Rank-two root-system classification

For a linear subspace $V\subseteq E$ with $\mathrm{\Phi }\cap V\mathrm{\ne }\mathrm{\varnothing }$, the set $\mathrm{\Phi }\cap V$ is a reduced crystallographic root system in $\mathrm{span}(\mathrm{\Phi }\cap V)$: it is finite, contains no zero vector, inherits integrality and reducedness, and $s_{\gamma }(\mathrm{\Phi }\cap V)\subseteq \mathrm{\Phi }\cap V$ for $\gamma \in \mathrm{\Phi }\cap V$ because $s_{\gamma }$ preserves $\mathrm{\Phi }$ and maps $V$ into $V$. In particular the plane subsystem $\mathrm{\Phi }\cap \mathrm{span}(\alpha ,\beta )$ is a rank-two reduced crystallographic root system.

Both $n_{\alpha \beta }$ and $n_{\beta \alpha }$ are integers, and $n_{\alpha \beta }{n}_{\beta \alpha }=\frac{4(\alpha ,\beta {)}^2}{(\alpha ,\alpha )(\beta ,\beta )}=4{\cos }^2\theta \mathrm{.}$ Since $\alpha ,\beta$ are nonproportional, Cauchy-Schwarz gives $\mathrm{\mid }(\alpha ,\beta )\mathrm{\mid }<\mathrm{\mid }\alpha \mathrm{\mid }\mathrm{\mid }\beta \mathrm{\mid }$, so $0\le 4{\cos }^2\theta <4$; being a product of integers, $4{\cos }^2\theta$ is a nonnegative integer, hence lies in $\{0,1,2,3\}$. If $\mathrm{\mid }\alpha \mathrm{\mid }\ge \mathrm{\mid }\beta \mathrm{\mid }$ then $\mathrm{\mid }{n}_{\alpha \beta }\mathrm{\mid }=2\mathrm{\mid }\cos \theta \mathrm{\mid }\frac{\mathrm{\mid }\beta \mathrm{\mid }}{\mathrm{\mid }\alpha \mathrm{\mid }}\le 2\mathrm{\mid }\cos \theta \mathrm{\mid }\frac{\mathrm{\mid }\alpha \mathrm{\mid }}{\mathrm{\mid }\beta \mathrm{\mid }}=\mathrm{\mid }{n}_{\beta \alpha }\mathrm{\mid },$ and both absolute values cannot be at least $2$, since then their product would be at least $4$. Hence $\mathrm{\mid }{n}_{\alpha \beta }\mathrm{\mid }\in \{0,1\}$; moreover $n_{\alpha \beta }\mathrm{\ne }0$ if and only if $n_{\beta \alpha }\mathrm{\ne }0$, and then $n_{\alpha \beta }=\pm 1$, $n_{\beta \alpha }=\pm (n_{\alpha \beta }{n}_{\beta \alpha })$ with the same sign, and $\mathrm{\mid }\alpha {\mathrm{\mid }}^2\mathrm{/}\mathrm{\mid }\beta {\mathrm{\mid }}^2=n_{\beta \alpha }\mathrm{/}{n}_{\alpha \beta }$, because $n_{\beta \alpha }\mathrm{/}{n}_{\alpha \beta }=(\alpha ,\alpha )\mathrm{/}(\beta ,\beta )$.

Fix a vector $v\in E$ with $(v,\gamma )\mathrm{\ne }0$ for every $\gamma \in \mathrm{\Phi }$, which exists because $\mathrm{\Phi }$ is finite and $E$ is not a finite union of the proper subspaces ${\gamma }^{\perp }$ [L5]. Call $\gamma$ positive when $(v,\gamma )>0$ and negative otherwise, so that $\mathrm{\Phi }$ is the disjoint union of its positive and negative roots and the negative roots are the negatives of the positive ones; call a positive root simple when it is not a sum of two positive roots. If a positive root is not simple, write it as a sum of two positive roots; the value of $v$ on each summand is strictly smaller than on the sum, so iterating the decomposition and always decomposing a summand that is not simple terminates after finitely many steps (the values of $v$ on positive roots form a finite set and strictly decrease along the iteration). The terminal summands are simple, so every positive root is a sum of simple roots.

