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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.

Choose the given short root $\alpha$ as a simple root, and let $\beta$ be the long simple root. This entails no loss of generality: the Weyl group is transitive on the short roots of the G2 root system. The relevant Cartan integers are $⟨\beta ,{\alpha }^{\vee }⟩=-3$ and $⟨\alpha ,{\beta }^{\vee }⟩=-1$. The positive roots areUnder ${\mathfrak{m}}_{\alpha }\cong {\mathfrak{sl}}_2$, the root space for $m\alpha +n\beta$ has $h_{\alpha }$ eigenvalue $2m-3n$. The subalgebra ${\mathfrak{m}}_{\alpha }$ itself is the three-dimensional irreducible $V(2)$ in the Adjoint representation.

The four root spaces along the root string $\beta ,\alpha +\beta ,2\alpha +\beta ,3\alpha +\beta$ have eigenvalues $-3,-1,1,3$. Consecutive spaces are connected by nonzero raising operators and lowering operators, so their sum is $V(3)$. The negative root string supplies another $V(3)$. The root spaces for $\pm (3\alpha +2\beta )$ have eigenvalue zero, and neither adding nor subtracting $\alpha$ gives a root; they are two copies of $V(0)$. Finally, the one-dimensional space $\ker \alpha \subset \mathfrak{t}$ commutes with ${\mathfrak{m}}_{\alpha }$, giving one more $V(0)$. We have accounted for all $14$ dimensions, and therefore the G2 adjoint branching to a short-root sl2 subalgebra isFor a nonzero root vector $x\in {\mathfrak{g}}_{\alpha }$, its Adjoint representation action is a nonzero scalar multiple of the raising operator on each summand. On each irreducible $V(n)$ this operator has a one-dimensional kernel, including $V(0)$. Since the Lie algebra centralizer is that kernel,

An automorphism of a root system outside the Weyl group is the sign changeIt preserves the set $\{\pm {\epsilon }_i\pm {\epsilon }_j\}$, fixes ${\alpha }_1$, and interchanges ${\alpha }_2$ and ${\alpha }_3$. It therefore realizes the nontrivial symmetry of the Dynkin diagram while preserving the chosen root basis.

To verify that $\tau \notin W$, note that $s_1$ swaps ${\epsilon }_1,{\epsilon }_2$, $s_2$ swaps ${\epsilon }_2,{\epsilon }_3$, and $s_3$ swaps ${\epsilon }_2,{\epsilon }_3$ while changing both their signs. Every product of these generators is a signed permutation with an even number of sign changes, whereas $\tau$ has one sign change. Thus it cannot be a Weyl group element. The parity here concerns sign changes, rather than the determinant of the permutation.

The formal character of a weight module is $\mathrm{ch}V={\sum }_{\mu }(\dim V_{\mu })e^{\mu }$, where $e^{\mu }$ are formal basis symbols in the group algebra of the weight lattice, satisfying $e^{\mu }{e}^{\nu }=e^{\mu +\nu }$. For a finite-dimensional irreducible highest-weight representation, the highest weight has weight multiplicity one. Another standard result is that weight multiplicities are invariant under the Weyl group: $\dim V_{w\mu }=\dim V_{\mu }$.

A minuscule representation has all its weights in a single Weyl group orbit. Since the highest weight $\lambda$ is present, that orbit is $W\lambda$, and the invariance just stated shows every point of it is present with multiplicity one. Consequently the formal character of a minuscule representation isThe sum is over distinct weights, rather than over all elements of $W$: summing $e^{w\lambda }$ over $w\in W$ would count each weight $|{\mathrm{Stab}}_W(\lambda )|$ times.

Use the coroot ${\alpha }^{\vee }$ in the pairing. The original PDF retains this mark, which is lost in the local TeX transcription. Fix any positive root $\alpha$, and put $n=⟨\lambda ,{\alpha }^{\vee }⟩$. For a dominant integral weight, $n$ is a nonnegative integer; positive coroots are nonnegative integral combinations of simple coroots.

Let $v_{\lambda }$ be a highest-weight vector, and restrict to the sl2 subalgebra associated with a root $\alpha$. It is killed by its raising operator $e_{\alpha }$ and has $h_{\alpha }$ eigenvalue $n$. If $n\ge 1$, its lowering operator produces a nonzero vector: otherwise $[e_{\alpha },f_{\alpha }]v_{\lambda }=n{v}_{\lambda }$ would vanish. Therefore $\lambda -\alpha$ is a weight.

Choose a Weyl group invariant positive-definite inner product on the real weight space. All weights of a minuscule representation have the same norm, since they belong to $W\lambda$. ButIf $n\ge 2$, this is negative, a contradiction. Thus for every positive root,When $n=1$, the lower weight is precisely the Weyl reflection $s_{\alpha }\lambda =\lambda -\alpha$, consistent with the orbit condition.

