Christoffel symbols for the Poincaré ball model
$begingroup$
The metric tensor $g_{ij}$ of the Poincaré ball model is
$$ g_{ij} = frac{delta_{ij}}{(1 - x_k x^k)^2} $$
where $delta_{ij}$ is the Kronecker delta and $x^k$ are the ambient Cartesian coordinates.
Hence the partial derivative of the metric tensor with respect to a coordinate $x^l$ is
$$ partial_l g_{ij} = partial_l frac{delta_{ij}}{(1 - x_k x^k)^2} = delta_{ij} partial_l (1 - x_k x^k)^{-2} = -2 delta_{ij} (1 - x_k x^k)^{-3} partial_l (1 - x_k x^k) = 2 delta_{ij} (1 - x_k x^k)^{-3} (x_k partial_l x^k + x^k partial_l x_k) = 2 delta_{ij} (1 - x_k x^k)^{-3} (x_k delta_l^k + x^k delta_{lk}) $$
In summary
$$ partial_l g_{ij} = 2 delta_{ij} (1 - x_k x^k)^{-3} (x_k delta_l^k + x^k delta_{lk}) = 4 delta_{ij} (1 - x_k x^k)^{-3} x_l $$
The Christoffel symbols are defined in terms of the partial derivatives of the metric tensor as
$$ Gamma^i_{jk} = frac{1}{2} g^{il} (partial_j g_{lk} + partial_k g_{lj} - partial_l g_{jk}) $$
Hence we substitute our expression for $ partial_l g_{ij} $ with the right indices.
Is this correct? I have not been able to find an online source to verify that these are the correct Christoffel symbols for the Poincaré ball model.
differential-geometry riemannian-geometry tensors hyperbolic-geometry connections
asked Dec 20 '15 at 16:32
user76284user76284
1,2031125
$endgroup$
add a comment |
$begingroup$
The metric tensor $g_{ij}$ of the Poincaré ball model is
$$ g_{ij} = frac{delta_{ij}}{(1 - x_k x^k)^2} $$
where $delta_{ij}$ is the Kronecker delta and $x^k$ are the ambient Cartesian coordinates.
Hence the partial derivative of the metric tensor with respect to a coordinate $x^l$ is
$$ partial_l g_{ij} = partial_l frac{delta_{ij}}{(1 - x_k x^k)^2} = delta_{ij} partial_l (1 - x_k x^k)^{-2} = -2 delta_{ij} (1 - x_k x^k)^{-3} partial_l (1 - x_k x^k) = 2 delta_{ij} (1 - x_k x^k)^{-3} (x_k partial_l x^k + x^k partial_l x_k) = 2 delta_{ij} (1 - x_k x^k)^{-3} (x_k delta_l^k + x^k delta_{lk}) $$
In summary
$$ partial_l g_{ij} = 2 delta_{ij} (1 - x_k x^k)^{-3} (x_k delta_l^k + x^k delta_{lk}) = 4 delta_{ij} (1 - x_k x^k)^{-3} x_l $$
The Christoffel symbols are defined in terms of the partial derivatives of the metric tensor as
$$ Gamma^i_{jk} = frac{1}{2} g^{il} (partial_j g_{lk} + partial_k g_{lj} - partial_l g_{jk}) $$
Hence we substitute our expression for $ partial_l g_{ij} $ with the right indices.
Is this correct? I have not been able to find an online source to verify that these are the correct Christoffel symbols for the Poincaré ball model.
differential-geometry riemannian-geometry tensors hyperbolic-geometry connections
asked Dec 20 '15 at 16:32
user76284user76284
1,2031125
$endgroup$
$begingroup$
@Winther Thanks! Fixed it.
$endgroup$
– user76284
Dec 20 '15 at 17:43
1
$begingroup$
Then it looks fine. For reference I get $Gamma^i_{jk} = frac{2}{(1 - x_mu x^mu)}[delta^i_kx_{j} + delta^i_jx_{k} - delta_{jk}x^i]$ as the final result. The only thing I can see to criticize is the use of the notation $x_kx^k$ (which I also use above:) This usually means $g^{ij}x_ix_j$, i.e. the lower indices are lowered using the metric so it could potentially be confused.
