What is $\nabla X$ in Riemannian geometry?

Question

What exactly is $\nabla X$, for a vector field $X$? Can we write it in coordinates?

I'm familiar with $\nabla f$, and also $\nabla_Y X$, where both $X$ and $Y$ are vector fields. However, what would $\nabla X$ be? Should we think of it as a map $TM\to TM$?

Answer 1

"nabla", ∇, can be interpreted as the "vector operator", $\frac{\partial}{\partial x}\vec{i}+ \frac{\partial}{\partial y}\vec{j}+ \frac{\partial}{\partial z}\vec{k}$.

There are three different kinds of vector multiplication- scalar product (between a scalar and a vector), dot product, and cross product (between two vectors)- so there are three different ways to use nabla:

"gradient": One takes the "grad" of a scalar valued function of x, y, and z and the result is a vector valued function.

$\nabla f= \left(\frac{\partial}{\partial x}\vec{i}+ \frac{\partial}{\partial y}\vec{j}+ \frac{\partial}{\partial z}\vec{k}\right) f(x, y, z)= \frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}+ \frac{\partial f}{\partial z}\vec{k}$.

That is the vector pointing in the direction of fastest increase of f. It's length is the rate of increase in that direction.

"divergence" One takes the "dot product" with a vector valued function and the result is a scalar.

$\nabla\cdot f\vec{i}+ g\vec{j}+ h\vec{k}= $$\left(\frac{\partial}{\partial x}\vec{i}+ \frac{\partial}{\partial y}\vec{j}+ \frac{\partial}{\partial z}\vec{k}\right)\cdot \left(f\vec{i}+ g\vec{j}+ h\vec{k}\right)=$$ \frac{\partial f}{\partial x}\vec{i}+ \frac{\partial g}{\partial y}\vec{j}+ \frac{\partial h}{\partial z}\vec{k}$.

If we think of the vector function as the velocity vector of a fluid at each point, that measures, roughly, how fast the fluid "spreads out".

"curl": One takes the "cross product" with a vector valued function and the result is another vector value function.

$\nabla \times \left(f\vec{i}+ g\vec{j}+ h\vec{k}\right)= \left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ f & g & h \end{array}\right|= \left(\frac{\partial h}{\partial y}- \frac{\partial g}{\partial z}\right)\vec{i}- \left(\frac{\partial h}{\partial x}- \frac{\partial f}{\partial z}\right)\vec{j}+ \left(\frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}\right)\vec{k}$

If we think of the vector function as the velocity vector of a fluid at each point, that measures, roughly, the rotation of the fluid.

Also of great importance in physics is the "Laplacian", the divergence of the gradient of a scalar valued function:

$\nabla^2 f= \nabla\cdot\left(\nabla f\right)= \frac{\partial^2 f}{\partial x^2}+ \frac{\partial^2 f}{\partial y^2}+ \frac{\partial^2 f}{\partial x^2}$

For some reason physicists tend to write that as "$\triangle f$"

Answer 2

I imagine this is in the context of differential geometry (rather than vector calculus, where I would not know what it stands for). Then it is very simply related to $\nabla_YX $ which you say you are familiar with: \begin{equation} \nabla X (Y) = \nabla _Y X. \end{equation} While the covariant derivative preserves a tensor field rank (that is, the covariant derivative of a vector is a vector, of a 1-form is a 1-form, and so on), the action of $\nabla$ itself increases the covariant (differential form) degree by one. Hence we can characterise it by saying how it acts on vector fields as above.

Answer 3

For a $C^k$ manifold $M$ and an associated linear connection $\nabla$, with a vector field (i.e the section of the tangent bundle) $X\in \mathfrak{X}(M)$, $\nabla X$ is a $C^k$-linear map from $\mathfrak{X}(M)$ to $\mathfrak{X}(M)$. This follows directly from definition. In the standard way, this can also be identified with a $(1,1)$-tensor field, as $\nabla X(Y,\alpha)=\alpha(\nabla_YX)$ (since it is linear in $Y$, this identification works).

As a $(1,1)$-tensor field, it has a local representative in terms of a coordinate frame; i.e. given any chart $(U,\kappa)$, define $(\nabla X)_{\restriction_U}(\partial_\mu,dx^\nu):=(\nabla X)_\mu{}^\nu \in C^k(U)$. Then, one has, $$(\nabla X)_{\restriction_U}=\Sigma_{\mu\nu}(\nabla X)_\mu{}^\nu\partial_\nu \otimes dx^\mu$$ (Due to Leibniz rule, a similar construction would fail if you chose $\nabla_X$ in the same way).

Further, this makes it easier to talk about the term "divergence", using the contraction from tensor fields, as $div(X)=tr(\nabla X) \in C^k(M)$.

For any further interpretation, more context is needed.