variations of objects: to be or not to be a tensor On math.stackexchange we find that a variation of the connection $\Gamma_{\nu\sigma}^\rho$ (not a tensor) is a tensor (it obeys the the tensor transformation equation). 
On physics.stackexchange we find that $\delta g_{\mu\nu} = -g_{\mu\rho}g_{\nu\sigma}\delta g^{\rho\sigma}$. So a variation of the metric (a tensor) is not a tensor (it does not obey the rule about raising and lowering indices).
This seems like a minefield. Do I have to check the tensor-ness of every object I vary? Or is there some rule?
 A: You have to handle everything on a case by case basis, but you don't always have to go back to the tensor transformational law. The trick is to remember a few results, such as that $\delta\nabla_aX=\nabla_a\delta X$ and $\delta\partial_aX=\partial_a\delta X$ (and that $\nabla$ sends tensors to tensors, as can be proven from their definition), $\delta(XY)=X\delta Y+(\delta X)Y$, and $\delta\delta_a^b=0$. Thus$$\Gamma_{ab}^cV_c=(\partial_a-\nabla_a)V_b\implies\delta\Gamma_{ab}^c\cdot V_c+\Gamma_{ab}^c\delta V_c=(\partial_a-\nabla_a)\delta V_b=\Gamma_{ab}^c\delta V_c\implies\delta\Gamma_{ab}^c\cdot V_c=0.$$Finally, this contraction with arbitrary vectors to form a (vanishing) rank-$2$ tensor implies $\delta\Gamma_{ab}^c$ transforms as a tensor (again, this is easily verified from the definition). By contrast, when no obvious proof a quantity transforms as a tensor can be obtained with such tricks, it's worth trying to prove the opposite from the definition, as you can with the second example you gave.
A: I have been very confused by all this!
It is not true that $\nabla_a\delta X=\delta\nabla_aX$ as J.G. says. It is easy to show by using the usual formula for the covariant derivative that, for example $$
\nabla_a\left({\delta V}^b\right)=\delta\left(\nabla_aV^b\right)+V^c\delta\Gamma_{ac}^b
$$It is also not true that $\delta\Gamma_{ac}^b=0$. J.G.'s proof of it relied on $\nabla_a\delta X=\delta\nabla_aX$.
It is also not true that ${\delta g}_{ab}\ ,\ {\delta g}^{ab}$ are not tensors as I said. They are tensors but they are different tensors. One is the variation of the metric, the other the variation of the inverse metric. For total clarity one might replace ${\delta g}^{ab}$ by ${\delta\bar{g}}^{ab}$ or even ${\delta\left(g^{-1}\right)}^{ab}$.
