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If you have a Lorentz tensor $T$ with components $T_{\mu\nu}$, it seems clear that $$ \frac{\partial T_{\mu\nu}}{\partial T_{\rho\sigma}} = \delta^\rho_\mu \delta^\sigma_\nu. \tag{1} $$ However, if $T$ is symmetric, so $T_{\mu\nu} = T_{\mu\nu}$, there are two ways this derivative could be nonzero, so $$ \frac{\partial T_{\mu\nu}}{\partial T_{\rho\sigma}} = \delta^\rho_\mu \delta^\sigma_\nu + \delta^\sigma_\mu \delta^\rho_\nu. \tag{2} $$ Does this mean the first expression is wrong? It seems that some information is lost if one naively uses the first expression with a symmetric matrix, because the result lacks the symmetry in $\mu$ and $\nu$ and in $\rho$ and $\sigma$ that the derivative must have.

Additionally, should there be a factor of $\frac{1}{2}$? If one sets $\mu = \nu = \rho = \sigma$, then the second expression yields 2. However, introducing this factor seems to then yield $\frac{1}{2}$ when $\mu = \rho$ and $\nu = \sigma$ but $\mu \neq \sigma$, when it should yield 1. What am I missing?

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    $\begingroup$ Possible duplicate: physics.stackexchange.com/q/149066/2451 $\endgroup$ – Qmechanic May 24 '17 at 19:54
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    $\begingroup$ The answer to that comment is No. It can be traced back to that the tensor components are not independent. $\endgroup$ – Qmechanic May 24 '17 at 20:00
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    $\begingroup$ $\uparrow$ Right. $\endgroup$ – Qmechanic May 24 '17 at 20:03
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    $\begingroup$ The two objects are as different as a 2D surface and a curve on $\mathbb{R}^3$ $\endgroup$ – user126422 May 24 '17 at 20:30
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    $\begingroup$ I meant, a tensor with independent components and a symmetric tensor. The analogy is only as seen from the point of view of a partial derivative. $\endgroup$ – user126422 May 24 '17 at 21:08