Geodesics and Distance in Hyperbolic SpaceExpression of the Hyperbolic Distance in the Upper Half PlaneDoes distance in hyperbolic space satisfy such properties which Euclidean distance have?Geodesics of this metricDistance of two hyperbolic linesQuestion on ideal triangle and hyperbolic distanceMetric tensor and Christoffel symbols of the hyperbolic n-spaceIsometry between two moldes for hyperbolic spaceFinding Geodesics on the Unit Cylinderdoes a proper variation through geodesics keep the length of geodesic fixed?Two distinct geodesics in hyperbolic Space
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Geodesics and Distance in Hyperbolic Space
Expression of the Hyperbolic Distance in the Upper Half PlaneDoes distance in hyperbolic space satisfy such properties which Euclidean distance have?Geodesics of this metricDistance of two hyperbolic linesQuestion on ideal triangle and hyperbolic distanceMetric tensor and Christoffel symbols of the hyperbolic n-spaceIsometry between two moldes for hyperbolic spaceFinding Geodesics on the Unit Cylinderdoes a proper variation through geodesics keep the length of geodesic fixed?Two distinct geodesics in hyperbolic Space
$begingroup$
I am trying to understand geodesics and distance in hyperbolic space $H^n$ (Inverse image of $1$ under $f(x_0,x_1,dots,x_n)=-x_0^2+x_1^2+dots+x_n^2$ and with the inherited metric). More precisely, I want to show that the Riemann distance between two points $p,q$ is $d(p,q)=arccosh(-pcdot q)$.
Easy I thought, just compute the geodesics between two points $p,q$ and evaluate the length.
However, I don't really understand how to do this and would appreciate some help. How do I get a nice analytical formula which I can just "evaluate" to obtain the distance? In all the books I've read there are only very geometric arguments which don't really produce formulas and I don't really understand how to go between the two...
differential-geometry riemannian-geometry
$endgroup$
add a comment |
$begingroup$
I am trying to understand geodesics and distance in hyperbolic space $H^n$ (Inverse image of $1$ under $f(x_0,x_1,dots,x_n)=-x_0^2+x_1^2+dots+x_n^2$ and with the inherited metric). More precisely, I want to show that the Riemann distance between two points $p,q$ is $d(p,q)=arccosh(-pcdot q)$.
Easy I thought, just compute the geodesics between two points $p,q$ and evaluate the length.
However, I don't really understand how to do this and would appreciate some help. How do I get a nice analytical formula which I can just "evaluate" to obtain the distance? In all the books I've read there are only very geometric arguments which don't really produce formulas and I don't really understand how to go between the two...
differential-geometry riemannian-geometry
$endgroup$
$begingroup$
The dot product here is the Lorenz metric right?
$endgroup$
– Elad
May 8 at 11:56
add a comment |
$begingroup$
I am trying to understand geodesics and distance in hyperbolic space $H^n$ (Inverse image of $1$ under $f(x_0,x_1,dots,x_n)=-x_0^2+x_1^2+dots+x_n^2$ and with the inherited metric). More precisely, I want to show that the Riemann distance between two points $p,q$ is $d(p,q)=arccosh(-pcdot q)$.
Easy I thought, just compute the geodesics between two points $p,q$ and evaluate the length.
However, I don't really understand how to do this and would appreciate some help. How do I get a nice analytical formula which I can just "evaluate" to obtain the distance? In all the books I've read there are only very geometric arguments which don't really produce formulas and I don't really understand how to go between the two...
differential-geometry riemannian-geometry
$endgroup$
I am trying to understand geodesics and distance in hyperbolic space $H^n$ (Inverse image of $1$ under $f(x_0,x_1,dots,x_n)=-x_0^2+x_1^2+dots+x_n^2$ and with the inherited metric). More precisely, I want to show that the Riemann distance between two points $p,q$ is $d(p,q)=arccosh(-pcdot q)$.
Easy I thought, just compute the geodesics between two points $p,q$ and evaluate the length.
