Why use two spacetime indices to label Lorentz generators?

Question

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.

Answer 1

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.

Answer 2

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 $M_{ij}$.