Why is the space of Riemannian metric on a manifold an open set with respect to the Sobolev class of...
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Let $M$ be a smooth manifold. Let $H^s(T^2_0M)$ be the Sobolev class of symmetric $0,2$ tensors defined by the class of $0,2$ tensors whose derivatives up to order $s$ exist in the sense of distributions and are square integrable. Since it is a vector space, it can be identified as $mathbb{R}^n$ for some $n$, equipped with usual topology.
Let $mathcal{M}^s$ be the subset of $H^s(T^2_0M)$ consisting of Riemannian metrics on $M$. How can I show that it is an open set in $H^s(T^2_0M)$? The usual approach is to construct continuous map whose preimage of some open set in the range of the function is $mathcal{M}^s$. However, I have no idea how to construct such a map.
differential-geometry riemannian-geometry
asked Jan 12 at 15:39
JerryJerry
422313
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add a comment |
$begingroup$
Let $M$ be a smooth manifold. Let $H^s(T^2_0M)$ be the Sobolev class of symmetric $0,2$ tensors defined by the class of $0,2$ tensors whose derivatives up to order $s$ exist in the sense of distributions and are square integrable. Since it is a vector space, it can be identified as $mathbb{R}^n$ for some $n$, equipped with usual topology.
Let $mathcal{M}^s$ be the subset of $H^s(T^2_0M)$ consisting of Riemannian metrics on $M$. How can I show that it is an open set in $H^s(T^2_0M)$? The usual approach is to construct continuous map whose preimage of some open set in the range of the function is $mathcal{M}^s$. However, I have no idea how to construct such a map.
differential-geometry riemannian-geometry
asked Jan 12 at 15:39
JerryJerry
422313
$endgroup$
1
$begingroup$
Two things: note that your Sobolev space is not a finite dimensional one, so it is not isomorphic to any $mathbb{R}^n$. Second, as far as I remember Riemannian metrics are smooth (or at least continuous), so I do not think there would be any Sobolev space where they would be open...
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– Mindlack
Jan 14 at 18:15
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Here Riemannian metrics are assumed to be in the Sobolev space (i.e. those positive definite elements in the space).
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– Jerry
Jan 14 at 18:44
$begingroup$
I think $s$ needs to satisfy at least $s>frac{n}2$. Also, is $M$ bounded?
$endgroup$
– timur
Jan 16 at 3:11
add a comment |
$begingroup$
Let $M$ be a smooth manifold. Let $H^s(T^2_0M)$ be the Sobolev class of symmetric $0,2$ tensors defined by the class of $0,2$ tensors whose derivatives up to order $s$ exist in the sense of distributions and are square integrable. Since it is a vector space, it can be identified as $mathbb{R}^n$ for some $n$, equipped with usual topology.
Let $mathcal{M}^s$ be the subset of $H^s(T^2_0M)$ consisting of Riemannian metrics on $M$. How can I show that it is an open set in $H^s(T^2_0M)$? The usual approach is to construct continuous map whose preimage of some open set in the range of the function is $mathcal{M}^s$. However, I have no idea how to construct such a map.
differential-geometry riemannian-geometry
asked Jan 12 at 15:39
JerryJerry
422313
$endgroup$
Let $M$ be a smooth manifold. Let $H^s(T^2_0M)$ be the Sobolev class of symmetric $0,2$ tensors defined by the class of $0,2$ tensors whose derivatives up to order $s$ exist in the sense of distributions and are square integrable. Since it is a vector space, it can be identified as $mathbb{R}^n$ for some $n$, equipped with usual topology.