$endgroup$
– Winther
Dec 20 '15 at 17:50
$begingroup$
@Winther Ah, of course. What would be a better notation?
$endgroup$
– user76284
Dec 20 '15 at 18:10
add a comment |
$begingroup$
The metric tensor $g_{ij}$ of the Poincaré ball model is
$$ g_{ij} = frac{delta_{ij}}{(1 - x_k x^k)^2} $$
where $delta_{ij}$ is the Kronecker delta and $x^k$ are the ambient Cartesian coordinates.
Hence the partial derivative of the metric tensor with respect to a coordinate $x^l$ is
$$ partial_l g_{ij} = partial_l frac{delta_{ij}}{(1 - x_k x^k)^2} = delta_{ij} partial_l (1 - x_k x^k)^{-2} = -2 delta_{ij} (1 - x_k x^k)^{-3} partial_l (1 - x_k x^k) = 2 delta_{ij} (1 - x_k x^k)^{-3} (x_k partial_l x^k + x^k partial_l x_k) = 2 delta_{ij} (1 - x_k x^k)^{-3} (x_k delta_l^k + x^k delta_{lk}) $$
In summary
$$ partial_l g_{ij} = 2 delta_{ij} (1 - x_k x^k)^{-3} (x_k delta_l^k + x^k delta_{lk}) = 4 delta_{ij} (1 - x_k x^k)^{-3} x_l $$
The Christoffel symbols are defined in terms of the partial derivatives of the metric tensor as
$$ Gamma^i_{jk} = frac{1}{2} g^{il} (partial_j g_{lk} + partial_k g_{lj} - partial_l g_{jk}) $$
Hence we substitute our expression for $ partial_l g_{ij} $ with the right indices.
Is this correct? I have not been able to find an online source to verify that these are the correct Christoffel symbols for the Poincaré ball model.
differential-geometry riemannian-geometry tensors hyperbolic-geometry connections
asked Dec 20 '15 at 16:32
user76284user76284
1,2031125
$endgroup$
The metric tensor $g_{ij}$ of the Poincaré ball model is
$$ g_{ij} = frac{delta_{ij}}{(1 - x_k x^k)^2} $$
where $delta_{ij}$ is the Kronecker delta and $x^k$ are the ambient Cartesian coordinates.
Hence the partial derivative of the metric tensor with respect to a coordinate $x^l$ is
$$ partial_l g_{ij} = partial_l frac{delta_{ij}}{(1 - x_k x^k)^2} = delta_{ij} partial_l (1 - x_k x^k)^{-2} = -2 delta_{ij} (1 - x_k x^k)^{-3} partial_l (1 - x_k x^k) = 2 delta_{ij} (1 - x_k x^k)^{-3} (x_k partial_l x^k + x^k partial_l x_k) = 2 delta_{ij} (1 - x_k x^k)^{-3} (x_k delta_l^k + x^k delta_{lk}) $$
In summary
$$ partial_l g_{ij} = 2 delta_{ij} (1 - x_k x^k)^{-3} (x_k delta_l^k + x^k delta_{lk}) = 4 delta_{ij} (1 - x_k x^k)^{-3} x_l $$
The Christoffel symbols are defined in terms of the partial derivatives of the metric tensor as
$$ Gamma^i_{jk} = frac{1}{2} g^{il} (partial_j g_{lk} + partial_k g_{lj} - partial_l g_{jk}) $$
Hence we substitute our expression for $ partial_l g_{ij} $ with the right indices.
Is this correct? I have not been able to find an online source to verify that these are the correct Christoffel symbols for the Poincaré ball model.
differential-geometry riemannian-geometry tensors hyperbolic-geometry connections
differential-geometry riemannian-geometry tensors hyperbolic-geometry connections
asked Dec 20 '15 at 16:32
user76284user76284
1,2031125
asked Dec 20 '15 at 16:32
user76284user76284
1,2031125
edited Dec 20 '15 at 17:43
user76284
asked Dec 20 '15 at 16:32
user76284user76284
1,2031125
asked Dec 20 '15 at 16:32
user76284user76284
1,2031125
asked Dec 20 '15 at 16:32
user76284user76284
1,2031125
1,2031125
$begingroup$
@Winther Thanks! Fixed it.
