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Distance function-inequality in unit sphere



The 2019 Stack Overflow Developer Survey Results Are InDecomposition of normal curvature of curve in $mathbbR^3$Each point of $M$ has a smooth coordinate neighborhood in which the coordinate frame is orthonormal iff $g$ is flatExistence of a local geodesic frameRelationship between euclidean metric in sphere of radius $r$ and the unit sphere.Is the unit bundle of a Finsler vector bundle a sphere bundle?What is the best approximate of points on a sphere?A constant curvature manifold is EinsteinGiven a basis of $T_pM (e_i)$. Extend this base to a local orthonormal frame $(E_i)$ with $nabla E_i (p) = 0 $Distance function on the complex projective spaceExponential map on Heisenberg group is a diffeomorphism.Is the Euclidean Distance and the Euclidean Norm the same thing?










0












$begingroup$


Assume that $d$ is a distance function (i.e. Riemannian metric) on a unit sphere $X$ of Euclidean space $mathbbE^3$ (That is, $d$ is an angle between two unit vectors). If $e_i$ is an orthonormal basis in $mathbbE^3$, i.e. $d(e_i,e_j)=fracpi2$ for all $ineq j$, then prove that




$$ sum_i=1^3
bigg|pi - d(e_i,u)-d(e_i,v) bigg| geq pi - d(u,v)
$$




Proof : When $f(x)=bigg|pi - T(x)bigg|$ and $T(x)=d(x,u)+d(x,v)$, then $T$ is a 2-Lipschitz function. And $f(x)=f(-x)$.










share|cite|improve this question











$endgroup$











  • $begingroup$
    What's $S^2(1)$?
    $endgroup$
    – Alex R.
    Apr 17 '18 at 3:51










  • $begingroup$
    It is a unit sphere in $mathbbR^3$ and $d$ is intrinsic metric on it.
    $endgroup$
    – HK Lee
    Apr 17 '18 at 3:52










  • $begingroup$
    Are u,v generic vectors of $mathbbR^3$?
    $endgroup$
    – big-lion
    Apr 17 '18 at 3:59










  • $begingroup$
    @big-lion : I edited.
    $endgroup$
    – HK Lee
    Apr 17 '18 at 4:03















0












$begingroup$


Assume that $d$ is a distance function (i.e. Riemannian metric) on a unit sphere $X$ of Euclidean space $mathbbE^3$ (That is, $d$ is an angle between two unit vectors). If $e_i$ is an orthonormal basis in $mathbbE^3$, i.e. $d(e_i,e_j)=fracpi2$ for all $ineq j$, then prove that




$$ sum_i=1^3
bigg|pi - d(e_i,u)-d(e_i,v) bigg| geq pi - d(u,v)
$$




Proof : When $f(x)=bigg|pi - T(x)bigg|$ and $T(x)=d(x,u)+d(x,v)$, then $T$ is a 2-Lipschitz function. And $f(x)=f(-x)$.










share|cite|improve this question











$endgroup$











  • $begingroup$
    What's $S^2(1)$?
    $endgroup$
    – Alex R.
    Apr 17 '18 at 3:51










  • $begingroup$
    It is a unit sphere in $mathbbR^3$ and $d$ is intrinsic metric on it.
    $endgroup$
    – HK Lee
    Apr 17 '18 at 3:52










  • $begingroup$
    Are u,v generic vectors of $mathbbR^3$?
    $endgroup$
    – big-lion
    Apr 17 '18 at 3:59










  • $begingroup$
    @big-lion : I edited.
    $endgroup$
    – HK Lee
    Apr 17 '18 at 4:03













0












0








0





$begingroup$


Assume that $d$ is a distance function (i.e. Riemannian metric) on a unit sphere $X$ of Euclidean space $mathbbE^3$ (That is, $d$ is an angle between two unit vectors). If $e_i$ is an orthonormal basis in $mathbbE^3$, i.e. $d(e_i,e_j)=fracpi2$ for all $ineq j$, then prove that




$$ sum_i=1^3
bigg|pi - d(e_i,u)-d(e_i,v) bigg| geq pi - d(u,v)
$$




Proof : When $f(x)=bigg|pi - T(x)bigg|$ and $T(x)=d(x,u)+d(x,v)$, then $T$ is a 2-Lipschitz function. And $f(x)=f(-x)$.










share|cite|improve this question











$endgroup$




Assume that $d$ is a distance function (i.e. Riemannian metric) on a unit sphere $X$ of Euclidean space $mathbbE^3$ (That is, $d$ is an angle between two unit vectors). If $e_i$ is an orthonormal basis in $mathbbE^3$, i.e. $d(e_i,e_j)=fracpi2$ for all $ineq j$, then prove that




$$ sum_i=1^3
bigg|pi - d(e_i,u)-d(e_i,v) bigg| geq pi - d(u,v)
$$




Proof : When $f(x)=bigg|pi - T(x)bigg|$ and $T(x)=d(x,u)+d(x,v)$, then $T$ is a 2-Lipschitz function. And $f(x)=f(-x)$.







euclidean-geometry riemannian-geometry geometric-inequalities






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 8 at 0:45







HK Lee

















asked Apr 17 '18 at 3:40









HK LeeHK Lee

14.1k52362




14.1k52362











  • $begingroup$
    What's $S^2(1)$?
    $endgroup$
    – Alex R.
    Apr 17 '18 at 3:51










  • $begingroup$
    It is a unit sphere in $mathbbR^3$ and $d$ is intrinsic metric on it.
    $endgroup$
    – HK Lee
    Apr 17 '18 at 3:52










  • $begingroup$
    Are u,v generic vectors of $mathbbR^3$?
    $endgroup$
    – big-lion
    Apr 17 '18 at 3:59










  • $begingroup$
    @big-lion : I edited.
    $endgroup$
    – HK Lee
    Apr 17 '18 at 4:03
















  • $begingroup$
    What's $S^2(1)$?
    $endgroup$
    – Alex R.
    Apr 17 '18 at 3:51










  • $begingroup$
    It is a unit sphere in $mathbbR^3$ and $d$ is intrinsic metric on it.
    $endgroup$
    – HK Lee
    Apr 17 '18 at 3:52










  • $begingroup$
    Are u,v generic vectors of $mathbbR^3$?
    $endgroup$
    – big-lion
    Apr 17 '18 at 3:59










  • $begingroup$
    @big-lion : I edited.
    $endgroup$
    – HK Lee
    Apr 17 '18 at 4:03















$begingroup$
What's $S^2(1)$?
$endgroup$
– Alex R.
Apr 17 '18 at 3:51




$begingroup$
What's $S^2(1)$?
$endgroup$
– Alex R.
Apr 17 '18 at 3:51












$begingroup$
It is a unit sphere in $mathbbR^3$ and $d$ is intrinsic metric on it.
$endgroup$
– HK Lee
Apr 17 '18 at 3:52




$begingroup$
It is a unit sphere in $mathbbR^3$ and $d$ is intrinsic metric on it.
$endgroup$
– HK Lee
Apr 17 '18 at 3:52












$begingroup$
Are u,v generic vectors of $mathbbR^3$?
$endgroup$
– big-lion
Apr 17 '18 at 3:59




$begingroup$
Are u,v generic vectors of $mathbbR^3$?
$endgroup$
– big-lion
Apr 17 '18 at 3:59












$begingroup$
@big-lion : I edited.
$endgroup$
– HK Lee
Apr 17 '18 at 4:03




$begingroup$
@big-lion : I edited.
$endgroup$
– HK Lee
Apr 17 '18 at 4:03










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