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*}

We have \(\alpha\neq \pm \beta\) since \(\Delta\) is a linearly independent set. By a previous lemma, if \(( \alpha, \beta) > 0\) then \(\beta- \alpha \in \Phi\) by a previous lemma. \(\contradiction\)

1. Claim: each root in \(\Phi^+( \gamma)\) is in \({\mathbf{Z}}_{\geq 0} \Delta( \gamma)\).

The proof: if not, pick \(\beta\in \Phi^+(\gamma)\) which cannot be written this way and choose it such that \((\beta, \gamma)\) is maximal (by finiteness). Since \(\beta\not\in \Delta( \gamma)\), it is decomposable as \(\beta = \beta_1 + \beta_2\) with \(\beta_i \in \Phi^+\). Now \((\beta, \gamma) = \sum (\beta_i, \gamma)\) is a sum of nonnegative numbers, so \((\beta_i, \gamma) < (\beta, \gamma)\) for \(i=1,2\). By minimality, \(\beta_i\in {\mathbf{Z}}_{\geq 0 } \Delta(\gamma)\), but then by adding them we get \(\beta\in {\mathbf{Z}}_{\geq 0 } \Delta(\gamma)\).

2. Claim: if \(\alpha, \beta\in \Delta( \gamma)\) with \(\alpha\neq \beta\) then \((\alpha, \beta) \leq 0\). Note \(\alpha\neq - \beta\) since \((\alpha, \gamma), (\beta, \gamma) > 0\). By lemma 9.4, if \(( \alpha, \beta) > 0\) then \(\alpha - \beta\in \Phi\) is a root. Then one of \(\alpha- \beta, \beta- \alpha\in \Phi^+( \gamma)\). In the first case, \(\beta + (\alpha- \beta ) = \alpha\), decomposing \(\alpha\). In the second, \(\alpha + (\beta- \alpha) = \beta\), again a contradiction.

3. Claim: \(\Delta\mathrel{\vcenter{:}}=\Delta( \gamma)\) is linearly independent. Suppose \(\sum_{ \alpha\in \Delta} r_ \alpha \alpha = 0\) for some \(r_ \alpha \in {\mathbf{R}}\). Separate the positive terms \(\alpha\in \Delta'\) and the remaining \(\alpha\in \Delta''\) to write \({\varepsilon}\mathrel{\vcenter{:}}=\sum_{ \alpha\in \Delta'}r_ \alpha \alpha = \sum_{ \beta\in \Delta''} t_\beta \beta\) where now \(r_ \alpha, t_\beta > 0\). Use the two expressions for \({\varepsilon}\) to write \begin{align*} ({\varepsilon}, {\varepsilon}) = \sum _{ \alpha\in \Delta', \beta\in \Delta''} r_ \alpha t_ \beta (\alpha, \beta) \leq 0 ,\end{align*} since \(r_ \alpha t_ \beta >0\) and \((\alpha, \beta) \leq 0\). So \({\varepsilon}= 0\), since \(({-}, {-})\) is an inner product. Write \(0 = (\gamma, {\varepsilon}) = \sum_{ \alpha\in \Delta'} r_\alpha (\alpha, \delta)\) where \(r_ \alpha > 0\) and \((\alpha, \gamma) > 0\), so it must be the case that \(\Delta' = \emptyset\). Similarly \(\Delta'' = \emptyset\), so \(r_\alpha = 0\) for all \(\alpha\in \Delta\).

4. Claim: \(\Delta( \gamma)\) is a base for \(\Phi\). Since \(\Phi = \Phi^+(\gamma){\textstyle\coprod}\Phi^-( \gamma)\), we have B2 by step 1. This also implies \(\Delta( \gamma)\) is a basis for \({\mathbb{E}}\), since we have linear independent by step 3. Thus \({\mathbf{Z}}\Delta ( \gamma) \supseteq\Phi\) and \({\mathbf{R}}\Phi = {\mathbb{E}}\).

5. Claim: every base of \(\Phi\) is \(\Delta( \gamma)\) for some regular \(\gamma\). Given \(\Delta\), choose \(\gamma\in {\mathbb{E}}\) such that \((\gamma, \alpha) > 0\) for all \(\alpha\in \Delta\). Then \(\gamma\) is regular by B2. Moreover \(\Phi^+ \subseteq \Phi^+( \gamma)\) and similarly \(\Phi^- \subseteq \Phi^-( \gamma)\), and taking disjoint unions yields \(\Phi\) for both the inner and outer sets, forcing them to be equal, i.e. \(\Phi^{\pm} = \Phi^{\pm}( \gamma)\). One can check that \(\Delta \subseteq \Delta( \gamma)\) using \(\Phi^+ = \Phi^+( \gamma)\) and linear independence of \(\Delta\) – but both sets are bases for \({\mathbb{E}}\) and thus have the same cardinality \(\ell = \dim {\mathbb{E}}\), making them equal.

If \(( \beta, \alpha)\leq 0\) for all \(\alpha\in \Delta\), then the proof of theorem 10.1 would show \(\beta = 0\) by taking \({\varepsilon}\mathrel{\vcenter{:}}=\beta\) in step 3. One can then find an \(\alpha\in \Delta\) with \(( \beta, \alpha) > 0\), where clearly \(\beta \neq \pm \alpha\). By lemma 9.4, \(\beta- \alpha\in \Phi\). Why is this positive? Note that \(\beta\) has at least one coefficient for a simple root (not the coefficient for \(\alpha\)) which is strictly positive, and thus all coefficients are \(\geq 0\). This coefficient stays the same in \(\beta- \alpha\), so its coefficients are all non-negative by axiom B2 and \(\beta-\alpha\in \Phi^+\).

Let \(\beta \in \Phi^+\setminus\left\{{ \alpha }\right\}\); if \(\beta = \sum _{\gamma\in \Delta} k_ \gamma \gamma\) with \(k_ \gamma \in {\mathbf{Z}}_{\geq 0}\), since \(\beta\neq \alpha\), some \(k_ \gamma > 0\) for some \(\gamma \neq \alpha\). Using the formula \(s_ \alpha( \beta) = \beta- {\left\langle {\beta },~{\alpha } \right\rangle}\alpha\) still has coefficient \(k_ \gamma\) for \(\gamma\). Thus \(s_ \alpha( \beta) \in \Phi^+\) and \(s_ \alpha( \beta)\neq \alpha\) since \(s_ \alpha(- \alpha) = \alpha\) and \(s_\alpha\) is bijective, and so \(s_\alpha(\beta)\in \Phi^+\setminus\left\{{ \alpha }\right\}\). As a result, \(s_\alpha\) permutes this set since it is invertible.

