Mixing
Ordinary and covariant derivative inequality: $ | u|_1 leq Cleft( | frac{Du}{dt} |_0 + |u|_0 right) $
$begingroup$
Let $I=[0,1]$.
Define:
$H_0 = L^2(I,mathbb{R}^3)$ with inner product $langle u,v rangle_0 = int_0^1 langle u(t), v(t) rangle text{ dt}$
$H_1 = W^{1,2}(I, mathbb{R}^3)$ with inner product $langle u,v rangle_1 = langle u,v rangle_0 + langle u',v' rangle_0$.
and the submanifold $Omega = {omega in H_1 mid |omega(t)| = 1 text{ and } w(0) = w(1) }$ of $H$.
I have trouble showing the following "elementary" inequality between covariant and ordinary derivative.
If $A$ is an $H_1$ bounded subset of $Omega$ then there exists a constant $C$ such that
$$ | u|_1 leq Cleft( | frac{Du}{dt} |_0 + |u|_0 right) $$
for any $w in A$ and any $u in H_1(w^*TS^2)$.
The covariant derivative for some $u in H_1(w^*TS^2)$ along a curve $w$ is that of the standard metric on $S^2$ and can be written as
$$ frac{Du}{partial t} = u'(t) - langle u'(t), w(t) rangle w(t).$$
I've tried to keep the outline of the specific setting as brief and precise as possible, if there is anything unclear about it, feel free to comment.
What I've tried: Squaring both sides of the inequality and trying to get the $| |_1$ and $| |_0$ terms of $w$ together, since subtracting would only leave the $L^2$-norm of the ordinary derivative, which might help. When squaring I also used that
$$ langle w(t), frac{Du}{dt}(t) rangle = 0$$
for any $t$ since $frac{Du}{dt}(t)$ lies in the tangential plane $S^2_{w(t)}$ and is thus orthogonal to $w(t)$.
I can't get any further though and I lack (forgot) some general things to look out for when showing such inequality.
Any help would be appreciated.
differential-geometry riemannian-geometry smooth-manifolds connections global-analysis
asked Jan 22 at 21:35
NhatNhat
1,0261017
$endgroup$
add a comment |
$begingroup$
Let $I=[0,1]$.
Define:
$H_0 = L^2(I,mathbb{R}^3)$ with inner product $langle u,v rangle_0 = int_0^1 langle u(t), v(t) rangle text{ dt}$
$H_1 = W^{1,2}(I, mathbb{R}^3)$ with inner product $langle u,v rangle_1 = langle u,v rangle_0 + langle u',v' rangle_0$.
and the submanifold $Omega = {omega in H_1 mid |omega(t)| = 1 text{ and } w(0) = w(1) }$ of $H$.
I have trouble showing the following "elementary" inequality between covariant and ordinary derivative.
If $A$ is an $H_1$ bounded subset of $Omega$ then there exists a constant $C$ such that
$$ | u|_1 leq Cleft( | frac{Du}{dt} |_0 + |u|_0 right) $$
for any $w in A$ and any $u in H_1(w^*TS^2)$.
The covariant derivative for some $u in H_1(w^*TS^2)$ along a curve $w$ is that of the standard metric on $S^2$ and can be written as
$$ frac{Du}{partial t} = u'(t) - langle u'(t), w(t) rangle w(t).$$
I've tried to keep the outline of the specific setting as brief and precise as possible, if there is anything unclear about it, feel free to comment.
What I've tried: Squaring both sides of the inequality and trying to get the $| |_1$ and $| |_0$ terms of $w$ together, since subtracting would only leave the $L^2$-norm of the ordinary derivative, which might help. When squaring I also used that
$$ langle w(t), frac{Du}{dt}(t) rangle = 0$$
for any $t$ since $frac{Du}{dt}(t)$ lies in the tangential plane $S^2_{w(t)}$ and is thus orthogonal to $w(t)$.
I can't get any further though and I lack (forgot) some general things to look out for when showing such inequality.
Any help would be appreciated.
differential-geometry riemannian-geometry smooth-manifolds connections global-analysis
asked Jan 22 at 21:35
NhatNhat
1,0261017
$endgroup$
add a comment |
$begingroup$
Let $I=[0,1]$.
