By induction on \(i\) – it STS that \([x[ab]] \in L^{(i)}\) for \(a,b \in L^{(i-1)}\) and \(x\in L\). Use the Jacobi identity and the induction hypothesis that \(L^{(i-1)} {~\trianglelefteq~}L\): \begin{align*} [x,[ab]] = [[xa]b] + [a[xb]] \in L^{(i-1)} + L^{(i-1)} \subseteq L^{(i)} .\end{align*}

\(\supseteq\): Centers are always solvable ideals, since it’s abelian and brackets are ideals, and the radical is sum of all solvable ideals.

\(\subseteq\): Suppose \(Z\subsetneq R\) is proper, then there is a non-scalar matrix \(x\in R\). Write \(x = aI_n + y\) for \(a = {\mathrm{tr}}(x)/n\) and \(0\neq y\in {\mathfrak{sl}}_n({\mathbf{C}})\) is traceless. Consider \(I = \left\langle{x}\right\rangle {~\trianglelefteq~}{\mathfrak{gl}}_n({\mathbf{C}})\), i.e. the span of all brackets \([zx]\) for \(z\in {\mathfrak{g}}\) and their iterated brackets containing \(x\), e.g. \([z_1[z_2x]]\). Note that \([zx]=[zy]\) since \(aI_n\) is central. Since \({\mathfrak{sl}}_n({\mathbf{C}})\) is simple, so \(\left\langle{y}\right\rangle_{{\mathfrak{sl}}_n({\mathbf{C}})} = {\mathfrak{sl}}_n({\mathbf{C}})\) and thus \({\mathfrak{sl}}_n({\mathbf{C}}) \subseteq I\). This containment must be proper, since \(I \subseteq \sqrt{(}{\mathfrak{g}})\) and the latter is solvable, so \(I\) must be solvable – but \({\mathfrak{sl}}_n({\mathbf{C}})\) is not solvable. We can thus choose \(x\in I\) such that \(x = aI_n + y\) with \(a\neq 0\) and \(0\neq y \in {\mathfrak{sl}}_n({\mathbf{C}})\), so \(x-y= aI \in I\) since \(y\in I\) because \({\mathfrak{sl}}_n({\mathbf{C}}) \subseteq I\). Since \(a\neq 0\), we must have \(I_n\in I\). Then \({\mathbf{C}}\cdot I_n \subseteq I\), forcing \(I = {\mathfrak{g}}\) since every matrix in \({\mathfrak{gl}}_n({\mathbf{C}})\) is a scalar multiple of the identity plus a traceless matrix. This contradicts that \(I\) is solvable, since \({\mathfrak{g}}^{(1)} \mathrel{\vcenter{:}}=[{\mathfrak{g}}{\mathfrak{g}}] = {\mathfrak{sl}}_n({\mathbf{C}})\). But \({\mathfrak{g}}^{(1)} = {\mathfrak{sl}}_n({\mathbf{C}})\), so the derived series never terminates. \(\contradiction\)

To be covered in an upcoming section.

Let \(\lambda_x, \rho_x \in { \operatorname{End} }({\mathfrak{gl}}(V))\) be left and right multiplication by \(x\), which are commuting nilpotent operators. The binomial theorem shows that if \(D_1, D_2\) are any two commuting nilpotent endomorphisms of a vector space, then \(D_1\pm D_2\) is again nilpotent. But then one can write \({ \operatorname{ad}}_x = \lambda_x - \rho_x\).

Proceed by induction on \(n = \dim L\) (assuming it holds for all vector spaces), where the \(n=1\) case is clear – the characteristic polynomial of such an operator is \(f(t) = t^n\), which has roots \(t=0\) and every field contains zero. Once one has an eigenvalue, there is at least one eigenvector.

For \(n > 1\), suppose \(K \leq L\) is a proper Lie subalgebra. By hypothesis, \(K\) consists of nilpotent elements in \({ \operatorname{End} }(V)\), so apply the previous lemma to see that \({ \operatorname{ad}}(K) \subseteq { \operatorname{End} }(L)\) acts by nilpotent endomorphisms of \(L\) since they are restrictions to \(L\) of nilpotent endomorphisms of \({\mathfrak{gl}}(V)\). Since \([KK] \subseteq K\), we can view \({ \operatorname{ad}}(K) \subseteq { \operatorname{End} }(L/K)\) where \(L/K\) is a vector space.

By the IH with \(V = L/K\), where \({ \operatorname{End} }(L/K)\) has smaller dimension, one can find a nonzero \(x+K \in L/K\) such that \({ \operatorname{ad}}(K)(x+K)=0\). Hence one can find an \(x\in L\setminus K\) such that for all \(y\in K\) one has \([yx] \in K\), so \(x\in N_L(K)\setminus K\). Thus \(K \subsetneq N_L(K)\) is a proper containment.

To be continued.

We’re inducting on \(\dim L\) (over all \(V\)). Assuming \(\dim V > 1\), we showed that proper subalgebras are strictly contained in their normalizers: \begin{align*} K \lneq L \implies K \subsetneq N_L(K) .\end{align*} Let \(K\) be a maximal proper subalgebra of \(L\), then \(N_L(K) = L\) by maximality and thus \(K\) is a proper ideal of \(L\). Then \(L/K\) is a Lie algebra of some dimension, which must be 1 – otherwise the preimage in \(L\) under \(L\twoheadrightarrow L/K\) would be a subalgebra of \(L\) properly between \(K\) and \(L\). Thus \(K\) is a codimension 1 ideal in \(L\).Minimal model program Choosing any \(z\in L\setminus K\) yields a decomposition \(L = K \bigoplus { \mathbf{F} }z\) as vector spaces. Let \(W \mathrel{\vcenter{:}}=\left\{{v\in V{~\mathrel{\Big\vert}~}Kv=0}\right\}\), then \(W\neq 0\) by the IH.

Since \(z\curvearrowright W\) nilpotently, choose an eigenvector \(v\) for \(z\) in \(W\) for the eigenvalue zero. Then \(z.v=0\), so \(Lv=0\).

Induct on \(\dim L\). Note that \({ \operatorname{ad}}(L) \leq {\mathfrak{gl}}(V)\) consists of nilpotent endomorphisms. Use the theorem to pick \(x\in L\) such that \({ \operatorname{ad}}(L).x = 0\), i.e. \([L, x] = 0\), i.e. \(x\in Z(L)\) and thus \(Z(L)\) is nonzero. Now \(\dim L/Z(L) < \dim L\), and a fortiori its elements are still ad-nilpotent so \(L/Z(L)\) is nilpotent. By proposition 3.2b, \(L\) is nilpotent.⁷

Induct on \(\dim V\). Use the theorem to pick a nonzero \(v_1\) with \(Lv_1=0\). Consider \(W\mathrel{\vcenter{:}}= V/{ \mathbf{F} }v_1\), and view \(L \subseteq { \operatorname{End} }(V)\) as a subspace of \({ \operatorname{End} }(W)\) – these are still nilpotent endomorphisms. By the IH, \(W\) has a basis \(\left\{{\overline{v}_i}\right\}_{2\leq i \leq n}\) with respect to the matrices in \(L\) (viewed as a subspace of \({ \operatorname{End} }(W)\)) are strictly upper triangular. Let \(\left\{{v_i}\right\} \subseteq V\) be their preimages in \(L\); this basis has the desired properties. This results in a matrix of the following form:

Induct on \(\dim L\). If \(\dim L = 1\), then \(L\) is spanned by 1 linearly operator \(x\) and over an algebraically closed field, \(x\) has at least one eigenvector. For \(\dim L > 1\), take the following strategy:

1. Identify \(K{~\trianglelefteq~}L\) proper of codimension 1,
2. By IH, find a common eigenvector for \(K\),
3. Verify that \(L\) stabilizes “the” subspace of all such common eigenvectors (much harder than before!)
4. In this subspace, find an eigenvector for some \(z\in L\setminus K\) with \(L = K \oplus { \mathbf{F} }z\).

