$frac{d}{dt}g(Y,Z) = g(D_tY,Z) + g(Y,D_tZ)$
This is Lemma 5.2 in Lee's Riemannian manifolds book. Given a Riemannian manifold $(M,g)$ with a connection that is compatible with $g$, I cannot show why the following holds for vector fields $Y,Z$ along any curve $gamma$: $$frac{d}{dt}g(Y,Z) = g(D_tY,Z) + g(Y,D_tZ)$$
Since $Y$ and $Z$ are vector fields along a curve, does this mean that when you evaluate them at some $t in I$ that you actually use $Y(gamma(t))$?
Because I thought you could just use that $$frac{d}{dt}g(Y,Z)(gamma(t)) = frac{d}{dt}(g(Y,Z) circ gamma)(t) = dot{gamma}(t)left(g(Y,Z)right) = nabla_{dot{gamma}(t)}g(Y,Z) = D_t g(Y,Z)$$
And the result would follow from $nabla$ being compatible with $g$.
Can someone please validate this? I'm not sure whether I have understood the definition of vector fields along curves correctly.
differential-geometry
edited 2 days ago
Kenny Wong
18k21438
asked 2 days ago
eager2learn
1,23211430
add a comment |
This is Lemma 5.2 in Lee's Riemannian manifolds book. Given a Riemannian manifold $(M,g)$ with a connection that is compatible with $g$, I cannot show why the following holds for vector fields $Y,Z$ along any curve $gamma$: $$frac{d}{dt}g(Y,Z) = g(D_tY,Z) + g(Y,D_tZ)$$
Since $Y$ and $Z$ are vector fields along a curve, does this mean that when you evaluate them at some $t in I$ that you actually use $Y(gamma(t))$?
Because I thought you could just use that $$frac{d}{dt}g(Y,Z)(gamma(t)) = frac{d}{dt}(g(Y,Z) circ gamma)(t) = dot{gamma}(t)left(g(Y,Z)right) = nabla_{dot{gamma}(t)}g(Y,Z) = D_t g(Y,Z)$$
And the result would follow from $nabla$ being compatible with $g$.
Can someone please validate this? I'm not sure whether I have understood the definition of vector fields along curves correctly.
differential-geometry
edited 2 days ago
Kenny Wong
18k21438
asked 2 days ago
eager2learn
1,23211430
1
Your second argument looks fine to me. The point is that $g(Y,Z)$ is a smooth function on $M$ (i.e. a tensor field of rank $(0,0)$), so ordinary derivatives and covariant derivatives are the same thing.
– Kenny Wong
2 days ago
1
That said, regarding your first interpretation, it is also a true statement that $g(Y, Z)|_{gamma(t)} = g|_{gamma(t)} (Y|_{gamma(t)}, Z|_{gamma(t)})$.
– Kenny Wong
2 days ago
add a comment |
This is Lemma 5.2 in Lee's Riemannian manifolds book. Given a Riemannian manifold $(M,g)$ with a connection that is compatible with $g$, I cannot show why the following holds for vector fields $Y,Z$ along any curve $gamma$: $$frac{d}{dt}g(Y,Z) = g(D_tY,Z) + g(Y,D_tZ)$$
Since $Y$ and $Z$ are vector fields along a curve, does this mean that when you evaluate them at some $t in I$ that you actually use $Y(gamma(t))$?
Because I thought you could just use that $$frac{d}{dt}g(Y,Z)(gamma(t)) = frac{d}{dt}(g(Y,Z) circ gamma)(t) = dot{gamma}(t)left(g(Y,Z)right) = nabla_{dot{gamma}(t)}g(Y,Z) = D_t g(Y,Z)$$
And the result would follow from $nabla$ being compatible with $g$.
