Mixing
Frenet Frame along a curve and Riemannian Curvature on $S^2$
$begingroup$
I would appreciate some help showing the following statement.
Let $omega: [0,1] rightarrow S^2$ be a smooth curve with velocity vector $V = omega'$, speed $v = |V|$ and Frenet frame ${T,N}$.
Then for all vector fields $W: S^2 rightarrow TS^2$ along the curve $omega$ we have:
$$R(W,V)V = v^2 langle W , N rangle N $$
where $R$ is the Riemannian curvature tensor on $S^2$.
If $omega$ were arc-length parametrized / unit-length, I suppose the terms $v^2$ could join the left side because of the normalization of $N = frac{|T'|}{T'}$, but this isn't the case here.
differential-geometry riemannian-geometry smooth-manifolds curves frenet-frame
asked Jan 29 at 16:15
NhatNhat
1,0261017
$endgroup$
add a comment |
$begingroup$
I would appreciate some help showing the following statement.
Let $omega: [0,1] rightarrow S^2$ be a smooth curve with velocity vector $V = omega'$, speed $v = |V|$ and Frenet frame ${T,N}$.
Then for all vector fields $W: S^2 rightarrow TS^2$ along the curve $omega$ we have:
$$R(W,V)V = v^2 langle W , N rangle N $$
where $R$ is the Riemannian curvature tensor on $S^2$.
If $omega$ were arc-length parametrized / unit-length, I suppose the terms $v^2$ could join the left side because of the normalization of $N = frac{|T'|}{T'}$, but this isn't the case here.
differential-geometry riemannian-geometry smooth-manifolds curves frenet-frame
asked Jan 29 at 16:15
NhatNhat
1,0261017
$endgroup$
$begingroup$
No, the $v^2$ is there because $V=vT$. What definition of the Frenet frame are they using? We're not thinking of $S^2subsetBbb R^3$ and of the Frenet frame of a curve in $Bbb R^3$, I presume?
$endgroup$
– Ted Shifrin
Jan 29 at 19:52
$begingroup$
@TedShifrin I think they are but I’m not entirely sure. If it helps: Later calculations are made representing the vectorfield $W$ (as above) in terms of $W = fT + gN$. Furthermore for $gamma $ with $gamma‘ = omega$ they get the Frenet Frame ${T_gamma, N_gamma, B_gamma }$ such that $T_gamma = omega, N_gamma = T, B_gamma$ as expected.
$endgroup$
– Nhat
Jan 30 at 12:42
add a comment |
$begingroup$
I would appreciate some help showing the following statement.
Let $omega: [0,1] rightarrow S^2$ be a smooth curve with velocity vector $V = omega'$, speed $v = |V|$ and Frenet frame ${T,N}$.
Then for all vector fields $W: S^2 rightarrow TS^2$ along the curve $omega$ we have:
$$R(W,V)V = v^2 langle W , N rangle N $$
where $R$ is the Riemannian curvature tensor on $S^2$.
If $omega$ were arc-length parametrized / unit-length, I suppose the terms $v^2$ could join the left side because of the normalization of $N = frac{|T'|}{T'}$, but this isn't the case here.
differential-geometry riemannian-geometry smooth-manifolds curves frenet-frame
asked Jan 29 at 16:15
NhatNhat
1,0261017
$endgroup$
I would appreciate some help showing the following statement.
Let $omega: [0,1] rightarrow S^2$ be a smooth curve with velocity vector $V = omega'$, speed $v = |V|$ and Frenet frame ${T,N}$.
Then for all vector fields $W: S^2 rightarrow TS^2$ along the curve $omega$ we have:
$$R(W,V)V = v^2 langle W , N rangle N $$
where $R$ is the Riemannian curvature tensor on $S^2$.