Off we go!

Step 1: Since \(L\) is solvable, we have \([LL]\) properly contained in \(L\). In \(L/[LL]\) choose any codimension 1 subspace – it is an ideal, which lifts to a codimension 1 ideal \(K \subset L\).

Step 2: Since subalgebras of solvable algebras are again solvable, \(K\) is solvable. By the IH, pick a common nonzero eigenvector \(v\) for \(K\). There exists a linear map \(\lambda: K\to { \mathbf{F} }\) such that \(x.v = \lambda(x) v\) for all \(x\in K\). Let \(W \mathrel{\vcenter{:}}=\left\{{v\in V {~\mathrel{\Big\vert}~}y.v = \lambda(y) v\,\,\forall y\in K}\right\}\), which is nonzero.

Step 3: Note \(L.W \subseteq W\). Let \(w\in W, x\in L, y\in K\); we WTS \(y.(x.w) = \lambda(y)x.w\). Write \begin{align*} y.(x.w) &= x.(y.w) - [xy].w \\ &= \lambda(y)(x.w) - \lambda([xy])w ,\end{align*} where the second line follows since \([xy]\in K\). We then need \(\lambda([xy]) = 0\) for all \(x\in L\) and \(y\in K\). Since \(\dim V < \infty\), choose \(n\) minimal such that \(\left\{{w, x.w, x^2.w,\cdots, x^n.w}\right\}\) is linearly dependent. Set \(W_i \mathrel{\vcenter{:}}=\mathop{\mathrm{span}}_{ \mathbf{F} }\left\{{w, x.w, \cdots, x^i.w}\right\}\), so \(W_0 = 0, W_1 = \mathop{\mathrm{span}}_{ \mathbf{F} }\left\{{w}\right\}\), and so on, noting that

• \(\dim W_n = n\),
• \(W_{n+k} = W_n\) for all \(k\geq 0\),
• \(x.W_n \subseteq W_n\).

To be continued!

Step 3: Fix \(x\in L, w\in W\) and \(n\) minimal such that \(\left\{{x^i w}\right\}_{i\leq n}\) is linearly dependent. For \(i\geq 0\) set \(W_i = { \mathbf{F} }\left\langle{w, xw, \cdots, x^{i-1}w}\right\rangle\). Then \(\dim W_n = n, W_n = W_{n+i}\) for \(i\geq 0\), and \(xW_n \subseteq W_n\).

Given this claim, for \(i=n\) this says that the matrices of any \(y\in K\) with respect to the basis \(\left\{{x^iw}\right\}_{0\leq i \leq n-1}\) is upper triangular with diagonal entries all equal to \(\lambda(y)\). Thus \({ \left.{{\operatorname{tr}(y)}} \right|_{{W_m}} } = n \lambda(y)\), and so \([xy]\curvearrowright W_n\) with trace \(n \lambda([xy])\). On the other hand, \(x,y\) both act on \(W_n\) (e.g. by the formula in the claim for \(yx^i.w\)) and so \begin{align*} { \left.{{[xy]}} \right|_{{W_n}} } = { \left.{{xy}} \right|_{{W_n}} } - { \left.{{yx}} \right|_{{W_n}} } ,\end{align*} thus \({ \left.{{ \operatorname{tr}([xy])}} \right|_{{W_n}} } = 0\). Since \({ \mathbf{F} }\) is characteristic zero, we have \(n \lambda([xy]) = 0 \implies \lambda([xy]) = 0\).

Step 4: By step 1, \(L = K \oplus { \mathbf{F} }z\) for some \(z\in L\setminus K\). Viewing \(z: W\to W\) and using \({ \mathbf{F} }= \overline{{ \mathbf{F} }}\), \(z\) has an eigenvector \(v\in W\). Since \(v\in W\), it is also a common eigenvector for \(K\) and thus an eigenvector for \(L\) by additivity.

Use the theorem and induct on \(n=\dim V\) as in Engel’s theorem – find a common eigenvector for \(V^1\), since \(L\) stabilizes one gets an action \(L\curvearrowright V^i/V^{i-1}\) which is smaller dimension. Then just lift through the quotient.

Consider \({ \operatorname{ad}}L \leq {\mathfrak{gl}}(L)\). Apply Lie’s theorem: \(( { \operatorname{ad}}L)L_i \subseteq L_i \iff [LL_i] \subseteq L_i\), making \(L_i{~\trianglelefteq~}L\) an ideal.

Find a flag of ideals by and let \(\left\{{x_1,\cdots, x_n}\right\}\) be a compatible basis. Then the matrices \(\left\{{ { \operatorname{ad}}_x{~\mathrel{\Big\vert}~}x\in L}\right\}\) are all upper triangular. If \(x\in [LL]\), without loss of generality \(x = [yz]\) for some \(y,z\in L\). Then \begin{align*} { \operatorname{ad}}_x = [ { \operatorname{ad}}_y { \operatorname{ad}}_z] = { \operatorname{ad}}_y { \operatorname{ad}}_z - { \operatorname{ad}}_z { \operatorname{ad}}_y \end{align*} will be strictly upper triangular (since these are upper triangular and the commutator cancels diagonals) and hence nilpotent.

Later!

Let \(z\in \mathop{\mathrm{Rad}}(\beta)\) and \(x,y\in L\). To see if \([zx]\in \mathop{\mathrm{Rad}}(\beta)\), check \begin{align*} \beta([zx], y) = \beta(z, [xy]) = 0 \end{align*} since \(z\in \mathop{\mathrm{Rad}}(\beta)\). Thus \([zx] \in \mathop{\mathrm{Rad}}(\beta)\).

If \(x\in {\mathfrak{gl}}(V)\) is semisimple then so is \({ \operatorname{ad}}_x\) since the eigenvalues for \({ \operatorname{ad}}_x\) are differences of eigenvalues of \(x\). I.e. if \(\left\{{v_1,\cdots, v_n}\right\}\) is an eigenbasis for \(V\) and \(x.v_i = a_i v_i\) in this bases, we have \([x e_{ij}] = (a_i - a_j) = e_{ij}\), so \(\left\{{e_{ij}}\right\}\) is an eigenbasis for \({ \operatorname{ad}}_x\). If \(x\) is nilpotent then \({ \operatorname{ad}}_x\) is nilpotent, since \({ \operatorname{ad}}_x(y) = \lambda_x(y) - \rho_x(y)\) where \(\lambda, \rho\) are left/right multiplication, and sums of nilpotents are nilpotent. One can check \([ { \operatorname{ad}}_{x_s} { \operatorname{ad}}_{x_n}] = { \operatorname{ad}}_{[x_s x_n} = 0\) since they commute.

Let \(\delta\in \mathop{\mathrm{Der}}(U)\) and write \(\delta = \sigma + v\) be the Jordan decomposition of \(\delta\) in \({ \operatorname{End} }(U)\). It STS \(\sigma\) is a derivation, so for \(a\in { \mathbf{F} }\) define \begin{align*} U_a \mathrel{\vcenter{:}}=\left\{{x\in U {~\mathrel{\Big\vert}~}(\delta - a)^k x = 0 \,\,\text{for some } k}\right\} .\end{align*} Note \(U = \bigoplus _{a\in \Lambda} U_a\) where \(\Lambda\) is the set of eigenvalues of \(\delta\), which are also the eigenvalues of \(\sigma\) – this is because \(\sigma, v\) are commuting operators, so eigenvalues of \(\delta\) are sums of eigenvalues of \(s\) and \(v\).

