Cartan Subalgebras and Root Space Decompositions — Examples

Killing-form computation: ${\mathrm{ad}}_h=\mathrm{diag}(0,2,-2)$ on the basis $(h,e,f)$, ${\mathrm{ad}}_e(h)=-2e$, ${\mathrm{ad}}_e(e)=0$, ${\mathrm{ad}}_e(f)=h$, and ${\mathrm{ad}}_f(h)=2f$, ${\mathrm{ad}}_f(e)=-h$, ${\mathrm{ad}}_f(f)=0$. Hence $B(h,h)=8$, $B(e,f)=B(f,e)=4$, and all other basis pairings vanish. The Killing matrix $\left(\begin{array}{ccc}8& 0& 0\\ 0& 0& 4\\ 0& 4& 0\end{array}\right)$ has determinant $-128\mathrm{\ne }0$, so [L1] proves that ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$ is semisimple.

The subspace $\mathfrak{h}=\mathbb{C}h$ is a Cartan subalgebra: it is one-dimensional, hence abelian and nilpotent, and $N_{\mathfrak{g}}(\mathfrak{h})=\{ah+be+cf:[ah+be+cf,h]\in \mathbb{C}h\}$ equals $\mathbb{C}h$, because $[h,h]=0$, $[e,h]=-2e$ and $[f,h]=2f$, so a normalizing element has $b=c=0$.

The eigenspaces of ${\mathrm{ad}}_h$ are $\mathbb{C}h$ with eigenvalue $0$, $\mathbb{C}e$ with eigenvalue $2$ and $\mathbb{C}f$ with eigenvalue $-2$. Defining $\alpha \in {\mathfrak{h}}^{\ast }$ by $\alpha (h)=2$ and using Root and root space, the nonzero eigenspaces are ${\mathfrak{g}}_{\alpha }=\mathbb{C}e$ and ${\mathfrak{g}}_{-\alpha }=\mathbb{C}f$, so $\mathrm{\Phi }=\{\pm \alpha \}$ and the root-space decomposition is the displayed one.

The dual vector $H_{\alpha }$ satisfies $B(H_{\alpha },h)=\alpha (h)=2$; writing $H_{\alpha }=th$ gives $8t=2$, so $t=\frac{1}{4}$ and $H_{\alpha }=\frac{1}{4}h$; then $\alpha (H_{\alpha })=\frac{1}{4}\cdot 2=\frac{1}{2}=B(H_{\alpha },H_{\alpha })$, so the coroot is $h_{\alpha }=2H_{\alpha }\mathrm{/}\alpha (H_{\alpha })=2\cdot \frac{1}{4}h\mathrm{/}\frac{1}{2}=h$.

The product formula $E_{ij}{E}_{kl}={\delta }_{jk}{E}_{il}$ is immediate from matrix multiplication: the product has the single nonzero entry $1$ in position $(i,l)$ exactly when the middle indices match. Hence $[E_{ij},E_{kl}]={\delta }_{jk}{E}_{il}-{\delta }_{li}{E}_{kj}$.

There are four cases. If $k=j$ and $l\mathrm{\ne }i$, then $[E_{ij},E_{jl}]=E_{il}$ and the root is $({\epsilon }_i-{\epsilon }_j)+({\epsilon }_j-{\epsilon }_l)={\epsilon }_i-{\epsilon }_l$. If $l=i$ and $k\mathrm{\ne }j$, then $[E_{ij},E_{ki}]=-E_{kj}$ and the root is $({\epsilon }_i-{\epsilon }_j)+({\epsilon }_k-{\epsilon }_i)={\epsilon }_k-{\epsilon }_j$. If $k=j$ and $l=i$, then $[E_{ij},E_{ji}]=E_{ii}-E_{jj}\in \mathfrak{h}={\mathfrak{g}}_0$ and the functional sum is zero. Finally, if $k\mathrm{\ne }j$ and $l\mathrm{\ne }i$, both Kronecker terms vanish; the functional sum has no cancellation producing a root or zero, so its root space is zero. Thus every case has the asserted bracket inclusion.

