Restricted root systems may be nonreduced

By [L2] the restriction $(\cdot ,\cdot )$ of $B_{\theta }$ to $\mathfrak{a}\subseteq {\mathfrak{p}}_0$ is a positive definite inner product; for $\lambda \in {\mathfrak{a}}^{\ast }$ the vector $H_{\lambda }\in \mathfrak{a}$ with $(H_{\lambda },H)=\lambda (H)$ for all $H\in \mathfrak{a}$ exists and is unique, and $⟨\lambda ,\mu ⟩=(H_{\lambda },H_{\mu })$ is an inner product on ${\mathfrak{a}}^{\ast }$ with $\mathrm{\mid }\lambda {\mathrm{\mid }}^2>0$ for $\lambda \mathrm{\ne }0$; also $\lambda (H_{\mu })=⟨\lambda ,\mu ⟩$; the reflection $s_{\lambda }(\mu )=\mu -2⟨\mu ,\lambda ⟩\mathrm{\mid }\lambda {\mathrm{\mid }}^{-2}\lambda$ is orthogonal, fixes ${\lambda }^{\perp }$ pointwise and sends $\lambda$ to $-\lambda$; put $H_{\lambda }^{\mathrm{\prime }}=2\mathrm{\mid }\lambda {\mathrm{\mid }}^{-2}{H}_{\lambda }$, so that $\lambda (H_{\lambda }^{\mathrm{\prime }})=2$.

Witness setup: let $J=\mathrm{diag}(1,1,-1)$ and ${\mathfrak{g}}_0=\mathfrak{s}\mathfrak{u}(2,1)=\{X\in M_3(\mathbb{C}):X^{\ast }J+JX=0,\mathrm{tr}X=0\}$ with $\theta (X)=-X^{\ast }$; then $\theta$ is an involutive automorphism of ${\mathfrak{g}}_0$, and $X\in M_3(\mathbb{C})$ lies in ${\mathfrak{g}}_0$ exactly when $X=\left(\begin{array}{cc}a& b\\ b^{\ast }& d\end{array}\right)$ with $a\in u(2)$, $b\in {\mathbb{C}}^2$ and $d=-\mathrm{tr}a\in i\mathbb{R}$, so that $k_0=\{X\in {\mathfrak{g}}_0:b=0\}$ and $p_0=\{X\in {\mathfrak{g}}_0:a=0,d=0\}=\{z{E}_{13}+w{E}_{23}+\bar{z}{E}_{31}+\bar{w}{E}_{32}:z,w\in \mathbb{C}\}$; write $a=\left(\begin{array}{cc}iA& \gamma \\ -\bar{\gamma }& iB\end{array}\right)$ with $A,B\in \mathbb{R}$, $\gamma \in \mathbb{C}$ and $b=\left(\begin{array}{c}p\\ q\end{array}\right)$; the element $H=E_{13}+E_{31}$ lies in $p_0$ and $\mathfrak{a}=\mathbb{R}H$ is a $1$-dimensional subspace of $p_0$; finally ${\mathfrak{g}}_0$ is semisimple, because ${\mathfrak{g}}_0^{\mathbb{C}}={\mathfrak{s}\mathfrak{l}}_3(\mathbb{C})$ has nondegenerate Killing form and ${\mathfrak{g}}_0$ is its real form.

$\mathrm{\Sigma }$ spans ${\mathfrak{a}}^{\ast }$: if $H\in \mathfrak{a}$ satisfies $\lambda (H)=0$ for every $\lambda \in \mathrm{\Sigma }$, then $[H,{\mathfrak{g}}_0^{\lambda }]=0$ for every $\lambda \in \mathrm{\Sigma }$ and also $[H,{\mathfrak{g}}_0^0]=0$ because $H\in \mathfrak{a}\subseteq {\mathfrak{g}}_0^0=Z_{{\mathfrak{g}}_0}(\mathfrak{a})$, so $[H,{\mathfrak{g}}_0]=0$ by [L1], i.e. $H\in Z({\mathfrak{g}}_0)=0$ because ${\mathfrak{g}}_0$ is semisimple; hence no nonzero $H\in \mathfrak{a}$ annihilates $\mathrm{\Sigma }$, and $\mathrm{\Sigma }$ spans ${\mathfrak{a}}^{\ast }$.