Assuming this, it STS \(\sigma(xy) = \sigma(x)y + x \sigma(y)\) when \(x\in U_a, y\in U_b\) where \(a,b\) are eigenvalues. Using that eigenvalues of \(\delta\) are also eigenvalues of \(\sigma\), since \(xy\in U_{a+b}\) by the claim, \(\sigma(xy) = (a+b)xy\) and thus \begin{align*} \sigma(x)y + x \sigma(y) = axy + xby = (a+b)xy .\end{align*} So \(\sigma\in \mathop{\mathrm{Der}}(U)\).

A sub-claim: \begin{align*} (\delta - (a+b) 1) (xy) = \sum_{0\leq i\leq n} {n\choose i} (\delta - aI)^{n-i}x (\delta- b 1)^i y .\end{align*}

To show \(L\) is solvable, it STS that \([LL]\) is nilpotent since the ideals used to check nilpotency are bigger than those to check solvability. By Engel’s theorem, it STS to show each \(w\in [LL]\) is ad-nilpotent. Since \(L \leq {\mathfrak{gl}}(V)\), it STS to show each \(w\in [LL]\) is a nilpotent endomorphism. As in the setup of the lemma, set \(B = L, A = [LL]\), then \begin{align*} M \mathrel{\vcenter{:}}=\left\{{z\in {\mathfrak{gl}}(V) {~\mathrel{\Big\vert}~}[zL] \subseteq [LL] }\right\} \supseteq L \supseteq[LL] .\end{align*} Let \(w\in [LL] \subseteq M\), then note that \(\operatorname{tr}(wz) = 0\) for all \(z\in L\), but we need to know this for all \(z\in M\). Letting \(z\in M\) be arbitrary; by linearity of the trace it STS \(\operatorname{tr}(wz) = 0\) on generators \(w = [xy]\) on \([LL]\) for \(x,y\in L\). We thus WTS \(\operatorname{tr}([xy]z) = 0\): \begin{align*} \operatorname{tr}([xy]z) &= \operatorname{tr}(x [yz] ) \\ &=\operatorname{tr}([yz] x) \qquad \in \operatorname{tr}(LMx) \subseteq \operatorname{tr}([LL]L) \\ &= 0 \end{align*} by assumption. By the lemma, \(w\) is nilpotent.

Use \({ \operatorname{ad}}: L\to {\mathfrak{gl}}(V)\), a morphism of Lie algebras. Its image is solvable by Cartan’s criterion above, and \(\ker { \operatorname{ad}}= Z(L)\) which is abelian and hence a solvable ideal.⁹ Therefore \(L\) is solvable.

Let \(w = s + n\) be the Jordan-Chevalley decomposition of \(w\). Choose a basis for \(V\) such that this is the JCF of \(w\), i.e. \(s = \operatorname{diag}(a_1,\cdots, a_n)\) and \(n\) is strictly upper triangular. Idea: show \(s=0\) by showing \(A\mathrel{\vcenter{:}}={\mathbf{Q}}\left\langle{a_1,\cdots, a_n}\right\rangle = 0\) by showing \(A {}^{ \vee }= 0\), i.e. any \({\mathbf{Q}}{\hbox{-}}\)linear functional \(f: A\to {\mathbf{Q}}\) is zero. If \(\sum a_i f(a_i) = 0\) then \begin{align*} 0 = f(\sum a_i f(a_i)) = \sum f(a_i)^2 \implies f(a_i) = 0 \,\,\forall i ,\end{align*} so we’ll show this. Let \(y = \operatorname{diag}( f(a_1), \cdots, f(a_n) )\), then \({ \operatorname{ad}}_y\) is a polynomial (explicitly constructed using Lagrange interpolation) in \({ \operatorname{ad}}_s\) without a constant term. So do this for \({ \operatorname{ad}}_y\) and \({ \operatorname{ad}}_s\) (see exercise). Since \({ \operatorname{ad}}_s\) is a polynomial in \({ \operatorname{ad}}_w\) with zero constant term, and since \({ \operatorname{ad}}_w: B\to A\), we have \({ \operatorname{ad}}_s(B) \subseteq A\) and the same is thus true for \({ \operatorname{ad}}_y\). So \(y\in M\) and \(w\in M\), and applying the trace condition in the lemma with \(z\mathrel{\vcenter{:}}= y\) we get \begin{align*} 0 = \operatorname{tr}(wy) = \sum a_i f(a_i) ,\end{align*} noting that \(w\) is upper triangular and \(y\) is diagonal. So \(s=0\) and \(w=n\) is nilpotent.

Let \(x\in I\), then \({ \operatorname{ad}}_x(L) \subseteq I\) since \(I\) is an ideal. Choosing a basis for \(I\) yields a matrix:

So if \(x,y\in I\), we have \begin{align*} \kappa(x,y) &= \operatorname{tr}( { \operatorname{ad}}_x \circ { \operatorname{ad}}_y) \\ &= \operatorname{tr}( { \operatorname{ad}}_{I, x} \circ { \operatorname{ad}}_{I, y}) \\ &= \kappa_I(x, y) .\end{align*}

Let \(S = \mathop{\mathrm{Rad}}(\kappa) {~\trianglelefteq~}L\), which is easy to check using “invariance” (associativity) of the form. Given \(s,s'\in S\), the restricted form \(\kappa_S(x, y) = \operatorname{tr}( { \operatorname{ad}}_{S, s} { \operatorname{ad}}_{S, s'}) = \operatorname{tr}( { \operatorname{ad}}_{L, s} { \operatorname{ad}}_{L, s'})\), which was proved in a previous lemma. But this is equal to \(\kappa(s, s') = 0\). In particular, we can take \(s\in [SS]\), so by (the corollary of) Cartan’s criterion for solvable Lie algebras, \(S\) is solvable as a Lie algebra and thus solvable as an ideal in \(L\).

\(\implies\): Since \(\mathop{\mathrm{Rad}}(L)\) is the sum of all solvable ideals, we have \(S \subseteq \mathop{\mathrm{Rad}}(L)\), but since \(L\) is semisimple \(\mathop{\mathrm{Rad}}(L) = 0\) and thus \(S=0\).

\(\impliedby\): Assume \(S=0\). If \(I {~\trianglelefteq~}L\) is a solvable ideal so \(I^{(n)} = 0\) for some \(n\geq 0\). If \(I^{(n-1)} \neq 0\), it is a nonzero abelian ideal – since we want to show \(\mathop{\mathrm{Rad}}(L) = 0\), we don’t want this to happen! Thus it STS every abelian ideal is contained in \(S\).

So let \(I {~\trianglelefteq~}L\) be an abelian ideal, \(x\in I\), \(y\in L\). Define an operator \begin{align*} A_{xy}^2 \mathrel{\vcenter{:}}=( { \operatorname{ad}}_x { \operatorname{ad}}_y)^2: L \xrightarrow{ { \operatorname{ad}}_y} L \xrightarrow{ { \operatorname{ad}}_x} I \xrightarrow{ { \operatorname{ad}}_y} I \xrightarrow{ { \operatorname{ad}}_x} 0 ,\end{align*} which is zero since \([II] =0\). Thus \(A_{xy}\) is a nilpotent endomorphism, which are always traceless, so \(0 = \operatorname{tr}( { \operatorname{ad}}_x { \operatorname{ad}}_y) = \kappa(x, y)\) for all \(y\in L\), and so \(x\in S\). Thus \(I \subseteq S\).

Let \(I {~\trianglelefteq~}L\) and define \begin{align*} I^\perp \mathrel{\vcenter{:}}=\left\{{x\in L {~\mathrel{\Big\vert}~}\kappa(x, I) = 0 }\right\} ,\end{align*} the orthogonal complement of \(I\) with respect to \(\kappa\). This is an ideal by the associativity of \(\kappa\). Set \(J\mathrel{\vcenter{:}}= I \cap I^\perp {~\trianglelefteq~}L\), then \(\kappa([JJ], J) = 0\) and by Cartan’s criterion \(J\) is a solvable ideal and thus \(J = 0\), making \(L\) semisimple.