(Reducible case) If $\mathrm{\Phi }$ is a reducible rank-two root system then $\mathrm{\Phi }={\mathrm{\Phi }}_1\bigsqcup {\mathrm{\Phi }}_2$ with ${\mathrm{\Phi }}_i$ root systems spanning pairwise orthogonal nonzero subspaces $E_i$ with $E=E_1\oplus E_2$ [L4]; hence $\dim E_i=1$, and a rank-one reduced crystallographic root system is $\{\pm \gamma \}$ for its unique positive root $\gamma$, since every root lies on the line $\mathbb{R}\gamma$ and reducedness excludes proper multiples. Thus the reducible rank-two system is $A_1\bigsqcup A_1$.

With $\mathrm{\mid }\alpha \mathrm{\mid }\ge \mathrm{\mid }\beta \mathrm{\mid }$, the identity of step 1.2 enumerates the possibilities. If $n_{\alpha \beta }{n}_{\beta \alpha }=0$ then $(n_{\alpha \beta },n_{\beta \alpha })=(0,0)$ and $\theta ={90}^{\circ }$. If the product is $1$ then $n_{\alpha \beta }=n_{\beta \alpha }=\pm 1$, so $\mathrm{\mid }\alpha {\mathrm{\mid }}^2=\mathrm{\mid }\beta {\mathrm{\mid }}^2$, $4{\cos }^2\theta =1$, and $\theta ={60}^{\circ }$ if $\cos \theta >0$, $\theta ={120}^{\circ }$ if $\cos \theta <0$. If the product is $2$ then $n_{\alpha \beta }=\pm 1$, $n_{\beta \alpha }=\pm 2$ with the same sign, $\mathrm{\mid }\alpha {\mathrm{\mid }}^2=2\mathrm{\mid }\beta {\mathrm{\mid }}^2$, $4{\cos }^2\theta =2$, and the angle is ${45}^{\circ }$ or ${135}^{\circ }$ according to the sign of $\cos \theta$. If the product is $3$ then $n_{\alpha \beta }=\pm 1$, $n_{\beta \alpha }=\pm 3$, $\mathrm{\mid }\alpha {\mathrm{\mid }}^2=3\mathrm{\mid }\beta {\mathrm{\mid }}^2$, and the angle is ${30}^{\circ }$ or ${150}^{\circ }$.

Assume $(\alpha ,\beta )>0$. Then both Cartan integers are positive, so by step 1.2 the one attached to the longer of the two roots equals $1$: if $\mathrm{\mid }\alpha \mathrm{\mid }\le \mathrm{\mid }\beta \mathrm{\mid }$ then $n_{\beta \alpha }=1$ and $s_{\beta }(\alpha )=\alpha -n_{\beta \alpha }\beta =\alpha -\beta \in \mathrm{\Phi }$; if $\mathrm{\mid }\alpha \mathrm{\mid }\ge \mathrm{\mid }\beta \mathrm{\mid }$ then $n_{\alpha \beta }=1$ and $s_{\alpha }(\beta )=\beta -n_{\alpha \beta }\alpha =\beta -\alpha \in \mathrm{\Phi }$, so that $\alpha -\beta =-(\beta -\alpha )\in \mathrm{\Phi }$. Applying this to $-\alpha$ gives the companion statement: if $(\alpha ,\beta )<0$ then $\alpha +\beta \in \mathrm{\Phi }$.

If $\alpha ,\beta$ are distinct simple roots and $(\alpha ,\beta )>0$, then $\alpha -\beta \in \mathrm{\Phi }$ by step 2.2, and this root is positive or negative: if it is positive then $\alpha =(\alpha -\beta )+\beta$ exhibits $\alpha$ as a sum of two positive roots, and if it is negative then $\beta =(\beta -\alpha )+\alpha$ exhibits $\beta$ as a sum of two positive roots, contradicting simplicity in either case. Hence $(\alpha ,\beta )\le 0$. The same reasoning shows $\alpha -\beta \mathrm{\notin }\mathrm{\Phi }$, since a root $\alpha -\beta$ is positive, giving the first contradiction, or negative, giving the second.