However, I don't really understand how to do this and would appreciate some help. How do I get a nice analytical formula which I can just "evaluate" to obtain the distance? In all the books I've read there are only very geometric arguments which don't really produce formulas and I don't really understand how to go between the two...
differential-geometry riemannian-geometry
differential-geometry riemannian-geometry
edited May 8 at 8:06
afightingchance
asked May 8 at 8:01
afightingchanceafightingchance
1106
1106
$begingroup$
The dot product here is the Lorenz metric right?
$endgroup$
– Elad
May 8 at 11:56
add a comment |
$begingroup$
The dot product here is the Lorenz metric right?
$endgroup$
– Elad
May 8 at 11:56
$begingroup$
The dot product here is the Lorenz metric right?
$endgroup$
– Elad
May 8 at 11:56
$begingroup$
The dot product here is the Lorenz metric right?
$endgroup$
– Elad
May 8 at 11:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I will write you a sketch. First start with geometric reasoning to determine the geodesics in the hyperboloid model of hyperbolic space.
Let $O(1, n)$ denote the subgroup of linear transformations of $mathbbR^n+1$ preserving the pseudo-Riemannian metric called the Lorentz metric $(cdot, cdot)$ of the . Let $O_+(1, n)$ denote the matrices $(a_ij)_i,j=0,dots, n$ in $O(1, n)$ with $a_00 > 0$. Show that
$O_+(1, n)$ is a subgroup of $O(1, n)$. Show that elements of $O_+(1, n)$
restrict to isometries of $H^n$.Determine the geodesics in the hyperboloid model of hyperbolic space:
Let $P$ be a two-dimensional linear subspace of $mathbbR^n+1$ containing the
point $p in H^n$. Show there is an element of $O_+(1, n)$ fixing $P$ and acting
non-trivially on each point of $mathbbR^n+1backslash P$. Conclude that all geodesics of $H^n$ through $p$ can be obtained as intersections $P cap H^n$ for some $P$ as
above.
You will need to use:
Let $(M,g)$ be a Riemannian and let $varphi:M to M$ be an isometry. Let $p in M$ such that $$varphi(p)=p$$
and let $v in T_p M$ such that:
$$dvarphi_p(v)=v$$
If $gamma(t)=exp_p(tv)$ (the geodesic with velocity $v$ that start at $p$) then:
$$varphi circ gamma(t)=gamma(t)$$
for all $t$. You will also use the fact that geodesic in a given deirection is unique.Klein disk model of hyperbolic space:
Let $y_1,dots, y_n$, be the standard coordinates on $D^n$. Equip $D^n$ with metric defined by
$$g_ij =
fracdelta_ij^2+fracy_i y_j(1-
$$
Define a map $h : H^n_1 to D^n$ by
$$y_i=fracx_ix_0$$
Show that $h$ is an isometry.
Show that the geodesics of part (2) are sent by $h$ to straight lines in
the Klein model. (The map $h$ is a stereographic projection along
lines through the origin, to $D^n subset x_0 = 1 subset L^n+1$).
Now calculate the length of a straight line paramtized in arc length between the two points with the Klien metric.
This is not the fastest way to see this but it connects the geometric reasoning to a formula and you can understant the arrcosh function. Maybe now you will even be satisfied with the $arrcosh$ formula without the calculation. Because you can understand what we are measuring.
$endgroup$
$begingroup$
An interesting side note, angles measured with respect to the Klein metric on the disk are not the same as with respect to the Euclidean metric on the disk. On the other hand, angles in the Poincare disk model are the same as in the Euclidean metric on the disk. Thus, we can portray angles (Poincare) or geodesics (Klein) intuitively, but not both.
$endgroup$
– Elad
May 8 at 11:44
add a comment |
Your Answer
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$begingroup$
I will write you a sketch. First start with geometric reasoning to determine the geodesics in the hyperboloid model of hyperbolic space.