Define:
$H_0 = L^2(I,mathbb{R}^3)$ with inner product $langle u,v rangle_0 = int_0^1 langle u(t), v(t) rangle text{ dt}$
$H_1 = W^{1,2}(I, mathbb{R}^3)$ with inner product $langle u,v rangle_1 = langle u,v rangle_0 + langle u',v' rangle_0$.
and the submanifold $Omega = {omega in H_1 mid |omega(t)| = 1 text{ and } w(0) = w(1) }$ of $H$.
I have trouble showing the following "elementary" inequality between covariant and ordinary derivative.
If $A$ is an $H_1$ bounded subset of $Omega$ then there exists a constant $C$ such that
$$ | u|_1 leq Cleft( | frac{Du}{dt} |_0 + |u|_0 right) $$
for any $w in A$ and any $u in H_1(w^*TS^2)$.
The covariant derivative for some $u in H_1(w^*TS^2)$ along a curve $w$ is that of the standard metric on $S^2$ and can be written as
$$ frac{Du}{partial t} = u'(t) - langle u'(t), w(t) rangle w(t).$$
I've tried to keep the outline of the specific setting as brief and precise as possible, if there is anything unclear about it, feel free to comment.
What I've tried: Squaring both sides of the inequality and trying to get the $| |_1$ and $| |_0$ terms of $w$ together, since subtracting would only leave the $L^2$-norm of the ordinary derivative, which might help. When squaring I also used that
$$ langle w(t), frac{Du}{dt}(t) rangle = 0$$
for any $t$ since $frac{Du}{dt}(t)$ lies in the tangential plane $S^2_{w(t)}$ and is thus orthogonal to $w(t)$.
I can't get any further though and I lack (forgot) some general things to look out for when showing such inequality.
Any help would be appreciated.
differential-geometry riemannian-geometry smooth-manifolds connections global-analysis
asked Jan 22 at 21:35
NhatNhat
1,0261017
$endgroup$
Let $I=[0,1]$.
Define:
$H_0 = L^2(I,mathbb{R}^3)$ with inner product $langle u,v rangle_0 = int_0^1 langle u(t), v(t) rangle text{ dt}$
$H_1 = W^{1,2}(I, mathbb{R}^3)$ with inner product $langle u,v rangle_1 = langle u,v rangle_0 + langle u',v' rangle_0$.
and the submanifold $Omega = {omega in H_1 mid |omega(t)| = 1 text{ and } w(0) = w(1) }$ of $H$.
I have trouble showing the following "elementary" inequality between covariant and ordinary derivative.
If $A$ is an $H_1$ bounded subset of $Omega$ then there exists a constant $C$ such that
$$ | u|_1 leq Cleft( | frac{Du}{dt} |_0 + |u|_0 right) $$
for any $w in A$ and any $u in H_1(w^*TS^2)$.
The covariant derivative for some $u in H_1(w^*TS^2)$ along a curve $w$ is that of the standard metric on $S^2$ and can be written as
$$ frac{Du}{partial t} = u'(t) - langle u'(t), w(t) rangle w(t).$$
I've tried to keep the outline of the specific setting as brief and precise as possible, if there is anything unclear about it, feel free to comment.
What I've tried: Squaring both sides of the inequality and trying to get the $| |_1$ and $| |_0$ terms of $w$ together, since subtracting would only leave the $L^2$-norm of the ordinary derivative, which might help. When squaring I also used that
$$ langle w(t), frac{Du}{dt}(t) rangle = 0$$
for any $t$ since $frac{Du}{dt}(t)$ lies in the tangential plane $S^2_{w(t)}$ and is thus orthogonal to $w(t)$.
I can't get any further though and I lack (forgot) some general things to look out for when showing such inequality.