For \(0\leq i\leq t-1\), let \(\beta_i \mathrel{\vcenter{:}}= s_{i+1} \cdots s_{t-1}( \alpha_t)\) and \(\beta_{t-1} \mathrel{\vcenter{:}}=\alpha_t\). Since \(\beta_0 \mathrel{\vcenter{:}}= s_1\cdots s_{t-1}( \alpha_t ) < 0\) and \(\beta_{t-1} = \alpha_t > 0\), there must be a smallest index \(u\) such that \(\beta_u > 0\). Note that \(u\geq 1\) since \(\beta_0\) is negative. Then \begin{align*} s_u( \beta_u) &= s_u s_{u+1} \cdots s_{t-1}( \alpha_t) \\ &= \beta_{u-1} \\ & < 0 \end{align*} by choice of \(u\). Noting that \(\beta_u = s_{u+1}\cdots s_{t-1} (\alpha_t)\), by lemma B, \(s_u = s_{\alpha_u}\) permutes roots other than \(\alpha_u\) since \(\beta_u > 0\) and \(s_{\alpha_u}( \beta_u) < 0\). By lemma 9.2, write \begin{align*} s_{\alpha_u} = s_{\beta_u} = s_{ s_{u+1}\cdots s_{t-1}( \alpha_t ) } = (s_{u+1} \cdots s_{t-1}) s_{\alpha_u} (s_{u+1}\cdots s_{t-1})^{-1} .\end{align*} Multiply both sides on the left by \((s_1\cdots s_u)\) and on the right by \((s_{u+1}\cdots s_{t-1})\) to obtain \begin{align*} (s_1 \cdots s_{u-1})(s_{u+1}\cdots s_{t-1}) = (s_1\cdots s_u)(s_{u+1}\cdots s_t), \qquad s_t \mathrel{\vcenter{:}}= s_{\alpha_t} .\end{align*}

Part c: Set \(W' \mathrel{\vcenter{:}}=\left\langle{s_ \alpha{~\mathrel{\Big\vert}~}\alpha\in \Delta}\right\rangle\), we’ll prove (c) with \(W\) replaced \(W'\), which is larger. First suppose \(\beta\in \Phi^+\) and consider \(W' \beta \cap\Phi^+\). This is nonempty since it includes \(\beta\) and is a finite set, so choose \(\gamma\) in it of minimal height. Claim: \(\operatorname{ht}(\gamma) = 1\), making \(\gamma\) simple. If not, supposing \(\operatorname{ht}( \gamma) > 1\), write \(\gamma = \sum_{ \alpha\in \Delta} k_ \alpha \alpha\) with \(k_ \alpha > 0\). Since \(\gamma\neq 0\), we have \((\gamma, \gamma) > 0\), so substitute to yield \begin{align*} 0 < (\gamma, \gamma) = (\gamma, \sum_{\alpha \in \Delta} k_ \alpha \alpha) = \sum_{\alpha\in \Delta} k_ \alpha (\gamma, \alpha) ,\end{align*} so \((\gamma, \alpha)>0\) for some \(\alpha \in \Delta\), and \(s_{ \alpha} \gamma = \gamma - {\left\langle {\gamma },~{\alpha } \right\rangle}\alpha\in \Phi^+\) is positive where \({\left\langle {\gamma },~{\alpha } \right\rangle}> 0\). This is a contradiction, since it has a smaller height. Note that if \(\beta\in \Phi^-\) then \(-\beta\in \Phi^+\) and there exists a \(\sigma\in W'\) such that \(\sigma( - \beta) = \alpha\in \Delta\). So \(\sigma( \beta) = - \alpha\), and \(s_ \alpha \sigma( \beta) = s_ \alpha( - \alpha) = \alpha \in \Delta\).

Part d: Given \(\beta\), pick \(\sigma\in W'\) such that \(\sigma^{-1}( \beta) = \alpha\in \Delta\). Then \begin{align*} s_ \beta = s_{ \sigma( \alpha)} = \sigma s_{ \alpha} \sigma^{-1}\in W' ,\end{align*} so \(W \leq W' \leq W\), making \(W = W'\).

Parts a and b: Recall \(\rho = {1\over 2}\sum _{ \beta\in \Phi^+}\) and choose \(\sigma\in W\) such that \(( \sigma(\delta), \rho)\) is maximal (picking from a finite set). Given \(\alpha\in \Delta\), we have \(s_\alpha \sigma\in W\), and so \begin{align*} ( \sigma(\delta), \rho) &\geq ( \sigma(\delta), \rho) \\ &= (\sigma(\delta), s_ \alpha \rho) \\ &= (\sigma(\delta), \rho - \alpha) \\ &= (\sigma(\delta), \rho ) - ( \sigma( \delta), \alpha) ,\end{align*} and so \(( \sigma( \delta), \alpha)\geq 0\) for all \(\alpha\in \Delta\). Importantly, \(\gamma\) is regular, so this inequality is structure for all \(\alpha\in \Delta\). So \(W\) acts transitively on the Weyl chambers, and consequently on simple systems (i.e. bases for \(\Phi\)) by the discussion at the end of \(\S 10.1\).2

Part e: Suppose \(\sigma( \Delta) = \Delta\) and \(\sigma \neq 1\), and write \(\sigma = \prod_{1\leq i \leq t} s_i\) with \(s_i \mathrel{\vcenter{:}}= s_{\alpha_i}\) for \(\alpha_i \in \Delta\) with \(t \geq 1\) minimal. Note \(\sigma( \Delta) = \Delta\) and \(\alpha_t \in \Delta\), we have \(\sigma( \alpha_t) > 0\) and \(\prod_{1\leq i\leq t}(\alpha_t) = \prod_{1\leq i \leq t-1}s_i (-\alpha_t)\) so \(\prod_{1\leq i\leq t-1} s_i(\alpha_t) < 0\). This fulfills the deletion condition, so \(\prod_{1\leq i \leq t} = s_1\cdots \widehat{s_u}\cdots \widehat{s_t}\) which is of smaller length.

Induct on \(\ell(\sigma)\): if zero, then \(\sigma = 1\) and \(n(1) = 0\) since it fixes all positive roots. If \(\ell( \sigma ) = 1\) then \(\sigma = s_{ \alpha}\) for some simple \(\alpha\), and we know from the last section that \(\sigma\) permutes \(\Phi^+\setminus\left\{{ \alpha }\right\}\) and \(\sigma( \alpha) = - \alpha\), so \(n( \sigma) = 1\).

We’re proving \(\ell( \sigma) = n(\sigma) \mathrel{\vcenter{:}}={\sharp}( \Phi^- \cap\sigma(\Phi^-))\) by induction on \(\ell(\sigma)\), where we already checked the zero case. Assume the result for all \(\tau\) with \(\ell( \tau) \leq \ell( \sigma)\) for \(\tau \in W\). Write \(\sigma = s_1\cdots s_t\) with \(s_i \mathrel{\vcenter{:}}= s_{ \alpha_i}, \alpha_i\in \Delta\) reduced. Set \(\tau \mathrel{\vcenter{:}}=\sigma s_t = s_1\cdots s_{t-1}\) which is again reduced with \(\ell(\tau) = \ell( \sigma) - 1\). By the deletion condition, \(s_1 \cdots s_{t-1}( \alpha_t) > 0\), so \(s_1\cdots s_{t-1}s_t (\alpha_t) = s_1 \cdots s_{t-1}(- \alpha_t) < 0\). Thus \(n(\tau) = n( \sigma) - 1\), since \(s_t\) permutes \(\Phi^+\setminus\left\{{ \alpha_t }\right\}\), so \begin{align*} \ell( \sigma) - 1 = \ell( \tau) = n( \tau) = n( \sigma) -1 \implies \ell( \sigma) = n( \sigma) .\end{align*}

\(\Phi\) reducible implies \(\Delta\) reducible: write \(\Phi = \Phi_1 {\textstyle\coprod}\Phi_2\) where \(( \Phi_1, \Phi_2) = 0\); this induces a similar partition of \(\Delta\). Then \((\Delta, \Phi_2) = 0 \implies ({\mathbb{E}}, \Phi_2) = 0 \implies {\mathbb{E}}= \emptyset\) using nondegeneracy of the bilinear form. \(\contradiction\)

Now \(\Delta\) reducible implies \(\Phi\) reducible: write \(\Delta =\Delta_1 {\textstyle\coprod}\Delta_2\) with \((\Delta_1, \Delta_2) = 0\). Let \(\Phi_i\) be the roots which are \(W{\hbox{-}}\)conjugate to an element of \(\Delta_i\). Then elements in \(\Phi_i\)are obtained from \(\Delta_i\) by adding and subtracting only elements of \(\Delta_i\), so \((\Phi_1, \Phi_2) = 0\) and \(\Phi = \Phi_1 \cup\Phi_2\) by a previous lemma that every \(\beta\in \Phi\) is conjugate to some \(\alpha\in \Delta\).

