The sl2 principal admissible reduction data at level minus one half

Direct multiplication gives $[h,e]=2e$, $[h,f]=-2f$, and $[e,f]=h$. Thus the three matrices are linearly independent and span the traceless $2\times 2$ matrices. The algebra is nonabelian.

Let $I$ be an ideal and $X=ae+bh+cf\in I$. Since $I$ is stable under $\mathrm{ad}h$, it is stable under polynomials in that operator. The eigenvalues on $e,h,f$ are $2,0,-2$; the polynomials $\frac{A(A+2)}{8},1-\frac{A^2}{4},\frac{A(A-2)}{8},A=\mathrm{ad}h,$ project onto the $e,h,f$ lines, respectively. If $I$ is nonzero, one of these projections puts a nonzero multiple of one of $e,h,f$ in $I$. Bracketing that element with the other two basis elements using step 1.1 puts all three in $I$. Therefore $I={\mathfrak{s}\mathfrak{l}}_2$, proving simplicity.

The brackets are $[f,e]=-h$, $[f,h]=2f$, and $[f,f]=0$, so $(\mathrm{ad}f{)}^3=0$. Hence $f$ is nilpotent by [F2]. The relations in step 1.1 also verify the chosen Jacobson-Morozov triple directly, without invoking its general existence theorem.

By step 1.1, $\mathrm{ad}h$ is diagonal on the basis $e,h,f$ with eigenvalues $2,0,-2$. Its zero eigenspace is $\mathbb{C}h$, so $\mathrm{ad}h$ is diagonalizable and the centralizer of $\mathbb{C}h$ is itself; hence $\mathbb{C}h$ is a Cartan subalgebra of rank $1$. The two nonzero weights are $\alpha$ and $-\alpha$ with $\alpha (h)=2$. For $X=ae+bh+cf$, $[f,X]=-ah+2bf$, so ${\mathfrak{g}}^f=\mathbb{C}f$ and has dimension $1$. Thus $f$ is principal under the stated convention.

Since $\mathrm{ad}(h\mathrm{/}2)$ has eigenvalues $1,0,-1$ on $e,h,f$, the displayed grading is a Lie grading. The only nonzero required map with $j\ge 1\mathrm{/}2$ is $\mathrm{ad}f:\mathbb{C}e\to \mathbb{C}h$, $e\mapsto -h$, which is injective. For $j\le 1\mathrm{/}2$, the nonzero target cases are $\mathbb{C}h\to \mathbb{C}f$, $h\mapsto 2f$, which is surjective, and $\mathbb{C}f\to 0$, which is also surjective. All other graded pieces are zero. Since ${\mathfrak{g}}_{1\mathrm{/}2}=0$, its skew form is nondegenerate on the zero space and the unique Lagrangian is $\mathrm{\ell }=0$. Thus the grading is good and $\mathfrak{m}=\mathbb{C}e$.

For $X=ae+bh+cf$, $[e,X]=-2be+ch$. This vanishes exactly when $b=c=0$, so ${\mathfrak{g}}^e=\mathbb{C}e$. By [F4], $S_f=f+\mathbb{C}e$.

The trace form has $(h\mathrm{\mid }h)=2$ and $(e\mathrm{\mid }f)=1$, so $\alpha$ has squared length $2$ and ${\alpha }^{\vee }=h$. Thus $\theta =\alpha$, $\rho =\alpha \mathrm{/}2$, $h=1+\mathrm{ht}(\alpha )=2$, and $h^{\vee }=1+\rho (h)=2$. The coroot root set is again $\{\alpha ,-\alpha \}$, so its dual datum has $h_{\tilde{\mathfrak{g}}}^{\vee }=2$; there are no off-diagonal Cartan entries, giving $r^{\vee }=1$.

With $p=3,q=2$, $\gcd (p,q)=1$ and $\gcd (q,r^{\vee })=\gcd (2,1)=1$. The first branch requires $p\ge h^{\vee }$ and $q\ge h$; both hold because $3\ge 2$ and $2\ge 2$. Thus $k=-h^{\vee }+p\mathrm{/}q=-2+3\mathrm{/}2=-1\mathrm{/}2$, and $k+h^{\vee }=3\mathrm{/}2>0$. This verifies the declared level condition and noncriticality.