Any help would be appreciated.
differential-geometry riemannian-geometry smooth-manifolds connections global-analysis
differential-geometry riemannian-geometry smooth-manifolds connections global-analysis
asked Jan 22 at 21:35
NhatNhat
1,0261017
asked Jan 22 at 21:35
NhatNhat
1,0261017
edited Jan 26 at 8:51
Nhat
asked Jan 22 at 21:35
NhatNhat
1,0261017
asked Jan 22 at 21:35
NhatNhat
1,0261017
asked Jan 22 at 21:35
NhatNhat
1,0261017
1,0261017
add a comment |
add a comment |
1 Answer
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Let $n(t)=langle u'(t),w(t)rangle w(t)$ so $u'(t)=Du/dt(t)+n(t).$
We want to bound $|u|_1^2=|u|_0^2+|Du/dt|_0^2+|n|_0^2$ by $C(|Du|_0+|u|_0)^2$ for some $C.$
The only troublesome term is $|n|_0^2.$
Differentiating $langle u,wrangle=0$ gives $n(t)=langle u'(t),w(t)rangle w(t)=-langle u(t),w'(t)rangle w(t),$ so
$$|n|_0^2=int_0^1langle u(t),w'(t)rangle^2dtleqint_0^1| u(t)|^2|w'(t)|^2dtleq C'|u|_infty^2$$
for some $C',$ using Cauchy-Schwarz and the assumption that $|w|_1$ is bounded.
Here $|u|_infty$ is the essential supremum of $|u|.$
Between any $0leq a<bleq 1,$ the change in $|u|^2$ is
$$int_a^bfrac{d}{dt}|u|^2dt=2int_a^blangle u(t),u'(t)rangle dt=2int_a^blangle u(t),Du/dt(t)rangle dt.$$
To justify the first expressions use a smooth approximation in $W^{1,2}.$
The second equality uses the fact that $n(t)$ is orthogonal to $u(t)$ (note $u(t)$ lies in the tangent space to $S^2$ at $w(t)$). The right-hand-side is at most $|Du/dt(t)|_0^2+|u|_0^2.$
And the average value of $|u|$ is $int u(t)dtleq |u|_0,$ so the essential supremum of $|u|$ is at most $C(|u|_0+|Du/dt|_0)$ for some $C.$
answered Jan 26 at 14:09
DapDap
17.9k841
$endgroup$
$begingroup$
Thanks for the detailed input. There are some inconsistencies in your answer, however, making it harder to follow: The definition of $n(t)$ your using for your calculations seem to vary from the one at the top. Also how is $langle u,w rangle = 0$? It’s the covariant derivative of $u$ that is orthogonal to $w$ isn’t it? Edit: I made a mistake looking at your calculations with $n$. The second point still stands, i will check the first one now.
$endgroup$
– Nhat
Jan 26 at 18:32
$begingroup$
@Nhat: I've corrected a couple of missing $w$'s where I used $n(t),$ and added "note $u(t)$ lies in the tangent plane to $S^2$ at $w(t)$" - at least that's how I understand it
$endgroup$
– Dap
Jan 26 at 18:37
$begingroup$
I’m not fluent in differential geometry enough so you may be right. Is the tangent plane at $w(t)$ the one with $w(t)$ at the origin? Or is it the plane with origin at $0$ that’s tangent to $S^2$ after translation? I feel like i’m misunderstanding the difference between tangent space and tangent plane.
$endgroup$
– Nhat
Jan 26 at 18:42
$begingroup$
@Nhat: sorry for the confusion, I meant tangent space not plane.
$endgroup$
– Dap
Jan 26 at 18:42
$begingroup$
So the picture i have in mind of a curve $u in H_1(w^*TS^2)$ being a curve that follows $w$ through its tangent planes is wrong then. What a shame - but it makes sense, I just checked the definition by the author (Curve Straightening Flow by Langer / Singer) and it says $u$ is closed at $0$ which only makes sense with your interpretation. I will check your calculations when I’m at home and then mark it as answered. Tyvm!
$endgroup$
– Nhat
Jan 26 at 18:47
add a comment |
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1 Answer
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$begingroup$
Let $n(t)=langle u'(t),w(t)rangle w(t)$ so $u'(t)=Du/dt(t)+n(t).$
We want to bound $|u|_1^2=|u|_0^2+|Du/dt|_0^2+|n|_0^2$ by $C(|Du|_0+|u|_0)^2$ for some $C.$
The only troublesome term is $|n|_0^2.$
Differentiating $langle u,wrangle=0$ gives $n(t)=langle u'(t),w(t)rangle w(t)=-langle u(t),w'(t)rangle w(t),$ so
$$|n|_0^2=int_0^1langle u(t),w'(t)rangle^2dtleqint_0^1| u(t)|^2|w'(t)|^2dtleq C'|u|_infty^2$$
for some $C',$ using Cauchy-Schwarz and the assumption that $|w|_1$ is bounded.
