If $\Lambda$ represents the Lorentz transformation matrix, then the transformation of contravariant components $x^\mu$ is given by $$x'^\mu=\Lambda^{\mu}{}_{\nu} x^\nu$$ and that of the covariant components is given by $$x^\prime_\mu =\Lambda_{\mu}{}^{\nu}x_\nu$$ we used $\eta_{\mu\nu}$ to raise and lower the indices of $\Lambda$. In this notation, we remember that the first index is always the row index and the second index is the column index irrespective of its position upstairs or downstairs. Now using th relation $$\eta_{\rho\sigma}=\eta_{\mu\nu}\Lambda^{\mu}{}_{\rho}\Lambda^{\nu}{}_{\sigma},$$ it can be shown that $$ (\Lambda^{-1})^{\sigma}{}_{\rho}=\Lambda_{\rho}{}^\sigma$$

Now, here are the questions.

  1. If we define $\Lambda^{\mu}{}_{\nu}$ to be the $\mu\nu-$th component of the Lorentz transformation matrix, then $=\Lambda_{\mu}{}^{\nu}$ is the $\mu\nu-$th component of of $\Lambda^{-1}$. Then what do the objects $\Lambda_{\mu\nu}$ or $\Lambda^{\mu\nu}$ represent?

  2. If we want to take, the matrix element of $\eta^{-1}\eta=I$, what should one write? Should it be: $(\eta^{-1})^{\mu}{}_{\sigma}(\eta)^{\sigma}{}_{\nu}$ or $(\eta^{-1})^{\mu\sigma}(\eta)^{\sigma\nu}=\delta^{\mu\nu}$, or $(\eta^{-1})_{\mu\sigma}(\eta)_{\sigma\nu}=\delta_{\mu\nu}$ or something else?

  3. What is the relation between $\eta^{\mu\nu}$ and $\eta_{\mu\nu}$ and how to establish that relation?

EDIT: This is an additional question related to the manipulation of indice. Since $\eta=\Lambda^T\eta\Lambda$, we have $\Lambda^{-1} =\eta^{-1}\Lambda^T \eta$ taking matrix elements on both sides we get, $$(\Lambda^{-1})^{\sigma}{}_{\rho}=(\eta^{-1}\Lambda^T \eta)^{\sigma}{}_{\rho}$$ $$=(\eta^{-1})^{\sigma\alpha}\Lambda^{\beta}{}_{\alpha}\eta_{\beta\rho}$$ $$=?$$ Starting from this how can we arrive at the relation $$ (\Lambda^{-1})^{\sigma}{}_{\rho}=\Lambda_{\rho}{}^\sigma$$ Since, I don't know the action of $(\eta^{-1})^{\sigma\alpha}$, I'm unable to proceed further (given $\Lambda^{-1}=\eta^{-1}\Lambda^T\eta$ as the starting point).


1 Answer 1

  1. $\Lambda_{\mu\nu} = {\Lambda_\mu}^\sigma\eta_{\sigma\nu}$. It doesn't "do" anything.

  2. $\delta_{\mu\nu}$ and $\delta^{\mu\nu}$ are not tensors, as I explain at length in this answer of mine. The matrix elements of the identity are $\delta_\mu^\nu$, which you could have determined by thinking about the fact that the identity must send vectors $v^\mu$ to other vectors, so it needs a lower index that can be contracted with the upper vector index, and it needs an upper index so that the result can still be a vector. Writing $\eta^{-1}\eta$ is sort of non-sensical because the metric is a (0,2)-tensor, not a matrix that has an inverse in the linear algebra sense. However:

  3. $\eta^{\mu\nu}$ and $\eta_{\mu\nu}$ are "inverses" of each other in the following sense: $\eta^{\mu\sigma}\eta_{\sigma\nu} = \delta^\mu_\nu$. This follows from the very definition of $\eta^{\mu\nu}$ - it is the object that raises indices, while $\eta_{\mu\nu}$ defines lowering indices. First lowering and then raising an index should be the identity, which is exactly what the equation $\eta^{\mu\sigma}\eta_{\sigma\nu} = \delta^\mu_\nu$ means.

  • $\begingroup$ @ACuriousMind-When you write $\delta^\nu_\mu$, you don't seem to distinguish between first or second index. Why is that? Question 3 was asked because I was having a problem in writing the equation $\Lambda^{-1}=\eta^{-1}\Lambda^T\eta$ (which is obtained from $\Lambda^T \eta\Lambda=\eta$) in a component form. $\endgroup$
    – SRS
    Commented Jun 27, 2016 at 13:25
  • 1
    $\begingroup$ @SRS: I don't distinguish because I'm lazy and because it doesn't matter - what counts is which index is upper and which is lower, the $\delta$ doesn't care about the order. $\endgroup$
    – ACuriousMind
    Commented Jun 27, 2016 at 13:28
  • $\begingroup$ Thank you. Now it has started to make sense. You said that $\delta^{\mu\nu}$ is not a tensor, but we know $T^{\mu\nu}$ or $F^{\mu\nu}$ are tensors. If I understood it correct then $\delta^{\mu\nu}$ do not transform the same way as $F^{\mu\nu}$ or $T^{\mu\nu}$ does? $\endgroup$
    – SRS
    Commented Jun 27, 2016 at 13:32
  • $\begingroup$ @SRS: I linked a post in my answer where I explain in what sense $\delta^{\mu\nu}$ is not a tensor but $\delta_\mu^\nu$ is - please read it. $\endgroup$
    – ACuriousMind
    Commented Jun 27, 2016 at 13:38

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