Can someone please validate this? I'm not sure whether I have understood the definition of vector fields along curves correctly.
differential-geometry
edited 2 days ago
Kenny Wong
18k21438
asked 2 days ago
eager2learn
1,23211430
This is Lemma 5.2 in Lee's Riemannian manifolds book. Given a Riemannian manifold $(M,g)$ with a connection that is compatible with $g$, I cannot show why the following holds for vector fields $Y,Z$ along any curve $gamma$: $$frac{d}{dt}g(Y,Z) = g(D_tY,Z) + g(Y,D_tZ)$$
Since $Y$ and $Z$ are vector fields along a curve, does this mean that when you evaluate them at some $t in I$ that you actually use $Y(gamma(t))$?
Because I thought you could just use that $$frac{d}{dt}g(Y,Z)(gamma(t)) = frac{d}{dt}(g(Y,Z) circ gamma)(t) = dot{gamma}(t)left(g(Y,Z)right) = nabla_{dot{gamma}(t)}g(Y,Z) = D_t g(Y,Z)$$
And the result would follow from $nabla$ being compatible with $g$.
Can someone please validate this? I'm not sure whether I have understood the definition of vector fields along curves correctly.
differential-geometry
differential-geometry
edited 2 days ago
Kenny Wong
18k21438
asked 2 days ago
eager2learn
1,23211430
edited 2 days ago
Kenny Wong
18k21438
asked 2 days ago
eager2learn
1,23211430
edited 2 days ago
Kenny Wong
18k21438
edited 2 days ago
Kenny Wong
18k21438
edited 2 days ago
Kenny Wong
18k21438
18k21438
asked 2 days ago
eager2learn
1,23211430
asked 2 days ago
eager2learn
1,23211430
asked 2 days ago
eager2learn
1,23211430
1,23211430
1
Your second argument looks fine to me. The point is that $g(Y,Z)$ is a smooth function on $M$ (i.e. a tensor field of rank $(0,0)$), so ordinary derivatives and covariant derivatives are the same thing.
– Kenny Wong
2 days ago
1
That said, regarding your first interpretation, it is also a true statement that $g(Y, Z)|_{gamma(t)} = g|_{gamma(t)} (Y|_{gamma(t)}, Z|_{gamma(t)})$.
– Kenny Wong
2 days ago
add a comment |
1
Your second argument looks fine to me. The point is that $g(Y,Z)$ is a smooth function on $M$ (i.e. a tensor field of rank $(0,0)$), so ordinary derivatives and covariant derivatives are the same thing.
– Kenny Wong
2 days ago
1
That said, regarding your first interpretation, it is also a true statement that $g(Y, Z)|_{gamma(t)} = g|_{gamma(t)} (Y|_{gamma(t)}, Z|_{gamma(t)})$.
– Kenny Wong
2 days ago
1
1
Your second argument looks fine to me. The point is that $g(Y,Z)$ is a smooth function on $M$ (i.e. a tensor field of rank $(0,0)$), so ordinary derivatives and covariant derivatives are the same thing.
– Kenny Wong
2 days ago
Your second argument looks fine to me. The point is that $g(Y,Z)$ is a smooth function on $M$ (i.e. a tensor field of rank $(0,0)$), so ordinary derivatives and covariant derivatives are the same thing.
– Kenny Wong
2 days ago
1
1
That said, regarding your first interpretation, it is also a true statement that $g(Y, Z)|_{gamma(t)} = g|_{gamma(t)} (Y|_{gamma(t)}, Z|_{gamma(t)})$.
– Kenny Wong
2 days ago
That said, regarding your first interpretation, it is also a true statement that $g(Y, Z)|_{gamma(t)} = g|_{gamma(t)} (Y|_{gamma(t)}, Z|_{gamma(t)})$.
– Kenny Wong
2 days ago
add a comment |
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1
Your second argument looks fine to me. The point is that $g(Y,Z)$ is a smooth function on $M$ (i.e. a tensor field of rank $(0,0)$), so ordinary derivatives and covariant derivatives are the same thing.
– Kenny Wong
2 days ago
1
That said, regarding your first interpretation, it is also a true statement that $g(Y, Z)|_{gamma(t)} = g|_{gamma(t)} (Y|_{gamma(t)}, Z|_{gamma(t)})$.
– Kenny Wong
2 days ago