Taylor expansion of $g^-1/2$ where $g$ is riemannian metric in normal coordinatesDiagonalization of Riemannian Metric and the Laplace Beltrami OperatorApproximate expression for the metric in normal coordinatesTaylor expansion of Riemannian exponential map and Jacobi fields?Taylor series representation for a Riemannian hypersurfaceIntuitive derivation of Taylor expansion?Compute the Riemannian metric induced on $mathbbT^2$ from $mathbbR^3$ in the local coordinates $(θ, ϕ)$.Trying to understand a proof of the existence of isothermal coordinates for $2$-dimensional Riemannian manifoldsRiemannian metric with specified totally geodesic submanifoldsThe identity $x_i = g^ijx_j$ in geodesic normal coordinatesTaylor Expansion in Normal Coordinates
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Taylor expansion of $g^-1/2$ where $g$ is riemannian metric in normal coordinates
Diagonalization of Riemannian Metric and the Laplace Beltrami OperatorApproximate expression for the metric in normal coordinatesTaylor expansion of Riemannian exponential map and Jacobi fields?Taylor series representation for a Riemannian hypersurfaceIntuitive derivation of Taylor expansion?Compute the Riemannian metric induced on $mathbbT^2$ from $mathbbR^3$ in the local coordinates $(θ, ϕ)$.Trying to understand a proof of the existence of isothermal coordinates for $2$-dimensional Riemannian manifoldsRiemannian metric with specified totally geodesic submanifoldsThe identity $x_i = g^ijx_j$ in geodesic normal coordinatesTaylor Expansion in Normal Coordinates
$begingroup$
Let $(M,g)$ be a riemannian manifold.
In normal coordinates for any $qin M$, there is a Taylor expansion of $g_q$ given by
$$(g_q)_ij(x)=delta_ij+frac13R_kijl(q)x^k x^l+O(|x|^3)$$
Now here is my question:
How does one derive from the above expression, that
$$left(sqrtg_q^-1right)^ij(x)
=delta^ij-frac16R_kijl(q)x^k x^l+O(|x|^3)$$
I know how one derives the Taylor expansion of $g$, using a geodesic variation.
Here $sqrtg_q$ denotes the positive square root of $g$ (as a matrix).
I am kinda clueless here, any help would be very much appreciated!
I read this in a paper, but there is no further explanation.
differential-geometry taylor-expansion riemannian-geometry
asked yesterday
Pink PantherPink Panther
12711
$endgroup$
add a comment |
$begingroup$
Let $(M,g)$ be a riemannian manifold.
In normal coordinates for any $qin M$, there is a Taylor expansion of $g_q$ given by
$$(g_q)_ij(x)=delta_ij+frac13R_kijl(q)x^k x^l+O(|x|^3)$$
Now here is my question:
How does one derive from the above expression, that
$$left(sqrtg_q^-1right)^ij(x)
=delta^ij-frac16R_kijl(q)x^k x^l+O(|x|^3)$$
I know how one derives the Taylor expansion of $g$, using a geodesic variation.
Here $sqrtg_q$ denotes the positive square root of $g$ (as a matrix).
I am kinda clueless here, any help would be very much appreciated!
I read this in a paper, but there is no further explanation.
differential-geometry taylor-expansion riemannian-geometry
asked yesterday
Pink PantherPink Panther
12711
$endgroup$
$begingroup$
Could you share the link to the paper? What have you tried?
$endgroup$
– Thibaut Dumont
yesterday
$begingroup$
I think the coefficient is wrong. Inverse of a matrix $1+ epsilon$ is (up to terms of order $epsilon^2$) equal to $1- epsilon$ for a small matrix $epsilon$. Square root of $1-epsilon$ is clearly $1 - frac12 epsilon$.
$endgroup$
– Blazej
15 hours ago
$begingroup$
@Blazej yes you are right, there should be a $frac13$ in the first expression
$endgroup$
– Pink Panther
10 hours ago
add a comment |
$begingroup$
Let $(M,g)$ be a riemannian manifold.
In normal coordinates for any $qin M$, there is a Taylor expansion of $g_q$ given by
$$(g_q)_ij(x)=delta_ij+frac13R_kijl(q)x^k x^l+O(|x|^3)$$
Now here is my question:
How does one derive from the above expression, that
$$left(sqrtg_q^-1right)^ij(x)
=delta^ij-frac16R_kijl(q)x^k x^l+O(|x|^3)$$
I know how one derives the Taylor expansion of $g$, using a geodesic variation.
Here $sqrtg_q$ denotes the positive square root of $g$ (as a matrix).
I am kinda clueless here, any help would be very much appreciated!
I read this in a paper, but there is no further explanation.
differential-geometry taylor-expansion riemannian-geometry
asked yesterday
Pink PantherPink Panther
12711
$endgroup$
Let $(M,g)$ be a riemannian manifold.
In normal coordinates for any $qin M$, there is a Taylor expansion of $g_q$ given by
$$(g_q)_ij(x)=delta_ij+frac13R_kijl(q)x^k x^l+O(|x|^3)$$
Now here is my question:
How does one derive from the above expression, that
$$left(sqrtg_q^-1right)^ij(x)
=delta^ij-frac16R_kijl(q)x^k x^l+O(|x|^3)$$
I know how one derives the Taylor expansion of $g$, using a geodesic variation.
