# The root space decomposition of a semisimple Lie algebra

*Written by GPT-6.1 Sol (OpenAI). Public domain (CC0).*

Commuting diagonalizable adjoint operators separate a Lie algebra into simultaneous eigenspaces. The bracket adds their eigenvalues. Each pair of opposite nonzero eigenspaces then supplies a copy of \(\mathfrak{sl}_2\). Its finite-dimensional representation theory forces the eigenspaces to be lines, arranges them in strings and supplies reflections. Finally, a trace calculation turns the resulting finite set of eigenvalues into a Euclidean root system.

Throughout, \(\mathfrak g\) is a finite-dimensional semisimple Lie algebra over \(\mathbb C\), and \(\kappa\) is its Killing form. The exact prerequisites are [RTF-001 (Engel)](RT-LIE-foundations.md#RTF-001), [RTF-002 (Lie)](RT-LIE-foundations.md#RTF-002), [RTF-003 (all finite rank-one modules)](RT-LIE-foundations.md#RTF-003), [RT-LIE-03, Theorem 3.3](RT-LIE-03.md#RTLIE03-T33) for the nondegenerate Killing form and zero centre, and [RT-LIE-03, Theorem 6.2](RT-LIE-03.md#RTLIE03-T62) for intrinsic Jordan decomposition. The rank-one decomposition is proved algebraically in the companion before root geometry; no compact real form or general Weyl theorem is assumed here.

The zero algebra has the empty decomposition and a zero-dimensional Euclidean space. Statements about a root concern the nonzero case. We postpone conjugacy of the chosen subalgebras to *Cartan subalgebras and their conjugacy*.

## 1. Find commuting diagonalizable adjoint operators

An element \(x\) is **semisimple** here when \(\operatorname{ad}x\) is diagonalizable. A **toral subalgebra** is an abelian subalgebra of such elements. Commuting diagonalizable operators are simultaneously diagonalizable: diagonalize one, restrict the others to its preserved eigenspaces, and continue with a finite spanning list of operators. Consequently every linear combination in an abelian family of semisimple elements is semisimple.

The abelian requirement is automatic if a subalgebra consists entirely of semisimple elements.

<a id="RTLIE07-L11"></a>
**Lemma 1.1.** A subalgebra \(\mathfrak t\subseteq\mathfrak g\) all of whose elements are semisimple is abelian.

**Proof.** For \(x\in\mathfrak t\), restriction of \(\operatorname{ad}x\) to the invariant space \(\mathfrak t\) is diagonalizable. If it had an eigenvector y of nonzero eigenvalue a, Jacobi would make \(\operatorname{ad}y\) shift every x-eigenvalue b on \(\mathfrak g\) to b+a. Repeated shifts eventually leave the finite spectrum; characteristic zero and \(a\ne0\) give a common finite bound. Thus \(\operatorname{ad}y\) is nilpotent. It is diagonalizable by hypothesis, hence zero. The zero-centre consequence of RT-LIE-03, Theorem 3.3, gives y=0, a contradiction. The restriction therefore has only zero eigenvalues, so x commutes with all of \(\mathfrak t\). This holds for every x. \(\square\)

This lemma does not say that the set of all semisimple elements is itself a subalgebra. For example, in \(\mathfrak{sl}_2\), \(h\) and \(h+e\) are semisimple, whereas their bracket \(2e\) is nonzero nilpotent.

Maximal toral subalgebras exist: start with zero and choose a toral subalgebra of largest dimension. There is a nonzero one if \(\mathfrak g\ne0\). Indeed, if every \(\operatorname{ad}x\) were nilpotent, RTF-001 would make \(\mathfrak g\) nilpotent, contradicting semisimplicity. Choose an \(x\) with a nonzero semisimple Jordan part \(x_s\). RT-LIE-03, Theorem 6.2, gives \(x_s\in\mathfrak g\) and \(\operatorname{ad}x_s=(\operatorname{ad}x)_s\); its span is a nonzero toral subalgebra.

Fix a maximal toral subalgebra \(\mathfrak h\). Simultaneous diagonalization gives
\[
\mathfrak g=\bigoplus_{\lambda\in\mathfrak h^*}\mathfrak g_\lambda,
\qquad
\mathfrak g_\lambda=\{v:[H,v]=\lambda(H)v
\text{ for every }H\in\mathfrak h\}.
\tag{1.1}
\]
Only finitely many summands are nonzero. Linearity of the adjoint action makes each eigenvalue assignment \(\lambda\) a linear functional. Write
\[
\Phi=\{\lambda\ne0:\mathfrak g_\lambda\ne0\}.
\]
These functionals are the **roots**, and their eigenspaces are **root spaces**. At this stage their dimensions are unknown.

Jacobi and invariance of the Killing form give
\[
[\mathfrak g_\lambda,\mathfrak g_\mu]
\subseteq\mathfrak g_{\lambda+\mu},
\qquad
\kappa(\mathfrak g_\lambda,\mathfrak g_\mu)=0
\quad\text{if }\lambda+\mu\ne0.
\tag{1.2}
\]
For the bracket, apply \([H,-]\) and use the derivation rule. For the pairing, if \(v,w\) have weights \(\lambda,\mu\), then
\[
0=\kappa([H,v],w)+\kappa(v,[H,w])
=(\lambda+\mu)(H)\kappa(v,w).
\]
Choose \(H\) on which the nonzero functional \(\lambda+\mu\) does not vanish.