Notation: $\mathrm{\Sigma }$ is reducible if there are nonzero orthogonal subspaces $E_1,E_2\subseteq {\mathfrak{a}}^{\ast }$ with $\mathrm{\Sigma }\subseteq E_1\cup E_2$ and $\mathrm{\Sigma }\cap E_i\mathrm{\ne }\mathrm{\varnothing }$ for $i=1,2$, and irreducible otherwise; ${\mathrm{\Sigma }}_s=\{\alpha \in \mathrm{\Sigma }:\alpha \mathrm{/}2\mathrm{\notin }\mathrm{\Sigma }\}$ and $\mathrm{\Psi }=\{\alpha \in {\mathrm{\Sigma }}_s:2\alpha \in \mathrm{\Sigma }\}$.

Let $\lambda \in \mathrm{\Sigma }$ and choose $0\mathrm{\ne }{E}_{\lambda }\in {\mathfrak{g}}_0^{\lambda }$; then $B(E_{\lambda },\theta E_{\lambda })=-B_{\theta }(E_{\lambda },E_{\lambda })<0$ by [L2], the bracket $[E_{\lambda },\theta E_{\lambda }]$ lies in ${\mathfrak{g}}_0^0$ by [L1] and satisfies $\theta [E_{\lambda },\theta E_{\lambda }]=-[E_{\lambda },\theta E_{\lambda }]$, hence lies in $\mathfrak{a}$, and for every $H\in \mathfrak{a}$ one has $B([E_{\lambda },\theta E_{\lambda }],H)=B(E_{\lambda },[\theta E_{\lambda },H])=\lambda (H)B(E_{\lambda },\theta E_{\lambda })=B(B(E_{\lambda },\theta E_{\lambda })H_{\lambda },H)$, so $[E_{\lambda },\theta E_{\lambda }]=B(E_{\lambda },\theta E_{\lambda })H_{\lambda }$; rescaling $E_{\lambda }$ by a positive real number we arrange $B_{\theta }(E_{\lambda },E_{\lambda })=2\mathrm{\mid }\lambda {\mathrm{\mid }}^{-2}$, that is $B(E_{\lambda },\theta E_{\lambda })=-2\mathrm{\mid }\lambda {\mathrm{\mid }}^{-2}$, and then $[E_{\lambda },\theta E_{\lambda }]=-H_{\lambda }^{\mathrm{\prime }}$.

Witness computation, part 1: for $X\in {\mathfrak{g}}_0$ with coordinates $(A,B,\gamma ,p,q)$ as in step 1.2, a direct computation of $[H,X]$ gives the new coordinates $A^{\mathrm{\prime }}={\displaystyle \frac{\bar{p}-p}{i}}$, $B^{\mathrm{\prime }}=0$, ${\gamma }^{\mathrm{\prime }}=\bar{q}$, $p^{\mathrm{\prime }}=-i(2A+B)$, $q^{\mathrm{\prime }}=\bar{\gamma }$; hence $[H,X]=0$ if and only if $p\in \mathbb{R}$, $q=\gamma =0$ and $B=-2A$; consequently $Z_{p_0}(H)=\{p(E_{13}+E_{31}):p\in \mathbb{R}\}=\mathfrak{a}$, and $\mathfrak{a}$ is maximal abelian in ${\mathfrak{p}}_0$, since every abelian subspace of ${\mathfrak{p}}_0$ containing $\mathfrak{a}$ is contained in $Z_{p_0}(H)=\mathfrak{a}$.

The normalization of step 2.1 gives the bracket relations $[H_{\lambda }^{\mathrm{\prime }},E_{\lambda }]=2E_{\lambda }$, $[H_{\lambda }^{\mathrm{\prime }},\theta E_{\lambda }]=-2\theta E_{\lambda }$ and $[E_{\lambda },\theta E_{\lambda }]=-H_{\lambda }^{\mathrm{\prime }}$, so the real span of $H_{\lambda }^{\mathrm{\prime }},E_{\lambda },\theta E_{\lambda }$ is a three-dimensional Lie subalgebra of ${\mathfrak{g}}_0$ isomorphic to ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ with $H_{\lambda }^{\mathrm{\prime }}$ corresponding to $\mathrm{diag}(1,-1)$; complexifying, its complexification is a copy of ${\mathfrak{s}\mathfrak{l}}_2$ in ${\mathfrak{g}}_0^{\mathbb{C}}$ acting on the finite-dimensional complex vector space ${\mathfrak{g}}_0^{\mathbb{C}}$.