(Root strings) Let $\alpha \in \mathrm{\Phi }$ and $\beta \in \mathrm{\Phi }$ with $\beta \mathrm{\ne }\pm \alpha$. Then the set of integers $k$ with $\beta +k\alpha \in \mathrm{\Phi }$ is a nonempty interval $\{-p,-p+1,\dots ,q\}$ of consecutive integers with $p\ge 0$, $q\ge 0$, no gaps, and $p-q=n_{\alpha \beta }$; moreover $p+q\le 3$. Indeed, the set is nonempty because $k=0$ occurs, and is invariant under $k\mapsto -n_{\alpha \beta }-k$ because $s_{\alpha }(\beta +k\alpha )=\beta -(n_{\alpha \beta }+k)\alpha \in \mathrm{\Phi }$, so it is finite and symmetric about $-n_{\alpha \beta }\mathrm{/}2$. If it had a gap, there would be $r<s-1$ with $\beta +r\alpha \in \mathrm{\Phi }$, $\beta +s\alpha \in \mathrm{\Phi }$ and $\beta +(r+1)\alpha ,\beta +(s-1)\alpha \mathrm{\notin }\mathrm{\Phi }$; then $(\beta +r\alpha ,\alpha )\ge 0$, since otherwise step 2.2 applied to $-\alpha$ would give $\beta +(r+1)\alpha \in \mathrm{\Phi }$, and similarly $(\beta +s\alpha ,\alpha )\le 0$; subtracting gives $(s-r)(\alpha ,\alpha )\le 0$, a contradiction. Hence there are no gaps, and the symmetry of an interval about $-n_{\alpha \beta }\mathrm{/}2$ gives $p-q=n_{\alpha \beta }$. Finally, replacing $\beta$ by $\beta +q\alpha$ reduces to the case $q=0$ and $p=n_{\alpha \beta }$, and $\mathrm{\mid }{n}_{\alpha \beta }\mathrm{\mid }\le 3$ by steps 1.2 and 2.1 applied to the nonproportional pair $(\beta +q\alpha ,\alpha )$; if that pair is proportional then reducedness gives at most three elements.

In a rank-two root system the simple roots are linearly independent, hence exactly two; explicitly, if ${\sum }_{i\in I}{c}_i{\alpha }_i={\sum }_{j\in J}{c}_j{\alpha }_j$ with disjoint finite index sets and positive real coefficients $c_i,c_j$ (which is the shape of every nontrivial linear relation), then for $\gamma ={\sum }_{i\in I}{c}_i{\alpha }_i\mathrm{\ne }0$ one computes $0<(\gamma ,\gamma )=({\sum }_{i\in I}{c}_i{\alpha }_i,{\sum }_{j\in J}{c}_j{\alpha }_j)={\sum }_{i\in I,j\in J}{c}_i{c}_j({\alpha }_i,{\alpha }_j)\le 0,$ because the two index sets are disjoint and distinct simple roots have nonpositive inner product by step 3.1. Therefore the simple roots are independent; since they span $E$ by step 1.3, a rank-two system has exactly two simple roots ${\alpha }_1,{\alpha }_2$, every root is $\pm (m{\alpha }_1+n{\alpha }_2)$ with $m,n\ge 0$ integers, and the Cartan matrix is one of $\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right),\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right),\left(\begin{array}{cc}2& -2\\ -1& 2\end{array}\right),\left(\begin{array}{cc}2& -1\\ -2& 2\end{array}\right),\left(\begin{array}{cc}2& -3\\ -1& 2\end{array}\right),\left(\begin{array}{cc}2& -1\\ -3& 2\end{array}\right),$ because both off-diagonal entries are nonpositive integers whose product is one of $0,1,2,3$.

(Descent and constraints for a base) Let ${\alpha }_1,{\alpha }_2$ be the simple roots of a rank-two system in the ordering of step 1.3, and let $\gamma =m{\alpha }_1+n{\alpha }_2$ be a positive root, $m,n\ge 0$ integers. Write $c=n_{{\alpha }_1{\alpha }_2}=2({\alpha }_1,{\alpha }_2)\mathrm{/}({\alpha }_1,{\alpha }_1)$ and $c^{\mathrm{\prime }}=n_{{\alpha }_2{\alpha }_1}=2({\alpha }_1,{\alpha }_2)\mathrm{/}({\alpha }_2,{\alpha }_2)$; both are nonpositive integers with product in $\{0,1,2,3\}$ by steps 3.1 and 1.2. Then: (a) $n_{{\alpha }_1\gamma }=2m+nc$ and $n_{{\alpha }_2\gamma }=2n+m{c}^{\mathrm{\prime }}$, and the reflected roots $s_{{\alpha }_1}\gamma =-(m+nc){\alpha }_1+n{\alpha }_2$ and $s_{{\alpha }_2}\gamma =m{\alpha }_1-(n+m{c}^{\mathrm{\prime }}){\alpha }_2$ again have coefficients of one sign; hence if $n\ge 1$ then $-(m+nc)\ge 0$, that is $m\le n\mathrm{\mid }c\mathrm{\mid }$, and if $m\ge 1$ then $n\le m\mathrm{\mid }{c}^{\mathrm{\prime }}\mathrm{\mid }$. (b) The string bounds of step 3.2 give $\mathrm{\mid }2m+nc\mathrm{\mid }\le 3$ and $\mathrm{\mid }2n+m{c}^{\mathrm{\prime }}\mathrm{\mid }\le 3$. (c) Reducedness gives: if $n=0$ then $m=1$, and if $m=0$ then $n=1$. (d) If $\gamma \mathrm{\ne }{\alpha }_1$ and $2m+nc\ge 1$, then the ${\alpha }_1$-string through $\gamma$ contains $(m-1){\alpha }_1+n{\alpha }_2$, so this vector lies in $\mathrm{\Phi }$ and $m\ge 1$; and if $\gamma \mathrm{\ne }{\alpha }_2$ and $2n+m{c}^{\mathrm{\prime }}\ge 1$, then $m{\alpha }_1+(n-1){\alpha }_2\in \mathrm{\Phi }$ and $n\ge 1$.