Nondegeneracy on \(\mathfrak g\) now gives a nondegenerate pairing between \(\mathfrak g_\lambda\) and \(\mathfrak g_{-\lambda}\). A vector orthogonal to that opposite space is orthogonal to every summand in (1.1), hence zero. In particular \(-\alpha\in\Phi\) whenever \(\alpha\in\Phi\), and \(\kappa\) restricts nondegenerately to \(\mathfrak g_0\).

## 2. Why the zero eigenspace is exactly the chosen subalgebra

<a id="RTLIE07-T21"></a>
**Theorem 2.1.** For the maximal toral \(\mathfrak h\),
\[
C_{\mathfrak g}(\mathfrak h)=\mathfrak h,\qquad
\mathfrak g=\mathfrak h\oplus\bigoplus_{\alpha\in\Phi}\mathfrak g_\alpha,
\qquad \kappa|_{\mathfrak h}\text{ is nondegenerate}.
\tag{2.1}
\]

**Proof.** Set \(C=\mathfrak g_0\). Section 1 has proved that the Killing restriction to C is nondegenerate. For \(x\in C\), \(\operatorname{ad}x\) commutes with every \(\operatorname{ad}H\), \(H\in\mathfrak h\). Its Jordan parts are polynomials in that operator, so also commute with these adjoints. RT-LIE-03, Theorem 6.2, and zero centre therefore put \(x_s,x_n\) in C. The commuting diagonalizable family \(\mathfrak h+\mathbb Cx_s\) is toral; maximality forces \(x_s\in\mathfrak h\). Its adjoint vanishes on C. Hence every \(\operatorname{ad}_C x=\operatorname{ad}_C x_n\) is nilpotent. RTF-001 makes C nilpotent, and therefore solvable.

Triangularize C's representation on all of \(\mathfrak g\) using RTF-002. Derived elements act strictly upper triangularly, and products with the represented C have zero trace. Thus \(\kappa(C,[C,C])=0\). The nondegenerate restriction gives \([C,C]=0\).

Now \(x_n\in C\) commutes with every y in C. The product of its nilpotent adjoint with \(\operatorname{ad}y\) is nilpotent: its Nth power factors as \((\operatorname{ad}x_n)^N(\operatorname{ad}y)^N\). Its trace is zero, so \(\kappa(x_n,C)=0\). Nondegeneracy gives \(x_n=0\). Every x in C is therefore its semisimple part in \(\mathfrak h\). This proves \(C=\mathfrak h\), and Section 1 gives the remaining assertions. \(\square\)

The roots span \(\mathfrak h^*\). Otherwise their common annihilator contains a nonzero \(H\). It commutes with \(\mathfrak h\) and every root space, so is central in \(\mathfrak g\), a contradiction.

For \(\lambda\in\mathfrak h^*\), let \(t_\lambda\in\mathfrak h\) be its unique Killing-dual vector:
\[
\kappa(t_\lambda,H)=\lambda(H)
\quad(H\in\mathfrak h).
\tag{2.2}
\]
This is a linear isomorphism \(\mathfrak h^*\to\mathfrak h\). The induced bilinear form on the dual is
\[
(\lambda,\mu)=\kappa(t_\lambda,t_\mu)
=\lambda(t_\mu).
\tag{2.3}
\]
It is complex bilinear at present; positivity has not yet been established.

For \(u\in\mathfrak g_\alpha\), \(v\in\mathfrak g_{-\alpha}\), the bracket belongs to \(\mathfrak h\), and
\[
\kappa([u,v],H)=\kappa(u,[v,H])
=\alpha(H)\kappa(u,v).
\]
Thus
\[
[u,v]=\kappa(u,v)t_\alpha,\qquad
[\mathfrak g_\alpha,\mathfrak g_{-\alpha}]
=\mathbb Ct_\alpha.
\tag{2.4}
\]
The equality follows because the opposite-space pairing is nondegenerate and \(t_\alpha\ne0\).

## 3. One rank-one algebra at every root

A nonzero vector for a complex bilinear form can have square zero. We must prove that this does not happen to a root.

<a id="RTLIE07-L31"></a>
**Lemma 3.1.** For every \(\alpha\in\Phi\), \(a_\alpha=(\alpha,\alpha)=\alpha(t_\alpha)\ne0\). There are \(e_\alpha\in\mathfrak g_\alpha\), \(f_\alpha\in\mathfrak g_{-\alpha}\) such that
\[
h_\alpha=\frac{2t_\alpha}{a_\alpha},\qquad
[h_\alpha,e_\alpha]=2e_\alpha,\quad
[h_\alpha,f_\alpha]=-2f_\alpha,\quad
[e_\alpha,f_\alpha]=h_\alpha.
\tag{3.1}
\]

