[proofplan] We replace the generalized Cartan matrix by its symmetric positive definite form, whose off-diagonal entries record the edge multiplicities of the Dynkin diagram. Positivity of this form forces strong restrictions on principal subdiagrams: there are no cycles, no multiple branching patterns, and no forbidden weighted paths. The simply-laced case reduces to classifying positive definite trees, giving $A_n$, $D_n$, and $E_6,E_7,E_8$. The non-simply-laced case is controlled by determinant recurrences for weighted paths, giving only $B_n,C_n,F_4,G_2$, and direct determinant checks prove that every diagram in the list is finite type. [/proofplan] [step:Pass from the Cartan matrix to a symmetric quadratic form] Let $I$ be the finite vertex set of $\Delta$, and let $A=(a_{ij})_{i,j\in I}$ be its indecomposable generalized Cartan matrix. For distinct vertices $i,j\in I$, define \begin{align*} m_{ij}:=a_{ij}a_{ji}\in \mathbb{Z}_{\geq 0}. \end{align*} Thus $m_{ij}=0$ means that $i$ and $j$ are not adjacent, while $m_{ij}=1,2,3$ corresponds respectively to a single, double, or triple edge in a finite-type Dynkin diagram. Define the symmetric matrix $S=(s_{ij})_{i,j\in I}$ by \begin{align*} s_{ii}&:=2,\\ s_{ij}&:=-\sqrt{m_{ij}}\quad \text{for }i\neq j. \end{align*} Define the quadratic form $Q:\mathbb{R}^{I}\to \mathbb{R}$ by \begin{align*} Q(x):=\sum_{i,j\in I}s_{ij}x_ix_j. \end{align*} In this proof, a connected Dynkin diagram is called finite type when the symmetric form $S$ attached to its generalized Cartan matrix is positive definite; equivalently, $Q(x)>0$ for every nonzero vector $x\in\mathbb{R}^{I}$. This is the standard finite-type criterion for symmetrizable generalized Cartan matrices, obtained by replacing the Cartan matrix by its congruent symmetric [bilinear form](/page/Bilinear%20Form). By the [positive definiteness criterion by principal submatrices](/page/Positive%20Definite%20Matrix), every principal submatrix of a positive definite matrix is positive definite, so every connected full subdiagram of a finite-type diagram is again finite type. For a two-vertex subdiagram with edge multiplicity $m$, the associated matrix is \begin{align*} \begin{pmatrix} 2&-\sqrt m\\ -\sqrt m&2 \end{pmatrix}, \end{align*} whose determinant is $4-m$. Positivity gives $4-m>0$, hence \begin{align*} m\in \{0,1,2,3\}. \end{align*} Thus no edge of multiplicity at least $4$ can occur. [/step] [step:Exclude cycles in the underlying graph] Let $\Gamma$ be the underlying unoriented graph of $\Delta$. Suppose that $\Gamma$ contains a cycle with distinct vertices $i_1,\dots,i_k$, where $k\geq 3$, and edges $i_\ell i_{\ell+1}$ for $1\leq \ell<k$ and $i_ki_1$. Define $x\in\mathbb{R}^{I}$ by \begin{align*} x_i:= \begin{cases} 1,& i\in \{i_1,\dots,i_k\},\\ 0,& i\notin \{i_1,\dots,i_k\}. \end{cases} \end{align*} Then \begin{align*} Q(x) &=2k-2\sum_{\ell=1}^{k}\sqrt{m_{i_\ell i_{\ell+1}}}. \end{align*} Here $i_{k+1}:=i_1$. Since each edge on the cycle has multiplicity at least $1$, each $\sqrt{m_{i_\ell i_{\ell+1}}}\geq 1$, and therefore \begin{align*} Q(x)\leq 2k-2k=0. \end{align*} This contradicts positive definiteness because $x\neq 0$. Hence the underlying graph of a finite-type connected Dynkin diagram is a tree. [guided] The first structural restriction is that positive definiteness forbids cycles. Let $\Gamma$ denote the underlying graph obtained from $\Delta$ by forgetting orientations but keeping adjacency. Assume that $\Gamma$ contains a cycle with distinct vertices $i_1,\dots,i_k$, where $k\geq 3$. We test the quadratic form on the vector that is constant on this cycle and zero elsewhere: \begin{align*} x_i:= \begin{cases} 1,& i\in \{i_1,\dots,i_k\},\\ 0,& i\notin \{i_1,\dots,i_k\}. \end{cases} \end{align*} Only vertices on the cycle and edges between