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Orientation-preserving homeomorphisms
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I am starting to study the Frechet distance, and the expression "orientation-preserving homeomorphisms" is very used, but I have not found a formal definition...For instance, for two given to parametrised surfaces $A,B:[0,1]^2longrightarrow X$, $(X,d)$ being a metric space, the Fréchet distance is defined as
$inf_phi,psiinmathrmAut([0,1]^2) max_xin[0,1]^2 d(A(phi(x),B(psi(x)))$,
where $mathrmAut([0,1]^2) $ denotes the set of orientation-preserving homeomorphisms of $[0,1]^2$. What means, in this context (or replacing $[0,1]^2$ by $[0,1]^d$), the expression "orientation-preserving homeomorphisms"?
Many thanks in advance for your comments.
linear-algebra computational-geometry
edited Jul 17 at 6:43
P Vanchinathan
13.9k12035
asked Jul 17 at 6:24
user123043
10010
add a comment |Â
up vote
0
down vote
favorite
I am starting to study the Frechet distance, and the expression "orientation-preserving homeomorphisms" is very used, but I have not found a formal definition...For instance, for two given to parametrised surfaces $A,B:[0,1]^2longrightarrow X$, $(X,d)$ being a metric space, the Fréchet distance is defined as
$inf_phi,psiinmathrmAut([0,1]^2) max_xin[0,1]^2 d(A(phi(x),B(psi(x)))$,
where $mathrmAut([0,1]^2) $ denotes the set of orientation-preserving homeomorphisms of $[0,1]^2$. What means, in this context (or replacing $[0,1]^2$ by $[0,1]^d$), the expression "orientation-preserving homeomorphisms"?
Many thanks in advance for your comments.
linear-algebra computational-geometry
edited Jul 17 at 6:43
P Vanchinathan
13.9k12035
asked Jul 17 at 6:24
user123043
10010
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am starting to study the Frechet distance, and the expression "orientation-preserving homeomorphisms" is very used, but I have not found a formal definition...For instance, for two given to parametrised surfaces $A,B:[0,1]^2longrightarrow X$, $(X,d)$ being a metric space, the Fréchet distance is defined as
$inf_phi,psiinmathrmAut([0,1]^2) max_xin[0,1]^2 d(A(phi(x),B(psi(x)))$,
where $mathrmAut([0,1]^2) $ denotes the set of orientation-preserving homeomorphisms of $[0,1]^2$. What means, in this context (or replacing $[0,1]^2$ by $[0,1]^d$), the expression "orientation-preserving homeomorphisms"?
Many thanks in advance for your comments.
linear-algebra computational-geometry
edited Jul 17 at 6:43
P Vanchinathan
13.9k12035
asked Jul 17 at 6:24
user123043
10010
I am starting to study the Frechet distance, and the expression "orientation-preserving homeomorphisms" is very used, but I have not found a formal definition...For instance, for two given to parametrised surfaces $A,B:[0,1]^2longrightarrow X$, $(X,d)$ being a metric space, the Fréchet distance is defined as
$inf_phi,psiinmathrmAut([0,1]^2) max_xin[0,1]^2 d(A(phi(x),B(psi(x)))$,
where $mathrmAut([0,1]^2) $ denotes the set of orientation-preserving homeomorphisms of $[0,1]^2$. What means, in this context (or replacing $[0,1]^2$ by $[0,1]^d$), the expression "orientation-preserving homeomorphisms"?