Write $R_i=\mathrm{rad}({\mathfrak{g}}_i)$ and $R=\mathrm{rad}(\mathfrak{g})$. The subspace $R_1\oplus R_2$ is a solvable ideal because brackets and every term of its derived series are computed componentwise, so $R_1\oplus R_2\subseteq R$. Conversely each projection ${\pi }_i(R)$ is an ideal of ${\mathfrak{g}}_i$ and is solvable: ${\pi }_i(R{)}^{(m)}={\pi }_i(R^{(m)})=0$ once $R^{(m)}=0$. Thus ${\pi }_i(R)\subseteq R_i$ and $R\subseteq R_1\oplus R_2$. Hence $R=R_1\oplus R_2=0$, proving that $\mathfrak{g}$ is semisimple before the Cartan/maximal-toral theorem is applied.

If each ${\mathfrak{h}}_i$ is a Cartan subalgebra of ${\mathfrak{g}}_i$, then ${\mathfrak{h}}_1\oplus {\mathfrak{h}}_2$ is nilpotent, being a direct sum of nilpotent algebras, and its normalizer is $N_{{\mathfrak{g}}_1}({\mathfrak{h}}_1)\oplus N_{{\mathfrak{g}}_2}({\mathfrak{h}}_2)={\mathfrak{h}}_1\oplus {\mathfrak{h}}_2$: an element $x=x_1+x_2$ normalizes ${\mathfrak{h}}_1\oplus {\mathfrak{h}}_2$ exactly when $[x_i,{\mathfrak{h}}_i]\subseteq {\mathfrak{h}}_i$ for $i=1,2$, because brackets in a direct sum are computed componentwise and mixed brackets vanish.

Conversely let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$. By step 1.1 and Cartan subalgebras are exactly maximal toral subalgebras it is maximal toral, hence abelian with all adjoint operators semisimple (Toral and maximal toral subalgebras). Let ${\mathfrak{h}}_i$ be the image of $\mathfrak{h}$ under the projection $\mathfrak{g}\to {\mathfrak{g}}_i$; each ${\mathfrak{h}}_i$ is abelian, since it is the image of an abelian subalgebra under a Lie-algebra homomorphism, and each of its elements is semisimple, because the adjoint operator of $x_1+x_2$ splits as the direct sum of the adjoint operators of $x_1$ and $x_2$, and a direct sum of endomorphisms is semisimple exactly when both summands are. Hence ${\mathfrak{h}}_1\oplus {\mathfrak{h}}_2$ is toral and contains $\mathfrak{h}$, so maximality gives $\mathfrak{h}={\mathfrak{h}}_1\oplus {\mathfrak{h}}_2$.

Each ${\mathfrak{h}}_i$ is maximal toral in ${\mathfrak{g}}_i$: if ${\mathfrak{t}}_i⊋{\mathfrak{h}}_i$ were toral in ${\mathfrak{g}}_i$, then replacing the $i$th summand of ${\mathfrak{h}}_1\oplus {\mathfrak{h}}_2$ by ${\mathfrak{t}}_i$ would give a toral subalgebra of $\mathfrak{g}$ strictly containing $\mathfrak{h}$, contradicting maximality. By Cartan subalgebras are exactly maximal toral subalgebras each ${\mathfrak{h}}_i$ is a Cartan subalgebra of ${\mathfrak{g}}_i$, which completes the first half. The dimension formula is additivity of dimensions over a direct sum.

For the root statement, the eigenvectors of ${\mathrm{ad}}_H$ for $H=H_1+H_2\in \mathfrak{h}$ are exactly the sums of eigenvectors in the two summands: a functional on $\mathfrak{h}$ that is nonzero on both summands occurs for no nonzero eigenvector, while the roots of $\mathfrak{g}$ are the union of the roots of ${\mathfrak{g}}_1$ with respect to ${\mathfrak{h}}_1$ and of ${\mathfrak{g}}_2$ with respect to ${\mathfrak{h}}_2$, extended by zero on the other summand. Hence the root systems form a disjoint union, as asserted.