two cycle vertices contribute to $Q(x)$. Edges leaving the cycle do not contribute because the outside endpoint has coordinate $0$. Therefore \begin{align*} Q(x) &=2k-2\sum_{\ell=1}^{k}\sqrt{m_{i_\ell i_{\ell+1}}}, \end{align*} where $i_{k+1}:=i_1$. Every edge on the cycle has multiplicity at least $1$, so each weight satisfies $\sqrt{m_{i_\ell i_{\ell+1}}}\geq 1$. Hence \begin{align*} Q(x)\leq 2k-2k=0. \end{align*} But positive definiteness requires $Q(x)>0$ for every nonzero $x$. Since this test vector is nonzero, we obtain a contradiction. Thus the underlying graph must be a tree. [/guided] [/step] [step:Classify the simply-laced positive definite trees] Assume first that every edge of $\Delta$ is a single edge. Then $S=2I-M$, where $M$ is the adjacency matrix of the tree $\Gamma$. [claim:There is at most one branching vertex, and every branching vertex has degree three] [/claim] [proof] For a finite path with $p\geq 1$ vertices, let $T_p$ denote the $p\times p$ symmetric matrix with diagonal entries $2$ and off-diagonal entries $-1$ between adjacent vertices. Its leading determinants $d_p:=\det T_p$ satisfy \begin{align*} d_0&:=1,\qquad d_1:=2,\qquad d_p=2d_{p-1}-d_{p-2}\quad (p\geq 2), \end{align*} so $d_p=p+1$. [Cramer's rule](/theorems/3305) applied to the first diagonal entry gives \begin{align*} (T_p^{-1})_{11}=\frac{d_{p-1}}{d_p}=\frac{p}{p+1}. \end{align*} Let $v$ be a vertex of degree $d\geq 3$. Choose the full subdiagram consisting of $v$ and the $d$ branches issuing from $v$, with branch lengths $p_1,\dots,p_d\in\mathbb{N}$. Ordering the vertices with $v$ first, its symmetric matrix has block form \begin{align*} S_v= \begin{pmatrix} 2&-e_1^\top&-e_1^\top&\cdots&-e_1^\top\\ -e_1&T_{p_1}&0&\cdots&0\\ -e_1&0&T_{p_2}&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ -e_1&0&0&\cdots&T_{p_d} \end{pmatrix}, \end{align*} where $e_1\in\mathbb{R}^{p_\ell}$ denotes the first standard basis vector in the corresponding branch block. Since each $T_{p_\ell}$ is positive definite, the Schur complement criterion gives that $S_v$ is positive definite exactly when \begin{align*} 2-\sum_{\ell=1}^{d} e_1^\top T_{p_\ell}^{-1}e_1 &=2-\sum_{\ell=1}^{d}\frac{p_\ell}{p_\ell+1}\\ &=2-d+\sum_{\ell=1}^{d}\frac{1}{p_\ell+1} >0. \end{align*} Thus positivity of this principal submatrix requires \begin{align*} \sum_{\ell=1}^{d}\frac{1}{p_\ell+1}>d-2. \end{align*} Since $p_\ell\geq 1$ for each $\ell$, the left-hand side is at most $d/2$. If $d\geq 4$, then $d/2\leq d-2$, with equality only when $d=4$ and all $p_\ell=1$. The strict inequality is therefore impossible. Hence every branching vertex has degree three. It remains to exclude two branching vertices. Suppose that $u$ and $v$ are distinct degree-three vertices, and let the path from $u$ to $v$ contain $t\geq 1$ edges. Choose two terminal branches at $u$ not containing $v$ and two terminal branches at $v$ not containing $u$; let their lengths be $p_1,p_2,q_1,q_2\in\mathbb{N}$. The full subdiagram on these four branches and the path from $u$ to $v$ is a tree with two degree-three vertices. Define $x\in\mathbb{R}^{I}$ as follows: put $x_u=x_v=1$, put value $1$ on every internal vertex of the path from $u$ to $v$, and on a terminal branch of length $p$ attached to either branching vertex put the values \begin{align*} 1-\frac{1}{p+1},\ 1-\frac{2}{p+1},\ \dots,\ 1-\frac{p}{p+1} \end{align*} away from the branching vertex. Put $x_i=0$ outside this full subdiagram. Using \begin{align*} Q(x)=\sum_{\{i,j\}\text{ edge}}(x_i-x_j)^2+\sum_{i\in I}(2-\deg i)x_i^2 \end{align*} for simply-laced trees, the path from $u$ to $v$ contributes $0$ to the edge-square sum, each of the four terminal branches contributes $1/(p+1)$ or $1/(q+1)$ after its endpoint term is included, and the two degree-three vertices contribute $-1$ each through $(2-\deg i)x_i^2$. Hence \begin{align*} Q(x)=\frac{1}{p_1+1}+\frac{1}{p_2+1}+\frac{1}{q_1+1}+\frac{1}{q_2+1}-2\leq 0, \end{align*} because each of the four fractions is at most $1/2$. This contradicts positive definiteness of the principal submatrix on the chosen full subdiagram. Therefore there is at most one branching vertex. [/proof] If there is no branching vertex, the tree is a path, hence the diagram is $A_n$ for some $n\geq 1$. If there is a branching vertex, it has degree three. Let $p,q,r\in\mathbb{N}$ be the three branch lengths, ordered so that \begin{align*} 1\leq p\leq q\leq r. \end{align*} The Schur complement computation in the claim, now with $d=3$, proves that the symmetric matrix is positive definite exactly when \begin{align*} \frac{1}{p+1}+\frac{1}{q+1}+\frac{1}{r+1}>1. \end{align*} Indeed, the branch matrices $T_p,T_q,T_r$ are positive definite, and the remaining Schur complement is precisely this displayed quantity. If $p=1$ and $q=1$, this inequality holds for every $r\geq 1$, producing the diagrams $D_n$ with $n=r+3\geq 4$. If $p=1$ and $q=2$, then \begin{align*} \frac12+\frac13+\frac{1}{r+1}>1, \end{align*} so $r+1<6$, and hence $r=2,3,4$, producing $E_6,E_7,E_8$. If $q\geq 3$, then \begin{align*} \frac{1}{p+1}+\frac{1}{q+1}+\frac{1}{r+1}\leq \frac12+\frac14+\frac14=1, \end{align*} which violates strict positivity. Thus the simply-laced finite-type connected diagrams are exactly \begin{align*} A_n,\quad D_n,\quad E_6,\quad E_7,\quad E_8. \end{align*} [/step] [step:Restrict the possible multiple edges] Now suppose that at least one edge has multiplicity $2$ or $3$. Since the underlying graph is a tree, every full subdiagram supported on a path is again a weighted path. For a weighted path on vertices $1,\dots,n$, define $m_k\in\{1,2,3\}$ to be the multiplicity of the edge between $k$ and $k+1$. Let $\Delta_k$ denote the determinant of the leading $k\times k$ principal submatrix of its symmetric matrix $S$. Then \begin{align*} \Delta_0&:=1,\\ \Delta_1&:=2,\\ \Delta_k&=2\Delta_{k-1}-m_{k-1}\Delta_{k-2}\quad (k\geq 2). \end{align*} Positive definiteness requires $\Delta_k>0$ for every $k$. If a path contains two multiple edges, choose the shortest subpath containing both and order its vertices so that the first and last edges of the subpath are precisely the two multiple edges and every intervening edge, if any, is a single edge. For this ordered subpath, the initial determinants are $\Delta_0=1$ and $\Delta_1=2$. If the first multiple edge is double, then \begin{align*} \Delta_2=2\Delta_1-2\Delta_0=2. \end{align*} Thus $\Delta_1=\Delta_2=2$, and each intervening single edge preserves equality of consecutive determinants because $\Delta_{k}=2\Delta_{k-1}-\Delta_{k-2}$. At the second multiple edge, a double edge gives $\Delta_k=2c-2c=0$ and a triple edge gives $\Delta_k=2c-3c=-c<0$, where $c>0$ is the common preceding determinant. If the first multiple edge is triple, then \begin{align*} \Delta_2=2\Delta_1-3\Delta_0=1. \end{align*} A following single edge gives $\Delta_3=2\cdot 1-2=0$, while an adjacent second multiple edge gives $\Delta_3=2\cdot 1-m_2\cdot 2<0$ for $m_2\in\{2,3\}$. In every case the chosen principal subpath has a nonpositive leading principal determinant, contradicting positive definiteness by [Sylvester's criterion](/page/Sylvester%27s%20Criterion). Hence a finite-type connected diagram contains at most one multiple edge, and a triple edge can occur only in rank two. The triple-edge case is therefore exactly $G_2$. It remains to consider one double edge. We first exclude branching. Suppose that a branching vertex $v$ occurs. Since the underlying graph is a tree and there is only one double edge, choose the minimal full subtree containing $v$ and the double edge, and choose two terminal arms