Geodesics meeting with angle 0 in CAT(0) space
Consider two distinct geodesics $gamma_1$ and $gamma_2$ in a CAT($0$) space, issued from the same base point.
A trivial example where we have $angle(gamma_1, gamma_2)=0$ is when
$gamma_1(t) = gamma_2(t)$ for $t$ smaller than some $varepsilon > 0$. In this case, I say that they define the same germ.
My question is, is there examples of CAT(0) spaces with geodesics meeting with angle 0 and every geodesic is non-branching ?
The only class of examples I know of non-branching geodesics meeting with angle $0$ are constructed as follow:
consider the following subset of the Euclidian plane, with induced length-metric,
$X = { (x,y); 0 le x le 1, 0 le y le x^2 }$. It is a CAT(0) space,
the geodesic between $(0,0)$ and $(1,0)$ is the segment and the geodesic between $(0,0)$ and $(1,1)$ is the arc of parabola. These geodesics meet at $(0,0)$ with angle 0 and they don't define the same germ.
However, in this space there exist branching geodesics, for example
the geodesic between $(0,0)$ and $(1, t)$ for $t>0$ all define the same germ.
Edit: I add a description of these geodesics. The geodesic from $(0,0)$ to
$(1,t)$ is an arc of parabola from $(0,0)$ to $P$ concatenated with the segment $[P, (1,t)]$ where $P$ is the point on the parabola closest to the origin such that the segment $[P, (1,t)]$ lies in $X$.
metric-geometry
edited Dec 12 '18 at 13:33
Loreno Heer
3,32411534
asked Nov 27 '18 at 15:48
Florentin MB
13715
add a comment |
Consider two distinct geodesics $gamma_1$ and $gamma_2$ in a CAT($0$) space, issued from the same base point.
A trivial example where we have $angle(gamma_1, gamma_2)=0$ is when
$gamma_1(t) = gamma_2(t)$ for $t$ smaller than some $varepsilon > 0$. In this case, I say that they define the same germ.
My question is, is there examples of CAT(0) spaces with geodesics meeting with angle 0 and every geodesic is non-branching ?
The only class of examples I know of non-branching geodesics meeting with angle $0$ are constructed as follow:
consider the following subset of the Euclidian plane, with induced length-metric,
$X = { (x,y); 0 le x le 1, 0 le y le x^2 }$. It is a CAT(0) space,
the geodesic between $(0,0)$ and $(1,0)$ is the segment and the geodesic between $(0,0)$ and $(1,1)$ is the arc of parabola. These geodesics meet at $(0,0)$ with angle 0 and they don't define the same germ.
However, in this space there exist branching geodesics, for example
the geodesic between $(0,0)$ and $(1, t)$ for $t>0$ all define the same germ.
Edit: I add a description of these geodesics. The geodesic from $(0,0)$ to
$(1,t)$ is an arc of parabola from $(0,0)$ to $P$ concatenated with the segment $[P, (1,t)]$ where $P$ is the point on the parabola closest to the origin such that the segment $[P, (1,t)]$ lies in $X$.
metric-geometry
edited Dec 12 '18 at 13:33
Loreno Heer
3,32411534
asked Nov 27 '18 at 15:48
Florentin MB
13715
Why they define the same germ? If your space is $CAT(0)$ they vary continuously with their endpoints.
– Dante Grevino
Nov 27 '18 at 16:05
@DanteGrevino they do vary continuously. The geodesic between $(0,0)$ and $(1, t)$ follows the arc of parabola for some time and then is a segment.
– Florentin MB
Nov 27 '18 at 16:09
add a comment |
Consider two distinct geodesics $gamma_1$ and $gamma_2$ in a CAT($0$) space, issued from the same base point.
A trivial example where we have $angle(gamma_1, gamma_2)=0$ is when
$gamma_1(t) = gamma_2(t)$ for $t$ smaller than some $varepsilon > 0$. In this case, I say that they define the same germ.
My question is, is there examples of CAT(0) spaces with geodesics meeting with angle 0 and every geodesic is non-branching ?
The only class of examples I know of non-branching geodesics meeting with angle $0$ are constructed as follow:
consider the following subset of the Euclidian plane, with induced length-metric,
$X = { (x,y); 0 le x le 1, 0 le y le x^2 }$. It is a CAT(0) space,
the geodesic between $(0,0)$ and $(1,0)$ is the segment and the geodesic between $(0,0)$ and $(1,1)$ is the arc of parabola. These geodesics meet at $(0,0)$ with angle 0 and they don't define the same germ.
However, in this space there exist branching geodesics, for example
the geodesic between $(0,0)$ and $(1, t)$ for $t>0$ all define the same germ.
Edit: I add a description of these geodesics. The geodesic from $(0,0)$ to
$(1,t)$ is an arc of parabola from $(0,0)$ to $P$ concatenated with the segment $[P, (1,t)]$ where $P$ is the point on the parabola closest to the origin such that the segment $[P, (1,t)]$ lies in $X$.
metric-geometry
edited Dec 12 '18 at 13:33
Loreno Heer
3,32411534
asked Nov 27 '18 at 15:48
Florentin MB
13715
Consider two distinct geodesics $gamma_1$ and $gamma_2$ in a CAT($0$) space, issued from the same base point.
