About the covariant derivative - is this correct?
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Let $S subset mathbbR^3$ be a regular surface, the image of a parametrization $X: U subset mathbbR^2 to S$ and $alpha:I to W subset S$ be a curve in $S$. We can write $alpha(t) = X(u(t), v(t))$, then the covariant derivative of $alpha'$ at some $t in I$ is given by:
$$beginalign
dfracsf Dalpha'rmdt &= rm proj_T_largealpha(t)S alpha''(t) \
&= alpha''(t) - langlealpha''(t), N(u(t), v(t)) rangle N(u(t), v(t))
endalign$$
Is this correct? If not, what am I getting wrong?
differential-geometry riemannian-geometry surfaces curves
asked Aug 9 at 16:12
Matheus Andrade
626214
add a comment |Â
up vote
2
down vote
favorite
Let $S subset mathbbR^3$ be a regular surface, the image of a parametrization $X: U subset mathbbR^2 to S$ and $alpha:I to W subset S$ be a curve in $S$. We can write $alpha(t) = X(u(t), v(t))$, then the covariant derivative of $alpha'$ at some $t in I$ is given by:
$$beginalign
dfracsf Dalpha'rmdt &= rm proj_T_largealpha(t)S alpha''(t) \
&= alpha''(t) - langlealpha''(t), N(u(t), v(t)) rangle N(u(t), v(t))
endalign$$
Is this correct? If not, what am I getting wrong?
differential-geometry riemannian-geometry surfaces curves
asked Aug 9 at 16:12
Matheus Andrade
626214
1
Yes it is correct.
– John Ma
Aug 10 at 14:07
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $S subset mathbbR^3$ be a regular surface, the image of a parametrization $X: U subset mathbbR^2 to S$ and $alpha:I to W subset S$ be a curve in $S$. We can write $alpha(t) = X(u(t), v(t))$, then the covariant derivative of $alpha'$ at some $t in I$ is given by:
$$beginalign
dfracsf Dalpha'rmdt &= rm proj_T_largealpha(t)S alpha''(t) \
&= alpha''(t) - langlealpha''(t), N(u(t), v(t)) rangle N(u(t), v(t))
endalign$$
Is this correct? If not, what am I getting wrong?
differential-geometry riemannian-geometry surfaces curves
asked Aug 9 at 16:12
Matheus Andrade
626214
Let $S subset mathbbR^3$ be a regular surface, the image of a parametrization $X: U subset mathbbR^2 to S$ and $alpha:I to W subset S$ be a curve in $S$. We can write $alpha(t) = X(u(t), v(t))$, then the covariant derivative of $alpha'$ at some $t in I$ is given by:
$$beginalign
dfracsf Dalpha'rmdt &= rm proj_T_largealpha(t)S alpha''(t) \
&= alpha''(t) - langlealpha''(t), N(u(t), v(t)) rangle N(u(t), v(t))
endalign$$
Is this correct? If not, what am I getting wrong?
differential-geometry riemannian-geometry surfaces curves
asked Aug 9 at 16:12
Matheus Andrade
626214
asked Aug 9 at 16:12
Matheus Andrade
626214
asked Aug 9 at 16:12
Matheus Andrade
626214
asked Aug 9 at 16:12
Matheus Andrade
626214
626214
1
Yes it is correct.
– John Ma
Aug 10 at 14:07
add a comment |Â
1
Yes it is correct.
– John Ma
Aug 10 at 14:07
1
1
Yes it is correct.
– John Ma
Aug 10 at 14:07
Yes it is correct.
– John Ma
Aug 10 at 14:07
add a comment |Â
1 Answer
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The expression I wrote is correct.
answered Aug 15 at 16:43
Matheus Andrade
626214
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The expression I wrote is correct.
answered Aug 15 at 16:43
Matheus Andrade
626214
add a comment |Â
up vote
0
down vote
accepted
The expression I wrote is correct.
answered Aug 15 at 16:43
Matheus Andrade
626214
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The expression I wrote is correct.
answered Aug 15 at 16:43
Matheus Andrade
626214
The expression I wrote is correct.
answered Aug 15 at 16:43
Matheus Andrade
626214
answered Aug 15 at 16:43
Matheus Andrade
626214
answered Aug 15 at 16:43
Matheus Andrade
626214
answered Aug 15 at 16:43
Matheus Andrade
626214
626214
add a comment |Â
add a comment |Â
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1
Yes it is correct.
– John Ma
Aug 10 at 14:07