Mixing
Geodesics meeting with angle $0$ in ${rm CAT}(0)$ space
$begingroup$
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 Feb 1 at 9:08
HK Lee
14.1k52362
asked Nov 27 '18 at 15:48
Florentin MBFlorentin MB
14215
$endgroup$
add a comment |
$begingroup$
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 Feb 1 at 9:08
HK Lee
14.1k52362
asked Nov 27 '18 at 15:48
Florentin MBFlorentin MB
14215
$endgroup$
$begingroup$
Why they define the same germ? If your space is $CAT(0)$ they vary continuously with their endpoints.
$endgroup$
– Dante Grevino
Nov 27 '18 at 16:05
$begingroup$
@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.
$endgroup$
– Florentin MB
Nov 27 '18 at 16:09
$begingroup$
@ Florentin MB : Could you explain "branching" ?
$endgroup$
– HK Lee
Feb 1 at 9:13
$begingroup$
@HKLee Two geodesics are branching when they are distinct but define the same germ
$endgroup$
– Florentin MB
Feb 4 at 15:40
add a comment |
$begingroup$
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 Feb 1 at 9:08
HK Lee
14.1k52362
asked Nov 27 '18 at 15:48
Florentin MBFlorentin MB
14215
$endgroup$
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 Feb 1 at 9:08
HK Lee
14.1k52362
asked Nov 27 '18 at 15:48
Florentin MBFlorentin MB
14215
edited Feb 1 at 9:08
HK Lee
14.1k52362
asked Nov 27 '18 at 15:48
Florentin MBFlorentin MB
14215
edited Feb 1 at 9:08
HK Lee
14.1k52362
edited Feb 1 at 9:08
HK Lee
14.1k52362
edited Feb 1 at 9:08
HK Lee
14.1k52362
14.1k52362
asked Nov 27 '18 at 15:48
Florentin MBFlorentin MB
14215
asked Nov 27 '18 at 15:48
Florentin MBFlorentin MB
14215
asked Nov 27 '18 at 15:48
Florentin MBFlorentin MB
14215
14215
$begingroup$
Why they define the same germ? If your space is $CAT(0)$ they vary continuously with their endpoints.
$endgroup$
– Dante Grevino
Nov 27 '18 at 16:05
$begingroup$
@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.
$endgroup$
– Florentin MB
Nov 27 '18 at 16:09
$begingroup$
@ Florentin MB : Could you explain "branching" ?
$endgroup$
– HK Lee
Feb 1 at 9:13
$begingroup$
@HKLee Two geodesics are branching when they are distinct but define the same germ
$endgroup$
– Florentin MB
Feb 4 at 15:40
add a comment |
$begingroup$
Why they define the same germ? If your space is $CAT(0)$ they vary continuously with their endpoints.
$endgroup$
– Dante Grevino
Nov 27 '18 at 16:05
$begingroup$
@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.
$endgroup$
– Florentin MB
Nov 27 '18 at 16:09
$begingroup$
@ Florentin MB : Could you explain "branching" ?
$endgroup$
– HK Lee
Feb 1 at 9:13
$begingroup$
@HKLee Two geodesics are branching when they are distinct but define the same germ
$endgroup$
– Florentin MB
Feb 4 at 15:40
$begingroup$
Why they define the same germ? If your space is $CAT(0)$ they vary continuously with their endpoints.
$endgroup$
– Dante Grevino
Nov 27 '18 at 16:05
$begingroup$
Why they define the same germ? If your space is $CAT(0)$ they vary continuously with their endpoints.
$endgroup$
– Dante Grevino
Nov 27 '18 at 16:05
$begingroup$
@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.
$endgroup$
– Florentin MB
Nov 27 '18 at 16:09
$begingroup$
@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.
$endgroup$
– Florentin MB
Nov 27 '18 at 16:09
$begingroup$
@ Florentin MB : Could you explain "branching" ?
$endgroup$
– HK Lee
Feb 1 at 9:13
$begingroup$
@ Florentin MB : Could you explain "branching" ?
