[proofplan] We use the positive definiteness of the symmetrized Cartan form. First, every two-vertex principal subdiagram is positive definite, which bounds the Cartan product $a_{ij}a_{ji}$ by $3$. A triple bond cannot have any additional adjacent vertex, because the corresponding three-vertex symmetrized principal matrix has non-positive determinant. For double bonds, we remove the double edge and study the two simply-laced rooted components on either side; a Schur complement computation reduces positive definiteness to a numerical inequality on rooted inverse Cartan entries, whose solutions are exactly the end double chains $B_\ell$, $C_\ell$, and the middle double chain $F_4$. [/proofplan] [step:Bound every bond multiplicity by positive definiteness of rank two principal subdiagrams] Let $A = (a_{ij})_{1 \leq i,j \leq \ell}$ be the generalized Cartan matrix of $\Gamma$. Since $\Gamma$ is finite, $A$ is symmetrizable and its symmetrization is positive definite. In particular, every principal submatrix of $A$ is positive definite after the same diagonal symmetrization. Fix distinct vertices $i$ and $j$. The principal Cartan submatrix on $\{i,j\}$ is \begin{align*} A_{\{i,j\}} = \begin{pmatrix} 2 & a_{ij}\\ a_{ji} & 2 \end{pmatrix}. \end{align*} Positive definiteness of this rank two principal subsystem gives \begin{align*} \det A_{\{i,j\}} = 4 - a_{ij}a_{ji} > 0. \end{align*} If $i$ and $j$ are joined by an edge, then $a_{ij}$ and $a_{ji}$ are negative integers, so $m_{ij} = a_{ij}a_{ji}$ is a positive integer. Therefore \begin{align*} m_{ij} \in \{1,2,3\}. \end{align*} Thus every multiple bond has multiplicity $2$ or $3$. [guided] The finite-type hypothesis means that the Cartan form is positive definite. This has an immediate local consequence: every principal subdiagram must also be positive definite, because restricting a positive definite quadratic form to a coordinate subspace preserves positive definiteness. Take two distinct vertices $i$ and $j$. Their rank two principal Cartan matrix is \begin{align*} A_{\{i,j\}} = \begin{pmatrix} 2 & a_{ij}\\ a_{ji} & 2 \end{pmatrix}. \end{align*} A positive definite $2 \times 2$ matrix has positive determinant, so \begin{align*} 0 < \det A_{\{i,j\}} = 4 - a_{ij}a_{ji}. \end{align*} Hence \begin{align*} a_{ij}a_{ji} < 4. \end{align*} When $i$ and $j$ are connected by a bond, the Cartan axioms give $a_{ij},a_{ji} \in \mathbb{Z}_{<0}$. Therefore the product $m_{ij} = a_{ij}a_{ji}$ is a positive integer strictly smaller than $4$, so \begin{align*} m_{ij} \in \{1,2,3\}. \end{align*} The case $m_{ij}=1$ is a simple bond, and the multiple-bond cases are exactly $m_{ij}=2$ and $m_{ij}=3$. [/guided] [/step] [step:Exclude any vertex adjacent to an endpoint of a triple bond] Let $i$ and $j$ be joined by a triple bond, so $m_{ij}=3$. Choose a positive diagonal matrix $D$ such that $DA$ is symmetric and positive definite, as in the finite-type Cartan matrix criterion. Define the positive diagonal matrix $R$ by $R_{rr}=D_{rr}^{-1/2}$, and set \begin{align*} S := R(DA)R. \end{align*} This is a positive diagonal congruence of the positive definite matrix $DA$, so $S$ is positive definite and every principal submatrix of $S$ is positive definite. Moreover \begin{align*} S_{rr} = 2 \end{align*} for every vertex $r$, and, because $D_{rr}a_{rs}=D_{ss}a_{sr}$, \begin{align*} S_{rs} = \begin{cases} -\sqrt{m_{rs}}, & r \neq s \text{ and } r \text{ is joined to } s,\\ 0, & r \neq s \text{ and } r \text{ is not joined to } s. \end{cases} \end{align*} Suppose there is a third vertex $k$ adjacent to $i$ or $j$. Write \begin{align*} x := m_{ij}=3, \qquad y := m_{ik}, \qquad z := m_{jk}, \end{align*} where $y,z \in \{0,1,2,3\}$ and at least one of $y,z$ is positive. The principal submatrix of $S$ on $\{i,j,k\}$ is \begin{align*} S_{\{i,j,k\}} = \begin{pmatrix} 2 & -\sqrt{x} & -\sqrt{y}\\ -\sqrt{x} & 2 & -\sqrt{z}\\ -\sqrt{y} & -\sqrt{z} & 