Question about indices and matrix This is essentially a trivial question, which can be answer probably immediately, but i have this doubt anyway.
If, say, $$\Lambda^{a}_{b} = \begin{pmatrix}
f & -fc\\ 
-fc & f
\end{pmatrix}$$
$$\Lambda^{b}_{a} = \begin{pmatrix}
f & -fc\\ 
-fc & f
\end{pmatrix}^T or \begin{pmatrix}
f & -fc\\ 
-fc & f
\end{pmatrix}^{-1}$$ ??
For example, suppose $$T^{c'd'}=T^{cd}\Lambda^{c'}_{c}\Lambda^{d'}_{d} = = \begin{pmatrix}
f & -fc\\ 
-fc & f
\end{pmatrix}T\begin{pmatrix}
f & -fc\\ 
-fc & f
\end{pmatrix}^T or \begin{pmatrix}
f & -fc\\ 
-fc & f
\end{pmatrix}T\begin{pmatrix}
f & -fc\\ 
-fc & f
\end{pmatrix}^{-1} $$
?
 A: Here is the rule that always works: always multiply matrices by keeping the indices which are being summed over touching each other. The upper/lower placement doesn't matter. Only the right-left.
The standard matrix multiplication rule is
$$
(AB)_{ik} = A_{ij} B_{jk}.
$$
Notice how the the two instances index that is being summed over, $j$, are adjacent to each other. This is what corresponds to normal matrix multiplication.
When it comes to raising/lowering, this doesn't affect the matrix multiplication rule.
However, if you have an expression like
$$
A_{ij} B_{kj}
$$
where the two instances of $j$ aren't adjacent, then you must take the transpose so that they are. For instance, the above expression corresponds to $$A B^T.$$
So for instance, say you have
$$
\eta^{\mu \nu} = \eta_{\mu \nu} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}
$$
and
$$
\Lambda^\mu_{\;\; \nu} = \begin{pmatrix} \gamma & \gamma v \\  \gamma v &  \gamma \end{pmatrix}
$$
which implies
\begin{align}
T^{\mu' \nu'} &= \Lambda^{\mu'}_{\;\; \mu} \Lambda^{\nu'}_{\;\; \nu} T^{\mu \nu} \\
&= \Lambda^{\mu'}_{\;\; \mu} T^{\mu \nu} \Lambda^{\nu'}_{\;\; \nu} \\
&= \begin{pmatrix} \gamma & \gamma v \\  \gamma v & \gamma \end{pmatrix} T \begin{pmatrix} \gamma & \gamma v \\ \gamma v & \gamma \end{pmatrix}^T.
\end{align}
Also notice that
\begin{align}
\Lambda_{\mu}^{\;\;\nu} &= \eta_{\mu \mu'} \eta^{\nu \nu'} \Lambda^{\mu'}_{\;\; \nu'} \\
&= \eta_{\mu \mu'} \Lambda^{\mu'}_{\;\; \nu'}  \eta^{\nu' \nu} \\
&= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} \gamma & \gamma v \\  \gamma v &  \gamma \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \\
&= \begin{pmatrix} \gamma & -\gamma v \\  -\gamma v & \gamma \end{pmatrix}
\end{align}
A: The tensor indices can represent matrix elements, but to do so you have to give them an order! Placing them one on top of each other is ambiguous. For example:
$
\Lambda_a \,^b
$
has as rows $\Lambda_0 \,^b$.
Then there is another issue. Transposing a matrix means interchanging its indices only if they are both contravariant or covariant. So what you have done there is lowered (or raised) one index, transposing the tensor, and then raising (or lowering) againg. In general, it doesn't have to be either the transpose or the inverse, although it can happen (as with the lorentz transformation tensors).
For example take this matric together with the minkowski space-like metric:
$
\begin{pmatrix}
1 & 0 & 0 & 1\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
\end{pmatrix}
= \Lambda^a \, _b
$
And see that it is neither the transpose or the inverse. (The edit was for this final sentece xD)
