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For $n\ge 2$, the finite Dynkin diagram has a chain of short roots ${\alpha }_1,\dots ,{\alpha }_{n-1}$ with single bonds, followed by a double bond to the long root ${\alpha }_n$, pointing toward ${\alpha }_{n-1}$. The Extended Dynkin diagram adds the long root ${\alpha }_0=-2{\epsilon }_1$, doubly bonded to ${\alpha }_1$ with arrow toward ${\alpha }_1$. All other bonds remain. For $n=1$, the finite diagram is a single vertex; the affine diagram has two equal-length vertices joined by a double bond without an arrow, with Cartan matrix $\left(\begin{array}{cc}2& -2\\ -2& 2\end{array}\right)$.

Write $\bar{i}=2n+1-i$, and let $E_{ab}$ denote a matrix unit. The diagonal Cartan subalgebra consists ofFor a diagonal matrix, $[t,E_{ab}]=(t_a-t_b)E_{ab}$. Substituting into the symplectic relation shows that the following are root vectors:The zero weight space is spanned by $H_i=E_{ii}-E_{\bar{i}\bar{i}}$. The symplectic equation pairs the remaining entries exactly as in this table; the displayed vectors and the $H_i$ are independent and span the algebra. This proves the root-space decompositionThere are $2n^2$ roots of a root system, each with one-dimensional root space, giving $\dim \mathfrak{g}=n(2n+1)$.

The upper triangular root vectors select preciselyThe simple roots of this Cn root system areIndeed, the positive roots of a root system have expansionsThese expansions also identify the highest root:Normalize the coordinate inner product by $({\epsilon }_i,{\epsilon }_j)={\delta }_{ij}$. The simple coroots are ${\alpha }_i^{\vee }={\epsilon }_i-{\epsilon }_{i+1}$ for $i<n$ and ${\alpha }_n^{\vee }={\epsilon }_n$. Solving $⟨{\omega }_j,{\alpha }_i^{\vee }⟩={\delta }_{ij}$ gives the fundamental weightsIn the sum of the positive roots of a root system, the pair $({\epsilon }_i-{\epsilon }_j)+({\epsilon }_i+{\epsilon }_j)$ contributes $2{\epsilon }_i$, and the long root contributes another $2{\epsilon }_i$. Therefore the Weyl vector isThese are the Positive-root data for Cn.

For $n\ge 2$, the finite Dynkin diagram is a chain of $n$ vertices. The bonds between ${\alpha }_i$ and ${\alpha }_{i+1}$ for $i<n-1$ are single. The last bond is double, with its arrow pointing from the long root ${\alpha }_n$ toward the short root ${\alpha }_{n-1}$. For the Extended Dynkin diagram, add ${\alpha }_0=-\theta =-2{\epsilon }_1$ at the other end, doubly bonded to ${\alpha }_1$, with arrow toward ${\alpha }_1$. The following Cn Dynkin diagrams include both ends and the low-rank cases.

At $n=1$, there is just the root ${\alpha }_1=2{\epsilon }_1$ and one finite vertex. The affine Cartan matrix is $\left(\begin{array}{cc}2& -2\\ -2& 2\end{array}\right)$; its two vertices have equal root length, so the double bond has no arrow.

An abstract root system, in the reduced crystallographic convention, is a finite subset $\Phi$ of a finite-dimensional real Euclidean space $E$ satisfying the following conditions:

The root system is irreducible if it is not a union of two nonempty mutually orthogonal root systems. The integrality condition is essential for the finite angle list below; arbitrary finite reflection systems need not satisfy it.

The possible angles are $0$, $\pi /6$, $\pi /4$, $\pi /3$, $\pi /2$, $2\pi /3$, $3\pi /4$, $5\pi /6$, and $\pi$. For roots $\alpha ,\beta$, putBoth are integers, they have the same sign when nonzero, andIf the roots are linearly independent, strict Cauchy-Schwarz inequality gives $0\le ab<4$. Hence $ab\in \{0,1,2,3\}$. These values give, respectively,For the last column, the possible absolute pairs $(|a|,|b|)$ are $(1,1)$, $(1,2)$, and $(1,3)$, up to interchange, and $a/b=(\beta ,\beta )/(\alpha ,\alpha )$. Parallel roots are equal or opposite by reducedness, giving $0$ or $\pi$. Thus the list is exhaustive, and the rank-two root systems realize each nonparallel possibility.

