Degrees of freedom of a metric up to coordinate changes (precise formulation)

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Let $M$ be a smooth $n$-dimensional manifold. I have heard that a Riemannian metric on $M$, depends locally on $ n(n+1) / 2 - n = n(n-1) /2$ "independent" functions up to coordinate changes.



I can roughly see the intuition for that: The symmetric matrix $g_ij
$ depends on $n(n+1) / 2$ functions, but you can "change the coordinates" which loses $n$ degrees of freedom.




Is there a precise formulation of this statement? Given an arbitrary metric, can you really specify $n$ out of the $ n(n+1)/2$ $g_ij$ functions as you wish? How can we see it's impossible specify more than $n$?




I heard that Cartan proved something like this, using the language of jets.
(Something about the dimension of the space of $k$-jets of metrics modulu diffeomorphisms).



Maybe there are other references; any pointer in the right direction is welcomed.



This question was supposedly answered here, but I find the answer there to be too vague. I am looking for a precise statement.




differential-geometry reference-request differential-topology riemannian-geometry jet-bundles




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  • This might be related to the fact that the sectional curvature completely determines the Riemann curvature, and the sectional planes (bivectors) have $n(n-1)/2$ degrees of freedom. But the curvature does not determine the metric... math.stackexchange.com/questions/1973957/…
    – mr_e_man
    Aug 22 at 3:04














up vote
5
down vote

favorite












Let $M$ be a smooth $n$-dimensional manifold. I have heard that a Riemannian metric on $M$, depends locally on $ n(n+1) / 2 - n = n(n-1) /2$ "independent" functions up to coordinate changes.



I can roughly see the intuition for that: The symmetric matrix $g_ij
$ depends on $n(n+1) / 2$ functions, but you can "change the coordinates" which loses $n$ degrees of freedom.




Is there a precise formulation of this statement? Given an arbitrary metric, can you really specify $n$ out of the $ n(n+1)/2$ $g_ij$ functions as you wish? How can we see it's impossible specify more than $n$?




I heard that Cartan proved something like this, using the language of jets.
(Something about the dimension of the space of $k$-jets of metrics modulu diffeomorphisms).



Maybe there are other references; any pointer in the right direction is welcomed.



This question was supposedly answered here, but I find the answer there to be too vague. I am looking for a precise statement.




differential-geometry reference-request differential-topology riemannian-geometry jet-bundles




share|cite|improve this question






















  • This might be related to the fact that the sectional curvature completely determines the Riemann curvature, and the sectional planes (bivectors) have $n(n-1)/2$ degrees of freedom. But the curvature does not determine the metric... math.stackexchange.com/questions/1973957/…
    – mr_e_man
    Aug 22 at 3:04












up vote
5
down vote

favorite









up vote
5
down vote

favorite











Let $M$ be a smooth $n$-dimensional manifold. I have heard that a Riemannian metric on $M$, depends locally on $ n(n+1) / 2 - n = n(n-1) /2$ "independent" functions up to coordinate changes.



I can roughly see the intuition for that: The symmetric matrix $g_ij
$ depends on $n(n+1) / 2$ functions, but you can "change the coordinates" which loses $n$ degrees of freedom.




Is there a precise formulation of this statement? Given an arbitrary metric, can you really specify $n$ out of the $ n(n+1)/2$ $g_ij$ functions as you wish? How can we see it's impossible specify more than $n$?




I heard that Cartan proved something like this, using the language of jets.
(Something about the dimension of the space of $k$-jets of metrics modulu diffeomorphisms).



Maybe there are other references; any pointer in the right direction is welcomed.



This question was supposedly answered here, but I find the answer there to be too vague. I am looking for a precise statement.




differential-geometry reference-request differential-topology riemannian-geometry jet-bundles




share|cite|improve this question














Let $M$ be a smooth $n$-dimensional manifold. I have heard that a Riemannian metric on $M$, depends locally on $ n(n+1) / 2 - n = n(n-1) /2$ "independent" functions up to coordinate changes.



I can roughly see the intuition for that: The symmetric matrix $g_ij
$ depends on $n(n+1) / 2$ functions, but you can "change the coordinates" which loses $n$ degrees of freedom.




Is there a precise formulation of this statement? Given an arbitrary metric, can you really specify $n$ out of the $ n(n+1)/2$ $g_ij$ functions as you wish? How can we see it's impossible specify more than $n$?




I heard that Cartan proved something like this, using the language of jets.
(Something about the dimension of the space of $k$-jets of metrics modulu diffeomorphisms).



Maybe there are other references; any pointer in the right direction is welcomed.



This question was supposedly answered here, but I find the answer there to be too vague. I am looking for a precise statement.





differential-geometry reference-request differential-topology riemannian-geometry jet-bundles





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edited Aug 21 at 14:09

























asked Jun 6 at 6:51









Asaf Shachar

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  • This might be related to the fact that the sectional curvature completely determines the Riemann curvature, and the sectional planes (bivectors) have $n(n-1)/2$ degrees of freedom. But the curvature does not determine the metric... math.stackexchange.com/questions/1973957/…
    – mr_e_man
    Aug 22 at 3:04
















  • This might be related to the fact that the sectional curvature completely determines the Riemann curvature, and the sectional planes (bivectors) have $n(n-1)/2$ degrees of freedom. But the curvature does not determine the metric... math.stackexchange.com/questions/1973957/…
    – mr_e_man
    Aug 22 at 3:04















This might be related to the fact that the sectional curvature completely determines the Riemann curvature, and the sectional planes (bivectors) have $n(n-1)/2$ degrees of freedom. But the curvature does not determine the metric... math.stackexchange.com/questions/1973957/…
– mr_e_man
Aug 22 at 3:04




This might be related to the fact that the sectional curvature completely determines the Riemann curvature, and the sectional planes (bivectors) have $n(n-1)/2$ degrees of freedom. But the curvature does not determine the metric... math.stackexchange.com/questions/1973957/…
– mr_e_man
Aug 22 at 3:04















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