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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,

Suppose first that $x\ne 0$, and let $E$ be the real span of the root system, of dimension equal to the rank of a semisimple Lie algebra $\ell$. The images of the roots span $E/\mathbb{R}\alpha$, so select roots ${\beta }_1,\dots ,{\beta }_{\ell -1}$ whose images form a basis of this quotient.

For each $i$, take the highest endpoint ${\gamma }_i^{+}$ of the $\alpha$ root string through ${\beta }_i$, and the highest endpoint ${\gamma }_i^{-}$ of the root string through $-{\beta }_i$. By construction, ${\gamma }_i^{\pm }+\alpha$ is not a root. Moreover ${\gamma }_i^{\pm }\ne -\alpha$, since their images in $E/\mathbb{R}\alpha$ are nonzero. The root-space decomposition and its bracket rule therefore giveThese $2(\ell -1)$ roots are distinct: their quotient images are the two signs of a basis, which are distinct. Their one-dimensional root spaces consequently contribute $2(\ell -1)$ independent vectors to the Lie algebra centralizer.

In addition, the $\ell -1$ dimensional subspace $\ker \alpha \subset \mathfrak{t}$ commutes with $x$, because $[h,x]=\alpha (h)x$. The line $\mathbb{C}x\subset {\mathfrak{g}}_{\alpha }$ also commutes with $x$. These contributions are independent by the root-space decomposition; in particular none of the selected endpoint roots is $\alpha$. Thus the centralizer lower bound for a root vector isIf $x=0$, then $Z_{\mathfrak{g}}(x)=\mathfrak{g}$. The root system contains the $2\ell$ distinct roots $\pm {\alpha }_i$, so $\dim \mathfrak{g}=\ell +|\Phi |\ge 3\ell$, which also proves the desired inequality. The argument includes rank one, where the list of ${\beta }_i$ is empty.

Write an element of the diagonal Cartan subalgebra asThe root system of this Special orthogonal Lie algebra isThus the maps requested are $H(t)\mapsto \pm t_i\pm t_j$, with the two signs independent. This is the D3 root system, with twelve roots.

For a direct check, the defining matrix identity gives the block form $X=\left(\begin{array}{cc}A& B\\ C& -A^T\end{array}\right)$ with $B^T=-B$ and $C^T=-C$. If $E_{ij}$ denotes a matrix unit, the root space for ${\epsilon }_i-{\epsilon }_j$ is spanned by $E_{ij}-E_{j+3,i+3}$; those for ${\epsilon }_i+{\epsilon }_j$ and $-{\epsilon }_i-{\epsilon }_j$ are spanned respectively by $E_{i,j+3}-E_{j,i+3}$ and $E_{i+3,j}-E_{j+3,i}$. Commuting these with $H(t)$ gives exactly the displayed maps.

Work over $\mathbb{R}$ or $\mathbb{C}$, in finite dimension and characteristic zero. The Killing form of a Lie algebra is the trace form of its Adjoint representation:It is bilinear and symmetric by the cyclic property of the trace. It is an invariant bilinear form on a Lie algebra, since the Jacobi identity gives ${\mathrm{ad}}_{[x,y]}=[{\mathrm{ad}}_x,{\mathrm{ad}}_y]$ and henceEquivalently, $\kappa ([z,x],y)+\kappa (x,[z,y])=0$. A Lie algebra automorphism preserves the form, because it conjugates the adjoint matrices. Its radical of a bilinear form ${\mathfrak{r}}_{\kappa }=\{x:\kappa (x,\mathfrak{g})=0\}$ is an ideal of a Lie algebra by invariance. The center of a Lie algebra lies in this radical, so the form vanishes for an abelian algebra. On a direct sum of ideals the summands are orthogonal and the form restricts to their own Killing forms.

A short argument proves the requested implication without assuming the Cartan criterion for semisimplicity. Let $\mathfrak{a}$ be an abelian ideal of a Lie algebra, $x\in \mathfrak{a}$, and $y\in \mathfrak{g}$. The map ${\mathrm{ad}}_x$ takes $\mathfrak{g}$ into $\mathfrak{a}$ and vanishes on $\mathfrak{a}$, while ${\mathrm{ad}}_y$ preserves $\mathfrak{a}$. Therefore ${\mathrm{ad}}_x{\mathrm{ad}}_y$ has image in $\mathfrak{a}$ and is zero on $\mathfrak{a}$. In a basis extending a basis of $\mathfrak{a}$, both diagonal blocks are zero, so its trace vanishes. This proves that abelian ideals lie in the radical of the Killing form.

If the solvable radical $\mathfrak{r}$ were nonzero, its derived series of a Lie algebra would have a last nonzero term $\mathfrak{a}$. The Jacobi identity makes every derived term an ideal of $\mathfrak{g}$, and the last one is abelian. Thus $0\ne \mathfrak{a}\subseteq {\mathfrak{r}}_{\kappa }$, contradicting nondegeneracy. We obtainThe converse holds as well in characteristic zero; together these implications are the Cartan criterion for semisimplicity.

For a complex simple Lie algebra, that criterion gives nondegeneracy on $\mathfrak{g}$. Let $\mathfrak{h}$ be a Cartan subalgebra. Use the standard root-space decompositionFor $h,h^{\prime }\in \mathfrak{h}$ and a root vector $e_{\alpha }$, choose $t\in \mathfrak{h}$ with $\alpha (t)\ne 0$. Invariance givesThus $\mathfrak{h}$ is orthogonal to every nonzero root space. If $h\in \mathfrak{h}$ is also orthogonal to $\mathfrak{h}$, it is orthogonal to all of $\mathfrak{g}$ and hence is zero. This establishes nondegeneracy of the Killing form on a Cartan subalgebra:The same invariance calculation shows that $\kappa ({\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta })=0$ unless $\alpha +\beta =0$. Opposite root spaces are therefore paired nondegenerately. Tracing the adjoint action on the root-space decomposition yieldssince the root spaces of a complex semisimple algebra are one-dimensional and its adjoint action on $\mathfrak{h}$ is zero.