Here $sqrtg_q$ denotes the positive square root of $g$ (as a matrix).
I am kinda clueless here, any help would be very much appreciated!
I read this in a paper, but there is no further explanation.
differential-geometry taylor-expansion riemannian-geometry
differential-geometry taylor-expansion riemannian-geometry
asked yesterday
Pink PantherPink Panther
12711
asked yesterday
Pink PantherPink Panther
12711
edited 10 hours ago
Pink Panther
asked yesterday
Pink PantherPink Panther
12711
asked yesterday
Pink PantherPink Panther
12711
asked yesterday
Pink PantherPink Panther
12711
12711
$begingroup$
Could you share the link to the paper? What have you tried?
$endgroup$
– Thibaut Dumont
yesterday
$begingroup$
I think the coefficient is wrong. Inverse of a matrix $1+ epsilon$ is (up to terms of order $epsilon^2$) equal to $1- epsilon$ for a small matrix $epsilon$. Square root of $1-epsilon$ is clearly $1 - frac12 epsilon$.
$endgroup$
– Blazej
15 hours ago
$begingroup$
@Blazej yes you are right, there should be a $frac13$ in the first expression
$endgroup$
– Pink Panther
10 hours ago
add a comment |
$begingroup$
Could you share the link to the paper? What have you tried?
$endgroup$
– Thibaut Dumont
yesterday
$begingroup$
I think the coefficient is wrong. Inverse of a matrix $1+ epsilon$ is (up to terms of order $epsilon^2$) equal to $1- epsilon$ for a small matrix $epsilon$. Square root of $1-epsilon$ is clearly $1 - frac12 epsilon$.
$endgroup$
– Blazej
15 hours ago
$begingroup$
@Blazej yes you are right, there should be a $frac13$ in the first expression
$endgroup$
– Pink Panther
10 hours ago
$begingroup$
Could you share the link to the paper? What have you tried?
$endgroup$
– Thibaut Dumont
yesterday
$begingroup$
Could you share the link to the paper? What have you tried?
$endgroup$
– Thibaut Dumont
yesterday
$begingroup$
I think the coefficient is wrong. Inverse of a matrix $1+ epsilon$ is (up to terms of order $epsilon^2$) equal to $1- epsilon$ for a small matrix $epsilon$. Square root of $1-epsilon$ is clearly $1 - frac12 epsilon$.
$endgroup$
– Blazej
15 hours ago
$begingroup$
I think the coefficient is wrong. Inverse of a matrix $1+ epsilon$ is (up to terms of order $epsilon^2$) equal to $1- epsilon$ for a small matrix $epsilon$. Square root of $1-epsilon$ is clearly $1 - frac12 epsilon$.
$endgroup$
– Blazej
15 hours ago
$begingroup$
@Blazej yes you are right, there should be a $frac13$ in the first expression
$endgroup$
– Pink Panther
10 hours ago
$begingroup$
@Blazej yes you are right, there should be a $frac13$ in the first expression
$endgroup$
– Pink Panther
10 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let us start with another description of the setting. We have $0in UsubsetmathbbR^n$ an open set and a map $g:UtomathrmSym^2left(mathbbR^nright),$ where $mathrmSym^2left(mathbbR^nright)$ denotes the space of symmetric matrices. The Taylor expansion you quote can also be written in terms of matrix-valued maps as $$g(x)=I+frac13R_kl(0)x^kx^l+Oleft(|x|^3right),$$ where $R_kl(0)$ is the matrix whose $ij$-entry is $R_kijl(0)$. By Taylor's theorem, this is equivalent to saying that the map $g$ is twice differentiable at $0$ and satisfies $$g(0)=I,quad fracpartial gpartial x^i=0,quadfracpartial^2gpartial x^kpartial x^l=frac23R_kl(0).$$ Finally, the map $sqrt;;:mathrmSym^2_+left(mathbbR^nright)tomathrmSym^2_+left(mathbbR^nright)$ is differentiable, and so is the inverse map for matrices. This means that the map $sqrtg^-1$ is also differentiable, and you can compute its derivatives (that is, its Taylor expansion) using the chain rule.
answered 15 hours ago
Amitai YuvalAmitai Yuval
15.5k11127
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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active
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votes
$begingroup$
Let us start with another description of the setting. We have $0in UsubsetmathbbR^n$ an open set and a map $g:UtomathrmSym^2left(mathbbR^nright),$ where $mathrmSym^2left(mathbbR^nright)$ denotes the space of symmetric matrices. The Taylor expansion you quote can also be written in terms of matrix-valued maps as $$g(x)=I+frac13R_kl(0)x^kx^l+Oleft(|x|^3right),$$ where $R_kl(0)$ is the matrix whose $ij$-entry is $R_kijl(0)$. By Taylor's theorem, this is equivalent to saying that the map $g$ is twice differentiable at $0$ and satisfies $$g(0)=I,quad fracpartial gpartial x^i=0,quadfracpartial^2gpartial x^kpartial x^l=frac23R_kl(0).$$ Finally, the map $sqrt;;:mathrmSym^2_+left(mathbbR^nright)tomathrmSym^2_+left(mathbbR^nright)$ is differentiable, and so is the inverse map for matrices. This means that the map $sqrtg^-1$ is also differentiable, and you can compute its derivatives (that is, its Taylor expansion) using the chain rule.
