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There is a very interesting paper from "The Pin Groups in Physics: C, P, and T" on improper and antichronous Lorentz transformations, but on page 5 I got quite confused as it states there

${L^\alpha}_\beta \in O(1,3)$ with the properties ${L^\alpha}_\beta {(L^{-1})^\beta}_\gamma =\delta^\alpha_\gamma$ and $L L^{T}=\mathbb{1}$.

How should I understand that? I always thought that Lorentz transformations would satisfy $L^T\eta L=\eta$ with $\eta$ as Minkowski metric.

EDIT: In order to make my doubt clearer: Wikipedia says on elements of "Indefinite orthogonal groups", using $g=\mathrm{diag}(\underbrace{1,1,\ldots, 1}_{p},\underbrace{-1,-1,\ldots,-1}_q)$, one can define $Q(p,q)$ as a group of matrices just as for the classical orthogonal group $O(p)$. More explicitly, $Q(p,q)$ consists of matrices $A$ such that

$$gA^{tr}g=A^{-1}$$

So it looks very like that $L$ is like $A$. So I cannot just think that $LL^{tr}=\mathbb{1}$.

Due to that I lost trust in the paper although I consider its topic interesting. I would appreciate if anybody could clarify what is meant there.

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  • $\begingroup$ $L^T\eta L=\eta$ and $LL^T=1$ are not incompatible. Rotations, for example, satisfy both. $\endgroup$ – AccidentalFourierTransform May 28 at 12:41
  • $\begingroup$ I see no problem with $L^{\alpha}_{\beta}(L^{-1})^{\beta}_{\gamma}=\delta^{\alpha}_{\gamma}$ and $LL^{T} = \mathbb{I}$, where $\mathbb{I}$ represents the 4 x 4 identity matrix. Keep in mind that det$(L^{T})=$det$(L)$ and that det($L)=\pm 1\,.$ The choice of sign for det$(L)$ gives, I believe, proper/improper Lorentz transformations (not taking other things into account.) Please note that I have not read the paper so I cannot comment on it. $\endgroup$ – Physics_Et_Al May 28 at 13:30
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The simplest explanation is that they are defining the transpose as $L^T = \eta L^t \eta$, where $L^t$ is what you would think of as the ordinary transpose. It's a definition that mostly has to do with the position of the indices; we have $(L^t)^\mu{}_\nu = L^\nu{}_\mu$, so $(L^T)^\mu{}_\nu = L_\nu{}^\mu$. As I've said before, since $\Lambda$ is not a tensor I find this notation dangerous and potentially confusing, but it does simplify things.

There is, however, a deeper mathematical reason for it. If you look at the formula

$$(L^t)^\mu{}_\nu = L^\nu{}_\mu,$$

the indices seem to be in the wrong position. This is because the transpose is really something that has to be defined with respect to an inner product, and when the metric for this inner product is not $\delta_{\mu\nu}$, one has to be careful. If we have an inner product $\langle \cdot, \cdot \rangle$, the "real" transpose (or adjoint) $L^T$ of $L$ satisfies

$$\langle L^T x, y \rangle = \langle x, L y \rangle$$

for all $x$ and $y$. Writing the inner product $\langle x, y\rangle$ as $x^t \eta y$, where again $^t$ is the usual transpose, this becomes

$$x^t (L^T)^t \eta y = x^t \eta L y,$$

from which we can deduce that $L^T = \eta L^t \eta$. This is still iffy, though, for the same reason as before: $L$ not being a tensor makes $\langle x, Ly\rangle$ not coordinate independent, so this derivation is a bit dubious.

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  • $\begingroup$ Thank you. Your answer looks like the proper explanation. But I still have to think about it. $\endgroup$ – Frederic Thomas May 28 at 15:00

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