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Schoen & Yau's proof of the positive-mass theorem: Why is the surface S homeomorphic to $mathbb{R}^2$?
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I'm currently reading through Schoen & Yau's 1979 proof of the positive-mass theorem and am trying to understand the following statement on p. 55 of the publication (page 11 of the proof / PDF):
Remark 2.1: The Cohn-Vossen inequality says that $int_S K leq 2pi chi(S)$, where [$K$ is the Gauss curvature and] $chi(S)$ is the Euler characteristic of $S$. Combining this with (2.18) we see immediately that $S$ is homeomorphic to $mathbb{R}^2$
Here, (2.18) refers to the inequality $int_S K > 0$ derived just before.
How can I see that the surface $S$ is homeomorphic to $mathbb{R}^2$? The immediate implication of (2.18) obviously is that $chi(S) > 0$, so $chi(S) geq 1$. Apart from that, I know that $S$ is non-compact but I'm unsure about other properties$^dagger$ of $S$ as my understanding of its construction is somewhat limited so far. Is it those properties that imply that the Betti numbers $b_i$ vanish for $i geq 1$?
--
$^dagger$ For instance, I suspect that $S$ is boundaryless, connected & simply connected and also a closed subset of the ambient manifold $N$ but since I only have a rough idea of how $S$ is constructed as a limit of surfaces $S_sigma$ (namely by representing the minimal surfaces $S_sigma$ locally as graphs in the tangent space (using normal coordinates) and using Arzelà-Ascoli for finding the limit), I'm having trouble coming up with rigorous proofs of any properties of $S$ that could help prove the claim.
differential-geometry algebraic-topology general-relativity
asked Jan 29 at 13:48
balubalu
10712
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I'm currently reading through Schoen & Yau's 1979 proof of the positive-mass theorem and am trying to understand the following statement on p. 55 of the publication (page 11 of the proof / PDF):
Remark 2.1: The Cohn-Vossen inequality says that $int_S K leq 2pi chi(S)$, where [$K$ is the Gauss curvature and] $chi(S)$ is the Euler characteristic of $S$. Combining this with (2.18) we see immediately that $S$ is homeomorphic to $mathbb{R}^2$
Here, (2.18) refers to the inequality $int_S K > 0$ derived just before.
How can I see that the surface $S$ is homeomorphic to $mathbb{R}^2$? The immediate implication of (2.18) obviously is that $chi(S) > 0$, so $chi(S) geq 1$. Apart from that, I know that $S$ is non-compact but I'm unsure about other properties$^dagger$ of $S$ as my understanding of its construction is somewhat limited so far. Is it those properties that imply that the Betti numbers $b_i$ vanish for $i geq 1$?
--
$^dagger$ For instance, I suspect that $S$ is boundaryless, connected & simply connected and also a closed subset of the ambient manifold $N$ but since I only have a rough idea of how $S$ is constructed as a limit of surfaces $S_sigma$ (namely by representing the minimal surfaces $S_sigma$ locally as graphs in the tangent space (using normal coordinates) and using Arzelà-Ascoli for finding the limit), I'm having trouble coming up with rigorous proofs of any properties of $S$ that could help prove the claim.
differential-geometry algebraic-topology general-relativity
asked Jan 29 at 13:48
balubalu
10712
$endgroup$
add a comment |
$begingroup$
I'm currently reading through Schoen & Yau's 1979 proof of the positive-mass theorem and am trying to understand the following statement on p. 55 of the publication (page 11 of the proof / PDF):
Remark 2.1: The Cohn-Vossen inequality says that $int_S K leq 2pi chi(S)$, where [$K$ is the Gauss curvature and] $chi(S)$ is the Euler characteristic of $S$. Combining this with (2.18) we see immediately that $S$ is homeomorphic to $mathbb{R}^2$
Here, (2.18) refers to the inequality $int_S K > 0$ derived just before.
How can I see that the surface $S$ is homeomorphic to $mathbb{R}^2$? The immediate implication of (2.18) obviously is that $chi(S) > 0$, so $chi(S) geq 1$. Apart from that, I know that $S$ is non-compact but I'm unsure about other properties$^dagger$ of $S$ as my understanding of its construction is somewhat limited so far. Is it those properties that imply that the Betti numbers $b_i$ vanish for $i geq 1$?
--
$^dagger$ For instance, I suspect that $S$ is boundaryless, connected & simply connected and also a closed subset of the ambient manifold $N$ but since I only have a rough idea of how $S$ is constructed as a limit of surfaces $S_sigma$ (namely by representing the minimal surfaces $S_sigma$ locally as graphs in the tangent space (using normal coordinates) and using Arzelà-Ascoli for finding the limit), I'm having trouble coming up with rigorous proofs of any properties of $S$ that could help prove the claim.
differential-geometry algebraic-topology general-relativity
asked Jan 29 at 13:48
balubalu
10712
$endgroup$
I'm currently reading through Schoen & Yau's 1979 proof of the positive-mass theorem and am trying to understand the following statement on p. 55 of the publication (page 11 of the proof / PDF):
Remark 2.1: The Cohn-Vossen inequality says that $int_S K leq 2pi chi(S)$, where [$K$ is the Gauss curvature and] $chi(S)$ is the Euler characteristic of $S$. Combining this with (2.18) we see immediately that $S$ is homeomorphic to $mathbb{R}^2$
Here, (2.18) refers to the inequality $int_S K > 0$ derived just before.
