$C^{k}$-manifolds: how and why?












33












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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


















33












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















33












33








33


14



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 31 '14 at 12:58







Pedro

















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
















  • 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












1 Answer
1






active

oldest

votes


















0












$begingroup$

@Pedro: As you know, any $C^k$-atlas is compatible with a $C^infty$-atlas. For Lie groups we have more: $C^1Longrightarrow C^omega$.




  1. 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:
    $$partial_{x_i}partial_{x_j}f=partial_{x_j}partial_{x_i}f, text{if} fin C^2(U)$$

  2. Differential topology deals with functions with less regularity (to use a most general form of Sard's theorem, Morse theory...)






share|cite|improve this answer









$endgroup$









  • 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











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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0












$begingroup$

@Pedro: As you know, any $C^k$-atlas is compatible with a $C^infty$-atlas. For Lie groups we have more: $C^1Longrightarrow C^omega$.




  1. 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:
    $$partial_{x_i}partial_{x_j}f=partial_{x_j}partial_{x_i}f, text{if} fin C^2(U)$$

  2. Differential topology deals with functions with less regularity (to use a most general form of Sard's theorem, Morse theory...)






share|cite|improve this answer









$endgroup$









  • 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
















0












$begingroup$

@Pedro: As you know, any $C^k$-atlas is compatible with a $C^infty$-atlas. For Lie groups we have more: $C^1Longrightarrow C^omega$.




  1. 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:
    $$partial_{x_i}partial_{x_j}f=partial_{x_j}partial_{x_i}f, text{if} fin C^2(U)$$

  2. Differential topology deals with functions with less regularity (to use a most general form of Sard's theorem, Morse theory...)






share|cite|improve this answer









$endgroup$









  • 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














0












0








0





$begingroup$

@Pedro: As you know, any $C^k$-atlas is compatible with a $C^infty$-atlas. For Lie groups we have more: $C^1Longrightarrow C^omega$.




  1. 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:
    $$partial_{x_i}partial_{x_j}f=partial_{x_j}partial_{x_i}f, text{if} fin C^2(U)$$

  2. Differential topology deals with functions with less regularity (to use a most general form of Sard's theorem, Morse theory...)






share|cite|improve this answer









$endgroup$



@Pedro: As you know, any $C^k$-atlas is compatible with a $C^infty$-atlas. For Lie groups we have more: $C^1Longrightarrow C^omega$.




  1. 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:
    $$partial_{x_i}partial_{x_j}f=partial_{x_j}partial_{x_i}f, text{if} fin C^2(U)$$

  2. Differential topology deals with functions with less regularity (to use a most general form of Sard's theorem, Morse theory...)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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














  • 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


















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.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
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