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Boundary of a product manifold 
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#1
Mar1112, 12:32 AM

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Is it true, that if [itex]A[/itex] and [itex]B[/itex] are oriented manifolds with boundary, having dimensions [itex]n[/itex] and [itex]m[/itex] respectivelly, then the boundary of [itex]A\times B[/itex] is
[itex]\partial(A\times B)=\partial A\times B + (1)^n A\times \partial B[/itex]? If not, then what can we say about the boundary of product manifolds? Could someone recommend a textbook or a lecture notes that discusses this? 


#2
Mar1212, 01:36 AM

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Yes, that's how one typically orients the boundary of AxB.



#3
Mar1212, 04:00 AM

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Okay, but orientation is only one aspect of my question. I'm also curious that why is
[itex]\partial(A\times B)=(\partial A\times B) \cup (A\times \partial B)[/itex] 


#4
Mar1212, 09:18 AM

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Boundary of a product manifold
This should not be hard to prove directly: Assume dimA=m, dimB=n. Take a point (a,b) in dA x B. Two cases:
i) b is not a boundary point ii) b is a boundary point In case i), a nbhd of (a,b) is of the form R^{m1} x R_{+} x R^{n} ≈ R^{m+n1} x R_{+}. Hence (a,b) is in d(AxB). In case (ii), a nbhd of (a,b) is of the form R^{m1} x R_{+} x R^{n1} x R_{+}. You just have to convince yourself now that R_{+} x R_{+} ≈ R x R_{+} to see that the above nbhd is just R^{m+n1} x R_{+} again. Hence (a,b) is in d(AxB) in this case also. Etc. 


#5
Mar1312, 12:52 AM

P: 230

let's [itex] \psi_1: (R_+\backslash\{0\})\times [0,\pi] \to (R\times R_+)\backslash\{(0,0)\}[/itex]:[itex](r,\varphi)\mapsto (\cos\varphi,\sin\varphi)[/itex], [itex]\psi_2: R^2\to R^2: (r,\varphi)\mapsto (r,2\varphi)[/itex], then the mapping from [itex]R_+\times R_+[/itex] to [itex]R\times R_+[/itex] that leaves (0,0) fixed and the other points maps to [itex]\psi_1\circ\psi_2\circ\psi_1^{1}[/itex] is a diffeomorphism from [itex]R_+\times R_+[/itex] to [itex]R\times R_+[/itex] Your proof proves only that [itex]\partial(A\times B)\supseteq (\partial A\times B)\cup(A\times\partial B)[/itex], but I think that the reverse direction goes similarly. We can prove with your method that if (a,b) is not in the RHS set,then it also isn't in the LHS set. So you thought? 


#6
Mar1312, 07:40 AM

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Latex is not showing up, but whatever your map is, I doubt it is a diffeomorphism. Because any homeomorphism F: R_{+} x R  R_{+} x R_{+} must send boundary to boundary and hence it is not differentiable at F^{1}(0), where 0 is the "corner point" of R_{+} x R_{+}.
What I meant by "R_{+} x R_{+} ≈ R x R_{+}" was only a homeomorphism... because the boundary of a manifold, after all, is a topological concept. So far, the argument I gave shows [itex](\partial A\times B) \subset \partial(A\times B)[/itex] But it goes without saying that we can get [itex](\partial B\times A) \subset \partial(A\times B)[/itex] in the same way. Hence we have so far [itex]\partial A\times B + (1)^n A\times \partial B \subset \partial(A\times B)[/itex] For the reverse inclusion, pick (a,b) in [itex]\partial(A\times B)[/itex]. Then it has a nbhd of the form R_{+} x R^{m+n1}. Find out which of a or b has one of its coordinates in R+. If it is a, then (a,b) is in dA x B. It if is b, then (a,b) is in A x dB. 


#7
Mar1412, 12:36 AM

P: 230

Dear Quasar987, I greatly enjoy your concise and expressive explanations, they are really lucid, thank you for them.
Also thank you for your remark, that my map isn't a diffeomorphism. Really, as far as I see, however it is differentiable, it's inverse isn't at (0,0). (by the way, is it really a good definition for a bijective map to be smooth, that is sends every smooth curve into a smooth curve?) Of course I screwed up my formula, I wanted to write [itex](r \cos\varphi, r \sin \varphi)[/itex] instead of [itex](\cos\varphi, \sin \varphi)[/itex], sorry. So, my map simply bends the y_{+} half axis to the x_{} half axis, and stretches the area between it and the x_{+} axis. And one more note. I understand your proof for the reverse inclusion and I like it, but I'd also like to know if my original thought was correct or not. I imagined an indirect proof, that is, reversing the direction of the inclusion sign an also reversing the statements. If (a,b) is not in [itex](\partial A\times B) \cup (A\times \partial B)[/itex], then it has a whole nbhd inside [itex]A\times B[/itex], that is it isnt in [itex]\partial(A\times B)[/itex]. Is it correct so? 


#8
Mar1412, 12:14 PM

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#9
Mar1412, 11:30 PM

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Now it is quite clear, thank you again!



#10
Mar1512, 11:25 AM

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I am thinking of the product of two closed intervals. This resulting rectangle has four corners. At the corners I wonder what happens. 


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