= Solution
For the intended <module>, index a <basis> by $v_k=a_{l-2k}$ for $0\le k\le l$, and use
$$
Hv_k=(l-2k)v_k,\qquad fv_k=v_{k+1},\qquad ev_k=k(l+1-k)v_{k-1},
$$
with $v_{-1}=v_{l+1}=0$. This makes the <basis> endpoints distinct, identifies the lowering operator, and includes its missing raising action at $k=l$. These are the necessary indexing corrections to the printed display. The same definition for $l=0$ supplies the trivial <module> needed for classification.
Directly on $v_k$,
$$
\begin{aligned}
[H,f]v_k&=-2fv_k,\\
[H,e]v_k&=2ev_k,\\
[e,f]v_k&=((k+1)(l-k)-k(l+1-k))v_k=(l-2k)v_k.
\end{aligned}
$$
These hold at the endpoints too, proving the three defining $\mathfrak{sl}_2$ relations. Finally
$$
\begin{aligned}
\Delta v_k
&=\bigl(4k(l+1-k)+(l-2k)^2+2(l-2k)\bigr)v_k\\
&=\boxed{l(l+2)v_k.}
\end{aligned}
$$
Thus the <Casimir element> acts by that scalar on the whole <module>.
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