$endgroup$
– user76284
Dec 20 '15 at 17:43
1
$begingroup$
Then it looks fine. For reference I get $Gamma^i_{jk} = frac{2}{(1 - x_mu x^mu)}[delta^i_kx_{j} + delta^i_jx_{k} - delta_{jk}x^i]$ as the final result. The only thing I can see to criticize is the use of the notation $x_kx^k$ (which I also use above:) This usually means $g^{ij}x_ix_j$, i.e. the lower indices are lowered using the metric so it could potentially be confused.
$endgroup$
– Winther
Dec 20 '15 at 17:50
$begingroup$
@Winther Ah, of course. What would be a better notation?
$endgroup$
– user76284
Dec 20 '15 at 18:10
add a comment |
$begingroup$
@Winther Thanks! Fixed it.
$endgroup$
– user76284
Dec 20 '15 at 17:43
1
$begingroup$
Then it looks fine. For reference I get $Gamma^i_{jk} = frac{2}{(1 - x_mu x^mu)}[delta^i_kx_{j} + delta^i_jx_{k} - delta_{jk}x^i]$ as the final result. The only thing I can see to criticize is the use of the notation $x_kx^k$ (which I also use above:) This usually means $g^{ij}x_ix_j$, i.e. the lower indices are lowered using the metric so it could potentially be confused.
$endgroup$
– Winther
Dec 20 '15 at 17:50
$begingroup$
@Winther Ah, of course. What would be a better notation?
$endgroup$
– user76284
Dec 20 '15 at 18:10
$begingroup$
@Winther Thanks! Fixed it.
$endgroup$
– user76284
Dec 20 '15 at 17:43
$begingroup$
@Winther Thanks! Fixed it.
$endgroup$
– user76284
Dec 20 '15 at 17:43
1
1
$begingroup$
Then it looks fine. For reference I get $Gamma^i_{jk} = frac{2}{(1 - x_mu x^mu)}[delta^i_kx_{j} + delta^i_jx_{k} - delta_{jk}x^i]$ as the final result. The only thing I can see to criticize is the use of the notation $x_kx^k$ (which I also use above:) This usually means $g^{ij}x_ix_j$, i.e. the lower indices are lowered using the metric so it could potentially be confused.
$endgroup$
– Winther
Dec 20 '15 at 17:50
$begingroup$
Then it looks fine. For reference I get $Gamma^i_{jk} = frac{2}{(1 - x_mu x^mu)}[delta^i_kx_{j} + delta^i_jx_{k} - delta_{jk}x^i]$ as the final result. The only thing I can see to criticize is the use of the notation $x_kx^k$ (which I also use above:) This usually means $g^{ij}x_ix_j$, i.e. the lower indices are lowered using the metric so it could potentially be confused.
$endgroup$
– Winther
Dec 20 '15 at 17:50
$begingroup$
@Winther Ah, of course. What would be a better notation?
$endgroup$
– user76284
Dec 20 '15 at 18:10
$begingroup$
@Winther Ah, of course. What would be a better notation?
$endgroup$
– user76284
Dec 20 '15 at 18:10
add a comment |
1 Answer
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For future reference: check pages 160 and 161 of (the first edition?) do Carmo's Riemannian Geometry book. For $$g_{ij} = frac{delta_{ij}}{F^2}$$he proves that $$Gamma_{ij}^k = -delta_{jk}f_i-delta_{ki}f_j+delta_{ij}f_k,$$where $log F = f$ and $f_i = partial f/partial x_i$. In this case, we have $F(x) = 1-|x|^2$, so that $$f(x) = log F(x) = log (1-|x|^2)implies frac{partial f}{partial x_i}(x) = frac{-2x_i}{1-|x|^2}.$$Putting everything together, one obtains $$Gamma_{ij}^k = frac{2(delta_{jk}x_i + delta_{ki}x_j-delta_{ij}x_k)}{1-|x|^2}.$$
answered Jan 10 at 5:43
Ivo TerekIvo Terek
45.4k952141
$endgroup$
$begingroup$
Thanks. Shouldn't some of your indices be raised?
