Inverse of anti-symmetric rank 4 tensors?

Question

I am trying to find an inverse of a tensor of the form $$M_{\mu\nu\rho\sigma}$$ such that $M$ is anti-symmetric in the $(\mu, \nu)$ exchange and $(\rho, \sigma)$ exchange. The inverse should be such that

$$M^{-1}_{\alpha\beta\mu\nu}M_{\mu\nu\rho\sigma} = \frac{1}{2}(\eta_{\alpha\rho}\eta_{\beta\sigma}-\eta_{\alpha\sigma}\eta_{\beta\rho})$$ where $\eta$ is the Minowski metric. Is this a well-known thing in some literature or does anyone have any ideas how to go about it? More specifically, I am interested in the case where $M_{\mu\nu\rho\sigma} = [J_{\mu\nu}, J_{\rho\sigma}]$ where $J$ is the generator of the Lorentz group. Any ideas or suggestions?

Answer 1

1. Consider an antisymmetric double-index $$ I=(i_1,i_2),\qquad i_1,i_2~\in~\{1,\ldots,n\},$$ i.e. the double-index $I$ can take $n^2$ values, but only $\frac{n(n-1)}{2}$ values (say when e.g. $i_1<i_2$) yield independent components of tensors, e.g. $$\begin{align} V^I~\equiv~&V^{i_1i_2}~=~-V^{i_2i_1}, \cr M^{I,J}~\equiv~&M^{i_1i_2,j_1j_2}~=~-M^{i_2i_1,j_1j_2}~=~-M^{i_1i_2,j_2j_1},\end{align}$$ and so forth.

2. Then one can define a corresponding Kronecker tensor that represents the identity on this antisymmetric double-index space $$\delta^I_J~=~\frac{1}{2}\left(\delta^{i_1}_{j_1}\delta^{i_2}_{j_2}-\delta^{i_2}_{j_1}\delta^{i_1}_{j_2}\right). $$ Note the half $\frac{1}{2}$, which e.g. needed for idempotency $$ \sum_J\delta^I_J \delta^J_K~=~\delta^I_K, \qquad \sum_J~=~\sum_{j_1,j_2=1}^n,$$ of the identity. For further motivation, see e.g. this and this Phys.SE posts.

3. Finally, we can define an inverse tensor $M^{-1}_{I,J}$ to $M^{I,J}$ so that $$ \sum_JM^{-1}_{I,J}M^{J,K}~=~\delta_I^K~=~\sum_JM^{K,J}M^{-1}_{J,I}. $$