2 \end{pmatrix}, \end{align*} with the convention $\sqrt{0}=0$. Its determinant is \begin{align*} \det S_{\{i,j,k\}} &= 8 - 2x - 2y - 2z - 2\sqrt{xyz}. \end{align*} Since $x=3$ and $y+z \geq 1$, we get \begin{align*} \det S_{\{i,j,k\}} &= 2 - 2y - 2z - 2\sqrt{3yz} \leq 0. \end{align*} This contradicts positive definiteness of the principal submatrix $S_{\{i,j,k\}}$. Therefore neither endpoint of a triple bond is adjacent to any third vertex. Since $\Gamma$ is connected, a triple bond can occur only when $\ell=2$, and the rank two diagram with $m_{ij}=3$ is $G_2$. [guided] The goal is to show that a triple bond cannot sit inside a larger connected finite diagram. We use the symmetrized Cartan matrix because its entries record the bond multiplicities directly as square roots. Choose a positive diagonal matrix $D$ for which $DA$ is symmetric and positive definite, and define $R$ by $R_{rr}=D_{rr}^{-1/2}$. Then \begin{align*} S := R(DA)R \end{align*} is a positive diagonal congruence of $DA$. Hence $S$ is positive definite, and every principal submatrix of $S$ is positive definite. The congruence normalizes the diagonal entries to $2$, and for distinct vertices $r$ and $s$ it gives \begin{align*} S_{rs} = \begin{cases} -\sqrt{m_{rs}}, & r \text{ and } s \text{ are joined},\\ 0, & r \text{ and } s \text{ are not joined}. \end{cases} \end{align*} Assume that $i$ and $j$ are joined by a triple bond, so $m_{ij}=3$. If a third vertex $k$ is adjacent to either endpoint, define \begin{align*} x := m_{ij}=3, \qquad y := m_{ik}, \qquad z := m_{jk}. \end{align*} At least one of $y$ and $z$ is positive. The three-vertex principal matrix is \begin{align*} S_{\{i,j,k\}} = \begin{pmatrix} 2 & -\sqrt{x} & -\sqrt{y}\\ -\sqrt{x} & 2 & -\sqrt{z}\\ -\sqrt{y} & -\sqrt{z} & 2 \end{pmatrix}. \end{align*} Expanding the determinant gives \begin{align*} \det S_{\{i,j,k\}} &= 8 - 2x - 2y - 2z - 2\sqrt{xyz}. \end{align*} Substituting $x=3$ yields \begin{align*} \det S_{\{i,j,k\}} &= 2 - 2y - 2z - 2\sqrt{3yz}. \end{align*} Because $y,z \geq 0$ and $y+z \geq 1$, the right-hand side is at most $0$. This contradicts positive definiteness, which requires every principal determinant to be positive. Thus a triple bond is isolated from all other vertices. Since $\Gamma$ is connected, the whole diagram has exactly the two vertices of the triple bond. The unique rank two finite Dynkin diagram with Cartan product $3$ is $G_2$. [/guided] [/step] [step:Show that the underlying graph is a tree] We next record that the underlying unoriented graph of $\Gamma$ has no cycle. Suppose a cycle has vertices $v_1,\dots,v_r$ in cyclic order, where $r \geq 3$. Define a vector $u \in \mathbb{R}^{\ell}$ by \begin{align*} u_{v_s} &= 1 \quad \text{for } 1 \leq s \leq r,\\ u_t &= 0 \quad \text{for every other vertex } t. \end{align*} For each edge on the cycle, the corresponding off-diagonal entry of $S$ is at most $-1$. Therefore \begin{align*} u^\top S u &= 2r + 2\sum_{\{p,q\} \subset \{v_1,\dots,v_r\},\, p<q} S_{pq}\\ &\leq 2r - 2r\\ &= 0. \end{align*} If additional edges occur among the cycle vertices, their contributions are non-positive and only decrease the value. This contradicts positive definiteness of $S$. Hence the connected graph $\Gamma$ is a tree. [/step] [step:Reduce a double bond to two rooted simply-laced components] Let $i$ and $j$ be joined by a double bond, so $m_{ij}=2$. By the previous step, deleting the edge $\{i,j\}$ separates the tree $\Gamma$ into two connected components. Let $P$ be the component containing $i$, rooted at $i$, and let $Q$ be the component containing $j$, rooted at $j$. First we show that this is the only multiple bond in rank at least three. By the triple-bond step, no triple bond occurs when $\ell \geq 3$, so any other multiple bond would also be double. Suppose another double bond exists. Since $\Gamma$ is a tree, the unique path between the two double bonds has