Let $O(1, n)$ denote the subgroup of linear transformations of $mathbbR^n+1$ preserving the pseudo-Riemannian metric called the Lorentz metric $(cdot, cdot)$ of the . Let $O_+(1, n)$ denote the matrices $(a_ij)_i,j=0,dots, n$ in $O(1, n)$ with $a_00 > 0$. Show that
$O_+(1, n)$ is a subgroup of $O(1, n)$. Show that elements of $O_+(1, n)$
restrict to isometries of $H^n$.Determine the geodesics in the hyperboloid model of hyperbolic space:
Let $P$ be a two-dimensional linear subspace of $mathbbR^n+1$ containing the
point $p in H^n$. Show there is an element of $O_+(1, n)$ fixing $P$ and acting
non-trivially on each point of $mathbbR^n+1backslash P$. Conclude that all geodesics of $H^n$ through $p$ can be obtained as intersections $P cap H^n$ for some $P$ as
above.
You will need to use:
Let $(M,g)$ be a Riemannian and let $varphi:M to M$ be an isometry. Let $p in M$ such that $$varphi(p)=p$$
and let $v in T_p M$ such that:
$$dvarphi_p(v)=v$$
If $gamma(t)=exp_p(tv)$ (the geodesic with velocity $v$ that start at $p$) then:
$$varphi circ gamma(t)=gamma(t)$$
for all $t$. You will also use the fact that geodesic in a given deirection is unique.Klein disk model of hyperbolic space:
Let $y_1,dots, y_n$, be the standard coordinates on $D^n$. Equip $D^n$ with metric defined by
$$g_ij =
fracdelta_ij^2+fracy_i y_j(1-
$$
Define a map $h : H^n_1 to D^n$ by
$$y_i=fracx_ix_0$$
Show that $h$ is an isometry.
Show that the geodesics of part (2) are sent by $h$ to straight lines in
the Klein model. (The map $h$ is a stereographic projection along
lines through the origin, to $D^n subset x_0 = 1 subset L^n+1$).
Now calculate the length of a straight line paramtized in arc length between the two points with the Klien metric.
This is not the fastest way to see this but it connects the geometric reasoning to a formula and you can understant the arrcosh function. Maybe now you will even be satisfied with the $arrcosh$ formula without the calculation. Because you can understand what we are measuring.
$endgroup$
$begingroup$
An interesting side note, angles measured with respect to the Klein metric on the disk are not the same as with respect to the Euclidean metric on the disk. On the other hand, angles in the Poincare disk model are the same as in the Euclidean metric on the disk. Thus, we can portray angles (Poincare) or geodesics (Klein) intuitively, but not both.
$endgroup$
– Elad
May 8 at 11:44
add a comment |
$begingroup$
I will write you a sketch. First start with geometric reasoning to determine the geodesics in the hyperboloid model of hyperbolic space.
Let $O(1, n)$ denote the subgroup of linear transformations of $mathbbR^n+1$ preserving the pseudo-Riemannian metric called the Lorentz metric $(cdot, cdot)$ of the . Let $O_+(1, n)$ denote the matrices $(a_ij)_i,j=0,dots, n$ in $O(1, n)$ with $a_00 > 0$. Show that
$O_+(1, n)$ is a subgroup of $O(1, n)$. Show that elements of $O_+(1, n)$
restrict to isometries of $H^n$.Determine the geodesics in the hyperboloid model of hyperbolic space:
Let $P$ be a two-dimensional linear subspace of $mathbbR^n+1$ containing the
point $p in H^n$. Show there is an element of $O_+(1, n)$ fixing $P$ and acting
non-trivially on each point of $mathbbR^n+1backslash P$. Conclude that all geodesics of $H^n$ through $p$ can be obtained as intersections $P cap H^n$ for some $P$ as
above.
