Both Srednicki [pg.16] and Weinberg I [pg.57] give the result:

$$ \Lambda^{-1} = \Lambda^{T}\tag{1}$$

Where $\Lambda$ is a member of the Lorentz group - i.e the group $O(3,1)$, the group of matrices which preserve the interval $\eta_{\mu\nu}x^{\mu}x^{\nu}$

(1) is easily derived by equating an interval in frame $S$ with one in frame $S'$. What I am confused about, is that the condition: $$MM^{T} = I$$ Is the definition for the group of rotations - in this case $O(4)$. (1) would therefore be the statement $$O(3,1) \cong O(4)$$ But it seems self-evident that there exist Lorentz transformations which aren't equivalent to rotations in $\mathbb{R}^{4}$?

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    $\begingroup$ (1) is certainly not true in the Lorentz group. $O(3,1)$ consists of matrices $M$ such that $M^T \eta M = \eta$. $\endgroup$ – childofsaturn Nov 18 '17 at 21:00
  • $\begingroup$ @childofsaturn I know, but I can't see what the caveat is - as far as I know what they've both done is: $$ x^{\mu}x^{\nu}\eta_{\mu\nu} = \eta_{\alpha\beta}x'^{\alpha}x'^{\beta} \implies \eta_{\mu \nu} = \eta_{\alpha\beta} \Lambda^{\alpha}_{\mu}\Lambda^{\beta}_{\nu} $$ And then multiplied through by say $\eta^{\mu \rho}$ on both sides, result follows. $\endgroup$ – thesundayscientist Nov 18 '17 at 21:08
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    $\begingroup$ The non-compact group $O(3,1)$ is not isomorphic to the compact group $O(4)$. $\endgroup$ – Qmechanic Nov 18 '17 at 21:14
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    $\begingroup$ That seems to be a misinterpretation of what Srednicki and Weinberg are saying because of indices raised and lowered by the metric tensor. Related: physics.stackexchange.com/q/158309/2451 , physics.stackexchange.com/q/255933/2451 , physics.stackexchange.com/q/322305/2451 and links therein. $\endgroup$ – Qmechanic Nov 18 '17 at 21:26
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    $\begingroup$ (1) is wrong. The correct formula is $\Lambda^{-1}=\eta \Lambda^T \eta$. $\endgroup$ – Valter Moretti Nov 18 '17 at 21:47

Equation $\Lambda^T=\Lambda^{-1}$ is not true for Lorentz group. The correct formula is $\Lambda^T\eta=\eta\Lambda^{-1}$.

The easiest way to see this is to match indices: If we choose $\Lambda\equiv\Lambda^\mu_{\;\;\nu}$, we have $(\Lambda^{-1})^\mu_{\;\;\nu}$ but $(\Lambda^T)_{\nu}^{\;\;\mu}$. So one needs to contract $\Lambda^T$ ($\Lambda^{-1}$) with the metric from right (left) so that one can match them. Otherwise covariance-contravariance do not match!

This is also why the correct formula is $M^T\eta M=\eta$ as you cannot directly contract $M^T$ with $M$.

In Euclidean signature, covariance-contravarince no longer matters hence we got our usual formulae back!

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