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Studying Algebraic Topology, I learned that Singular Homology (over de integers) it's supposed to represent the number of "holes" the space has. The main examples are $$H(S^1,\Bbb Z)=(\Bbb Z,\Bbb Z,0,0,...) \quad \quad H(S^2,\Bbb Z)=(\Bbb Z,0,\Bbb Z,0,0,...) \quad \quad H(S_1 \times S_1)=(\Bbb Z,\Bbb Z^2,\Bbb Z,0,0,...)$$ And they all (kind of) make sense. Now, the first wierd example is the Proyective Space with $$H(\Bbb P^2(\Bbb R),\Bbb Z)=(\Bbb Z,\, \Bbb Z/2 \Bbb Z \, ,\Bbb Z,0,0,...)$$ Which makes sense, and at the same time it doesn't. What I find "wierd" it's that there is a (differential) embedding of $\Bbb P^2(\Bbb R)$ into $\Bbb R^4$ so there is a reglular submanifold of $\Bbb R^4$ with a... $\Bbb Z/2 \Bbb Z$-hole?

In my goal of finding intuition of what on earth could that mean, I started wondering if it exists a regular differential submanifold of $\Bbb R^3$ who has an singular homology group with torsion. In that case, I could visualize how a $\Bbb Z/n \Bbb Z$-hole looks like. So the question is, does such a Regular Sub-Manifold of $\Bbb R^3$ exists?

By Regular Sub-Manifold of $\Bbb R^3$, I mean a $C^{\infty}$ real manifold contained in $\Bbb R^3$ with the subspace topology and an inmersive inclusion. The manifold could have boundary and need not to be compact nor orientable.

The Reason I'm interested in $C^{\infty}$ real manifold is because I find them more intuitive and less pathological than an arbitrary Topological subspace of $\Bbb R^3$. I want the Manifold to be a sub-manifold of $\Bbb R^3$ so I can visualize it.

symplectomorphic
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Marcos Martínez Wagner
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    https://math.stackexchange.com/questions/1170056/intuition-of-non-free-homology-groups?rq=1 This is a question about non-free homology groups, which is related to your question. The answer to that question actually explains that $\Bbb{R}P^2$ is an ideal example of understanding torsion despite the fact that it can't be embedded in $\Bbb{R}^3$. – Kevin.S Dec 30 '20 at 11:38
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    This is not possible for submanifolds of dimensions 0,1 or 2 (even if you only require them to be topological manifolds, though that makes the proof harder). This leaves open the case of submanifolds of dimension 3, i.e. open subsets of $\mathbb{R}^3$, to which I don't know the answer. – Thorgott Dec 30 '20 at 11:41
  • @Thorgott For 3-manifolds embedded in $\Bbb{R}^3$, I think even if there exists such spaces, they wouldn't be ideal for understanding torsion (OP's need), because of Alexander duality (probably). – Kevin.S Dec 30 '20 at 11:46
  • Ah right, Alexander duality actually rules open subsets of $\mathbb{R}^3$ out completely, so no such submanifold can exist in $\mathbb{R}^3$. – Thorgott Dec 30 '20 at 12:22
  • I know it's not possible for $0$ or $1$ dimensional manifolds. For 2-dimensional manifolds, if the manifold is compact and with no boundary, it's difeomorphic to a Torus with $g$ handles so it's homology is free, but what about not compact manifolds or manifolds with boundary? Also, I don't understand how to use the Alexander duality for three dimensional submanifolds in $\Bbb R^3$, aka open sets. (Thanks for the interest!) – Marcos Martínez Wagner Dec 30 '20 at 19:12

1 Answers1

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For the first question regarding the intuition behind elements of homology groups that admits finite order. This link is certainly helpful.

For your second question in the comment, I'll answer by parts under the assumption of finitely generated homology groups.

What about compact $2$ (resp. $3$) dimensional manifolds with boundary?

Suppose $X$ is any compact $2$-manifold or $3$-manifold that can be embedded in $\Bbb{R}^3$. Then Alexander Duality tells us that $$H^2(X)\cong \tilde{H}_0(S^3-X;\Bbb{Z})$$ which must be torsion-free. Then, by UCT (Universal Coefficient Theorem), $$ H^2(X)\cong\operatorname{Hom}(H_2(X);\Bbb{Z})\oplus\operatorname{Ext}(H_1(X);\Bbb{Z})\\ \implies H_1(X) \text{ torsion-free} $$ Besides, one may notice that if $\dim X=2$ and $X$ non-compact, we still have torsion-free $H_1$ by the same argument. (The only difference is that $H_2(X)\cong 0$ in that case, but it doesn't matter)

What about those cases when $X$ is a non-compact $3$ manifold? (This is the intriguing part of your question.)

