Can $$ \nabla_{\nu} A^{ \mu\nu} $$ represent a covariant derivative with respect to $\nu$? If not what can it be?
I'm reading a textbook on General Relativity, and such operations appear without any explanation. The above (unidentified) notation appears, which I suspect represents covariant differentiation, but so does the standard nortation (with a semicolon), that's what makes it confusing.
For example, can $$ \nabla_{\nu} T^{ \mu\nu} $$ be the cocariant derivative of the strss-energy-momentum tensor?
Indeed the ''nabla'' symbol is usually meant to be the covariant derivative. There are actually quiet a lot of notations for that derivative. It really depends on literature.
For a Tensor of rank zero, witch is just a scalar, the covariant derivative becomes the partial derivative. For a contravariant tensor of rank one, witch is a four-vector or a Riemann-vector you will get
$\nabla_\alpha j^{\mu}=\partial_\alpha j^\mu+\Gamma^\mu_{\alpha \beta}j^\beta$.
If you have to equal indices, its meant to be a contraction of the Tensor $\textbf{j}$ with the covariant derivative $\textbf{$\nabla$}$. In case you don't know it yet, you would have to some over equal indices. This could be for instance the continuity equation of the four current in electrodynamics, taking general relativity into Account: $\nabla_\mu j^{\mu}=\partial_\mu j^\mu+\Gamma^\mu_{\mu \beta}j^\beta=0$.
However, you asked if $\textbf{T}$ could be the electromagnetic stress energy momentum tensor. Yes it is usual to choose $\textbf{T}$ as the character for this symmetric second rank tensor. And again the "nabla" is probably meant to be the covariant derivative. The covariant derivative of a second rank contravariant Tensor is given, with respect to some observer by
$\nabla_\alpha T^{\mu \nu}= \partial_\alpha T^{\mu \nu}+\Gamma^\mu_{\alpha \beta}T^{\beta \nu}+\Gamma^\nu_{\alpha \beta} T^{\mu \beta}$.
And again, if the indices coincide, it is a contraction of the tensor. For instance, in your case that would be the contraction of the EM S E M tensor with the covariant derivative. This is nothing else then force density of the "stuff" witch is described by the EMSEM tensor. If you take a perfect fluid this turns out to be the Euler equations.
Cheers buddy, Sam.