Witness computation, part 2: solving the coordinate equations of step 2.2 for eigenvectors of $\mathrm{ad}H$ shows that the eigenvalues are $0$ with multiplicity $2$, $\pm 1$ with multiplicity $2$ each, and $\pm 2$ with multiplicity $1$ each; explicitly $[H,X]=2X$ for $X=i(E_{33}-E_{11}+E_{13}-E_{31})$, the elements $E_{12}-E_{21}+E_{23}+E_{32}$ and $i(E_{12}+E_{21}-E_{23}+E_{32})$ of ${\mathfrak{g}}_0$ satisfy $[H,X]=X$ and are linearly independent, hence span the eigenspace for the functional $f$ with $f(H)=1$, and $\ker (\mathrm{ad}H)={\mathfrak{g}}_0^0$ is spanned by $H$ and $i(E_{11}-2E_{22}+E_{33})$.

Let $k_{\lambda }=\exp (\frac{\pi }{2}(E_{\lambda }+\theta E_{\lambda }))$; since $\theta (E_{\lambda }+\theta E_{\lambda })=E_{\lambda }+\theta E_{\lambda }$ one has $E_{\lambda }+\theta E_{\lambda }\in {\mathfrak{k}}_0$, so $k_{\lambda }\in K=\exp ({\mathfrak{k}}_0)$; moreover $(\mathrm{ad}(E_{\lambda }+\theta E_{\lambda }))H=0$ for $H\in \ker \lambda$ and, with $X=\frac{\pi }{2}(E_{\lambda }+\theta E_{\lambda })$, $(\mathrm{ad}X)H_{\lambda }^{\mathrm{\prime }}=\pi (\theta E_{\lambda }-E_{\lambda })$ and $(\mathrm{ad}X{)}^2{H}_{\lambda }^{\mathrm{\prime }}=-{\pi }^2{H}_{\lambda }^{\mathrm{\prime }}$, so the exponential series gives $\mathrm{Ad}(k_{\lambda })H_{\lambda }^{\mathrm{\prime }}=\cos (\pi )H_{\lambda }^{\mathrm{\prime }}+\sin (\pi )(\theta E_{\lambda }-E_{\lambda })=-H_{\lambda }^{\mathrm{\prime }}$; hence $\mathrm{Ad}(k_{\lambda })$ preserves $\mathfrak{a}=\ker \lambda \oplus \mathbb{R}{H}_{\lambda }^{\mathrm{\prime }}$, acts as the identity on $\ker \lambda$ and as $-1$ on $H_{\lambda }^{\mathrm{\prime }}$, i.e. acts on $\mathfrak{a}$ by the reflection with fixed hyperplane $\ker \lambda$.

Witness computation, part 3: by step 3.2 the restricted-root system of the pair $({\mathfrak{g}}_0,\mathfrak{a})$ of step 1.2 is $\mathrm{\Sigma }=\{\pm f,\pm 2f\}$ with $f(H)=1$, the spaces ${\mathfrak{g}}_0^{\pm f}$ being the two $2$-dimensional eigenspaces and ${\mathfrak{g}}_0^{\pm 2f}$ the two $1$-dimensional eigenspaces; hence $m_{\pm f}=2$ and $m_{\pm 2f}=1$, and this restricted-root system is not reduced, since both $f$ and $2f$ occur.

Integrality: let $\mu ,\lambda \in \mathrm{\Sigma }$; by step 3.1 the copy of ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$ spanned by $H_{\lambda }^{\mathrm{\prime }},E_{\lambda },\theta E_{\lambda }$ acts on ${\mathfrak{g}}_0^{\mathbb{C}}$, and $H_{\lambda }^{\mathrm{\prime }}$ is its standard diagonal element; by [L6] the operator $\mathrm{ad}{H}_{\lambda }^{\mathrm{\prime }}$ is diagonalisable with integer eigenvalues on ${\mathfrak{g}}_0^{\mathbb{C}}$; since ${\mathfrak{g}}_0^{\mu }$ lies in the eigenspace of $\mathrm{ad}{H}_{\lambda }^{\mathrm{\prime }}$ for the eigenvalue $\mu (H_{\lambda }^{\mathrm{\prime }})=2⟨\mu ,\lambda ⟩\mathrm{\mid }\lambda {\mathrm{\mid }}^{-2}$, this number is an integer.