From the Endman-Wildon lemma in the appendix (posted on ELC, lemma 16.11), \(\dim L = \dim I + \dim I^\perp\) and \(L = I \oplus I^\perp\), so now induct on \(\dim L\) to get the decomposition when \(L\) is not simple. These are semisimple ideals since solvable ideals in the \(I, I^\perp\) remain solvable in \(L\). Finally let \(I{~\trianglelefteq~}L\) be simple, then \([I, L] \subseteq L\) is a ideal (in both \(L\) and \(I\)), which is nonzero since \(Z(L) = 0\). Since \(I\) is simple, this forces \([I, L] = I\). Writing \(L = \bigoplus I_i\) as a sum of simple ideals, we have \begin{align*} I = [I, L] = [I, \bigoplus I_i] = \bigoplus [I, I_i] ,\end{align*} and by simplicity only one term can be nonzero, so \(I = [I, I_j]\) for some \(j\). Since \(I_j\) is an ideal, \([I, I_j] \subseteq I_j\), and by simplicity of \(I_j\) we have \(I = I_j\).

Take the canonical decomposition \(L = \bigoplus I_i\) and check \begin{align*} [L, L] = [\bigoplus I_i, \bigoplus I_j] = \bigoplus [I_i, I_i] = \bigoplus I_i ,\end{align*} where in the last step we’ve used that the \(I_i\) are (not?) abelian. Let \(J {~\trianglelefteq~}L\) to write \(L = J \bigoplus J^\perp\), both of which are semisimple as Lie algebras. In particular, if \(\phi: L\to L'\), set \(J \mathrel{\vcenter{:}}=\ker \phi {~\trianglelefteq~}L\). Then \(\operatorname{im}\phi = L/J \cong J^\perp\) as Lie algebras, using the orthogonal decomposition, so \(\operatorname{im}\phi\) is semisimple. Finally if \(J {~\trianglelefteq~}L\) then \(L = J \oplus J^\perp\) with \(J\) semisimple, so by the previous theorem \(J\) decomposes as \(J = \oplus K_i\) with \(K_i\) (simple) ideals in \(J\) – but these are (simple) ideals in \(L\) as well since it’s a direct sum. Thus the \(K_i\) are a subset of the \(I_j\), since these are the only simple ideals of \(L\).

We know \({ \operatorname{ad}}(L) \leq \mathop{\mathrm{Der}}L\) is a subalgebra, and \(L\) semisimple implies \(0 = Z(L) = \ker { \operatorname{ad}}\), so \({ \operatorname{ad}}: L { \, \xrightarrow{\sim}\, } { \operatorname{ad}}(L)\) is an isomorphism and \({ \operatorname{ad}}(L)\) is semisimple. The Killing form of a semisimple Lie algebra is always nondegenerate, so \(\kappa_{ { \operatorname{ad}}(L)}\) is nondegenerate, while \(\kappa_{\mathop{\mathrm{Der}}L}\) may be degenerate. Recall that \({ \operatorname{ad}}(L) {~\trianglelefteq~}\mathop{\mathrm{Der}}L\), so \([\mathop{\mathrm{Der}}L, { \operatorname{ad}}(L)] \subseteq { \operatorname{ad}}(L)\). Define \({ \operatorname{ad}}(L)^\perp {~\trianglelefteq~}\mathop{\mathrm{Der}}L\) to be the orthogonal complement in \(\mathop{\mathrm{Der}}(L)\) with respect to \(\kappa_{\mathop{\mathrm{Der}}L}\), which is an ideal by the associative property.

Claim: \({ \operatorname{ad}}(L) \cap { \operatorname{ad}}(L)^\perp = 0\), This follows readily from the fact that \(\kappa_{ { \operatorname{ad}}(L)}\) is nondegenerate and so \(\mathop{\mathrm{Rad}}(\kappa_{ { \operatorname{ad}}(L)}) = 0\).

Note that \({ \operatorname{ad}}(L), { \operatorname{ad}}(L)^\perp\) are both ideals, so \([ { \operatorname{ad}}(L), { \operatorname{ad}}(L)^\perp] \subseteq { \operatorname{ad}}(L) \cap { \operatorname{ad}}(L)^\perp = 0\). Let \(\delta \in { \operatorname{ad}}(L)^\perp\) and \(x\in L\), then \(0 = [\delta, { \operatorname{ad}}_x] = { \operatorname{ad}}_{ \delta(x) }\) where the last equality follows from an earlier exercise. Since \({ \operatorname{ad}}\) is injective, \(\delta(x) = 0\) and so \(\delta = 0\), thus \({ \operatorname{ad}}(L)^\perp = 0\). So we have \(\mathop{\mathrm{Rad}}\kappa_{\mathop{\mathrm{Der}}L} \subseteq { \operatorname{ad}}(L)^\perp = 0\) since any derivation orthogonal to all derivations is in particular orthogonal to inner derivations, and thus \(\kappa_{\mathop{\mathrm{Der}}L}\) is nondegenerate. Finally, we can write \(\mathop{\mathrm{Der}}L = { \operatorname{ad}}(L) \oplus { \operatorname{ad}}(L)^\perp = { \operatorname{ad}}(L) \oplus 0 = { \operatorname{ad}}(L)\).

If \(V\) is irreducible then every \(f\in {}_{L}{\mathsf{Mod}}(V, V)\) is either zero or an isomorphism since \(f(V) \leq V\) is a submodule. Thus \({ \operatorname{End} }_L(V)\) is a division algebra over \({ \mathbf{F} }\), but the only such algebra is \({ \mathbf{F} }\) since \({ \mathbf{F} }= \overline{{ \mathbf{F} }}\).

Letting \(\phi\) be as above, it has an eigenvalue \(\lambda \in { \mathbf{F} }\), again since \({ \mathbf{F} }= \overline{{ \mathbf{F} }}\). Then \(\phi - \lambda I \in { \operatorname{End} }_L(V)\) has a nontrivial kernel, the \(\lambda{\hbox{-}}\)eigenspace. So \(\phi - \lambda I = 0 \implies \varphi = \lambda I\).

Case I: \(\dim V/W = 1\).

Case 1: \(W\) is reducible. We got \(0 < W' < W < V\) (proper submodules), represented schematically by a triangle. We showed \(V/W' \cong W/W' \oplus \tilde W/W'\), since

• \(\tilde W \cap W \subseteq W'\),
• \(V= W + \tilde W + W' = W + \tilde W\) since \(W' \subseteq W\).
• \(\tilde W = W' \oplus X\) for some submodule \(X \leq_L \tilde W \leq_L V\).

Thus replacing \(\tilde W\) in the second point yields \(V = W + \tilde W = W + W' +X = W + X\); we want to prove this sum is direct. Since \(X\) is contained in \(\tilde W\), we can write \begin{align*} X \cap W &= (X \cap\tilde W) \cap W \\ &= X \cap(\tilde W \cap W) \qquad\text{by 1}\\ &\subseteq X \cap W' = 0 \qquad \text{by 2} ,\end{align*} so \(V = W \oplus X\).

