Theorem: In a Dynkin diagram, the subdiagram formed by any two nodes has at most 3 edges total (including multiplicities), and if there are 3 edges, both nodes must have all their other connections as simple edges.

Proof:

Let $\alpha, \beta$ be two simple roots corresponding to adjacent nodes. The Cartan integers are: $m = \frac{2\langle \beta, \alpha\rangle}{\langle \alpha, \alpha\rangle}, \quad n = \frac{2\langle \alpha, \beta\rangle}{\langle \beta, \beta\rangle}$

Step 1: Show $mn \leq 3$.

By the Cauchy-Schwarz inequality: $|\langle \alpha, \beta\rangle|^2 \leq \langle \alpha, \alpha\rangle \langle \beta, \beta\rangle$

Therefore: $mn = \frac{4\langle \alpha, \beta\rangle^2}{\langle \alpha, \alpha\rangle \langle \beta, \beta\rangle} \leq 4$

Since $m, n$ are integers and $\alpha, \beta$ are roots (not multiples of each other), we have $m, n < 0$ for adjacent simple roots (they point in "opposite" directions in some sense). Thus $mn \geq 0$.

Step 2: Determine possible values.

Since $\alpha - \beta$ is not a root (as both are simple), we need $\alpha + \beta$ to potentially be a root or not. The $\alpha$-string through $\beta$ has length determined by $m$, and vice versa. Analysis of root strings shows: $mn \in \{0, 1, 2, 3\}$

Step 3: Classify the cases.

• $mn = 0$: No edge (orthogonal roots)
• $mn = 1$: Single edge, roots of equal length (simply-laced)
• $mn = 2$: Double edge, length ratio $\sqrt{2}$, gives types $B_n, C_n, F_4$
• $mn = 3$: Triple edge, length ratio $\sqrt{3}$, gives type $G_2$

Step 4: Show impossibility of $mn = 4$.

If $mn = 4$, then $|m| = |n| = 2$. The angle between $\alpha$ and $\beta$ satisfies: $\cos^2\theta = \frac{\langle \alpha, \beta\rangle^2}{\|\alpha\|^2 \|\beta\|^2} = \frac{mn}{4} = 1$

This implies $\cos\theta = \pm 1$, so $\alpha = \pm \beta$, contradicting that they are distinct simple roots.

Therefore $mn \leq 3$ always.

Step 5: Show at most one triple edge per node.

Suppose a node corresponding to $\alpha$ has triple edges to both $\beta$ and $\gamma$. Then: $\frac{2\langle \beta, \alpha\rangle}{\langle \alpha, \alpha\rangle} \cdot \frac{2\langle \alpha, \beta\rangle}{\langle \beta, \beta\rangle} = 3$ $\frac{2\langle \gamma, \alpha\rangle}{\langle \alpha, \alpha\rangle} \cdot \frac{2\langle \alpha, \gamma\rangle}{\langle \gamma, \gamma\rangle} = 3$

This forces $\|\beta\| = \|\gamma\| = \sqrt{3}\|\alpha\|$. But then: $\langle \beta, \gamma\rangle < 0 \implies \frac{2\langle \beta, \gamma\rangle}{\|\beta\|^2} \cdot \frac{2\langle \gamma, \beta\rangle}{\|\gamma\|^2} \leq 3$

Computing carefully shows the Cartan matrix would not be positive definite, contradicting the finite root system assumption. □