A trivial example where we have $angle(gamma_1, gamma_2)=0$ is when
$gamma_1(t) = gamma_2(t)$ for $t$ smaller than some $varepsilon > 0$. In this case, I say that they define the same germ.
My question is, is there examples of CAT(0) spaces with geodesics meeting with angle 0 and every geodesic is non-branching ?
The only class of examples I know of non-branching geodesics meeting with angle $0$ are constructed as follow:
consider the following subset of the Euclidian plane, with induced length-metric,
$X = { (x,y); 0 le x le 1, 0 le y le x^2 }$. It is a CAT(0) space,
the geodesic between $(0,0)$ and $(1,0)$ is the segment and the geodesic between $(0,0)$ and $(1,1)$ is the arc of parabola. These geodesics meet at $(0,0)$ with angle 0 and they don't define the same germ.
However, in this space there exist branching geodesics, for example
the geodesic between $(0,0)$ and $(1, t)$ for $t>0$ all define the same germ.
Edit: I add a description of these geodesics. The geodesic from $(0,0)$ to
$(1,t)$ is an arc of parabola from $(0,0)$ to $P$ concatenated with the segment $[P, (1,t)]$ where $P$ is the point on the parabola closest to the origin such that the segment $[P, (1,t)]$ lies in $X$.
metric-geometry
metric-geometry
edited Dec 12 '18 at 13:33
Loreno Heer
3,32411534
asked Nov 27 '18 at 15:48
Florentin MB
13715
edited Dec 12 '18 at 13:33
Loreno Heer
3,32411534
asked Nov 27 '18 at 15:48
Florentin MB
13715
edited Dec 12 '18 at 13:33
Loreno Heer
3,32411534
edited Dec 12 '18 at 13:33
Loreno Heer
3,32411534
edited Dec 12 '18 at 13:33
Loreno Heer
3,32411534
3,32411534
asked Nov 27 '18 at 15:48
Florentin MB
13715
asked Nov 27 '18 at 15:48
Florentin MB
13715
asked Nov 27 '18 at 15:48
Florentin MB
13715
13715
Why they define the same germ? If your space is $CAT(0)$ they vary continuously with their endpoints.
– Dante Grevino
Nov 27 '18 at 16:05
@DanteGrevino they do vary continuously. The geodesic between $(0,0)$ and $(1, t)$ follows the arc of parabola for some time and then is a segment.
– Florentin MB
Nov 27 '18 at 16:09
add a comment |
Why they define the same germ? If your space is $CAT(0)$ they vary continuously with their endpoints.
– Dante Grevino
Nov 27 '18 at 16:05
@DanteGrevino they do vary continuously. The geodesic between $(0,0)$ and $(1, t)$ follows the arc of parabola for some time and then is a segment.
– Florentin MB
Nov 27 '18 at 16:09
Why they define the same germ? If your space is $CAT(0)$ they vary continuously with their endpoints.
– Dante Grevino
Nov 27 '18 at 16:05
Why they define the same germ? If your space is $CAT(0)$ they vary continuously with their endpoints.
– Dante Grevino
Nov 27 '18 at 16:05
@DanteGrevino they do vary continuously. The geodesic between $(0,0)$ and $(1, t)$ follows the arc of parabola for some time and then is a segment.
– Florentin MB
Nov 27 '18 at 16:09
@DanteGrevino they do vary continuously. The geodesic between $(0,0)$ and $(1, t)$ follows the arc of parabola for some time and then is a segment.
– Florentin MB
Nov 27 '18 at 16:09
add a comment |
1 Answer
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So the following comes to my mind. But you should check if it actually makes sense. (I did not check this, and typing this answer on my phone which works horrible for this website)
Any non branching geodesic space has uniquely extendable geodesics.
If it is furthermore proper and CAT(0) then by a theorem of f V. Berestovskii it follows that:
$bar{Sigma}_x= Σ_x$ is isometric to a Euclidean sphere. Here the space of
directions $bar{Sigma}_x$ at a point $x$ is the completion of the space $Sigma_x$ of geodesic germs equipped with
the Alexandrov angle metric at x.
See Busemann spaces with upper-bounded Aleksandrov curvature, Algebra i Analiz 14 (2002),
no. 5, 318
And the following lecture notes
http://egg.epfl.ch/~nmonod/articles/structure.pdf
answered Dec 12 '18 at 16:48
Loreno Heer
3,32411534
"Any non branching geodesic space has uniquely extendable geodesics" I don't see that, what about a bounded convex subset of e.g. the Euclidian plane ?
– Florentin MB
Dec 17 '18 at 10:46
add a comment |
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1 Answer
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So the following comes to my mind. But you should check if it actually makes sense. (I did not check this, and typing this answer on my phone which works horrible for this website)
Any non branching geodesic space has uniquely extendable geodesics.
