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Boundary of a product manifold

by mma
Tags: boundary, manifold, product
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mma
#1
Mar11-12, 12:32 AM
P: 230
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?
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morphism
#2
Mar12-12, 01:36 AM
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Yes, that's how one typically orients the boundary of AxB.
mma
#3
Mar12-12, 04:00 AM
P: 230
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]

quasar987
#4
Mar12-12, 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 Rm-1 x R+ x Rn ≈ Rm+n-1 x R+. Hence (a,b) is in d(AxB).
In case (ii), a nbhd of (a,b) is of the form Rm-1 x R+ x Rn-1 x R+. You just have to convince yourself now that R+ x R+ ≈ R x R+ to see that the above nbhd is just Rm+n-1 x R+ again. Hence (a,b) is in d(AxB) in this case also.
Etc.
mma
#5
Mar13-12, 12:52 AM
P: 230
Quote Quote by quasar987 View Post
You just have to convince yourself now that R+ x R+ ≈ R x R+
I think I can:

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?
quasar987
#6
Mar13-12, 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 Rm+n-1. 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.
mma
#7
Mar14-12, 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?
quasar987
#8
Mar14-12, 12:14 PM
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Quote Quote by mma View Post
Dear Quasar987, I greatly enjoy your concise and expressive explanations, they are really lucid, thank you for them.
I am touched by your words of thanks mma. I also enjoy discussing with you :)

Quote Quote by mma View Post
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).
At the time I wrote my previous post, I could not see the latex code of your post, so I just said "whatever your homeomorphism is, the direction that goes from R+ x R into R+ x R+ , call it F, won't be differentiable at F-1(0)." The argument I had in mind is the following. For the sake of clarity, suppose that F-1(0)=0. Then, the limit defining ∂Fi/∂x at (0,0) will be different as you approach from the right and from the left.

Quote Quote by mma View Post
(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?)
I doubt that it is, but it is a good criterion for deciding if a map is non-smooth :)

Quote Quote by mma View Post
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?
Yes, it seems to be essentially the same as the argument I wrote.
mma
#9
Mar14-12, 11:30 PM
P: 230
Now it is quite clear, thank you again!
lavinia
#10
Mar15-12, 11:25 AM
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Quote Quote by mma View Post
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?
If you have two manifolds with boundary I think the product will be a manifold with corners.

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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