$endgroup$
– user76284
Jan 10 at 6:08
$begingroup$
If you want to respect the index balance required by Einstein's convention, yes. The only reason I did not raise those indices was to keep the notation the same notation as do Carmo's (he does not use Einstein's convention), in case anyone reading actually wants to look at the book.
$endgroup$
– Ivo Terek
Jan 11 at 2:35
add a comment |
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$begingroup$
For future reference: check pages 160 and 161 of (the first edition?) do Carmo's Riemannian Geometry book. For $$g_{ij} = frac{delta_{ij}}{F^2}$$he proves that $$Gamma_{ij}^k = -delta_{jk}f_i-delta_{ki}f_j+delta_{ij}f_k,$$where $log F = f$ and $f_i = partial f/partial x_i$. In this case, we have $F(x) = 1-|x|^2$, so that $$f(x) = log F(x) = log (1-|x|^2)implies frac{partial f}{partial x_i}(x) = frac{-2x_i}{1-|x|^2}.$$Putting everything together, one obtains $$Gamma_{ij}^k = frac{2(delta_{jk}x_i + delta_{ki}x_j-delta_{ij}x_k)}{1-|x|^2}.$$
answered Jan 10 at 5:43
Ivo TerekIvo Terek
45.4k952141
$endgroup$
$begingroup$
Thanks. Shouldn't some of your indices be raised?
$endgroup$
– user76284
Jan 10 at 6:08
$begingroup$
If you want to respect the index balance required by Einstein's convention, yes. The only reason I did not raise those indices was to keep the notation the same notation as do Carmo's (he does not use Einstein's convention), in case anyone reading actually wants to look at the book.
$endgroup$
– Ivo Terek
Jan 11 at 2:35
add a comment |
$begingroup$
For future reference: check pages 160 and 161 of (the first edition?) do Carmo's Riemannian Geometry book. For $$g_{ij} = frac{delta_{ij}}{F^2}$$he proves that $$Gamma_{ij}^k = -delta_{jk}f_i-delta_{ki}f_j+delta_{ij}f_k,$$where $log F = f$ and $f_i = partial f/partial x_i$. In this case, we have $F(x) = 1-|x|^2$, so that $$f(x) = log F(x) = log (1-|x|^2)implies frac{partial f}{partial x_i}(x) = frac{-2x_i}{1-|x|^2}.$$Putting everything together, one obtains $$Gamma_{ij}^k = frac{2(delta_{jk}x_i + delta_{ki}x_j-delta_{ij}x_k)}{1-|x|^2}.$$
answered Jan 10 at 5:43
Ivo TerekIvo Terek
45.4k952141
$endgroup$
$begingroup$
Thanks. Shouldn't some of your indices be raised?
$endgroup$
– user76284
Jan 10 at 6:08
$begingroup$
If you want to respect the index balance required by Einstein's convention, yes. The only reason I did not raise those indices was to keep the notation the same notation as do Carmo's (he does not use Einstein's convention), in case anyone reading actually wants to look at the book.