Case 2: Let \(c_\phi\) be the Casimir element of \(\phi\), and note that \(c_\phi(W) \subseteq W\) since \(c_\phi\) is built out of endomorphisms in \(\phi(L)\) sending \(W\) to \(W\) (since \(W\) is a submodule). In fact, \(\phi(L)(V) = W\) since \(V/W\) is a 1-dimensional representation of the semisimple Lie algebra \(L\), hence trivial. Thus \(c_\phi(V) \subseteq W\) and thus \(\ker c_\phi \neq 0\) since \(W < V\) is proper. Note also that \(c_\phi\) commutes with anything in \(c_\phi(L)\) on \(V\), and so defines a morphism \(c_\phi \in {}_{L}{\mathsf{Mod}}(V, V)\) and \(\ker c_\phi \leq_L V\). On the other hand, \(c_\phi\) induces an element of \({ \operatorname{End} }_{L}(W)\), and since \(W\) is irreducible, \({ \left.{{c_\phi}} \right|_{{W}} } = \lambda \operatorname{id}_W\) for some scalar \(\lambda\). This can’t be zero, since \(\operatorname{tr}({ \left.{{c_\phi}} \right|_{{W}} }) = {\dim L \over \dim W} > 0\), so \(\ker c_\phi \cap W = 0\). Since \(\operatorname{codim}_V W = 1\), i.e. \(\dim W = \dim V - 1\), we must have \(\dim \ker c_\phi = 1\) and we have a direct sum decomposition \(V = W \oplus \ker c_\phi\).

Use of the Casimir element in basic Lie theory: producing a complement to an irreducible submodule.

Case 2: Suppose \(0 < W < V\) with \(W\) any nontrivial \(L{\hbox{-}}\)submodule; there is a SES \(W \hookrightarrow V \twoheadrightarrow V/W\) Consider \(H \mathrel{\vcenter{:}}=\hom_{\mathbf{C}}(V, W)\), then \(H \in {\mathsf{L}{\hbox{-}}\mathsf{Mod}}\) by \((x.f)(v) \mathrel{\vcenter{:}}= x.(f(v)) - f(x.v)\) for \(f\in H, x\in L, v\in V\). Let \({\mathcal{V}}\mathrel{\vcenter{:}}=\left\{{f {~\mathrel{\Big\vert}~}H {~\mathrel{\Big\vert}~}{ \left.{{f}} \right|_{{W}} } = \alpha \operatorname{id}_W \,\text{ for some } \alpha \in {\mathbf{C}}}\right\} \subseteq H\). For \(f\in V\) and \(w\in W\), we have \begin{align*} (x.f)(w) = x.f(w) - f(x.w) = \alpha x.w - \alpha x.w = 0 .\end{align*} So let \({\mathcal{W}}\mathrel{\vcenter{:}}=\left\{{f\in {\mathcal{V}}{~\mathrel{\Big\vert}~}f(W) = 0}\right\} \subseteq {\mathcal{V}}\), then we’ve shown that \(L.{\mathcal{V}}\subseteq {\mathcal{W}}\). Now roughly, the complement is completely determined by the scalar. Rigorously, since \(\dim {\mathcal{V}}/{\mathcal{W}}= 1\), any \(f\in {\mathcal{V}}\) is determined by the scalar \({ \left.{{f}} \right|_{{W}} } = \alpha \operatorname{id}_W\): we have \(f-\alpha \chi_W \in {\mathcal{W}}\) where \(\chi_W\) is any extension of \(\operatorname{id}_W\) to \(V\) which is zero on \(V/W\), e.g. by extending a basis and having it act by zero on the new basis elements.

Now \({\mathcal{W}}\hookrightarrow{\mathcal{V}}\twoheadrightarrow{\mathcal{V}}/{\mathcal{W}}\in {\mathsf{L}{\hbox{-}}\mathsf{Mod}}\) with \(\operatorname{codim}_{\mathcal{V}}{\mathcal{W}}= 1\). By Case I, \({\mathcal{V}}= {\mathcal{W}}\oplus {\mathcal{W}}''\) for some complement \(L{\hbox{-}}\)submodule \({\mathcal{W}}''\). Let \(f: V\to W\) span \({\mathcal{W}}''\), then \({ \left.{{f}} \right|_{{W}} }\) is a nonzero scalar – a scalar since it’s in \({\mathcal{V}}\), and nonzero since it’s in the complement of \({\mathcal{W}}\). By rescaling, we can assume the scalar is 1, so \(\operatorname{im}f = W\) and by rank-nullity \(\dim \ker f = \dim V - \dim W\). Thus \(\ker f\) has the right dimension to be the desired complement. It is an \(L{\hbox{-}}\)submodule, since \(L.f \subseteq {\mathcal{W}}'' \cap{\mathcal{W}}= 0\) since \({\mathcal{W}}''\) is an \(L{\hbox{-}}\)submodule and \(f\in {\mathcal{W}}\) since \(L.{\mathcal{V}}\subseteq {\mathcal{W}}\). Noting that if \((x.f) = 0\) then \(x.(f(v)) = f(x.v)\), making \(f\) an \({\mathsf{L}{\hbox{-}}\mathsf{Mod}}\) morphism. Thus \(W'' \mathrel{\vcenter{:}}=\ker f \leq_L V\), and \(W \cap W'' = 0\) so \({ \left.{{f}} \right|_{{W}} } = \operatorname{id}_W\). Since the dimensions add up correctly, we get \(V = W \oplus W''\).

The proof is technical, but here’s a sketch:

• Construct a subspace \(L \leq L' \leq_L {\mathfrak{gl}}(V)\) such that \(l'\) contains the semisimple and nilpotent parts of all elements where \(L\curvearrowright{\mathfrak{gl}}(V)\) by \({ \operatorname{ad}}: L\to {\mathfrak{gl}}({\mathfrak{gl}}(V))\).
• Check \(L' {\mathfrak{q}}N_{{\mathfrak{gl}}(V)}(L)\) (the normalizer), so \([LL'] \subseteq L\).
• Show \(L' = L\):
  • If \(L'\neq L\), use Weyl’s theorem to get a complement with \(L' = L \oplus M\).
  • Check \([LM] \subseteq [LL'] \subseteq L\) and \([LM] \subseteq M\) since \(M\leq_L M\), forcing \([LM] \subseteq L \cap M = 0\).
  • Use Weyl’s theorem on all of \(V\) splits it into sums of irreducibles, bracket against the irreducibles, and use specific properties of this \(L'\) to show \(M = 0\).
• Since \(s, n\in L\) when \(x=s+n\) and \({ \operatorname{ad}}_x = { \operatorname{ad}}_s + { \operatorname{ad}}_n = { \operatorname{ad}}_{x_s} + { \operatorname{ad}}_{x_n}\), the result follows from uniqueness of the abstract JCF that \(s=x_s, n=x_n\) (using that \({ \operatorname{ad}}\) is injective when \(L\) is semisimple since \(Z(L) = 0\)).

Consider \({ \operatorname{ad}}_{\phi(L)}\phi(s)\) and \({ \operatorname{ad}}_{\phi(L)}\phi(n)\), which are semisimple (acts on a vector space and decomposes into a direct sum of eigenspaces) and nilpotent respectively and commute, yielding the abstract Jordan decomposition of \({ \operatorname{ad}}_{\phi(x)}\). Now apply the theorem.

In parts:

1. By the lemma and induction on \(i\).

2. Clear!

3. Follows from \(i x.v_i = x.(y.v_{i-1}) = y.(x.v_{i-1}) + [xy].v_{i-1}\) and induction on \(i\).

Let \(T \leq L\) be toral and let \(x\in T\) be a basis element. Since \(x\) is semisimple, it STS \({ \operatorname{ad}}_{T, x} = 0\). Semisimplicity of \(x\) implies \({ \operatorname{ad}}_{L, x}\) diagonalizable, so we want to show \({ \operatorname{ad}}_{T, x}\) has no non-zero eigenvalues. Suppose that there exists a nonzero \(y\in T\) such that \({ \operatorname{ad}}_{T, x}(y) = \lambda y\) for \(\lambda \neq 0\). Then \({ \operatorname{ad}}_{T, y}(x) = [yx] = -[xy] = - { \operatorname{ad}}_{T, x} = - \lambda y\neq 0\), and since \({ \operatorname{ad}}_{T, y}(y) = -[yy] = 0\), \(y\) is an eigenvector with eigenvalue zero. Since \({ \operatorname{ad}}_{T, y}\) is also diagonalizable and \(x\in T\), write \(x\) as a linear combination of eigenvectors for it, say \(x = \sum a_i v_i\). Then \({ \operatorname{ad}}_{T, y}(x) = \sum \lambda_i a_i v_i\) and the terms with \(\lambda_i = 0\) vanish, and the remaining element is a sum of eigenvectors for \({ \operatorname{ad}}_{T, y}\) with nonzero eigenvalues. \(\contradiction\)

1. Follows from the Jacobi identity.

2. Follows from (1), that \(\dim L < \infty\), and the root space decomposition. This is because \(L_{\alpha}\neq L_{\alpha + \beta}\neq L_{\alpha + 2\beta} \neq \cdots\) and there are only finitely many \(\beta\) to consider.