The Killing form of ${\mathfrak{s}\mathfrak{l}}_n(\mathbb{C})$ is $B(X,Y)=2n\mathrm{tr}(XY)$ on traceless matrices: on the basis of matrix units, ${\mathrm{ad}}_{E_{ab}}(E_{cd})={\delta }_{bc}{E}_{ad}-{\delta }_{da}{E}_{cb}$, and summing the diagonal contributions of ${\mathrm{ad}}_{E_{ab}}{\mathrm{ad}}_{E_{cd}}$ over the basis gives $2n{\delta }_{ad}{\delta }_{bc}-2{\delta }_{ab}{\delta }_{cd}$, which is $2n\mathrm{tr}(E_{ab}{E}_{cd})$ on traceless elements because the correction term vanishes there.

With this form, $B(\frac{1}{2n}(E_{ii}-E_{jj}),H)=\mathrm{tr}((E_{ii}-E_{jj})H)=x_i-x_j=\alpha (H)$ for $H=\mathrm{diag}(x_1,\dots ,x_n)\in \mathfrak{h}$, so $H_{\alpha }=\frac{1}{2n}(E_{ii}-E_{jj})$. Since $H_{\alpha }$ is a multiple of the traceless diagonal matrix $E_{ii}-E_{jj}$, we get $\alpha (H_{\alpha })=\frac{1}{2n}\alpha (E_{ii}-E_{jj})=\frac{2}{2n}=\frac{1}{n}$, and therefore $2H_{\alpha }\mathrm{/}\alpha (H_{\alpha })=2n{H}_{\alpha }=E_{ii}-E_{jj}$.

The bracket relations are matrix multiplications: $[E_{ij},E_{ji}]=E_{ii}-E_{jj}$, $[E_{ii}-E_{jj},E_{ij}]=2E_{ij}$ and $[E_{ii}-E_{jj},E_{ji}]=-2E_{ji}$. Hence the displayed triple satisfies exactly the relations of The special linear Lie algebra sl_2 with $h_{\alpha }$ in the role of $h$, which is the claim of The root sl_2 triple realized concretely.

The roots of ${\mathfrak{s}\mathfrak{l}}_3(\mathbb{C})$ are the six functionals ${\epsilon }_i-{\epsilon }_j$, $i\mathrm{\ne }j$, so $\beta$, $\beta +\alpha ={\epsilon }_1-{\epsilon }_3$, $\alpha$ and $-\alpha$ are roots while $2\alpha =2{\epsilon }_1-2{\epsilon }_2$ is not, by the reducedness statement that the only scalar multiples of a root that are roots are $\pm$ themselves.

The Cartan integer evaluates as $\beta (h_{\alpha })=({\epsilon }_2-{\epsilon }_3)(E_{11}-E_{22})$: the diagonal matrix $E_{11}-E_{22}$ has coordinate vector $(1,-1,0)$, so ${\epsilon }_2-{\epsilon }_3$ gives $(-1)-0=-1$, matching $p-q=0-1=-1$.

For the $\alpha$-string through $\beta$: the indices $k\in \mathbb{Z}$ with $\beta +k\alpha$ a root or $0$ are $k=0,1$; indeed $\beta +\alpha ={\epsilon }_1-{\epsilon }_3$ is a root, while $\beta -\alpha =2{\epsilon }_2-{\epsilon }_1-{\epsilon }_3$ and $\beta +2\alpha =2{\epsilon }_1-{\epsilon }_2-{\epsilon }_3$ are not among the six roots and are nonzero. Hence $p=0$, $q=1$.

