Parallel Transport Along Radial Geodesics Yields a Smooth Vector Field?
$begingroup$
Let $M$ be a Riemannian manifold and $p$ be a point on $M$. Let $U$ be a normal neighborhood about $p$ (that is, the exponential map $exp_p$ maps a neighborhood of the origin in $T_pM$ diffeomorphically onto $U$). Fix a vector $v_pin T_pM$.
For each $qin U$, let $v_q$ be the vector in $T_qM$ obtained by parallel transporting $v_p$ along the radial geodesic joining $p$ to $q$. So we get a map $X:Uto TU$ which takes $q$ to $v_q$. So $X$ is a vector field on $U$.
Question. Is $X$ necessarily smooth?
What I thought is that by definition of $X$, we have $nabla_{partial/partial r}X=0$ at all point of $Usetminus{p}$, where $partial/partial r$ is the radial vector field (which is defined at all points of $U$ except $p$). So $X$ is a solution of a system of partial differential equations. I do not know if this guarantees that $X$ is smooth on $Usetminus{p}$.
riemannian-geometry
asked Jun 2 '17 at 18:09
caffeinemachinecaffeinemachine
6,56721351
$endgroup$
add a comment |
$begingroup$
Let $M$ be a Riemannian manifold and $p$ be a point on $M$. Let $U$ be a normal neighborhood about $p$ (that is, the exponential map $exp_p$ maps a neighborhood of the origin in $T_pM$ diffeomorphically onto $U$). Fix a vector $v_pin T_pM$.
For each $qin U$, let $v_q$ be the vector in $T_qM$ obtained by parallel transporting $v_p$ along the radial geodesic joining $p$ to $q$. So we get a map $X:Uto TU$ which takes $q$ to $v_q$. So $X$ is a vector field on $U$.
Question. Is $X$ necessarily smooth?
What I thought is that by definition of $X$, we have $nabla_{partial/partial r}X=0$ at all point of $Usetminus{p}$, where $partial/partial r$ is the radial vector field (which is defined at all points of $U$ except $p$). So $X$ is a solution of a system of partial differential equations. I do not know if this guarantees that $X$ is smooth on $Usetminus{p}$.
riemannian-geometry
asked Jun 2 '17 at 18:09
caffeinemachinecaffeinemachine
6,56721351
$endgroup$
2
$begingroup$
My first instinct would be to look for a counterexample in the Heisenberg group (3-dimensional simply-connected nilmanifold).
$endgroup$
– Neal
Jun 2 '17 at 18:13
add a comment |
$begingroup$
Let $M$ be a Riemannian manifold and $p$ be a point on $M$. Let $U$ be a normal neighborhood about $p$ (that is, the exponential map $exp_p$ maps a neighborhood of the origin in $T_pM$ diffeomorphically onto $U$). Fix a vector $v_pin T_pM$.
For each $qin U$, let $v_q$ be the vector in $T_qM$ obtained by parallel transporting $v_p$ along the radial geodesic joining $p$ to $q$. So we get a map $X:Uto TU$ which takes $q$ to $v_q$. So $X$ is a vector field on $U$.
Question. Is $X$ necessarily smooth?
What I thought is that by definition of $X$, we have $nabla_{partial/partial r}X=0$ at all point of $Usetminus{p}$, where $partial/partial r$ is the radial vector field (which is defined at all points of $U$ except $p$). So $X$ is a solution of a system of partial differential equations. I do not know if this guarantees that $X$ is smooth on $Usetminus{p}$.
riemannian-geometry
asked Jun 2 '17 at 18:09
caffeinemachinecaffeinemachine
6,56721351
$endgroup$
Let $M$ be a Riemannian manifold and $p$ be a point on $M$. Let $U$ be a normal neighborhood about $p$ (that is, the exponential map $exp_p$ maps a neighborhood of the origin in $T_pM$ diffeomorphically onto $U$). Fix a vector $v_pin T_pM$.
For each $qin U$, let $v_q$ be the vector in $T_qM$ obtained by parallel transporting $v_p$ along the radial geodesic joining $p$ to $q$. So we get a map $X:Uto TU$ which takes $q$ to $v_q$. So $X$ is a vector field on $U$.
Question. Is $X$ necessarily smooth?
What I thought is that by definition of $X$, we have $nabla_{partial/partial r}X=0$ at all point of $Usetminus{p}$, where $partial/partial r$ is the radial vector field (which is defined at all points of $U$ except $p$). So $X$ is a solution of a system of partial differential equations. I do not know if this guarantees that $X$ is smooth on $Usetminus{p}$.
riemannian-geometry
riemannian-geometry
asked Jun 2 '17 at 18:09
caffeinemachinecaffeinemachine
6,56721351
asked Jun 2 '17 at 18:09
caffeinemachinecaffeinemachine
6,56721351
asked Jun 2 '17 at 18:09
caffeinemachinecaffeinemachine
6,56721351
asked Jun 2 '17 at 18:09
caffeinemachinecaffeinemachine
6,56721351
asked Jun 2 '17 at 18:09
caffeinemachinecaffeinemachine
6,56721351
6,56721351
2
$begingroup$
My first instinct would be to look for a counterexample in the Heisenberg group (3-dimensional simply-connected nilmanifold).