Here $|u|_infty$ is the essential supremum of $|u|.$
Between any $0leq a<bleq 1,$ the change in $|u|^2$ is
$$int_a^bfrac{d}{dt}|u|^2dt=2int_a^blangle u(t),u'(t)rangle dt=2int_a^blangle u(t),Du/dt(t)rangle dt.$$
To justify the first expressions use a smooth approximation in $W^{1,2}.$
The second equality uses the fact that $n(t)$ is orthogonal to $u(t)$ (note $u(t)$ lies in the tangent space to $S^2$ at $w(t)$). The right-hand-side is at most $|Du/dt(t)|_0^2+|u|_0^2.$
And the average value of $|u|$ is $int u(t)dtleq |u|_0,$ so the essential supremum of $|u|$ is at most $C(|u|_0+|Du/dt|_0)$ for some $C.$
answered Jan 26 at 14:09
DapDap
17.9k841
$endgroup$
$begingroup$
Thanks for the detailed input. There are some inconsistencies in your answer, however, making it harder to follow: The definition of $n(t)$ your using for your calculations seem to vary from the one at the top. Also how is $langle u,w rangle = 0$? It’s the covariant derivative of $u$ that is orthogonal to $w$ isn’t it? Edit: I made a mistake looking at your calculations with $n$. The second point still stands, i will check the first one now.
$endgroup$
– Nhat
Jan 26 at 18:32
$begingroup$
@Nhat: I've corrected a couple of missing $w$'s where I used $n(t),$ and added "note $u(t)$ lies in the tangent plane to $S^2$ at $w(t)$" - at least that's how I understand it
$endgroup$
– Dap
Jan 26 at 18:37
$begingroup$
I’m not fluent in differential geometry enough so you may be right. Is the tangent plane at $w(t)$ the one with $w(t)$ at the origin? Or is it the plane with origin at $0$ that’s tangent to $S^2$ after translation? I feel like i’m misunderstanding the difference between tangent space and tangent plane.
$endgroup$
– Nhat
Jan 26 at 18:42
$begingroup$
@Nhat: sorry for the confusion, I meant tangent space not plane.
$endgroup$
– Dap
Jan 26 at 18:42
$begingroup$
So the picture i have in mind of a curve $u in H_1(w^*TS^2)$ being a curve that follows $w$ through its tangent planes is wrong then. What a shame - but it makes sense, I just checked the definition by the author (Curve Straightening Flow by Langer / Singer) and it says $u$ is closed at $0$ which only makes sense with your interpretation. I will check your calculations when I’m at home and then mark it as answered. Tyvm!
$endgroup$
– Nhat
Jan 26 at 18:47
add a comment |
$begingroup$
Let $n(t)=langle u'(t),w(t)rangle w(t)$ so $u'(t)=Du/dt(t)+n(t).$
We want to bound $|u|_1^2=|u|_0^2+|Du/dt|_0^2+|n|_0^2$ by $C(|Du|_0+|u|_0)^2$ for some $C.$
The only troublesome term is $|n|_0^2.$
Differentiating $langle u,wrangle=0$ gives $n(t)=langle u'(t),w(t)rangle w(t)=-langle u(t),w'(t)rangle w(t),$ so
$$|n|_0^2=int_0^1langle u(t),w'(t)rangle^2dtleqint_0^1| u(t)|^2|w'(t)|^2dtleq C'|u|_infty^2$$
for some $C',$ using Cauchy-Schwarz and the assumption that $|w|_1$ is bounded.