We first prove the useful lemma that dominant root-lattice highest weights have zero weight. Write $Q=\mathbb{Z}\Phi$ for the root lattice, and let a dominant integral weight $\lambda \in Q$ be the highest weight of a finite-dimensional irreducible representation.

First, $\lambda$ is a nonnegative integral combination of simple roots. Indeed, write $\lambda ={\lambda }_{+}-{\lambda }_{-}$, separating its positive and negative coefficients in the root basis. The two parts have disjoint supports, and distinct simple roots have nonpositive inner product, so $({\lambda }_{+},{\lambda }_{-})\le 0$. If ${\lambda }_{-}\ne 0$, thenOn the other hand, dominance gives $(\lambda ,{\alpha }_i)\ge 0$ for each simple root, hence $(\lambda ,{\lambda }_{-})\ge 0$. This contradiction proves ${\lambda }_{-}=0$.

Now suppose a nonzero weight $\mu ={\sum }_i{m}_i{\alpha }_i$ has all $m_i\in {\mathbb{Z}}_{\ge 0}$. The identityshows that some $i$ satisfies $m_i>0$ and $⟨\mu ,{\alpha }_i^{\vee }⟩>0$. We use the standard sl2 Lie algebra fact that its lowering operator is injective on any positive $h$ eigenspace in a finite-dimensional representation. This follows from the classification of finite-dimensional sl2 representations: in each irreducible $V(n)$, the only weight killed by the lowering operator is the lowest weight $-n\le 0$.

Consequently a nonzero vector of weight $\mu$ lowers to a nonzero vector of weight $\mu -{\alpha }_i$. Its simple-root coefficients remain nonnegative and their sum decreases by one. Starting at $\lambda$, repeated lowering must therefore reach the zero weight. Notice that intermediate weights need not remain dominant; positivity of the chosen coroot pairing is enough at each step.

For a minuscule representation, every weight belongs to $W\lambda$, so the zero weight just obtained lies in this orbit. Every Weyl group element is invertible, and $w\lambda =0$ forces $\lambda =0$. This proves that minuscule weights in the root lattice are zero. Under the assumption $X=Q$, every possible highest weight lies in $Q$, henceHere $V(0)$ is the one-dimensional trivial Lie algebra representation, by the classification of finite-dimensional irreducible highest-weight representations.

Use the wiki's Cartan matrix conventionLabel the long simple root first when the lengths differ. The matrix of an irreducible rank-two root system has the form $A=\left(\begin{array}{cc}2& -a\\ -b& 2\end{array}\right)$, where $a,b$ are positive integers. Integrality and the signs follow from the Cartan integers for distinct simple roots, and irreducibility rules out a zero off-diagonal pair. Its positive-definite symmetrization gives $ab<4$, equivalently $ab=4{\cos }^2\theta <4$. Moreover $a/b=\Vert {\alpha }_1{\Vert }^2/\Vert {\alpha }_2{\Vert }^2\ge 1$. Thus $(a,b)=(1,1),(2,1),(3,1)$, giving the entire classification of rank-two root systems relevant to a simple algebra:These correspond to ${\mathfrak{sl}}_3(\mathbb{C})$, ${\mathfrak{so}}_5(\mathbb{C})\cong {\mathfrak{sp}}_4(\mathbb{C})$, and the exceptional algebra ${\mathfrak{g}}_2$. Here $B_2=C_2$ denotes the Isomorphism between so5 and sp4, not an additional case. The disconnected $A_1\times A_1$ system corresponds to a semisimple but nonsimple algebra and is excluded.

The Dynkin diagrams, with nodes in the same order as the matrices, areThe second and third diagrams have two and three bonds respectively, with the arrowhead pointing to the short root ${\alpha }_2$. In the last diagram $\equiv$ draws the three bonds and $>$ their arrowhead.

To enumerate roots, use the standard root string theorem: for $\beta \ne \pm \alpha$, the roots $\beta +n\alpha$ are consecutive from $\beta -p\alpha$ through $\beta +q\alpha$, with $p-q=⟨\beta ,{\alpha }^{\vee }⟩$. Also use the standard results that the root system is reduced, every root is conjugate under the Weyl group to a simple root, and each root space has dimension one.

