Variation of Maxwell action with respect to the vierbein - Einstein-Cartan Theory

I understand the notation in the book this way \begin{eqnarray} e^a \wedge e^b \wedge \star |_e F &\equiv& e^a \wedge e^b \wedge \star F\;\\ &=& F \wedge \star (e^a \wedge e^b)\;\\ &=& \star (e^a \wedge e^b) \wedge F\;. \end{eqnarray} Now in the more accurate definition of Hodge dual (in the proof from definition there is a step of flipping the indices for more symmetric looking result, but we not flip them here, at first) we have \begin{equation}\boxed{ \star (f^a \wedge e^b) =\frac 1 2 \epsilon^{abcd} f_c \wedge e_d }\;. \end{equation} In the case of $\star(e \wedge e)$, there are no problems for flipping the indices \begin{equation} \frac 1 2 \epsilon^{abcd} e_c \wedge e_d = \frac 1 2 \epsilon^{ab}{}_{cd}\, e^c \wedge e^d \;. \end{equation} But in the case of $\star(f \wedge e)$ we have \begin{eqnarray} \star (f^a \wedge e^b) &=& \frac 1 2 \epsilon^{abcd} f_c \wedge e_d \\ &=& \frac 1 2 \epsilon^{ab}{}_{ef} \eta^{ce} \eta^{fd} f_c \wedge e_d \\ &=& \frac 1 2 \epsilon^{ab}{}_{ef} \eta^{ce} f_c \wedge e^f \\ &=& - \frac 1 2 \epsilon^{ab}{}_{ef} f^e \wedge e^f\;. \end{eqnarray} This because $e$ obeys the transformation laws \begin{eqnarray} e^a &\longmapsto& e^a+f^a \\ e_a &\longmapsto& e_a-f_a \\ && \eta^{ab}f_a=-f^b\;. \end{eqnarray} With \begin{equation} f^a \wedge e^b \wedge \star |_e F = \star(f^a \wedge e^b) \wedge F =- \frac 1 2 \epsilon^{ab}{}_{cd} f^c \wedge e^d \wedge F \end{equation} we can obtained the same result as shown in the reference.

Sorry I have to dredge this thread up, but I've got the same problem as the author had. Let's summarize: user148471 was trying to say, that

1. We have to make the substitution $e^a \mapsto e^a + f^a $, and THEN compute the Hodge dual, not conversely.
2. We've got $$\star (e^a \wedge e^b)= \frac{1}{2} \epsilon ^{ab}_{\quad cd}e^c \wedge e^d$$ BUT $$\star (f^a \wedge e^b) = -\frac{1}{2} \epsilon ^{ab}_{\quad cd}f^c \wedge e^d.$$
3. We define $f_a$ as $$f_a:=-g_{ab}f^b=-\eta_{ab}f^b,$$ hence $$\eta^{ab}f_a=-\eta^{ab}\eta_{ac}f^c=-\delta ^b _{\ c}f^c=-f^b$$

Concerning 2. The formula $$\star (f^a \wedge e^b) = -\frac{1}{2} \epsilon ^{ab}_{\quad cd}f^c \wedge e^d.$$ would solve the problem. But is it correct? The formula seems not to make sense since $f^a,\ a=0,1,2,3$ are any variations of 1-forms of the basis. What if we take $$f^a :=e^a,\ a=0,1,2,3?$$ Then we'll have $$\star (f^a \wedge e^b)=\star (e^a \wedge e^b)$$ and $$-\frac{1}{2} \epsilon ^{ab}_{\quad cd}e^c \wedge e^d=-\frac{1}{2} \epsilon ^{ab}_{\quad cd}f^c \wedge e^d=\star(f^a \wedge e^b)=\star (e^a \wedge e^b)=\frac{1}{2} \epsilon ^{ab}_{\quad cd}e^c \wedge e^d.$$ This tells us that $$\star (e^a \wedge e^b)=0,$$ but why should it equal 0? When $a=b$ then $$\star (e^a \wedge e^b)=0,$$ but $a$ and $b$ don't have to be equal.