$endgroup$
– Neal
Jun 2 '17 at 18:13
add a comment |
2
$begingroup$
My first instinct would be to look for a counterexample in the Heisenberg group (3-dimensional simply-connected nilmanifold).
$endgroup$
– Neal
Jun 2 '17 at 18:13
2
2
$begingroup$
My first instinct would be to look for a counterexample in the Heisenberg group (3-dimensional simply-connected nilmanifold).
$endgroup$
– Neal
Jun 2 '17 at 18:13
$begingroup$
My first instinct would be to look for a counterexample in the Heisenberg group (3-dimensional simply-connected nilmanifold).
$endgroup$
– Neal
Jun 2 '17 at 18:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes.
There's kind of a trick to it. Let $(x^i)$ be normal coordinates centered at $p$, so the radial geodesics are given by $gamma(t) = (tx^1,dots, tx^n)$. The differential equation satisfied by $v = v^k partial_k$ along such a geodesic is
$$
dot v^k(t) = - Gamma_{ij}^k(tx^1,dots,tx^n)x^i v^j(t),tag{$*$}
$$
with initial condition $v^k(0) = a^k$ (some arbitrary constants). (I'm using the summation convention here.)
The trick is to replace the $x^i$'s by new dependent variables $w^i$, and write this as a system of ODEs for the $2n$ functions $(v^i,dots,v^n,w^1,dots,w^n)$:
begin{align*}
dot v^k(t) &= - Gamma_{ij}^k(tw^1(t),dots,tw^n(t))w^i(t) v^j(t),\
dot w^k(t) &= 0,
end{align*}
with initial conditions
begin{align*}
v^k(0) &= a^k,\
w^k(0) &= x^k.
end{align*}
Because solutions to smooth ODEs depend smoothly on initial conditions as well as time, the solutions to this system can be written as smooth functions $v^k(t,a,x)$ and $w^k(t,a,x)$. It follows immediately from the form of the equation that $w^k$ is constant, $w^k equiv x^k$, and therefore the vector field you're interested in is
$$
v(x) = v^k(1,a,x) partial_k,
$$
which depends smoothly on $x$.
answered Jun 2 '17 at 22:01
Jack LeeJack Lee
27.4k54767
$endgroup$
$begingroup$
Thank you. This is a very neat proof.
$endgroup$
– caffeinemachine
Jun 3 '17 at 18:23
add a comment |
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$begingroup$
Yes.
There's kind of a trick to it. Let $(x^i)$ be normal coordinates centered at $p$, so the radial geodesics are given by $gamma(t) = (tx^1,dots, tx^n)$. The differential equation satisfied by $v = v^k partial_k$ along such a geodesic is
$$
dot v^k(t) = - Gamma_{ij}^k(tx^1,dots,tx^n)x^i v^j(t),tag{$*$}
$$
with initial condition $v^k(0) = a^k$ (some arbitrary constants). (I'm using the summation convention here.)
The trick is to replace the $x^i$'s by new dependent variables $w^i$, and write this as a system of ODEs for the $2n$ functions $(v^i,dots,v^n,w^1,dots,w^n)$:
begin{align*}
dot v^k(t) &= - Gamma_{ij}^k(tw^1(t),dots,tw^n(t))w^i(t) v^j(t),\
dot w^k(t) &= 0,
end{align*}
with initial conditions
begin{align*}
v^k(0) &= a^k,\
w^k(0) &= x^k.
end{align*}
Because solutions to smooth ODEs depend smoothly on initial conditions as well as time, the solutions to this system can be written as smooth functions $v^k(t,a,x)$ and $w^k(t,a,x)$. It follows immediately from the form of the equation that $w^k$ is constant, $w^k equiv x^k$, and therefore the vector field you're interested in is
$$
v(x) = v^k(1,a,x) partial_k,
$$
which depends smoothly on $x$.
answered Jun 2 '17 at 22:01
Jack LeeJack Lee
27.4k54767
$endgroup$
$begingroup$
Thank you. This is a very neat proof.
$endgroup$
– caffeinemachine
Jun 3 '17 at 18:23
add a comment |
$begingroup$
Yes.
There's kind of a trick to it. Let $(x^i)$ be normal coordinates centered at $p$, so the radial geodesics are given by $gamma(t) = (tx^1,dots, tx^n)$. The differential equation satisfied by $v = v^k partial_k$ along such a geodesic is
$$
dot v^k(t) = - Gamma_{ij}^k(tx^1,dots,tx^n)x^i v^j(t),tag{$*$}
$$
with initial condition $v^k(0) = a^k$ (some arbitrary constants). (I'm using the summation convention here.)