Since your goal of finding such manifold is to understand the torsion of homology groups, I'll show that "normal" examples don't satisfy your criteria.

For simplicity, assume that $X=S^3-Y$, where $Y$ is a finite simplicial complex, then by Alexander Duality, we can still obtain the usual isomorphism relationship: $H_1(S^3-Y)\cong H^1(Y)$. Then by UCT, $$H^1(Y)\cong\operatorname{Hom}(H_1(Y);\Bbb{Z})\cong\operatorname{Hom}(\Bbb{Z}^\alpha\oplus T;\Bbb{Z})\cong \Bbb{Z}^\alpha$$ which is torsion-free. This result actually eliminate most common $3$ dimensional non-compact manifolds without boundary. Probably, there are some ugly spaces that admits torsion in their first homology group, but I think they wouldn't be suitable for visualizing elements of finite orders...

Besides the clear answer in the link, I'd like to discuss the cycles with finite order in terms of cellular structure (at least it's imaginable). During the process of constructing $\Bbb{R}P^2$, we often attach a $2$-cell via $f:\partial D^2\to S^1\to\Bbb{R}P^{n-1}$ (fill the circle with a disk and identify antipodes on $S^1$). Alternatively, this can be viewed as gluing the boundary of the disk to a circle via a degree $2$ map, which wraps around $S^1$ twice. By definition, a $1$-cycle must have an image $\partial(1\text{-cycle})=e^0-e^0=0$. In particular, only multiples of $e^1$ ($1$-cell) can be candidates.

  • For $m$ odd, $\partial(me^1)=0$ and $\not\exists n\in\Bbb{Z}$ s.t. $\partial(ne^2)=me^1$ since going around $S^1$ one time only goes half way around the only $2$-cell, which is not an element of $\Bbb{Z}$.
  • For $m$ even, then we see that it wraps around the $2$-cell $\frac{m}{2}\in\Bbb{Z}$ times because of the attaching map (degree $2$). Hence it is homologous to $0\implies$ triviality.

You may also connect this to the deformation of closed paths.

I hope this response answers your question. :)

Kevin.S
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    There's a more intrinsic way to deal with non-closed surfaces: every connected surface which is not closed (i.e. either not compact, or has nonempty boundary) deformation retracts to a 1-complex, hence its homology in all ranks is free abelian. – Lee Mosher Dec 31 '20 at 20:02
  • @LeeMosher Oh, thank you, I didn't catch that. But still, the case of $3$-manifold is partially open, cuz there might exist some strange space that admits torsion in its 1st homology group, and I don't know any intuitive example of that... – Kevin.S Jan 01 '21 at 00:35
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  • A manifold with boundary is homotopy equivalent to its interior, so we may assume boundaryless WLOG. 2. A compact manifold automatically has f.g. homology, so your argument takes care of those (alternatively, a compact 2-manifold embedding in R^3 is orientable, then quote classification). 3. Non-compact surfaces are dealt with by Lee Mosher's comment.
    • – Thorgott Jan 01 '21 at 15:36
  • Non-compact boundaryless submanifolds of R^3 are open subsets. I don't immediately see why you can apply Alexander duality directly. Is the complement necessarily locally contractible at the boundary? This isn't clear to me. OTOH, if you replace singular with Cech cohomology, this always works and tells us H_1 and H_2 are torsionfree, because Cech cohomology is always torsion-free in degrees 1 and 0. H_0 is free in any case and H_3 vanishes, because a noncompact (smooth) 3-manifold deformation retracts to a 2-dimensional CW-complex. So this takes care of everything.
    • – Thorgott Jan 01 '21 at 15:36
  • @Thorgott I understand. The reason I didn't replace singular with Cech cohomology is because the space $Y$ is assumed to be a "nice" simplicial complex (actually a polyhedron in $\Bbb{R}^3$). In this case Cech cohomology is isomorphic to simplicial and singular. Also, I think there's nothing wrong with $H^1(Y)\cong H_1(S^3-Y)$ because there's no assumption on $S^3-Y$, which is the complement that you mentioned. In general, the duality does not hold if Cech cohomology is simply replaced by singular cohomology. – Kevin.S Jan 02 '21 at 02:37
  • For Alexander duality to hold with singular cohomology, you want an assumption such as $Y$ being locally contractible and I don't see why that would be satisfied a priori. Of course, if $Y$ is a simplical complex, this is satisfied, but you mention the isomorphism before you mention the assumption on $Y$. Anyhow, the point of my comment was simply to ascertain that there are no open cases left. – Thorgott Jan 02 '21 at 02:54
  • @Thorgott Oh indeed, I should have mentioned $Y$ being a simpliciale complex before claiming the isomorphism. Thank you. – Kevin.S Jan 02 '21 at 02:57