You will need to use:
Let $(M,g)$ be a Riemannian and let $varphi:M to M$ be an isometry. Let $p in M$ such that $$varphi(p)=p$$
and let $v in T_p M$ such that:
$$dvarphi_p(v)=v$$
If $gamma(t)=exp_p(tv)$ (the geodesic with velocity $v$ that start at $p$) then:
$$varphi circ gamma(t)=gamma(t)$$
for all $t$. You will also use the fact that geodesic in a given deirection is unique.Klein disk model of hyperbolic space:
Let $y_1,dots, y_n$, be the standard coordinates on $D^n$. Equip $D^n$ with metric defined by
$$g_ij =
fracdelta_ij^2+fracy_i y_j(1-
$$
Define a map $h : H^n_1 to D^n$ by
$$y_i=fracx_ix_0$$
Show that $h$ is an isometry.
Show that the geodesics of part (2) are sent by $h$ to straight lines in
the Klein model. (The map $h$ is a stereographic projection along
lines through the origin, to $D^n subset x_0 = 1 subset L^n+1$).
Now calculate the length of a straight line paramtized in arc length between the two points with the Klien metric.
This is not the fastest way to see this but it connects the geometric reasoning to a formula and you can understant the arrcosh function. Maybe now you will even be satisfied with the $arrcosh$ formula without the calculation. Because you can understand what we are measuring.
$endgroup$
$begingroup$
An interesting side note, angles measured with respect to the Klein metric on the disk are not the same as with respect to the Euclidean metric on the disk. On the other hand, angles in the Poincare disk model are the same as in the Euclidean metric on the disk. Thus, we can portray angles (Poincare) or geodesics (Klein) intuitively, but not both.
$endgroup$
– Elad
May 8 at 11:44
add a comment |
$begingroup$
I will write you a sketch. First start with geometric reasoning to determine the geodesics in the hyperboloid model of hyperbolic space.
Let $O(1, n)$ denote the subgroup of linear transformations of $mathbbR^n+1$ preserving the pseudo-Riemannian metric called the Lorentz metric $(cdot, cdot)$ of the . Let $O_+(1, n)$ denote the matrices $(a_ij)_i,j=0,dots, n$ in $O(1, n)$ with $a_00 > 0$. Show that
$O_+(1, n)$ is a subgroup of $O(1, n)$. Show that elements of $O_+(1, n)$
restrict to isometries of $H^n$.Determine the geodesics in the hyperboloid model of hyperbolic space:
Let $P$ be a two-dimensional linear subspace of $mathbbR^n+1$ containing the
point $p in H^n$. Show there is an element of $O_+(1, n)$ fixing $P$ and acting
non-trivially on each point of $mathbbR^n+1backslash P$. Conclude that all geodesics of $H^n$ through $p$ can be obtained as intersections $P cap H^n$ for some $P$ as
above.
You will need to use:
Let $(M,g)$ be a Riemannian and let $varphi:M to M$ be an isometry. Let $p in M$ such that $$varphi(p)=p$$
and let $v in T_p M$ such that:
$$dvarphi_p(v)=v$$
If $gamma(t)=exp_p(tv)$ (the geodesic with velocity $v$ that start at $p$) then:
$$varphi circ gamma(t)=gamma(t)$$
for all $t$. You will also use the fact that geodesic in a given deirection is unique.Klein disk model of hyperbolic space:
Let $y_1,dots, y_n$, be the standard coordinates on $D^n$. Equip $D^n$ with metric defined by
$$g_ij =
fracdelta_ij^2+fracy_i y_j(1-
$$
Define a map $h : H^n_1 to D^n$ by
$$y_i=fracx_ix_0$$
Show that $h$ is an isometry.
Show that the geodesics of part (2) are sent by $h$ to straight lines in
the Klein model. (The map $h$ is a stereographic projection along
lines through the origin, to $D^n subset x_0 = 1 subset L^n+1$).
Now calculate the length of a straight line paramtized in arc length between the two points with the Klien metric.
This is not the fastest way to see this but it connects the geometric reasoning to a formula and you can understant the arrcosh function. Maybe now you will even be satisfied with the $arrcosh$ formula without the calculation. Because you can understand what we are measuring.
$endgroup$
I will write you a sketch. First start with geometric reasoning to determine the geodesics in the hyperboloid model of hyperbolic space.