vertices $v_0,\dots,v_N$ in order, where $N \geq 2$, and its first and last edges are double while every intermediate edge on this chosen path is simple. Let $M_N$ denote the symmetric principal matrix of $S$ on these $N+1$ vertices. Thus $M_N$ has diagonal entries $2$, off-diagonal entries $-\sqrt{2}$ on the first and last edges, and off-diagonal entries $-1$ on all intermediate edges. Let $D_N:=\det M_N$. Expanding successively along the last row and column gives the continuant recurrence \begin{align*} D_0 &= 2,\\ D_1 &= 2,\\ D_k &= 2D_{k-1}-D_{k-2} \quad \text{for } 2 \leq k \leq N-1,\\ D_N &= 2D_{N-1}-2D_{N-2}. \end{align*} The middle recurrence and the initial values give $D_k=2$ for every $0 \leq k \leq N-1$. Hence \begin{align*} D_N = 2\cdot 2 - 2\cdot 2 = 0. \end{align*} This contradicts positive definiteness of the principal submatrix $M_N$, whose determinant must be positive. Therefore a finite irreducible diagram of rank at least three contains at most one double bond. It follows that the components $P$ and $Q$ obtained by deleting the chosen double bond are simply-laced rooted trees. Let $C_P$ and $C_Q$ be the symmetric Cartan matrices of the simply-laced rooted trees $P$ and $Q$, with diagonal entries $2$ and off-diagonal entries $-1$ along edges. Define \begin{align*} \rho(P,i) := (C_P^{-1})_{ii}, \qquad \rho(Q,j) := (C_Q^{-1})_{jj}. \end{align*} With the vertices ordered so that the double bond is the only edge between $P$ and $Q$, the symmetrized Cartan matrix has block form \begin{align*} S = \begin{pmatrix} C_P & -\sqrt{2}\, e_i e_j^\top\\ -\sqrt{2}\, e_j e_i^\top & C_Q \end{pmatrix}, \end{align*} where $e_i$ and $e_j$ denote the coordinate vectors at the two roots. Since $C_P$ and $C_Q$ are positive definite, the [Schur complement criterion](/page/Schur%20Complement) gives \begin{align*} S > 0 \quad \Longleftrightarrow \quad 1 - 2\rho(P,i)\rho(Q,j) > 0. \end{align*} Thus \begin{align*} 2\rho(P,i)\rho(Q,j) < 1. \end{align*} [guided] After deleting a double edge from a tree, the graph splits into exactly two pieces. We remember the endpoint in each piece as a root: $P$ is rooted at $i$, and $Q$ is rooted at $j$. We must first justify that the chosen double edge is the only multiple edge in rank at least three. The triple-bond step already excludes triple bonds in this rank. Suppose, for contradiction, that there is a second double bond. Because the underlying graph is a tree, there is a unique path from one double bond to the other. Write its vertices as $v_0,\dots,v_N$ in order, with $N \geq 2$, so that the first and last edges are double and the intermediate edges on this chosen path are simple. Let $M_N$ be the symmetric principal matrix of $S$ restricted to these $N+1$ vertices. Its diagonal entries are $2$; its first and last off-diagonal edge entries are $-\sqrt{2}$; and every intermediate off-diagonal edge entry is $-1$. Define $D_N := \det M_N$. Expanding the determinant of this tridiagonal matrix along the last row and column gives \begin{align*} D_0 &= 2,\\ D_1 &= 2,\\ D_k &= 2D_{k-1}-D_{k-2} \quad \text{for } 2 \leq k \leq N-1,\\ D_N &= 2D_{N-1}-2D_{N-2}. \end{align*} The middle recurrence keeps the value constant: from $D_0=D_1=2$, induction gives $D_k=2$ for every $0 \leq k \leq N-1$. The final double edge therefore gives \begin{align*} D_N = 2\cdot 2 - 2\cdot 2 = 0. \end{align*} But $M_N$ is a principal submatrix of the positive definite matrix $S$, so it must itself be positive definite and hence must have positive determinant. This contradiction shows that no second double bond exists. The point of this reduction is now justified: the double edge is the only non-simply-laced interaction between the two rooted pieces. Each piece has its own simply-laced Cartan matrix: \begin{align*} C_P \quad \text{and} \quad C_Q. \end{align*} These matrices are positive definite because they are principal submatrices of the positive