Now suppose two simple roots $\alpha ,\beta$ have angle $5\pi /6$. Distinct simple roots have nonpositive inner product. Here is a proof from their defining sign property: if $(\delta ,ϵ)>0$, then $s_{\delta }(ϵ)=ϵ-n\delta$ with a positive Cartan integer $n$. This would be a root whose coefficients in the root basis include both a positive and a negative coefficient, which is impossible. Therefore, after normalizing the simple roots to unit length, any nonzero off-diagonal inner product is at most $-1/2$, by the angle list.

Let $u=\alpha /\Vert \alpha \Vert$, $v=\beta /\Vert \beta \Vert$, so $(u,v)=-\sqrt{3}/2$. If there were a third simple root with unit vector $w$, write $r=(u,w)\le 0$ and $s=(v,w)\le 0$. Independence of the simple roots makes their Gram matrix positive definite, soIf either $r$ or $s$ were nonzero, its absolute value would be at least $1/2$. Since $rs\ge 0$, the displayed determinant would then be nonpositive. Thus $r=s=0$. Every other simple root is orthogonal to both $\alpha$ and $\beta$: a triple bond isolates a G2 component.

Indeed, the root reflections in $\alpha ,\beta$ preserve their plane and fix its orthogonal complement; the root reflections in the other simple roots preserve that complement and fix the plane. The given facts that these reflections generate the Weyl group and that every root is Weyl group-conjugate to a simple root imply that every root lies entirely in one of those two orthogonal subspaces. If there were any other simple root, this would make $\Phi$ reducible. Therefore irreducibility forces rank two.

Up to interchanging the simple roots, take $\alpha$ short and $\beta$ long. The squared-length ratio is $3$. Choose a scale and orthonormal coordinates so thatThe Cartan integers are $⟨\beta ,{\alpha }^{\vee }⟩=-3$ and $⟨\alpha ,{\beta }^{\vee }⟩=-1$. In coefficient coordinates $m\alpha +n\beta$, the two root reflections areTheir orbits of $\alpha$ and $\beta$ are, respectively,These sets are closed under both displayed reflections; for example $s_{\beta }(\alpha )=\alpha +\beta$, $s_{\alpha }(\alpha +\beta )=2\alpha +\beta$, $s_{\alpha }(\beta )=3\alpha +\beta$, and $s_{\beta }(3\alpha +\beta )=3\alpha +2\beta$. The assumption that every root is conjugate to a simple root now determines all twelve roots, with no further ones possible.

For completeness, this twelve-element set really is a root system. It is finite, spans the plane, and is reduced. The six short roots have length $1$ and directions spaced by $\pi /3$; the six long roots have length $\sqrt{3}$ and interleaved directions, also spaced by $\pi /3$. Their mutual Cartan integers are integral: within either length class they are $0$, $\pm 1$, or $\pm 2$ as applicable, and between the two length classes they are $0$, $\pm 1$, or $\pm 3$. Closure under all root reflections follows by conjugating the two simple reflections, since every listed root lies in a simple-root orbit. The nonorthogonal simple roots make it irreducible. We have therefore proved existence as well as uniqueness, up to isometry and overall scale:

The triple bond in the original Dynkin diagram gives the Cartan integerswhere $\alpha$ is the long root and $\beta$ the short root. Their product is $4{\cos }^2\theta =3$, and the inner product of distinct simple roots is nonpositive. The ratio of the two integers gives the squared-length ratio. ThereforeIt is convenient to normalize $(\alpha ,\alpha )=6$, $(\beta ,\beta )=2$ and $(\alpha ,\beta )=-3$. Scaling the inner product does not affect the root system or the fundamental-weight relations.

