Meaningfulness of metric and covariant derivative induced by spherical coordinations












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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







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  • 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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













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edited Jan 1 at 0:17









Bernard

123k741117




123k741117










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














  • 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










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 $S^2$.






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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 $S^2$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      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 $S^2$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        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 $S^2$.






        share|cite|improve this answer









        $endgroup$



        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 $S^2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 1 at 16:05









        Alex ProvostAlex Provost

        15.6k22351




        15.6k22351






























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