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
asked Jun 6 at 6:51
Asaf Shachar
4,7043835
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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
asked Jun 6 at 6:51
Asaf Shachar
4,7043835
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
add a comment |Â
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
asked Jun 6 at 6:51
Asaf Shachar
4,7043835
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
asked Jun 6 at 6:51
Asaf Shachar
4,7043835
edited Aug 21 at 14:09
asked Jun 6 at 6:51
Asaf Shachar
4,7043835
asked Jun 6 at 6:51
Asaf Shachar
4,7043835
asked Jun 6 at 6:51
Asaf Shachar
4,7043835
4,7043835
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
add a comment |Â
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
add a comment |Â
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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