(The three irreducible cases) Let $\mathrm{\Phi }$ be an irreducible rank-two root system with simple roots ${\alpha }_1,{\alpha }_2$ and Cartan matrix as in step 4.1; exclude the first matrix, which gives a reducible system by step 1.4 (no positive root has both coefficients nonzero by (a) of step 4.2). For the five remaining cases define $K=\{m{\alpha }_1+n{\alpha }_2:(m,n)\in S\},S=\{\begin{array}{ll}\{(1,0),(0,1),(1,1)\},& (c,c^{\mathrm{\prime }})=(-1,-1),\\ \{(1,0),(0,1),(1,1),(2,1)\},& (c,c^{\mathrm{\prime }})=(-2,-1),\\ \{(1,0),(0,1),(1,1),(1,2)\},& (c,c^{\mathrm{\prime }})=(-1,-2),\\ \{(1,0),(0,1),(1,1),(2,1),(3,1),(3,2)\},& (c,c^{\mathrm{\prime }})=(-3,-1),\\ \{(1,0),(0,1),(1,1),(1,2),(1,3),(2,3)\},& (c,c^{\mathrm{\prime }})=(-1,-3)\mathrm{.}\end{array}$ Every element of $K$ is a root: ${\alpha }_1,{\alpha }_2$ are simple; ${\alpha }_1+{\alpha }_2$, ${\alpha }_1+2{\alpha }_2$, ${\alpha }_1+3{\alpha }_2$ are the images of ${\alpha }_1$ under the reflections $s_{{\alpha }_2}$, and ${\alpha }_1+{\alpha }_2$, $2{\alpha }_1+{\alpha }_2$, $3{\alpha }_1+{\alpha }_2$ are the images of ${\alpha }_2$ under $s_{{\alpha }_1}$; for $(c,c^{\mathrm{\prime }})=(-3,-1)$ one has $2{\alpha }_1+{\alpha }_2=s_{{\alpha }_1}({\alpha }_1+{\alpha }_2)$ and $3{\alpha }_1+2{\alpha }_2=s_{{\alpha }_2}(3{\alpha }_1+{\alpha }_2)$, and the last case is its mirror image. In each case $K$ has $3$, $4$, $4$, $6$, $6$ elements and spans $E$.