For the $\alpha$-string through $\beta =\alpha$, the terms are $\beta +k\alpha =(k+1)\alpha$. They are roots or zero exactly for $k\in \{-2,-1,0\}$, giving the terms $-\alpha ,0,\alpha$ and hence $p=2$, $q=0$, and $p-q=2$. Also $\alpha (h_{\alpha })=({\epsilon }_1-{\epsilon }_2)(E_{11}-E_{22})=1-(-1)=2$, so the string identity $p-q=\beta (h_{\alpha })$ holds.

The equations defining both matrix spaces are closed under commutators, and [L2] makes the resulting Lie algebras semisimple. Their displayed diagonal subalgebras are abelian. Choose $H_B=\mathrm{diag}(1,2,0,-2,-1)$ and $H_C=\mathrm{diag}(1,2,-1,-2)$; each has pairwise distinct diagonal entries. If $X$ normalizes the corresponding diagonal subalgebra, then $[X,H_B]$ or $[X,H_C]$ is diagonal, but a commutator with a diagonal matrix has zero diagonal and therefore vanishes. The matrix-unit bracket formula then makes $X$ diagonal, and the defining form equation places it in ${\mathfrak{h}}_B$ or ${\mathfrak{h}}_C$. Thus each displayed subalgebra is nilpotent and self-normalizing, hence Cartan by [L1].

For ${\mathfrak{s}\mathfrak{o}}_5(\mathbb{C})$, put $a^{\mathrm{\prime }}=6-a$. Its zero-weight space is ${\mathfrak{h}}_B$, and its nonzero weight spaces are spanned by the nonzero vectors $E_{ab}-E_{b^{\mathrm{\prime }}{a}^{\mathrm{\prime }}}$ with $a\mathrm{\ne }b$ and $b\mathrm{\ne }{a}^{\mathrm{\prime }}$. Each is an ${\mathrm{ad}}_H$-eigenvector of weight $H_{aa}-H_{bb}$ because $H_{a^{\mathrm{\prime }}{a}^{\mathrm{\prime }}}=-H_{aa}$ and $H_{b^{\mathrm{\prime }}{b}^{\mathrm{\prime }}}=-H_{bb}$. Listing these weights gives $\pm {\epsilon }_1,\pm {\epsilon }_2,\pm ({\epsilon }_1+{\epsilon }_2),\pm ({\epsilon }_1-{\epsilon }_2)$; the would-be weights $\pm 2{\epsilon }_i$ correspond to $b=a^{\mathrm{\prime }}$, where the displayed vector is zero. Thus there are exactly eight roots, the set $B_2$.

For ${\mathfrak{s}\mathfrak{p}}_4(\mathbb{C})$, the form-compatible root vectors in the $a$-blocks are $E_{12}-E_{43}$ and $E_{21}-E_{34}$, of weights $\pm ({\eta }_1-{\eta }_2)$. The symmetric $b$-block gives $E_{13},E_{24},E_{14}+E_{23}$ of weights $2{\eta }_1,2{\eta }_2,{\eta }_1+{\eta }_2$, and the symmetric $c$-block gives their three negative weights. Hence the roots are $\pm 2{\eta }_1,\pm 2{\eta }_2,\pm ({\eta }_1+{\eta }_2),\pm ({\eta }_1-{\eta }_2)$ — exactly eight roots, the set $C_2$.

By [L3], the two root sets computed in steps 2.1 and 2.2 are reduced crystallographic root systems. The linear map $\phi ({\epsilon }_1)={\eta }_1+{\eta }_2$, $\phi ({\epsilon }_2)={\eta }_1-{\eta }_2$ carries the eight vectors of ${\mathrm{\Phi }}_B$ bijectively onto those of ${\mathrm{\Phi }}_C$. Its two image basis vectors are orthogonal of squared length $2$, so $(\phi u,\phi v)=2(u,v)$ for all $u,v$; the common factor cancels from every Cartan integer. Thus [L3] makes $\phi$ a root-system isomorphism $B_2\cong C_2$.