Here $|u|_infty$ is the essential supremum of $|u|.$
Between any $0leq a<bleq 1,$ the change in $|u|^2$ is
$$int_a^bfrac{d}{dt}|u|^2dt=2int_a^blangle u(t),u'(t)rangle dt=2int_a^blangle u(t),Du/dt(t)rangle dt.$$
To justify the first expressions use a smooth approximation in $W^{1,2}.$
The second equality uses the fact that $n(t)$ is orthogonal to $u(t)$ (note $u(t)$ lies in the tangent space to $S^2$ at $w(t)$). The right-hand-side is at most $|Du/dt(t)|_0^2+|u|_0^2.$
And the average value of $|u|$ is $int u(t)dtleq |u|_0,$ so the essential supremum of $|u|$ is at most $C(|u|_0+|Du/dt|_0)$ for some $C.$
answered Jan 26 at 14:09
DapDap
17.9k841
$endgroup$
$begingroup$
Thanks for the detailed input. There are some inconsistencies in your answer, however, making it harder to follow: The definition of $n(t)$ your using for your calculations seem to vary from the one at the top. Also how is $langle u,w rangle = 0$? It’s the covariant derivative of $u$ that is orthogonal to $w$ isn’t it? Edit: I made a mistake looking at your calculations with $n$. The second point still stands, i will check the first one now.
$endgroup$
– Nhat
Jan 26 at 18:32
$begingroup$
@Nhat: I've corrected a couple of missing $w$'s where I used $n(t),$ and added "note $u(t)$ lies in the tangent plane to $S^2$ at $w(t)$" - at least that's how I understand it
$endgroup$
– Dap
Jan 26 at 18:37
$begingroup$
I’m not fluent in differential geometry enough so you may be right. Is the tangent plane at $w(t)$ the one with $w(t)$ at the origin? Or is it the plane with origin at $0$ that’s tangent to $S^2$ after translation? I feel like i’m misunderstanding the difference between tangent space and tangent plane.
$endgroup$
– Nhat
Jan 26 at 18:42
$begingroup$
@Nhat: sorry for the confusion, I meant tangent space not plane.
$endgroup$
– Dap
Jan 26 at 18:42
$begingroup$
So the picture i have in mind of a curve $u in H_1(w^*TS^2)$ being a curve that follows $w$ through its tangent planes is wrong then. What a shame - but it makes sense, I just checked the definition by the author (Curve Straightening Flow by Langer / Singer) and it says $u$ is closed at $0$ which only makes sense with your interpretation. I will check your calculations when I’m at home and then mark it as answered. Tyvm!
$endgroup$
– Nhat
Jan 26 at 18:47
add a comment |
$begingroup$
Let $n(t)=langle u'(t),w(t)rangle w(t)$ so $u'(t)=Du/dt(t)+n(t).$
We want to bound $|u|_1^2=|u|_0^2+|Du/dt|_0^2+|n|_0^2$ by $C(|Du|_0+|u|_0)^2$ for some $C.$
The only troublesome term is $|n|_0^2.$
Differentiating $langle u,wrangle=0$ gives $n(t)=langle u'(t),w(t)rangle w(t)=-langle u(t),w'(t)rangle w(t),$ so
$$|n|_0^2=int_0^1langle u(t),w'(t)rangle^2dtleqint_0^1| u(t)|^2|w'(t)|^2dtleq C'|u|_infty^2$$
for some $C',$ using Cauchy-Schwarz and the assumption that $|w|_1$ is bounded.
Here $|u|_infty$ is the essential supremum of $|u|.$
Between any $0leq a<bleq 1,$ the change in $|u|^2$ is
$$int_a^bfrac{d}{dt}|u|^2dt=2int_a^blangle u(t),u'(t)rangle dt=2int_a^blangle u(t),Du/dt(t)rangle dt.$$
To justify the first expressions use a smooth approximation in $W^{1,2}.$
The second equality uses the fact that $n(t)$ is orthogonal to $u(t)$ (note $u(t)$ lies in the tangent space to $S^2$ at $w(t)$). The right-hand-side is at most $|Du/dt(t)|_0^2+|u|_0^2.$
And the average value of $|u|$ is $int u(t)dtleq |u|_0,$ so the essential supremum of $|u|$ is at most $C(|u|_0+|Du/dt|_0)$ for some $C.$
answered Jan 26 at 14:09
DapDap
17.9k841
$endgroup$
Let $n(t)=langle u'(t),w(t)rangle w(t)$ so $u'(t)=Du/dt(t)+n(t).$
We want to bound $|u|_1^2=|u|_0^2+|Du/dt|_0^2+|n|_0^2$ by $C(|Du|_0+|u|_0)^2$ for some $C.$
The only troublesome term is $|n|_0^2.$
Differentiating $langle u,wrangle=0$ gives $n(t)=langle u'(t),w(t)rangle w(t)=-langle u(t),w'(t)rangle w(t),$ so
$$|n|_0^2=int_0^1langle u(t),w'(t)rangle^2dtleqint_0^1| u(t)|^2|w'(t)|^2dtleq C'|u|_infty^2$$
for some $C',$ using Cauchy-Schwarz and the assumption that $|w|_1$ is bounded.
