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I've seen (e.g. in Srednicki) the following notation for the connection between a Lorentz transformation $\Lambda$ and the Lorentz generators $M^{\mu\nu}$: \begin{equation} {\Lambda^\mu}_\nu = {\left( \exp \left( \frac{\text{i}}{2} \, \omega_{\alpha\beta} M^{\alpha\beta} \right)\right)^\mu}_\nu , \tag{1} \label{1} \end{equation} where—as far as I understand—the parameters $\omega_{\alpha\beta}$ are antisymmetric in $\alpha, \beta$; while the generators $(M^{\alpha\beta})^{\mu\nu}$ (note the raised $\nu$!) are antisymmetric in both $\alpha, \beta$ and $\mu, \nu$. Obviously, for any specific $\alpha, \beta$, the matrices $\Lambda$ and $M^{\alpha\beta}$ belong to the same vector space (to make my question clearer, I have here considered the ordinary spacetime representation of the Lorentz group).

The antisymmetry in $\alpha, \beta$ gives e.g. $\omega_{10} M^{10} = - \omega_{01} M^{10} = \omega_{01} M^{01}$, whereby \begin{equation} \omega_{\alpha\beta} M^{\alpha\beta} = 2 \sum_{\alpha<\beta} \, \omega_{\alpha\beta} M^{\alpha\beta} , \tag{2} \label{2} \end{equation} so it is easy to see where the factor $1/2$ in eq. \eqref{1} comes from. However, what is not clear to me is the following:

  1. Why the imaginary factor? Obviously it does no harm, since it can be accounted for when defining the $\omega$-s, but why include it in the first place?

  2. Why use two four-indices (!) in the product between parameters and generators? Surely an expression like \begin{equation} {\Lambda^\mu}_\nu = {\left(\exp \omega^i M_i \right)^\mu}_\nu \tag{3} \label{3} \end{equation} would be far less likely to cause confusion, especially when antisymmetry of the generators (by some authors, at least) is derived from considering infinitesimal Lorentz transformations on the form ${\Lambda^\mu}_\nu = \delta^\mu_\nu + {\omega^\mu}_\nu$ (c.f. this question and the aforementioned Srednicki)?

Question number 2 is what puzzles me the most, as I guess no. 1 is linked to unitarity.

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  • $\begingroup$ Do you mean "for any specific $\alpha,\beta$"? $\endgroup$ Commented Aug 7, 2020 at 16:12
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    $\begingroup$ Also people do frequently translate between $\omega_{\alpha\beta}$ and $J_i$, $K_i$ (see the Wikipedia article on Lorentz transformations as an example), so people do use one-index objects and it is mostly a matter of what notation you prefer. But it is similar to the case of angular momentum in classical mechanics; angular momentum is naturally an antisymmetric two-index tensor, but we often contract it with $\epsilon_{ijk}$ to make it a pseudovector for convenience. $\endgroup$ Commented Aug 7, 2020 at 16:18
  • $\begingroup$ @JahanClaes: Yes, good catch! I've edited the question accordingly. As to your second comment: Do you mean that people translate between $M^{\alpha\beta}$ and $J_i$, $K_i$? If so, then yes, I'm aware, and my question becomes why it is "natural" to consider $M^{\alpha\beta}$ as an antisymmetric four-index tensor. If not, then I am even more confused ... $\endgroup$
    – B. Bergtun
    Commented Aug 7, 2020 at 16:29
  • $\begingroup$ Yes, apologies, that's what I meant! $\endgroup$ Commented Aug 7, 2020 at 16:30

2 Answers 2

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  1. More generally, let there be given a finite-dimensional vector space $V$ over a field $\mathbb{F}$ and equipped with a (not necessarily positive definite) non-degenerate $\mathbb{F}$-bilinear form $\eta:V\times V\to \mathbb{F}$. The Lie algebra $$so(V)~=~\left\{\Lambda\in{\rm End}(V)\mid \forall v,w\in V:~\eta(\Lambda v,w)=-\eta(v,\Lambda w) \right\} ~\cong~ \bigwedge\!{}^2V$$ of pseudo-orthogonal transformations is isomorphic to the exterior tensor product $\bigwedge^2V \equiv V\wedge V$.

