Mixing
$exp_q^{-1}(exp_p(tX)) = exp_q^{-1}(p)+tGamma_{pto q}(X)+O(t^2)$ as $tto 0$?
$begingroup$
Let $(M,g)$ be a Riemannian manifold with induced metric $d$ and injectivity radius $r>0$. Let $p, q$ be two points in $M$ such that $d(p,q)<r$. It is easy to see that $p$ and $q$ can be connected by an unique minimizing geodesic $gamma={gamma(t):0le tle 1 }$ with $gamma(0)=p$ and $gamma(1)=q$. Denote by $Gamma_{pto q}:T_p(M)to T_q(M)$ the parallel transport along the geodesic $gamma$ from $p$ to $q$.
My question is: for a tangent vector $Xin T_p(M)$, does the following holds
$$exp_q^{-1}(exp_p(tX)) = exp_q^{-1}(p)+tGamma_{pto q}(X)+O(t^2), quad tto 0?$$
Any comments or hints will be appreciated. TIA.
differential-geometry manifolds differential-topology riemannian-geometry smooth-manifolds
asked Jan 28 at 6:45
Q. HuangQ. Huang
603213
$endgroup$
add a comment |
$begingroup$
Let $(M,g)$ be a Riemannian manifold with induced metric $d$ and injectivity radius $r>0$. Let $p, q$ be two points in $M$ such that $d(p,q)<r$. It is easy to see that $p$ and $q$ can be connected by an unique minimizing geodesic $gamma={gamma(t):0le tle 1 }$ with $gamma(0)=p$ and $gamma(1)=q$. Denote by $Gamma_{pto q}:T_p(M)to T_q(M)$ the parallel transport along the geodesic $gamma$ from $p$ to $q$.
My question is: for a tangent vector $Xin T_p(M)$, does the following holds
$$exp_q^{-1}(exp_p(tX)) = exp_q^{-1}(p)+tGamma_{pto q}(X)+O(t^2), quad tto 0?$$
Any comments or hints will be appreciated. TIA.
differential-geometry manifolds differential-topology riemannian-geometry smooth-manifolds
asked Jan 28 at 6:45
Q. HuangQ. Huang
603213
$endgroup$
add a comment |
$begingroup$
Let $(M,g)$ be a Riemannian manifold with induced metric $d$ and injectivity radius $r>0$. Let $p, q$ be two points in $M$ such that $d(p,q)<r$. It is easy to see that $p$ and $q$ can be connected by an unique minimizing geodesic $gamma={gamma(t):0le tle 1 }$ with $gamma(0)=p$ and $gamma(1)=q$. Denote by $Gamma_{pto q}:T_p(M)to T_q(M)$ the parallel transport along the geodesic $gamma$ from $p$ to $q$.
My question is: for a tangent vector $Xin T_p(M)$, does the following holds
$$exp_q^{-1}(exp_p(tX)) = exp_q^{-1}(p)+tGamma_{pto q}(X)+O(t^2), quad tto 0?$$
Any comments or hints will be appreciated. TIA.
differential-geometry manifolds differential-topology riemannian-geometry smooth-manifolds
asked Jan 28 at 6:45
Q. HuangQ. Huang
603213
$endgroup$
Let $(M,g)$ be a Riemannian manifold with induced metric $d$ and injectivity radius $r>0$. Let $p, q$ be two points in $M$ such that $d(p,q)<r$. It is easy to see that $p$ and $q$ can be connected by an unique minimizing geodesic $gamma={gamma(t):0le tle 1 }$ with $gamma(0)=p$ and $gamma(1)=q$. Denote by $Gamma_{pto q}:T_p(M)to T_q(M)$ the parallel transport along the geodesic $gamma$ from $p$ to $q$.
