= Solution
For the <A3 root system>, take <simple roots> $\alpha_i=e_i-e_{i+1}$ for $i=1,2,3$ in the hyperplane $\sum_{i=1}^4x_i=0$ of an orthonormal $\mathbb R^4$. Its <Dynkin diagram> is the single-bond chain $1-2-3$. The <Cartan matrix> has entries $C_{ij}=2(\alpha_i,\alpha_j)/(\alpha_j,\alpha_j)$, giving
$$
\boxed{C_{A_3}=\begin{pmatrix}2&-1&0\\-1&2&-1\\0&-1&2\end{pmatrix}.}
$$
For a <weight> $\mu$, use <Dynkin labels> $w_i=2(\mu,\alpha_i)/(\alpha_i,\alpha_i)$. The four <weights> $\mu_j=e_j-\tfrac14\sum_ke_k$ of the defining <representation> have labels
$$
\boxed{F=\{(1,0,0),(-1,1,0),(0,-1,1),(0,0,-1)\}.}
$$
They form the <weight system> with <highest weight> $(1,0,0)$. Equivalently, subtract successive rows of $C_{A_3}$ from that <highest weight>. The <dual representation> negates every <weight>, so the <weight system> with <highest weight> $(0,0,1)$ is
$$
\boxed{\overline F=-F=\{(0,0,1),(0,1,-1),(1,-1,0),(-1,0,0)\}.}
$$
All these <weights> have multiplicity one. The numbered <Dynkin diagrams> used in this solution and in <4/b/solution> are shown below.
\Image[/past-exam-of-the-groups-and-representations-course-of-the-university-of-oxford/2023-dynkin-diagrams.png]
{title=Numbered <Dynkin diagrams> for the <A3 root system> and <D5 root system>, with deletion from the <Extended Dynkin diagram>}
{description=All bonds are single. In the <Extended Dynkin diagram> of the <D5 root system>, deleting node 2 separates nodes 0 and 1 and leaves the three-node chain 4-3-5, giving the <root subsystem> $A_1\oplus A_1\oplus A_3$.}
Back to article page