Existence: Let \(\tilde \alpha\) be any maximal root in the ordering. Given \(\alpha \in \Delta, (\tilde \alpha, \alpha) \geq 0\) – otherwise \(s_ \alpha(\tilde \alpha)= \tilde \alpha-{\left\langle { \tilde \alpha},~{\alpha} \right\rangle} \alpha > \alpha\), a contradiction. \(\contradiction\) Write \(\tilde \alpha = \sum_{\alpha\in \Delta} k_ \alpha \alpha\) with \(k_ \alpha \in {\mathbf{Z}}_{\geq 0}\), where it’s easy to see these are all non-negative. Suppose some \(k_\gamma = 0\), then \((\tilde \alpha, \gamma)\leq 0\) – otherwise \(s_\gamma( \tilde \alpha) = \tilde \alpha - {\left\langle {\tilde \alpha},~{\gamma} \right\rangle} \gamma\) has both positive and negative coefficients, which is not possible. Since \((\tilde \alpha, \alpha) \geq 0\), we must have \(( \tilde \alpha, \gamma) = 0\). So write \begin{align*} 0 = (\tilde \alpha, \gamma) = \sum_{ \alpha\in \Delta} k_ \alpha( \alpha, \gamma) \leq 0 ,\end{align*} so \(( \alpha, \gamma)= 0\) whenever \(k_\alpha \neq 0\), otherwise this expression would be strictly \(< 0\). Thus \(\Delta = \Delta_1 {\textstyle\coprod}\Delta_2\) where \(\Delta_1 = \left\{{\alpha\in \Delta {~\mathrel{\Big\vert}~}K_ \alpha\neq 0}\right\}\) and \(\Delta_2 = \left\{{ \alpha\in \Delta{~\mathrel{\Big\vert}~}k_ \alpha = 0 }\right\}\). This is an orthogonal decomposition of \(\Delta\), since any \(\gamma \in \Delta_2\) is orthogonal to any \(\alpha\in \Delta_1\). Note that \(\Delta_1\neq \empty\) since \(\tilde\alpha \neq 0\), and if \(\Delta_2\neq \empty\) then this is a contradiction, so \(\Delta_2\) must be empty. So no such \(\gamma\) exists.

Uniqueness: let \(\tilde \alpha\) be any maximal root in the ordering and let \(\tilde \alpha'\) be another such root. Then \((\tilde \alpha, \tilde \alpha') = \sum_{\alpha\in \Delta} k_ \alpha (\alpha, \tilde \alpha')\) with \(k_ \alpha > 0\) and \((\alpha, \tilde \alpha') \geq 0\). So \((\tilde \alpha, \tilde \alpha ') > 0\) since \(\Delta\) is a basis for \({\mathbb{E}}\) and anything orthogonal to a basis is zero by nondegeneracy of the form. Since \(\tilde \alpha \neq 0\), it is not orthogonal to everything. By Lemma 9.4, either \(\tilde \alpha, \tilde \alpha '\) are proportional (which was excluded in the lemma), in which case they are equal since they’re both positive, or otherwise \(a \mathrel{\vcenter{:}}=\tilde \alpha - \tilde \alpha' \in \Phi\) is a root. In the latter case, \(a > 0 \implies \tilde \alpha > \tilde \alpha'\) or \(a< 0 \implies \tilde \alpha < \tilde \alpha'\), both contradicting maximality.

Omitted.

Omitted.

Omitted.

An easy 10 steps:

1. If some \({\varepsilon}_i\) are discarded, the remaining ones still form an admissible set in \({\mathbb{E}}\) whose graph is obtained from \(\Gamma\) by omitting the corresponding discarded vertices.
2. The number of pairs of vertices in \(\Gamma\) connected by at least one edge is strictly less than \(n\).

Proof: Set \({\varepsilon}\mathrel{\vcenter{:}}=\sum_{i=1}^n {\varepsilon}_i\), which is nonzero by linear independence. Then \(0 < ({\varepsilon}, {\varepsilon}) = n + \sum_{i<j} 2 ({\varepsilon}_i, {\varepsilon}_j)\), and if \(i< j\) are joined, so \(({\varepsilon}_i, {\varepsilon}_j)\neq 0\), then \(4({\varepsilon}_i, {\varepsilon}_j)^2 = 1,2,3\) and so \(2({\varepsilon}_i, {\varepsilon}_j)\leq -1\). However, since the sum is positive, there are \(\leq n-1\) such pairs.

3. \(\Gamma\) contains no cycles.

Proof: A cycle is subgraph that corresponds to an admissible subset \(A' \subseteq A\) with \(m\) vertices and \(m\) edges corresponding to \(m\) connected pairs. This contradicts (2).

4. No more than 3 edges can be incident to any vertex of \(\Gamma\).

Let \({\varepsilon}\in A\) be a vertex, and suppose \({ \eta _1, \cdots, \eta _{k} }\) are the vertices connected to \({\varepsilon}\):

By (3), no two \(\eta_i, \eta_j\) are connected, so \(( \eta_i, \eta_j) = 0\) for all \(i,j\). Apply Gram-Schmidt to \(\left\{{{ \eta _1, \cdots, \eta _{k} } , {\varepsilon}}\right\}\) only involves modifying \({\varepsilon}\) since \(\left\{{{ \eta _1, \cdots, \eta _{k} }}\right\}\) are already orthonormal. Call the new vector \(\eta_0\) and let \(\left\{{{ \eta _0, \cdots, \eta _{k} }}\right\}\) be the resulting orthonormal set. One can then write \({\varepsilon}= \sum_{i=0}^k ({\varepsilon}, \eta_i)\eta_i\), which implies \(({\varepsilon}, \eta_0)\neq 0\) by linear independence. Then \(1 = ({\varepsilon}, {\varepsilon}) = \sum_{i=0}^k ({\varepsilon}, \eta_i)^2\), but \(({\varepsilon}, \eta_0)^2 > 0\) so \(\sum_{i=1}^k ({\varepsilon}, \eta_i)^2 < 1\) and thus \(\sum_{i=1}^k 4 ({\varepsilon}, \eta_i)^2 < 4\). But this sum is the number of edges incident to \({\varepsilon}\) in \(\Gamma\).

5. The only connected graph which contains a triple edge is the Coxeter graph of \(G_2\) by (4), since the triple edge forces each vertex to already have 3 incident edges.