3. If \(\alpha + \beta\neq 0\) then \(\exists h\in H\) such that \((\alpha + \beta)(h) \neq 0\). For \(x\in L_\alpha, y\in L_ \beta\), \begin{align*} \alpha(h) \kappa(x,y) &= \kappa([hx], y) \\ &= -\kappa([xh], y) \\ &= -\kappa(x, [hy]) \\ &= -\beta(h)\kappa(x,y) \\ &\implies ( \alpha + \beta)(h) \kappa(x,y) = 0 \\ &\implies \kappa(x,y)=0 \text{ since } (\alpha + \beta)(h) \neq 0 .\end{align*}

Skipped, about 1 page of dense text broken into 7 steps. Uses the last corollary along with Engel’s theorem.

1. If it does not span, choose \(h \in H\setminus\left\{{0}\right\}\) with \(\alpha(h) = 0\) for all \(\alpha\in \Phi\). Then \([h, L_ \alpha] = 0\) for all \(\alpha\), but \([h H] = 0\) since \(H\) is abelian. Using the root space decomposition, \([h L] =0\) and so \(h\in Z(L) = 0\) since \(L\) is semisimple. \(\contradiction\)

2. Follows from proposition 8.2 and \(\kappa(L_ \alpha, L_ \beta) = 0\) when \(\beta\neq \alpha\).

3. Let \(h\in H\), then \begin{align*} \kappa(h, [xy]) &= \kappa([hx], y) \\ &= \alpha(h) \kappa(x, y)\\ &= \kappa(t_\alpha, h) \kappa(x, y) \\ &= \kappa( \kappa(x, y) t_ \alpha, h) \\ &= \kappa( h, \kappa(x, y) t_ \alpha) \\ &\implies \kappa(h, [xy] - \kappa(x,y)t_\alpha) = 0 \\ &\implies [xy] = \kappa (x,y) t_ \alpha ,\end{align*} where we’ve used that \([xy]\in H\) and \(\kappa\) is nondegenerate on \(H\) and \([L_{ \alpha}, L_{ - \alpha}] \subseteq L_0 = H\).

4. 3. shows \([L_{ \alpha}, L_{ - \alpha}]\) is spanned by \(t_ \alpha\) if it is nonzero. Let \(x\in L_ \alpha\setminus\left\{{0}\right\}\), then if \(\kappa(x, L_{ - \alpha}) = 0\) then \(\kappa\) would have to be degenerate, a contradiction. So there is some \(y\in L_{ - \alpha}\) with \(\kappa(x, y) \neq 0\). Moreover \(t_\alpha\neq 0\) since \(\alpha\neq 0\) and \(\alpha\mapsto t_\alpha\) is an isomorphism. Thus \([xy] = \kappa(x,y) t_ \alpha\).

5. Suppose \(\alpha(t_\alpha) = \kappa(t_{ \alpha}, t_{ \alpha}) = 0\), then for \(x\in L_{\alpha}, y\in L_{ - \alpha}\), we have \([t_ \alpha, x] = \alpha(t_ \alpha)x = 0\) and similarly \([t_ \alpha, y] = 0\). As before, find \(x\in L_{ \alpha}, y\in L_{ - \alpha}\) with \(\kappa(x,y)\neq 0\) and scale one so that \(\kappa(x, y) = 1\). Then by (c), \([x, y] = t_ \alpha\), so combining this with the previous formula yields that \(S \mathrel{\vcenter{:}}=\left\langle{x, y, t_ \alpha}\right\rangle\) is a 3-dimensional solvable subalgebra.¹⁵ Taking \({ \operatorname{ad}}: L\hookrightarrow{\mathfrak{gl}}(L)\), which is injective by semisimplicity, and similarly \({ \left.{{ { \operatorname{ad}}}} \right|_{{S}} }: S\hookrightarrow { \operatorname{ad}}(S) \leq {\mathfrak{gl}}(L)\). We’ll use Lie’s theorem to show everything here is a commutator of upper triangular, thus strictly upper triangular, thus nilpotent and reach a contradiction.

Part e: We have \(\alpha(t_ \alpha ) = \kappa(t_ \alpha, t_ \alpha)\), so suppose this is zero. Pick \(x\in L_{ \alpha}, y\in L_{ - \alpha}\) such that \(\kappa(x, y) = 1\), then

• \([t_ \alpha, x] = 0\),
• \([t_ \alpha, y] = 0\),
• \([x, y] = t_ \alpha\).

Set \(S \mathrel{\vcenter{:}}={\mathfrak{sl}}( \alpha) \mathrel{\vcenter{:}}={\mathbf{C}}\left\langle{x,y,t_ \alpha}\right\rangle\) and restrict \({ \operatorname{ad}}: L\hookrightarrow{\mathfrak{gl}}(L)\) to \(S\). Then \({ \operatorname{ad}}(S) \cong S\) by injectivity, and this is a solvable linear subalgebra of \({\mathfrak{gl}}(L)\). Apply Lie’s theorem to choose a basis for \(L\) such that the matrices for \({ \operatorname{ad}}(S)\) are upper triangular. Then use that \({ \operatorname{ad}}_L([SS]) = [ { \operatorname{ad}}_L(S) { \operatorname{ad}}_L(S)]\), which is strictly upper triangular and thus nilpotent. In particular, \({ \operatorname{ad}}_L (t_ \alpha)\) is nilpotent, but since \(t_\alpha\in H\) which is semisimple, so \({ \operatorname{ad}}_L( t_ \alpha)\) is semisimple. The only things that are semisimple and nilpotent are zero, so \({ \operatorname{ad}}_L( t_ \alpha) = 0 \implies t_\alpha = 0\). This contradicts that \(\alpha\in H {}^{ \vee }\setminus\left\{{0}\right\}\). \(\contradiction\)

Part f: Given \(x_ \alpha\in L_ \alpha\setminus\left\{{0}\right\}\), choose \(y_ \alpha \in L_{ - \alpha}\) and rescale it so that \begin{align*} \kappa(x_ \alpha, y _{\alpha} ) = {2\over \kappa(t_ \alpha, t_ \alpha)} .\end{align*} Set \(h_ \alpha \mathrel{\vcenter{:}}={2t_ \alpha\over \kappa(t_ \alpha, t_ \alpha) }\), then by (c), \([x_ \alpha, y_ \alpha] = \kappa( x_ \alpha, y_ \alpha) t_ \alpha = h_ \alpha\). So \begin{align*} [ h_ \alpha, x_ \alpha] = {2\over \alpha(t_ \alpha) }[ t_ \alpha, x_ \alpha] = {2\over \alpha(t_ \alpha)} \alpha(t_ \alpha) x_ \alpha = 2x_ \alpha ,\end{align*} and similarly \([h_ \alpha, y_ \alpha] = -2 y_ \alpha\). Now the span \(\left\langle{x_ \alpha, h_ \alpha, y_ \alpha}\right\rangle \leq L\) is a subalgebra with the same multiplication table as \({\mathfrak{sl}}_2({\mathbf{C}})\), so \(S \cong {\mathfrak{sl}}_2({\mathbf{C}})\).

