$C^{k}$-manifolds: how and why?
$begingroup$
First of all, I have a specific question. Suppose $M$ is an $m$-dimensional $C^k$-manifold, for $1 leq k < infty$. Is the tangent space to a point defined as the space of $C^k$ derivations on the germs of $C^k$ functions near that point? If so, is it $m$-dimensional? Bredon's book Topology and Geometry comments that only in the $C^infty$ case can one prove that every derivation is given by a tangent vector to a curve. If so, this would suggest that (if indeed given this definition), the tangent space to a $C^k$-manifold would be bigger in the case $k < infty$. Additionally, out of curiosity, would anybody have an example of a derivation that is not a tangent vector to a curve?
Secondly, it would seem to me that a fair share of the things I learned about smooth manifolds should fail or at least require more elaborate proofs in the $C^k$ case. We only used higher derivatives in proving Sard's theorem, but all the time we used the identification that the tangent space is given by tangent vectors to curves; the tubular neighborhood theorem comes to mind. What are the standard facts of smooth manifolds that do fail in the $C^k$ case?
Thirdly, are they really important? It seems a lot of books deal only with smooth manifolds, but a fair share also seem to deal with $C^k$-manifolds; Hirsch's Differential Topology deals with them all throughout, and Duistermaat's book on Lie groups defines them as $C^2$-manifolds. Should I, as a student of topology / geometry, be paying close attention to $C^k$-manifolds and the distinctions with the smooth case?
differential-geometry manifolds differential-topology
asked Aug 31 '14 at 11:02
PedroPedro
2,79131834
$endgroup$
|
show 2 more comments
$begingroup$
First of all, I have a specific question. Suppose $M$ is an $m$-dimensional $C^k$-manifold, for $1 leq k < infty$. Is the tangent space to a point defined as the space of $C^k$ derivations on the germs of $C^k$ functions near that point? If so, is it $m$-dimensional? Bredon's book Topology and Geometry comments that only in the $C^infty$ case can one prove that every derivation is given by a tangent vector to a curve. If so, this would suggest that (if indeed given this definition), the tangent space to a $C^k$-manifold would be bigger in the case $k < infty$. Additionally, out of curiosity, would anybody have an example of a derivation that is not a tangent vector to a curve?
Secondly, it would seem to me that a fair share of the things I learned about smooth manifolds should fail or at least require more elaborate proofs in the $C^k$ case. We only used higher derivatives in proving Sard's theorem, but all the time we used the identification that the tangent space is given by tangent vectors to curves; the tubular neighborhood theorem comes to mind. What are the standard facts of smooth manifolds that do fail in the $C^k$ case?
Thirdly, are they really important? It seems a lot of books deal only with smooth manifolds, but a fair share also seem to deal with $C^k$-manifolds; Hirsch's Differential Topology deals with them all throughout, and Duistermaat's book on Lie groups defines them as $C^2$-manifolds. Should I, as a student of topology / geometry, be paying close attention to $C^k$-manifolds and the distinctions with the smooth case?
differential-geometry manifolds differential-topology
asked Aug 31 '14 at 11:02
PedroPedro
2,79131834
$endgroup$
3
$begingroup$
I remember losing a little bit of sleep about exactly those questions. My understanding is that every maximal $C^k (kgeq 1)$ atlas contains a $C^infty$ atlas, which is a relief.
$endgroup$
– Tim kinsella
Aug 31 '14 at 11:07
2
$begingroup$
Duistermaat defines Lie groups as $C^2$-manifolds because that suffices to prove they are analytic (better than $C^{infty}$). Nice exposition at terrytao.wordpress.com/2011/06/21/…
$endgroup$
– Colin McLarty
Aug 31 '14 at 11:20
1
$begingroup$
@Lolman: Very interesting! How is it defined then? It'd be lovely if someone could get a reference for showing it's infinite-dimensional in the derivations definition.
$endgroup$
– Pedro
Aug 31 '14 at 11:27
3
$begingroup$
Much of this question is answered at math.stackexchange.com/questions/73677/…
$endgroup$
– Colin McLarty
Aug 31 '14 at 11:57
2
$begingroup$
As a general comment, you shouldn't use the word "tangent space" to refer to the space of all derivations. Even in the $C^k$ context, the word "tangent space" means the (finite-dimensional) space of all tangent vectors. Books that restrict to the $C^infty$ case sometimes define the tangent space using derivations, but that's not the primary definition. In particular, books that consider the $C^k$ case generally ignore derivations completely in favor of some other formalism.
