= Solution
The <dual Thurston polytope> is the polar of the <Thurston norm> unit ball. More intrinsically, in the real dual of $H_2(Y,\partial Y)$ it is
$$
B_T(Y)=\{\alpha:|\alpha(u)|\leq\|u\|_T\text{ for every }u\}.
$$
The pairing can be regarded as evaluation of $H^2(Y,\partial Y;\mathbb R)$ on <relative homology>. Its definition remains valid when the <Thurston norm> has a kernel: the polytope then lies in the annihilator of that kernel.
Identify $Y$ with the <pair of pants> product $P\times S^1$. A regular <Seifert fiber> has homology class $h=\mu_1+\mu_2+\mu_3$. Let $e_i$ be the relative <homology class> corresponding by <Poincare-Lefschetz duality> to the homomorphism taking the $i$th meridian to one and the other two to zero. A spanning disk for $L_1$ punctured once by each of $L_2,L_3$ is an embedded <pair of pants> $F$ representing $e_1$, with $c(F)=1$.
Take two arcs in $P$, one joining boundary one to boundary two, the other joining boundary one to boundary three. Their products with $S^1$ are embedded <vertical surfaces in a Seifert fibered space>, namely <annuli> $A_{12},A_{13}$. Orient them so that their relative classes are $e_1-e_2$ and $e_1-e_3$. They cost zero. Thus the three required inequalities, with $\alpha_i=\alpha(e_i)$, are
$$
|\alpha_1|\leq1,\qquad |\alpha_1-\alpha_2|\leq0,\qquad |\alpha_1-\alpha_3|\leq0.
$$
For completeness they give the whole polytope. Oriented cut-and-paste of copies of $F,A_{12},A_{13}$ gives the upper bound $\|a e_1+b e_2+c e_3\|_T\leq|a+b+c|$. For the reverse bound, compress a minimizing surface and use the <classification of incompressible surfaces in Seifert fibered spaces>. Its horizontal components cover $P$ and have negative <Euler characteristic> equal to their unsigned covering degree; its vertical components have zero cost and zero intersection with a regular <Seifert fiber>. The total signed horizontal degree is $a+b+c$, so its cost is at least $|a+b+c|$. Hence
$$
\boxed{\|a e_1+b e_2+c e_3\|_T=|a+b+c|,\qquad B_T(Y)=\{(s,s,s):-1\leq s\leq1\}.}
$$
It is a line segment, because this <Thurston norm> has a two-dimensional kernel.
Back to article page