$endgroup$
– HK Lee
Feb 1 at 9:13
$begingroup$
@HKLee Two geodesics are branching when they are distinct but define the same germ
$endgroup$
– Florentin MB
Feb 4 at 15:40
$begingroup$
@HKLee Two geodesics are branching when they are distinct but define the same germ
$endgroup$
– Florentin MB
Feb 4 at 15:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider an arc $x=0, (y-1)^2+z^2=1, 0<yleq 1, z>0$. When we
rotate it wrt $z$-zxis, then we have $X$, homeomorphic to a disc
with a hole.
Note that it has a nonpositive
curvature. A completion of universal cover of $X$ is homeomorphic to
$bigcup_{tin mathbb{R}} [oc(t)]$ where $c:mathbb{R}rightarrow
mathbb{S}^2$ is an immersion, $o$ is origin, and $[ab]$ is a line segment in
$mathbb{E}^3$.
Geodesics with Angle $0$ : Assume that $$p=(0,1-cos theta,ast),
q=(0,-1+cos theta
,ast)in X$$
Now consider a completion of $X$ : $|p-o|=theta =|q-o|$. And $|p-q|
< pi(1-cos theta )$.
answered Feb 5 at 7:04
HK LeeHK Lee
14.1k52362
$endgroup$
add a comment |
$begingroup$
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 HeerLoreno Heer
3,35411534
$endgroup$
$begingroup$
"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 ?
$endgroup$
– Florentin MB
Dec 17 '18 at 10:46
$begingroup$
I think your Berestovskii Theorem works just for manifolds and it fails for graphs
$endgroup$
– Dante Grevino
Jan 21 at 4:54
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider an arc $x=0, (y-1)^2+z^2=1, 0<yleq 1, z>0$. When we
rotate it wrt $z$-zxis, then we have $X$, homeomorphic to a disc
with a hole.
Note that it has a nonpositive
curvature. A completion of universal cover of $X$ is homeomorphic to
$bigcup_{tin mathbb{R}} [oc(t)]$ where $c:mathbb{R}rightarrow
mathbb{S}^2$ is an immersion, $o$ is origin, and $[ab]$ is a line segment in
$mathbb{E}^3$.
Geodesics with Angle $0$ : Assume that $$p=(0,1-cos theta,ast),
q=(0,-1+cos theta
,ast)in X$$
Now consider a completion of $X$ : $|p-o|=theta =|q-o|$. And $|p-q|
< pi(1-cos theta )$.
answered Feb 5 at 7:04
HK LeeHK Lee
14.1k52362
$endgroup$
add a comment |
$begingroup$
Consider an arc $x=0, (y-1)^2+z^2=1, 0<yleq 1, z>0$. When we
rotate it wrt $z$-zxis, then we have $X$, homeomorphic to a disc
with a hole.
Note that it has a nonpositive
curvature. A completion of universal cover of $X$ is homeomorphic to
$bigcup_{tin mathbb{R}} [oc(t)]$ where $c:mathbb{R}rightarrow
mathbb{S}^2$ is an immersion, $o$ is origin, and $[ab]$ is a line segment in
$mathbb{E}^3$.
Geodesics with Angle $0$ : Assume that $$p=(0,1-cos theta,ast),
q=(0,-1+cos theta
,ast)in X$$
Now consider a completion of $X$ : $|p-o|=theta =|q-o|$. And $|p-q|
< pi(1-cos theta )$.
answered Feb 5 at 7:04
HK LeeHK Lee
14.1k52362
$endgroup$
add a comment |
$begingroup$
Consider an arc $x=0, (y-1)^2+z^2=1, 0<yleq 1, z>0$. When we
rotate it wrt $z$-zxis, then we have $X$, homeomorphic to a disc
with a hole.
Note that it has a nonpositive
curvature. A completion of universal cover of $X$ is homeomorphic to
$bigcup_{tin mathbb{R}} [oc(t)]$ where $c:mathbb{R}rightarrow
mathbb{S}^2$ is an immersion, $o$ is origin, and $[ab]$ is a line segment in
$mathbb{E}^3$.
