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Let us denote a 1 form on manifold M with $\eta$ which in a chart looks like $\eta=\eta_{\mu}dx^{\mu}$ where $\eta_{\mu}$ are smooth functions on M. Now given the coordinate vector fields $\frac{\partial}{\partial x^{\mu}}$, $$\nabla_{\nu}\eta\equiv\nabla_{\frac{\partial}{\partial x^{\nu}}}\eta$$ is a (0,1) tensor field, i.e, another 1 form. so, it makes sense to talk about $(\nabla_{\nu}\eta)_{\mu}$ and I can see that (after acting $\nabla_{\nu}\eta$ on a coordinate vector field) the following holds $$(\nabla_{\nu}\eta)_{\mu}=\nabla_{\nu}\eta_{\mu}-\Gamma^{\rho}_{\mu\nu}\eta_{\rho}.$$ Note that since $\eta_{\mu}$ are smooth functions on M, $\nabla_{\nu}\eta_{\mu}=\eta_{\mu,\nu}$ are ordinary partial derivatives of $\eta_{\mu}$. The above equation can be rewritten as: $$\eta_{\mu;\nu}=\eta_{\mu,\nu}-\Gamma^{\rho}_{\mu\nu}\eta_{\rho}.$$ My question is why in the physics/general relativity literature then $\nabla_{\nu}\eta_{\mu}$ denotes the covariant derivative of the 1 form $\eta$ along the coordinate vector field $\frac{\partial}{\partial x^{\nu}}$? The covariant derivative of $\eta$ along $\frac{\partial}{\partial x^{\nu}}$, denoted by $\nabla_{\nu}\eta$ is a (0,1) tensor field whose components are denoted by $(\nabla_{\nu}\eta)_{\mu}$ (the left hand side of the second equation above) where as $\nabla_{\nu}\eta_{\mu}$ are mere partial derivatives of the component functions $\eta_{\mu}$.

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This use of notation is not limited to physics - it's very popular amongst differential geometers too. The reason it's useful to interpret things this way is because it makes it easier to write down coordinate-invariant expressions (which are those that are geometrically or physically meaningful) while retaining the power of index notation to clearly express complicated tensor products and contractions. We're much more likely to care about $(\nabla_\nu \eta)_\mu$ than we are $\partial_\nu \eta_\mu$, so why not make the former easier to write down?

If you don't adopt this convention, then you either have to clutter your calculations with extra parentheses as in your $(\nabla_\nu \eta)_\mu$ (becomes extremely unwieldy once you start taking iterated derivatives), or forget about covariance entirely and work with partial derivatives and Christoffel symbols (ugly!).

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  • $\begingroup$ So, is $(\nabla_{\nu}(\nabla\eta))_{\rho\mu}$ denoted by $\nabla_{\nu}\nabla_{\mu}\eta_{\rho}$? The problem, as you can see, is that it's hard to realize what your final object is from this notation. From $(\nabla_{\nu}(\nabla\eta))_{\rho\mu}$ one can easily detect that the final object is a (0,2) tensor field while the later is less informative. $\endgroup$ – user78032 Oct 14 '17 at 14:42
  • $\begingroup$ @user78032: In abstract index notation, $\nabla_\nu \nabla_\mu \eta_\rho$ denotes the 3-tensor $(\nabla \nabla \eta)_{\nu \mu \rho}.$ If you're interpreting $\nabla_\nu(\nabla \eta)_{\rho \mu}$ as a $2$-tensor (so $\nu$ is fixed) then this is not a coordinate-invariant expression (since $x^\nu$ is some distinguished coordinate), so it's not going to be well-adapted to abstract index notation. $\endgroup$ – Anthony Carapetis Oct 14 '17 at 14:48
  • $\begingroup$ The idiomatic way to do this would be to make the coordinate vector field $X = \partial_\nu$ an object in its own right, and write $X^\alpha \nabla_\alpha \nabla_\mu \eta_\rho$, which is clearly a 2-tensor. $\endgroup$ – Anthony Carapetis Oct 14 '17 at 14:49
  • $\begingroup$ OK, thanks, I get it, so $\nabla_{\nu}\nabla_{\mu}\eta_{\rho}$ is the (0,3) tensor field whose components are $(\nabla\nabla\eta)_{\nu\mu\rho}$. $\endgroup$ – user78032 Oct 14 '17 at 14:56
  • $\begingroup$ Yeah, that's right. $\endgroup$ – Anthony Carapetis Oct 14 '17 at 14:58

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