**Proof.** Choose opposite-weight u,v with \(\kappa(u,v)=1\), possible by Section 1's pairing. Their bracket is \(t_\alpha\) by (2.4). If \(\alpha(t_\alpha)=0\), the span of u,v,\(t_\alpha\) is a solvable algebra whose derived algebra is the central line \(\mathbb Ct_\alpha\). RTF-002 makes \(\operatorname{ad}t_\alpha\) nilpotent in its action on \(\mathfrak g\). But it is diagonalizable since \(t_\alpha\in\mathfrak h\). Thus it is zero, and zero centre implies \(t_\alpha=0\), contradicting its defining nonzero functional \(\alpha\). The scalar \(a_\alpha\) is consequently nonzero. Set \(e_\alpha=u\), \(f_\alpha=2v/a_\alpha\), \(h_\alpha=2t_\alpha/a_\alpha\); their weight equations and (2.4) give (3.1). Distinct weights \(\alpha,0,-\alpha\) make the three elements independent. \(\square\)

The vectors have distinct weights \(\alpha,0,-\alpha\), so are independent and their span is isomorphic to \(\mathfrak{sl}_2\). \(\square\)

The vector \(h_\alpha\) is the **coroot**, with \(\alpha(h_\alpha)=2\). The algebra \(\mathfrak g\), under the adjoint action of this rank-one subalgebra, is a finite-dimensional \(\mathfrak{sl}_2\)-module. RTF-003 gives
\[
\beta(h_\alpha)\in\mathbb Z
\quad\text{for every }\beta\in\Phi.
\tag{3.2}
\]

We next prove the root-space dimension and reducedness before using any Euclidean geometry.

<a id="RTLIE07-P32"></a>
**Proposition 3.2.** For every root \(\alpha\),
\[
\dim\mathfrak g_\alpha=1,\qquad
\mathbb C\alpha\cap\Phi=\{\alpha,-\alpha\}.
\tag{3.3}
\]

**Proof.** Take the rank-one module
\[
M=\mathfrak h\oplus\bigoplus_{c\alpha\in\Phi}\mathfrak g_{c\alpha}.
\]
The bracket-addition and opposite-bracket formulas make it invariant. Its \(h_\alpha\)-weight-zero space is precisely \(\mathfrak h\), and raising there is \(H\mapsto-\alpha(H)e_\alpha\), of rank one. RTF-003 decomposes M into irreducibles. Each nontrivial even string \(V(2m)\) contributes one to this rank; an odd string has no zero weight, and a trivial string contributes zero. The contained rank-one algebra is already the irreducible string \(V(2)\). Its highest vector e lies in the kernel of raising at weight two. Within an even string this kernel occurs at weight two only for \(V(2)\); hence at least one \(V(2)\) summand occurs. The total rank one forces it to be the only nontrivial even string.

Weight two in M is exactly \(\mathfrak g_\alpha\); it is therefore one-dimensional. Weight four is absent, so \(2\alpha\) is not a root. The same even-string argument excludes every integer multiple other than \(\pm\alpha\).

If \(c\alpha\) is a root, integrality (3.2) gives \(2c\in\mathbb Z\). Duality gives \(t_{c\alpha}=ct_\alpha\) and \(h_{c\alpha}=h_\alpha/c\), so the reverse pairing gives \(2/c\in\mathbb Z\). These nonzero integers have product four, hence \(c\in\{\pm\tfrac12,\pm1,\pm2\}\). The double cases have already been excluded. A half case would make twice the root \(c\alpha\) a root, excluded by the same argument for that root. Only \(c=\pm1\) remains. \(\square\)

Scaling the Killing form by a nonzero scalar \(z\) changes \(t_\alpha\) to \(t_\alpha/z\) and its induced root square to \(a_\alpha/z\). Their ratio in (3.1) is unchanged. Thus the coroot is independent of that scaling, while its unnormalized dual vector is not.

## 4. Strings, brackets and reflections

<a id="RTLIE07-T41"></a>
**Theorem 4.1.** Fix \(\alpha\in\Phi\). If \(\beta\in\Phi\) is not proportional to \(\alpha\), there are integers \(r,q\ge0\) such that
\[
\{j\in\mathbb Z:\beta+j\alpha\in\Phi\}
=\{-r,-r+1,\ldots,q\},\qquad
r-q=\beta(h_\alpha).
\tag{4.1}
\]
For all roots \(\beta\), including \(\pm\alpha\),
\[
s_\alpha(\beta)=\beta-\beta(h_\alpha)\alpha\in\Phi.
\tag{4.2}
\]
Whenever \(\alpha+\beta\in\Phi\),
\[
[\mathfrak g_\alpha,\mathfrak g_\beta]
=\mathfrak g_{\alpha+\beta}.
\tag{4.3}
\]

**Proof.** For nonproportional \(\beta\), the finite module
\(M=\bigoplus_j\mathfrak g_{\beta+j\alpha}\)
has no zero-root summand and is stable under the rank-one raising and lowering operators. Its \(h_\alpha\)-weights are the integers \(\beta(h_\alpha)+2j\), each of multiplicity at most one by Proposition 3.2. RTF-003 makes it a sum of simple strings, all of the same parity. Two even strings would share weight zero, and two odd strings would share weight one; either would violate that multiplicity. Thus M is one string \(V(n)\).

