vertex, say $e = \overrightarrow{u_1u_2}$ and $e' = \overrightarrow {u_1 u_3}$, the contribution to the sum is $1/3$. This is also true if $e$ and $e'$ are both in-edges of a single vertex. Lastly, if $e$ is an out-edge of a vertex $v$ and $e'$ is an in-edge then the contribution to the sum is $-1/3$. By counting the non-zero contributions of pairs $\{e,e'\}$ according to the vertex of $V(H)$ which intersect in, by the coefficient of $x^2$ in ${q_H}(x)$ is $$\begin{aligned}
            \sum _{u \in V(H)} \Big ( \frac{1}{3} \binom {d^+_H(u)}{2}  
            + \frac {1}{3}\binom {d^-_H(u)}{2} 
            -\frac {1}{3} {d^+_H(u)}{d^-_H(u)} \Big ) 
            & =
            \sum _{u \in V(H)} \frac {(d^+_H(u) - d^-_H(u))^2}{6} 
            - \frac {h(h-1)}{6}.
        \end{aligned}$$ as claimed.

Finally, if fails, ${q_H}(\varepsilon ) = 1 + a \varepsilon ^2 + O(\varepsilon ^4)$ with $a > 0$, where we used Lemma \[lemma: properties of counting poly\] *(ii)* to get ${q_H}(0)=1$ and *(iii)* to get that ${q_H}$ is even. This implies that ${q_H}(\varepsilon ) > 1$ for sufficiently small $\varepsilon >0$. As $H$ is not transitive, Lemma \[lemma: properties of counting poly\] *(ii)* gives ${q_H}(1) = 0$, and so by the intermediate value theorem ${q_H}(\varepsilon ) = 1$ for some $\varepsilon \in (0,1)$. But 