Shape restrictions on Dynkin diagrams

The graph has no cycle: if $i_1,\dots ,i_m$ with $m\ge 3$ formed a cycle, set $x_{i_k}=1$ and $x_j=0$ otherwise; then ${\sum }_j{x}_j^2=m$ and the adjacency sum equals $m$, since each of the $m$ cycle edges contributes $1$, so $\sum x^2\le {\sum }_{i\sim j}{x}_i{x}_j$, contradicting [L2] (a multiple edge in the cycle only increases the right side). Since the graph is connected by [L3], it is a tree by [L4].

No vertex has four neighbours: if $v$ had distinct neighbours $n_1,\dots ,n_4$, set $x_v=2$, $x_{n_k}=1$ and $x_j=0$ otherwise; then $\sum x^2=4+4=8$ and the four edges at $v$ contribute at least $4\cdot (2\cdot 1)=8$, so $\sum x^2\le {\sum }_{i\sim j}{x}_i{x}_j$, contradicting [L2].

(Simply-laced trivalent case.) Suppose all edges are simple and $\delta$ is the unique trivalent vertex, its arms having $p-1,q-1,r-1$ edges with $p,q,r\ge 2$; all simple roots then have a common squared length $d^2$ by [L4], and adjacent simple roots have inner product $-\frac{1}{2}{d}^2$. Let $\alpha ={\sum }_{i=1}^{p-1}i{\alpha }_i$, where ${\alpha }_1,\dots ,{\alpha }_{p-1}$ are the roots of the first arm ordered from its free end toward $\delta$, and define $\beta ,\gamma$ similarly for the other two arms; the three vectors are mutually orthogonal because their supports are disjoint, and direct expansion using the adjacent inner products gives $\mathrm{\mid }\alpha {\mathrm{\mid }}^2=\frac{1}{2}p(p-1)d^2$, $\mathrm{\mid }\beta {\mathrm{\mid }}^2=\frac{1}{2}q(q-1)d^2$, $\mathrm{\mid }\gamma {\mathrm{\mid }}^2=\frac{1}{2}r(r-1)d^2$ and $(\alpha ,\delta )=-\frac{1}{2}(p-1)d^2$, with the analogous formulas for $\beta ,\gamma$. The set $\{\alpha ,\beta ,\gamma \}$ is orthogonal, and $\delta$ is not in its span (the supports are disjoint from $\delta$), so Bessel's inequality with the nonzero residual component gives $\mathrm{\mid }\delta {\mathrm{\mid }}^2>{\sum }_{u\in \{\alpha ,\beta ,\gamma \}}\frac{(u,\delta {)}^2}{\mathrm{\mid }u{\mathrm{\mid }}^2}=\frac{(p-1)d^2}{2p}+\frac{(q-1)d^2}{2q}+\frac{(r-1)d^2}{2r}$; dividing by $d^2=\mathrm{\mid }\delta {\mathrm{\mid }}^2$ and multiplying by $2$ gives $2>3-(\frac{1}{p}+\frac{1}{q}+\frac{1}{r})$, that is $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}>1$.

(Path with a unique double edge.) Suppose the underlying graph is a path with exactly one multiple edge, that edge is double, and its deletion splits the vertices into arms of $p$ and $q$ vertices. Every edge within either arm is then simple, so the roots on an arm have one common length by [L4]. Let $\alpha ={\sum }_{i=1}^{p}i{\alpha }_i$, $\beta ={\sum }_{j=1}^{q}j{\beta }_j$ with the vertices ordered from the free ends toward the double edge. From the double edge one has $a_{{\alpha }_p{\beta }_q}{a}_{{\beta }_q{\alpha }_p}=2$, so $2({\alpha }_p,{\beta }_q{)}^2=\mathrm{\mid }{\alpha }_p{\mathrm{\mid }}^2\mathrm{\mid }{\beta }_q{\mathrm{\mid }}^2$, while the simple-arm expansions give $\mathrm{\mid }\alpha {\mathrm{\mid }}^2=\frac{1}{2}p(p+1)\mathrm{\mid }{\alpha }_p{\mathrm{\mid }}^2$, $\mathrm{\mid }\beta {\mathrm{\mid }}^2=\frac{1}{2}q(q+1)\mathrm{\mid }{\beta }_q{\mathrm{\mid }}^2$ and $(\alpha ,\beta )=pq({\alpha }_p,{\beta }_q)$. Substituting into the strict Schwarz inequality $(\alpha ,\beta {)}^2<\mathrm{\mid }\alpha {\mathrm{\mid }}^2\mathrm{\mid }\beta {\mathrm{\mid }}^2$ for the nonproportional vectors $\alpha ,\beta$ gives $\frac{1}{2}{p}^2{q}^2<\frac{1}{4}p(p+1)q(q+1)$, hence $2pq<(p+1)(q+1)$ and $(p-1)(q-1)<2$. Therefore either $p=1$ or $q=1$, giving a double edge at an end of the path, or $p=q=2$, giving a four-vertex path with central double edge.