Its index interval is \(-r\le j\le q\), with r,q nonnegative because index zero occurs. The endpoint weights satisfy \(\beta(h_\alpha)-2r=-n\) and \(\beta(h_\alpha)+2q=n\). Their sum proves \(\beta(h_\alpha)=r-q\), and every intermediate index occurs. Reflection has index \(q-r\), which lies in this interval. The proportional cases are \(\pm\alpha\), interchanged by reflection.

If \(\alpha+\beta\) is a root, reducedness ensures nonproportionality. Its existence means that raising is an interior step of M and hence nonzero by the explicit formulas in RTF-003. The target is a line, as is \(\mathfrak g_\alpha=\mathbb Ce_\alpha\). The nonzero bracket therefore fills the target. \(\square\)

The nonproportional condition in (4.1) is necessary. In \(\{\alpha,-\alpha\}\), the translates of \(\beta=\alpha\) that are roots have indices \(0,-2\); index \(-1\) gives zero and is absent. The rank-one module has a zero-weight coroot between those two root spaces, but that coroot is not a root.

For an \(A_2\) example take \(\alpha=\epsilon_1-\epsilon_2\) and \(\beta=\epsilon_2-\epsilon_3\). The string is \(\beta,\beta+\alpha\), so \(r=0,q=1\) and \(\beta(h_\alpha)=-1\). The bracket of the corresponding matrix units is
\([E_{12},E_{23}]=E_{13}\), realizing the nonzero step in (4.3).

## 5. Recover a rational and positive Euclidean form

The complex bilinear form (2.3) becomes positive on a particular real space. That conclusion requires a real form; it is not a positivity claim on all of \(\mathfrak h^*\).

<a id="RTLIE07-T51"></a>
**Theorem 5.1.** Put
\[
\mathfrak h_{\mathbb Q}=\operatorname{span}_{\mathbb Q}
\{h_\alpha:\alpha\in\Phi\},\quad
\mathfrak h_{\mathbb R}=\operatorname{span}_{\mathbb R}
\{h_\alpha:\alpha\in\Phi\},\quad
E=\operatorname{span}_{\mathbb R}\Phi.
\]
Then \(\mathfrak h_{\mathbb R}\) is a real form of \(\mathfrak h\), the restriction of \(\kappa\) to it is real and positive definite, and \(E\) is its real dual under evaluation. The induced form (2.3) is positive definite on \(E\) and rational on \(\operatorname{span}_{\mathbb Q}\Phi\). With this form, \(\Phi\) is a reduced root system.

**Proof.** Roots span the complex dual, as proved after Theorem 2.1; their Killing-dual vectors and thus their coroots span \(\mathfrak h\). Choose a complex basis \(H_j\) among the coroots and a complex dual-space basis \(\gamma_i\) among the roots. The evaluation matrix \(A_{ij}=\gamma_i(H_j)\) is invertible and integral by (3.2). For every coroot \(h_\beta=\sum c_jH_j\), its evaluation column is integral, so \(c=A^{-1}(\gamma_i(h_\beta))\) is rational. This proves
\[
\mathfrak h_{\mathbb Q}=\bigoplus_j\mathbb QH_j,\qquad
\mathfrak h_{\mathbb R}=\bigoplus_j\mathbb RH_j,\qquad
\mathfrak h=\mathfrak h_{\mathbb R}\oplus i\mathfrak h_{\mathbb R}.
\tag{5.1}
\]
Every root evaluates really on this real form. Since every root space is a line, tracing the diagonal adjoint action gives
\[
\kappa(H,K)=\sum_{\gamma\in\Phi}\gamma(H)\gamma(K).
\tag{5.2}
\]
For real H its square is a sum of real squares, positive unless every root vanishes on H. Their common annihilator is zero. Thus the restriction is real and positive definite. The invertible real evaluation matrix also shows that the \(\gamma_i\) are a basis of the real dual; consequently E is that dual under complex-linear extension.

The Gram matrix \(G_{ij}=\kappa(H_i,H_j)\) is integral by (5.2) and positive definite. The evaluations of a root on the \(H_i\) are integers, so solving the duality equation with \(G^{-1}\) puts its Killing-dual vector in \(\mathfrak h_{\mathbb Q}\). The dual form is rational on the rational root span and positive definite on E, being the inverse positive form. Finally
\[
\beta(h_\alpha)=\frac{2(\beta,\alpha)}{(\alpha,\alpha)}\in\mathbb Z.
\tag{5.3}
\]
Section 1 gives a finite nonzero spanning root set, Proposition 3.2 gives reducedness, and Theorem 4.1 gives reflection stability. The map \(s_\alpha(\lambda)=\lambda-2(\lambda,\alpha)\alpha/(\alpha,\alpha)\) fixes the perpendicular hyperplane and negates \(\alpha\); decomposing along these two spaces proves it orthogonal. All Euclidean reduced crystallographic root-system axioms follow. \(\square\)

For the zero algebra this argument is the empty-basis calculation, with \(E=0\) and \(\Phi=\varnothing\). If a definition requires root systems to be nonempty, this case is recorded separately.