How can I see that the surface $S$ is homeomorphic to $mathbb{R}^2$? The immediate implication of (2.18) obviously is that $chi(S) > 0$, so $chi(S) geq 1$. Apart from that, I know that $S$ is non-compact but I'm unsure about other properties$^dagger$ of $S$ as my understanding of its construction is somewhat limited so far. Is it those properties that imply that the Betti numbers $b_i$ vanish for $i geq 1$?
--
$^dagger$ For instance, I suspect that $S$ is boundaryless, connected & simply connected and also a closed subset of the ambient manifold $N$ but since I only have a rough idea of how $S$ is constructed as a limit of surfaces $S_sigma$ (namely by representing the minimal surfaces $S_sigma$ locally as graphs in the tangent space (using normal coordinates) and using Arzelà-Ascoli for finding the limit), I'm having trouble coming up with rigorous proofs of any properties of $S$ that could help prove the claim.
differential-geometry algebraic-topology general-relativity
differential-geometry algebraic-topology general-relativity
asked Jan 29 at 13:48
balubalu
10712
asked Jan 29 at 13:48
balubalu
10712
edited Jan 29 at 15:35
balu
asked Jan 29 at 13:48
balubalu
10712
asked Jan 29 at 13:48
balubalu
10712
asked Jan 29 at 13:48
balubalu
10712
10712
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A noncompact surface has no homology groups above dimension 1. So $chi(S) = 1 - dim H_1(S)$. If $chi(S) = 1$, then in fact $H_1(S) = 0$.
A noncompact surface with finitely generated homology groups is diffeomorphic to some $S_{g,n}$, the genus $g$ surface with $n>0$ points removed. This is proved using the similar classification of compact surfaces, and using a compact exhaustion of your manifold. I wrote something about the simplest case in the second part of the post here but it's not hard to extrapolate to the general case.
Now $H_1(Sigma_{g,n}) = Bbb Z^{2g+n-1}$ so long as $n > 0$. If this is zero (so that the Euler characteristic is 1), then necessarily $g=0$ and $n=1$. Puncturing the 2-sphere leaves you with $Bbb R^2$, as desired.
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1 Answer
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1 Answer
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$begingroup$
A noncompact surface has no homology groups above dimension 1. So $chi(S) = 1 - dim H_1(S)$. If $chi(S) = 1$, then in fact $H_1(S) = 0$.
A noncompact surface with finitely generated homology groups is diffeomorphic to some $S_{g,n}$, the genus $g$ surface with $n>0$ points removed. This is proved using the similar classification of compact surfaces, and using a compact exhaustion of your manifold. I wrote something about the simplest case in the second part of the post here but it's not hard to extrapolate to the general case.
Now $H_1(Sigma_{g,n}) = Bbb Z^{2g+n-1}$ so long as $n > 0$. If this is zero (so that the Euler characteristic is 1), then necessarily $g=0$ and $n=1$. Puncturing the 2-sphere leaves you with $Bbb R^2$, as desired.
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add a comment |
$begingroup$
A noncompact surface has no homology groups above dimension 1. So $chi(S) = 1 - dim H_1(S)$. If $chi(S) = 1$, then in fact $H_1(S) = 0$.
A noncompact surface with finitely generated homology groups is diffeomorphic to some $S_{g,n}$, the genus $g$ surface with $n>0$ points removed. This is proved using the similar classification of compact surfaces, and using a compact exhaustion of your manifold. I wrote something about the simplest case in the second part of the post here but it's not hard to extrapolate to the general case.
Now $H_1(Sigma_{g,n}) = Bbb Z^{2g+n-1}$ so long as $n > 0$. If this is zero (so that the Euler characteristic is 1), then necessarily $g=0$ and $n=1$. Puncturing the 2-sphere leaves you with $Bbb R^2$, as desired.
$endgroup$
add a comment |
$begingroup$
A noncompact surface has no homology groups above dimension 1. So $chi(S) = 1 - dim H_1(S)$. If $chi(S) = 1$, then in fact $H_1(S) = 0$.
A noncompact surface with finitely generated homology groups is diffeomorphic to some $S_{g,n}$, the genus $g$ surface with $n>0$ points removed. This is proved using the similar classification of compact surfaces, and using a compact exhaustion of your manifold. I wrote something about the simplest case in the second part of the post here but it's not hard to extrapolate to the general case.
Now $H_1(Sigma_{g,n}) = Bbb Z^{2g+n-1}$ so long as $n > 0$. If this is zero (so that the Euler characteristic is 1), then necessarily $g=0$ and $n=1$. Puncturing the 2-sphere leaves you with $Bbb R^2$, as desired.
$endgroup$
A noncompact surface has no homology groups above dimension 1. So $chi(S) = 1 - dim H_1(S)$. If $chi(S) = 1$, then in fact $H_1(S) = 0$.
A noncompact surface with finitely generated homology groups is diffeomorphic to some $S_{g,n}$, the genus $g$ surface with $n>0$ points removed. This is proved using the similar classification of compact surfaces, and using a compact exhaustion of your manifold. I wrote something about the simplest case in the second part of the post here but it's not hard to extrapolate to the general case.
Now $H_1(Sigma_{g,n}) = Bbb Z^{2g+n-1}$ so long as $n > 0$. If this is zero (so that the Euler characteristic is 1), then necessarily $g=0$ and $n=1$. Puncturing the 2-sphere leaves you with $Bbb R^2$, as desired.
answered Jan 29 at 15:39
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