The reflection $s_{\lambda }$ is realised in $N_K(\mathfrak{a})$ and permutes $\mathrm{\Sigma }$: by step 4.1 the element $k_{\lambda }\in K$ satisfies $\mathrm{Ad}(k_{\lambda })\mathfrak{a}=\mathfrak{a}$ and acts on ${\mathfrak{a}}^{\ast }$ by $s_{\lambda }$, so $k_{\lambda }\in N_K(\mathfrak{a})$; and for $\mu \in \mathrm{\Sigma }$ and $0\mathrm{\ne }X\in {\mathfrak{g}}_0^{\mu }$ one has $[H,\mathrm{Ad}(k_{\lambda })X]=\mathrm{Ad}(k_{\lambda })[\mathrm{Ad}(k_{\lambda }{)}^{-1}H,X]=(s_{\lambda }\mu )(H)\mathrm{Ad}(k_{\lambda })X$ for all $H\in \mathfrak{a}$, so $\mathrm{Ad}(k_{\lambda }){\mathfrak{g}}_0^{\mu }={\mathfrak{g}}_0^{s_{\lambda }\mu }$ and $s_{\lambda }\mu \in \mathrm{\Sigma }$; hence $s_{\lambda }(\mathrm{\Sigma })=\mathrm{\Sigma }$ for every $\lambda \in \mathrm{\Sigma }$.

Multiples of a restricted root: let $\lambda \in \mathrm{\Sigma }$ and let $C=\{c>0:c\lambda \in \mathrm{\Sigma }\}$, a finite nonempty set of positive reals with minimum $a$; by step 4.3 one has $2c\mathrm{/}d\in \mathbb{Z}$ for all $c,d\in C$, so with $c=a$ and $d=\max C$ one gets $2a\mathrm{/}\max C\in \{1,2\}$ and hence $\max C\in \{a,2a\}$; moreover no $c\in C$ satisfies $a<c<2a$, since then $2a\mathrm{/}c$ would lie strictly between $1$ and $2$; therefore $C=\{a\}$ or $C=\{a,2a\}$, and since $1\in C$ we have $a=1$ or $a=1\mathrm{/}2$; hence every element of $\mathrm{\Sigma }$ is either indivisible or twice an indivisible root, that is $\mathrm{\Sigma }={\mathrm{\Sigma }}_s\bigsqcup 2\mathrm{\Psi }$, and for every $\beta \in {\mathrm{\Sigma }}_s$ the only positive multiples of $\beta$ in $\mathrm{\Sigma }$ are $\beta$ and possibly $2\beta$.

${\mathrm{\Sigma }}_s$ is a reduced crystallographic root system in ${\mathfrak{a}}^{\ast }$: it is finite by [L1]; it spans ${\mathfrak{a}}^{\ast }$ because $\mathrm{\Sigma }\subseteq \mathrm{span}{\mathrm{\Sigma }}_s$ by step 5.2 and $\mathrm{\Sigma }$ spans ${\mathfrak{a}}^{\ast }$ by step 1.3; it satisfies $s_{\alpha }({\mathrm{\Sigma }}_s)={\mathrm{\Sigma }}_s$ for $\alpha \in {\mathrm{\Sigma }}_s$ because $s_{\alpha }(\mathrm{\Sigma })=\mathrm{\Sigma }$ by step 5.1 and $s_{\alpha }$ preserves the property $\alpha \mathrm{/}2\mathrm{\notin }\mathrm{\Sigma }$, since $s_{\alpha }(\beta )\mathrm{/}2=s_{\alpha }(\beta \mathrm{/}2)$; it satisfies the integrality condition by step 4.3; and it is reduced because $\mathbb{R}\beta \cap {\mathrm{\Sigma }}_s=\{\pm \beta \}$ for $\beta \in {\mathrm{\Sigma }}_s$, which is exactly the statement proved in step 5.2.

$\mathrm{\Psi }$ is invariant under the Weyl group $W({\mathrm{\Sigma }}_s)$: if $\alpha \in \mathrm{\Psi }$ and $\beta \in {\mathrm{\Sigma }}_s$, then $2s_{\beta }(\alpha )=s_{\beta }(2\alpha )\in \mathrm{\Sigma }$ by step 5.1, and $s_{\beta }(\alpha )\in {\mathrm{\Sigma }}_s$ by step 6.1, so $s_{\beta }(\alpha )\in \mathrm{\Psi }$.

