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Suppose you are given two oriented manifolds with boundary $M$ say $B, B'$ and $\partial B = M = \partial B'$. Identify the boundaries and form $C = B \sqcup_{Id: M \to M} B'$. I want to see why, with the orientation induced by being submanifolds of $C$, $B$ and $B'$ induce opposite orientations to their boundary $M$. I'm particularly interested in the way the fundamental classes of $C,B,B'$ and $M$ behave. Thanks a lot!

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  • $\begingroup$ The phrase "induce opposite orientations to their boundary" is a little confusing to me. I guess this means that if you give the boundary the standard outward normal orientation, it will be opposite depending on whether you give it induced from $B$ or $B'$? This seems kind of intuitive, right? The outward normals are opposite, since you glued along $M$... $\endgroup$ – Matt Feb 4 '11 at 17:09
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I'm assuming you're JuanOS from MO, and this question corresponds to the MO question: https://mathoverflow.net/questions/54278/orientation-of-a-glued-manifold

Here's how I interpret your question. You have an oriented manifold $C$ which is compact without boundary, and you've decomposed it into the union of two submanifolds $B$ and $B'$ with $B \cap B' = M$, $M$ a compact manifold. Let $n=dim(C)$, so $n-1=dim(M)$. The global orientation class for $C$ is a generator $\mu_C \in H_n C$. There are restriction maps $H_n C \to H_n(B,M)$ and $H_n C \to H_n(B',M)$ which give the corresponding global orientations $\mu_B$ for $B$ and $\mu_{B'}$ for $B'$ respectively. Then there is the pairs $M \to B \to (B,M)$ and $M \to B' \to (B',M)$ and you want to know how the two generators of $H_n(B,M)$ and $H_n(B',M)$ compare when mapped to elements of $H_{n-1}M$ via the two connecting maps for the above pairs, specifically you want to show that $\partial \mu_{B} + \partial \mu_{B'} = 0$. Moreover, you want an argument that's fairly generic, in particular not specific to triangulated smooth manifolds or anything like that.

The above "restriction maps" are formally induced by inclusion $C \to (C, C \setminus int(B') ) \leftarrow (B,M)$, one being an excision inclusion, the other just an inclusion.

I don't believe this is as complicated as Kuperberg makes out -- the complication comes when attempting to bridge the gap between the smooth or simplicial views with the strictly homological view, especially say in the singular homology setting. But the above formulation side-steps those complications as your question is phrased entirely in terms of Mayer-Vietoris type constructions.

Okay, so here's a cheap way to check that it's true. Since $M$ is a submanifold of $C$, given a point $p \in M$ you can find an orientation-preserving degree 1 map $f : C \to S^n$ such that $f(M) \subset S^{n-1} \times \{0\} \subset S^n$. Moreover, you can ensure $f$ when restricted to $M$ is an isomorphism on the top homology groups of $M$ and $S^{n-1}$ respectively, and that $f$ sends $B$ to the top hemi-sphere, and $B'$ to the bottom hemi-sphere. So by naturality, you've reduced your problem to the case $D^n \sqcup_{S^{n-1}} D^n = S^n$, i.e. $C= S^n$, and $B$ and $B'$ both discs, which one way or another boils down to a cellular homology computation (we are using singular homology but what I'm saying is this is singular homology of a CW-complex to effectively cellular homology). This is the equivalent step to using outward pointing normals and determinants for smooth manifolds.

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  • $\begingroup$ Thanks, yet again, Ryan! Maybe this is too much but how might one prove that that the result of the composition $H_n(C) \to H_n(C,C\setminus int(B')) \cong H_n(B,M)$ is actually the fundamental class of $(B,M)$? and is the connecting homomorphism of the long exact sequence associated to $M \to B \to (B,M)$ the same as the connecting homomorphism of the Mayer-Vietoris sequence of the triple (C;B,B')? $\endgroup$ – Jsos88 Feb 6 '11 at 5:58
  • $\begingroup$ Oh, this is relatively easy. A local orientation at $p$ is a generator of $H_n(C,C\setminus \{p\})$. A global orientation of a manifold $C$ is a class in $H_n(C)$ that restricts to local orientations at all $p$. So the map $H_n(C) \to H_n(B,M)$ gives little commutative diagrams of maps to $H_n(C, C\setminus\{p\}) = H_n(B, B \setminus \{p\})$ where we identify these two objects via excision inclusions. $\endgroup$ – Ryan Budney Feb 6 '11 at 7:53
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I don't know if this answers what you meant by "why", but to my mind this is why the orientations are opposite: Given a point $P$ on the boundary, consider a neighbourhood $U$ of $P$ in $C$ which is homeomorphic to an open ball in $\mathbb{R}^n$, with the boundary mapped to a hyperplane bisecting the ball. An orientation of $C$ determines an orientation of this ball. For the ball, it is clear that its two halves induce opposite orientations on the hyperplane bisecting it. Then this has to be true also of the intersections of $M$, $B$ and $B'$ with $U$, and hence, since $P$ was arbitrary, of $M$, $B$ and $B'$ in general.

If this wasn't the sort of answer you were looking for, please elaborate.

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  • $\begingroup$ Thanks joriki, Ryan's answer is closer to the one i seek. $\endgroup$ – Jsos88 Feb 6 '11 at 5:59
  • $\begingroup$ OK -- you would have saved me the work of writing this up if you'd written here what you wrote on MO, or provided a link to that discussion :-) $\endgroup$ – joriki Feb 6 '11 at 9:31
  • $\begingroup$ Sorry! Will do in the future $\endgroup$ – Jsos88 Feb 6 '11 at 18:40

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