6. Let \(\left\{{{\varepsilon}_1,\cdots, {\varepsilon}_k}\right\} \subseteq A\) have a simple chain \(\cdot \to \cdot \to \cdots \to \cdot\) as a subgraph. If \(A' \mathrel{\vcenter{:}}=\left\{{A\setminus\left\{{ { {\varepsilon}_1, \cdots, {\varepsilon}_{k} } }\right\}}\right\} \cup\left\{{{\varepsilon}}\right\}\) where \({\varepsilon}\mathrel{\vcenter{:}}=\sum_{i=1}^k {\varepsilon}_i\), then \(A'\) is admissible. The corresponding graph \(\Gamma'\) is obtained by shrinking the chain to a point, where any edge that was incident to any vertex in the chain is now incident to \({\varepsilon}\), with the same multiplicity.

Proof: number the vertices in the chain \(1,\cdots, k\). Linear independence of \(A'\) is clear. Note \(4({\varepsilon}_i, {\varepsilon}_{i+1})^2 = 1\implies 2({\varepsilon}_i, {\varepsilon}_{i+1}) \implies ({\varepsilon}, {\varepsilon}) = k + 2 \sum_{i< j} ({\varepsilon}_i, {\varepsilon}_j) = k + (-1)(k-1) = 1\). Any \(\eta\in A\setminus\left\{{ { {\varepsilon}_1, \cdots, {\varepsilon}_{k} } }\right\}\) is connected to at most one of \({ {\varepsilon}_1, \cdots, {\varepsilon}_{k} }\) since this would otherwise form a cycle, so \((\eta, {\varepsilon}) = (\eta, {\varepsilon}_i)\) for a single \(i\). So \(4(\eta, {\varepsilon})^2 = 4(\eta, {\varepsilon}_i)^2 \in \left\{{0,1,2,3}\right\}\) and \((\eta, {\varepsilon}) = (\eta, {\varepsilon}_i) \leq 0\), which verifies all of the admissibility criteria.

7. \(\Gamma\) contains no graphs of the following forms:

Proof: collapsing the chain in the middle produces a vertex with 4 incident edges.

The collection of such monomials of length exactly \(n\) forms a basis for \(S^m(L)\), and via \(w\), a basis for \(G^m(L)\). In particular these monomials form a linearly independent in \(U_m/U_{m-1}\), since taking quotients can only increase linear dependencies, and hence these are linearly independent in \(U_m\) and \(U(L)\). By induction on \(m\), \(U_{m-1}\) has a basis consisting of all monomials of length \(\leq m-1\). We can then get a basis of \(U_m\) by adjoining to this basis of \(U_{m-1}\) any preimages in \(U_m\) of basis elements for the quotient \(U_m/U_{m-1}\). So a basis for \(U_m\) is all ordered monomials of length \(\leq m\). Since \(U(L) = \cup_{m\geq 0} U_m\), taking the union of bases over all \(m\) yields the result.

Recall that differences of simple roots are never roots, since the coefficients have mixed signs. Since \(\alpha_i - \alpha_j \not\in \Phi\), we have \([x_i y_j] = 0\) for \(i\neq j\) since it would have to be in \(L_{\alpha_i - \alpha_j}\). Consider the \(\alpha_i\) root string through \(\alpha_j\): we have \(r=0\) from above, and the string is \begin{align*} \alpha, \alpha+ \alpha_i, \cdots, \alpha_j - {\left\langle {\alpha_j},~{\alpha_i} \right\rangle} \alpha_i \end{align*} since \(L_\beta = 0\) for \(\beta \mathrel{\vcenter{:}}=\alpha_j - \qty{ {\left\langle {\alpha_j},~{ \alpha_i} \right\rangle} + 1} \alpha_i\). The relations for \(S_{ij}^\pm\) follow similarly.

Straightforward but tedious checking of all relations, e.g.  \begin{align*} \widehat{\phi}(\widehat{h}_i) \circ \widehat{\phi}(\widehat{x}_j) - \widehat{\phi}(\widehat{x}_j )\widehat{\phi}(\widehat{h}_i) = {\left\langle { \alpha_j},~{\alpha_i} \right\rangle} \widehat{\phi}(\widehat{x}_j) .\end{align*}

Steps 1 and 2: :::{.claim} \(\pi(\widehat{H}) = H\) is \(\ell{\hbox{-}}\)dimensional.

Case 1: \(k\neq i\).

In this case, \([x_k y_i] = 0\) and thus \begin{align*} ( { \operatorname{ad}}_{x_k}) ( { \operatorname{ad}}_{y_i})^{- {\left\langle { \alpha_j },~{ ga_i } \right\rangle} +1 }(y_j) = ( { \operatorname{ad}}_{y_i})^{- {\left\langle { \alpha_j },~{ \alpha_i } \right\rangle} +1 } ( { \operatorname{ad}}_{x_k})(y_j) .\end{align*}

• Case i, \(k\neq j\): then \(( { \operatorname{ad}}_{x_k})(y_j) = 0\).
• Case ii, \(k=j\): then \(( { \operatorname{ad}}_{x_j})(y_j) = h_j\) and \(( { \operatorname{ad}}_{y_i})(h_j) = {\left\langle { \alpha_i },~{ \alpha_j } \right\rangle} y_i\).
  • Case a, \({\left\langle { \alpha_j },~{ \alpha_i } \right\rangle} \neq 0\), then \begin{align*} ( { \operatorname{ad}}_{y_i})^{- {\left\langle { \alpha_j },~{ \alpha_i } \right\rangle} +1 } ( { \operatorname{ad}}_{x_j}) (y_j) = {\left\langle { \alpha_i },~{ \alpha_j } \right\rangle} ( { \operatorname{ad}}_{y_i})^{- {\left\langle { \alpha_j },~{ \alpha_i } \right\rangle} }(y_i) = 0 .\end{align*}
  • Case b, \({\left\langle { \alpha_j },~{ \alpha_i } \right\rangle} = 0\), then \({\left\langle { \alpha_i },~{ \alpha_j } \right\rangle} =0\). In this case we have \(( { \operatorname{ad}}_{y_i})^1 (h_j) = {\left\langle { \alpha_i },~{ \alpha_j } \right\rangle} y_i = 0\).

Case 2: \(k=i\).

In this case, we saw that for any fixed \(i\), \(\left\{{x_i, h_i, y_i}\right\}\) spans a standard \({\mathfrak{sl}}_2\) triple in \(L_0\), so consider the \({\mathfrak{sl}}_2{\hbox{-}}\)submodule of \(Y_J \leq L_0\) generated by \(y_j\). Since \(i\neq j\), we know \([x_i y_j] = 0\), so \(y_j\) is a maximal vector for \(Y_j\) with weight \(m \mathrel{\vcenter{:}}=- {\left\langle { \alpha_j },~{ \alpha_i } \right\rangle}\).

One can show by induction on \(t\) that the following formula holds: \begin{align*} ( { \operatorname{ad}}_{x_i}) ( { \operatorname{ad}}_{y_i})^t (y_j) = t (m-t+1) ( { \operatorname{ad}}_{y_i})^{t-1} (y_j) \qquad t\geq 1 .\end{align*} So in particular \(( { \operatorname{ad}}_{x_i}) ( { \operatorname{ad}}_{y_i})^{m+1}(y_j) = 0\), and the LHS is \(y_{ij}\).

Omitted. See \(\S 14.3\) and \(\S 2.3\) for a very similar calculation.