$endgroup$
– Jim Belk
Jun 4 '15 at 16:48
|
show 2 more comments
$begingroup$
First of all, I have a specific question. Suppose $M$ is an $m$-dimensional $C^k$-manifold, for $1 leq k < infty$. Is the tangent space to a point defined as the space of $C^k$ derivations on the germs of $C^k$ functions near that point? If so, is it $m$-dimensional? Bredon's book Topology and Geometry comments that only in the $C^infty$ case can one prove that every derivation is given by a tangent vector to a curve. If so, this would suggest that (if indeed given this definition), the tangent space to a $C^k$-manifold would be bigger in the case $k < infty$. Additionally, out of curiosity, would anybody have an example of a derivation that is not a tangent vector to a curve?
Secondly, it would seem to me that a fair share of the things I learned about smooth manifolds should fail or at least require more elaborate proofs in the $C^k$ case. We only used higher derivatives in proving Sard's theorem, but all the time we used the identification that the tangent space is given by tangent vectors to curves; the tubular neighborhood theorem comes to mind. What are the standard facts of smooth manifolds that do fail in the $C^k$ case?
Thirdly, are they really important? It seems a lot of books deal only with smooth manifolds, but a fair share also seem to deal with $C^k$-manifolds; Hirsch's Differential Topology deals with them all throughout, and Duistermaat's book on Lie groups defines them as $C^2$-manifolds. Should I, as a student of topology / geometry, be paying close attention to $C^k$-manifolds and the distinctions with the smooth case?
differential-geometry manifolds differential-topology
asked Aug 31 '14 at 11:02
PedroPedro
2,79131834
$endgroup$
First of all, I have a specific question. Suppose $M$ is an $m$-dimensional $C^k$-manifold, for $1 leq k < infty$. Is the tangent space to a point defined as the space of $C^k$ derivations on the germs of $C^k$ functions near that point? If so, is it $m$-dimensional? Bredon's book Topology and Geometry comments that only in the $C^infty$ case can one prove that every derivation is given by a tangent vector to a curve. If so, this would suggest that (if indeed given this definition), the tangent space to a $C^k$-manifold would be bigger in the case $k < infty$. Additionally, out of curiosity, would anybody have an example of a derivation that is not a tangent vector to a curve?
Secondly, it would seem to me that a fair share of the things I learned about smooth manifolds should fail or at least require more elaborate proofs in the $C^k$ case. We only used higher derivatives in proving Sard's theorem, but all the time we used the identification that the tangent space is given by tangent vectors to curves; the tubular neighborhood theorem comes to mind. What are the standard facts of smooth manifolds that do fail in the $C^k$ case?
Thirdly, are they really important? It seems a lot of books deal only with smooth manifolds, but a fair share also seem to deal with $C^k$-manifolds; Hirsch's Differential Topology deals with them all throughout, and Duistermaat's book on Lie groups defines them as $C^2$-manifolds. Should I, as a student of topology / geometry, be paying close attention to $C^k$-manifolds and the distinctions with the smooth case?
differential-geometry manifolds differential-topology
differential-geometry manifolds differential-topology
asked Aug 31 '14 at 11:02
PedroPedro
2,79131834
asked Aug 31 '14 at 11:02
PedroPedro
2,79131834
edited Aug 31 '14 at 12:58
Pedro
asked Aug 31 '14 at 11:02
PedroPedro
2,79131834
asked Aug 31 '14 at 11:02
PedroPedro
2,79131834
asked Aug 31 '14 at 11:02
PedroPedro
2,79131834
2,79131834
3
$begingroup$
I remember losing a little bit of sleep about exactly those questions. My understanding is that every maximal $C^k (kgeq 1)$ atlas contains a $C^infty$ atlas, which is a relief.