## 6. Read the classical roots directly from matrices

Write \(E_{ij}\) for matrix units. The following calculations use convenient forms with paired coordinate blocks. Reversing the order in the last block makes the symmetric paired form anti-diagonal. These are changes of basis, so they do not change the Lie algebra.

For \(\mathfrak{sl}_n\), \(n\ge2\), take diagonal traceless
\(H=\operatorname{diag}(t_1,\ldots,t_n)\) and define \(\epsilon_i(H)=t_i\). Then
\[
[H,E_{ij}]=(t_i-t_j)E_{ij},\qquad
\Phi=\{\epsilon_i-\epsilon_j:i\ne j\},\qquad
\mathfrak g_{\epsilon_i-\epsilon_j}=\mathbb CE_{ij}.
\tag{6.1}
\]
The diagonal zero-weight space has dimension \(n-1\), and \(\sum_i\epsilon_i=0\). The root space is the real hyperplane \(\sum_i x_i=0\) in \(\mathbb R^n\), with the indicated difference vectors. There are \(n(n-1)\) roots.

For even orthogonal and symplectic algebras use, respectively,
\[
Q=\begin{pmatrix}0&I\\I&0\end{pmatrix},
\qquad
J=\begin{pmatrix}0&I\\-I&0\end{pmatrix}.
\]
Solving \(X^{\mathsf t}Q+QX=0\) or
\(X^{\mathsf t}J+JX=0\) gives
\[
X=\begin{pmatrix}A&B\\C&-A^{\mathsf t}\end{pmatrix},
\tag{6.2}
\]
where \(B,C\) are skew-symmetric for \(Q\) and symmetric for \(J\). Indeed the lower-right block is forced by the upper-left one; the other two equations are exactly the displayed symmetry conditions.

In both cases choose
\(H(t)=\operatorname{diag}(t_1,\ldots,t_n,-t_1,\ldots,-t_n)\).
Write
\[
A(F)=\begin{pmatrix}F&0\\0&-F^{\mathsf t}\end{pmatrix},\qquad
B(F)=\begin{pmatrix}0&F\\0&0\end{pmatrix},\qquad
C(F)=\begin{pmatrix}0&0\\F&0\end{pmatrix}.
\]
These satisfy (6.2) when \(F\) has the required symmetry in the latter two cases. Put \(\epsilon_i(H(t))=t_i\). The root vectors and weights are
\[
\begin{array}{c|c}
\text{matrix}&\text{weight}\\ \hline
A(E_{ij}),\ i\ne j&\epsilon_i-\epsilon_j\\
B(E_{ij}-E_{ji}),\ i<j&\epsilon_i+\epsilon_j\\
C(E_{ij}-E_{ji}),\ i<j&-\epsilon_i-\epsilon_j
\end{array}
\quad\text{for }\mathfrak{so}_{2n},
\tag{6.3}
\]
whereas for \(\mathfrak{sp}_{2n}\) replace the two minus signs inside the \(B,C\) matrices by plus signs and also include
\[
B(E_{ii})\text{ of weight }2\epsilon_i,\qquad
C(E_{ii})\text{ of weight }-2\epsilon_i.
\tag{6.4}
\]
Each weight is verified by multiplying the diagonal \(H(t)\) on the left and right. The free entries in (6.2) show that these vectors together with \(A(E_{ii})\) form a basis. Hence the complete root lists are
\[
\begin{aligned}
D_n:\quad&\{\pm\epsilon_i\pm\epsilon_j:i<j\},
&&n\ge2,\\
C_n:\quad&\{\pm\epsilon_i\pm\epsilon_j:i<j\}
\ \cup\ \{\pm2\epsilon_i\},
&&n\ge1.
\end{aligned}
\tag{6.5}
\]
The two signs in a pair are independent. The counts are \(2n(n-1)\) and \(2n^2\), respectively.

For odd orthogonal matrices order the coordinates in blocks of sizes \(n,1,n\), and take
\[
Q_o=\begin{pmatrix}0&0&I\\0&1&0\\I&0&0\end{pmatrix}.
\]
Solving the defining equation gives every matrix uniquely as
\[
X=\begin{pmatrix}
A&u&B\\
v^{\mathsf t}&0&-u^{\mathsf t}\\
C&-v&-A^{\mathsf t}
\end{pmatrix},
\qquad B^{\mathsf t}=-B,\quad C^{\mathsf t}=-C.
\tag{6.6}
\]
For example the upper-middle equation forces the lower-middle block to be \(-v\), the middle-right equation forces \(-u^{\mathsf t}\), and the middle diagonal equation forces zero. Choose
\(H(t)=\operatorname{diag}(t_1,\ldots,t_n,0,-t_1,\ldots,-t_n)\).
The \(A,B,C\) entries give the \(D_n\) roots. A single \(u_i\) entry has weight \(\epsilon_i\), and a single \(v_i\) entry has weight \(-\epsilon_i\). These supply the additional roots
\[
B_n:\quad
\{\pm\epsilon_i\pm\epsilon_j:i<j\}
\ \cup\ \{\pm\epsilon_i\},
\qquad n\ge1.
\tag{6.7}
\]
There are \(2n^2\) roots. In all three block models the zero-weight space is precisely the chosen diagonal algebra. Thus it centralizes only itself and is maximal toral: a larger toral algebra would commute with it and lie in that same zero space.

