Raising and lowering indices of the Levi-Civita epsilon symbol in two dimensions In two dimensions, what is the relation between $\epsilon^a{}_b$ and $\epsilon_{ab}$ where $a, b$ take the values $\{1,2\}$?
By that I mean, how does the sign change in that case? In four dimensions for example I know that $ \epsilon_{0123} = - \epsilon^{0123}$.
 A: $\epsilon_{ab}$ are the components of a tensor, so you can raise and lower indices with the metric:
\begin{align}
\epsilon^a{}_b & = g^{ac} \epsilon_{cb} \\
\epsilon_b{}^a & = g^{ac} \epsilon_{bc}.
\end{align}
Note that order matters: $\epsilon^a{}_b = -\epsilon_b{}^a$.

Since it seems you are just relying on heuristics for when signs get introduced, here is the full story. The symbol (not tensor) $\tilde{\epsilon}_{ab\cdots} \equiv [a\ b\ \cdots]$ is defined to be $+1$ if the indices are an even permutation of the $n$ available indices ($n$ being the dimension of the space), $-1$ if they are an odd permutation, and $0$ otherwise. Sometimes you see an upper-indexed symbol $\tilde{\epsilon}^{ab\cdots} \equiv \mathrm{sgn}(g) \tilde{\epsilon}_{ab\cdots}$.
The tensors are related to these symbols:
\begin{align}
\epsilon_{ab\cdots} & = \lvert g \rvert^{1/2} \tilde{\epsilon}_{ab\cdots} \\
\epsilon^{ab\cdots} & = \lvert g \rvert^{-1/2} \tilde{\epsilon}^{ab\cdots}.
\end{align}
In special relativity, much of the detail is lost, since $g = -1$. This means the tensors are not distinct from their similarly-indexed symbols, but you still get the sign difference in going from lower to upper.
In 2D Cartesian space, $g = 1$. Here again the distinction between symbol and tensor is lost, and in fact there is no sign change between all upper indices and all lower indices.
