Following this question that asserts that $\Lambda$ (the transformation matrix in Lorentz group) is not a tensor, then if $\Lambda^\mu_{~~~\nu}$ is THE Lorentz transformation matrix, what is the meaning of $\Lambda_{\mu \nu}$, $\Lambda_\mu^{~~~\nu}$ and $\Lambda^{\mu\nu}$ ?

I know how are they related to $\Lambda^\mu_{~~~\nu}$, for example: $$\Lambda_{\mu\nu} = \eta_{\mu\sigma} \Lambda^\sigma_{~~~\nu}.$$ Considering the fact that $\eta$ is indeed a tensor and $\Lambda$ is "just a number" (well, just a matrix), does this mean that $\Lambda_{\mu\nu}$ is a tensor in the same way that if $\vec{r}$ is an ordinary 3D vector and $a\in\mathbb{R}$ just a number then $\vec{a} = a \vec{r}$ is a vector?

  • 1
    $\begingroup$ possible duplicate of physics.stackexchange.com/q/255933/84967 and links therein $\endgroup$ May 1, 2018 at 16:37
  • $\begingroup$ I have seen those links before asking. There it is explained that $\Lambda _\mu^\nu$ is the inverse, but I still don't know what are all the other combinations. I added $\Lambda _\mu^\nu$ into my question just for completion. $\endgroup$
    – user171780
    May 1, 2018 at 16:47

1 Answer 1


Even though $\Lambda$ isn't a tensor, you can still lower and raise indices with the metric. You are just multiplying matrices after all. They are useful because they let you express the inverse of $\Lambda$ in a convenient way. To be explicit, let's define "the" transformation matrices (the ones that transform vectors) with one index up and one down:

$$V'^\mu = \Lambda^\mu{}_\nu V^\nu.$$

A covariant vector transforms with the inverse:

$$A'_\mu = A_\nu (\Lambda^{-1})^\nu{}_\mu.$$

Note that $\Lambda^{-1}$ has the same index positioning as $\Lambda$, since the inverse of a linear transformation (which is what a 1-1 tensor is) is also a linear transformation. Now, one defining property of a Lorentz transformation is that $\eta_{\mu\nu} = \Lambda^\alpha{}_\mu \Lambda^\beta{}_\nu \eta_{\alpha\beta}$. Multiplying both sides by $\eta^\nu{}_\rho (\Lambda^{-1})^\mu{}_\lambda$, this is equivalent to

$$(\Lambda^{-1})^\rho{}_\lambda = \eta_{\lambda\beta} \eta^{\nu\rho} \Lambda^\beta{}_\nu = \Lambda_\lambda{}^\rho$$

which kind of says that $\Lambda^{-1} = \Lambda^T$ as long as you're careful with the index positioning. Indeed, this just says that the Lorentz matrices are orthogonal matrices with respect to the Minkowski inner product. It also lets us remember the covariant transformation law as

$$A'_\mu = \Lambda_\mu{}^\nu A_\nu$$

which is kind of like the contravariant version but with the indices reversed. I personally get confused, though, and prefer to not use any of this and just write $\Lambda^{-1}$ whenever I need to.

  • $\begingroup$ Thanks for your answer, although I was looking for an interpretation for $\Lambda_{\mu\nu}$ and $\Lambda^{\mu\nu}$. And also if these are tensors or remain as "just matrices". $\endgroup$
    – user171780
    May 1, 2018 at 18:04
  • $\begingroup$ @user171780 None of them are tensors, so there isn't much to say about them since the versions with two indices up or two down don't show up very often. The only interesting fact I can think of is that they are antisymmetric, which might (or might not) help you remember the explicit form of a Lorentz matrix. $\endgroup$
    – Javier
    May 1, 2018 at 18:07

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