The lists also verify semisimplicity of these classical models without assuming the classification. Taking the trace of \(\operatorname{ad}H(t)\operatorname{ad}H(u)\) in the displayed matrix bases gives
\[
\begin{array}{c|c|c}
\text{algebra}&\kappa(H(t),H(u))&
\text{dual coordinate metric}\\ \hline
\mathfrak{sl}_n&2n\sum_i t_i u_i,\ \sum t_i=\sum u_i=0
&\frac1{2n}\text{ times the hyperplane dot product}\\
\mathfrak{so}_{2n+1}&(4n-2)\sum_i t_i u_i&
\frac1{4n-2}\text{ times the dot product}\\
\mathfrak{sp}_{2n}&4(n+1)\sum_i t_i u_i&
\frac1{4(n+1)}\text{ times the dot product}\\
\mathfrak{so}_{2n}&4(n-1)\sum_i t_i u_i&
\frac1{4(n-1)}\text{ times the dot product}
\end{array}
\tag{6.8}
\]
For the difference roots of \(\mathfrak{sl}_n\), expanding
\(\sum_{i\ne j}(t_i-t_j)(u_i-u_j)\) gives the first row. For the other rows each four-root pair
\(\{\pm\epsilon_i\pm\epsilon_j\}\) contributes
\(4(t_i u_i+t_j u_j)\); the short pair \(\pm\epsilon_i\) contributes \(2t_i u_i\), and the long pair \(\pm2\epsilon_i\) contributes \(8t_i u_i\). Summing gives every coefficient in (6.8).

To check nondegeneracy on the whole algebra, not just its diagonal part, opposite root matrices have a bracket \(K\) of the following forms: \(H_i-H_j\) for difference roots, \(H_i+H_j\) for pair-sum roots, and \(H_i\) for the short odd-orthogonal or long symplectic roots, where \(H_i=A(E_{ii})\). In \(\mathfrak{sl}_n\) use \(K=E_{ii}-E_{jj}\). For skew pair-sum matrices take the negative of the \(C\) matrix as the opposite vector; for symmetric pair-sum matrices take the \(C\) matrix itself. Direct multiplication gives these brackets. The \(u_i,v_i\) matrices in (6.6) also bracket to \(H_i\). Their respective root evaluations \(\alpha(K)\) are \(2,2,1,2\), all nonzero.

Invariance yields
\(\kappa(K,K)=\alpha(K)\kappa(X_\alpha,X_{-\alpha})\).
The left side is positive by the diagonal formulas, so every opposite-root pairing is nonzero. All other weight pairings are zero by the argument in (1.2), which uses invariance alone. The diagonal part is nondegenerate by (6.8), and each opposite pair of root lines is nondegenerate. Thus the entire Killing form is nondegenerate and the Killing criterion proves semisimplicity.

The low-dimensional boundaries matter. \(\mathfrak{sl}_1=0\); \(\mathfrak{sp}_2\) and \(\mathfrak{so}_3\) have rank one. The even orthogonal formula starts at \(n=2\): \(\mathfrak{so}_2\) is one-dimensional abelian, its Killing form is zero and it is not a semisimple \(D_1\) example. At \(n=2\), \(D_2\) is a reducible root system.

## 7. Three low-rank pictures

For \(\mathfrak{sl}_3\), the roots are the six vectors
\(\epsilon_i-\epsilon_j\) in the plane \(x_1+x_2+x_3=0\).
Use the ambient orthonormal axes
\[
a=\frac{(1,-1,0)}{\sqrt2},\qquad
b=\frac{(1,1,-2)}{\sqrt6}.
\]
The positive difference vectors have coordinates
\[
\epsilon_1-\epsilon_2=(\sqrt2,0),\quad
\epsilon_2-\epsilon_3=(-1/\sqrt2,\sqrt{3/2}),\quad
\epsilon_1-\epsilon_3=(1/\sqrt2,\sqrt{3/2}),
\tag{7.1}
\]
and the other roots are their negatives. These are six equally spaced points on a circle. The Killing-dual metric is \(1/6\) of the ambient dot product, so each root has squared length \(1/3\). This is the regular hexagon called \(A_2\).

For \(\mathfrak{so}_5\) and \(\mathfrak{sp}_4\), write their coordinate roots as \(B_2\) in \(\epsilon_i\) and \(C_2\) in \(\delta_i\). Their sets are
\[
\begin{aligned}
B_2&=\{\pm\epsilon_1,\pm\epsilon_2,
\pm\epsilon_1\pm\epsilon_2\},\\
C_2&=\{\pm2\delta_1,\pm2\delta_2,
\pm\delta_1\pm\delta_2\}.
\end{aligned}
\tag{7.2}
\]
The map
\[
T(\epsilon_1)=\delta_1+\delta_2,\qquad
T(\epsilon_2)=\delta_1-\delta_2
\tag{7.3}
\]
sends the first root set bijectively to the second: it sends the axis roots to diagonal roots, and the diagonal roots to doubled axis roots. In the usual unscaled coordinates it is a rotation or reflection combined with a dilation by \(\sqrt2\). In the actual Killing-dual metrics it is an isometry, because those matrices are \(I/6\) for \(B_2\) and \(I/12\) for \(C_2\), and \(T^{\mathsf t}T=2I\). Thus the two root configurations agree with their intrinsic metrics. The construction of the corresponding Lie algebra isomorphism belongs to the later classical-model lesson.

