Classification of irreducible root systems

Suppose first that all edges of $\mathrm{\Gamma }$ are simple and there is no trivalent vertex. Then by [L1] the graph is a path on $n\ge 1$ vertices, and the Cartan matrix is the $A_n$ matrix $a_{ii}=2$, $a_{i,i+1}=a_{i+1,i}=-1$, $a_{ij}=0$ for $\mathrm{\mid }i-j\mathrm{\mid }\ge 2$; the corresponding root system is $A_n$.

If $\mathrm{\Gamma }$ is simply laced with exactly one trivalent vertex, write its arms as $p-1,q-1,r-1$ edges with $2\le p\le q\le r$; by [L1] $1\mathrm{/}p+1\mathrm{/}q+1\mathrm{/}r>1$. If $p\ge 3$ then $1\mathrm{/}p+1\mathrm{/}q+1\mathrm{/}r\le 3\cdot \frac{1}{3}=1$, so $p=2$; then $1\mathrm{/}q+1\mathrm{/}r>1\mathrm{/}2$, so $q\le 3$: for $q=2$ every $r\ge 2$ occurs, giving the diagrams $D_{r+2}$ with arms $1,1,r-1$, and for $q=3$ the condition $1\mathrm{/}r>1\mathrm{/}6$ gives $r=3,4,5$, the diagrams $E_6,E_7,E_8$. No other simply-laced trivalent diagrams occur.

If $\mathrm{\Gamma }$ has a multiple edge, then by [L1] its underlying graph is a path and the possibilities are: a double edge at an end, which gives the two orientation choices $B_n$ and $C_n$ on $n\ge 2$ vertices (the double edge being the end edge of the path); a central double edge on exactly four vertices, which is $F_4$; or a two-vertex triple edge, which is $G_2$.

The diagram types are distinguished by rank, edge multiplicities and positions, and, for a simply-laced branch, the unordered arm lengths. At rank $n\ge 4$ the $D_n$ arms $(1,1,n-3)$ differ from the exceptional arms, which all have just one arm of length $1$. Reversing the path interchanges the two orientations for $F_4$ and for $G_2$, so these introduce no extra types. To check unbased uniqueness it remains to explain why isomorphisms preserve diagram types. In particular $B_n$ and $C_n$ for $n\ge 3$ are non-isomorphic even as unbased root systems. If an isomorphism $\phi :B_n\to C_n$ existed, the image of a chosen base ${\mathrm{\Delta }}_B$ would be a base. Indeed it is a basis of roots, every root has integral coefficients of one sign relative to it because this is true relative to ${\mathrm{\Delta }}_B$ and $\phi$ is linear, and a vector pairing positively with every member of $\phi ({\mathrm{\Delta }}_B)$ therefore defines the corresponding positive system, whose indecomposable roots are precisely those basis vectors. By [L4] a Weyl element of $C_n$ carries $\phi ({\mathrm{\Delta }}_B)$ to the standard base ${\mathrm{\Delta }}_C$. After ordering the bases, the composite based isomorphism would identify their Cartan matrices up to a simultaneous row-and-column permutation, because it preserves every Cartan integer. But for $n\ge 3$ the $B_n$ and $C_n$ matrices are transposes and no vertex permutation identifies them: the unique double edge fixes its end of the path, while its arrow is reversed. This contradiction proves non-isomorphism. For $n=2$ the two orientations of the single double edge are interchanged by permuting the two vertices, so [L2] gives $B_2\cong C_2$; and for $n=1$ the unique reduced rank-one system $\{\pm \alpha \}$ is simultaneously $A_1$, $B_1$ and $C_1$.

For the low-rank $D$ coincidences, use the coordinate set $D_n=\{\pm e_i\pm e_j:i<j\}$ at $n=2,3$. These are root systems: a reflection in $e_i-e_j$ exchanges coordinates $i,j$, and one in $e_i+e_j$ exchanges them and negates both, preserving this set. Every root has squared norm $2$, pairwise inner products are integers, reducedness is immediate, and the displayed roots span. For $D_3$ put $a=e_1-e_2$, $b=e_2-e_3$, $c=e_2+e_3$. Its roots are exactly the positives and negatives of $a,b,c,a+b,a+c,a+b+c\mathrm{.}$ The vector $(2,1,0)$ pairs positively with all three basis vectors, and this list shows they are exactly the simple positive roots. Their squared lengths are $2$, with $(a,b)=(a,c)=-1$ and $(b,c)=0$, so the diagram is the three-vertex path $b-a-c$, giving $D_3\cong A_3$ by [L2]. In $D_2$, the two root lines generated by $e_1+e_2$ and $e_1-e_2$ are orthogonal, each containing just a pair of opposite roots, giving $A_1\bigsqcup A_1$. In particular arms $(1,1,1)$ describe $D_4$, not $D_3$.

Combining steps 1.1, 1.2 and 1.3, every irreducible system has one of the listed diagrams, and by [L2] its isomorphism class is determined by the Cartan matrix, hence by that diagram; so the classification list is complete, and steps 1.4–1.5 give the stated low-rank coincidences and uniqueness. Finally, if $\mathrm{\Phi }=\mathrm{\varnothing }$ then $E=0$ by [L5], and no decomposition into two nonzero orthogonal spaces exists, so it is the additional irreducible rank-zero case under the local definition. It is not one of the positive-rank types in the display.