Part g: Note that we would have \(h_{ - \alpha} = {2 t_{ - \alpha} \over \kappa( t_ { - \alpha}, t_{- \alpha} ) } = - h_ \alpha\) if \(t_{ \alpha} = t_{ - \alpha}\). This follows from the fact that \(H {}^{ \vee } { \, \xrightarrow{\sim}\, }H\) sends \(\alpha\mapsto t_\alpha, -\alpha\mapsto t_{- \alpha}\), but by linearity \(-\alpha\mapsto -t_{ \alpha}\).

It STS \(H \subseteq \left\langle{\left\{{L_\alpha}\right\}_{\alpha\in \Phi}}\right\rangle\). Given \(\alpha\in \Phi\), \begin{align*} \exists x_\alpha\in L_ \alpha, y_ \alpha\in L_{- \alpha} \quad\text{such that}\quad \left\langle{x_ \alpha, y_ \alpha, h_ \alpha\mathrel{\vcenter{:}}=[x_ \alpha, y_ \alpha] }\right\rangle \cong {\mathfrak{sl}}_2({\mathbf{C}}) .\end{align*} Note any \(h_\alpha\in {\mathbf{C}}^{\times}t_ \alpha\) corresponds via \(\kappa\) to some \(\alpha\in H {}^{ \vee }\). By (a), \(\Phi\) spans \(H {}^{ \vee }\), so \(\left\{{t_\alpha}\right\}_{\alpha\in \Phi}\) spans \(H\).

Apply Weyl’s theorem to decompose \(M\) into a finite direct sum of irreducibles in \({}_{ {\mathfrak{sl}}_2({\mathbf{C}}) }{\mathsf{Mod}}\). The weights of \(h_\alpha\) are of the form \(m, m-2,\cdots, -m+2, -m\in {\mathbf{Z}}\).¹⁶

Note \(L_\alpha\) can only pair with \(L_{- \alpha}\) to give a nondegenerate Killing form. Set \begin{align*} M \mathrel{\vcenter{:}}=\bigoplus _{c\in {\mathbf{C}}} L_{c \alpha} = H \oplus \bigoplus _{c \alpha\in \Phi} L_{c \alpha} .\end{align*} By Weyl’s theorem, this decomposes into irreducibles. This allows us to take a complement of the decomposition from before to write \(M = H \bigoplus {\mathfrak{sl}}(\alpha) \oplus W\), and we WTS \(W = 0\) since this contains all \(L_{c \alpha}\) where \(c\neq \pm 1\). Since \(H \subseteq K \oplus {\mathfrak{sl}}( \alpha)\), we have \(W \cap H = 0\). If \(c_ \alpha\) is a root of \(L\), then \(h_ \alpha\) has \((c \alpha)(h_ \alpha) = 2c\) as an eigenvalue, which must be an integer by a previous lemma. So \(c\in {\mathbf{Z}}\) or \(c\in {\mathbf{Z}}+ {1\over 2}\).

Suppose \(W\neq 0\), and let \(V(s)\) (or \(L(s)\) in modern notation) be an irreducible \({\mathfrak{sl}}( \alpha){\hbox{-}}\)submodule of \(W\) for \(s\in {\mathbf{Z}}_{\geq 0}\). If \(s\) is even, \(V(s)\) contains an eigenvector \(w\) for \(h_ \alpha\) of eigenvalue zero by applying \(y_\alpha\) \(s/2\) times. We can then write \(w = \sum_{c\in {\mathbf{C}}} v_{c \alpha}\) with \(v_{ ca } \in L_{ c \alpha}\), and by finiteness of direct sums we have \(v_{c \alpha} = 0\) for almost every \(c\in {\mathbf{C}}\). Then \begin{align*} 0 &= [h_ \alpha, w] \\ &= \sum_{c\in {\mathbf{C}}} [h_ \alpha, v_{ c \alpha} ] \\ &= \sum_{c\in {\mathbf{C}}} (c\alpha)( h _{\alpha} )v_{c \alpha} \\ &= \sum 2c v_{c \alpha} \\ &\implies v_{c \alpha} = 0 \text{ when } c\neq 0 ,\end{align*} forcing \(w\in H\), the zero eigenspace. But \(w\in W\), so \(w\in W \cap H = 0\). \(\contradiction\)

Consider \(M \mathrel{\vcenter{:}}=\bigoplus \bigoplus _{ c\in {\mathbf{C}}} L_{c \alpha} \leq L\in {}_{{\mathfrak{sl}}( \alpha)}{\mathsf{Mod}}\). Write \({\mathfrak{sl}}(\alpha) = \left\langle{ x_ \alpha, h_ \alpha, y_ \alpha}\right\rangle\), we decomposed \(M = K \oplus {\mathfrak{sl}}(\alpha) \oplus W\) where \(\ker \alpha \leq H\) and \(W \cap H = 0\). WTS: \(W = 0\). So far, we’ve shown that if \(L(s) \subseteq W\) for \(s\in {\mathbf{Z}}_{\geq 0}\) (which guarantees finite dimensionality), then \(s\) can’t be even – otherwise it has a weight zero eigenvector, forcing it to be in \(H\), but \(W \cap H = 0\).

Aside: \(\alpha\in \Phi \implies 2\alpha\not\in \Phi\), since it would have weight \((2\alpha)(h_ \alpha) = 2\alpha h_ \alpha = 4\), but weights in irreducible modules have the same parity as the highest weight and no such weights exist in \(M\) (only \(0, \pm 2\) in \(K \oplus {\mathfrak{sl}}_2(\alpha)\) and only odd in \(W\)). Suppose \(L(s) \subseteq W\) and \(s\geq 1\) is odd. Then \(L(s)\) has a weight vector for \(h_ \alpha\) of weight 1. This must come from \(c=1/2\), since \((1/2) \alpha (h_ \alpha) = (1/2) 2 = 1\), so this is in \(L_{\alpha/2}\). However, by the aside, if \(\alpha\in \Phi\) then \(\alpha/2\not\in\Phi\).

Thus it \(W\) can’t contain any odd roots or even roots, so \(W = 0\). Note also that \(L_{\pm\alpha}\not\subset K \oplus W\), forcing it to be in \({\mathfrak{sl}}( \alpha)\), so \(L_{ \alpha} = \left\langle{x_ \alpha}\right\rangle\) and \(L_{- \alpha} = \left\langle{y _{\alpha} }\right\rangle\).

Consider \begin{align*} M \mathrel{\vcenter{:}}=\bigoplus _{k\in {\mathbf{Z}}} L_{ \beta + k \alpha} \leq L \quad\in {}_{{\mathfrak{sl}}( \alpha)}{\mathsf{Mod}} .\end{align*}

1. \(\beta(h _{\alpha} )\) is the eigenvalues of \(h_ \alpha\) acting on \(L_ \beta\). But by the lemma, \(\beta(h_ \alpha)\in {\mathbf{Z}}\).

2. By the previous proposition, \(\dim L_{ \beta+ k \alpha} = 1\) if nonzero, and the weight of \(h_\alpha\) acting on it is \(\beta( h _{\alpha} ) + 2k\) all different for distinct \(k\). By \({\mathfrak{sl}}_2{\hbox{-}}\)representation theory, we know the multiplicities of various weight spaces as the sum of dimensions of the zero and one weight spaces, and thus \(M\) is a single irreducible \({\mathfrak{sl}}(\alpha){\hbox{-}}\)module. So write \(M - L(d)\) for some \(d\in {\mathbf{Z}}_{\geq 0}\), then \(h_ \alpha\curvearrowright M\) with eigenvalues \(\left\{{d,d-2,\cdots, -d+2, -d}\right\}\). But \(h_ \alpha\curvearrowright M\) with eigenvalues \(\beta( h_ \alpha) + 2k\) for those \(k\in {\mathbf{Z}}\) with \(L_{\beta + k \alpha}\neq 0\). Since the first list is an unbroken string of integers of the same parity, thus the \(k\) that appear must also be an unbroken string. Define \(r\) and \(q\) by setting \(d = \beta(h_\alpha) + 2q\) and \(-d =\beta( h_ \alpha ) - 2r\) to obtain \([-r, q]\). Adding these yields \(0 = 2\beta( h_ \alpha) + 2q-2r\) and \(r-q = \beta(h_ \alpha)\).