For \(\mathfrak{so}_4\), the roots are
\[
\pm(\epsilon_1-\epsilon_2),\qquad
\pm(\epsilon_1+\epsilon_2).
\tag{7.4}
\]
The two pairs are orthogonal. We can see the two Lie algebra factors explicitly in the block model (6.2). Put
\[
K=\begin{pmatrix}0&1\\-1&0\end{pmatrix}.
\]
For \(\alpha=\epsilon_1-\epsilon_2\), take
\[
e_\alpha=A(E_{12}),\quad f_\alpha=A(E_{21}),\quad
h_\alpha=A(\operatorname{diag}(1,-1)).
\]
For \(\beta=\epsilon_1+\epsilon_2\), take
\[
e_\beta=B(K),\quad f_\beta=C(-K),\quad h_\beta=A(I).
\]
Since \(K^2=-I\), multiplying the blocks gives
\([B(K),C(-K)]=A(I)\). The other brackets in each triple are
\([h,e]=2e,[h,f]=-2f\); the first triple also has
\([A(E_{12}),A(E_{21})]=A(\operatorname{diag}(1,-1))\).
Across the two triples, the coroot evaluations are zero and the sums or differences of their root weights are not roots or zero. Hence every cross bracket vanishes by (1.2). The six displayed matrices span \(\mathfrak{so}_4\): the four root lines and two independent diagonal directions are its whole basis. Consequently
\[
\mathfrak{so}_4\simeq
\mathfrak{sl}_2\oplus\mathfrak{sl}_2,
\qquad D_2=A_1\sqcup A_1.
\tag{7.5}
\]
This proof uses no compact group or conjugacy theorem.

## 8. Exercises with complete solutions

**Exercise 8.1 (easy).** Compute the root decomposition of \(\mathfrak{sl}_3\), including its zero-weight space and Killing-dual metric.

**Solution.** Let
\(\mathfrak h=\{\operatorname{diag}(t_1,t_2,t_3):t_1+t_2+t_3=0\}\).
For \(i\ne j\),
\([H,E_{ij}]=(t_i-t_j)E_{ij}\). Thus the six nonzero weights are \(\epsilon_i-\epsilon_j\) and each root space is the span of that matrix unit. The diagonal trace-zero matrices are exactly the zero-weight space; they have dimension two. Therefore
\[
\mathfrak{sl}_3=\mathfrak h\oplus
\mathbb CE_{12}\oplus\mathbb CE_{21}\oplus
\mathbb CE_{13}\oplus\mathbb CE_{31}\oplus
\mathbb CE_{23}\oplus\mathbb CE_{32}.
\]
This accounts for all eight dimensions. The root trace sum is
\(\sum_{i\ne j}(t_i-t_j)(u_i-u_j)=6\sum_i t_i u_i\),
because both coordinate sums are zero. Its inverse metric on the root plane is the ambient dot product divided by six. For example the dual vector for \(\epsilon_i-\epsilon_j\) is
\(t_\alpha=(E_{ii}-E_{jj})/6\), its squared length is \(1/3\), and (3.1) gives \(h_\alpha=E_{ii}-E_{jj}\). The six coordinates in (7.1) and their negatives give the hexagon.

**Exercise 8.2 (medium).** Compute the roots of \(\mathfrak{sp}_4\) and \(\mathfrak{so}_5\), and compare them.

**Solution.** For the symplectic algebra set \(n=2\) in (6.2). Its \(A\)-block is arbitrary and its \(B,C\)-blocks are symmetric. The two diagonal \(A\)-matrices give a two-dimensional zero-weight space. The off-diagonal \(A\)-matrices give \(\pm(\delta_1-\delta_2)\); the off-diagonal symmetric \(B,C\)-matrices give \(\pm(\delta_1+\delta_2)\); their diagonal entries give \(\pm2\delta_1,\pm2\delta_2\). Each entry direction is one-dimensional. These eight root directions and two diagonal directions account for \(4+3+3=10\) dimensions.

For the odd orthogonal algebra use the blocks of sizes \(2,1,2\) in (6.6). The off-diagonal \(A\)-directions give \(\pm(\epsilon_1-\epsilon_2)\), the skew \(B,C\)-directions give \(\pm(\epsilon_1+\epsilon_2)\), and \(u_1,u_2,v_1,v_2\) give \(\pm\epsilon_1,\pm\epsilon_2\). The two diagonal directions are the zero-weight space, and the total dimension is \(4+1+1+2+2=10\).

The eight-root sets are (7.2). Apply (7.3): the images of \(\epsilon_1,\epsilon_2,\epsilon_1+\epsilon_2,\epsilon_1-\epsilon_2\) are respectively
\(\delta_1+\delta_2,\delta_1-\delta_2,2\delta_1,2\delta_2\).
Including negatives proves a bijection. The Killing trace sums give metric matrices \(I/6,I/12\), so \(T^{\mathsf t}(I/12)T=I/6\). This establishes the precise root-system isometry.