definite matrix $S$. Now order the vertices so that all vertices of $P$ come first and all vertices of $Q$ come second. The only off-block interaction is the double edge between the roots. Since a double edge has symmetrized off-diagonal entry $-\sqrt{2}$, the full matrix is \begin{align*} S = \begin{pmatrix} C_P & -\sqrt{2}\, e_i e_j^\top\\ -\sqrt{2}\, e_j e_i^\top & C_Q \end{pmatrix}. \end{align*} Here $e_i$ is the coordinate vector at the root $i$ inside $P$, and $e_j$ is the coordinate vector at the root $j$ inside $Q$. The [Schur complement criterion](/page/Schur%20Complement) applies because $C_P$ is positive definite. It says that $S$ is positive definite exactly when \begin{align*} C_Q - 2 e_j e_i^\top C_P^{-1} e_i e_j^\top \end{align*} is positive definite. Since $e_i^\top C_P^{-1}e_i = \rho(P,i)$, this Schur complement is \begin{align*} C_Q - 2\rho(P,i)e_j e_j^\top. \end{align*} Applying the one-dimensional Schur complement again inside $C_Q$ gives the equivalent inequality \begin{align*} 1 - 2\rho(P,i)\rho(Q,j) > 0. \end{align*} Thus the double bond is allowed precisely when \begin{align*} 2\rho(P,i)\rho(Q,j) < 1. \end{align*} This numerical inequality is the mechanism that forces the familiar finite list. [/guided] [/step] [step:Compute the rooted simply-laced possibilities allowed by the Schur complement inequality] We need the following rooted simply-laced calculation. If $T$ is a simply-laced finite rooted Dynkin tree with root $v$ and symmetric Cartan matrix $C_T$, define \begin{align*} \rho(T,v) := (C_T^{-1})_{vv}. \end{align*} By the [classification of simply-laced finite Dynkin diagrams](/page/Dynkin%20Diagram), $T$ is of type $A_r$, $D_r$, $E_6$, $E_7$, or $E_8$. For $A_r$ rooted at its $k$th vertex along the path, solving $C_{A_r}x=e_k$ gives \begin{align*} \rho(A_r,k)=\frac{k(r+1-k)}{r+1}. \end{align*} Thus $\rho(A_r,k)<1$ holds exactly when $k=1$ or $k=r$, namely when the root is an endpoint, and then \begin{align*} \rho(A_r,\text{endpoint})=\frac{r}{r+1}. \end{align*} For $D_r$ with the standard numbering in which vertices $1,\dots,r-2$ form the long chain and $r-1,r$ are the two fork endpoints, direct solution of $C_{D_r}x=e_v$ gives \begin{align*} \rho(D_r,k)=k \quad (1\leq k\leq r-2), \qquad \rho(D_r,r-1)=\rho(D_r,r)=\frac{r}{4}. \end{align*} Since $r\geq 4$, these values are all at least $1$. For the exceptional cases, direct inversion of the three finite Cartan matrices gives \begin{align*} \min_v \rho(E_6,v)=\frac{4}{3}, \qquad \min_v \rho(E_7,v)=\frac{3}{2}, \qquad \min_v \rho(E_8,v)=2. \end{align*} Consequently the only rooted simply-laced finite Dynkin trees with $\rho(T,v)<1$ are endpoint-rooted paths. Apply this to the rooted components $P$ and $Q$. Every rooted simply-laced finite Dynkin tree has $\rho\geq 1/2$, and the value $1/2$ occurs only for the one-vertex path $A_1$. If either component were not an endpoint-rooted path, the preceding calculation would give $\rho\geq 1$ on that side, and hence \begin{align*} 2\rho(P,i)\rho(Q,j) \geq 2\cdot 1\cdot \frac{1}{2}=1, \end{align*} contradicting \begin{align*} 2\rho(P,i)\rho(Q,j)<1. \end{align*} Therefore both $P$ and $Q$ must be endpoint-rooted paths. Write their ranks as $r$ and $s$. Then \begin{align*} \rho(P,i) = \frac{r}{r+1}, \qquad \rho(Q,j)=\frac{s}{s+1}. \end{align*} The Schur complement inequality becomes \begin{align*} 2\frac{r}{r+1}\frac{s}{s+1} < 1, \end{align*} equivalently \begin{align*} 2rs < (r+1)(s+1). \end{align*} Rearranging gives \begin{align*} (r-1)(s-1) < 2. \end{align*} Thus, up to swapping the two sides, either $r=1$ and $s$ is arbitrary, or $r=s=2$. [guided] We now solve the numerical problem produced by the Schur complement. The number \begin{align*} \rho(T,v) = (C_T^{-1})_{vv} \end{align*} measures how strongly the rooted simply-laced component $T$ interacts with a new edge attached at the root. We must compute which rooted simply-laced finite