If $omega$ were arc-length parametrized / unit-length, I suppose the terms $v^2$ could join the left side because of the normalization of $N = frac{|T'|}{T'}$, but this isn't the case here.
differential-geometry riemannian-geometry smooth-manifolds curves frenet-frame
differential-geometry riemannian-geometry smooth-manifolds curves frenet-frame
asked Jan 29 at 16:15
NhatNhat
1,0261017
asked Jan 29 at 16:15
NhatNhat
1,0261017
asked Jan 29 at 16:15
NhatNhat
1,0261017
asked Jan 29 at 16:15
NhatNhat
1,0261017
asked Jan 29 at 16:15
NhatNhat
1,0261017
1,0261017
$begingroup$
No, the $v^2$ is there because $V=vT$. What definition of the Frenet frame are they using? We're not thinking of $S^2subsetBbb R^3$ and of the Frenet frame of a curve in $Bbb R^3$, I presume?
$endgroup$
– Ted Shifrin
Jan 29 at 19:52
$begingroup$
@TedShifrin I think they are but I’m not entirely sure. If it helps: Later calculations are made representing the vectorfield $W$ (as above) in terms of $W = fT + gN$. Furthermore for $gamma $ with $gamma‘ = omega$ they get the Frenet Frame ${T_gamma, N_gamma, B_gamma }$ such that $T_gamma = omega, N_gamma = T, B_gamma$ as expected.
$endgroup$
– Nhat
Jan 30 at 12:42
add a comment |
$begingroup$
No, the $v^2$ is there because $V=vT$. What definition of the Frenet frame are they using? We're not thinking of $S^2subsetBbb R^3$ and of the Frenet frame of a curve in $Bbb R^3$, I presume?
$endgroup$
– Ted Shifrin
Jan 29 at 19:52
$begingroup$
@TedShifrin I think they are but I’m not entirely sure. If it helps: Later calculations are made representing the vectorfield $W$ (as above) in terms of $W = fT + gN$. Furthermore for $gamma $ with $gamma‘ = omega$ they get the Frenet Frame ${T_gamma, N_gamma, B_gamma }$ such that $T_gamma = omega, N_gamma = T, B_gamma$ as expected.
$endgroup$
– Nhat
Jan 30 at 12:42
$begingroup$
No, the $v^2$ is there because $V=vT$. What definition of the Frenet frame are they using? We're not thinking of $S^2subsetBbb R^3$ and of the Frenet frame of a curve in $Bbb R^3$, I presume?
$endgroup$
– Ted Shifrin
Jan 29 at 19:52
$begingroup$
No, the $v^2$ is there because $V=vT$. What definition of the Frenet frame are they using? We're not thinking of $S^2subsetBbb R^3$ and of the Frenet frame of a curve in $Bbb R^3$, I presume?
$endgroup$
– Ted Shifrin
Jan 29 at 19:52
$begingroup$
@TedShifrin I think they are but I’m not entirely sure. If it helps: Later calculations are made representing the vectorfield $W$ (as above) in terms of $W = fT + gN$. Furthermore for $gamma $ with $gamma‘ = omega$ they get the Frenet Frame ${T_gamma, N_gamma, B_gamma }$ such that $T_gamma = omega, N_gamma = T, B_gamma$ as expected.
$endgroup$
– Nhat
Jan 30 at 12:42
$begingroup$
@TedShifrin I think they are but I’m not entirely sure. If it helps: Later calculations are made representing the vectorfield $W$ (as above) in terms of $W = fT + gN$. Furthermore for $gamma $ with $gamma‘ = omega$ they get the Frenet Frame ${T_gamma, N_gamma, B_gamma }$ such that $T_gamma = omega, N_gamma = T, B_gamma$ as expected.
$endgroup$
– Nhat
Jan 30 at 12:42
add a comment |
1 Answer
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Yes, I think the authors' use of the term Frenet frame is a bit misleading — I personally would call this a Darboux frame. Regardless, by $N$ they mean the normalized covariant derivative of $T$.