$\mathrm{\Sigma }$ is irreducible if and only if ${\mathrm{\Sigma }}_s$ is: if ${\mathrm{\Sigma }}_s={\mathrm{\Phi }}_1\bigsqcup {\mathrm{\Phi }}_2$ with $({\mathrm{\Phi }}_1,{\mathrm{\Phi }}_2)=0$ and both parts nonempty, then $\mathrm{\Sigma }=({\mathrm{\Phi }}_1\cup 2(\mathrm{\Psi }\cap {\mathrm{\Phi }}_1))\bigsqcup ({\mathrm{\Phi }}_2\cup 2(\mathrm{\Psi }\cap {\mathrm{\Phi }}_2))$ by step 5.2, and both parts are nonempty and mutually orthogonal, so $\mathrm{\Sigma }$ is reducible; conversely if $\mathrm{\Sigma }={\mathrm{\Sigma }}_1\bigsqcup {\mathrm{\Sigma }}_2$ with $({\mathrm{\Sigma }}_1,{\mathrm{\Sigma }}_2)=0$ and both parts nonempty, then every $\lambda \in {\mathrm{\Sigma }}_i$ is $\alpha$ or $2\alpha$ with $\alpha \in {\mathrm{\Sigma }}_s\cap {\mathrm{\Sigma }}_i$ (if $\alpha \in {\mathrm{\Sigma }}_j$, $j\mathrm{\ne }i$, then $(\alpha ,2\alpha )=0$, impossible), so ${\mathrm{\Sigma }}_s=({\mathrm{\Sigma }}_s\cap {\mathrm{\Sigma }}_1)\bigsqcup ({\mathrm{\Sigma }}_s\cap {\mathrm{\Sigma }}_2)$ with both parts nonempty and orthogonal, and ${\mathrm{\Sigma }}_s$ is reducible.

Every root of ${\mathrm{\Sigma }}_s$ is a $W({\mathrm{\Sigma }}_s)$-conjugate of a simple root: it suffices to treat $\gamma \in {\mathrm{\Sigma }}_s^{+}$ and to induct on the height $h=\mathrm{ht}(\gamma )$, the negative case following from $-\delta =s_{\delta }(\delta )$; if $\gamma \in {\mathrm{\Delta }}_s$ there is nothing to prove, so suppose $\gamma \mathrm{\notin }{\mathrm{\Delta }}_s$, write $\gamma ={\sum }_i{n}_i{\alpha }_i$ over ${\mathrm{\Delta }}_s$ with $n_i\ge 0$ integers and $h={\sum }_i{n}_i\ge 2$ by [L4], and note that $0<(\gamma ,\gamma )={\sum }_i{n}_i(\gamma ,{\alpha }_i)$ gives an index $i$ with $(\gamma ,{\alpha }_i)>0$, which forces $n_i\ge 1$ because $({\alpha }_j,{\alpha }_i)\le 0$ for $j\mathrm{\ne }i$ by [L4]; with $n=n(\gamma ,{\alpha }_i)=2(\gamma ,{\alpha }_i)({\alpha }_i,{\alpha }_i{)}^{-1}\ge 1$ one has $n\le 2n_i\le 2h$ and $n=2n_i=2h$ would force $\gamma =n_i{\alpha }_i$ with $n_i=h$, hence $h=1$ by reducedness of ${\mathrm{\Sigma }}_s$ and $\gamma ={\alpha }_i\in {\mathrm{\Delta }}_s$, a contradiction, so $\mathrm{\mid }h-n\mathrm{\mid }<h$; the root $\beta =s_{{\alpha }_i}(\gamma )=\gamma -n{\alpha }_i$ has height $h-n$, so either $\beta$ or $-\beta$ is a positive root of smaller height, which by induction is a $W({\mathrm{\Sigma }}_s)$-conjugate of a simple root, and since $\gamma =s_{{\alpha }_i}(\beta )$ and $\gamma =s_{{\alpha }_i}{s}_{-\beta }(-\beta )$ in the respective cases, $\gamma$ is such a conjugate too.

For $\alpha \in \mathrm{\Psi }$ and $\beta \in {\mathrm{\Sigma }}_s$ one has $⟨\beta ,\alpha ⟩⟨\alpha ,\alpha {⟩}^{-1}\in \mathbb{Z}$: indeed the pair $(\beta ,2\alpha )$ consists of two elements of $\mathrm{\Sigma }$, so step 4.3 applied to it gives $2⟨\beta ,2\alpha ⟩\mathrm{\mid }⟨2\alpha ,2\alpha ⟩{\mathrm{\mid }}^{-1}\in \mathbb{Z}$, which is the displayed number.