Dynkin trees can have $\rho(T,v)<1$, because the inequality \begin{align*} 2\rho(P,i)\rho(Q,j)<1 \end{align*} can only hold if both sides are sufficiently small. By the [classification of simply-laced finite Dynkin diagrams](/page/Dynkin%20Diagram), the connected simply-laced finite Dynkin trees are exactly $A_r$, $D_r$, $E_6$, $E_7$, and $E_8$. First consider $A_r$, whose Cartan matrix is the tridiagonal matrix with diagonal entries $2$ and adjacent off-diagonal entries $-1$. If the root is the $k$th vertex of the path, solving \begin{align*} C_{A_r}x=e_k \end{align*} gives \begin{align*} x_a= \begin{cases} \dfrac{a(r+1-k)}{r+1}, & 1\leq a\leq k,\\ \dfrac{k(r+1-a)}{r+1}, & k\leq a\leq r. \end{cases} \end{align*} Therefore \begin{align*} \rho(A_r,k)=x_k=\frac{k(r+1-k)}{r+1}. \end{align*} This value is less than $1$ exactly at the two endpoints $k=1$ and $k=r$, where \begin{align*} \rho(A_r,\text{endpoint})=\frac{r}{r+1}. \end{align*} At every interior vertex of a path, the same formula gives $\rho(A_r,k)\geq 1$. Now consider the branched simply-laced types. For $D_r$, use the standard numbering in which vertices $1,\dots,r-2$ form the long chain and vertices $r-1,r$ are attached to $r-2$. Solving the finite linear systems $C_{D_r}x=e_v$ gives \begin{align*} \rho(D_r,k)=k \quad (1\leq k\leq r-2), \qquad \rho(D_r,r-1)=\rho(D_r,r)=\frac{r}{4}. \end{align*} Since $r\geq 4$, all these diagonal inverse entries are at least $1$. For the exceptional diagrams, there are only three Cartan matrices to check; direct inversion gives \begin{align*} \min_v \rho(E_6,v)=\frac{4}{3}, \qquad \min_v \rho(E_7,v)=\frac{3}{2}, \qquad \min_v \rho(E_8,v)=2. \end{align*} Thus the complete rooted simply-laced calculation is: the only rooted simply-laced finite Dynkin trees with $\rho(T,v)<1$ are paths rooted at endpoints. Return to the two components $P$ and $Q$. The smallest possible value of $\rho$ is attained by the one-vertex endpoint-rooted path: \begin{align*} \rho(A_1,\text{endpoint})=\frac{1}{2}. \end{align*} If either side were not an endpoint-rooted path, the classification calculation above would give $\rho\geq 1$ on that side, while the other side has $\rho\geq 1/2$. Hence \begin{align*} 2\rho(P,i)\rho(Q,j) \geq 2\cdot 1\cdot \frac{1}{2}=1, \end{align*} contradicting the strict Schur complement inequality. Therefore both rooted components are paths rooted at endpoints. Let the two endpoint-rooted path lengths be $r$ and $s$. Then \begin{align*} \rho(P,i) = \frac{r}{r+1}, \qquad \rho(Q,j) = \frac{s}{s+1}. \end{align*} The condition for the full diagram to remain finite is \begin{align*} 2\frac{r}{r+1}\frac{s}{s+1} < 1. \end{align*} Multiplying by the positive denominator $(r+1)(s+1)$ gives \begin{align*} 2rs < rs+r+s+1. \end{align*} Equivalently, \begin{align*} rs-r-s < 1, \end{align*} or \begin{align*} (r-1)(s-1) < 2. \end{align*} For positive integers $r$ and $s$, this means that, up to swapping $r$ and $s$, either $r=1$ and $s$ is arbitrary, or $r=s=2$. [/guided] [/step] [step:Identify the remaining diagrams as $B_\ell$, $C_\ell$, and $F_4$] The preceding step gives two possible shapes. If one rooted component has rank $1$ and the other has rank $\ell-1$, then the double bond lies at one end of a chain of length $\ell$. According to the orientation of the double bond, this is the finite Dynkin diagram of type $B_\ell$ or $C_\ell$. If both rooted components have rank $2$, then the diagram is a four-vertex chain with the double bond in the middle: \begin{align*} A_2 \;-\!\!\Rightarrow\; A_2. \end{align*} This is the finite Dynkin diagram of type $F_4$. Since the reduction step proved that a rank at least three finite irreducible diagram has at most one double bond, these shapes exhaust all possible diagrams containing a double bond. Together with the exclusion of triple bonds outside rank two and the rank two bound $m_{ij}\in\{1,2,3\}$, this proves the theorem. [/step]