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We know that a Lorentz boost can be written as $$ \begin{aligned} x_0^{\prime} &=\gamma\left(x_0-\beta x\right) \\ x^{\prime} &=\gamma\left(x-\beta x_0\right) \\ y^{\prime} &=y \\ z^{\prime} &= z, \end{aligned} $$ symmetric between X and t.

However, infinitesimally, it is included in $$ \Lambda_{~~~\nu}^\mu=\delta^\mu{ }_\nu+\omega^\mu{ }_\nu, $$ whose infinitesimal transformations amount to $$ x^{\prime \mu}=x^\mu+\omega^\mu{ }_\nu x^\nu. $$ Here $$ \omega_{\mu\nu}=-\omega_{\nu\mu}, $$ antisymmetric.

Question: how is a symmetric boost transformation quantified by infinitesimal antisymmetric parameters?

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It's in the funny Minkowski metric. In point of fact, as a matrix, for a boost, $$ \omega^\mu_{~~\nu} = \omega^\nu_{~~\mu}, $$ so it is symmetric, unlike the antisymmetric covariant object, $$ \eta_{\mu\kappa} \omega^\kappa_{~~\nu} ~~~~~~~~\leadsto \\ \omega_{\mu\nu}= - \omega_{\nu\mu}, $$ as the lowering of the space-like indices pick up a sign w.r.t. the timelike index.

So, leaving the irrelevant y,z inert directions alone, your infinitesimal boost (~to lowest order in β) is but $$ \begin{pmatrix}x^0 \\ x^1 \end{pmatrix} '= \begin{pmatrix}1&-\beta\\ -\beta & 1 \end{pmatrix}\begin{pmatrix}x^0\\ x^1 \end{pmatrix} =\left (I+ \begin{pmatrix}0& \omega^0_{~~1}\\ \omega^1_{~~0} & 0 \end{pmatrix}\right )\begin{pmatrix}x^0\\x^1 \end{pmatrix} , $$ since $\omega^0_{~~0}=0=\omega^1_{~~1}$.

To be sure, this mismatch miracle does not occur for rotations, which entail only spacelike indices, so the mixed tensor has the same antisymmetry as the covariant one.

  • In conclusion, the antisymmetry of the covariant tensor $\omega_{\mu\nu}$ elegantly unifies rotations with boosts (hyperbolic rotations) by dint of the Minkowski metric. Neat, huh?

Clarification to comment question

Indeed, you don't understand the notation: The mixed tensor (one covariant and one contravariant index) is not always symmetric: only for the boost, but not for rotations. So, for the boost, $$ \omega_{0~1}=\eta_{0\kappa} \omega^ \kappa_{~~1}=\omega^ 0_{~~1}= \omega^ 1_{~~0}= -\omega_{1~0}\equiv b, $$ but for a rotation, $$ \omega_{2~1}=\eta_{2\kappa} \omega^ \kappa_{~~1}=-\omega^ 2_{~~1}= \omega^ 1_{~~2}= -\omega_{1~2}\equiv a. $$

If we take $\omega_{0~2}=0$, and ignore the z direction, we have the mixed-symmetry mixed-tensor matrix , $$ \omega^ \mu _{~~\nu} = \begin{pmatrix}0 & b&0 \\ b&0 & a \\ 0& - a&0 \end{pmatrix}, $$ with the standard structure of the boost and rotation generators.

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  • $\begingroup$ thank you, it clarifies a bit the situation. But I want to understand it better, so we are multiplying 2 symmetric tensors ($\eta$ and $\omega$) and we are getting an antisymmetric tensor? I tried to multiply them in Walfram and I didn't get an antisymmetric tensor. I think that the problem is that I don't understand the notation $\omega^\mu{ }_\nu=\omega_\mu{ }^\nu$. Why do you say that it is symmetric if it includes rotations that are antisymmetric. Can you please write these transformations in explicit matrix form? $\endgroup$ Jan 6, 2023 at 11:50
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    $\begingroup$ O, maybe I understand! $\omega^\mu{}_\nu$ contains the symmetric and antisymmetric part, and Minkiwsky's metric makes it fully antisymmetric? $\endgroup$ Jan 6, 2023 at 12:19
  • $\begingroup$ Wait, can you please clarify the notation ω_ν^μ=ω^μ_ν from the previous comment? How we are changing the indices? $\endgroup$ Jan 6, 2023 at 12:24
  • $\begingroup$ Thank you for the clarification! One last question what can we say about the matrix of tensor that has a property $\omega^\mu{}_\nu=\omega^\nu{}_\mu$? $\endgroup$ Jan 6, 2023 at 13:37
  • $\begingroup$ As I illustrated, that matrix is symmetric for one spacelike and one timeline index. But you need a minus sign when two indices are space like, as you also see in the conventional generators linked in WP! $\endgroup$ Jan 6, 2023 at 14:47

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