By definition, \(L \mathrel{\vcenter{:}}= L_0/K\) where \(K{~\trianglelefteq~}L_0\) is generated by the elements \(x_{ij}, y_{ij}\) where \(i\neq j\). Recall that \(X, Y\leq L_0\) are the subalgebras generated by the \(x_i\) and \(y_i\) respectively, so let \(I\) (resp. \(J\)) be the ideal in \(X\) (resp. \(Y\)) generated by the \(x_{ij}\) (resp. \(y_{ij}\)) for \(i\neq j\). Clearly \(I, J \subseteq K\).

Recall that we have \begin{align*} L = N^- \oplus H \oplus N^+ \mathrel{\vcenter{:}}= Y/\left\langle{(S_{ij}^-)}\right\rangle \oplus {\mathbf{C}}\left\langle{h_1,\cdots, h_\ell}\right\rangle \oplus X/\left\langle{ (S_{ij}^+) }\right\rangle .\end{align*}

Most of this follows from Theorem 10.3, exercise 10.14, lemma 10.3B, and corollary 10.2C, along with induction on \(\ell(w)\). We’ll omit the details.

Let \(\lambda, \mu\in \Lambda^+\) and write \(\lambda - \mu\) as a nonnegative integer linear combination of simple roots. Note \begin{align*} 0 \leq (\lambda+ \mu, \lambda- \mu ) = (\lambda, \lambda ) - (\mu, \mu) = {\left\lVert { \lambda} \right\rVert}^2 - {\left\lVert {\mu} \right\rVert}^2 ,\end{align*} so \(\mu\) lies in the compact set of vectors whose length is \({\left\lVert {\lambda} \right\rVert}\) and also in the discrete set \(\Lambda^+\). The intersection of a compact set and a discrete set is always finite.

If \(\alpha, \beta\in \Phi^+\) then \([L_{ \alpha}, L_{ \beta}] = L_{\alpha + \beta}\) where \(\alpha + \beta\in \Phi^+\) (if this is still a root), so \(H\leq L\). One has \(B^{(i)} \mathrel{\vcenter{:}}=[B^{(i-1)}, B^{(i-1)}] \subseteq \sum_{\operatorname{ht}( \beta) \geq 2^{i-1} } L_{\beta}\), since bracketing elements of \(H\) together will vanish (since \(H\) is abelian) and bracketing height 1 roots yields height 2, bracketing height 2 yields height 4, and so on. Thus \(B\) is a solvable subalgebra, since the height is uniformly bounded above by a finite number. To see that its maximal, note that any subalgebra \(B' \leq L\) containing \(B\), it must also contain some \(L_{ - \alpha}\) for some \(\alpha\in \Phi^+\). But then \(B' \supseteq{\mathfrak{sl}}_2({\mathbf{C}}) = L_{ - \alpha} \oplus [ L_{ - \alpha }, L_{ \alpha}] \oplus L_{ \alpha}\) which is not solvable, so \(B'\) can not be solvable.

1. Use the PBW theorem and extend a basis for any \(B\leq L\) to a basis where the \(B\) basis elements come second. Writing \(L = N^ \oplus B\), one can decompose \(U(L) = U(N^-) \otimes_{\mathbf{C}}U(B)\) and get \(U(L) v^+ = U(N^-)U(B)v^+ = U(N^-)U(H \oplus N^+)v^+\)

2. Writing the \(\beta\) in terms of \(\alpha\) yields this expression.

3. Clear.

We’ll finish the rest next time.

Use that \({ \operatorname{ad}}\) acts by derivations: \begin{align*} [x_i, y_i^{k+1}] &= x_i y_i^{k+1} - y_i^{k+1} x_i \\ &= x_i y_i y_i^k - y_i x_i y_i^k + y_i x_i y_i^k - y_i y_i^k x_i \\ &= [x_i y_i] y_i^k + y_i[x_i y_i^k] \\ &= h_i y_i^k + y_i [x_i y_i^k] \\ &= \qty{ y_i^k h_i - k \alpha_i (h_i) y_i^k } + y_i \qty{k y_i^{k-1} (h_i - (k-1)\cdot 1) } \qquad\text{by (b) and induction} \\ &= y_i^k h_i - 2k y_i^k + ky_i^k (h_i - (k-1)\cdot 1) \\ &= (k+1)y_i^k h_i - (2k + k(k-1)) y_i^k \\ &= (k+1) y_i^k h_i - k(k+1) y_i^k \\ &= (k+1)y_i^k (h_i - k\cdot 1) .\end{align*}

The main work is showing the last part involving equality of dimensions. It STS this for a simple reflection \(s_i \mathrel{\vcenter{:}}= s_{\alpha_i}\) since \(\sigma\) is a product of such reflections. Write \(\phi: L\to {\mathfrak{gl}}(V)\) be the representation associated to \(V\) – the strategy is to show that \(\phi(x_i)\) and \(\phi(y_i)\) are locally nilpotent endomorphisms of \(V\). Let \(v^+ \in V_ \lambda\setminus\left\{{0}\right\}\) be a fixed maximal vector and set \(m_i \mathrel{\vcenter{:}}=\lambda(h_i)\) so \(h_i.v^+ = m_i v^+\).

1. Set \(w \mathrel{\vcenter{:}}= y_i^{m_i+1}. v^+\), then the claim is that \(w=0\). Supposing not, we’ll show \(w\) is a maximal vector of weight not equal to \(\lambda\), and thus not a scalar multiple of \(v^+\). We have \({\operatorname{wt}}( w) = \lambda- (m_i +1) \alpha_i < \lambda\) (a strict inequality). If \(j\neq i\) then \(x_j.w = x_j y_i^{m_i + 1} v^+ = y_i^{m_i + 1} x_j v\) by part (a) of the lemma above, and this is zero since \(v^+\) is highest weight and thus maximal (recalling that these are distinct notions). Otherwise \begin{align*} x_i w &= x_i y_i^{m_i + 1} v^+ \\ &= y_i^{m_i+1} x_i v^+ + (m_i + 1) y_i^{m_i} (h_i - m_i\cdot 1)v^+ \\ &= 0 + (m_i+1)y_i^{m_i}(m_i - m_i) v^+ = 0 .\end{align*} So \(w\) is a maximal vector of weight distinct from \(\lambda\), contradicting corollary 20.2 since this would generate a proper submodule. \(\contradiction\)

2. Let \(S_i = \left\langle{x_i, y_i, h_i}\right\rangle \cong {\mathfrak{sl}}_2\), then the claim is that \(v^+, y_i v^+, \cdots, y_i^{m_i} v^+\) span a nonzero finite-dimensional \(S_i{\hbox{-}}\)submodule of \(V\). The span is closed under (the action of) \(h_i\) since all of these are eigenvectors for \(h_i\), and is closed under \(y_i\) since \(y_i\) raises generators and annihilates \(y_i^{m_i} v^+\), and is closed under \(x_i\) by part (c) of the lemma (since it lowers generators).

