Spivak & Dimension of Manifold

  • #1
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1. Homework Statement

I'm taking a swing at Spivak's Differential Geometry, and a question that Spivak asks his reader to show is that if ##x\in M## for ##M## a manifold and there is a neighborhood (Note that Spivak requires neighborhoods to be sets which contain an open set containing the point of interest) of ##x## such that this neighborhood is homeomorphic to ##\mathbb{R}^n##, then this ##n## is fixed for any other such neighborhood of ##x##.

If I could prove this, then I could prove his other assertion that if ##U\subseteq M## is a neighborhood of some ##x\in M##, and is homeomorphic to ##\mathbb{R}^n## (for any ##n##), then ##U## must be open; via the Invariance of Domain, the proof follows almost immediately from what I'm proving here.



2. The attempt at a solution

I'm a little far removed from topology so I'm not entirely confident in my proof. In particular, when I tried to find some sort of verification online, I found this which doesn't look anything like my argument.

I more or less mimic my proof that ##\mathbb{R}^n## is not homeomorphic to ##\mathbb{R}^m## if ##n\ne m## by the invariance of domain in my proof:


Spivak has shown that one can choose, for any neighborhood ##U## of ##x\in M##, a subset ##V\subseteq U## such that ##V## is open with respect to the topology of ##M## and homeomorphic to the same ##\mathbb{R}^n## as ##U##. This is straight-forward.

So if I let ##x\in M##, and ##N_1##, ##N_2## be two neighborhoods of ##x## homeomorphic, respectively, to ##\mathbb{R}^n## and ##\mathbb{R}^m##. Let's denote by ##f\colon N_1\to\mathbb{R}^n## and ##g\colon N_{2}\to\mathbb{R}^m## the homeomorphisms. WLOG we can assume ##n<m##, attempting to obtain a contradiction.

I can widdle ##N_1## and ##N_2## down into subsets ##V_1## and ##V_2## open in ##M## as I noted above. Let ##V=V_1\cap V_2##. Then, of course, ##V\subseteq N_{1}## and ##V\subseteq N_2##; moreover ##V## is still open in ##M##. Since ##g## and ##f## are homeomorphisms, the image of ##V## is open in ##\mathbb{R}^n## under ##f## and the image of ##V## under ##g## is open in ##\mathbb{R}^m##. Denote by ##\pi## the embedding of ##\mathbb{R}^n## into ##\mathbb{R}^m##. Then the map ##h:=\pi\circ f\circ g^{-1}## passes ##g(V)## to ##f(V)\subseteq\mathbb{R}^n## viewed as a subset of ##\mathbb{R}^m##. But in the metric topology of ##\mathbb{R}^m## (where, again, ##n<m##), any subset of ##\mathbb{R}^n## is closed. So this contradicts the invariance of domain, since ##h## defines a continuous injection of an open subset of ##\mathbb{R}^m## into ##\mathbb{R}^m##.

Does this look alright? Any pointers?
 
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Answers and Replies

  • #2
Samy_A
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Homework Statement


I'm taking a swing at Spivak's Differential Geometry, and a question that Spivak asks his reader to show is that if ##x\in M## for ##M## a manifold and there is a neighborhood (Note that Spivak requires neighborhoods to be sets which contain an open set containing the point of interest) of ##x## such that this neighborhood is homeomorphic to ##\mathbb{R}^n##, then this ##n## is fixed for any other such neighborhood of ##x##.

If I could prove this, then I could prove his other assertion that if ##U\subseteq M## is a neighborhood of some ##x\in M##, and is homeomorphic to ##\mathbb{R}^n## (for any ##n##), then ##U## must be open; via the Invariance of Domain, the proof follows almost immediately from what I'm proving here.



2. The attempt at a solution

I'm a little far removed from topology so I'm not entirely confident in my proof. In particular, when I tried to find some sort of verification online, I found this which doesn't look anything like my argument.

I more or less mimic my proof that ##\mathbb{R}^n## is not homeomorphic to ##\mathbb{R}^m## if ##n\ne m## by the invariance of domain in my proof:


Spivak has shown that one can choose, for any neighborhood ##U## of ##x\in M##, a subset ##V\subseteq U## such that ##V## is open with respect to the topology of ##M## and homeomorphic to the same ##\mathbb{R}^n## as ##U##. This is straight-forward.