Suppose $\mathrm{\Sigma }$ is irreducible and nonreduced, so that ${\mathrm{\Sigma }}_s$ is irreducible by step 7.2 and $\mathrm{\Psi }\mathrm{\ne }\mathrm{\varnothing }$ by step 5.2; fix a positive system of ${\mathrm{\Sigma }}_s$ with simple roots ${\mathrm{\Delta }}_s$; choosing $\alpha \in \mathrm{\Psi }$ and applying step 7.3 to the roots of ${\mathrm{\Sigma }}_s$ shows that some $\beta \in {\mathrm{\Delta }}_s$ satisfies $2\beta \in \mathrm{\Sigma }$, that is $\beta \in \mathrm{\Psi }\cap {\mathrm{\Delta }}_s$, because $\mathrm{\Psi }$ is $W({\mathrm{\Sigma }}_s)$-invariant by step 7.1.

The rank-one case: if $\mathfrak{a}$ is one-dimensional, then ${\mathrm{\Sigma }}_s=\{\pm \alpha \}$ for some $\alpha$ by step 5.2 and step 6.1, so $W({\mathrm{\Sigma }}_s)=\{1,s_{\alpha }\}$ acts transitively on ${\mathrm{\Sigma }}_s$; by step 7.1 the nonempty set $\mathrm{\Psi }$ is $W({\mathrm{\Sigma }}_s)$-invariant, hence $\mathrm{\Psi }={\mathrm{\Sigma }}_s$ and $\mathrm{\Sigma }=\{\pm \alpha ,\pm 2\alpha \}$ by step 5.2; then the linear map sending $\alpha$ to $e_1$ and $2\alpha$ to $2e_1$ carries $\mathrm{\Sigma }$ onto $B{C}_1=\{\pm e_1,\pm 2e_1\}$ and preserves all Cartan integers, since $2⟨e_1,2e_1⟩\mathrm{\mid }2e_1{\mathrm{\mid }}^{-2}=1$ and $2⟨2e_1,e_1⟩\mathrm{\mid }{e}_1{\mathrm{\mid }}^{-2}=4$ match the corresponding numbers computed in ${\mathfrak{a}}^{\ast }$.

Suppose from now on that $\dim \mathfrak{a}\ge 2$ and let $\beta \in \mathrm{\Psi }\cap {\mathrm{\Delta }}_s$ be as in step 8.1; if $\gamma \in {\mathrm{\Delta }}_s$, $\gamma \mathrm{\ne }\beta$, satisfies $(\gamma ,\beta )\mathrm{\ne }0$, then $(\gamma ,\beta )<0$ and, in the convention $n_{\alpha \delta }=2(\delta ,\alpha )\mathrm{/}(\alpha ,\alpha )$ of [L4], one has $n_{\beta \gamma }=2(\gamma ,\beta )(\beta ,\beta {)}^{-1}\in \{-1,-2,-3\}$. This integer is even by step 7.4, so $n_{\beta \gamma }=-2$ and [L4] gives $n_{\gamma \beta }=-1$ and $(\gamma ,\gamma )=2(\beta ,\beta )$; hence every neighbour of $\beta$ in the Dynkin diagram is longer than $\beta$ and is joined to $\beta$ by a double edge, and since the diagram has at most one multiple edge by [L5] the vertex $\beta$ has degree one; applying [L5] once more, the diagram of ${\mathrm{\Sigma }}_s$ is a path with exactly one multiple edge, which is the double edge at the end $\beta$, all other edges being simple, and the double edge carries the arrow pointing to the shorter root $\beta$: indeed the remaining configurations of [L5] with a multiple edge are a double edge in the middle of a four-vertex path and a two-vertex triple edge, and neither can occur here because every edge at $\beta$ is a double edge while $\beta$ has degree one.

