Let $W$ be a vector field of constant length and let $v$ be a vector. Show that the covariant derivative, $\nabla_vW$ and $W$ are orthogonal

Question

I know that the covariant derivative of $W$ with respect to $v$ is the tangent vector

$\nabla_vW=W(p+tv)′(0)$

at the point $p$

We want to show that the dot product of $W$ and $\nabla_vW$ is zero which would imply orthogonality

So if $||W||=L$, $L$ is constant, would it be true that our vector $v=(a,b,c)$, is a vector of just coefficients without any variable $x,y,z$?

So any $v$ in the tangent space is constant?

and...

I want to write a clear proof of the above claim, but am not sure where to go from here

Answer 1

You need Koszul's formula. It tells you that:

$\langle\nabla_XY,Z\rangle = \frac{1}{2}(X\langle Y,Z\rangle+Y\langle X,Z\rangle-Z\langle X,Y\rangle+\langle[X,Y],Z\rangle-\langle[X,Z],Y\rangle-\langle[Y,Z],X\rangle)$

Then we substitute $X=V$, $Y=Z=W$ and we use the fact that $ V\langle W,W\rangle=0$ since $W$ has a constant norm and it's derivative by any vector field (in this case $V$) is zero. Explicitly, we have the following:

$\langle\nabla_V W,W\rangle = \frac{1}{2}(V\langle W,W\rangle+W\langle V,W\rangle-W\langle V,W\rangle+\langle[V,W],W\rangle-\langle[V,W],W\rangle-\langle[W,W],V\rangle)=0$

Answer 2

The usual proof of this fact given in differential and/or Riemannian geometry relies on the Leibniz product rule for the connection $\nabla$ associated with the metric tensor

$\mathbf g(\cdot, \cdot) = \langle \cdot, \cdot \rangle; \tag 1$

we have in general for vector fields $X$, $Y$, $Z$:

$Z[\langle X, Y \rangle] = \nabla_Z \langle X, Y \rangle = \langle \nabla_Z X, Y \rangle + \langle X, \nabla_Z Y \rangle; \tag 2$

then if

$\Vert W \Vert = \langle W, W \rangle^{1/2} = L = \text{a constant}, \tag 3$

we also have

$\Vert W \Vert^2 = \langle W, W \rangle = L^2 = \text{constant}, \tag 4$

and from this

$\nabla_v \langle W, W \rangle = v[L^2] = 0; \tag 5$

now from (2),

$\nabla_v \langle W, W \rangle = \langle \nabla_v W, W \rangle + \langle W, \nabla_v W \rangle$ $= \langle \nabla_v W, W \rangle + \langle \nabla_v W, W \rangle = 2\langle \nabla_v W, W \rangle, \tag 6$

since $\langle \cdot, \cdot \rangle$ is symmetric, that is

$\langle Y, Z \rangle = \langle Z, Y \rangle \tag 7$

for all vectors $Y$, $Z$; combining (5) and (6) yields

$2\langle \nabla_v W, W \rangle = 0, \tag 8$

whence

$\langle \nabla_v W, W \rangle = 0, \tag 9$

the desired result.