# Tensor index in special relativity?

I'm studying special relativity and I have some difficulties with tensor index.

Take for example the Lorentz matrix, whose elements are written as $$\Lambda^\mu{}_\nu$$.

$$\Lambda^\mu{}_\nu(v) = \begin{bmatrix} \gamma & -\gamma \beta & 0 & 0 \\ -\gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0& 1 \end{bmatrix} \ \Lambda_\mu{}^\nu(v) = \begin{bmatrix} \gamma & \gamma \beta & 0 & 0 \\ \gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0& 1 \end{bmatrix}$$

Now I know that the $$u$$ is the index that is linked with rows. This is ok and it is ok how we can write the multiplication of vectors and matrix in this way.

But I have seen for example this equation

$$g_\alpha{}^\beta = \Lambda^\mu{}_\alpha\Lambda_\mu{}^\beta$$

where $$g$$ is identity matrix. I see that both $$\mu$$ represents rows. So it is not a usual matrix multiplication. How can he tell that $$\alpha$$ represents row and $$\beta$$ the column in $$g$$? (Ok $$\Lambda$$ is symmetric but if we don't take a symmetric matrix i don't know)

Here $$\Lambda^\mu{}_\alpha\Lambda_\mu{}^\beta$$ represent matrix multiplication. But in matrix multiplication elements in a row should be multiplied with elements in a column, and added. Here the index notation represent, each element of a column of first matrix is multiplied with the corresponding elements in another column of the second matrix, and added, means two matrix are multiplied by taking transpose of first matrix. The transpose of first matrix is $$(\Lambda^\mu{}_\alpha)^T= \Lambda_\alpha{}^\mu.$$ Then it become by index cancellation $$\Lambda_\alpha{}^\mu\Lambda_\mu{}^\beta=g_\alpha{}^\beta.$$ Thus it is obvious that $$\alpha$$ represent row and $$\beta$$ represent column.
You can verify these formulae satisfy $$\Lambda_\mu^{\:\nu}=\eta_{\mu\rho}\eta^{\nu\sigma}\Lambda_{\:\sigma}^\rho$$, so the equation you ask about can be rewritten as $$\delta_\alpha^\beta=\Lambda^\mu_{\:\alpha}\eta_{\mu\rho}\eta^{\beta\sigma}\Lambda^{\:\rho}_\sigma$$, which uses only the first version of the Lorentz transformation. You could similarly switch to only using the second viz. $$\Lambda^\mu_{\:\nu}=\eta^{\mu\rho}\eta_{\nu\sigma}\Lambda_\rho^{\:\sigma}$$. The two conversions are equivalent because$$\eta^{\mu\rho}\eta_{\nu\sigma}\Lambda^{\:\sigma}_\rho=\eta^{\mu\rho}\eta_{\nu\sigma}\eta_{\rho\kappa}\eta^{\sigma\lambda}\Lambda_{\:\lambda}^\kappa=\delta^\mu_\kappa\eta_\nu^\lambda\Lambda_{\:\lambda}^\kappa=\Lambda_{\:\nu}^\mu.$$