Many thanks in advance for your comments.
linear-algebra computational-geometry
edited Jul 17 at 6:43
P Vanchinathan
13.9k12035
asked Jul 17 at 6:24
user123043
10010
edited Jul 17 at 6:43
P Vanchinathan
13.9k12035
edited Jul 17 at 6:43
P Vanchinathan
13.9k12035
edited Jul 17 at 6:43
P Vanchinathan
13.9k12035
13.9k12035
asked Jul 17 at 6:24
user123043
10010
asked Jul 17 at 6:24
user123043
10010
asked Jul 17 at 6:24
user123043
10010
10010
add a comment |Â
add a comment |Â
1 Answer
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The concept of "orientation" is too subtle to give a comprehensive explanation in this answer. You certainly know it from linear algebra (an isomorphism $f : mathbbR^n to mathbbR^n$ is orientation preserving if $det(f) > 0$). In the context of your question we deal with homeomorphims $h$ between manifolds (with boundary). Locally a manifold looks like Euclidean space $mathbbR^n$ (or, at the boundary, like a half-space $[0,infty) times mathbbR^n-1$) and it is possible to give a meaning to "$h$ is locally orientation preserving".
In your special case an orientation preserving homeomorphism $h : [0,1] to [0,1]$ is one such that $h(0) = 0 , h(1) = 1$; if $h(0) = 1, h(1) = 0$ it is orientation reversing. On $[0,1]^d$ it is not that easy to explain, but if $h$ is continously differentiable on the interior of $[0,1]^d$ (i.e. on $(0,1)^d$), then it means that the Jacobian matrix of $h$ has a positive determinant for all $x_0 in (0,1)^d$. Intuitively, an orientation preserving homeomorphism is obtained by "deforming the identity", an orientation reversing homeomorphism by "deforming a reflection of $[0,1]^d$ at same hyperplane $H_k = x = (x_1,..,,x_n) in mathbbR^n mid x_k = 1/2 $".
answered Jul 17 at 8:10
Paul Frost
3,703420
Many thanks for your kind answer. Now I have the concept a bit clearer,
– user123043
Jul 17 at 9:57
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The concept of "orientation" is too subtle to give a comprehensive explanation in this answer. You certainly know it from linear algebra (an isomorphism $f : mathbbR^n to mathbbR^n$ is orientation preserving if $det(f) > 0$). In the context of your question we deal with homeomorphims $h$ between manifolds (with boundary). Locally a manifold looks like Euclidean space $mathbbR^n$ (or, at the boundary, like a half-space $[0,infty) times mathbbR^n-1$) and it is possible to give a meaning to "$h$ is locally orientation preserving".
In your special case an orientation preserving homeomorphism $h : [0,1] to [0,1]$ is one such that $h(0) = 0 , h(1) = 1$; if $h(0) = 1, h(1) = 0$ it is orientation reversing. On $[0,1]^d$ it is not that easy to explain, but if $h$ is continously differentiable on the interior of $[0,1]^d$ (i.e. on $(0,1)^d$), then it means that the Jacobian matrix of $h$ has a positive determinant for all $x_0 in (0,1)^d$. Intuitively, an orientation preserving homeomorphism is obtained by "deforming the identity", an orientation reversing homeomorphism by "deforming a reflection of $[0,1]^d$ at same hyperplane $H_k = x = (x_1,..,,x_n) in mathbbR^n mid x_k = 1/2 $".
answered Jul 17 at 8:10
Paul Frost
3,703420
Many thanks for your kind answer. Now I have the concept a bit clearer,
– user123043
Jul 17 at 9:57
add a comment |Â
up vote
1
down vote
accepted
The concept of "orientation" is too subtle to give a comprehensive explanation in this answer. You certainly know it from linear algebra (an isomorphism $f : mathbbR^n to mathbbR^n$ is orientation preserving if $det(f) > 0$). In the context of your question we deal with homeomorphims $h$ between manifolds (with boundary). Locally a manifold looks like Euclidean space $mathbbR^n$ (or, at the boundary, like a half-space $[0,infty) times mathbbR^n-1$) and it is possible to give a meaning to "$h$ is locally orientation preserving".
