Meaningfulness of metric and covariant derivative induced by spherical coordinations
$begingroup$
On $S^2$ have the spherical parametrization
$f:(theta,phi)rightarrow (sin(theta) cos(phi), sin(theta) sin(phi), cos(theta))$. Is it meaningful to talk about the Riemannian metric induced by this only parameterisation? As far as I know we can define a Riemannian metric on any manifold induced by all parametrizations with partition of unity, not only with one parametrization.
riemannian-geometry
edited Jan 1 at 0:17
Bernard
123k741117
asked Dec 31 '18 at 23:33
User12239User12239
343216
$endgroup$
add a comment |
$begingroup$
On $S^2$ have the spherical parametrization
$f:(theta,phi)rightarrow (sin(theta) cos(phi), sin(theta) sin(phi), cos(theta))$. Is it meaningful to talk about the Riemannian metric induced by this only parameterisation? As far as I know we can define a Riemannian metric on any manifold induced by all parametrizations with partition of unity, not only with one parametrization.
riemannian-geometry
edited Jan 1 at 0:17
Bernard
123k741117
asked Dec 31 '18 at 23:33
User12239User12239
343216
$endgroup$
3
$begingroup$
This $f$ is an embedding of $(0,pi) times (0, 2pi)$ into $S^2$. You can use this to transport the Euclidean metric to the image of $f$ in $S^2$, but this doesn't give you a metric on all of $S^2$.
$endgroup$
– Alex Provost
Dec 31 '18 at 23:56
$begingroup$
Thanks, that solved my problem
$endgroup$
– User12239
Jan 1 at 8:32
$begingroup$
In that case I have made the comment into an answer, in order to remove it from the unanswered list.
$endgroup$
– Alex Provost
Jan 1 at 16:06
add a comment |
$begingroup$
On $S^2$ have the spherical parametrization
$f:(theta,phi)rightarrow (sin(theta) cos(phi), sin(theta) sin(phi), cos(theta))$. Is it meaningful to talk about the Riemannian metric induced by this only parameterisation? As far as I know we can define a Riemannian metric on any manifold induced by all parametrizations with partition of unity, not only with one parametrization.
riemannian-geometry
edited Jan 1 at 0:17
Bernard
123k741117
asked Dec 31 '18 at 23:33
User12239User12239
343216
$endgroup$
On $S^2$ have the spherical parametrization
$f:(theta,phi)rightarrow (sin(theta) cos(phi), sin(theta) sin(phi), cos(theta))$. Is it meaningful to talk about the Riemannian metric induced by this only parameterisation? As far as I know we can define a Riemannian metric on any manifold induced by all parametrizations with partition of unity, not only with one parametrization.
riemannian-geometry
riemannian-geometry
edited Jan 1 at 0:17
Bernard
123k741117
asked Dec 31 '18 at 23:33
User12239User12239
343216
edited Jan 1 at 0:17
Bernard
123k741117
asked Dec 31 '18 at 23:33
User12239User12239
343216
edited Jan 1 at 0:17
Bernard
123k741117
edited Jan 1 at 0:17
Bernard
123k741117
edited Jan 1 at 0:17
Bernard
123k741117
123k741117
asked Dec 31 '18 at 23:33
User12239User12239
343216
asked Dec 31 '18 at 23:33
User12239User12239
343216
asked Dec 31 '18 at 23:33
User12239User12239
343216
343216
3
$begingroup$
This $f$ is an embedding of $(0,pi) times (0, 2pi)$ into $S^2$. You can use this to transport the Euclidean metric to the image of $f$ in $S^2$, but this doesn't give you a metric on all of $S^2$.
$endgroup$
– Alex Provost
Dec 31 '18 at 23:56
$begingroup$
Thanks, that solved my problem
$endgroup$
– User12239
Jan 1 at 8:32
$begingroup$
In that case I have made the comment into an answer, in order to remove it from the unanswered list.
$endgroup$
– Alex Provost
Jan 1 at 16:06
add a comment |
3
$begingroup$
This $f$ is an embedding of $(0,pi) times (0, 2pi)$ into $S^2$. You can use this to transport the Euclidean metric to the image of $f$ in $S^2$, but this doesn't give you a metric on all of $S^2$.
