Let $\Sigma$ be a connected oriented surface, and $[-,-]\colon \Bbb Z\big[\widehat\pi(\Sigma)\big]\times Z\big[\widehat\pi(\Sigma)\big]\to Z\big[\widehat\pi(\Sigma)\big]$ be the Goldman Bracket. Note that Goldman Bracket is a bi-linear skew-symmetric map satisfying the Jacobi Identity. Here $\widehat\pi(\Sigma)$ is the set of all free homotopy classes of loops in $\Sigma$, and $Z\big[\widehat\pi(\Sigma)\big]$ denotes the free module of $\widehat\pi(\Sigma)$ over the ring of integers.
One of the exciting properties of the Goldman bracket is the following:
Theorem: If $\alpha$ is a simple closed curve representing $x\in \widehat\pi(\Sigma)$, then, for $y\in \widehat \pi(\Sigma)$, $[x, y] = 0$ if and only if $y$ can be represented by a closed curve $\beta$ such that $\beta\cap \alpha=\varnothing$.
Now, in practice, you can not always expect $x$ has a representation by a simple closed curve, but after a homotopy, we have a representation of $x$ by a self-transverse immersion $\Bbb S^1\to \Sigma$ without any triple point. So, my question is the following:
To what extent can we relax the condition of having a simple-closed representative? For example, let's say $x$ has a representation by the figure-eight curve $\alpha$ but $x$ has no simple closed representative, then can we say that $[x, y] = 0$ if and only if $y$ can be represented by a closed curve $\beta$ such that $\beta\cap \alpha=\varnothing$
