# Converting an invariant matrix to a non-invariant tensor

I'm working on the following problem:

In 4-dimensional notations, given a transformation matrix

Calculate the matrices $$\Lambda_{\mu\nu}$$, $$\Lambda_\mu^\nu$$ and $$\Lambda^{\mu\nu}$$

The matrix $$\Lambda_\mu^\nu$$ seems simple enough, as it should simply be the transpose of $$\Lambda_\nu^\mu$$, which in this case is identical s.t. $$\Lambda_\mu^\nu = \Lambda_\nu^\mu$$.

As I understand it however, $$\Lambda_\mu^\nu$$ and its transpose are invariant matrices, but $$\Lambda_{\mu\nu}$$ and its transpose are only tensors (not invariant, not matrices).

I've looked for other examples of such a transformation, but so far I only get tensors with both lower or upper indices that are used as a notation for matrix dot products.

Is this a trick question? Or is there a way to solve this?

This seems like a relativity problem with a non-Euclidean metric tensor. If I am correct you need to raise and lower indices by applying the metric tensor. If you are dealing with a flat space-time you have either diag(-1, 1, 1, 1) or diag(1, -1, -1, -1) depending on signature.

Lambda_mu_nu = g_mu_alpha*Lambda^alpha_nu, summed over alpha.

Pardon my archaic notation ^ is an upper index, _ a lower index.

For example the (0,0) entry of the new tensor would be:

Lambda_0_0 = g_0_0 * Lambda^0_0 + g_0_1 * Lambda^1_0 + g_0_2 * Lambda^2_0 + g_3_0 * Lambda^3_0

• You are right, sorry. I'm doing a course that is dealing with both electrodynamics and minkowski space - I put the wrong tag. – JadaLovelace Dec 17 '18 at 14:17
• I didn't even see the tag. Does my answer help? – ggcg Dec 17 '18 at 14:20
• I'm trying to work it out explicitly now. This notation is new to me so I'm a bit slow with it. $\Lambda_{\mu\nu} = g_{\mu\alpha}*\Lambda^\alpha_\nu = \begin{bmatrix}g_{00}\Lambda^0_0 & g_{01}\Lambda^1_0 & g_{02}\Lambda^2_0 & g_{30}\Lambda^0_3\\g_{10}\Lambda^0_1 & g_{11}\Lambda^1_1 & g_{12}\Lambda^2_1 & g_{31}\Lambda^1_3\\g_{20}\Lambda^0_2 & g_{21}\Lambda^1_2 & g_{22}\Lambda^2_2 & g_{32}\Lambda^2_3\\g_{30}\Lambda^0_3 & g_{31}\Lambda^1_3 & g_{32}\Lambda^2_3 & g_{33}\Lambda^3_3\end{bmatrix}$ right? – JadaLovelace Dec 17 '18 at 14:42
• We are using diag[1,-1,-1,-1] in class. So my result would then be a matrix $\begin{bmatrix}cosh(\theta) & 0 & 0 & -sinh(\theta)\\0 & -1 & 0 & 0\\0 & 0 & -1 & 0\\sinh(\theta) & 0 & 0 & -cosh(\theta)\end{bmatrix}$ – JadaLovelace Dec 17 '18 at 14:51
• that last column in my first comment has wrong indices, let's pretend they are the other way around :P – JadaLovelace Dec 17 '18 at 15:09