# Why use two spacetime indices to label Lorentz generators?

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.

• Do you mean "for any specific $\alpha,\beta$"? Aug 7, 2020 at 16:12
• 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. Aug 7, 2020 at 16:18
• @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 ... Aug 7, 2020 at 16:29
• Yes, apologies, that's what I meant! Aug 7, 2020 at 16:30

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.

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

• Could you expand on this answer? I do not know the notation $\wedge^2 V$, nor do I understand it from context. Aug 7, 2020 at 17:20
• 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? Aug 7, 2020 at 17:50
• Cont.: And to be clear, this is not meant as a negative critique of your answer, simply a clarification of my currently limited understanding! Aug 7, 2020 at 17:53
• 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?". Aug 7, 2020 at 17:59
• I updated the answer. Aug 8, 2020 at 14:18

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 $$M_{ij}$$.