\section{Computations for the degree $8$ free curve}
\label{section-degree-8}

\noindent Let $\varphi:\mathbb{P}^1\to\mathbb{P}^5$ be a morphism given by the 6-tuple
\begin{align*}
&G_0 = S^7T \\
&G_1 = S^4T^4 + S^3T^3 \\
&G_2 = S^4T^4 + S^3T^5 + T^8 \\
&G_3 = S^7T + S^6T^2 + S^5T^3 + S^4T^4 + S^3T^5 \\
&G_4 = S^8 + S^7T + S^6T^2 + S^5T^3 + S^4T^4 + S^3T^5 + T^8 \\
&G_5 = S^8 + S^7T + S^6T^2 + S^5T^3 + S^4T^4 + S^3T^5 + S^2T^6 + ST^7.
\end{align*}

\noindent One can check by computer or by hand that this curve lies on the Fermat hypersurface $X\subset\mathbb{P}^5$.

\medskip
\noindent Due to twisting, the domain of the map $\tilde{\varphi}:R(-8)^{\oplus 6}\to R$ has its first nontrivial graded piece in dimension $8$. The $G_i$ are linearly independent over $k$, hence the kernel is trivial in dimension $8$. The matrix for the map $\tilde{\varphi}_9:R(-8)_{9}^{\oplus 6}\to R_9$ is
$$
\begin{pmatrix}
0 &0 &0 &0 &0 &0 &0 &0 &1 &0 &1 &0 \\
1 &0 &0 &0 &0 &0 &1 &0 &1 &1 &1 &1 \\
0 &1 &0 &0 &0 &0 &1 &1 &1 &1 &1 &1 \\
0 &0 &0 &0 &0 &0 &1 &1 &1 &1 &1 &1 \\
0 &0 &1 &0 &1 &0 &1 &1 &1 &1 &1 &1 \\
0 &0 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 \\
0 &0 &0 &1 &0 &1 &0 &1 &0 &1 &1 &1 \\
0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &1 &1 \\
0 &0 &0 &0 &1 &0 &0 &0 &1 &0 &0 &1 \\
0 &0 &0 &0 &0 &1 &0 &0 &0 &1 &0 &0
\end{pmatrix}
$$

\noindent where each direct summand of the domain has a basis $\left\{(S,0),(0,T)\right\}$, of which we take six copies (for total dimension $12$), and the range has basis given by the degree $9$ monomials in $S$ and $T$, ordered by increasing $T$-degree (for total dimension $10$). This matrix has rank $10$, which means that the map in degree $9$ is surjective. By rank-nullity, two dimensions of the kernel live in degree $9$; denote the generators by $x_1,x_2$. Surjectivity of $\tilde{\varphi}$ in degree $9$ implies surjectivity in all higher degrees. A second application of rank-nullity gives $\dim_k\Omega(\varphi)_{10}=7$. Four of the generators are inherited from the previous degree, taking the forms
$$
x_1S,x_2S,x_1T,x_2T.
$$

\noindent We conclude that there are three additional generators in degree $10$. Therefore, the splitting type of $\Omega_X(\varphi)$ is $e_1,...,e_5=-10,-10,-10,-9,-9$, which translates to a splitting type for $E_X(\varphi)$ of $f_1,...,f_5=0,0,0,4,4$, hence the curve is free.