Let $O(1, n)$ denote the subgroup of linear transformations of $mathbbR^n+1$ preserving the pseudo-Riemannian metric called the Lorentz metric $(cdot, cdot)$ of the . Let $O_+(1, n)$ denote the matrices $(a_ij)_i,j=0,dots, n$ in $O(1, n)$ with $a_00 > 0$. Show that
$O_+(1, n)$ is a subgroup of $O(1, n)$. Show that elements of $O_+(1, n)$
restrict to isometries of $H^n$.Determine the geodesics in the hyperboloid model of hyperbolic space:
Let $P$ be a two-dimensional linear subspace of $mathbbR^n+1$ containing the
point $p in H^n$. Show there is an element of $O_+(1, n)$ fixing $P$ and acting
non-trivially on each point of $mathbbR^n+1backslash P$. Conclude that all geodesics of $H^n$ through $p$ can be obtained as intersections $P cap H^n$ for some $P$ as
above.
You will need to use:
Let $(M,g)$ be a Riemannian and let $varphi:M to M$ be an isometry. Let $p in M$ such that $$varphi(p)=p$$
and let $v in T_p M$ such that:
$$dvarphi_p(v)=v$$
If $gamma(t)=exp_p(tv)$ (the geodesic with velocity $v$ that start at $p$) then:
$$varphi circ gamma(t)=gamma(t)$$
for all $t$. You will also use the fact that geodesic in a given deirection is unique.Klein disk model of hyperbolic space:
Let $y_1,dots, y_n$, be the standard coordinates on $D^n$. Equip $D^n$ with metric defined by
$$g_ij =
fracdelta_ij^2+fracy_i y_j(1-
$$
Define a map $h : H^n_1 to D^n$ by
$$y_i=fracx_ix_0$$
Show that $h$ is an isometry.
Show that the geodesics of part (2) are sent by $h$ to straight lines in
the Klein model. (The map $h$ is a stereographic projection along
lines through the origin, to $D^n subset x_0 = 1 subset L^n+1$).
Now calculate the length of a straight line paramtized in arc length between the two points with the Klien metric.
This is not the fastest way to see this but it connects the geometric reasoning to a formula and you can understant the arrcosh function. Maybe now you will even be satisfied with the $arrcosh$ formula without the calculation. Because you can understand what we are measuring.
edited May 8 at 10:24
answered May 8 at 9:16
EladElad
699313
699313
$begingroup$
An interesting side note, angles measured with respect to the Klein metric on the disk are not the same as with respect to the Euclidean metric on the disk. On the other hand, angles in the Poincare disk model are the same as in the Euclidean metric on the disk. Thus, we can portray angles (Poincare) or geodesics (Klein) intuitively, but not both.
$endgroup$
– Elad
May 8 at 11:44
add a comment |
$begingroup$
An interesting side note, angles measured with respect to the Klein metric on the disk are not the same as with respect to the Euclidean metric on the disk. On the other hand, angles in the Poincare disk model are the same as in the Euclidean metric on the disk. Thus, we can portray angles (Poincare) or geodesics (Klein) intuitively, but not both.
$endgroup$
– Elad
May 8 at 11:44
$begingroup$
An interesting side note, angles measured with respect to the Klein metric on the disk are not the same as with respect to the Euclidean metric on the disk. On the other hand, angles in the Poincare disk model are the same as in the Euclidean metric on the disk. Thus, we can portray angles (Poincare) or geodesics (Klein) intuitively, but not both.
$endgroup$
– Elad
May 8 at 11:44
$begingroup$
An interesting side note, angles measured with respect to the Klein metric on the disk are not the same as with respect to the Euclidean metric on the disk. On the other hand, angles in the Poincare disk model are the same as in the Euclidean metric on the disk. Thus, we can portray angles (Poincare) or geodesics (Klein) intuitively, but not both.
$endgroup$
– Elad
May 8 at 11:44
add a comment |
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$begingroup$
The dot product here is the Lorenz metric right?
$endgroup$
– Elad
May 8 at 11:56