Geodesics with Angle $0$ : Assume that $$p=(0,1-cos theta,ast),
q=(0,-1+cos theta
,ast)in X$$
Now consider a completion of $X$ : $|p-o|=theta =|q-o|$. And $|p-q|
< pi(1-cos theta )$.
answered Feb 5 at 7:04
HK LeeHK Lee
14.1k52362
$endgroup$
Consider an arc $x=0, (y-1)^2+z^2=1, 0<yleq 1, z>0$. When we
rotate it wrt $z$-zxis, then we have $X$, homeomorphic to a disc
with a hole.
Note that it has a nonpositive
curvature. A completion of universal cover of $X$ is homeomorphic to
$bigcup_{tin mathbb{R}} [oc(t)]$ where $c:mathbb{R}rightarrow
mathbb{S}^2$ is an immersion, $o$ is origin, and $[ab]$ is a line segment in
$mathbb{E}^3$.
Geodesics with Angle $0$ : Assume that $$p=(0,1-cos theta,ast),
q=(0,-1+cos theta
,ast)in X$$
Now consider a completion of $X$ : $|p-o|=theta =|q-o|$. And $|p-q|
< pi(1-cos theta )$.
answered Feb 5 at 7:04
HK LeeHK Lee
14.1k52362
answered Feb 5 at 7:04
HK LeeHK Lee
14.1k52362
answered Feb 5 at 7:04
HK LeeHK Lee
14.1k52362
answered Feb 5 at 7:04
HK LeeHK Lee
14.1k52362
14.1k52362
add a comment |
add a comment |
$begingroup$
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 HeerLoreno Heer
3,35411534
$endgroup$
$begingroup$
"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 ?
$endgroup$
– Florentin MB
Dec 17 '18 at 10:46
$begingroup$
I think your Berestovskii Theorem works just for manifolds and it fails for graphs
$endgroup$
– Dante Grevino
Jan 21 at 4:54
add a comment |
$begingroup$
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 HeerLoreno Heer
3,35411534
$endgroup$
$begingroup$
"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 ?
$endgroup$
– Florentin MB
Dec 17 '18 at 10:46
$begingroup$
I think your Berestovskii Theorem works just for manifolds and it fails for graphs
$endgroup$
– Dante Grevino
Jan 21 at 4:54
add a comment |
$begingroup$
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 HeerLoreno Heer
3,35411534
$endgroup$
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 HeerLoreno Heer
3,35411534
edited Dec 13 '18 at 10:25
answered Dec 12 '18 at 16:48
Loreno HeerLoreno Heer
3,35411534
answered Dec 12 '18 at 16:48
Loreno HeerLoreno Heer
3,35411534
answered Dec 12 '18 at 16:48
Loreno HeerLoreno Heer
3,35411534
3,35411534
$begingroup$
"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 ?
$endgroup$
– Florentin MB
Dec 17 '18 at 10:46
$begingroup$
I think your Berestovskii Theorem works just for manifolds and it fails for graphs
$endgroup$
– Dante Grevino
Jan 21 at 4:54
add a comment |
$begingroup$
"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 ?
$endgroup$
– Florentin MB
Dec 17 '18 at 10:46
$begingroup$
I think your Berestovskii Theorem works just for manifolds and it fails for graphs
$endgroup$
– Dante Grevino
Jan 21 at 4:54
$begingroup$
"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 ?
$endgroup$
– Florentin MB
Dec 17 '18 at 10:46
$begingroup$
"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 ?
$endgroup$
– Florentin MB
Dec 17 '18 at 10:46
$begingroup$
I think your Berestovskii Theorem works just for manifolds and it fails for graphs
$endgroup$
– Dante Grevino
Jan 21 at 4:54
$begingroup$
I think your Berestovskii Theorem works just for manifolds and it fails for graphs
$endgroup$
– Dante Grevino
Jan 21 at 4:54
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.
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– Dante Grevino
Nov 27 '18 at 16:05
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@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.
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– Florentin MB
Nov 27 '18 at 16:09
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@ Florentin MB : Could you explain "branching" ?
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– HK Lee
Feb 1 at 9:13
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@HKLee Two geodesics are branching when they are distinct but define the same germ
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– Florentin MB
Feb 4 at 15:40