My question is: for a tangent vector $Xin T_p(M)$, does the following holds
$$exp_q^{-1}(exp_p(tX)) = exp_q^{-1}(p)+tGamma_{pto q}(X)+O(t^2), quad tto 0?$$
Any comments or hints will be appreciated. TIA.
differential-geometry manifolds differential-topology riemannian-geometry smooth-manifolds
differential-geometry manifolds differential-topology riemannian-geometry smooth-manifolds
asked Jan 28 at 6:45
Q. HuangQ. Huang
603213
asked Jan 28 at 6:45
Q. HuangQ. Huang
603213
edited Jan 28 at 6:53
Q. Huang
asked Jan 28 at 6:45
Q. HuangQ. Huang
603213
asked Jan 28 at 6:45
Q. HuangQ. Huang
603213
asked Jan 28 at 6:45
Q. HuangQ. Huang
603213
603213
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
It does not hold : On $2$-dimensional unit sphere, consider two points of $frac{pi}{2}$ distance. And parallel vector field on the geodesic between two points is perpendicular to the geodesic (cf. Parallel transportion for Alexandrov space with curvature bounded below - Petrunin) :
When $M=mathbb{S}^2, |p-q|=frac{pi}{2}$, then $c$ is unit speed
geodesic from $p$ to $q$. Assume that $X(t)$ is a unit parallel
vector field along $c(t)$ s.t. $X(0)perp c'(0)$.
Then $$A=0, B=exp_p^{-1} q, C=exp_p^{-1} exp_q tX(pi/2)$$ is
a triangle in $T_pM$ s.t. $angle BAC = t$.
When $A, B, C'$ is triangle with $angle ABC'=pi/2$ and
$|B-C'|=t$, then we find $|C-C'|$ : By cosine law, begin{align*}
|C-C'|^2&=t^2+(pi sin frac{t}{2})^2-2t(pi sin
frac{t}{2})cos frac{t}{2} \& =( t-pi sin frac{t}{2})^2+ O(t^4)
\ |C-C'| &=t-pi sin frac{t}{2}+ O(t^2)end{align*}
answered Jan 29 at 2:39
HK LeeHK Lee
14.1k52361
$endgroup$
1
$begingroup$
How does your example negate that equality? Can you explain more? Thanks...
$endgroup$
– Q. Huang
Jan 29 at 3:32
$begingroup$
I think I understand your exmaple and it is really a good counterexample. But I derive the last two equations in your answer as $$|C-C'|^2 = (t- frac{pi}{2} sin t)^2 + (frac{pi}{2} - frac{pi}{2} cos t)^2 = t^2 - pi t sin t + frac{pi^2}{2} (1-cos t) = (1-pi+ frac{pi^2}{4}) t^2 + O(t^4),$$ $$|C-C'| = (1-frac{pi}{2}) t + O(t^2).$$ But, anyway, this negate that equality in my question. Thanks.
$endgroup$
– Q. Huang
Jan 29 at 6:11
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
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oldest
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$begingroup$
It does not hold : On $2$-dimensional unit sphere, consider two points of $frac{pi}{2}$ distance. And parallel vector field on the geodesic between two points is perpendicular to the geodesic (cf. Parallel transportion for Alexandrov space with curvature bounded below - Petrunin) :
When $M=mathbb{S}^2, |p-q|=frac{pi}{2}$, then $c$ is unit speed
geodesic from $p$ to $q$. Assume that $X(t)$ is a unit parallel
vector field along $c(t)$ s.t. $X(0)perp c'(0)$.
Then $$A=0, B=exp_p^{-1} q, C=exp_p^{-1} exp_q tX(pi/2)$$ is
a triangle in $T_pM$ s.t. $angle BAC = t$.
When $A, B, C'$ is triangle with $angle ABC'=pi/2$ and
$|B-C'|=t$, then we find $|C-C'|$ : By cosine law, begin{align*}
|C-C'|^2&=t^2+(pi sin frac{t}{2})^2-2t(pi sin
frac{t}{2})cos frac{t}{2} \& =( t-pi sin frac{t}{2})^2+ O(t^4)
\ |C-C'| &=t-pi sin frac{t}{2}+ O(t^2)end{align*}
answered Jan 29 at 2:39
HK LeeHK Lee
14.1k52361
$endgroup$
1
$begingroup$
How does your example negate that equality? Can you explain more? Thanks...
$endgroup$
– Q. Huang
Jan 29 at 3:32
$begingroup$
I think I understand your exmaple and it is really a good counterexample. But I derive the last two equations in your answer as $$|C-C'|^2 = (t- frac{pi}{2} sin t)^2 + (frac{pi}{2} - frac{pi}{2} cos t)^2 = t^2 - pi t sin t + frac{pi^2}{2} (1-cos t) = (1-pi+ frac{pi^2}{4}) t^2 + O(t^4),$$ $$|C-C'| = (1-frac{pi}{2}) t + O(t^2).$$ But, anyway, this negate that equality in my question. Thanks.