All simple roots of ${\mathrm{\Sigma }}_s$ other than $\beta$ have length $(\gamma ,\gamma )=2(\beta ,\beta )$: the neighbour $\gamma$ of $\beta$ does by step 9.1, and every further vertex is reached along a path of simple edges, along which the length ratio is $1$ by [L4]; consequently the Cartan matrix of ${\mathrm{\Sigma }}_s$ with respect to the base ${\mathrm{\Delta }}_s$ is that of the standard type-$B_r$ base ${\alpha }_i=e_i-e_{i+1}$ $(i<r)$, ${\alpha }_r=e_r$, as computed in [L5]: diagonal entries $2$, entries $-1$ between consecutive non-final simple roots, entry $n_{{\alpha }_{r-1}{\alpha }_r}=-1$ and $n_{{\alpha }_r{\alpha }_{r-1}}=-2$ along the double edge, and $0$ otherwise; since a based root system is determined up to isomorphism by its Cartan matrix by [L5], the linear isomorphism carrying ${\mathrm{\Delta }}_s$ to that base is an isomorphism of ${\mathrm{\Sigma }}_s$ onto $B_r$.

$\mathrm{\Psi }$ is exactly the set of short roots of ${\mathrm{\Sigma }}_s$: $\mathrm{\Psi }$ is $W({\mathrm{\Sigma }}_s)$-invariant by step 7.1, and under the isomorphism of step 10.1 the roots of ${\mathrm{\Sigma }}_s$ correspond to $B_r=\{\pm e_i\}\cup \{\pm e_i\pm e_j\}$ with $\beta$ corresponding to a short root $e_r$; since $W(B_r)$ is the group of signed permutations and acts transitively on $\{\pm e_i\}$ by [L5], $\mathrm{\Psi }$ contains the whole short class; conversely no long root $e_i\pm e_j$ lies in $\mathrm{\Psi }$, because for such a root $\alpha$, the short root $e_i$ satisfies $(\alpha ,\alpha )=2(e_i,e_i)$ and $\mathrm{\mid }(e_i,\alpha )\mathrm{\mid }=(e_i,e_i)$, so $(e_i,\alpha )(\alpha ,\alpha {)}^{-1}=\pm 1\mathrm{/}2\mathrm{\notin }\mathbb{Z}$, contradicting step 7.4.

Conclusion of (c): by steps 10.1 and 11.1 there is a linear isomorphism $\phi$ of ${\mathfrak{a}}^{\ast }$ onto ${\mathbb{R}}^r$ carrying ${\mathrm{\Sigma }}_s$ onto $B_r=\{\pm e_i\}\cup \{\pm e_i\pm e_j\}$ and $\mathrm{\Psi }$ onto $\{\pm e_i\}$; therefore $\phi$ carries $\mathrm{\Sigma }={\mathrm{\Sigma }}_s\bigsqcup 2\mathrm{\Psi }$ onto $B_r\cup 2\{\pm e_i\}=B{C}_r$; and $\phi$ preserves Cartan integers: on pairs of roots of ${\mathrm{\Sigma }}_s$ this holds by the definition of a root-system isomorphism, and for pairs involving a doubled root $2\alpha$ with $\alpha \in \mathrm{\Psi }$ one computes $2⟨2\alpha ,\beta ⟩\mathrm{\mid }\beta {\mathrm{\mid }}^{-2}=2\cdot 2⟨\alpha ,\beta ⟩\mathrm{\mid }\beta {\mathrm{\mid }}^{-2}$ and $2⟨2\alpha ,2\beta ⟩\mathrm{\mid }2\beta {\mathrm{\mid }}^{-2}=2⟨\alpha ,\beta ⟩\mathrm{\mid }\beta {\mathrm{\mid }}^{-2}$, which are the corresponding Cartan integers in $B{C}_r$ because $\alpha ,\beta$ correspond to short roots $\pm e_i$ while $2\alpha ,2\beta$ correspond to the doubled roots $\pm 2e_i$; combined with the rank-one case of step 8.2 this proves (c) for all $\dim \mathfrak{a}\ge 1$.

Statements (a), (b) and (c) are now proved: finiteness, spanning and the reflection and integrality properties of $\mathrm{\Sigma }$ in steps 1.3, 4.3 and 5.1, together with the realisation of $s_{\lambda }$ in $N_K(\mathfrak{a})$ in step 5.1, give (a); the explicit nonreduced restricted-root system of the pair $(\mathfrak{s}\mathfrak{u}(2,1),\mathbb{R}H)$ in step 4.2 gives (b); and the classification of the irreducible nonreduced case in step 12.1 completes (c). The Axiom of Choice was used only through the representation theory of [L6] in step 4.3.