This is a question I came across looking at special relativity and tensor products.
For example, we have the metric tensor and its corresponding matrix representation $$ g_{\mu\nu} = g^{\mu\nu} =\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{bmatrix}$$ which looks the same in co- and contravariant form. Another example would be the Lorentz transformation along the $x$-axis given by: $$ {\Lambda^\mu}_{\nu} =\begin{bmatrix} \gamma & -\gamma\beta & 0 & 0 \\ -\gamma\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$$ and the electromagnetic field: $$ F^{\mu\nu} = \begin{bmatrix} 0 & E_x & E_y & E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \end{bmatrix} $$
If we want to calculate the form of $F^{\mu\nu}$ in a moving coordinate frame we have to transform the elelectromagnetic field tensor: $$F'^{\mu\nu} = {\Lambda^\mu}_{\alpha} {\Lambda^\nu}_{\beta} F^{\alpha\beta}$$
This is sometimes calculated as a matrix product $$ F' = \Lambda F \Lambda $$
My question is now whether there is a simple way to see this product from the above notation using the indices. I think the right matrix $\Lambda$ should actually be a transpose (it does not matter for the given example). Is there a generalization for this?
In index notation, matrix multiplication takes the following form: $$ (AB)_{i}^j=A_{i}^kB_{k}^j, $$ which for three matrices (relevant to the case at hand) looks like this: $$ (ABC)_{i}^j=A_{i}^kB_{k}^lC_l^j. $$ Thus, we may write $$ (\Lambda F\Lambda)_{\mu}^{\nu}=\Lambda_\mu^\alpha F_\alpha^\beta\Lambda_\beta^\nu. $$ We can rewrite this as $$ \Lambda_\mu^\alpha \Lambda_\beta^\nu F_\alpha^\beta. $$ Finally, note that $\Lambda_\mu^\alpha F_\alpha^\beta = \Lambda^\mu_\alpha F^{\alpha\beta}$ (since $\Lambda$ is a symmetric matrix), giving the expression $$ \Lambda_\alpha^\mu \Lambda_\beta^\nu F^{\alpha\beta}. $$