At most one vertex is trivalent: if $u\mathrm{\ne }v$ both had degree at least three, let $v_0=u,v_1,\dots ,v_k=v$ be the unique path between them (existing by step 1.1 and [L4]) and set $x=2$ at the path vertices and $x=1$ at every other neighbour of $u$ or $v$; the numbers $e_u=\mathrm{deg}(u)-1\ge 2$ and $e_v=\mathrm{deg}(v)-1\ge 2$ of such extra neighbours satisfy $\sum x^2=4(k+1)+e_u+e_v$ and ${\sum }_{i\sim j}{x}_i{x}_j\ge 4k+2(e_u+e_v)$ (the $k$ path edges contribute $4$ each, the edges from $u,v$ to the extra neighbours contribute $2$ each, and the edge $uv$, when $k=1$, contributes $4$), so $\sum x^2=4k+4+e_u+e_v\le 4k+2(e_u+e_v)\le {\sum }_{i\sim j}{x}_i{x}_j$ because $e_u+e_v\ge 4$; this contradicts [L2].

(Multiple edges and the conclusion.) First exclude two multiple edges. If $\mathrm{\Gamma }$ had two multiple edges, choose such a pair joined by a path with the fewest edges; every internal edge of that path is then simple, by minimality. Label only the vertices of that path and give $0$ to every other vertex: every edge of $\mathrm{\Gamma }$ not on the path then has a vertex labelled $0$ and contributes nothing to either side, so a violation of [L2] on the labelled sub-path is a violation for $\mathrm{\Gamma }$. Let the path be $y_0,\dots ,y_n$, with the multiple edges $\{y_0,y_1\}$ and $\{y_{n-1},y_n\}$ and with $\sqrt{a_{y_0{y}_1}{a}_{y_1{y}_0}},\sqrt{a_{y_{n-1}{y}_n}{a}_{y_n{y}_{n-1}}}\in \{\sqrt{2},\sqrt{3}\}$. If $n=2$, take $x_{y_1}=1$, $x_{y_0}=x_{y_2}=\frac{\sqrt{2}}{2}$ when both factors are $\sqrt{2}$, and $x=1$ at all three vertices as soon as one factor is $\sqrt{3}$. If $n\ge 3$, take $x_{y_0}=x_{y_n}=\frac{\sqrt{2}}{2}$, $x=1$ at the remaining path vertices when both factors are $\sqrt{2}$, and $x=1$ at all path vertices as soon as one factor is $\sqrt{3}$. In the two double-edge cases both sides of [L2] equal $n$ (with $n=2$ in the first case), and in the mixed and triple cases ${\sum }_i{x}_i^2=n+1<\sqrt{2}+\sqrt{3}+n-2\le {\sum }_{i\sim j}\sqrt{a_{ij}{a}_{ji}}{x}_i{x}_j$; either way [L2] fails for a nonzero label vector. Hence $\mathrm{\Gamma }$ has at most one multiple edge. If $\mathrm{\Gamma }$ has a multiple edge $\{u,v\}$ and is not a path, then it has exactly one trivalent vertex $w$ by steps 1.1, 1.2 and 2.1, and the path $w=w_0,w_1,\dots ,w_k=u$ to the endpoint $u$ of that edge consists of simple edges. Label $x_w=1$, $x_b=x_c=\frac{1}{2}$ at the two neighbours $b,c$ of $w$ outside that path, $x_{w_i}=1$ for $1\le i\le k$, $x_v=t$, and $x=0$ at every other vertex. Then ${\sum }_i{x}_i^2=k+\frac{3}{2}+t^2$ and ${\sum }_{i\sim j}\sqrt{a_{ij}{a}_{ji}}{x}_i{x}_j=k+1+\sqrt{m}t$ with $m=a_{uv}{a}_{vu}\in \{2,3\}$, so the difference of the two sides is $t^2-\sqrt{m}t+\frac{1}{2}$; this vanishes at $t=\frac{\sqrt{2}}{2}$ for $m=2$ and equals $-\frac{1}{4}$ at $t=\frac{\sqrt{3}}{2}$ for $m=3$. Again [L2] fails, so $\mathrm{\Gamma }$ is a path. Finally, a path with a multiple edge has exactly one such edge. If its multiplicity is $3$ and the path has a third vertex adjacent to the triple edge, the label vector $\frac{\sqrt{3}}{2},1,\frac{1}{2}$ on the far endpoint of the triple edge, its other endpoint and that third vertex satisfies ${\sum }_i{x}_i^2={\sum }_{i\sim j}\sqrt{a_{ij}{a}_{ji}}{x}_i{x}_j$, contradicting [L2]; so a triple edge fills the whole path, which is then the two-vertex system $G_2$. If the multiple edge is double, step 1.4 gives $(p-1)(q-1)<2$ for the two arms of $p$ and $q$ vertices, so either one arm is a single vertex (a double edge at an end of the path) or $p=q=2$ (a four-vertex path with central double edge). This completes the verification of all five assertions.