Here $|u|_infty$ is the essential supremum of $|u|.$
Between any $0leq a<bleq 1,$ the change in $|u|^2$ is
$$int_a^bfrac{d}{dt}|u|^2dt=2int_a^blangle u(t),u'(t)rangle dt=2int_a^blangle u(t),Du/dt(t)rangle dt.$$
To justify the first expressions use a smooth approximation in $W^{1,2}.$
The second equality uses the fact that $n(t)$ is orthogonal to $u(t)$ (note $u(t)$ lies in the tangent space to $S^2$ at $w(t)$). The right-hand-side is at most $|Du/dt(t)|_0^2+|u|_0^2.$
And the average value of $|u|$ is $int u(t)dtleq |u|_0,$ so the essential supremum of $|u|$ is at most $C(|u|_0+|Du/dt|_0)$ for some $C.$
answered Jan 26 at 14:09
DapDap
17.9k841
edited Jan 26 at 18:59
answered Jan 26 at 14:09
DapDap
17.9k841
answered Jan 26 at 14:09
DapDap
17.9k841
answered Jan 26 at 14:09
DapDap
17.9k841
17.9k841
$begingroup$
Thanks for the detailed input. There are some inconsistencies in your answer, however, making it harder to follow: The definition of $n(t)$ your using for your calculations seem to vary from the one at the top. Also how is $langle u,w rangle = 0$? It’s the covariant derivative of $u$ that is orthogonal to $w$ isn’t it? Edit: I made a mistake looking at your calculations with $n$. The second point still stands, i will check the first one now.
$endgroup$
– Nhat
Jan 26 at 18:32
$begingroup$
@Nhat: I've corrected a couple of missing $w$'s where I used $n(t),$ and added "note $u(t)$ lies in the tangent plane to $S^2$ at $w(t)$" - at least that's how I understand it
$endgroup$
– Dap
Jan 26 at 18:37
$begingroup$
I’m not fluent in differential geometry enough so you may be right. Is the tangent plane at $w(t)$ the one with $w(t)$ at the origin? Or is it the plane with origin at $0$ that’s tangent to $S^2$ after translation? I feel like i’m misunderstanding the difference between tangent space and tangent plane.
$endgroup$
– Nhat
Jan 26 at 18:42
$begingroup$
@Nhat: sorry for the confusion, I meant tangent space not plane.
$endgroup$
– Dap
Jan 26 at 18:42
$begingroup$
So the picture i have in mind of a curve $u in H_1(w^*TS^2)$ being a curve that follows $w$ through its tangent planes is wrong then. What a shame - but it makes sense, I just checked the definition by the author (Curve Straightening Flow by Langer / Singer) and it says $u$ is closed at $0$ which only makes sense with your interpretation. I will check your calculations when I’m at home and then mark it as answered. Tyvm!
$endgroup$
– Nhat
Jan 26 at 18:47
add a comment |
$begingroup$
Thanks for the detailed input. There are some inconsistencies in your answer, however, making it harder to follow: The definition of $n(t)$ your using for your calculations seem to vary from the one at the top. Also how is $langle u,w rangle = 0$? It’s the covariant derivative of $u$ that is orthogonal to $w$ isn’t it? Edit: I made a mistake looking at your calculations with $n$. The second point still stands, i will check the first one now.
$endgroup$
– Nhat
Jan 26 at 18:32
$begingroup$
@Nhat: I've corrected a couple of missing $w$'s where I used $n(t),$ and added "note $u(t)$ lies in the tangent plane to $S^2$ at $w(t)$" - at least that's how I understand it
$endgroup$
– Dap
Jan 26 at 18:37
$begingroup$
I’m not fluent in differential geometry enough so you may be right. Is the tangent plane at $w(t)$ the one with $w(t)$ at the origin? Or is it the plane with origin at $0$ that’s tangent to $S^2$ after translation? I feel like i’m misunderstanding the difference between tangent space and tangent plane.