Let $mathcal{M}^s$ be the subset of $H^s(T^2_0M)$ consisting of Riemannian metrics on $M$. How can I show that it is an open set in $H^s(T^2_0M)$? The usual approach is to construct continuous map whose preimage of some open set in the range of the function is $mathcal{M}^s$. However, I have no idea how to construct such a map.
differential-geometry riemannian-geometry
differential-geometry riemannian-geometry
asked Jan 12 at 15:39
JerryJerry
422313
asked Jan 12 at 15:39
JerryJerry
422313
asked Jan 12 at 15:39
JerryJerry
422313
asked Jan 12 at 15:39
JerryJerry
422313
asked Jan 12 at 15:39
JerryJerry
422313
422313
1
$begingroup$
Two things: note that your Sobolev space is not a finite dimensional one, so it is not isomorphic to any $mathbb{R}^n$. Second, as far as I remember Riemannian metrics are smooth (or at least continuous), so I do not think there would be any Sobolev space where they would be open...
$endgroup$
– Mindlack
Jan 14 at 18:15
$begingroup$
Here Riemannian metrics are assumed to be in the Sobolev space (i.e. those positive definite elements in the space).
$endgroup$
– Jerry
Jan 14 at 18:44
$begingroup$
I think $s$ needs to satisfy at least $s>frac{n}2$. Also, is $M$ bounded?
$endgroup$
– timur
Jan 16 at 3:11
add a comment |
1
$begingroup$
Two things: note that your Sobolev space is not a finite dimensional one, so it is not isomorphic to any $mathbb{R}^n$. Second, as far as I remember Riemannian metrics are smooth (or at least continuous), so I do not think there would be any Sobolev space where they would be open...
$endgroup$
– Mindlack
Jan 14 at 18:15
$begingroup$
Here Riemannian metrics are assumed to be in the Sobolev space (i.e. those positive definite elements in the space).
$endgroup$
– Jerry
Jan 14 at 18:44
$begingroup$
I think $s$ needs to satisfy at least $s>frac{n}2$. Also, is $M$ bounded?
$endgroup$
– timur
Jan 16 at 3:11
1
1
$begingroup$
Two things: note that your Sobolev space is not a finite dimensional one, so it is not isomorphic to any $mathbb{R}^n$. Second, as far as I remember Riemannian metrics are smooth (or at least continuous), so I do not think there would be any Sobolev space where they would be open...
$endgroup$
– Mindlack
Jan 14 at 18:15
$begingroup$
Two things: note that your Sobolev space is not a finite dimensional one, so it is not isomorphic to any $mathbb{R}^n$. Second, as far as I remember Riemannian metrics are smooth (or at least continuous), so I do not think there would be any Sobolev space where they would be open...
$endgroup$
– Mindlack
Jan 14 at 18:15
$begingroup$
Here Riemannian metrics are assumed to be in the Sobolev space (i.e. those positive definite elements in the space).
$endgroup$
– Jerry
Jan 14 at 18:44
$begingroup$
Here Riemannian metrics are assumed to be in the Sobolev space (i.e. those positive definite elements in the space).
$endgroup$
– Jerry
Jan 14 at 18:44
$begingroup$
I think $s$ needs to satisfy at least $s>frac{n}2$. Also, is $M$ bounded?
$endgroup$
– timur
Jan 16 at 3:11
$begingroup$
I think $s$ needs to satisfy at least $s>frac{n}2$. Also, is $M$ bounded?
$endgroup$
– timur
Jan 16 at 3:11
add a comment |
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$begingroup$
Two things: note that your Sobolev space is not a finite dimensional one, so it is not isomorphic to any $mathbb{R}^n$. Second, as far as I remember Riemannian metrics are smooth (or at least continuous), so I do not think there would be any Sobolev space where they would be open...
$endgroup$
– Mindlack
Jan 14 at 18:15
$begingroup$
Here Riemannian metrics are assumed to be in the Sobolev space (i.e. those positive definite elements in the space).
$endgroup$
– Jerry
Jan 14 at 18:44
$begingroup$
I think $s$ needs to satisfy at least $s>frac{n}2$. Also, is $M$ bounded?
$endgroup$
– timur
Jan 16 at 3:11