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?
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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?
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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
$endgroup$
add a comment |
$begingroup$
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 $f(x)=f(-x)$.
euclidean-geometry riemannian-geometry geometric-inequalities
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$begingroup$
What's $S^2(1)$?
$endgroup$
– Alex R.
Apr 17 '18 at 3:51
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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
add a comment |
$begingroup$
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 $f(x)=f(-x)$.
euclidean-geometry riemannian-geometry geometric-inequalities
$endgroup$
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
euclidean-geometry riemannian-geometry geometric-inequalities
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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