then the new object is a codimemsion 0 submanifold homotopy equivalent to L, so everything works.
– user98602
Commented Aug 7, 2018 at 13:59