$endgroup$
– Tim kinsella
Aug 31 '14 at 11:07
2
$begingroup$
Duistermaat defines Lie groups as $C^2$-manifolds because that suffices to prove they are analytic (better than $C^{infty}$). Nice exposition at terrytao.wordpress.com/2011/06/21/…
$endgroup$
– Colin McLarty
Aug 31 '14 at 11:20
1
$begingroup$
@Lolman: Very interesting! How is it defined then? It'd be lovely if someone could get a reference for showing it's infinite-dimensional in the derivations definition.
$endgroup$
– Pedro
Aug 31 '14 at 11:27
3
$begingroup$
Much of this question is answered at math.stackexchange.com/questions/73677/…
$endgroup$
– Colin McLarty
Aug 31 '14 at 11:57
2
$begingroup$
As a general comment, you shouldn't use the word "tangent space" to refer to the space of all derivations. Even in the $C^k$ context, the word "tangent space" means the (finite-dimensional) space of all tangent vectors. Books that restrict to the $C^infty$ case sometimes define the tangent space using derivations, but that's not the primary definition. In particular, books that consider the $C^k$ case generally ignore derivations completely in favor of some other formalism.
$endgroup$
– Jim Belk
Jun 4 '15 at 16:48
|
show 2 more comments
3
$begingroup$
I remember losing a little bit of sleep about exactly those questions. My understanding is that every maximal $C^k (kgeq 1)$ atlas contains a $C^infty$ atlas, which is a relief.
$endgroup$
– Tim kinsella
Aug 31 '14 at 11:07
2
$begingroup$
Duistermaat defines Lie groups as $C^2$-manifolds because that suffices to prove they are analytic (better than $C^{infty}$). Nice exposition at terrytao.wordpress.com/2011/06/21/…
$endgroup$
– Colin McLarty
Aug 31 '14 at 11:20
1
$begingroup$
@Lolman: Very interesting! How is it defined then? It'd be lovely if someone could get a reference for showing it's infinite-dimensional in the derivations definition.
$endgroup$
– Pedro
Aug 31 '14 at 11:27
3
$begingroup$
Much of this question is answered at math.stackexchange.com/questions/73677/…
$endgroup$
– Colin McLarty
Aug 31 '14 at 11:57
2
$begingroup$
As a general comment, you shouldn't use the word "tangent space" to refer to the space of all derivations. Even in the $C^k$ context, the word "tangent space" means the (finite-dimensional) space of all tangent vectors. Books that restrict to the $C^infty$ case sometimes define the tangent space using derivations, but that's not the primary definition. In particular, books that consider the $C^k$ case generally ignore derivations completely in favor of some other formalism.
$endgroup$
– Jim Belk
Jun 4 '15 at 16:48
3
3
$begingroup$
I remember losing a little bit of sleep about exactly those questions. My understanding is that every maximal $C^k (kgeq 1)$ atlas contains a $C^infty$ atlas, which is a relief.
$endgroup$
– Tim kinsella
Aug 31 '14 at 11:07
$begingroup$
I remember losing a little bit of sleep about exactly those questions. My understanding is that every maximal $C^k (kgeq 1)$ atlas contains a $C^infty$ atlas, which is a relief.
$endgroup$
– Tim kinsella
Aug 31 '14 at 11:07
2
2
$begingroup$
Duistermaat defines Lie groups as $C^2$-manifolds because that suffices to prove they are analytic (better than $C^{infty}$). Nice exposition at terrytao.wordpress.com/2011/06/21/…
$endgroup$
– Colin McLarty
Aug 31 '14 at 11:20
$begingroup$
Duistermaat defines Lie groups as $C^2$-manifolds because that suffices to prove they are analytic (better than $C^{infty}$). Nice exposition at terrytao.wordpress.com/2011/06/21/…
$endgroup$
– Colin McLarty
Aug 31 '14 at 11:20
1
1
$begingroup$
@Lolman: Very interesting! How is it defined then? It'd be lovely if someone could get a reference for showing it's infinite-dimensional in the derivations definition.
$endgroup$
– Pedro
Aug 31 '14 at 11:27
$begingroup$
@Lolman: Very interesting! How is it defined then? It'd be lovely if someone could get a reference for showing it's infinite-dimensional in the derivations definition.
