Lets say after some calculations I arrive at the following expression
$ \int\mathrm{d}^4x\sqrt{-g} (T_{\mu\nu}+T_{\nu\mu})\delta g^{\mu\nu} = 0, $
where $T_{\mu\nu}$ is non-symmetric and $g^{\mu\nu}$ are the components of the metric. As it is, this implies that
$ T_{\mu\nu} + T_{\nu\mu}= 0. $
But, since the metric is symmetric, one could relabel the dummy indices and get
$ 2\int\mathrm{d}^4x\sqrt{-g} T_{\mu\nu}\delta g^{\mu\nu} = 0, $
which implies $T_{\mu\nu}=0$. So apparently there's something wrong that leads to different results. What is the source of this mistake? I feel that the answer is to do with Langrange multipliers, but I am not quite sure. Should I assume the metric is not symmetric while taking the variation and only in the end that I take it as being symmetric? Why there is no apparent problem when $T_{\mu\nu}$ is symmetric?
If $T_{\mu\nu}$ is an anti-symmetric tensor and metric is symmetric, then after the result that multiplication of an anti-symmetric tensor with a symmetric tensor gives zero, ( i.e.
$$ T_{\mu\nu} \delta g^{\mu\nu}=- T_{\nu\mu} \delta g^{\mu\nu}=- T_{\nu\mu} \delta g^{\nu\mu}=- T_{\mu\nu} \delta g^{\mu\nu} \\ \Rightarrow \quad T_{\mu\nu} \delta g^{\mu\nu}=0 \, ) $$
the integrand would be automatically zero, if $T_{\mu\nu}$ is anti-symmetric.
And, remember if $T_{\mu\nu}$ is symmetric, then you get the same expression.
And $T_{\mu\nu}=0$ is a special result of $T_{\mu\nu}+T_{\nu\mu}=0$, since the former satisfies the latter. So, why do you think the result is wrong?