3. Let \(M\cong L(d) \in {}_{{\mathfrak{sl}}(\alpha)}{\mathsf{Mod}}\) and \(x_\beta \in L_ \beta\setminus\left\{{0}\right\}\subseteq M\) with \(x_ \alpha\in L_{ \alpha}\). If \([x_ \alpha x_ \beta] = 0\) then \(x_ \beta\) is a maximal \({\mathfrak{sl}}(\alpha){\hbox{-}}\)vector in \(L(d)\) and thus \(d = \beta(h_ \alpha)\). But \(\alpha + \beta\in \Phi \implies \beta)(h_ \alpha) + 2\) is a weight in \(M\) bigger than \(d\), a contradiction. Thus \(\alpha + \beta\in \Phi \implies [x_ \alpha x_ \beta] \neq 0\). Since this bracket spans and \(\dim L_{ \alpha + \beta} = 1\), so \([L_ \alpha L_ \beta] = L_{ \alpha + \beta}\).

4. Use that \(q\geq 0, r\geq 0\) to write \(-r \leq -r+q \leq q\). Then \begin{align*} \beta - \beta(h_ \alpha) \alpha = \beta - (r-q) \alpha = \beta + (-r+q \alpha)\mathrel{\vcenter{:}}=\beta + \ell\alpha \end{align*} where \(\ell\in [-r, q]\). Thus \(\beta + \ell\alpha\in \Phi\) is an unbroken string by (b).

Write \(( \beta, \alpha_j) = \sum c_i (\alpha_i, \alpha_j)\) and m \begin{align*} {2 (\beta, \alpha_j) \over (\alpha_j, \alpha_j) } = \sum c_i {2 (\alpha_i, \alpha_j) \over (\alpha_j, \alpha_j) } ,\end{align*} where the LHS is in \({\mathbf{Z}}\), as is \(2( \alpha_i, \alpha_j) \over (\alpha_j, \alpha_j)\). On the other hand \begin{align*} {2 (\beta, \alpha_j) \over (\alpha_j, \alpha_j) } = {2 (t_ \beta, t_{\alpha_j} ) \over \kappa(t_{ \alpha_j}, \kappa_{ \alpha_j} ) } = \kappa(t_ \beta, h_{\alpha_j} ) = \beta(h_{ \alpha_j}) \end{align*} using that \(( \alpha_j, \alpha_j) = \kappa( t_{ \alpha_j}, t_{ \alpha_j} )\neq 0\) from before.¹⁷ Since \(\left\{{ \alpha_i}\right\}\) is a basis for \(H {}^{ \vee }\) and \(({-}, {-})\) is nondegenerate, the matrix \([ ( \alpha_i, \alpha _j) ]_{1\leq i, j\leq n}\) is invertible. Thus so is \(\left[ 2 ( \alpha_i, \alpha_j) \over (\alpha_j, \alpha_j ) \right]_{1\leq i,j \leq n}\), since it’s given by multiplying each column as a nonzero scalar, and one can solve for the \(c_i\) by inverting it. This involves denominators coming from the determinant, which is an integer, yielding entries in \({\mathbf{Q}}\).

If \(\lambda\in {\mathbb{E}}_{\mathbf{Q}}, ( \lambda, \mu) =0 \forall \mu\in {\mathbb{E}}_{\mathbf{Q}}\), then \(( \lambda, \alpha_i) = 0 \forall i \implies (\lambda, \nu) = 0 \forall \nu\in H {}^{ \vee }\implies \lambda= 0\).

Let \(\tau = \sigma s_ \alpha =\sigma s_{ \alpha}^{-1}\in \operatorname{GL}({\mathbb{E}})\), noting that every \(s_\alpha\) is order 2. Then \(\tau( \Phi) = \Phi\) and \(\tau( \alpha) = \alpha\), so \(\tau\) acts as the identity on the subspace \({\mathbf{R}}\alpha\) and the quotient space \({\mathbb{E}}/{\mathbf{R}}\alpha\) since there are two decompositions \({\mathbb{E}}= P_ \alpha \oplus {\mathbf{R}}\alpha = P \oplus {\mathbf{R}}\alpha\) using \(s_\alpha\) and \(\sigma\) respectively. So \(\tau - \operatorname{id}\) acts as zero on \({\mathbb{E}}/{\mathbf{R}}\alpha\), and so maps \({\mathbb{E}}\) into \({\mathbf{R}}\alpha\) and \({\mathbf{R}}\alpha\) to zero, s \((\tau - \operatorname{id})^2 = 0\) on \({\mathbb{E}}\) and its minimal polynomial \(m_\tau(t)\) divides \(f(t) \mathrel{\vcenter{:}}=(t-1)^2\).

Note that \(\Phi\) is finite, so the vectors \(\beta, \tau \beta, \tau^2 \beta, \tau^3 \beta,\cdots\) can not all be distinct. Since \(\tau\) is invertible we can assume \(\tau^k \beta = \beta\) for some particular \(k\). Taking the least common multiple of all such \(k\) yields a uniform \(k\) that works for all \(\beta\) simultaneously, so \(\tau^k \beta = \beta\) for all \(\beta \in \Phi\). Since \({\mathbf{R}}\Phi = {\mathbb{E}}, \tau^k\) acts as \(\operatorname{id}\) on all of \({\mathbb{E}}\), so \(\tau^k - 1 = 0\) and so \(m_\tau(t) \divides t^k - 1\) for some \(k\). Therefore \(m_\tau(t) \divides \gcd( (t-1)^2, t^k-1 ) = t-1\), forcing \(\tau = \operatorname{id}\) and \(\sigma = s_ \alpha\) and \(P = P_\alpha\).

Note that \(\sigma s_ \alpha \sigma^{-1}\) sends \(\sigma( \alpha)\) to its negative and fixes pointwise the hyperplane \(\sigma(P_\alpha)\). If \(\beta \in \Phi\) then \(s_{ \alpha}( \beta) \in \Phi\), so \(\sigma s_ \alpha ( \beta) \in \Phi\) and \begin{align*} (\sigma s_ \alpha \sigma^{-1}) ( \sigma( \beta)) = \sigma s_ \alpha(\beta) \in \sigma\Phi ,\end{align*} so \(\sigma s_ \alpha \sigma^{-1}\) leaves invariant the set \(\left\{{ \sigma( \beta) {~\mathrel{\Big\vert}~}\beta\in \Phi}\right\} = \Phi\). By the previous lemma, it must equal \(s_{ \sigma( \alpha)}\), and so \begin{align*} ( \sigma( \beta), \sigma( \alpha) {}^{ \vee }) = (\beta, \alpha {}^{ \vee }) \end{align*} by applying both sides to \(\sigma(\beta)\).

Note that (2) follows from (1) by replacing \(\beta\) with \(-\beta\). Assume \((\alpha, \beta) > 0\), then by the chart \({\left\langle {\alpha },~{\beta } \right\rangle}=1\) or \({\left\langle {\beta },~{\alpha } \right\rangle}= 1\). In the former case, \begin{align*} \Phi\ni s_{ \beta}( \alpha) = \alpha - {\left\langle {\alpha },~{\beta } \right\rangle}\beta = \alpha- \beta .\end{align*} In the latter, \begin{align*} s_{ \alpha}(\beta) = \beta- \alpha \in \Phi\implies - (\beta- \alpha) = \alpha- \beta\in \Phi .\end{align*}