$endgroup$
– Pedro
Aug 31 '14 at 11:27
3
3
$begingroup$
Much of this question is answered at math.stackexchange.com/questions/73677/…
$endgroup$
– Colin McLarty
Aug 31 '14 at 11:57
$begingroup$
Much of this question is answered at math.stackexchange.com/questions/73677/…
$endgroup$
– Colin McLarty
Aug 31 '14 at 11:57
2
2
$begingroup$
As a general comment, you shouldn't use the word "tangent space" to refer to the space of all derivations. Even in the $C^k$ context, the word "tangent space" means the (finite-dimensional) space of all tangent vectors. Books that restrict to the $C^infty$ case sometimes define the tangent space using derivations, but that's not the primary definition. In particular, books that consider the $C^k$ case generally ignore derivations completely in favor of some other formalism.
$endgroup$
– Jim Belk
Jun 4 '15 at 16:48
$begingroup$
As a general comment, you shouldn't use the word "tangent space" to refer to the space of all derivations. Even in the $C^k$ context, the word "tangent space" means the (finite-dimensional) space of all tangent vectors. Books that restrict to the $C^infty$ case sometimes define the tangent space using derivations, but that's not the primary definition. In particular, books that consider the $C^k$ case generally ignore derivations completely in favor of some other formalism.
$endgroup$
– Jim Belk
Jun 4 '15 at 16:48
|
show 2 more comments
1 Answer
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active
oldest
votes
$begingroup$
@Pedro: As you know, any $C^k$-atlas is compatible with a $C^infty$-atlas. For Lie groups we have more: $C^1Longrightarrow C^omega$.
- Differential geometry deals only with smooth atlas (in order to identify a tangent vector with a derivation, to work with vector fields as derivations on the algebra of functions $C^infty(M)$...Note that the Lie bracket of two vector fields is a vector field fails to be true if you restrict to $C^1$ functions:
$$partial_{x_i}partial_{x_j}f=partial_{x_j}partial_{x_i}f, text{if} fin C^2(U)$$ - Differential topology deals with functions with less regularity (to use a most general form of Sard's theorem, Morse theory...)
answered Nov 4 '14 at 19:02
amineamine
672516
$endgroup$
2
$begingroup$
The existence of a smooth structure may not be much help. The level set of $C^k$ function in Euclidean space cannot be given a smooth structure that is compatible with the smooth structure of Euclidean space.
$endgroup$
– shuhalo
Oct 24 '17 at 4:56
add a comment |
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$begingroup$
@Pedro: As you know, any $C^k$-atlas is compatible with a $C^infty$-atlas. For Lie groups we have more: $C^1Longrightarrow C^omega$.
- Differential geometry deals only with smooth atlas (in order to identify a tangent vector with a derivation, to work with vector fields as derivations on the algebra of functions $C^infty(M)$...Note that the Lie bracket of two vector fields is a vector field fails to be true if you restrict to $C^1$ functions:
$$partial_{x_i}partial_{x_j}f=partial_{x_j}partial_{x_i}f, text{if} fin C^2(U)$$ - Differential topology deals with functions with less regularity (to use a most general form of Sard's theorem, Morse theory...)
answered Nov 4 '14 at 19:02
amineamine
672516
$endgroup$
2
$begingroup$
The existence of a smooth structure may not be much help. The level set of $C^k$ function in Euclidean space cannot be given a smooth structure that is compatible with the smooth structure of Euclidean space.
$endgroup$
– shuhalo
Oct 24 '17 at 4:56
add a comment |
$begingroup$
@Pedro: As you know, any $C^k$-atlas is compatible with a $C^infty$-atlas. For Lie groups we have more: $C^1Longrightarrow C^omega$.
- Differential geometry deals only with smooth atlas (in order to identify a tangent vector with a derivation, to work with vector fields as derivations on the algebra of functions $C^infty(M)$...Note that the Lie bracket of two vector fields is a vector field fails to be true if you restrict to $C^1$ functions:
$$partial_{x_i}partial_{x_j}f=partial_{x_j}partial_{x_i}f, text{if} fin C^2(U)$$ - Differential topology deals with functions with less regularity (to use a most general form of Sard's theorem, Morse theory...)
answered Nov 4 '14 at 19:02
amineamine
672516
$endgroup$
2
$begingroup$
The existence of a smooth structure may not be much help. The level set of $C^k$ function in Euclidean space cannot be given a smooth structure that is compatible with the smooth structure of Euclidean space.