$endgroup$
– Ivo Terek
Jan 11 at 2:35
add a comment |
$begingroup$
For future reference: check pages 160 and 161 of (the first edition?) do Carmo's Riemannian Geometry book. For $$g_{ij} = frac{delta_{ij}}{F^2}$$he proves that $$Gamma_{ij}^k = -delta_{jk}f_i-delta_{ki}f_j+delta_{ij}f_k,$$where $log F = f$ and $f_i = partial f/partial x_i$. In this case, we have $F(x) = 1-|x|^2$, so that $$f(x) = log F(x) = log (1-|x|^2)implies frac{partial f}{partial x_i}(x) = frac{-2x_i}{1-|x|^2}.$$Putting everything together, one obtains $$Gamma_{ij}^k = frac{2(delta_{jk}x_i + delta_{ki}x_j-delta_{ij}x_k)}{1-|x|^2}.$$
answered Jan 10 at 5:43
Ivo TerekIvo Terek
45.4k952141
$endgroup$
For future reference: check pages 160 and 161 of (the first edition?) do Carmo's Riemannian Geometry book. For $$g_{ij} = frac{delta_{ij}}{F^2}$$he proves that $$Gamma_{ij}^k = -delta_{jk}f_i-delta_{ki}f_j+delta_{ij}f_k,$$where $log F = f$ and $f_i = partial f/partial x_i$. In this case, we have $F(x) = 1-|x|^2$, so that $$f(x) = log F(x) = log (1-|x|^2)implies frac{partial f}{partial x_i}(x) = frac{-2x_i}{1-|x|^2}.$$Putting everything together, one obtains $$Gamma_{ij}^k = frac{2(delta_{jk}x_i + delta_{ki}x_j-delta_{ij}x_k)}{1-|x|^2}.$$
answered Jan 10 at 5:43
Ivo TerekIvo Terek
45.4k952141
answered Jan 10 at 5:43
Ivo TerekIvo Terek
45.4k952141
answered Jan 10 at 5:43
Ivo TerekIvo Terek
45.4k952141
answered Jan 10 at 5:43
Ivo TerekIvo Terek
45.4k952141
45.4k952141
$begingroup$
Thanks. Shouldn't some of your indices be raised?
$endgroup$
– user76284
Jan 10 at 6:08
$begingroup$
If you want to respect the index balance required by Einstein's convention, yes. The only reason I did not raise those indices was to keep the notation the same notation as do Carmo's (he does not use Einstein's convention), in case anyone reading actually wants to look at the book.
$endgroup$
– Ivo Terek
Jan 11 at 2:35
add a comment |
$begingroup$
Thanks. Shouldn't some of your indices be raised?
$endgroup$
– user76284
Jan 10 at 6:08
$begingroup$
If you want to respect the index balance required by Einstein's convention, yes. The only reason I did not raise those indices was to keep the notation the same notation as do Carmo's (he does not use Einstein's convention), in case anyone reading actually wants to look at the book.
$endgroup$
– Ivo Terek
Jan 11 at 2:35
$begingroup$
Thanks. Shouldn't some of your indices be raised?
$endgroup$
– user76284
Jan 10 at 6:08
$begingroup$
Thanks. Shouldn't some of your indices be raised?
$endgroup$
– user76284
Jan 10 at 6:08
$begingroup$
If you want to respect the index balance required by Einstein's convention, yes. The only reason I did not raise those indices was to keep the notation the same notation as do Carmo's (he does not use Einstein's convention), in case anyone reading actually wants to look at the book.
$endgroup$
– Ivo Terek
Jan 11 at 2:35
$begingroup$
If you want to respect the index balance required by Einstein's convention, yes. The only reason I did not raise those indices was to keep the notation the same notation as do Carmo's (he does not use Einstein's convention), in case anyone reading actually wants to look at the book.
$endgroup$
– Ivo Terek
Jan 11 at 2:35
add a comment |
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$begingroup$
@Winther Thanks! Fixed it.
$endgroup$
– user76284
Dec 20 '15 at 17:43
1
$begingroup$
Then it looks fine. For reference I get $Gamma^i_{jk} = frac{2}{(1 - x_mu x^mu)}[delta^i_kx_{j} + delta^i_jx_{k} - delta_{jk}x^i]$ as the final result. The only thing I can see to criticize is the use of the notation $x_kx^k$ (which I also use above:) This usually means $g^{ij}x_ix_j$, i.e. the lower indices are lowered using the metric so it could potentially be confused.
$endgroup$
– Winther
Dec 20 '15 at 17:50
$begingroup$
@Winther Ah, of course. What would be a better notation?
$endgroup$
– user76284
Dec 20 '15 at 18:10