$endgroup$
– Nhat
Jan 26 at 18:42
$begingroup$
@Nhat: sorry for the confusion, I meant tangent space not plane.
$endgroup$
– Dap
Jan 26 at 18:42
$begingroup$
So the picture i have in mind of a curve $u in H_1(w^*TS^2)$ being a curve that follows $w$ through its tangent planes is wrong then. What a shame - but it makes sense, I just checked the definition by the author (Curve Straightening Flow by Langer / Singer) and it says $u$ is closed at $0$ which only makes sense with your interpretation. I will check your calculations when I’m at home and then mark it as answered. Tyvm!
$endgroup$
– Nhat
Jan 26 at 18:47
$begingroup$
Thanks for the detailed input. There are some inconsistencies in your answer, however, making it harder to follow: The definition of $n(t)$ your using for your calculations seem to vary from the one at the top. Also how is $langle u,w rangle = 0$? It’s the covariant derivative of $u$ that is orthogonal to $w$ isn’t it? Edit: I made a mistake looking at your calculations with $n$. The second point still stands, i will check the first one now.
$endgroup$
– Nhat
Jan 26 at 18:32
$begingroup$
Thanks for the detailed input. There are some inconsistencies in your answer, however, making it harder to follow: The definition of $n(t)$ your using for your calculations seem to vary from the one at the top. Also how is $langle u,w rangle = 0$? It’s the covariant derivative of $u$ that is orthogonal to $w$ isn’t it? Edit: I made a mistake looking at your calculations with $n$. The second point still stands, i will check the first one now.
$endgroup$
– Nhat
Jan 26 at 18:32
$begingroup$
@Nhat: I've corrected a couple of missing $w$'s where I used $n(t),$ and added "note $u(t)$ lies in the tangent plane to $S^2$ at $w(t)$" - at least that's how I understand it
$endgroup$
– Dap
Jan 26 at 18:37
$begingroup$
@Nhat: I've corrected a couple of missing $w$'s where I used $n(t),$ and added "note $u(t)$ lies in the tangent plane to $S^2$ at $w(t)$" - at least that's how I understand it
$endgroup$
– Dap
Jan 26 at 18:37
$begingroup$
I’m not fluent in differential geometry enough so you may be right. Is the tangent plane at $w(t)$ the one with $w(t)$ at the origin? Or is it the plane with origin at $0$ that’s tangent to $S^2$ after translation? I feel like i’m misunderstanding the difference between tangent space and tangent plane.
$endgroup$
– Nhat
Jan 26 at 18:42
$begingroup$
I’m not fluent in differential geometry enough so you may be right. Is the tangent plane at $w(t)$ the one with $w(t)$ at the origin? Or is it the plane with origin at $0$ that’s tangent to $S^2$ after translation? I feel like i’m misunderstanding the difference between tangent space and tangent plane.
$endgroup$
– Nhat
Jan 26 at 18:42
$begingroup$
@Nhat: sorry for the confusion, I meant tangent space not plane.
$endgroup$
– Dap
Jan 26 at 18:42
$begingroup$
@Nhat: sorry for the confusion, I meant tangent space not plane.
$endgroup$
– Dap
Jan 26 at 18:42
$begingroup$
So the picture i have in mind of a curve $u in H_1(w^*TS^2)$ being a curve that follows $w$ through its tangent planes is wrong then. What a shame - but it makes sense, I just checked the definition by the author (Curve Straightening Flow by Langer / Singer) and it says $u$ is closed at $0$ which only makes sense with your interpretation. I will check your calculations when I’m at home and then mark it as answered. Tyvm!
$endgroup$
– Nhat
Jan 26 at 18:47
$begingroup$
So the picture i have in mind of a curve $u in H_1(w^*TS^2)$ being a curve that follows $w$ through its tangent planes is wrong then. What a shame - but it makes sense, I just checked the definition by the author (Curve Straightening Flow by Langer / Singer) and it says $u$ is closed at $0$ which only makes sense with your interpretation. I will check your calculations when I’m at home and then mark it as answered. Tyvm!
$endgroup$
– Nhat
Jan 26 at 18:47
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