Euclidean Connection and Constant Vector fields
$begingroup$
Let $nabla$ be the Euclidean connection on a manifold $M = mathbb{R}^n$. The definition I'm following is if $X, Y$ is are smooth vector fields on $M$ with $Y$ given by:
$$
Y = sum_{i}^{n} Y^i frac{partial}{partial x_i}
$$
Then the Euclidean connection is defined as:
$$
nabla_Y X = sum_{i}^{n} X(Y^i)frac{partial}{partial x_i}
$$
I want to show that given a curve $gamma: I rightarrow M $, a vector field whose euclidean connection with $gamma'$ is $0$ has to be constant.
What I don't understand is how we can act on $gamma^i$ with $X$ when $gamma ^i$ is a map from $I subset mathbb{R}$ into $M$, unlike $Y^i$ which is $C^infty(M)$
differential-geometry manifolds smooth-manifolds
asked Dec 6 '18 at 1:55
David FengDavid Feng
929
$endgroup$
add a comment |
$begingroup$
Let $nabla$ be the Euclidean connection on a manifold $M = mathbb{R}^n$. The definition I'm following is if $X, Y$ is are smooth vector fields on $M$ with $Y$ given by:
$$
Y = sum_{i}^{n} Y^i frac{partial}{partial x_i}
$$
Then the Euclidean connection is defined as:
$$
nabla_Y X = sum_{i}^{n} X(Y^i)frac{partial}{partial x_i}
$$
I want to show that given a curve $gamma: I rightarrow M $, a vector field whose euclidean connection with $gamma'$ is $0$ has to be constant.
What I don't understand is how we can act on $gamma^i$ with $X$ when $gamma ^i$ is a map from $I subset mathbb{R}$ into $M$, unlike $Y^i$ which is $C^infty(M)$
differential-geometry manifolds smooth-manifolds
asked Dec 6 '18 at 1:55
David FengDavid Feng
929
$endgroup$
1
$begingroup$
When you define $nabla_YX$, shouldn't that be $nabla_XY$? Usually a connection is $C^infty(M)$-linear in the subscript. This would also eliminate the issue your having.
$endgroup$
– Aweygan
Dec 6 '18 at 2:01
$begingroup$
You're totally right, thanks. That was silly on my part
$endgroup$
– David Feng
Dec 6 '18 at 2:02
$begingroup$
@Aweygan If you'd like to post your comment I could mark it as the answer, I think I've figured out the rest :)
$endgroup$
– David Feng
Dec 6 '18 at 2:05
add a comment |
$begingroup$
Let $nabla$ be the Euclidean connection on a manifold $M = mathbb{R}^n$. The definition I'm following is if $X, Y$ is are smooth vector fields on $M$ with $Y$ given by:
$$
Y = sum_{i}^{n} Y^i frac{partial}{partial x_i}
$$
Then the Euclidean connection is defined as:
$$
nabla_Y X = sum_{i}^{n} X(Y^i)frac{partial}{partial x_i}
$$
I want to show that given a curve $gamma: I rightarrow M $, a vector field whose euclidean connection with $gamma'$ is $0$ has to be constant.
What I don't understand is how we can act on $gamma^i$ with $X$ when $gamma ^i$ is a map from $I subset mathbb{R}$ into $M$, unlike $Y^i$ which is $C^infty(M)$
differential-geometry manifolds smooth-manifolds
asked Dec 6 '18 at 1:55
David FengDavid Feng
929
$endgroup$
Let $nabla$ be the Euclidean connection on a manifold $M = mathbb{R}^n$. The definition I'm following is if $X, Y$ is are smooth vector fields on $M$ with $Y$ given by:
$$
Y = sum_{i}^{n} Y^i frac{partial}{partial x_i}
$$
Then the Euclidean connection is defined as:
$$
nabla_Y X = sum_{i}^{n} X(Y^i)frac{partial}{partial x_i}
$$
I want to show that given a curve $gamma: I rightarrow M $, a vector field whose euclidean connection with $gamma'$ is $0$ has to be constant.
What I don't understand is how we can act on $gamma^i$ with $X$ when $gamma ^i$ is a map from $I subset mathbb{R}$ into $M$, unlike $Y^i$ which is $C^infty(M)$
differential-geometry manifolds smooth-manifolds
differential-geometry manifolds smooth-manifolds
asked Dec 6 '18 at 1:55
David FengDavid Feng
929
asked Dec 6 '18 at 1:55
David FengDavid Feng
929
asked Dec 6 '18 at 1:55
David FengDavid Feng
929
asked Dec 6 '18 at 1:55
David FengDavid Feng
929
asked Dec 6 '18 at 1:55
David FengDavid Feng
929
929
1
$begingroup$
When you define $nabla_YX$, shouldn't that be $nabla_XY$? Usually a connection is $C^infty(M)$-linear in the subscript. This would also eliminate the issue your having.
