I have to prove that
$$ \omega_{\mu \nu} = \epsilon_{\mu \nu \lambda \kappa} \omega^\lambda u^\kappa $$
given the relations:
$$ u_k u^k = -1 $$ $$ \omega_{\mu \nu} u^\nu = 0 $$ $$ \omega^\mu = \frac{1}{2} \epsilon^{\mu \nu \alpha \beta} \omega_{\alpha \beta} u_\nu $$
I'm triying to solve this by multipliying at both sides by $u^{\nu}$ and then by $\epsilon_{\mu \nu \alpha \beta}$ but i'm not sure how to handle correctly the resulting expression:
$$\omega^\mu u^{\nu} \epsilon_{\mu \nu \alpha \beta}= \frac{1}{2}\epsilon_{\mu \nu \alpha \beta} \epsilon^{\mu \nu \alpha \beta} \omega_{\alpha \beta} u_\nu u^{\nu}$$
I'm sure that I'm not doing this correctly because I'm mixing inner contractions. Some ideas about how this can be solved?
This works if $\omega_{\alpha\beta}$ is antisymmetric (obviously, $\omega^\mu$ only contains info about the antisymmetric part, so that is the only part that can be recovered). Use the properties of the Levi-Civita symbols and generalised Kronecker delta symbols. Starting from:
$$\omega_{\mu \nu} = \epsilon_{\mu \nu \lambda \kappa} \omega^\lambda u^\kappa$$
Substitute in the definition of $\omega^\lambda$: $$\omega_{\mu \nu} = \epsilon_{\mu \nu \lambda \kappa}\left( \frac{1}{2} \epsilon^{\lambda \gamma \alpha \beta} \omega_{\alpha \beta} u_\gamma \right) u^\kappa$$
Group the Levi-Civita symbols, after a couple even permutations of their indices, and form the generalized Kronecker delta (it picks up a negative sign because of the Lorentzian signature): $$\omega_{\mu \nu} = \frac{1}{2}\omega_{\alpha \beta} u_\gamma u^\kappa \left( \epsilon_{\lambda \mu \nu \kappa} \epsilon^{\lambda \alpha \beta \gamma} \right)$$
$$\omega_{\mu \nu} = \frac{1}{2}\omega_{\alpha \beta} u_\gamma u^\kappa \left( -\delta_{\mu \nu \kappa}^{\alpha \beta \gamma} \right)$$
Expand the delta symbol (this comes from Laplace's formula for evaluating a determinant): $$\omega_{\mu \nu} = \frac{1}{2}\omega_{\alpha \beta} u_\gamma u^\kappa \left( -\delta^\gamma_\kappa \delta_{\mu \nu}^{\alpha \beta} +\delta^\gamma_\mu\delta^{\alpha\beta}_{\kappa\nu} +\delta^\gamma_\nu\delta^{\alpha\beta}_{\mu\kappa} \right)$$
$$\omega_{\mu \nu} = -\frac{1}{2}\omega_{\alpha \beta} u_\gamma u^\kappa \delta^\gamma_\kappa \delta_{\mu \nu}^{\alpha \beta} +\frac{1}{2}\omega_{\alpha \beta} u_\gamma u^\kappa\delta^\gamma_\mu\delta^{\alpha\beta}_{\kappa\nu} +\frac{1}{2}\omega_{\alpha \beta} u_\gamma u^\kappa\delta^\gamma_\nu\delta^{\alpha\beta}_{\mu\kappa} $$
The rank-2 delta changes the index of $u_\gamma$ and the rank-4 delta absorbs the $1/2$ and changes the indices of $\omega_{\alpha\beta}$ (and antisymmetrizes it, but it is already antisymmetric): $$\omega_{\mu \nu} = -\omega_{\mu \nu} u_\kappa u^\kappa +\omega_{\kappa \nu} u_\mu u^\kappa +\omega_{\mu \kappa} u_\nu u^\kappa$$
Then equality is shown by $u_\kappa u^\kappa=-1$, $\omega_{\mu\nu}u^\nu=0$, and the antisymmetry of $\omega_{\mu\nu}$.