3. The sum of two finite-dimensional \(S_i{\hbox{-}}\)submodules of \(V\) is again a finite-dimensional \(S_i{\hbox{-}}\)submodule, so let \(V_i\) be the sum of all finite-dimensional \(S_i{\hbox{-}}\)submodules of \(V\) (which is not obviously finite-dimensional, since we don’t yet know if \(V\) is finite-dimensional). The claim is \(V'\) is a nonzero \(L{\hbox{-}}\)submodule of \(V\). Let \(w\in V'\), then \(w\) is a finite sum and there exists a finite-dimensional \(S_i{\hbox{-}}\)submodule \(W\) of \(V\) with \(w\in W\). Construct \(U \mathrel{\vcenter{:}}=\sum_{ \alpha \in \Phi} x_ \alpha . W\) where \(x_ \alpha \mathrel{\vcenter{:}}= y_{- \alpha}\) if \(\alpha\in \Phi^-\), which is a finite-dimensional vector subspace of \(V\). Check that \begin{align*} h_i (x_ \beta . W) &= x_ \beta(h_i.W) + [h_i x_ \beta].W \subseteq x_ \beta.W \subseteq U \\ x_i (x_ \beta.W) &= x_ \beta(x_i.W) + [x_i x_ \beta] .W \subseteq x_ \beta .W + x _{ \beta + \alpha_i}.W \subseteq U \\ y_i(x_ \beta. W) &= x_{ \beta}(y_i.W) + [y_i x_ \beta].W \subseteq x_ \beta W + x_{ \beta - \alpha_i}.W \subseteq U ,\end{align*} and so \(U\) is a finite-dimensional \(S_i{\hbox{-}}\)submodule of \(V\) and thus \(U \subseteq V'\). So if \(w\in V'\) then \(x_ \alpha .w \in V'\) for all \(\alpha\in \Phi\) and \(V'\) is stable under a set of generators for \(L\), making \(V' \leq V\) an \(L{\hbox{-}}\)submodule. Since \(V'\neq 0\) (since it contains at least the highest weight space), it must be all of \(V\) since \(V\) is irreducible.

4. Given an arbitrary \(v\in V\), apply the argument for \(w\) in step 3 to show that there exists a finite-dimensional \(S_i{\hbox{-}}\)submodule \(W \subseteq V\) with \(v\in W\). The elements \(x_i, y_i\) act nilpotently on any finite-dimensional \(S_i{\hbox{-}}\)module, and so in particular they act nilpotently on \(v\) and we get local nilpotence.

5. Now \(\tau_i \mathrel{\vcenter{:}}= e^{\phi(x_i)} \circ e^{\phi(-y_i)} \circ e^{\phi(x_i)}\) is well-defined. As seen before, \(\tau_i(V_ \mu) = V_{s_i \mu}\), i.e. \(\tau_i\) is an automorphism that behaves like \(s_i\), and so \(\dim V_ \mu = \dim V_{ \sigma \mu}\) for all \(\mu\in \Pi(V)\) and \(\sigma\in W\). Now any \(\mu\in \Pi(V)\) is conjugate under \(W\) to a unique dominant weight \(\mu^+\), and by (4) \(\mu^+\in \Pi(V)\) and since the weights in \(V = L(\lambda)\) has only weights smaller than \(\lambda\), we have \(\mu^+ \leq \lambda\). Note \(\lambda \in \Lambda^+\) is dominant, and so by 13.2B there are only finitely many such weights \(\mu^+\). Now there are only finitely many conjugates of the finitely many possibilities for \(\mu^+\), so \({\sharp}\Pi(V) < \infty\). By the general theory of highest weight modules, all weight spaces \(V_{\mu} \leq L( \lambda)\) are finite-dimensional. Since \(\dim V_\mu < \infty\) for all \(\mu\in\Pi(V)\), we have \(\dim V < \infty\).

Take dimensions in the formula \((V\otimes W)_ \sigma = \sum_{ \lambda+ \mu = \sigma} V_ \lambda\otimes W_ \mu\).

Recall every \(\lambda\in \Lambda\) is conjugate to a unique \(\lambda\in \Lambda^+\). Since \(f\) is \(W{\hbox{-}}\)invariant, we can write it as \begin{align*} f = \sum_{ \lambda\in \Lambda^+} c( \lambda) \qty{\sum_{w\in W} e( w \lambda)} \end{align*} where \(c( \lambda)\in {\mathbf{Z}}\) and almost all are zero. Given \(\lambda\in \Lambda^+\) with \(c( \lambda)\neq 0\), a previous lemma (13.2B) shows that \({\sharp}\left\{{ \mu\in \Lambda^+ {~\mathrel{\Big\vert}~}\mu\leq \lambda}\right\} < \infty\) by a compactness argument. Set \begin{align*} M_f \mathrel{\vcenter{:}}=\bigcup_{c( \lambda) \neq 0, \lambda\in \Lambda^+} \left\{{ \mu\in \Lambda^+ {~\mathrel{\Big\vert}~}\mu\leq \lambda}\right\} ,\end{align*} all of the possible weights that could appear, then \({\sharp}M_f < \infty\) since this is a finite union of finite sets. Choose \(\lambda\in \Lambda^+\) maximal with respect to the property that \(c( \lambda)\neq 0\), and set \(f' \mathrel{\vcenter{:}}= f- c( \lambda) \operatorname{ch}_ \lambda\). Note that \(f'\) is again \(W{\hbox{-}}\)invariant, since \(f\) and \(\operatorname{ch}_ \lambda\) are both \(W{\hbox{-}}\)invariant, and \(M_{ \operatorname{ch}_ \lambda} \subseteq M_f\). However \(\lambda \not\in M_{f'}\) by maximality, since we’ve subtracted \(c( \lambda) e(\lambda)\) off, so \({\sharp}M_{f'} < {\sharp}M_f\). Inducting on \({\sharp}M_f\), \(f'\) is a \({\mathbf{Z}}{\hbox{-}}\)linear combination of \(\operatorname{ch}_{\lambda'}\) for \(\lambda' \in \Lambda\), and thus so is \(f\). One checks the base case \(L(0) \cong {\mathbf{C}}\) where everything acts with weight zero. Uniqueness is relegated to exercise 22.8.

The proof that \(w\) is a maximal vector is step 1 in theorem 21.2, which showed \(\dim L( \lambda) < \infty\) (using lemma 21.2 and commutator relations in \({\mathcal{U}}(L)\)). Then check that \begin{align*} { \operatorname{weight} }(w) = \lambda - (m+1) \alpha = \lambda- \qty{ {\left\langle {\lambda },~{\alpha } \right\rangle}+ 1} \alpha .\end{align*} In fact, for any \(\lambda\in H {}^{ \vee }, \alpha\in \Delta\) we can define \begin{align*} \mu \mathrel{\vcenter{:}}=\lambda- \qty{ {\left\langle {\lambda },~{\alpha } \right\rangle}+ 1} \alpha ,\end{align*} so in our case \({ \operatorname{weight} }(w) = \mu\). Note that \begin{align*} \mu &= \lambda- {\left\langle {\lambda},~{\alpha } \right\rangle}\alpha- \alpha \\ &= s_ \alpha( \lambda) + (s_ \alpha( \rho) - \rho) \\ &= s_ \alpha(\lambda+ \rho) - \rho\\ &= s_ \alpha \cdot \lambda .\end{align*} Now \(w\) generates a highest weight module \(W\leq M( \lambda)\) of highest weight \(\mu = s_ \alpha \cdot \lambda\). Note that \(M( \mu)\) is the universal highest weight module with highest weight \(\mu\), i.e. \(\exists! M(\mu) \twoheadrightarrow W\). This yields a \(B{\hbox{-}}\)module morphism \begin{align*} {\mathbf{C}}_ \mu &\to W \\ 1 &\mapsto w ,\end{align*} which yields an \(L{\hbox{-}}\)module morphism \({\mathcal{U}}(L) \otimes_{{\mathcal{U}}(B)} {\mathbf{C}}_ \mu \twoheadrightarrow W\). So \(W\) is a nonzero quotient of \(M( \mu)\) and \({\mathcal{Z}}(L)\curvearrowright W\) by \(\chi_ \mu\). On the other hand \(W\leq M( \lambda)\) and so \({\mathcal{Z}}(L)\curvearrowright W\) by \(\chi_ \lambda\), yielding \(\chi_ \lambda= \chi_ \mu\).