If it is furthermore proper and CAT(0) then by a theorem of f V. Berestovskii it follows that:
$bar{Sigma}_x= Σ_x$ is isometric to a Euclidean sphere. Here the space of
directions $bar{Sigma}_x$ at a point $x$ is the completion of the space $Sigma_x$ of geodesic germs equipped with
the Alexandrov angle metric at x.
See Busemann spaces with upper-bounded Aleksandrov curvature, Algebra i Analiz 14 (2002),
no. 5, 318
And the following lecture notes
http://egg.epfl.ch/~nmonod/articles/structure.pdf
answered Dec 12 '18 at 16:48
Loreno Heer
3,32411534
"Any non branching geodesic space has uniquely extendable geodesics" I don't see that, what about a bounded convex subset of e.g. the Euclidian plane ?
– Florentin MB
Dec 17 '18 at 10:46
add a comment |
So the following comes to my mind. But you should check if it actually makes sense. (I did not check this, and typing this answer on my phone which works horrible for this website)
Any non branching geodesic space has uniquely extendable geodesics.
If it is furthermore proper and CAT(0) then by a theorem of f V. Berestovskii it follows that:
$bar{Sigma}_x= Σ_x$ is isometric to a Euclidean sphere. Here the space of
directions $bar{Sigma}_x$ at a point $x$ is the completion of the space $Sigma_x$ of geodesic germs equipped with
the Alexandrov angle metric at x.
See Busemann spaces with upper-bounded Aleksandrov curvature, Algebra i Analiz 14 (2002),
no. 5, 318
And the following lecture notes
http://egg.epfl.ch/~nmonod/articles/structure.pdf
answered Dec 12 '18 at 16:48
Loreno Heer
3,32411534
"Any non branching geodesic space has uniquely extendable geodesics" I don't see that, what about a bounded convex subset of e.g. the Euclidian plane ?
– Florentin MB
Dec 17 '18 at 10:46
add a comment |
So the following comes to my mind. But you should check if it actually makes sense. (I did not check this, and typing this answer on my phone which works horrible for this website)
Any non branching geodesic space has uniquely extendable geodesics.
If it is furthermore proper and CAT(0) then by a theorem of f V. Berestovskii it follows that:
$bar{Sigma}_x= Σ_x$ is isometric to a Euclidean sphere. Here the space of
directions $bar{Sigma}_x$ at a point $x$ is the completion of the space $Sigma_x$ of geodesic germs equipped with
the Alexandrov angle metric at x.
See Busemann spaces with upper-bounded Aleksandrov curvature, Algebra i Analiz 14 (2002),
no. 5, 318
And the following lecture notes
http://egg.epfl.ch/~nmonod/articles/structure.pdf
answered Dec 12 '18 at 16:48
Loreno Heer
3,32411534
So the following comes to my mind. But you should check if it actually makes sense. (I did not check this, and typing this answer on my phone which works horrible for this website)
Any non branching geodesic space has uniquely extendable geodesics.
If it is furthermore proper and CAT(0) then by a theorem of f V. Berestovskii it follows that:
$bar{Sigma}_x= Σ_x$ is isometric to a Euclidean sphere. Here the space of
directions $bar{Sigma}_x$ at a point $x$ is the completion of the space $Sigma_x$ of geodesic germs equipped with
the Alexandrov angle metric at x.
See Busemann spaces with upper-bounded Aleksandrov curvature, Algebra i Analiz 14 (2002),
no. 5, 318
And the following lecture notes
http://egg.epfl.ch/~nmonod/articles/structure.pdf
answered Dec 12 '18 at 16:48
Loreno Heer
3,32411534
edited Dec 13 '18 at 10:25
answered Dec 12 '18 at 16:48
Loreno Heer
3,32411534
answered Dec 12 '18 at 16:48
Loreno Heer
3,32411534
answered Dec 12 '18 at 16:48
Loreno Heer
3,32411534
3,32411534
"Any non branching geodesic space has uniquely extendable geodesics" I don't see that, what about a bounded convex subset of e.g. the Euclidian plane ?
– Florentin MB
Dec 17 '18 at 10:46
add a comment |
"Any non branching geodesic space has uniquely extendable geodesics" I don't see that, what about a bounded convex subset of e.g. the Euclidian plane ?
– Florentin MB
Dec 17 '18 at 10:46
"Any non branching geodesic space has uniquely extendable geodesics" I don't see that, what about a bounded convex subset of e.g. the Euclidian plane ?
– Florentin MB
Dec 17 '18 at 10:46
"Any non branching geodesic space has uniquely extendable geodesics" I don't see that, what about a bounded convex subset of e.g. the Euclidian plane ?
– Florentin MB
Dec 17 '18 at 10:46
add a comment |
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Why they define the same germ? If your space is $CAT(0)$ they vary continuously with their endpoints.
– Dante Grevino
Nov 27 '18 at 16:05
@DanteGrevino they do vary continuously. The geodesic between $(0,0)$ and $(1, t)$ follows the arc of parabola for some time and then is a segment.
– Florentin MB
Nov 27 '18 at 16:09