$endgroup$
– Q. Huang
Jan 29 at 6:11
add a comment |
$begingroup$
It does not hold : On $2$-dimensional unit sphere, consider two points of $frac{pi}{2}$ distance. And parallel vector field on the geodesic between two points is perpendicular to the geodesic (cf. Parallel transportion for Alexandrov space with curvature bounded below - Petrunin) :
When $M=mathbb{S}^2, |p-q|=frac{pi}{2}$, then $c$ is unit speed
geodesic from $p$ to $q$. Assume that $X(t)$ is a unit parallel
vector field along $c(t)$ s.t. $X(0)perp c'(0)$.
Then $$A=0, B=exp_p^{-1} q, C=exp_p^{-1} exp_q tX(pi/2)$$ is
a triangle in $T_pM$ s.t. $angle BAC = t$.
When $A, B, C'$ is triangle with $angle ABC'=pi/2$ and
$|B-C'|=t$, then we find $|C-C'|$ : By cosine law, begin{align*}
|C-C'|^2&=t^2+(pi sin frac{t}{2})^2-2t(pi sin
frac{t}{2})cos frac{t}{2} \& =( t-pi sin frac{t}{2})^2+ O(t^4)
\ |C-C'| &=t-pi sin frac{t}{2}+ O(t^2)end{align*}
answered Jan 29 at 2:39
HK LeeHK Lee
14.1k52361
$endgroup$
1
$begingroup$
How does your example negate that equality? Can you explain more? Thanks...
$endgroup$
– Q. Huang
Jan 29 at 3:32
$begingroup$
I think I understand your exmaple and it is really a good counterexample. But I derive the last two equations in your answer as $$|C-C'|^2 = (t- frac{pi}{2} sin t)^2 + (frac{pi}{2} - frac{pi}{2} cos t)^2 = t^2 - pi t sin t + frac{pi^2}{2} (1-cos t) = (1-pi+ frac{pi^2}{4}) t^2 + O(t^4),$$ $$|C-C'| = (1-frac{pi}{2}) t + O(t^2).$$ But, anyway, this negate that equality in my question. Thanks.
$endgroup$
– Q. Huang
Jan 29 at 6:11
add a comment |
$begingroup$
It does not hold : On $2$-dimensional unit sphere, consider two points of $frac{pi}{2}$ distance. And parallel vector field on the geodesic between two points is perpendicular to the geodesic (cf. Parallel transportion for Alexandrov space with curvature bounded below - Petrunin) :
When $M=mathbb{S}^2, |p-q|=frac{pi}{2}$, then $c$ is unit speed
geodesic from $p$ to $q$. Assume that $X(t)$ is a unit parallel
vector field along $c(t)$ s.t. $X(0)perp c'(0)$.
Then $$A=0, B=exp_p^{-1} q, C=exp_p^{-1} exp_q tX(pi/2)$$ is
a triangle in $T_pM$ s.t. $angle BAC = t$.
When $A, B, C'$ is triangle with $angle ABC'=pi/2$ and
$|B-C'|=t$, then we find $|C-C'|$ : By cosine law, begin{align*}
|C-C'|^2&=t^2+(pi sin frac{t}{2})^2-2t(pi sin
frac{t}{2})cos frac{t}{2} \& =( t-pi sin frac{t}{2})^2+ O(t^4)
\ |C-C'| &=t-pi sin frac{t}{2}+ O(t^2)end{align*}
answered Jan 29 at 2:39
HK LeeHK Lee
14.1k52361
$endgroup$
It does not hold : On $2$-dimensional unit sphere, consider two points of $frac{pi}{2}$ distance. And parallel vector field on the geodesic between two points is perpendicular to the geodesic (cf. Parallel transportion for Alexandrov space with curvature bounded below - Petrunin) :
When $M=mathbb{S}^2, |p-q|=frac{pi}{2}$, then $c$ is unit speed
geodesic from $p$ to $q$. Assume that $X(t)$ is a unit parallel
vector field along $c(t)$ s.t. $X(0)perp c'(0)$.