$endgroup$
– shuhalo
Oct 24 '17 at 4:56
add a comment |
$begingroup$
@Pedro: As you know, any $C^k$-atlas is compatible with a $C^infty$-atlas. For Lie groups we have more: $C^1Longrightarrow C^omega$.
- Differential geometry deals only with smooth atlas (in order to identify a tangent vector with a derivation, to work with vector fields as derivations on the algebra of functions $C^infty(M)$...Note that the Lie bracket of two vector fields is a vector field fails to be true if you restrict to $C^1$ functions:
$$partial_{x_i}partial_{x_j}f=partial_{x_j}partial_{x_i}f, text{if} fin C^2(U)$$ - Differential topology deals with functions with less regularity (to use a most general form of Sard's theorem, Morse theory...)
answered Nov 4 '14 at 19:02
amineamine
672516
$endgroup$
@Pedro: As you know, any $C^k$-atlas is compatible with a $C^infty$-atlas. For Lie groups we have more: $C^1Longrightarrow C^omega$.
- Differential geometry deals only with smooth atlas (in order to identify a tangent vector with a derivation, to work with vector fields as derivations on the algebra of functions $C^infty(M)$...Note that the Lie bracket of two vector fields is a vector field fails to be true if you restrict to $C^1$ functions:
$$partial_{x_i}partial_{x_j}f=partial_{x_j}partial_{x_i}f, text{if} fin C^2(U)$$ - Differential topology deals with functions with less regularity (to use a most general form of Sard's theorem, Morse theory...)
answered Nov 4 '14 at 19:02
amineamine
672516
answered Nov 4 '14 at 19:02
amineamine
672516
answered Nov 4 '14 at 19:02
amineamine
672516
answered Nov 4 '14 at 19:02
amineamine
672516
672516
2
$begingroup$
The existence of a smooth structure may not be much help. The level set of $C^k$ function in Euclidean space cannot be given a smooth structure that is compatible with the smooth structure of Euclidean space.
$endgroup$
– shuhalo
Oct 24 '17 at 4:56
add a comment |
2
$begingroup$
The existence of a smooth structure may not be much help. The level set of $C^k$ function in Euclidean space cannot be given a smooth structure that is compatible with the smooth structure of Euclidean space.
$endgroup$
– shuhalo
Oct 24 '17 at 4:56
2
2
$begingroup$
The existence of a smooth structure may not be much help. The level set of $C^k$ function in Euclidean space cannot be given a smooth structure that is compatible with the smooth structure of Euclidean space.
$endgroup$
– shuhalo
Oct 24 '17 at 4:56
$begingroup$
The existence of a smooth structure may not be much help. The level set of $C^k$ function in Euclidean space cannot be given a smooth structure that is compatible with the smooth structure of Euclidean space.
$endgroup$
– shuhalo
Oct 24 '17 at 4:56
add a comment |
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$begingroup$
I remember losing a little bit of sleep about exactly those questions. My understanding is that every maximal $C^k (kgeq 1)$ atlas contains a $C^infty$ atlas, which is a relief.
$endgroup$
– Tim kinsella
Aug 31 '14 at 11:07
2
$begingroup$
Duistermaat defines Lie groups as $C^2$-manifolds because that suffices to prove they are analytic (better than $C^{infty}$). Nice exposition at terrytao.wordpress.com/2011/06/21/…
$endgroup$
– Colin McLarty
Aug 31 '14 at 11:20
1
$begingroup$
@Lolman: Very interesting! How is it defined then? It'd be lovely if someone could get a reference for showing it's infinite-dimensional in the derivations definition.
$endgroup$
– Pedro
Aug 31 '14 at 11:27
3
$begingroup$
Much of this question is answered at math.stackexchange.com/questions/73677/…
$endgroup$
– Colin McLarty
Aug 31 '14 at 11:57
2
$begingroup$
As a general comment, you shouldn't use the word "tangent space" to refer to the space of all derivations. Even in the $C^k$ context, the word "tangent space" means the (finite-dimensional) space of all tangent vectors. Books that restrict to the $C^infty$ case sometimes define the tangent space using derivations, but that's not the primary definition. In particular, books that consider the $C^k$ case generally ignore derivations completely in favor of some other formalism.
$endgroup$
– Jim Belk
Jun 4 '15 at 16:48