answered 15 hours ago
Amitai YuvalAmitai Yuval
15.5k11127
$endgroup$
add a comment |
$begingroup$
Let us start with another description of the setting. We have $0in UsubsetmathbbR^n$ an open set and a map $g:UtomathrmSym^2left(mathbbR^nright),$ where $mathrmSym^2left(mathbbR^nright)$ denotes the space of symmetric matrices. The Taylor expansion you quote can also be written in terms of matrix-valued maps as $$g(x)=I+frac13R_kl(0)x^kx^l+Oleft(|x|^3right),$$ where $R_kl(0)$ is the matrix whose $ij$-entry is $R_kijl(0)$. By Taylor's theorem, this is equivalent to saying that the map $g$ is twice differentiable at $0$ and satisfies $$g(0)=I,quad fracpartial gpartial x^i=0,quadfracpartial^2gpartial x^kpartial x^l=frac23R_kl(0).$$ Finally, the map $sqrt;;:mathrmSym^2_+left(mathbbR^nright)tomathrmSym^2_+left(mathbbR^nright)$ is differentiable, and so is the inverse map for matrices. This means that the map $sqrtg^-1$ is also differentiable, and you can compute its derivatives (that is, its Taylor expansion) using the chain rule.
answered 15 hours ago
Amitai YuvalAmitai Yuval
15.5k11127
$endgroup$
add a comment |
$begingroup$
Let us start with another description of the setting. We have $0in UsubsetmathbbR^n$ an open set and a map $g:UtomathrmSym^2left(mathbbR^nright),$ where $mathrmSym^2left(mathbbR^nright)$ denotes the space of symmetric matrices. The Taylor expansion you quote can also be written in terms of matrix-valued maps as $$g(x)=I+frac13R_kl(0)x^kx^l+Oleft(|x|^3right),$$ where $R_kl(0)$ is the matrix whose $ij$-entry is $R_kijl(0)$. By Taylor's theorem, this is equivalent to saying that the map $g$ is twice differentiable at $0$ and satisfies $$g(0)=I,quad fracpartial gpartial x^i=0,quadfracpartial^2gpartial x^kpartial x^l=frac23R_kl(0).$$ Finally, the map $sqrt;;:mathrmSym^2_+left(mathbbR^nright)tomathrmSym^2_+left(mathbbR^nright)$ is differentiable, and so is the inverse map for matrices. This means that the map $sqrtg^-1$ is also differentiable, and you can compute its derivatives (that is, its Taylor expansion) using the chain rule.
answered 15 hours ago
Amitai YuvalAmitai Yuval
15.5k11127
$endgroup$
Let us start with another description of the setting. We have $0in UsubsetmathbbR^n$ an open set and a map $g:UtomathrmSym^2left(mathbbR^nright),$ where $mathrmSym^2left(mathbbR^nright)$ denotes the space of symmetric matrices. The Taylor expansion you quote can also be written in terms of matrix-valued maps as $$g(x)=I+frac13R_kl(0)x^kx^l+Oleft(|x|^3right),$$ where $R_kl(0)$ is the matrix whose $ij$-entry is $R_kijl(0)$. By Taylor's theorem, this is equivalent to saying that the map $g$ is twice differentiable at $0$ and satisfies $$g(0)=I,quad fracpartial gpartial x^i=0,quadfracpartial^2gpartial x^kpartial x^l=frac23R_kl(0).$$ Finally, the map $sqrt;;:mathrmSym^2_+left(mathbbR^nright)tomathrmSym^2_+left(mathbbR^nright)$ is differentiable, and so is the inverse map for matrices. This means that the map $sqrtg^-1$ is also differentiable, and you can compute its derivatives (that is, its Taylor expansion) using the chain rule.
answered 15 hours ago
Amitai YuvalAmitai Yuval
15.5k11127
edited 10 hours ago
answered 15 hours ago
Amitai YuvalAmitai Yuval
15.5k11127
answered 15 hours ago
Amitai YuvalAmitai Yuval
15.5k11127
answered 15 hours ago
Amitai YuvalAmitai Yuval
15.5k11127
15.5k11127
add a comment |
add a comment |
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$begingroup$
Could you share the link to the paper? What have you tried?
$endgroup$
– Thibaut Dumont
yesterday
$begingroup$
I think the coefficient is wrong. Inverse of a matrix $1+ epsilon$ is (up to terms of order $epsilon^2$) equal to $1- epsilon$ for a small matrix $epsilon$. Square root of $1-epsilon$ is clearly $1 - frac12 epsilon$.
$endgroup$
– Blazej
15 hours ago
$begingroup$
@Blazej yes you are right, there should be a $frac13$ in the first expression
$endgroup$
– Pink Panther
10 hours ago