Tensors Differentiation I know that $\frac{\partial x^{\mu}}{\partial x^{\nu}}=\delta^{\mu}_{\nu}$ but a few days back, I read somewhere that $\frac{\partial x_{\mu}}{\partial x^{\nu}}=\eta_{\mu\nu}$. Can someone help me explaining these both expressions in an intuitive way?
 A: First one is the derivative of four-position in contravariant form with respect to contravariant vector (think about $\frac{\partial x}{\partial x} =1$ yet $\frac{\partial x}{\partial y} =0$). Therefore
$$\frac{\partial x^{\mu}}{\partial x^{\nu}} = \delta_{\nu}^{\mu}$$
where kronecker delta $\delta_{\nu}^{\mu}$ is $1$ when $\mu = \nu$ and $0$ otherwise.
For the latter, I assume you work on flat background where you can express the covariant vector as $x_{\mu}=\eta_{\mu \nu}x^{\nu}$ thus
$$\frac{\partial x_{\mu}}{\partial x^{\nu}}=\frac{\partial \eta_{\mu \nu} x^{\nu}}{\partial x^{\nu}}=\frac{\partial \eta_{\mu \nu}}{\partial x^{\nu}}x^{\nu}+\eta_{\mu \nu}=\eta_{\mu \nu}$$
since $\frac{\partial \eta_{\mu \nu}}{\partial x^{\nu}}=0$ because $\eta_{\mu \nu}$ is the Minkowski metric.
A: That's because the $dx^{\nu}$'s and the $dx_{\mu}$'s are switched one to the other by means of an $\eta$ metric tensor:
$$
dx_{\mu}=\eta_{\mu\alpha}dx^{\alpha}
$$
So that,
$$
\frac{\partial x_{\mu}}{\partial x^{\nu}}=\eta_{\mu\alpha}\frac{\partial x^{\alpha}}{\partial x^{\nu}}=\eta_{\mu\alpha}\delta_{\nu}^{\alpha}=\eta_{\mu\nu}
$$
while there's no "metric switching" from covariant to contravariant in,
$$
\frac{\partial x^{\mu}}{\partial x^{\nu}}
$$
It's just the identity.
