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I was studying variational methods in theoretical physics and I got stuck with a few simple questions. I have possible answers but I cannot see clearly and rigorously if they are correct.

Suppose we have an action $S$ that depends on two fields: an antisymmetric tensor field $T_{\mu \nu}$ and the spacetime metric $g_{\mu \nu}$. Now we vary the action to obtain the equations of motion:

$$ \delta S = \int \left[ \frac{\delta S}{\delta g_{\mu \nu}}\delta g_{\mu \nu} +\frac{\delta S}{\delta T_{\mu \nu}}\delta T_{\mu \nu} \right] d^D x . \tag{1} $$

I know that, due to the symmetry and antisymmetry of $g_{\mu \nu}$ and $T_{\mu \nu}$ respectively, the equations of motion should be symmetrized/antisymmetrized properly:

$$ \frac{\delta S}{\delta g_{(\mu \nu)}}=0 , \qquad \frac{\delta S}{\delta T_{[\mu \nu]}}=0 \tag{2} $$

But I don't see these symmetrizations (in books, articles, etc.) explicitly. I have always believed that the objects $\frac{\delta S}{\delta g_{\mu \nu}} $ and $\frac{\delta S}{\delta T_{\mu \nu}}$ do not have any particular (explicit) symmetry, and the symmetrizations in the equations of motion come from the $\delta g_{\mu \nu}$ and the $\delta T_{\mu \nu}$ that are multiplying in the variation. Am I wrong? (Question 1)

Suppose that for some reason I prefer $T_\mu {}^\nu$ to be the "fundamental" field. Then,

$$ \delta S = \int \left[ \frac{\delta S}{\delta g_{\mu \nu}}\delta g_{\mu \nu} +\frac{\delta S}{\delta T_\mu {}^\nu}\delta T_\mu {}^\nu \right] d^D x . \tag{3} $$

The antisymmetrization in the second term of $(3)$ is not explicit now. The indices of $T$ are one up and the other down and I cannot do things like:

$$ \frac{\delta S}{\delta T_\mu {}^\nu}\delta T_\mu {}^\nu = \frac{\delta S}{\delta T_{\mu \rho} } g_{\rho \nu} \delta T_\mu {}^\nu, \tag{4}$$

$$ \frac{\delta S}{\delta T_\mu {}^\nu} \delta (g^{\rho \nu} T_{\mu \rho}) = \frac{\delta S}{\delta T_\mu {}^\nu} g^{\rho \nu} \delta T_{\mu \rho} , \tag{5}$$

Because: in $(4)$ this metric would have to be affected by partial derivatives that are within the variation $\frac{\delta S}{\delta T_\mu {}^\nu}$; and, in $(5)$, we are forgetting a term ~$\delta g^{\rho \nu}$. Correct me if I am wrong (Question 2).

I think I should use the constraint:

$$ T_\mu {}^\nu = - T^\nu{} _\mu = - g^{\nu \rho} g_{\mu \tau} T_\rho {}^\tau .\tag{6}$$

But I do not know how. The confusing fact for me is that it depends on the other field, the metric. Any ideas? (Question 3)

Thanks!

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    $\begingroup$ For symmetric tensors $\frac{ \delta g_{\mu\nu}(x)}{ \delta g_{\rho\sigma}(y) } = ( \delta^\rho_\mu \delta^\sigma_\nu + \delta^\rho_\nu \delta^\sigma_\mu ) \delta^D(x-y)$. $\endgroup$ – Prahar Mar 19 '18 at 11:07
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The variational formulas for a symmetric tensor field $g_{\mu\nu}(x)\equiv g_{\nu\mu}(x)$ and an antisymmetric tensor field $F_{\mu\nu}(x)\equiv-F_{\nu\mu}(x)$ become

$$\frac{\delta g_{\mu\nu}(x)}{\delta g_{\alpha\beta}(y)} ~=~\frac{1}{2}\left( \delta_{\mu}^{\alpha}\delta_{\nu}^{\beta} + \delta_{\nu}^{\alpha}\delta_{\mu}^{\beta}\right)\delta^n(x-y),\tag{S} $$

and

$$\frac{\delta F_{\mu\nu}(x)}{\delta F_{\alpha\beta}(y)} ~=~\frac{1}{2}\left( \delta_{\mu}^{\alpha}\delta_{\nu}^{\beta} - \delta_{\nu}^{\alpha}\delta_{\mu}^{\beta}\right)\delta^n(x-y),\tag{A} $$

respectively. See also this related Phys.SE posts. Similarly, for a (1,1) tensor field such that $F^{\mu}{}_{\nu}(x)\equiv-g_{\nu\lambda}(x)F^{\lambda}{}_{\kappa}(x)g^{\mu\kappa}(x)$, the variational formula becomes

$$\frac{\delta F^{\mu}{}_{\nu}(x)}{\delta F^{\alpha}{}_{\beta}(y)} ~=~\frac{1}{2}\left( \delta^{\mu}_{\alpha}\delta_{\nu}^{\beta} - g_{\alpha\nu}(x)g^{\mu\beta}(x)\right)\delta^n(x-y).\tag{A'} $$

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