12:30 AM
@ACuriousMind And also some of them are conjugate-transposed, even with the factor of i
Interesting fact: Suppose $\mathrm{Cliff}(4) \to \mathrm{End}(S)$ is the spinor representation, so that the spinor space is $S = \Bbb C^4$. The map sends $\mathbf{e}_i$ to $\gamma^i$, $0 \leq i \leq 3$. The above is an algebra isomorphism, hence in particular a graded vector space isomorphism $\Lambda^* \Bbb R^4 \to \mathrm{End}(S)$. There is a chiral decomposition $S = S^+ \oplus S^-$, so we might as what $\mathrm{End}(S^+) \subset \mathrm{End}(S)$ under the inverse of that isomorphism.
I claim it is $\Lambda^0 \Bbb R^4 \oplus \Lambda^{2, +}\Bbb R^4$, where $\Lambda^{2, +}$ denotes the self-dual $2$-forms.
Proof: The space of self-dual 2-forms is generated by $\mathbf{e}_0 \wedge \mathbf{e}_1 + \mathbf{e}_2 \wedge \mathbf{e}_3$, and various permutations of the indices. Let's see what the matrix under this representation is: $\omega = \gamma^0 \gamma^1 + \gamma^2 \gamma^3$. Suppose $\gamma^5 = \gamma^0 \gamma^1 \gamma^2 \gamma^3$ is the chirality operator.
Suppose $\psi$ is a spinor such that $\gamma^5 \psi = \psi$. Then, $\omega \psi = \gamma^0 \gamma^1 \psi + \gamma^2 \gamma^3 \psi = \gamma^0 \gamma^1 \gamma^5 \psi + \gamma^2 \gamma^3 \psi = -\gamma^2 \gamma^3 \psi + \gamma^2 \gamma^3 \psi = 0$.
So it seems to me that $\omega$ vanishes on $S^{-}$, i.e., $\omega \in \mathrm{End}(S^+)$.
($\gamma^5 = i \gamma^0 \gamma^1 \gamma^2 \gamma^3$ in -+++ signature because $(\gamma^0)^2 = 1$, but in ++++ the $i$ isn't there. The eigenvalues are indeed $\pm 1$, I suppose it's a matter of convention to call which one as $S^+$ and which one as $S^-$, above I am calling $S^-$ to be the eigenspace with eigenvalue $+1$, it doesn't really matter)
After verifying all even permutations of indices (just also permute the order of the gamma matrices in $\gamma^5$; the operator doesn't change as it's an even permutation), this proves that $\Lambda^{2, +} \Bbb R^4 \to \mathrm{End}(S)$ maps into $\mathrm{End}(S^+)$. The image consists of skew-symmetric guys.
The domain is 4C2/2 = 3-dimensional, the the range is 2^2 = 4-dimensional. What we're missing is a 1-dimensional, I suppose these are just the scalars, which you identify with $\Lambda^0\Bbb R^4$
Bonus: Is there a physical way of thinking about this? This basically says a self-dual form is more or less a pair of positive chirality spinors. Is there a way to think about a self-dual form as a particle or something?
This is a good question, and could be very useful to people, so I would advise re-structuring it a bit e.g. removing the attempted answers to the questions, maybe adding them as a separate answer instead (I will remove this sentence if this advice is taken).
There are two nice ways to understand "...