$endgroup$
– Aweygan
Dec 6 '18 at 2:01
$begingroup$
You're totally right, thanks. That was silly on my part
$endgroup$
– David Feng
Dec 6 '18 at 2:02
$begingroup$
@Aweygan If you'd like to post your comment I could mark it as the answer, I think I've figured out the rest :)
$endgroup$
– David Feng
Dec 6 '18 at 2:05
add a comment |
1
$begingroup$
When you define $nabla_YX$, shouldn't that be $nabla_XY$? Usually a connection is $C^infty(M)$-linear in the subscript. This would also eliminate the issue your having.
$endgroup$
– Aweygan
Dec 6 '18 at 2:01
$begingroup$
You're totally right, thanks. That was silly on my part
$endgroup$
– David Feng
Dec 6 '18 at 2:02
$begingroup$
@Aweygan If you'd like to post your comment I could mark it as the answer, I think I've figured out the rest :)
$endgroup$
– David Feng
Dec 6 '18 at 2:05
1
1
$begingroup$
When you define $nabla_YX$, shouldn't that be $nabla_XY$? Usually a connection is $C^infty(M)$-linear in the subscript. This would also eliminate the issue your having.
$endgroup$
– Aweygan
Dec 6 '18 at 2:01
$begingroup$
When you define $nabla_YX$, shouldn't that be $nabla_XY$? Usually a connection is $C^infty(M)$-linear in the subscript. This would also eliminate the issue your having.
$endgroup$
– Aweygan
Dec 6 '18 at 2:01
$begingroup$
You're totally right, thanks. That was silly on my part
$endgroup$
– David Feng
Dec 6 '18 at 2:02
$begingroup$
You're totally right, thanks. That was silly on my part
$endgroup$
– David Feng
Dec 6 '18 at 2:02
$begingroup$
@Aweygan If you'd like to post your comment I could mark it as the answer, I think I've figured out the rest :)
$endgroup$
– David Feng
Dec 6 '18 at 2:05
$begingroup$
@Aweygan If you'd like to post your comment I could mark it as the answer, I think I've figured out the rest :)
$endgroup$
– David Feng
Dec 6 '18 at 2:05
add a comment |
1 Answer
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$begingroup$
The definition of the Euclidean connection should be
$$nabla_XY=sum_{i}^{n} X(Y^i)frac{partial}{partial x_i}.$$
(This makes $nabla$ satisfy the standard definition of a connection.) Using this, it should be easy to define $nabla_{gamma'}Y$ for a curve $gamma$.
answered Dec 6 '18 at 2:08
AweyganAweygan
14.1k21441
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add a comment |
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$begingroup$
The definition of the Euclidean connection should be
$$nabla_XY=sum_{i}^{n} X(Y^i)frac{partial}{partial x_i}.$$
(This makes $nabla$ satisfy the standard definition of a connection.) Using this, it should be easy to define $nabla_{gamma'}Y$ for a curve $gamma$.
answered Dec 6 '18 at 2:08
AweyganAweygan
14.1k21441
$endgroup$
add a comment |
$begingroup$
The definition of the Euclidean connection should be
$$nabla_XY=sum_{i}^{n} X(Y^i)frac{partial}{partial x_i}.$$
(This makes $nabla$ satisfy the standard definition of a connection.) Using this, it should be easy to define $nabla_{gamma'}Y$ for a curve $gamma$.
answered Dec 6 '18 at 2:08
AweyganAweygan
14.1k21441
$endgroup$
add a comment |
$begingroup$
The definition of the Euclidean connection should be
$$nabla_XY=sum_{i}^{n} X(Y^i)frac{partial}{partial x_i}.$$
(This makes $nabla$ satisfy the standard definition of a connection.) Using this, it should be easy to define $nabla_{gamma'}Y$ for a curve $gamma$.
answered Dec 6 '18 at 2:08
AweyganAweygan
14.1k21441
$endgroup$
The definition of the Euclidean connection should be
$$nabla_XY=sum_{i}^{n} X(Y^i)frac{partial}{partial x_i}.$$
(This makes $nabla$ satisfy the standard definition of a connection.) Using this, it should be easy to define $nabla_{gamma'}Y$ for a curve $gamma$.
answered Dec 6 '18 at 2:08
AweyganAweygan
14.1k21441
answered Dec 6 '18 at 2:08
AweyganAweygan
14.1k21441
answered Dec 6 '18 at 2:08
AweyganAweygan
14.1k21441
answered Dec 6 '18 at 2:08
AweyganAweygan
14.1k21441
14.1k21441
add a comment |
add a comment |
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1
$begingroup$
When you define $nabla_YX$, shouldn't that be $nabla_XY$? Usually a connection is $C^infty(M)$-linear in the subscript. This would also eliminate the issue your having.
$endgroup$
– Aweygan
Dec 6 '18 at 2:01
$begingroup$
You're totally right, thanks. That was silly on my part
$endgroup$
– David Feng
Dec 6 '18 at 2:02
$begingroup$
@Aweygan If you'd like to post your comment I could mark it as the answer, I think I've figured out the rest :)
$endgroup$
– David Feng
Dec 6 '18 at 2:05