The relevant Euclidean subspace of a Cartan subalgebra iswhere the $h_{{\alpha }_i}$ are the simple coroots, normalized by ${\alpha }_i(h_{{\alpha }_i})=2$. The standard Euclidean property is that $\kappa$ is real and positive definite on this space. The displayed trace formula explains it: $\kappa (h,h)$ is a sum of real squares, and the roots span ${\mathfrak{h}}^{\ast }$, so all squares vanish only for $h=0$. Via the nondegenerate form, the roots can consequently be regarded as vectors in a real Euclidean normed vector space. The induced dual inner product makes root reflections orthogonal and allows root lengths and angles to be encoded by Cartan integers and the Dynkin diagram.

This positivity is on a specified real subspace; the complex Killing form is a bilinear form, not a positive Hermitian inner product. On a compact real form the Killing form is instead negative definite. For example the Killing form of the special linear Lie algebra gives $\kappa (X,Y)=4\mathrm{tr}(XY)$ for ${\mathfrak{sl}}_2$, so $\kappa (L_0,L_0)=8$ but $\kappa (i{L}_0,i{L}_0)=-8$.

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.

Put $h_0=\psi (h)$ and $e_0=\psi (e)={\sum }_{i=1}^{\ell }{e}_i$, where $0\ne e_i\in {\mathfrak{g}}_{{\alpha }_i}$. Since $[h_0,e_0]=2e_0$ and the root spaces are a direct sum, ${\alpha }_i(h_0)=2$ for every simple root. Every root has simple-root coefficients of one sign, so no root vanishes on $h_0$. The Regular element criterion in a Cartan subalgebra therefore shows that $h_0$ is regular.

Restrict the Adjoint representation of $\mathfrak{g}$ along $\psi$. By complete reducibility of semisimple Lie algebra representations, it is a direct sum of finite-dimensional sl2 Lie algebra modules. The $h_0$-eigenvalues on a root space are twice the heights of the roots, so they are all even. Each irreducible summand consequently has even highest weight, contains exactly one zero-weight vector, and has a one-dimensional kernel for the raising operator $e_0$ by the classification of finite-dimensional sl2 representations.

Because $h_0$ is regular, its zero-weight space in $\mathfrak{g}$ is precisely $\mathfrak{t}$ and has dimension $\ell$. There are therefore exactly $\ell$ irreducible summands, whenceThus $e_0$ is regular; equivalently it is a principal nilpotent element in the given Principal sl2 subalgebra.

For root spaces ${\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta }$, the Jacobi identity gives $[{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta }]\subset {\mathfrak{g}}_{\alpha +\beta }$ when $\alpha +\beta \ne 0$. If the bracket is nonzero, that nonzero weight must be a root. When $\beta =-\alpha$, the bracket instead lies in the Cartan subalgebra and can be nonzero; zero is not a root. Thus the common nonzero-bracket implication requires the opposite-root exception.

A root is an individual nonzero vector in a root system. For a complex semisimple Lie algebra, it is a nonzero weight of the Adjoint representation restricted to a Cartan subalgebra; its root space consists of vectors $x$ satisfying $[h,x]=\alpha (h)x$ for all $h$ in that subalgebra.

For nonproportional roots $\gamma ,\delta$ of a reduced crystallographic root system, $S_{\gamma ,\delta }=\{\delta -p\gamma ,\dots ,\delta +q\gamma \}$ is consecutive, and $p-q=⟨\delta ,{\gamma }^{\vee }⟩$. Realize the crystallographic root system as the roots of a complex semisimple Lie algebra. The sum of the one-dimensional root spaces in this string is a module for the sl2 subalgebra associated with a root. Its weights are $⟨\delta ,{\gamma }^{\vee }⟩+2n$; they have one fixed parity. Complete reducibility gives a sum of irreducible representations of the sl2 Lie algebra. Two summands of this parity would overlap at weight zero when the parity is even, or at weights $\pm 1$ when it is odd, contradicting the one-dimensional root spaces. Thus there is one irreducible string, whose opposite endpoint weights sum to zero. This proves the relation. For distinct simple roots, $p=0$ and the cardinality is $1-⟨\delta ,{\gamma }^{\vee }⟩$. The nonproportional hypothesis excludes $\delta =\pm \gamma$, where zero is not a root and the consecutive-string statement fails.

A root vector is a nonzero element of a root space ${\mathfrak{g}}_{\alpha }$ in the root-space decomposition of a semisimple Lie algebra.

Choosing a positive system of a root system for a complex semisimple Lie algebra divides its root spaces into negative and positive ones. Their respective sums ${\mathfrak{n}}^{-}$ and ${\mathfrak{n}}^{+}$, together with the Cartan subalgebra $\mathfrak{h}$, give this direct sum of vector spaces. Both ${\mathfrak{n}}^{\pm }$ are nilpotent Lie algebras, and $\mathfrak{h}\oplus {\mathfrak{n}}^{+}$ is a Borel subalgebra. The Poincaré-Birkhoff-Witt theorem turns this into an ordered vector-space decomposition of the universal enveloping algebra, useful for building Verma modules.