$endgroup$
– Alex Provost
Dec 31 '18 at 23:56
$begingroup$
Thanks, that solved my problem
$endgroup$
– User12239
Jan 1 at 8:32
$begingroup$
In that case I have made the comment into an answer, in order to remove it from the unanswered list.
$endgroup$
– Alex Provost
Jan 1 at 16:06
3
3
$begingroup$
This $f$ is an embedding of $(0,pi) times (0, 2pi)$ into $S^2$. You can use this to transport the Euclidean metric to the image of $f$ in $S^2$, but this doesn't give you a metric on all of $S^2$.
$endgroup$
– Alex Provost
Dec 31 '18 at 23:56
$begingroup$
This $f$ is an embedding of $(0,pi) times (0, 2pi)$ into $S^2$. You can use this to transport the Euclidean metric to the image of $f$ in $S^2$, but this doesn't give you a metric on all of $S^2$.
$endgroup$
– Alex Provost
Dec 31 '18 at 23:56
$begingroup$
Thanks, that solved my problem
$endgroup$
– User12239
Jan 1 at 8:32
$begingroup$
Thanks, that solved my problem
$endgroup$
– User12239
Jan 1 at 8:32
$begingroup$
In that case I have made the comment into an answer, in order to remove it from the unanswered list.
$endgroup$
– Alex Provost
Jan 1 at 16:06
$begingroup$
In that case I have made the comment into an answer, in order to remove it from the unanswered list.
$endgroup$
– Alex Provost
Jan 1 at 16:06
add a comment |
1 Answer
1
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oldest
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$begingroup$
This $f$ is an embedding of $(0,pi) times (0,2pi)$ into $S^2$. You can use this to transport the Euclidean metric to the image of $f$ in $S^2$, which is the sphere minus a meridian. But this does not give you a metric on all of $S^2$.
answered Jan 1 at 16:05
Alex ProvostAlex Provost
15.6k22351
$endgroup$
add a comment |
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1 Answer
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$begingroup$
This $f$ is an embedding of $(0,pi) times (0,2pi)$ into $S^2$. You can use this to transport the Euclidean metric to the image of $f$ in $S^2$, which is the sphere minus a meridian. But this does not give you a metric on all of $S^2$.
answered Jan 1 at 16:05
Alex ProvostAlex Provost
15.6k22351
$endgroup$
add a comment |
$begingroup$
This $f$ is an embedding of $(0,pi) times (0,2pi)$ into $S^2$. You can use this to transport the Euclidean metric to the image of $f$ in $S^2$, which is the sphere minus a meridian. But this does not give you a metric on all of $S^2$.
answered Jan 1 at 16:05
Alex ProvostAlex Provost
15.6k22351
$endgroup$
add a comment |
$begingroup$
This $f$ is an embedding of $(0,pi) times (0,2pi)$ into $S^2$. You can use this to transport the Euclidean metric to the image of $f$ in $S^2$, which is the sphere minus a meridian. But this does not give you a metric on all of $S^2$.
answered Jan 1 at 16:05
Alex ProvostAlex Provost
15.6k22351
$endgroup$
This $f$ is an embedding of $(0,pi) times (0,2pi)$ into $S^2$. You can use this to transport the Euclidean metric to the image of $f$ in $S^2$, which is the sphere minus a meridian. But this does not give you a metric on all of $S^2$.
answered Jan 1 at 16:05
Alex ProvostAlex Provost
15.6k22351
answered Jan 1 at 16:05
Alex ProvostAlex Provost
15.6k22351
answered Jan 1 at 16:05
Alex ProvostAlex Provost
15.6k22351
answered Jan 1 at 16:05
Alex ProvostAlex Provost
15.6k22351
15.6k22351
add a comment |
add a comment |
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$begingroup$
This $f$ is an embedding of $(0,pi) times (0, 2pi)$ into $S^2$. You can use this to transport the Euclidean metric to the image of $f$ in $S^2$, but this doesn't give you a metric on all of $S^2$.
$endgroup$
– Alex Provost
Dec 31 '18 at 23:56
$begingroup$
Thanks, that solved my problem
$endgroup$
– User12239
Jan 1 at 8:32
$begingroup$
In that case I have made the comment into an answer, in order to remove it from the unanswered list.
$endgroup$
– Alex Provost
Jan 1 at 16:06