For nonproportional roots $\gamma ,\delta$, the root-string theorem states that the root string is consecutive:The endpoints are maximal, and reflection in $\gamma$ reverses the string. The Cartan integer determines $p-q$, not in general the total $p+q+1$ by itself. This result follows by restricting the Adjoint representation to the sl2 subalgebra associated with a root. If the roots are distinct simple roots, $\delta -\gamma$ cannot be a root: its simple-root coefficients have opposite signs. Thus $p=0$ andHere length means number of roots; the number of intervals between them is one less. Distinctness matters. If $\gamma =\delta$, a reduced root system gives the set $\{\gamma ,-\gamma \}$, with a missing zero between them, so the consecutive-string theorem and the displayed simple-root formula do not apply.

In the G2 root system, the initial strings areFor $\delta =\alpha +3\beta$, $⟨\delta ,{\alpha }^{\vee }⟩=-1$. Since $\delta -\alpha =3\beta$ is not a root in a reduced root system, $p=0$ and $q=1$: this generates $2\alpha +3\beta$. The remaining strings explain why the construction stops. The $\alpha$-strings through $\beta$ and $\alpha +\beta$ are the same two-element string; the one through $\alpha +2\beta$ is a singleton since subtracting $\alpha$ gives $2\beta$, and its Cartan integer is zero; the one through $2\alpha +3\beta$ is the string $\{\alpha +3\beta ,2\alpha +3\beta \}$. The $\beta$-strings through $\alpha$, $\alpha +\beta$, $\alpha +2\beta$ and $\alpha +3\beta$ are the initial four-element string. Finally, $2\alpha +3\beta$ is orthogonal to $\beta$; its $\beta$-string is a singleton because $2\alpha +2\beta =2(\alpha +\beta )$ is not a root. Apply the same reasoning to negatives. Using the permitted completeness of this procedure givesThe short positive roots are $\beta ,\alpha +\beta ,\alpha +2\beta$, of squared length two; the other three are long, of squared length six. Each root space is one-dimensional and the Cartan subalgebra has dimension two, soHere the dimension refers to the Lie algebra, with one Cartan generator per rank, not just the number of roots.

Write a prospective weight as $w=a\alpha +b\beta$. The pairings with simple coroots areThe fundamental weights are dual to those coroots. Solving the two linear systems gives, in the long-root-first numbering of this paper,The representation with Dynkin labels $(0,1)$ has highest weight $\lambda ={\omega }_2$, a short root. Numbering the short root first, as some references do, would call this the $(1,0)$ representation instead; the representation itself is unchanged.

The weight set of a finite-dimensional irreducible highest-weight representation is invariant under the Weyl group and lies in the convex hull of the orbit of its highest weight. All weights also differ from the highest weight by an element of the root lattice. The orbit of $\lambda$ comprises the six short roots, so all six are weights. The lowering operators give the chainIn particular zero occurs: at weight $\beta$, its pairing with ${\beta }^{\vee }$ is two, so the lowering operator is nonzero by the finite-dimensional sl2 Lie algebra representation theory. The $\beta$-string through it is the usual three-weight string $\beta ,0,-\beta$; higher weight $2\beta$ would lie outside the highest-weight convex hull.

There can be no further weights. Every point of that convex hull has squared norm at most $\Vert \lambda {\Vert }^2=2$, while a root-lattice point hasThe integer solutions of the bound $\Vert w{\Vert }^2\le 2$ are precisely zero and the six short roots: $a=0$ gives $b=-1,0,1$; $a=1$ gives $b=1,2$; and $a=-1$ gives $b=-1,-2$. ThusThe final dimension uses the stipulated nondegeneracy of the weights. It also agrees with the Weyl dimension formula without that stipulation. The zero weight is not in the Weyl orbit of the nonzero weights, so this is not a minuscule representation.

In the diagram above, a coordinate label $(a,b)$ means $a\alpha +b\beta$. The left panel contains all twelve roots of a root system; the right panel contains the six short-root weights and the zero weight. The long-root-first convention is the same as in the calculations.