So we conclude that if \(\mu = s_ \alpha \cdot \lambda\) with \({\left\langle {\lambda },~{\alpha} \right\rangle}\in {\mathbf{Z}}_{\geq 0}\), then \(\chi_ \mu = \chi_ \lambda\).

\(\mu = s_ \alpha \cdot \lambda = \lambda- (m+1) \alpha\) where \(m \mathrel{\vcenter{:}}={\left\langle {\lambda },~{\alpha} \right\rangle}\).

• Case 1: \(m=-1\), then \(\mu = \lambda\) and we’re done.
• Case 2: \(m\geq 0\), then \(\chi_ \lambda= \chi_ \mu\) by the proposition.
• Case 3: \(m\leq -2\), then \begin{align*} {\left\langle {\mu },~{\alpha } \right\rangle} &= {\left\langle { \lambda- (m+1) \alpha},~{\alpha} \right\rangle} \\ &= m-2(m+1) \\ &= -m-2 \geq 0 .\end{align*} We also have \(\mu = s_ \alpha\dot \lambda\implies s_ \alpha\cdot \mu= s_ \alpha^2 \cdot \lambda= \lambda\) by applying the action on both sides. By the proposition, swapping \(\mu, \lambda\) to get submodules of \(M( \mu)\) of highest weight \(\lambda\) and conclude \(\chi_ \lambda = \chi_ \mu\).

Say \(\mu = w\cdot \lambda\) for \(w\in W\), then write \(w = s_{i_1}\cdots s_{i_t}\) and use induction on \(t\), where the base case is the previous corollary.

Part a: The coefficient of \(e_\mu\) in \(\prod_{i=1}^m f_{ \beta_j}\) is the convolution \begin{align*} \sum_{i_1,\cdots, i_m \in {\mathbf{Z}}_{\geq 0}, -\sum i_j \beta_j = \mu } f_{\beta_1}(-i_1 \beta_1) \cdots f_{ \beta_m }(-i_m \beta_m) = p(\mu) .\end{align*}

Part b: \begin{align*} (e_0 - e_{- \alpha}) \ast(e_0 + e_{- \alpha} + e_{-2 \alpha} + \cdots) = e_0 + e_{ - \alpha} - e_{ - \alpha}+ e_{ -2 \alpha}- e_{ -2 \alpha} +\cdots = e_0 ,\end{align*} noting the telescoping. This can be checked rigorously by regarding these as functions instead of series.

Part c: Recall \(\rho = \sum_{ \alpha\in \Phi^+} {1\over 2}\alpha\), so \(e_\rho = \prod_{ \alpha\in \Phi^+} e_{\alpha\over 2}\). Thus the RHS is \begin{align*} \prod_{ \alpha\in \Phi^+} \qty{e_{\alpha\over 2} \ast(e_0 - e_{- \alpha}) } = \prod_{\alpha\in \Phi^+}(e_{\alpha\over 2} - e_{- \alpha\over 2} ) \mathrel{\vcenter{:}}= q .\end{align*} Note that \(q\neq 0\) since \(q(\rho) = 1\).

ETS for \(\alpha \in \Delta\). Recall \(s_ \alpha\) permutes \(\Phi^+ \setminus\left\{{ \alpha }\right\}\) and \(s_ \alpha(\alpha) = - \alpha\), so \(s_ \alpha q\) permutes the factors \((e_{ \beta\over 2} - e_{-{\beta\over 2}})\) for \(\beta\neq \alpha\) and negates \(e_{\alpha\over 2} - e_{- \alpha\over 2}\).

Use lemma A: \begin{align*} q\ast\rho \ast e_{-\rho} &= e_\rho \ast\qty{ \prod_{ \alpha\in \Phi^+} (e_0 - e_{ - \alpha} ) } \ast p \ast e_{ - \rho} \qquad \text{by C} \\ &= \qty{ \prod_{ \alpha\in \Phi^+} (e_0 - e_{ - \alpha} ) } \ast p \\ &= \qty{ \prod_{ \alpha\in \Phi^+} (e_0 - e_{ - \alpha}) \ast f_{ \alpha} } \qquad \text{by A}\\ &= \prod_{ \alpha\in \Phi^+} e_0 \qquad \text{by B}\\ &= e_0 .\end{align*}

\(M( \lambda)\) has basis \(y_{\beta_1}^{i_1}\cdots y_{ \beta_m}^{i_m} \cdot v^+\) where \(v^+\) is a highest weight vector of weight \(\lambda\). Note that \begin{align*} \mu = \lambda- \sum_{j=1}^M i_j \beta_j \iff \mu- \lambda= - \sum_{i=1}^m i_j \beta_j ,\end{align*} and \(\dim M(\lambda)_ \mu = p( \mu- \lambda)\). Now check \((p\ast e_ \lambda)( \mu) \mathrel{\vcenter{:}}= p( \mu - \lambda) e_ \lambda(\lambda) = p( \mu- \lambda)\).

\begin{align*} \text{LHS} \overset{D}{=} q \ast p \ast e_ \lambda \overset{C}{=} e_ \rho \ast e_ \lambda = e_{ \lambda+ \rho} .\end{align*}

By (3), the weights of \(V\) lie in a finite number of cones \(\lambda_i - {\mathbf{Z}}_{\geq 0} \Phi^+\). So if \(\mu\) is a weight of \(V\) and \(\alpha\in \Phi^+\), then \(\mu + k \alpha\) is not a weight of \(V\) for \(k\gg 0\). Iterating this for the finitely many positive weights, there exists a weight \(\mu\) such that \(\mu + \alpha\) is not a weight for any \(\alpha\in \Phi^+\). Then any \(0\neq v\in V_ \mu\) is a maximal vector.

By induction on the number of maximal vectors (up to scalar multiples). If \(M( \lambda)\) is irreducible then it’s an irreducible highest weight module, and these are unique up to isomorphism and so \(M( \lambda) = L( \lambda)\) and we’re done. Otherwise \(M( \lambda)\) has a proper irreducible submodule \(V\), and \(V\in {\mathcal{M}}_ \lambda\) by closure under submodules. By the lemma, \(V\) has a maximal vector of some weight \(\mu\) which must be strictly less than \(\lambda\), i.e. \(\mu < \lambda\). As before, \(\chi_{ \mu} = \chi_ \lambda\) and thus \(\mu\in \Theta( \lambda)\). Consider \(V\) and \(M( \lambda)/ V\) – each lies in \({\mathcal{M}}_ \lambda\) and either has fewer weights linked to \(\lambda\) than \(M( \lambda)\) has, or else it must have exactly the same set of weights linked to \(\lambda\), just with smaller multiplicities. By induction each of these has a composition series, and these can be pieced together into a series for \(M\) since they fit into a SES.

Part b: Each composition factor of \(M( \lambda)\) is in \({\mathcal{M}}_ \lambda\), hence by the lemma has a maximal vector. Since it’s irreducible, it is a highest weight module \(L( \mu)\) for some \(\mu\in \theta( \lambda)\).

