Geometric way to view affine connection and parallel transport.
$begingroup$
Given a parametrized curve $gamma$ on a manifold $M$ with metric $g$ and some affine connection on it, we can transport vectors of tangent spaces $T_{p} M$ and $T_{q}M$ to each other (when $p, q in gamma$). Then scalar product of vectors (w.r.t. $g$) is preserved under this transport and we can see it algebraically just because of Leibniz rule.
The question is: Can we see the fact that scalar product is preserved in a more geometric way?
Any notes on geodesics will also be appreciated.
riemannian-geometry smooth-manifolds curves connections
asked Jan 21 at 23:11
Rybin DmitryRybin Dmitry
1236
$endgroup$
add a comment |
$begingroup$
Given a parametrized curve $gamma$ on a manifold $M$ with metric $g$ and some affine connection on it, we can transport vectors of tangent spaces $T_{p} M$ and $T_{q}M$ to each other (when $p, q in gamma$). Then scalar product of vectors (w.r.t. $g$) is preserved under this transport and we can see it algebraically just because of Leibniz rule.
The question is: Can we see the fact that scalar product is preserved in a more geometric way?
Any notes on geodesics will also be appreciated.
riemannian-geometry smooth-manifolds curves connections
asked Jan 21 at 23:11
Rybin DmitryRybin Dmitry
1236
$endgroup$
add a comment |
$begingroup$
Given a parametrized curve $gamma$ on a manifold $M$ with metric $g$ and some affine connection on it, we can transport vectors of tangent spaces $T_{p} M$ and $T_{q}M$ to each other (when $p, q in gamma$). Then scalar product of vectors (w.r.t. $g$) is preserved under this transport and we can see it algebraically just because of Leibniz rule.
The question is: Can we see the fact that scalar product is preserved in a more geometric way?
Any notes on geodesics will also be appreciated.
riemannian-geometry smooth-manifolds curves connections
asked Jan 21 at 23:11
Rybin DmitryRybin Dmitry
1236
$endgroup$
Given a parametrized curve $gamma$ on a manifold $M$ with metric $g$ and some affine connection on it, we can transport vectors of tangent spaces $T_{p} M$ and $T_{q}M$ to each other (when $p, q in gamma$). Then scalar product of vectors (w.r.t. $g$) is preserved under this transport and we can see it algebraically just because of Leibniz rule.
The question is: Can we see the fact that scalar product is preserved in a more geometric way?
Any notes on geodesics will also be appreciated.
riemannian-geometry smooth-manifolds curves connections
riemannian-geometry smooth-manifolds curves connections
asked Jan 21 at 23:11
Rybin DmitryRybin Dmitry
1236
asked Jan 21 at 23:11
Rybin DmitryRybin Dmitry
1236
asked Jan 21 at 23:11
Rybin DmitryRybin Dmitry
1236
asked Jan 21 at 23:11
Rybin DmitryRybin Dmitry
1236
asked Jan 21 at 23:11
Rybin DmitryRybin Dmitry
1236
1236
add a comment |
add a comment |
1 Answer
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$begingroup$
Not all connections preserve the scalar product (the "metric"). We need to impose this condition - if, furthermore, we also ask the connection to be torsion-free, then we get the Levi-Civita connection, which is likely the one you have in mind.
One may consider $M$ as a submanifold in some ${mathbb R}^n$ and assume $X, Y$ are two vector fields on $M$. Then the Levi-Civita connection (at $xin M$) $nabla_XY$ is just taking the ordinary Euclidean directional derivative for $Y$ in the $X$ direction, then project to the tangent space $T_x M$. Euclidean derivative preserves the scalar product (by calculus), and this is still true after projecting to $T_x M$.
answered Jan 24 at 17:26
Yu DingYu Ding
3435
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Not all connections preserve the scalar product (the "metric"). We need to impose this condition - if, furthermore, we also ask the connection to be torsion-free, then we get the Levi-Civita connection, which is likely the one you have in mind.
One may consider $M$ as a submanifold in some ${mathbb R}^n$ and assume $X, Y$ are two vector fields on $M$. Then the Levi-Civita connection (at $xin M$) $nabla_XY$ is just taking the ordinary Euclidean directional derivative for $Y$ in the $X$ direction, then project to the tangent space $T_x M$. Euclidean derivative preserves the scalar product (by calculus), and this is still true after projecting to $T_x M$.
answered Jan 24 at 17:26
Yu DingYu Ding
3435
$endgroup$
add a comment |
$begingroup$
Not all connections preserve the scalar product (the "metric"). We need to impose this condition - if, furthermore, we also ask the connection to be torsion-free, then we get the Levi-Civita connection, which is likely the one you have in mind.
One may consider $M$ as a submanifold in some ${mathbb R}^n$ and assume $X, Y$ are two vector fields on $M$. Then the Levi-Civita connection (at $xin M$) $nabla_XY$ is just taking the ordinary Euclidean directional derivative for $Y$ in the $X$ direction, then project to the tangent space $T_x M$. Euclidean derivative preserves the scalar product (by calculus), and this is still true after projecting to $T_x M$.
answered Jan 24 at 17:26
Yu DingYu Ding
3435
$endgroup$
add a comment |
$begingroup$
Not all connections preserve the scalar product (the "metric"). We need to impose this condition - if, furthermore, we also ask the connection to be torsion-free, then we get the Levi-Civita connection, which is likely the one you have in mind.
One may consider $M$ as a submanifold in some ${mathbb R}^n$ and assume $X, Y$ are two vector fields on $M$. Then the Levi-Civita connection (at $xin M$) $nabla_XY$ is just taking the ordinary Euclidean directional derivative for $Y$ in the $X$ direction, then project to the tangent space $T_x M$. Euclidean derivative preserves the scalar product (by calculus), and this is still true after projecting to $T_x M$.
answered Jan 24 at 17:26
Yu DingYu Ding
3435
$endgroup$
Not all connections preserve the scalar product (the "metric"). We need to impose this condition - if, furthermore, we also ask the connection to be torsion-free, then we get the Levi-Civita connection, which is likely the one you have in mind.
One may consider $M$ as a submanifold in some ${mathbb R}^n$ and assume $X, Y$ are two vector fields on $M$. Then the Levi-Civita connection (at $xin M$) $nabla_XY$ is just taking the ordinary Euclidean directional derivative for $Y$ in the $X$ direction, then project to the tangent space $T_x M$. Euclidean derivative preserves the scalar product (by calculus), and this is still true after projecting to $T_x M$.
answered Jan 24 at 17:26
Yu DingYu Ding
3435
answered Jan 24 at 17:26
Yu DingYu Ding
3435
answered Jan 24 at 17:26
Yu DingYu Ding
3435
answered Jan 24 at 17:26
Yu DingYu Ding
3435
3435
add a comment |
add a comment |
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