= Filling pair on a closed genus-two surface
A concrete <filling set of curves> consisting of two <simple closed curves> can be built from four unit squares. Label the squares $1,2,3,4$. Glue the right side of square $i$ to the left side of square $h(i)$ and its top side to the bottom side of square $v(i)$ by <translations>, where
$$
h=(1\ 2\ 3\ 4),\qquad v=(1\ 2\ 4\ 3).
$$
This gives a connected oriented <square-tiled surface>. The horizontal midlines form one <simple closed curve> $\alpha$, because $h$ has one <cycle of a permutation> of length four; the vertical midlines form one <simple closed curve> $\beta$, because $v$ has one <cycle of a permutation> of length four. They meet once at each square center.
The corner identifications have two classes, corresponding to the <cycles of a permutation> of
$$
hvh^{-1}v^{-1}=(2\ 4\ 3),
$$
with $1$ fixed. Thus the square <CW complex> has $V=2$, $E=8$, $F=4$, and <Euler characteristic> $2-8+4=-2$. The <classification of compact surfaces> gives <genus> two. The complement of $\alpha\cup\beta$ consists of the two open <discs> around these vertex classes, so the pair fills.
Orient both curves in their positive coordinate directions. All four crossings have the same local sign, giving <algebraic intersection number of curves on an oriented surface> $\alpha\cdot\beta=4$. Therefore $i(\alpha,\beta)\geq4$, and the displayed representatives attain equality. In particular both curves are nonseparating; a separating curve could not have this nonzero signed pairing.
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