As I already commented, it suffices to show that $R(W,T)T = langle W,Nrangle N$ (since $V = vT$). Note that $R(T,T)T = 0$, so, writing $W=aT+bN$, we have $R(W,T)T = bR(N,T)T overset{(*)}{=} bN = langle W,Nrangle N$, as required. The equality (*) follows from the fact that the sphere has constant sectional curvature $1$.
answered Jan 31 at 17:39
Ted ShifrinTed Shifrin
64.7k44692
$endgroup$
add a comment |
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$begingroup$
Yes, I think the authors' use of the term Frenet frame is a bit misleading — I personally would call this a Darboux frame. Regardless, by $N$ they mean the normalized covariant derivative of $T$.
As I already commented, it suffices to show that $R(W,T)T = langle W,Nrangle N$ (since $V = vT$). Note that $R(T,T)T = 0$, so, writing $W=aT+bN$, we have $R(W,T)T = bR(N,T)T overset{(*)}{=} bN = langle W,Nrangle N$, as required. The equality (*) follows from the fact that the sphere has constant sectional curvature $1$.
answered Jan 31 at 17:39
Ted ShifrinTed Shifrin
64.7k44692
$endgroup$
add a comment |
$begingroup$
Yes, I think the authors' use of the term Frenet frame is a bit misleading — I personally would call this a Darboux frame. Regardless, by $N$ they mean the normalized covariant derivative of $T$.
As I already commented, it suffices to show that $R(W,T)T = langle W,Nrangle N$ (since $V = vT$). Note that $R(T,T)T = 0$, so, writing $W=aT+bN$, we have $R(W,T)T = bR(N,T)T overset{(*)}{=} bN = langle W,Nrangle N$, as required. The equality (*) follows from the fact that the sphere has constant sectional curvature $1$.
answered Jan 31 at 17:39
Ted ShifrinTed Shifrin
64.7k44692
$endgroup$
add a comment |
$begingroup$
Yes, I think the authors' use of the term Frenet frame is a bit misleading — I personally would call this a Darboux frame. Regardless, by $N$ they mean the normalized covariant derivative of $T$.
As I already commented, it suffices to show that $R(W,T)T = langle W,Nrangle N$ (since $V = vT$). Note that $R(T,T)T = 0$, so, writing $W=aT+bN$, we have $R(W,T)T = bR(N,T)T overset{(*)}{=} bN = langle W,Nrangle N$, as required. The equality (*) follows from the fact that the sphere has constant sectional curvature $1$.
answered Jan 31 at 17:39
Ted ShifrinTed Shifrin
64.7k44692
$endgroup$
Yes, I think the authors' use of the term Frenet frame is a bit misleading — I personally would call this a Darboux frame. Regardless, by $N$ they mean the normalized covariant derivative of $T$.
As I already commented, it suffices to show that $R(W,T)T = langle W,Nrangle N$ (since $V = vT$). Note that $R(T,T)T = 0$, so, writing $W=aT+bN$, we have $R(W,T)T = bR(N,T)T overset{(*)}{=} bN = langle W,Nrangle N$, as required. The equality (*) follows from the fact that the sphere has constant sectional curvature $1$.
answered Jan 31 at 17:39
Ted ShifrinTed Shifrin
64.7k44692
answered Jan 31 at 17:39
Ted ShifrinTed Shifrin
64.7k44692
answered Jan 31 at 17:39
Ted ShifrinTed Shifrin
64.7k44692
answered Jan 31 at 17:39
Ted ShifrinTed Shifrin
64.7k44692
64.7k44692
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$begingroup$
No, the $v^2$ is there because $V=vT$. What definition of the Frenet frame are they using? We're not thinking of $S^2subsetBbb R^3$ and of the Frenet frame of a curve in $Bbb R^3$, I presume?
$endgroup$
– Ted Shifrin
Jan 29 at 19:52
$begingroup$
@TedShifrin I think they are but I’m not entirely sure. If it helps: Later calculations are made representing the vectorfield $W$ (as above) in terms of $W = fT + gN$. Furthermore for $gamma $ with $gamma‘ = omega$ they get the Frenet Frame ${T_gamma, N_gamma, B_gamma }$ such that $T_gamma = omega, N_gamma = T, B_gamma$ as expected.
$endgroup$
– Nhat
Jan 30 at 12:42