Product of Lorentz Transformation with metric tensor and inverse metric tensor with different indexes

Question

I am trying to understand the following product: $$\eta_{\mu\lambda}\eta^{\nu\rho}\Lambda^{\lambda}_\rho.$$ I understand that the first metric lowers the $\lambda$ and changes it for a $\mu$, while the second one raises $\rho$ and changes it for $\nu$, giving as a result $\Lambda_\mu^\nu$. I also know that that matrix is the inverse of $\Lambda_\mu^\nu$. However, I want to do the multiplication to actually arrive at the result. For example, starting with $$\Lambda^{\mu}_\nu=\left(\begin{matrix} \cosh\omega & -\sinh\omega & 0 & 0 \\ -\sinh\omega & \cosh\omega & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 &1 \end{matrix}\right)$$ And applying the metric twice: $$\eta_{\lambda\sigma}=\eta^{\lambda\sigma}=\left(\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{matrix}\right)$$

But multiplying that matrix twice gives the identity matrix, which would leave the Lorentz transformation intact, Clearly I must be understanding something about the metric wrong, probably regarding the different indexes, but I don't know what it is.

Answer 1

Side note: You need to be careful with index notation. Instead of writing $\Lambda^{\mu}_\nu$ it is better to write $\Lambda^{\mu}{}_\nu$ to make clear what is the first index (i.e. the row index) and what is the second index (i.e. the column index).

Below we will use the definition of matrix multiplication several times: $$\begin{align} & (AB)^i{}_j=A^i{}_kB^k{}_j \\ \text{or}\quad & (AB)^{ij}=A^i{}_kB^{kj} \\ \text{or}\quad & (AB)_{ij}=A_{ik}B^k{}_j \\ \text{or}\quad & (AB)_i{}^j=A_{ik}B^{kj} \\ \text{or}\quad & ... \end{align}$$

(Mnemonic trick: The summing indices (here $k$) need to be side by side.)

We start with the Lorentz matrix: $$\begin{align} \Lambda^{\mu}{}_\nu=\begin{pmatrix} \cosh\omega & -\sinh\omega & 0 & 0 \\ -\sinh\omega & \cosh\omega & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 &1 \end{pmatrix}\end{align}$$

Now let us calculate the actual product. $$\begin{align} \Lambda_\mu{}^\nu =&\ \eta_{\mu\lambda}\ \eta^{\nu\rho}\Lambda^\lambda{}_\rho \quad\quad\quad \text{(reorder to get the $\lambda$ indices side by side)} \\ =&\ \eta_{\mu\lambda}\ \Lambda^\lambda{}_\rho\ \eta^{\nu\rho} \quad\quad\quad \text{(use the transposed matrix $\eta^T$ to get the $\rho$ indices side by side)} \\ =&\ \eta_{\mu\lambda}\ \Lambda^\lambda{}_\rho\ (\eta^T)^{\rho\nu} \quad\quad \text{(now we can interpret this as matrix products)} \\ =&\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} \cosh\omega&-\sinh\omega&0&0\\ -\sinh\omega&\cosh\omega&0&0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} \\ =&\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} \cosh\omega & \sinh\omega & 0 & 0 \\ -\sinh\omega & -\cosh\omega & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} \\ =&\begin{pmatrix} \cosh\omega & \sinh\omega & 0 & 0 \\ \sinh\omega & \cosh\omega & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \\ =&\begin{pmatrix} \cosh\omega & -\sinh\omega & 0 & 0 \\ -\sinh\omega & \cosh\omega & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}^{-1} \end{align}$$ So we finally got the inverse matrix of $\Lambda^\mu{}_\nu$, as it should be.

Answer 2

You are reading the expression wrongly. Recall that while those expressions look like matrices, they are not. For example, your expression does not involve multiplying $\eta$ by itself. The reason is that the indices of the $\eta$'s do not match, and hence they are not contracted.

If you really want to treat these objects as if they were matrices (which should work here, but not in all expressions involving tensors), notice which indices match which. Your expression can be written as $$\eta_{\mu\lambda} \Lambda^\lambda_\rho \eta^{\rho \nu},$$ which now can be written as a matrix multiplication $\eta \Lambda \eta$. The reason is that $\eta_{\mu\lambda} \Lambda^\lambda_\rho \eta^{\rho \nu}$ is an expression involving components, which are just real numbers and commute. There are hidden summations ensuring the tensor contractions are working right. Notice that, writing the summations explicitly and holding $\mu$ and $\nu$ fixed, $$\sum_{\lambda, \rho} \eta_{\mu\lambda} \Lambda^\lambda_\rho \eta^{\rho \nu} = \sum_{\lambda, \rho} \eta_{\mu\lambda} \eta^{\rho \nu} \Lambda^\lambda_\rho.$$ However, when writing this expression in terms of matrices, you do need to be more careful, since the matrices do not commute. It is then convenient to write the expression in "contraction order" (i.e., matching the contracting indices) and only then changing notation.