A two-dimensional matrix model for the p equals two endpoint conventions

The normalized operators satisfy $e(a)=b$, $e(b)=0$, $f(a)=0$, $f(b)=a$, $h(a)=-a$, $h(b)=b$. Hence $[e,f](a)=-a=h(a)$ and $[e,f](b)=b=h(b)$. Also $[h,e](a)=2b=2e(a)$ and $[h,e](b)=0$, while $[h,f](a)=0$ and $[h,f](b)=-2a=-2f(b)$. Equality on the basis proves the three operator identities. Direct matrix multiplication gives the same three brackets for $e_{\ast },f_{\ast },h_{\ast }$. The bracket is bilinear and alternating, so checking all three distinct basis pairs proves that the stipulated linear map preserves every bracket. Thus it is a representation of the explicitly defined matrix Lie algebra.

Let $R\subset U$ be a nonzero invariant subspace and choose $u=xa+yb\in R$, $u\mathrm{\ne }0$. Since $R$ is invariant under $h$ and the identity, it contains $(I-h)u\mathrm{/}2=xa$ and $(I+h)u\mathrm{/}2=yb$. At least one is nonzero. If $a\in R$, then $b=e(a)\in R$; if $b\in R$, then $a=f(b)\in R$. In either case $R=U$, proving irreducibility. This argument uses the two distinct diagonal eigenvalues and the explicitly nonzero connecting maps.

For any nonzero $u\in U$, let $C(u)$ be the span of all finite words in $e,f,h$ applied to $u$. It is a nonzero invariant subspace by prepending any of the three operators to each word. Step 2.1 gives $C(u)=U$. Since $E=-8e$, $F=4f$ and $H=4h$, multiplication by these nonzero scalars in either direction shows that the finite-word spans for the two operator triples coincide. Every nonzero vector is therefore cyclic for both triples.