    The proof essentially follows from the fact that ${\rm End}(V)\cong V\otimes V^{\ast}$ and use of the musical isomorphism. $\Box$

    Therefore we can label the generators $M^{\mu\nu}$ with two anti-symmetric vector-indices.

    In particular if $V$ is $(n\!+\!1)$-dimensional Minkowski spacetime, then $M^{\mu\nu}$ consist of $n(n\!-\!1)/2$ angular momentum generators and $n$ boost generators.

    See also this & this related Phys.SE posts.

  2. Concerning factors of the imaginary unit $i$, see footnote 1 in my Phys.SE answer here.

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  • $\begingroup$ Could you expand on this answer? I do not know the notation $\wedge^2 V$, nor do I understand it from context. $\endgroup$
    – B. Bergtun
    Commented Aug 7, 2020 at 17:20
  • $\begingroup$ I deleted my second comment, as I think I understood why the appropriate vector indices are spacetime indices: Simply because the vector space under consideration is Minkowski spacetime. However, I do not see any good reason why your general result should be true (possibly related: I am only barely able to parse it, though the latest edits help), except for the logically unsatisfying answer "well, it works when V is Minkowski spacetime, since then $M^{\mu\nu}$ must consist of angular momentum and boost generators" ... Does there exist an "intuitive" explanation of this result? $\endgroup$
    – B. Bergtun
    Commented Aug 7, 2020 at 17:50
  • $\begingroup$ Cont.: And to be clear, this is not meant as a negative critique of your answer, simply a clarification of my currently limited understanding! $\endgroup$
    – B. Bergtun
    Commented Aug 7, 2020 at 17:53
  • $\begingroup$ Further clarification: By asking for an "intuitive" explanation, I simply meant to ask for some hints for where the result comes from, i.e. "What causes this to be so?". $\endgroup$
    – B. Bergtun
    Commented Aug 7, 2020 at 17:59
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Aug 8, 2020 at 14:18
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This is just fleshing out the first point in Qmechanic's answer, but it's too long for a comment. Specifically, I want to give an example of the isomorphism $\mathfrak{so}(V) \simeq V \wedge V$. Since this holds whether we consider definite or indefinite signature and regardless of dimension, I will do the simple example of $\mathfrak{so}(2)$ acting on $\mathbf{R}^2$. Apologies to the mathematicians for butchering the nice mathematics.

We can represent an element $M\in \mathfrak{so}(2)$ as a $2\times2$ skew-symmetric matrix $$\begin{pmatrix}0&-\theta\\\theta&0\end{pmatrix}.$$ Its action on a vector $\mathbf{x}\in \mathbf{R}^2$ is \begin{align}M \mathbf{x}&= \begin{pmatrix}0&-\theta\\\theta&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}\\ &=\theta\begin{pmatrix}-x_2\\x_1\end{pmatrix}.\end{align}

Now let the action on $\mathbf{R}^2$ of the exterior product $\mathbf{v} \wedge \mathbf{w} \in \mathbf{R}^2\wedge \mathbf{R}^2$ be $$(\mathbf{v} \wedge \mathbf{w})\ast\mathbf{x}=\left<\mathbf{v},\mathbf{x}\right>\mathbf{w} - \left<\mathbf{w},\mathbf{x}\right>\mathbf{v}.$$ This gives \begin{align} (\mathbf{v} \wedge \mathbf{w})\ast\mathbf{x} = (v_1 w_2 - v_2 w_1)\begin{pmatrix}-x_2\\x_1\end{pmatrix}, \end{align} which is the same as the above with $\theta = v_1 w_2 - v_2 w_1$. In other words, we can identify $M\in \mathfrak{so}(2)$ with the two-index, antisymmetric bilinear $(\mathbf{v} \wedge \mathbf{w})_{ij}$, and so write $M_{ij}$.

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