(Exhaustiveness) In each of the five cases of step 5.1, every positive root lies in $K$. Suppose not, and choose a positive root $\gamma =m{\alpha }_1+n{\alpha }_2$ of least height among the positive roots outside $K$; it is not simple, so by step 1.3 it is a sum of two positive roots of smaller heights, and by minimality of $h=m+n$ both summands lie in $K$. Hence $\gamma$ is a sum of two elements of $K$, and each such sum is either an element of $K$, or violates one of conditions (a)-(c) of step 4.2, or descends by (d) to such a sum, or is excluded by reducedness [L1]; the following complete lists of the coordinate pairs of the sums of two elements of $K$ verify this case by case. For $(c,c^{\mathrm{\prime }})=(-1,-1)$ the sums are $(2,0),(1,1),(2,1),(0,2),(1,2),(2,2)$: $(2,0)$ and $(0,2)$ violate (c), $(2,1)$ violates $m\le n\mathrm{\mid }c\mathrm{\mid }=n$ and $(1,2)$ violates $n\le m\mathrm{\mid }{c}^{\mathrm{\prime }}\mathrm{\mid }=m$ in (a), and $(2,2)=2({\alpha }_1+{\alpha }_2)$ is excluded by reducedness because ${\alpha }_1+{\alpha }_2\in K$ is a root. For $(c,c^{\mathrm{\prime }})=(-2,-1)$ the sums with $n\ge 1$ are $(2,0),(1,1),(2,1),(3,1),(0,2),(1,2),(2,2),(3,2),(4,2)$: $(2,0)$ and $(0,2)$ violate (c), $(3,1)$ violates $m\le 2n$ and $(1,2)$ violates $n\le m$ in (a), $(2,2)=2({\alpha }_1+{\alpha }_2)$ and $(4,2)=2(2{\alpha }_1+{\alpha }_2)$ are excluded by reducedness, and $(3,2)$ satisfies (a)-(c) but (d) applied to ${\alpha }_1$ gives $(2,2)\in \mathrm{\Phi }$, already excluded. The case $(c,c^{\mathrm{\prime }})=(-1,-2)$ is the mirror image with the two coordinates and the two simple roots interchanged. For $(c,c^{\mathrm{\prime }})=(-3,-1)$ the sums of two elements of $K$ are $(2,0),(1,1),(2,1),(3,1),(4,1),(4,2),(0,2),(1,2),(2,2),(3,2),(3,3),(4,3),(5,2),(5,3),(6,2),(6,3),(6,4)$: $(2,0)$ and $(0,2)$ violate (c), $(4,1)$ violates $m\le 3n$ and $(1,2)$ violates $n\le m$ in (a), $(5,2)$ and $(6,2)$ violate $\mathrm{\mid }2m-3n\mathrm{\mid }\le 3$ in (b), $(3,3)=3({\alpha }_1+{\alpha }_2)$, $(2,2)=2({\alpha }_1+{\alpha }_2)$, $(4,2)=2(2{\alpha }_1+{\alpha }_2)$ and $(6,3)=3(2{\alpha }_1+{\alpha }_2)$ are excluded by reducedness because ${\alpha }_1+{\alpha }_2$ and $2{\alpha }_1+{\alpha }_2$ lie in $K$ and are roots, $(5,3)$ descends by (d) applied to ${\alpha }_1$ to $(4,3)$, and $(4,3)$ and $(6,4)$ descend by (d) applied to ${\alpha }_2$ to $(4,2)$ and $(6,3)$, all already excluded, while $(1,1),(2,1),(3,1),(3,2)$ lie in $K$. The mirror case $(c,c^{\mathrm{\prime }})=(-1,-3)$ is handled by the same interchange of coordinates and simple roots. Thus no positive root lies outside $K$, so $\mathrm{\Phi }=K\bigsqcup (-K)$ in each of the five cases, and the irreducible rank-two root systems are exactly the systems with $3$, $4$, $4$, $6$, $6$ positive roots.

The systems of 3, 4 and 6 positive roots are the root systems traditionally called $A_2$, $B_2\cong C_2$ and $G_2$: for $(c,c^{\mathrm{\prime }})=(-1,-1)$ the roots are $\pm {\alpha }_1,\pm {\alpha }_2,\pm ({\alpha }_1+{\alpha }_2)$ with $\mathrm{\mid }{\alpha }_1\mathrm{\mid }=\mathrm{\mid }{\alpha }_2\mathrm{\mid }$ and angle ${120}^{\circ }$; for $(c,c^{\mathrm{\prime }})=(-2,-1)$ they are $\pm {\alpha }_1,\pm {\alpha }_2,\pm ({\alpha }_1+{\alpha }_2),\pm (2{\alpha }_1+{\alpha }_2)$ with $\mathrm{\mid }{\alpha }_2{\mathrm{\mid }}^2=2\mathrm{\mid }{\alpha }_1{\mathrm{\mid }}^2$ and angle ${135}^{\circ }$; and for $(c,c^{\mathrm{\prime }})=(-3,-1)$ they are the six positive roots listed in $K$ with $\mathrm{\mid }{\alpha }_2{\mathrm{\mid }}^2=3\mathrm{\mid }{\alpha }_1{\mathrm{\mid }}^2$ and angle ${150}^{\circ }$. The two middle cases are isomorphic as root systems: the linear map that rotates the plane by ${45}^{\circ }$ and then rescales uniformly sends the four short root directions and the four long root directions of the $(-2,-1)$ system onto those of the $(-1,-2)$ system, and Cartan integers are unchanged by a uniform rescaling. Combining with steps 2.1, 3.1, 1.4 and 6.1 gives the full rank-two classification, and the angle statement of (iv) is step 3.1.