How do slanted indices work in special relativity? What is the difference between $T^{\mu}{}_{\nu}$ and $T_{\nu}{}^{\mu}$  where $T$ is a tensor?
 A: In some sense, it is basically a transpose. To be more specific, to get from the first to the second, you can raise the $\nu$ index using the inverse metric tensor
$$T^{\nu\mu}=g^{\nu\alpha}T^{\ \ \mu}_{\alpha}$$
Then transpose $T$
$$T^{\mu\nu}=(T^{\nu\mu})^T$$
and then lower the $\nu$ index with the metric tensor
$$T^{\mu}_{\ \ \ \nu}=g_{\nu\alpha}T^{\mu\alpha}$$
So you see that if $T$ was a symmetric tensor ($T^{\mu\nu}=T^{\nu\mu}$) then they are the same thing, but in general, the transposition step is what makes them differ.
A: If $T$ is defined as a $(1,1)$ tensor, the order of the indices is unimportant as they "live" in different spaces. Meaning that one transforms as vectors do and the other as covectors do.
However, sometimes $T$ could be originally e $(2,0)$ tensor (or $(0,2)$, let's consider the former for concreteness). Then the tensor with one index up and one down is defined as
$$
T^{\mu}_{\phantom{\mu\,}\nu} = \eta^{\mu\rho}\, T_{\rho\nu}\,,\qquad
T_{\mu}^{\phantom{\mu}\nu} = \eta^{\rho\nu}\, T_{\mu\rho}
\,.
$$
So, if $T$ is not symmetric, the order matters. Keep in mind though that it is not meaningful to symmetrize or antisymmetrize indices when they are not at the same height (i.e. both down or both up).