at $v$ not lying on the path from $v$ to the double edge. After truncating at the double edge, this full subtree has a central vertex $v$ and three arms: two simply-laced terminal arms of lengths $p,q\in\mathbb{N}$, and one arm ending in the unique double edge. Let $T_p$ be the simply-laced path matrix on $p$ vertices, so $(T_p^{-1})_{11}=p/(p+1)$ as computed above. For the weighted arm, let $R_r$ denote the tridiagonal matrix on the $r\in\mathbb{N}$ vertices of that arm after removing $v$, with diagonal entries $2$, all off-diagonal entries $-1$ except for the unique double-edge entry $-\sqrt2$ if the double edge is internal to the arm. If the double edge directly joins $v$ to the first vertex of the arm, the coupling vector has squared length $2$ and $R_1=(2)$, so its Schur-complement contribution is $1$. If the double edge is not incident to $v$, the coupling vector is the first standard basis vector and the determinant recurrence from the terminal end gives \begin{align*} \det R_r=2,\qquad \det R_r^{(1)}=2, \end{align*} where $R_r^{(1)}$ is the principal submatrix obtained by deleting the first row and first column. By Cramer's rule, the weighted arm contributes \begin{align*} e_1^\top R_r^{-1}e_1=\frac{\det R_r^{(1)}}{\det R_r}=1. \end{align*} Thus the [Schur complement criterion](/page/Schur%20Complement) at $v$ gives the scalar complement \begin{align*} 2-\frac{p}{p+1}-\frac{q}{q+1}-1 &=\frac{1}{p+1}+\frac{1}{q+1}-1\leq 0, \end{align*} because $p,q\geq 1$. This contradicts positive definiteness of the chosen principal submatrix. Hence no branching vertex can coexist with a double edge, and the underlying graph is a path. Let the double edge separate $a$ vertices on one side from $b$ vertices on the other side, with $a,b\geq 1$. Ordering the path from the side with $a$ vertices first, the determinant recurrence gives \begin{align*} \det S=2b-(b-1)(a+1). \end{align*} By symmetry the same condition holds after exchanging $a$ and $b$. Positivity is therefore possible exactly in the following cases: \begin{align*} \min\{a,b\}=1, \end{align*} or \begin{align*} a=b=2. \end{align*} The first case is a chain with the double edge at an end; its two orientations give $B_n$ and $C_n$. The second case is the four-vertex chain with the double edge in the middle, namely $F_4$. [guided] The weighted path recurrence is the main bookkeeping device in the non-simply-laced case. For a path with edge multiplicities $m_1,\dots,m_{n-1}$, the symmetric matrix is tridiagonal with diagonal entries $2$ and off-diagonal entries $-\sqrt{m_k}$. Expanding the leading $k\times k$ determinant along the last row gives \begin{align*} \Delta_0&:=1,\\ \Delta_1&:=2,\\ \Delta_k&=2\Delta_{k-1}-m_{k-1}\Delta_{k-2}. \end{align*} Because positive definiteness passes to principal submatrices, each $\Delta_k$ must be positive. This recurrence rules out two multiple edges once the shortest subpath is ordered correctly. Choose the shortest subpath containing both multiple edges, and order it so that the first and last edges are those multiple edges. Then the initial determinants are $\Delta_0=1$ and $\Delta_1=2$. If the first multiple edge is double, then $\Delta_2=2$, so the two consecutive determinants are equal; every intervening single edge preserves this equality by the recurrence $\Delta_k=2\Delta_{k-1}-\Delta_{k-2}$. When the second multiple edge is reached, a second double edge gives determinant $0$, and a triple edge gives a negative determinant. If the first multiple edge is triple, then $\Delta_2=1$; the next single edge gives determinant $0$, while an adjacent second multiple edge gives a negative determinant. By [Sylvester's criterion](/page/Sylvester%27s%20Criterion), a positive definite matrix cannot have such a principal path submatrix. Therefore a triple edge cannot