Suppose we are considering the Levi-Civita connection $\nabla$ on a Riemannian manifold $M$. By definition, $\nabla$ is a map
$$\nabla:\Gamma(TM)\to\Gamma(TM^*\otimes TM)$$
satisfying certain conditions. If $X,Y$ are vector fields on $M$, I can understand that $\nabla Y\in\Gamma(TM^*\otimes TM)$, and $\nabla_XY\in\Gamma(TM)$.
However, I've seen $\nabla$ used in the following way: Let $\xi^j\frac{\partial}{\partial x^j}$ be a tangent vector field. Then $\sum_{j}\nabla_j\xi^j$ equals the divergence of $\xi^j\frac{\partial}{\partial x^j}$. See here.
Questions:
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What does it mean to write $\nabla_j\xi^j$? $\xi^j$ itself is not a tensor, but a component of a tensor. But by definition, the input of $\nabla_j$ should be a tensor. Is it a shorthand or something else?
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I think using local coordinates, we have $\nabla_if=\frac{\partial f}{\partial x^i}$ and $\nabla_i\nabla_jf=\frac{\partial^2f}{\partial x^i\partial x^j}-\Gamma_{ij}^k\frac{\partial f}{\partial x^k}$. Is this the definition? What about higher derivatives?
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Even though I am fine with local coordinates, I hope someone can explain this to me in an intrinsic way.
Best Answer
For $V=V^\mu\partial_\mu$ a vector field, when physicists write something like $\nabla_\mu V^\nu$ they actually mean: $$\nabla_\mu V^\nu=\partial_\mu V^\nu+V^\rho\Gamma^\nu_{\mu\rho}$$ So this is indeed not just $\nabla_{\partial_\mu}$ applied to the function $V^\nu$. To see what it is, let also $X=X^\mu\partial_\mu$ be a vector field. In coordinates we have: $$\begin{array}{} \nabla_X V & =X^\mu\nabla_{\partial_\mu} V \\ &=X^\mu(\nabla_{\partial_\mu} V^\nu)\partial_\nu+X^\mu V^\nu\nabla_{\partial_\mu}\partial_\nu\\ &=X^\mu(\partial_\mu V^\nu)\partial_\nu+X^\mu V^\nu\Gamma^{\rho}_{\mu\nu}\partial_\rho \\ &=X^\mu(\nabla_\mu V^\nu)\partial_\nu \end{array}$$ This means that: $$\nabla_\mu V^\nu=dx^\nu (\nabla_{\partial_\mu} V)$$ I.e. the $\nabla_\mu V^\nu$ are the coefficients of the map $X\mapsto\nabla_X V$. In particular the quantity you are looking for is the trace of this map: $$\nabla_\mu V^\mu=Tr(X\mapsto\nabla_X V)$$ (Note that the covariant derivative $\nabla_X V$ is linear in $X$, hence this map is a well defined vector bundle homomorphism, i.e. linear on fibres, and it makes sense to take a trace.)