The roots are ${\epsilon }_i-{\epsilon }_j$ with $i\mathrm{\ne }j$, so $({\epsilon }_i-{\epsilon }_j)(H)=x_i-x_j$; hence a root vanishes at $H$ exactly when $x_i=x_j$ for the corresponding pair.

By Centralizer dimension from vanishing roots the centralizer of $H$ is $\mathfrak{h}\oplus {⨁}_{\alpha (H)=0}{\mathfrak{g}}_{\alpha }$, and each root space is one-dimensional, so $\dim {\mathfrak{s}\mathfrak{l}}_n(\mathbb{C}{)}^H=(n-1)+\mathrm{\#}\{i\mathrm{\ne }j:x_i=x_j\}$.

Therefore $H$ is regular in $\mathfrak{h}$, equivalently ${\mathfrak{s}\mathfrak{l}}_n(\mathbb{C}{)}^H=\mathfrak{h}$, exactly when no root vanishes at $H$, that is, when all the $x_i$ are distinct; this matches the general description of Regular elements form a dense Zariski-open subset of a Cartan subalgebra, whose regular set is the complement of the hyperplanes $x_i=x_j$.

The eigenvalue condition is intrinsic to the diagonal matrix: $\mathrm{diag}(x_1,\dots ,x_n)$ has the $x_i$ as eigenvalues with multiplicity, so pairwise distinct coordinates are exactly pairwise distinct eigenvalues. The stated dimension formula and the identification of the regular case with centralizer dimension $n-1$ follow.

The one-dimensional subalgebra $\mathbb{C}Y$ is abelian and maximal abelian: no abelian subalgebra can strictly contain it, because $\mathfrak{g}$ itself is not abelian, as $[X,Y]=Y\mathrm{\ne }0$. It is also a nonzero ideal, since both $[X,Y]$ and $[Y,Y]$ lie in $\mathbb{C}Y$. Being abelian it is solvable, so $0\mathrm{\ne }\mathbb{C}Y\subseteq \mathrm{rad}(\mathfrak{g})$; consequently $\mathfrak{g}$ is not semisimple.

However $N_{\mathfrak{g}}(\mathbb{C}Y)=\mathfrak{g}$: for $aX+bY$ one has $[aX+bY,Y]=aY\in \mathbb{C}Y$ and $[Y,Y]=0$, so every element of $\mathfrak{g}$ normalizes $\mathbb{C}Y$, while $\mathbb{C}Y\mathrm{\ne }\mathfrak{g}$. Hence $\mathbb{C}Y$ is not a Cartan subalgebra, since a Cartan subalgebra must equal its normalizer.

The notion is therefore strictly finer than maximal abelianness in this nonsemisimple algebra: nonsemisimplicity was proved in step 1.1, and $\mathbb{C}X$ is a Cartan subalgebra, as $[aX+bY,X]=-bY\in \mathbb{C}X$ forces $b=0$, so $N_{\mathfrak{g}}(\mathbb{C}X)=\mathbb{C}X$ and $\mathbb{C}X$ is abelian, whereas $\mathbb{C}Y$ is maximal abelian but not Cartan. This witnesses the failure of the claimed equivalence.

Since $\mathfrak{h}=\mathbb{C}h$ is one-dimensional, so is ${\mathfrak{h}}^{\ast }$; it is spanned by $\alpha$ with $\alpha (h)=2$. The reflection formula gives $s_{\alpha }(\alpha )=\alpha -\alpha (h_{\alpha })\alpha =\alpha -2\alpha =-\alpha$, so $s_{\alpha }=-\mathrm{id}$ on the whole line.

The element $W=e^e{e}^{-f}{e}^e$ is the product of the three matrix exponentials $e^e=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$, $e^{-f}=\left(\begin{array}{cc}1& 0\\ -1& 1\end{array}\right)$, $e^e=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$, which multiplies to $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$; it lies in $S{L}_2(\mathbb{C})$ and satisfies $W^2=-I$, $Wh{W}^{-1}=-h$.

