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Let $M$ be a compact, connected $n$-manifold. Consider the homology groups $H_n(M)$ with coefficients in $\mathbb{Z}$.

It is well known that if $M$ is not $\mathbb{Z}$-orientable, then we have $H_n(M) =0$ and $H_{n-1}(M) = \mathbb{Z}/2 \oplus \mathbb{Z}^i$ for some $i \ge 0$.

The proofs are clear for me (remark: the main instruments used in the proof are the Universal Coefficients Theorem and the existence of an orientable double cover of $M$), but I'm quite curious if there exists a geometric/intuitive explanation for the torsion summand $ \mathbb{Z}/2$ of $H_{n-1}(M)$.

Can this phenomenon be visualized in the case of a non-orientable $3$-manifold or is this a purely algebraic result?

Descartes Before the Horse
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user267839
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    This is a great question. If nobody else does, I'll write an answer later, though I only see how to do it using Poincare duality and cohomology rings. The general model is the Klein bottle, thought of as $S^1 \times [0,1]$ with the two sides glued by a reflection. Then $S^1 \times {1/2}$ is a closed submanifold which is homologically nontrivial. However, if you have two copies of those, eg $S^1 \times {.49, .51}$, then you could consider the cylinder wrapping around (but not including the bit between $.49$ and $.51$) as an oriented manifold they are the boundary of. –  Sep 22 '18 at 01:39
  • It is crucial here that we get an oriented manifold bounding the two sides! And in fact, the trick is that as you wrap around the two sides, the orientations on the circles swap. That's why this doesn't work on the torus (instead, one is essentially arguing that $[S^1 \times {1/2}] - [S^1 \times {1/2}]$ is zero in homology, which is obvious). –  Sep 22 '18 at 01:40
  • @Mike Miller: Ok, if I understand your construction of Klein bottle correctly then firstly you consider a cylinder as an orientated $2$-manifold with boundary consisting of the two disjunct circles $S^1 \times {0.49}$ and , $S^1 \times {0.51}$ and now you want to connect the boundary in a non orientable way as one expects for Klein bottle. But up to now I don’t see how this geometrical construction provides the mentioned $\mathcal{Z}/2$-torsion. Do you mean by the crucial point that the boundary here is not connected? – user267839 Sep 22 '18 at 03:13
  • The point is that if a closed oriented submanifold bounds an oriented submanifold with boundary, then it is zero in homology. I wanted to apply that to two copies of $S^1 \times {1/2}$. I replaced that with $S^1 \times {0.49, 0.51}$, which is homologically equivalent to two copies of $S^1 \times {0.5}$. I only wrote it like that to make the picture clearer (instead of thinking of two copies of the same circle). There were supposed to be two "crucial points": one is that the cylinder is oriented, and the other that the two boundary components are oriented the same. –  Sep 22 '18 at 04:17
  • That means this cylinder gives a null-bordism of $2 [S^1 \times {1/2}]$. So $[S^1 \times {1/2}]$ gives you a homology class, which is 2-torsion. –  Sep 22 '18 at 04:18

1 Answers1

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Here is an extended version of my, now deleted, comment.

Let $M$ be connected nonorientable a compact triangulated manifold (every topological 3-manifold admits a triangulation). A similar argument works for a manifold equipped with a CW complex structure, but it is less geometric in this case. I will work with simplicial homology.

Let $c\in C_n(M; {\mathbb Z})$ denote the chain equal to the sum of all top-dimensional simplices. The boundary of this chain is a cycle $b=\partial c$ with even coefficients. Therefore, $a=\frac{1}{2}b$ is still a cycle with integer coefficients. Since $M$ is unorientable, $a$ defines a nontrivial element of $H_{n-1}(M; {\mathbb Z})$. In order to prove this, you find a 1-cycle $e\in Z_1(M; {\mathbb Z}/2)$ (in the 1-skeleton of the dual triangulation) which has nonzero algebraic intersection number with $a$: Take $e$ which reverses orientation. From this, you see that $[a]\ne 0$ in $H_{n-1}(M; {\mathbb Z})$ as well. On the other hand, clearly, $2[a]=[b]=0$. Hence, $[a]$ generates ${\mathbb Z}_2$ in $H_{n-1}(M; {\mathbb Z})$.

I do not know how to see geometrically that $[a]$ generates a direct summand of $H_{n-1}(M; {\mathbb Z})$.

Moishe Kohan
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  • Hi, thank you for the answer. Two points aren't clear for me: Firstly, why the boundary of $c$ must have even coefficients? Do you mean that in the sense that each boundary of a $n$-cell is a intersection of two $n$-cells? And the second one: What is an algebraic intersection number in context with CW complexes? – user267839 Sep 22 '18 at 17:26
  • @KarlPeter: Yes, since every codimension 1 face f is the intersection of exactly two distinct facets $g_1, g_2$ , after taking boundary, the face $f$ appears with an even coefficient (no matter how $g_1, g_2$ are oriented). As for the second: The algebraic intersection number for general CW complexes is undefined. For CW complexes underlying manifolds, it takes some effort to define (involving approximating the attaching maps of cells), this is one reason I decided to restrict to simplicial complexes. – Moishe Kohan Sep 22 '18 at 17:31
  • ...and which definition of the a i n do you using in the case of simplicial complexes? I only know it from algebraic geometry / Bézout's theorem. – user267839 Sep 22 '18 at 17:53
  • @KarlPeter: If you have two mod 2 cycles in a triangulated manifold which intersect transversally then you simply count (mod 2) the number of intersection points. See the book of Guillemin and Pollack "Differential topology" in the case of smooth manifolds. – Moishe Kohan Sep 22 '18 at 18:48