Part c: \([M( \lambda) : L( \lambda)] = 1\) since \(\dim M( \lambda)_ \lambda = 1\) and all other weights are strictly less than \(\lambda\).

Take \(\lambda = 0\) in \(\star_2\), and use that \(\operatorname{ch}_0 = e_0\) where \(L(0) \cong {\mathbf{C}}\).

Apply \(\star_2\) and the lemma.

Note \(\operatorname{ch}_{ \lambda} = \sum _{\mu} \dim L( \lambda)_ \mu e_ \mu\in {\mathbf{Z}}[ \Lambda]\), and \(\dim L( \lambda) = \sum_{ \mu\in \Lambda} m_ \lambda( \mu)\). Viewing \(\operatorname{ch}_ \lambda: \Lambda\to {\mathbf{Z}}\) as a restriction of a function \(H {}^{ \vee }\to {\mathbf{C}}\), \(\dim L( \lambda)\) is the sum of all values of \(\operatorname{ch}_ \lambda\). Work in the \({\mathbf{C}}{\hbox{-}}\)subalgebra \({\mathcal{X}}_0\) of \({\mathcal{X}}\) generated by the characteristic functions \(S \mathrel{\vcenter{:}}=\left\{{e_ \mu{~\mathrel{\Big\vert}~}\mu\in \Lambda}\right\}\); this equals the span of \(S\) since \(e_{ \mu} \ast e_{\nu} = e_{ \mu+ \nu}\). We have a map \begin{align*} v: {\mathcal{X}}_0 &\to {\mathbf{C}}\\ f &\mapsto \sum_{ \mu\in \Lambda} f( \mu) ,\end{align*} which makes sense since \({\sharp}\mathop{\mathrm{supp}}f < \infty\). This function sums the values we’re after, so the goal is to compute \(v( \operatorname{ch}_ \lambda)\). By the exercise, attempting to apply this directly to the numerator and denominator yields \(0/0\), and we get around this by using a variant of L’Hopital’s rule. Define \({\partial}_ \alpha (e_ \mu) = ( \mu, \alpha)e_ \mu\), extended linearly to \({\mathcal{X}}_0\). In the basis \(S\) this operator is diagonal, and this is a derivation relative to convolution: \begin{align*} {\partial}_ \alpha \qty{ e_ \mu \ast e_{ \nu} } &= {\partial}_ \alpha( e_{ \mu+ \nu} ) \\ &= ( \mu+ \nu, \alpha)e_{ \mu+ \nu} \\ &= \qty{ (\mu, \alpha) e_ \mu} \ast e_ \nu + e_ \mu \ast\qty{ (\nu, \alpha) e_ \nu} \\ &= ({\partial}_ \alpha e_ \mu) \ast e_ \nu + e_ \mu \ast({\partial}_{\alpha} e _{\nu}) .\end{align*} Moreover they commute, i.e. \({\partial}_ \alpha {\partial}_ \beta = {\partial}_{\beta} {\partial}_{\beta}\). Set \({\partial}\mathrel{\vcenter{:}}=\prod_{ \alpha\in \Phi^+} {\partial}_ \alpha\) where the product here is composition, and view \({\partial}\) as an \(m\)th order differential operator. Write \(\omega( \lambda+ \rho)\mathrel{\vcenter{:}}=\sum_{w\in W} (-1)^{\ell(w)} e_{w (\lambda+ \rho)}\) for \(\lambda\in \Lambda^+\), so \(q = \omega ( \rho)\). Rewriting the WCF we have \begin{align*} \omega( \rho) \ast\operatorname{ch}_ \lambda = \omega( \lambda+ \rho) \qquad \star_1 ,\end{align*} and \begin{align*} \prod_{ \alpha\in \Phi^+}\qty{e_{ \alpha\over 2} - e_{- {\alpha\over 2}}} \ast\operatorname{ch}_ \lambda = \omega( \lambda+ \rho) .\end{align*} We now try to apply \({\partial}\) to both sides, followed by \(v\). Note that if any two factors of \({\partial}\) hit the same factor on the LHS, then noting that \(v(e_{\alpha\over 2} - e_{-{ \alpha\over 2}}) = 0\), such terms will vanish. So the total result will be zero unless all of the factors of \({\partial}\) are applied to the \(q\) factor in the LHS.

So apply \(v\circ {\partial}\) to \(\star_1\) to get \begin{align*} v( {\partial}\omega( \rho)) v( \operatorname{ch}_ \lambda) = v( {\partial}\omega( \lambda+ \rho)) .\end{align*} We can compute \begin{align*} v( {\partial}\omega ( \rho)) = v\qty{ {\partial}\sum_{w\in W} (-1)^{\ell(w)} e_{ w \rho} } = \sum_{w\in W} (-1)^{\ell(w)} v({\partial}( e_{ w \rho} ) ) .\end{align*} We have \begin{align*} v ( {\partial}( e_{ w \rho})) &= v\qty{ \qty{ \prod_{\alpha\in \Phi^+} {\partial}_\alpha } e_{ w \rho} } \\ &= v\qty{\prod_{ \alpha\in \Phi^+} (w \rho, \alpha) e_{ w \rho} } \\ &= \prod_{ \alpha\in \Phi^+} ( w \rho, \alpha) \\ &= \prod_{ \alpha\in \Phi^+} ( \rho, w^{-1}\alpha) \qquad \star_2 .\end{align*} Note that \(w^{-1}\cdot \Phi^+\) is a permutation of \(\Phi^+\), just potentially with some signs changed – in fact, exactly \(n(w^{-1})\), the number of positive roots sent to negative roots, and \(n(w^{-1}) = \ell(w^{-1})\). Thus the above is equal to \begin{align*} (-1)^{\ell(w)} \prod_{ \alpha\in \Phi^+}( \rho, \alpha) .\end{align*} Continuing \(\star_2\), we have \begin{align*} v( {\partial}( e_{ w \rho})) &= \sum_{w\in W} (-1)^{\ell(w)} (-1)^{\ell(w)} \prod_{ \alpha\in \Phi^+}( \rho, \alpha) \\ &= {\sharp}W \prod_{ \alpha\in \Phi^+}( \rho, \alpha) ,\end{align*} which is the LHS. Similarly for the RHS, \begin{align*} v({\partial}( \lambda+ \rho)) = {\sharp}W \prod_{ \alpha\in \Phi^+} (\lambda+ \rho, \alpha) .\end{align*}

Taking the quotient yields \begin{align*} \dim L(\lambda) = {{\sharp}W \prod_{ \alpha\in \Phi^+}( \rho, \alpha) \over {\sharp}W \prod_{ \alpha\in \Phi^+} (\lambda+ \rho, \alpha) } = {\prod_{ \alpha\in \Phi^+}( \rho, \alpha) \over \prod_{ \alpha\in \Phi^+} (\lambda+ \rho, \alpha) } .\end{align*} Multiplying the numerator and denominator by \(\prod_{ \alpha\in \Phi^+} {2\over (\alpha, \alpha)}\) yields \begin{align*} \prod_{ \alpha\in \Phi^+} {\left\langle {\rho},~{\alpha} \right\rangle} \over \prod_{ \alpha\in \Phi^+} {\left\langle { \lambda+ \rho},~{\alpha} \right\rangle} .\end{align*}