The trick is to replace the $x^i$'s by new dependent variables $w^i$, and write this as a system of ODEs for the $2n$ functions $(v^i,dots,v^n,w^1,dots,w^n)$:
begin{align*}
dot v^k(t) &= - Gamma_{ij}^k(tw^1(t),dots,tw^n(t))w^i(t) v^j(t),\
dot w^k(t) &= 0,
end{align*}
with initial conditions
begin{align*}
v^k(0) &= a^k,\
w^k(0) &= x^k.
end{align*}
Because solutions to smooth ODEs depend smoothly on initial conditions as well as time, the solutions to this system can be written as smooth functions $v^k(t,a,x)$ and $w^k(t,a,x)$. It follows immediately from the form of the equation that $w^k$ is constant, $w^k equiv x^k$, and therefore the vector field you're interested in is
$$
v(x) = v^k(1,a,x) partial_k,
$$
which depends smoothly on $x$.
answered Jun 2 '17 at 22:01
Jack LeeJack Lee
27.4k54767
$endgroup$
$begingroup$
Thank you. This is a very neat proof.
$endgroup$
– caffeinemachine
Jun 3 '17 at 18:23
add a comment |
$begingroup$
Yes.
There's kind of a trick to it. Let $(x^i)$ be normal coordinates centered at $p$, so the radial geodesics are given by $gamma(t) = (tx^1,dots, tx^n)$. The differential equation satisfied by $v = v^k partial_k$ along such a geodesic is
$$
dot v^k(t) = - Gamma_{ij}^k(tx^1,dots,tx^n)x^i v^j(t),tag{$*$}
$$
with initial condition $v^k(0) = a^k$ (some arbitrary constants). (I'm using the summation convention here.)
The trick is to replace the $x^i$'s by new dependent variables $w^i$, and write this as a system of ODEs for the $2n$ functions $(v^i,dots,v^n,w^1,dots,w^n)$:
begin{align*}
dot v^k(t) &= - Gamma_{ij}^k(tw^1(t),dots,tw^n(t))w^i(t) v^j(t),\
dot w^k(t) &= 0,
end{align*}
with initial conditions
begin{align*}
v^k(0) &= a^k,\
w^k(0) &= x^k.
end{align*}
Because solutions to smooth ODEs depend smoothly on initial conditions as well as time, the solutions to this system can be written as smooth functions $v^k(t,a,x)$ and $w^k(t,a,x)$. It follows immediately from the form of the equation that $w^k$ is constant, $w^k equiv x^k$, and therefore the vector field you're interested in is
$$
v(x) = v^k(1,a,x) partial_k,
$$
which depends smoothly on $x$.
answered Jun 2 '17 at 22:01
Jack LeeJack Lee
27.4k54767
$endgroup$
Yes.
There's kind of a trick to it. Let $(x^i)$ be normal coordinates centered at $p$, so the radial geodesics are given by $gamma(t) = (tx^1,dots, tx^n)$. The differential equation satisfied by $v = v^k partial_k$ along such a geodesic is
$$
dot v^k(t) = - Gamma_{ij}^k(tx^1,dots,tx^n)x^i v^j(t),tag{$*$}
$$
with initial condition $v^k(0) = a^k$ (some arbitrary constants). (I'm using the summation convention here.)
The trick is to replace the $x^i$'s by new dependent variables $w^i$, and write this as a system of ODEs for the $2n$ functions $(v^i,dots,v^n,w^1,dots,w^n)$:
begin{align*}
dot v^k(t) &= - Gamma_{ij}^k(tw^1(t),dots,tw^n(t))w^i(t) v^j(t),\
dot w^k(t) &= 0,
end{align*}
with initial conditions
begin{align*}
v^k(0) &= a^k,\
w^k(0) &= x^k.
end{align*}
Because solutions to smooth ODEs depend smoothly on initial conditions as well as time, the solutions to this system can be written as smooth functions $v^k(t,a,x)$ and $w^k(t,a,x)$. It follows immediately from the form of the equation that $w^k$ is constant, $w^k equiv x^k$, and therefore the vector field you're interested in is
$$
v(x) = v^k(1,a,x) partial_k,
$$
which depends smoothly on $x$.
answered Jun 2 '17 at 22:01
Jack LeeJack Lee
27.4k54767
answered Jun 2 '17 at 22:01
Jack LeeJack Lee
27.4k54767
answered Jun 2 '17 at 22:01
Jack LeeJack Lee
27.4k54767
answered Jun 2 '17 at 22:01
Jack LeeJack Lee
27.4k54767
27.4k54767
$begingroup$
Thank you. This is a very neat proof.
$endgroup$
– caffeinemachine
Jun 3 '17 at 18:23
add a comment |
$begingroup$
Thank you. This is a very neat proof.
$endgroup$
– caffeinemachine
Jun 3 '17 at 18:23
$begingroup$
Thank you. This is a very neat proof.
$endgroup$
– caffeinemachine
Jun 3 '17 at 18:23
$begingroup$
Thank you. This is a very neat proof.
$endgroup$
– caffeinemachine
Jun 3 '17 at 18:23
add a comment |
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My first instinct would be to look for a counterexample in the Heisenberg group (3-dimensional simply-connected nilmanifold).
$endgroup$
– Neal
Jun 2 '17 at 18:13