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Exponential map on Heisenberg group is a diffeomorphism.
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Assume that a Riemannian manifold $M$ is
simply connected. In further, assume that there is an global
orthnormal frame $e_i$. If $f_i$ is flow of $e_i$, and $M$ is
homeomorphic to $mathbbR^n$, then define $$ F(x)= f_n(x_n,cdots
f_2(x_2,f_1(x_1,o))cdots ) $$ where $o$ is a fixed point. Then $F$
is a diffeomorphism from $mathbbR^n$ to $M$. Hence an example of
such manifold is a Lie group homeomorphic to $mathbbR^n$, i.e.,
Heisenberg group $mathbbH^3$.
Question : If Riemannian metric on Heisenberg group $M=mathbbH^3$ has global orthonormal frame, then an exponential map $rm exp$ is a diffeomorphism on $T_oM$, where $o$ is any fixed point.
Proof : We have a claim that between two points, there is
unique
geodesic, i.e. geodesic space.
Try 1 : At each point in $M$, sectional curvatures have
both signs. If we have a point of negative sectional curvature, then
it is helpful : There is no subset $A$ of measure $0$ in canonical
sphere $S$ s.t. a connected set $S-A$ is a geodesic. But in some
torus in
$mathbbE^3$ there is such property.
Try 2 : $Mrightarrow mathbbE^2$ is a Riemannian
submersion. So
$exp_p tv, exp_p tw$ do not meet when $v, w$ are horizontal.
How can we finish the proof ?
riemannian-geometry heisenberg-group
edited Aug 1 at 12:33
John Ma
37.5k93669
asked Aug 1 at 5:21
HK Lee
13.5k31855
add a comment |Â
up vote
0
down vote
favorite
Assume that a Riemannian manifold $M$ is
simply connected. In further, assume that there is an global
orthnormal frame $e_i$. If $f_i$ is flow of $e_i$, and $M$ is
homeomorphic to $mathbbR^n$, then define $$ F(x)= f_n(x_n,cdots
f_2(x_2,f_1(x_1,o))cdots ) $$ where $o$ is a fixed point. Then $F$
is a diffeomorphism from $mathbbR^n$ to $M$. Hence an example of
such manifold is a Lie group homeomorphic to $mathbbR^n$, i.e.,
Heisenberg group $mathbbH^3$.
Question : If Riemannian metric on Heisenberg group $M=mathbbH^3$ has global orthonormal frame, then an exponential map $rm exp$ is a diffeomorphism on $T_oM$, where $o$ is any fixed point.
Proof : We have a claim that between two points, there is
unique
geodesic, i.e. geodesic space.
Try 1 : At each point in $M$, sectional curvatures have
both signs. If we have a point of negative sectional curvature, then
it is helpful : There is no subset $A$ of measure $0$ in canonical
sphere $S$ s.t. a connected set $S-A$ is a geodesic. But in some
torus in
$mathbbE^3$ there is such property.
Try 2 : $Mrightarrow mathbbE^2$ is a Riemannian
submersion. So
$exp_p tv, exp_p tw$ do not meet when $v, w$ are horizontal.
How can we finish the proof ?
riemannian-geometry heisenberg-group
edited Aug 1 at 12:33
John Ma
37.5k93669
asked Aug 1 at 5:21
HK Lee
13.5k31855
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Assume that a Riemannian manifold $M$ is
simply connected. In further, assume that there is an global
orthnormal frame $e_i$. If $f_i$ is flow of $e_i$, and $M$ is
homeomorphic to $mathbbR^n$, then define $$ F(x)= f_n(x_n,cdots
f_2(x_2,f_1(x_1,o))cdots ) $$ where $o$ is a fixed point. Then $F$
is a diffeomorphism from $mathbbR^n$ to $M$. Hence an example of
such manifold is a Lie group homeomorphic to $mathbbR^n$, i.e.,
Heisenberg group $mathbbH^3$.
Question : If Riemannian metric on Heisenberg group $M=mathbbH^3$ has global orthonormal frame, then an exponential map $rm exp$ is a diffeomorphism on $T_oM$, where $o$ is any fixed point.
Proof : We have a claim that between two points, there is
unique
geodesic, i.e. geodesic space.
Try 1 : At each point in $M$, sectional curvatures have
both signs. If we have a point of negative sectional curvature, then
it is helpful : There is no subset $A$ of measure $0$ in canonical
sphere $S$ s.t. a connected set $S-A$ is a geodesic. But in some
torus in
$mathbbE^3$ there is such property.
Try 2 : $Mrightarrow mathbbE^2$ is a Riemannian
submersion. So
$exp_p tv, exp_p tw$ do not meet when $v, w$ are horizontal.
How can we finish the proof ?
riemannian-geometry heisenberg-group
edited Aug 1 at 12:33
John Ma
37.5k93669
asked Aug 1 at 5:21
HK Lee
13.5k31855
Assume that a Riemannian manifold $M$ is
simply connected. In further, assume that there is an global
orthnormal frame $e_i$. If $f_i$ is flow of $e_i$, and $M$ is
homeomorphic to $mathbbR^n$, then define $$ F(x)= f_n(x_n,cdots
f_2(x_2,f_1(x_1,o))cdots ) $$ where $o$ is a fixed point. Then $F$
is a diffeomorphism from $mathbbR^n$ to $M$. Hence an example of
such manifold is a Lie group homeomorphic to $mathbbR^n$, i.e.,
Heisenberg group $mathbbH^3$.
Question : If Riemannian metric on Heisenberg group $M=mathbbH^3$ has global orthonormal frame, then an exponential map $rm exp$ is a diffeomorphism on $T_oM$, where $o$ is any fixed point.
Proof : We have a claim that between two points, there is
unique
geodesic, i.e. geodesic space.
Try 1 : At each point in $M$, sectional curvatures have
both signs. If we have a point of negative sectional curvature, then
it is helpful : There is no subset $A$ of measure $0$ in canonical
sphere $S$ s.t. a connected set $S-A$ is a geodesic. But in some
torus in
$mathbbE^3$ there is such property.
Try 2 : $Mrightarrow mathbbE^2$ is a Riemannian
submersion. So
$exp_p tv, exp_p tw$ do not meet when $v, w$ are horizontal.
How can we finish the proof ?
riemannian-geometry heisenberg-group
edited Aug 1 at 12:33
John Ma
37.5k93669
asked Aug 1 at 5:21
HK Lee
13.5k31855
edited Aug 1 at 12:33
John Ma
37.5k93669
edited Aug 1 at 12:33
John Ma
37.5k93669
edited Aug 1 at 12:33
John Ma
37.5k93669
37.5k93669
asked Aug 1 at 5:21
HK Lee
13.5k31855
asked Aug 1 at 5:21
HK Lee
13.5k31855
asked Aug 1 at 5:21
HK Lee
13.5k31855
13.5k31855
add a comment |Â
add a comment |Â
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