**Exercise 8.3 (medium).** Prove \(\dim\mathfrak g_\alpha=1\) using the module formed from root spaces on the line through \(\alpha\).

**Solution.** Include the zero-weight space: take
\(M=\mathfrak h\oplus\bigoplus_{c\alpha\in\Phi}\mathfrak g_{c\alpha}\).
It is a finite module for \(e_\alpha,h_\alpha,f_\alpha\) because raising and lowering shift \(c\) and the bracket of opposite root spaces is in \(\mathfrak h\). The weight-zero space is \(\mathfrak h\), and raising on it is \(H\mapsto-\alpha(H)e_\alpha\), which has rank one.

In a decomposition into \(V(n)\), a nontrivial even \(V(n)\) contributes exactly one to this rank: its zero-weight vector raises to a nonzero weight-two vector. Trivial summands contribute zero and odd summands have no weight zero. The adjoint span \(\mathbb Cf_\alpha\oplus\mathbb Ch_\alpha\oplus\mathbb Ce_\alpha\) is a \(V(2)\) summand by complete reducibility. It already contributes the full rank one, so no other nontrivial even summand occurs. Weight two can occur only in such an even summand. Thus the weight-two space of \(M\) is precisely \(\mathbb Ce_\alpha\); by its definition that space is \(\mathfrak g_\alpha\). Its dimension is one. Leaving out the zero-weight space would not give a submodule, since \([e_\alpha,f_\alpha]=h_\alpha\).

**Exercise 8.4 (hard).** Prove that the Killing-dual form is positive definite on the real span of the roots. Include the rationality assertion.

**Solution.** Roots span \(\mathfrak h^*\), so coroots span \(\mathfrak h\). Choose complex bases \(H_j\) from the coroots and \(\gamma_i\) from the roots. All entries \(\gamma_i(H_j)\) are integers, and their matrix is nonsingular. For any coroot \(h_\beta=\sum c_jH_j\), evaluation by all \(\gamma_i\) gives a nonsingular integer linear system with integer right side. Hence the coefficients \(c_j\) are rational. The real coroot span is therefore \(\bigoplus_j\mathbb RH_j\), a real form of \(\mathfrak h\), and each root takes real values on it.

The root-space dimension is one, so the trace of two diagonal adjoint operators is
\(\kappa(H,K)=\sum_{\alpha\in\Phi}\alpha(H)\alpha(K)\).
For real nonzero \(H\), this gives a sum of nonnegative real squares, and at least one is positive because roots have zero common annihilator. Thus the restriction is positive definite.

The chosen roots have nonsingular real evaluation matrix, so their real span is the whole real dual. Its induced Killing-dual form is the inverse positive definite form and hence positive definite. More explicitly, the Gram matrix \(G=(\kappa(H_i,H_j))\) has integer entries by the trace sum. It is nonsingular, so \(G^{-1}\) has rational entries. Every root evaluation column is integer; dual root pairings are obtained from these columns and \(G^{-1}\), proving rationality on their rational span. This proves both assertions without assuming a compact real form.

## What this lesson does not prove

Engel, Lie, the Killing criterion, zero centre, Jordan decomposition and the finite \(\mathfrak{sl}_2\)-module classification are the proved prerequisites linked at the start. No classification theorem for semisimple Lie algebras or abstract root systems is used.

Conjugacy of maximal toral subalgebras, and their equivalence with the general nilpotent self-normalizing definition of Cartan subalgebra, are treated in *Cartan subalgebras and their conjugacy*. The Lie algebra isomorphism \(\mathfrak{sp}_4\simeq\mathfrak{so}_5\) is not inferred from a picture here; its root-system comparison is proved, and the Lie algebra construction belongs to *The simple Lie algebras: classical models and the exceptional algebras*. Exceptional root systems and their classification are addressed in the subsequent root-system lessons.

The root decomposition, opposite pairings, rank-one triples, one-dimensional root spaces, reducedness, all Cartan integers and reflections, nonproportional root strings, bracket equality, rationality and positivity are all proved in this lesson. The classical matrix decompositions and their semisimplicity checks are included.

## References

- P. Etingof, [*18.745: Lie Groups and Lie Algebras, I*, author-hosted notes](https://math.mit.edu/~etingof/lnlg.pdf), Sections 16–17, especially Theorem 17.3, Lemmas 17.8–17.9, Corollary 17.11, Theorem 17.12 and Corollary 17.13. These are the locators in the actual checked 223-page edition.
- A. Kirillov, Jr., [*Introduction to Lie Groups and Lie Algebras*, author-hosted notes](https://math.stonybrook.edu/~kirillov/liegroups/liegroups.pdf), Sections 6.4–6.6, especially Theorems 6.35, 6.44 and 6.45, for comparison with the root decomposition and positivity proofs.
- J. S. Milne, [*Lie Algebras, Algebraic Groups, and Lie Groups*, version 2.00 (2013)](https://www.jmilne.org/math/CourseNotes/LAG.pdf), Chapter I, §§3–4, for the Killing form used in the receiving prerequisite RT-LIE-03.