So if I let ##x\in M##, and ##N_1##, ##N_2## be two neighborhoods of ##x## homeomorphic, respectively, to ##\mathbb{R}^n## and ##\mathbb{R}^m##. Let's denote by ##f\colon N_1\to\mathbb{R}^n## and ##g\colon N_{2}\to\mathbb{R}^m## the homeomorphisms. WLOG we can assume ##n<m##, attempting to obtain a contradiction.

I can widdle ##N_1## and ##N_2## down into subsets ##V_1## and ##V_2## open in ##M## as I noted above. Let ##V=V_1\cap V_2##. Then, of course, ##V\subseteq N_{1}## and ##V\subseteq N_2##; moreover ##V## is still open in ##M##. Since ##g## and ##f## are homeomorphisms, the image of ##V## is open in ##\mathbb{R}^n## under ##f## and the image of ##V## under ##g## is open in ##\mathbb{R}^m##. Denote by ##\pi## the embedding of ##\mathbb{R}^n## into ##\mathbb{R}^m##. Then the map ##h:=\pi\circ f\circ g^{-1}## passes ##g(V)## to ##f(V)\subseteq\mathbb{R}^n## viewed as a subset of ##\mathbb{R}^m##. But in the metric topology of ##\mathbb{R}^m## (where, again, ##n<m##), any subset of ##\mathbb{R}^n## is closed. So this contradicts the invariance of domain, since ##h## defines a continuous injection of an open subset of ##\mathbb{R}^m## into ##\mathbb{R}^m##.

Does this look alright? Any pointers?
(color added)
I don't know where the statement I colored in red comes from. I think it is wrong.
Remove it, and your proof looks correct, as ##f(V) \subseteq\mathbb R^n## and ##g(V) \subseteq \mathbb R^m## would be homeomorphic non-empty open sets, and that is not possible if
##m \neq n##.
 
  • #3
Fredrik
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I don't know where the statement I colored in red comes from. I think it is wrong.
By "subset of ##\mathbb R^n##", I assume that he means "subset of ##\{x\in\mathbb R^m|x_{n+1}=x_{n+2}=\cdots =x_m=0\}##". This set is closed because the limit of a convergent sequence in that set is in that set.
 
  • #4
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By "subset of ##\mathbb R^n##", I assume that he means "subset of ##\{x\in\mathbb R^m|x_{n+1}=x_{n+2}=\cdots =x_m=0\}##". This set is closed because the limit of a convergent sequence in that set is in that set.

In that interpretation, it's not wrong that ##\mathbb{R}^n## is closed in ##\mathbb{R}^m##. But it is definitely not true in that interpretation that any subset of ##\mathbb{R}^n## is closed in ##\mathbb{R}^m##.
 
  • #5
Samy_A
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By "subset of ##\mathbb R^n##", I assume that he means "subset of ##\{x\in\mathbb R^m|x_{n+1}=x_{n+2}=\cdots =x_m=0\}##". This set is closed because the limit of a convergent sequence in that set is in that set.
In that interpretation, it's not wrong that ##\mathbb{R}^n## is closed in ##\mathbb{R}^m##. But it is definitely not true in that interpretation that any subset of ##\mathbb{R}^n## is closed in ##\mathbb{R}^m##.
Thought so too.
If by "any subset" he meant "any closed subset" (in the ##\mathbb R^n## topology), then I agree.

Anyway, it is a red herring, as he doesn't really use the statement.
 
  • #6
Fredrik
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In that interpretation, it's not wrong that ##\mathbb{R}^n## is closed in ##\mathbb{R}^m##. But it is definitely not true in that interpretation that any subset of ##\mathbb{R}^n## is closed in ##\mathbb{R}^m##.
Ah, of course. Didn't really think about that part. Thanks.
 
  • #7
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Hey all, thanks for the replies and comments and I apologize for getting back a little late.

Not really sure why I threw that statement in there, there are plenty of counter-examples.

I appreciate the help, thanks!
 

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