Define ${\tau }_{\alpha }={\mathrm{Ad}}_W$ on the displayed matrix algebra. Since ${\mathrm{Ad}}_A{\mathrm{Ad}}_B={\mathrm{Ad}}_{AB}$ for invertible matrices, step 1.2 gives ${\tau }_{\alpha }={\mathrm{Ad}}_{e^e}{\mathrm{Ad}}_{e^{-f}}{\mathrm{Ad}}_{e^e}$ for the explicitly chosen standard triple. Direct conjugation acts on its basis by $h\mapsto -h$, $e\mapsto -f$, $f\mapsto -e$: indeed $We{W}^{-1}=\left(\begin{array}{cc}0& 0\\ -1& 0\end{array}\right)=-f$ and $Wf{W}^{-1}=\left(\begin{array}{cc}0& -1\\ 0& 0\end{array}\right)=-e$. Hence ${\tau }_{\alpha }$ maps ${\mathfrak{g}}_{\alpha }=\mathbb{C}e$ to ${\mathfrak{g}}_{-\alpha }=\mathbb{C}f$ and back, and its action on $\mathfrak{h}$ is $-\mathrm{id}$.

Since ${\tau }_{\alpha }$ acts on $\mathfrak{h}$ as $-1$, its induced action on ${\mathfrak{h}}^{\ast }$ is also $-1$: for $\lambda \in {\mathfrak{h}}^{\ast }$ the induced functional is $\lambda \circ {\tau }_{\alpha }^{-1}=\lambda \circ (-\mathrm{id})=-\lambda$. This equals $s_{\alpha }$ by step 1.1, so the Weyl reflection of the root $\alpha$ is realized by the inner automorphism ${\tau }_{\alpha }$ and swaps the roots $\pm \alpha$.

The Killing form of ${\mathfrak{s}\mathfrak{l}}_n(\mathbb{C})$ is $B(X,Y)=2n\mathrm{tr}(XY)$, so for $H_{\alpha }=\frac{1}{2n}(E_{ii}-E_{jj})$ and a diagonal traceless $H$ with coordinates $x_k$ one gets $B(H_{\alpha },H)=\mathrm{tr}((E_{ii}-E_{jj})H)=x_i-x_j=\alpha (H)$. Hence $H_{\alpha }$ is exactly the Killing-dual vector of $\alpha$ from Killing-dual vector of a root, and the map $H\mapsto B(H,\cdot )$ is an isomorphism onto ${\mathfrak{h}}^{\ast }$ because $B{\mathrm{\mid }}_{\mathfrak{h}}$ is nondegenerate by Orthogonality of root spaces and nondegeneracy on the Cartan subalgebra.

Since $E_{ii}-E_{jj}$ is traceless diagonal, $\alpha (E_{ii}-E_{jj})=2$, and therefore $\alpha (H_{\alpha })=\frac{1}{2n}\cdot 2=\frac{1}{n}$, which is nonzero as required by The Killing length of a root is nonzero and agrees with the direct computation $B(H_{\alpha },H_{\alpha })=2n\mathrm{tr}\left(\frac{(E_{ii}-E_{jj}{)}^2}{4n^2}\right)=\frac{1}{n}$.

Hence $h_{\alpha }=2H_{\alpha }\mathrm{/}\alpha (H_{\alpha })=2n{H}_{\alpha }=E_{ii}-E_{jj}$ by Coroot of a Lie-algebra root, and the coroot triple $e_{\alpha }=E_{ij}$, $f_{\alpha }=E_{ji}$, $h_{\alpha }=E_{ii}-E_{jj}$ is the one computed in The root sl_2 triple inside sl_n; in particular the coroot direction is the dual direction of the root under the Killing form, and the normalization is exactly $\alpha (h_{\alpha })=2$.