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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 $TM\to TM$?

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  • $\begingroup$ Gradient of vector $\endgroup$ – Winther Feb 25 at 12:22
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    $\begingroup$ Many people coming to this question seem to be ignoring the tags. I know next to nothing about Riemannian geometry, but I suspect the answer may not be as simple as anything from 3D calculus or related. Or even if it is, it's not restricted to the 3D case at least. The other notation Anju refers to might be related to the covariant derivative and since the question is tagged with "connections", maybe the exterior covariant derivative is relevant as well. $\endgroup$ – Mark S. Feb 25 at 12:35
  • $\begingroup$ Anju, can you provide some more context? Is there a particular book, paper, theorem, or course that you encountered this in? $\endgroup$ – Mark S. Feb 25 at 12:36
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    $\begingroup$ Yes, for any point $x\in M$, $(\nabla X)_x$ is in $\text{End}(T_xM).$ You can write $\nabla_Y X$ in coordonates. You can also think at $\nabla X$ as an 1-form with values in $TM.$ Or as a diferential operator of order 1 from $C^{\infty}(M,TM)$ to $C^{\infty}(M,\Lambda^1\bigotimes \text{End}(TM))$ $\endgroup$ – Hurjui Ionut Feb 25 at 12:41
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    $\begingroup$ I don't think it's a duplicate. At least the link added there has nothing to do with this question. As @MarkS. said, it has more to do with exterior covariant derivative than as a gradient of a vector field. $\endgroup$ – Sandesh Jr Feb 27 at 11:48
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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.

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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.

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  • $\begingroup$ There is more that can be said about why talking about curl, divergence and gradient works only in 3 dimensions. But it is not directly related, so let me know if you want to know more. $\endgroup$ – Sandesh Jr Mar 1 at 10:29
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"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 "$\triangle f$"

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    $\begingroup$ I have downvoted as this appears to ignore both the tags on the question and the notation that Anju said they understand. $\endgroup$ – Mark S. Feb 25 at 13:21
  • $\begingroup$ On a completely unrelated note, I would encourage you to check out Meta, if only to gain insight on how the site works and how people think you should use your rep privileges. $\endgroup$ – Mark S. Feb 25 at 13:21
  • $\begingroup$ Downvoted too. Essentially for the Sane reason, and while it is written that the divergence is a scalar, the result displayed incorrectly shows a vector. $\endgroup$ – GFR Feb 26 at 12:52

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