Then $$A=0, B=exp_p^{-1} q, C=exp_p^{-1} exp_q tX(pi/2)$$ is
a triangle in $T_pM$ s.t. $angle BAC = t$.
When $A, B, C'$ is triangle with $angle ABC'=pi/2$ and
$|B-C'|=t$, then we find $|C-C'|$ : By cosine law, begin{align*}
|C-C'|^2&=t^2+(pi sin frac{t}{2})^2-2t(pi sin
frac{t}{2})cos frac{t}{2} \& =( t-pi sin frac{t}{2})^2+ O(t^4)
\ |C-C'| &=t-pi sin frac{t}{2}+ O(t^2)end{align*}
answered Jan 29 at 2:39
HK LeeHK Lee
14.1k52361
edited Jan 29 at 3:40
answered Jan 29 at 2:39
HK LeeHK Lee
14.1k52361
answered Jan 29 at 2:39
HK LeeHK Lee
14.1k52361
answered Jan 29 at 2:39
HK LeeHK Lee
14.1k52361
14.1k52361
1
$begingroup$
How does your example negate that equality? Can you explain more? Thanks...
$endgroup$
– Q. Huang
Jan 29 at 3:32
$begingroup$
I think I understand your exmaple and it is really a good counterexample. But I derive the last two equations in your answer as $$|C-C'|^2 = (t- frac{pi}{2} sin t)^2 + (frac{pi}{2} - frac{pi}{2} cos t)^2 = t^2 - pi t sin t + frac{pi^2}{2} (1-cos t) = (1-pi+ frac{pi^2}{4}) t^2 + O(t^4),$$ $$|C-C'| = (1-frac{pi}{2}) t + O(t^2).$$ But, anyway, this negate that equality in my question. Thanks.
$endgroup$
– Q. Huang
Jan 29 at 6:11
add a comment |
1
$begingroup$
How does your example negate that equality? Can you explain more? Thanks...
$endgroup$
– Q. Huang
Jan 29 at 3:32
$begingroup$
I think I understand your exmaple and it is really a good counterexample. But I derive the last two equations in your answer as $$|C-C'|^2 = (t- frac{pi}{2} sin t)^2 + (frac{pi}{2} - frac{pi}{2} cos t)^2 = t^2 - pi t sin t + frac{pi^2}{2} (1-cos t) = (1-pi+ frac{pi^2}{4}) t^2 + O(t^4),$$ $$|C-C'| = (1-frac{pi}{2}) t + O(t^2).$$ But, anyway, this negate that equality in my question. Thanks.
$endgroup$
– Q. Huang
Jan 29 at 6:11
1
1
$begingroup$
How does your example negate that equality? Can you explain more? Thanks...
$endgroup$
– Q. Huang
Jan 29 at 3:32
$begingroup$
How does your example negate that equality? Can you explain more? Thanks...
$endgroup$
– Q. Huang
Jan 29 at 3:32
$begingroup$
I think I understand your exmaple and it is really a good counterexample. But I derive the last two equations in your answer as $$|C-C'|^2 = (t- frac{pi}{2} sin t)^2 + (frac{pi}{2} - frac{pi}{2} cos t)^2 = t^2 - pi t sin t + frac{pi^2}{2} (1-cos t) = (1-pi+ frac{pi^2}{4}) t^2 + O(t^4),$$ $$|C-C'| = (1-frac{pi}{2}) t + O(t^2).$$ But, anyway, this negate that equality in my question. Thanks.
$endgroup$
– Q. Huang
Jan 29 at 6:11
$begingroup$
I think I understand your exmaple and it is really a good counterexample. But I derive the last two equations in your answer as $$|C-C'|^2 = (t- frac{pi}{2} sin t)^2 + (frac{pi}{2} - frac{pi}{2} cos t)^2 = t^2 - pi t sin t + frac{pi^2}{2} (1-cos t) = (1-pi+ frac{pi^2}{4}) t^2 + O(t^4),$$ $$|C-C'| = (1-frac{pi}{2}) t + O(t^2).$$ But, anyway, this negate that equality in my question. Thanks.
$endgroup$
– Q. Huang
Jan 29 at 6:11
add a comment |
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