How to calculate the Hodge dual of a two-form defined in self-dual BF theory?

Question

I am trying to calculate the Hodge dual of the two form \begin{align}\Sigma^i=e^0\wedge e^i+\epsilon^i{}_{jk}e^j\wedge e^k,\quad (i,j,k \in \{1,2,3\}),\end{align} where Hodge dual with respect to internal indices has been given as \begin{align}\star A_{ab}=\frac{1}{2} \epsilon_{ab}{}^{cd}A_{cd},\quad (a,b,c,d \in \{0,1,2,3\}=\{0,i\}).\end{align} The reference that I am using is Quantum gravity with a positive cosmological constant by Lee Smolin (More precisely eq. (31) page 14).

For the first term: \begin{align} \star e^0\wedge e^i=\frac{1}{2} \epsilon^{0i}{}_{cd}e^c\wedge e^d. \end{align} Here, I understand that we can write j,k instead of c,d because of the properties of the Levi-Civita symbol. But, I dont understand how the term $\frac{1}{2}\epsilon^{0i}{}_{jk}$ becomes $\epsilon^{i}{}_{jk}$. Also for the second term, there is a Levi-Civita symbol with only three indices and I do not know how should I evaluate the Hodge dual in this term.

I would appreciate any help.

Answer 1

I found the solution thanks to the comments. There seems to be a factor of $\frac{1}{2}$ missing from the second term in the reference as well. Adding this factor will resolve the issue. \begin{align}\Sigma^i=e^0\wedge e^i+\frac{1}{2}\epsilon^i{}_{jk}e^j\wedge e^k,\quad (i,j,k \in \{1,2,3\}),\end{align}