
#1
Nov2811, 02:41 PM

P: 39

Hi,
I want to show that the set of boundary points on a manifold with boundary is well defined, i.e the image of a point on a manifold with boundary can not be both the interior point and boundary point on upper half space. To do this, it is enough to show that R^n can be homeomorphic to upper half space. How can i show this fact? Best regards 



#2
Nov2811, 03:07 PM

Mentor
P: 4,499

I would suggest starting by just proving R is homeomorphic to the positive real numbers, and seeing how you can generalize that result




#3
Nov2811, 03:31 PM

P: 39

Sorry,
This is not true. It will be R^n can not be homeomorphic to upper half plane. Note that upper half plane is {x=(x_1,,,,,x_n) \in R^n : x_n =>0}. 



#4
Nov2811, 05:41 PM

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PF Gold
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Hi,I want to show that the set of boundary points on a manifold
Try this: if you remove a point on the boundary of the upper half plane, show using MayerVietoris that you do not change its homology, but if you remove a point from eucidean space, this becomes a homotopy sphere and so the homology changes.




#5
Nov2811, 05:54 PM

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P: 1,168

How about invariance of dimension? Interior points have 'hoods homeomorphic to ℝ^{n1} , while interior points have 'hoods homeomorphic to ℝ^{n}. That would then produce a homeomorphism between ℝ^{n} and ℝ^{n1}.
Maybe now homeomorphic can write a post titled 'seydunas' 



#6
Nov2811, 07:10 PM

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#7
Nov2811, 07:16 PM

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In the case of a smooth manifold a differentiable ray starting on the boundary can only leave the boundary in one direction. That is, its tangent vector can only point inwards. In an open set it can point in both directions.




#8
Nov2811, 09:05 PM

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Hm, well, if there is a homeomorphism f btw H^n and R^n, then removing a point p on the boundary of H^n and removing f(p) in R^n we get a homeomorphism btw H^n  {p} and R^n  {f(p)}. But this is impossible because H^n  {p} is contractible, but R^n  {f(p)} is not.




#9
Nov2811, 09:08 PM

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The invariance of domain thm say if there is a homeo U>V with U open in R^n and V in R^m, then V is open and m=n. So yeah, H^n is not open so there is no homeo btw R^n and H^n. 



#10
Nov2811, 09:16 PM

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P: 1,168

Quasar987 (ism):
I meant that if a bdry. pt p was part of both the interior of M and the boundary of M, there would be an open set U containing p that is part of both the interior and the boundary. This set U would then be homeomorphic to R^n (as part of the manifold), and homeo to R^(n1), as part of the boundary. Isn't this correct? Bacle2(ism) 



#11
Nov2911, 07:33 AM

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I don't get how a 'hood of a point on the boundary of H^n is homeomorphic to R^(n1). A 'hood of a point on the boundary of H² is kind of a halfdisk and is thus certainly not homeomorphic to R.




#12
Nov3011, 09:17 AM

P: 120

You don't need more fancier staff than the knowledge of tangent spaces and diffeomorphisms I think.
Hint: consider the dimension of tangent spaces of boundary of upper halfplane and a interior point of half plane. notice that composition of charts belonging to same differentiable structure when composed gives diffeomorphisms. correction: if it is not smooth we will have to use homeomorphisms and by the correction quasar has made, tangent space arguement is not valid so we don't need tangent spaces. Indeed consider two charts ψ and ρ (from the manifold to Euclidean space and belonging to the differentiable structure of our manifold M) st for a point p in the boundary of our manifold ψ(p)=b is in the boundary of H^{n} and ρ(p)=i is an interior point of H^{n}. Now consider the map taking i to b which is ψ°ρ^{1}(i) = p. Since these two charts belong to the same differentiable structure this map is a diffeomorphism so it should be an isomorphism of the tangent spaces. However tangent space of b is n1 dimensional and that of i is n dimensional. So contradiction. 



#13
Nov3011, 09:43 AM

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The tangent space at b is not (n1)dimensional.




#14
Nov3011, 10:18 AM

P: 120

Okay this is where geometric intiution has failed me so I refine my answer. Again take the same construction with ψ°ρ^{1}(b) = i (there is a mistake also p should be b). This is a diffeomorphism and around b we can find a small enough nbd U such that ψ°ρ^{1}(U) = V is in the interior of H^{n} which is locally R^{n} and note tha U is locally H^{n}. This can not be because U is itself a manifold with boundary and it can not be locally modelled by R^{n} (using the definition of manifold with boundary, actually I know this is a heuristic arguement, actually I am thinking something for this now). Well okay my last arguement has already been made above so I don't need to refine it more. If it is not a smooth manifold replace diffeomorphism with homeomorphism. So I see I have not added anything new to the topic really :p




#15
Nov3011, 04:48 PM

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I think the 'hood needs to be a subspace one; what I'm using here is that the boundary of an nmanifold M with boundary BD(M) is an (n1)manifold (where I think the topology used in BD(M) is the subspace topology of BD(M) on M ). e.g., if we consider the closed disk D={(x,y) x^{2}+y^{2}≤1} , then the boundary (both manifold and topological, here) is the set of points BdS:={(x,y): x^{2}+y^{2}=1} . This is a manifold (without bdry, since bdry. of a mfld. with bdry is empty) , but with topology given by open sets from the subspace inherited from D, i.e., open arcs in D. 



#16
Nov3011, 05:47 PM

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P: 1,168

Never mind my previous post; I mean, it is true that the boundary of an nmfld with boundary is an nmanifold, is an (n1)manifold with subspace charts, but this won't help proof what Seydunas wants.




#17
Dec211, 03:44 PM

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P: 1,168

Quasar987 wrote, in part: (Sorry, my quote function is not working)
"Originally Posted by Bacle2 View Post How about invariance of dimension? Interior points have 'hoods homeomorphic to ℝn1 , while interior points have 'hoods homeomorphic to ℝn. That would then produce a homeomorphism between ℝn and ℝn1. Maybe now homeomorphic can write a post titled 'seydunas' Or seydunasism! :/ The invariance of domain thm say if there is a homeo U>V with U open in R^n and V in R^m, then V is open and m=n. So yeah, H^n is not open so there is no homeo btw R^n and H^n. " That was invariance of dimension; an invariance of domain argument would go like this: Assume there is a homeomorphism h between R^n and H^n , and then consider the inclusion map i: H^n>R^n i(x,y)=(x,y) between H^n and R^n . Then the composition ioh: R^n>R^n is a continuous injection between R^n and itself, with image H^n, so that H^n is open in R^n which it is not. 


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