Although my question is about some specific stuff about gravity, one does not need to know anything about it, because the main point of my question is a purely calculational issue, regarding condraction of indices.
Let me give you a little bit of context. The action of 3-dimensional gravity can be written as
$$S[e,\omega]\propto\frac{1}{2}\int_{\mathcal{M}}\varepsilon_{abc}e^{a}\wedge F^{bc}[\omega]$$
where $e^{a}=e^{a}_{\mu}\mathrm{d}x^{\mu}$ is a $1$-form, called the triad and where $\omega^{ab}=\omega^{ab}_{\mu}\mathrm{d}x^{\mu}$ is a $1$-form, called the spin connection. $F^{bc}[\omega]=\mathrm{d}\omega^{bc}+ {\omega^{b}}_{d}\wedge\omega^{dc}$ denotes the curvature corresponding to $\omega$.
Now in the case of $3$-dimensional gravity, the action is often written in another form, namely as $$S[e,\omega]\propto\int_{\mathcal{M}}e^{a}\wedge \bigg (\mathrm{d}\omega_{a}+\frac{1}{2}\varepsilon_{abc}\omega^{b}\wedge\omega^{c} \bigg )$$
where $\omega_{a}:=\frac{1}{2}\varepsilon_{abc}\omega^{bc}$ is the dualized object. So I would like to show that they are equivalent. I guess that it is not so hard, but I don't see where my error is...
So first of all, we have that
$$\frac{1}{2}\varepsilon_{abc} e^{a}\wedge F^{bc}[\omega]=\frac{1}{2}\varepsilon_{abc} e^{a}\wedge \bigg (\mathrm{d}\omega^{bc}+ {\omega^{b}}_{d}\wedge\omega^{dc}\bigg )=e^{a}\wedge\bigg (\mathrm{d}\omega_{a}+\frac{1}{2}\varepsilon_{abc}{\omega^{b}}_{d}\wedge\omega^{dc}\bigg )\stackrel{!}{=}e^{a}\wedge \bigg (\mathrm{d}\omega_{a}+\frac{1}{2}\varepsilon_{abc}\omega^{b}\wedge\omega^{c} \bigg )$$
Therefore, in order to show equivalence, I have to show that $${\omega^{b}}_{d}\wedge\omega^{dc} \stackrel{!}{=}\omega^{b}\wedge\omega^{c}$$
So far so good. I started like this:
$$\omega^{b}\wedge\omega^{c}=\frac{1}{4}\varepsilon^{bde}\varepsilon^{cfg}\omega_{de}\wedge\omega_{fg}$$ As a next step, I used the identity that $\varepsilon^{bde}\varepsilon^{cfg}$ can be written as a sum of $6$ terms consisting of a product of $3$-delta symbols each. Ignoring all combinations with delta functions like $\delta_{ab}$ (all deltas with two indices in $\{a,b,c\}$), because such terms cancel, since in the end we also have the antisymmetrc $\varepsilon^{abc}$ in front, we get the following: $$\omega^{b}\wedge\omega^{c}=\frac{1}{4}(\delta^{dg}\omega_{dc}\wedge\omega_{bg}-\delta^{df}\omega_{dc}\wedge\omega_{bg})=\frac{1}{2}\delta^{dg}\omega_{cd}\wedge\omega_{gb}$$
where I used that $\omega_{ab}$ is antisymmetry in the last step. Now I am stuck, because I can't see how
$$\frac{1}{2}\delta^{dg}\omega_{cd}\wedge\omega_{gb}\stackrel{!}{=}\omega^{b}\wedge\omega^{c}$$
First of all, we have this factor of $1/2$ in front and secondly, the indices of the spin connections $\omega_{ab}$ are raised and lowered with the Minkowski metric $\eta=\operatorname{diag}(1,-1,-1)$ and not with the Kronecker delta...
Any comments are welcome!