Write $m=a\in \{1,2,3\}$. Since a difference of simple roots is not a root, the ${\alpha }_2$-string starting at ${\alpha }_1$ has $p=0$ and $q=m$. It produces ${\alpha }_1,{\alpha }_1+{\alpha }_2,\dots ,{\alpha }_1+m{\alpha }_2$. The ${\alpha }_1$-string starting at ${\alpha }_2$ has $q=1$. In type $G_2$, the further string through $\beta ={\alpha }_1+3{\alpha }_2$ hasand $\beta -{\alpha }_1=3{\alpha }_2$ is not a root, so it also produces $2{\alpha }_1+3{\alpha }_2$. The resulting positive roots areAll roots are these and their negatives. To check completeness, the simple root reflections act on coordinates byEach listed set together with its negatives is stable under both reflections. It contains the simple roots and consists of roots already forced by strings. Since every root lies in a Weyl orbit of a simple root, no other roots can occur. This proves the lists for the A2 root system, B2 root system, and G2 root system. The root-space decomposition then gives

For the restriction to a sl2 subalgebra associated with a root, normalize its Cartan element to be $h_{\alpha }={\alpha }^{\vee }$. On a root vector of root $\beta$, its eigenvalue is $⟨\beta ,{\alpha }^{\vee }⟩$. The classification of finite-dimensional sl2 representations says that $R(\Lambda )$ has weights $\Lambda ,\Lambda -2,\dots ,-\Lambda$ and dimension $\Lambda +1$. A root string of $\Lambda +1$ nonparallel roots therefore supplies one such irreducible module: the raising and lowering brackets connect its consecutive root spaces.

For clarity, the positive-side strings for each simple-root direction are listed below. Brackets denote the whole consecutive string; a single listed root is a string of length one. In every row also include each negative string in reverse order, and separately the triple formed by the root ${\alpha }_i$, its negative, and $h_{{\alpha }_i}$.The root's own triple is always $R(2)$. The one-dimensional subspace of $\mathfrak{h}$ annihilated by ${\alpha }_i$ commutes with this subalgebra and gives $R(0)$. Each singleton root in the table, and its negative, contributes another $R(0)$. Reading the other string lengths now gives the adjoint branching to root sl2 subalgebras in rank two:Their dimensions are respectively $8,10,10,14,14$. The last row agrees with G2 adjoint branching to a short-root sl2 subalgebra. All labels in the table are highest weights, rather than dimensions.

Choose a positive system of a root system in which $\alpha$ is a simple root; this is possible after applying an element of the Weyl group. Let $\theta$ be the highest root. Since the rank is greater than one, $\theta \ne \alpha$, and the maximality of $\theta$ implies that $\theta +\alpha$ is not a root.

For $0\ne x\in {\mathfrak{g}}_{\alpha }$, the $(\ell -1)$-dimensional space $\ker \alpha \subset \mathfrak{t}$ centralizes $x$. The line $\mathbb{C}x$ also centralizes $x$, and $[x,{\mathfrak{g}}_{\theta }]\subseteq {\mathfrak{g}}_{\alpha +\theta }=0$. These independent spaces giveHence a nonzero simple-root vector is not regular when $\ell >1$.

The short-root Weyl reflection $s_{\alpha }$ interchanges ${\omega }_1$ and ${\omega }_2$, because $\alpha ={\omega }_2-{\omega }_1$ and ${\alpha }^{\vee }={\delta }_2^{\vee }-{\delta }_1^{\vee }$. HenceThe Weyl group orbit of the highest weight occurs in $U$ with a one-dimensional extremal weight space. Let $v$ be a nonzero vector of weight $s_{\alpha }\lambda$. The same reflection exchanges the long simple roots: $s_{\alpha }{\delta }_1={\delta }_2$ and $s_{\alpha }{\delta }_2={\delta }_1$. If an $A_2$ raising operator did not annihilate $v$, then $s_{\alpha }\lambda +{\delta }_i$ would be a weight of $U$; applying $s_{\alpha }$ would make $\lambda +{\delta }_j$ a weight, contradicting the fact that $\lambda$ is the highest weight.

Thus $v$ is an $A_2$ highest-weight vector of weight $b{\omega }_1+a{\omega }_2$. The complete reducibility of semisimple Lie algebra representations then supplies the corresponding irreducible summand, so

A simply-laced positive-definite Coxeter graph has no multiple edges and has positive-definite symmetric Cartan matrix, with diagonal entries $2$ and entry $-1$ precisely across an edge. Its connected components are exactly the ADE classification

The graph underlying $1\to 2$ is $A_2$. For simple roots ${\alpha }_1,{\alpha }_2$ with $({\alpha }_i,{\alpha }_i)=2$ and $({\alpha }_1,{\alpha }_2)=-1$, its roots areThere is one base for each Weyl chamber, hence six bases. The Weyl group is generated by the two root reflections withso $W\simeq S_3$. Relative to a chosen base, the two Coxeter elements are $s_1{s}_2$ and $s_2{s}_1$; both have order three.