In your special case an orientation preserving homeomorphism $h : [0,1] to [0,1]$ is one such that $h(0) = 0 , h(1) = 1$; if $h(0) = 1, h(1) = 0$ it is orientation reversing. On $[0,1]^d$ it is not that easy to explain, but if $h$ is continously differentiable on the interior of $[0,1]^d$ (i.e. on $(0,1)^d$), then it means that the Jacobian matrix of $h$ has a positive determinant for all $x_0 in (0,1)^d$. Intuitively, an orientation preserving homeomorphism is obtained by "deforming the identity", an orientation reversing homeomorphism by "deforming a reflection of $[0,1]^d$ at same hyperplane $H_k = x = (x_1,..,,x_n) in mathbbR^n mid x_k = 1/2 $".
answered Jul 17 at 8:10
Paul Frost
3,703420
Many thanks for your kind answer. Now I have the concept a bit clearer,
– user123043
Jul 17 at 9:57
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The concept of "orientation" is too subtle to give a comprehensive explanation in this answer. You certainly know it from linear algebra (an isomorphism $f : mathbbR^n to mathbbR^n$ is orientation preserving if $det(f) > 0$). In the context of your question we deal with homeomorphims $h$ between manifolds (with boundary). Locally a manifold looks like Euclidean space $mathbbR^n$ (or, at the boundary, like a half-space $[0,infty) times mathbbR^n-1$) and it is possible to give a meaning to "$h$ is locally orientation preserving".
In your special case an orientation preserving homeomorphism $h : [0,1] to [0,1]$ is one such that $h(0) = 0 , h(1) = 1$; if $h(0) = 1, h(1) = 0$ it is orientation reversing. On $[0,1]^d$ it is not that easy to explain, but if $h$ is continously differentiable on the interior of $[0,1]^d$ (i.e. on $(0,1)^d$), then it means that the Jacobian matrix of $h$ has a positive determinant for all $x_0 in (0,1)^d$. Intuitively, an orientation preserving homeomorphism is obtained by "deforming the identity", an orientation reversing homeomorphism by "deforming a reflection of $[0,1]^d$ at same hyperplane $H_k = x = (x_1,..,,x_n) in mathbbR^n mid x_k = 1/2 $".
answered Jul 17 at 8:10
Paul Frost
3,703420
The concept of "orientation" is too subtle to give a comprehensive explanation in this answer. You certainly know it from linear algebra (an isomorphism $f : mathbbR^n to mathbbR^n$ is orientation preserving if $det(f) > 0$). In the context of your question we deal with homeomorphims $h$ between manifolds (with boundary). Locally a manifold looks like Euclidean space $mathbbR^n$ (or, at the boundary, like a half-space $[0,infty) times mathbbR^n-1$) and it is possible to give a meaning to "$h$ is locally orientation preserving".
In your special case an orientation preserving homeomorphism $h : [0,1] to [0,1]$ is one such that $h(0) = 0 , h(1) = 1$; if $h(0) = 1, h(1) = 0$ it is orientation reversing. On $[0,1]^d$ it is not that easy to explain, but if $h$ is continously differentiable on the interior of $[0,1]^d$ (i.e. on $(0,1)^d$), then it means that the Jacobian matrix of $h$ has a positive determinant for all $x_0 in (0,1)^d$. Intuitively, an orientation preserving homeomorphism is obtained by "deforming the identity", an orientation reversing homeomorphism by "deforming a reflection of $[0,1]^d$ at same hyperplane $H_k = x = (x_1,..,,x_n) in mathbbR^n mid x_k = 1/2 $".
answered Jul 17 at 8:10
Paul Frost
3,703420
answered Jul 17 at 8:10
Paul Frost
3,703420
answered Jul 17 at 8:10
Paul Frost
3,703420
answered Jul 17 at 8:10
Paul Frost
3,703420
3,703420
Many thanks for your kind answer. Now I have the concept a bit clearer,
– user123043
Jul 17 at 9:57
add a comment |Â
Many thanks for your kind answer. Now I have the concept a bit clearer,
– user123043
Jul 17 at 9:57
Many thanks for your kind answer. Now I have the concept a bit clearer,
– user123043
Jul 17 at 9:57
Many thanks for your kind answer. Now I have the concept a bit clearer,
– user123043
Jul 17 at 9:57
add a comment |Â
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