be attached to anything else, so the only connected triple-edge diagram is the rank-two diagram $G_2$. Before treating paths with one double edge, we must rule out branching. Suppose a double edge and a branching vertex $v$ coexist. Take the full subtree consisting of $v$, two terminal simple arms at $v$, and the arm from $v$ toward the double edge, truncated at the double edge. If the two simple arm lengths are $p$ and $q$, then their Schur-complement contributions at $v$ are $p/(p+1)$ and $q/(q+1)$. The weighted arm contributes exactly $1$: if the double edge is incident to $v$, this is the contribution $(\sqrt2)^2/2=1$; otherwise the arm matrix has determinant $2$ and the cofactor obtained by deleting the vertex adjacent to $v$ also has determinant $2$, so Cramer's rule gives contribution $2/2=1$. Hence the scalar Schur complement at $v$ is \begin{align*} 2-\frac{p}{p+1}-\frac{q}{q+1}-1 &=\frac{1}{p+1}+\frac{1}{q+1}-1\leq 0. \end{align*} The [Schur complement criterion](/page/Schur%20Complement) then contradicts positive definiteness of this principal subdiagram. Hence a diagram with a double edge has no branching vertex, so its underlying graph is a path. For a single double edge in a path, let $a$ and $b$ be the numbers of vertices on the two sides of the double edge. Along the first $a$ vertices joined by single edges, the determinant is $a+1$. Crossing the double edge changes the next determinant to \begin{align*} 2(a+1)-2a=2. \end{align*} Each remaining single edge then follows the linear recurrence $\Delta_k=2\Delta_{k-1}-\Delta_{k-2}$. After the $b$ vertices on the other side, this gives \begin{align*} \det S=2b-(b-1)(a+1). \end{align*} For this determinant to be positive, and also positive after exchanging the two sides, the only possibilities are $\min\{a,b\}=1$ or $a=b=2$. These are precisely the end-double-edge chains $B_n,C_n$ and the middle-double-edge four-chain $F_4$. [/guided] [/step] [step:Verify that every diagram in the list is positive definite] For $A_n$, the symmetric matrix is the tridiagonal matrix with diagonal entries $2$ and off-diagonal entries $-1$. Its leading determinants satisfy \begin{align*} \Delta_0=1,\qquad \Delta_1=2,\qquad \Delta_k=2\Delta_{k-1}-\Delta_{k-2}, \end{align*} so $\Delta_k=k+1>0$ for every $k$. Thus $A_n$ is finite type. For the simply-laced diagrams with one branch point, the branch-length criterion proved above is also sufficient. The triples \begin{align*} (1,1,r),\qquad (1,2,2),\qquad (1,2,3),\qquad (1,2,4) \end{align*} satisfy \begin{align*} \frac{1}{p+1}+\frac{1}{q+1}+\frac{1}{r+1}>1, \end{align*} and therefore give positive definite matrices. These are precisely $D_n,E_6,E_7,E_8$. For $B_n$ and $C_n$, the underlying weighted path has one double edge at an end. Taking $a=1$ in the determinant formula gives \begin{align*} \det S=2>0, \end{align*} and every proper connected principal subdiagram is either an $A_k$ path or a shorter end-double-edge path, so all principal determinants are positive. Hence $B_n$ and $C_n$ are finite type. For $F_4$, the weighted path has $a=b=2$, and the determinant formula gives \begin{align*} \det S=2\cdot 2-(2-1)(2+1)=1>0. \end{align*} Its proper connected principal subdiagrams are of type $A_k$, $B_k$, or $C_k$, already shown positive definite. Hence $F_4$ is finite type. For $G_2$, the symmetric matrix is \begin{align*} \begin{pmatrix} 2&-\sqrt3\\ -\sqrt3&2 \end{pmatrix}, \end{align*} whose determinant is \begin{align*} 4-3=1>0. \end{align*} Thus $G_2$ is finite type. Combining the exhaustion steps with these positivity checks proves that the connected finite-type Dynkin diagrams are exactly \begin{align*} A_n,\ B_n,\ C_n,\ D_n,\ E_6,\ E_7,\ E_8,\ F_4,\ G_2, \end{align*} with the stated ranges of $n$. This proves both directions of the theorem. [/step]