A representation of the quiver is one linear map $f:V_1\to V_2$. Choosing bases that put $f$ in rank normal form decomposes it into copies ofThese are indecomposable and have dimension vectors $(1,0)$, $(0,1)$, and $(1,1)$, the three positive roots of $A_2$. More generally, Gabriel theorem says that for any orientation of a simply-laced positive-definite Dynkin graph, indecomposable quiver representations are in bijection with its positive roots. The finite ADE root system therefore gives finitely many indecomposables.

Starting from the highest weight ${\mu }_1=2{\alpha }_1+{\alpha }_2$, lowering with the simple-root operators and closing under the Weyl group givesEvery weight has multiplicity one. The diagram is the hexagon of short roots with one central weight, so this is the seven-dimensional fundamental irreducible representation of $G_2$.

A finite root system in a real Euclidean vector space $E$ is a finite spanning set $\Phi \subset E\setminus \{0\}$ such that, for every $\alpha \in \Phi$, the root reflectionpreserves $\Phi$, and the Cartan integer $2(\beta ,\alpha )/(\alpha ,\alpha )$ is an integer for all $\alpha ,\beta \in \Phi$. It is reduced when the only scalar multiples of $\alpha$ in $\Phi$ are $\alpha$ and $-\alpha$. Its Weyl group is the subgroup of the orthogonal group generated by the reflections $s_{\alpha }$. A base of a root system $\Delta$ is a basis of $E$ such that every root is an integer combination of elements of $\Delta$ whose nonzero coefficients all have the same sign.

The coroot of $\alpha$ isLet $C$ be the Weyl chamber determined by $\Delta$:The roots $\alpha$ and ${\alpha }^{\vee }$ are positive scalar multiples, so their reflecting hyperplanes and their positive half-spaces are identical. The same chamber $C$ therefore defines positivity in the coroot system ${\Phi }^{\vee }$. Its walls correspond exactly to the rays ${\mathbb{R}}_{>0}{\alpha }^{\vee }$ for $\alpha \in \Delta$. Hence its simple roots arewhich is therefore a base of ${\Phi }^{\vee }$.

For an arbitrary finite-dimensional complex Lie algebra $L$, a Cartan subalgebra is a nilpotent Lie subalgebra $H$ equal to its own normalizer. When $L$ is semisimple, this is equivalently a maximal abelian subalgebra consisting of elements that act semisimply in the Adjoint representation.

Choose such an $H$. Simultaneous diagonalization gives the root-space decompositionThe nonzero weights $\alpha \in H^{\ast }$ are the roots. The restriction of the Killing form $B$ to $H$ is a nondegenerate bilinear form, so each $\alpha$ corresponds to a unique $t_{\alpha }\in H$ with $\alpha (h)=B(t_{\alpha },h)$. On the real span of these $t_{\alpha }$, the restriction of $B$ supplies a positive-definite inner product after choosing the standard real form. The sl2 subalgebra associated with a root givesand shows that these reflections preserve the finite set $\Phi$. Thus the roots form a finite reduced crystallographic root system, whose Weyl group is generated by these reflections.

For every root, the Weyl reflection isFor $\alpha ={\epsilon }_i-{\epsilon }_j$, it swaps the $i$th and $j$th coordinates. For $\alpha ={\epsilon }_i+{\epsilon }_j$, it sendsand fixes all other coordinates. The Weyl group of $D_n$ is therefore the group of signed permutations with an even number of sign changes,

For a dominant integral weight $\lambda$, the Weyl character formula isHere $W$ is the Weyl group, $\ell (w)$ its Coxeter length, $\rho$ the Weyl vector, and $\mathrm{ch}$ the formal character of a weight module. Taking the limit at the identity gives the Weyl dimension formula

The linear mappreserves $\Phi$ and exchanges ${\alpha }_2$ with ${\alpha }_3$ while fixing ${\alpha }_1$. It is not in the $D_3$ Weyl group, whose signed permutations change an even number of signs. Thus it is an outer automorphism of the root system.

For a dominant integral weight $\lambda$, the Weyl character formula iswhere $W$ is the Weyl group, $\ell$ is Coxeter length, and $\rho$ is the half-sum of positive roots. The Weyl denominator formula is

Set $\lambda =k\rho$. Apply the denominator identity after replacing every formal exponential $e^{\mu }$ by $e^{(k+1)\mu }$:Dividing this by the ordinary denominator gives

Root bases correspond bijectively to Weyl chambers: the walls of a chamber determine its inward simple roots. The Weyl group acts transitively on the chambers. One proof joins interior points of two chambers by a generic line segment. Each time the segment crosses one reflecting hyperplane, reflect the remaining segment across that wall; the resulting product of root reflections sends the first chamber to the second. It consequently sends the first root basis to the second. Thus $W$ acts transitively on root bases.