Locally Confused

Solutions to Humphreys Chap. 3

last edited around 2020-07-07

\(\DeclareMathOperator{\ad}{ad}\) (The assignments are omitted, because it's probably not a wise idea to copy them word-for-word on a public site)

  1. By jacobi, the commutator of two ideals is again an ideal.
  2. \(L/I\) is abelian iff I contains \([LL]\), so \(\Rightarrow\) is clear. OTOH, \(L_i/L_{i+1}\) abelian implies \(L_{i+1}\geq [L_i L_i]\), implying this squashes the derived series.
  3. The generators \(h,x,y\) satisfy \([xy]=h\) and \([hx]=2x=0=[hy]\), so \({(L_i)}_i=(L, \langle h \rangle, 0)\).
  4. Solvability: note that \(L\) is an extension of \(\ad L\) by the center. Nilpotency: \(\ad\) is a hom, so termination of \({(\ad L)}_i=\ad L_i\) implies \(L_i\leq Z(L)\) eventually.
  5. For \(\mathfrak h\) with \([xy]=x\), the derived series is \((L, \langle x \rangle, 0)\), whereas the LCS stagnates at \(\langle x \rangle\). For the other one with \([xy]=z,[xz]=y,[yz]=0\), evidently, every ideal must contain \(I := \langle y, z \rangle\). We proceed to find \((L, I, 0)\) as the DS and \((L, I, I, \ldots)\) as the LCS.
  6. \([L(A+B)]\leq [LA]+[LB]\), so \({(I+J)}_i\) is bounded by \({(I)}_i+{(J)}_i\). For \(\mathfrak h\) this is \(\langle x \rangle\), for the other one its center \(\langle y, z \rangle\).
  7. The lie algebra induced by \(k\mapsto \alpha_k\colon (l+K\mapsto [kl]+K)\) consists of nilpotent endomorphisms, so Engel guarantees a nonzero \(l+K\) in every kernel. This translates to \(l\) normalizing \(K\).
  8. Take \(K\leq L\) maximal and \(l\) as above. Normalization turns \(K+ \mathsf{F} \cdot l\) into a subalgebra, necessarily \(L\). Also, \(N_L(K)\geq K+\mathsf{F}l=L\).
  9. First, \(z\in C_L(K)\) (nonempty because we have \(Z(L)\neq 0\)) ensures \(\delta \in \operatorname{Der}(L)\). If we had \(\delta=\ad_y\) with \(y\in L_i\), then \([yK]=0\) – so \(y\in C_L(K)\) – and \([yx]=z\) – so \(z\in L_{i-1}\). Therefore, taking \(z\) to be of minimal “depth” \(i\) prevents such a \(y\in C_L(K)\) from existing.
  10. Projection onto an ideal is a hom, so \({(L/K)}_i = L_i / K\). Ergo, \(L_i\leq K\) eventually. There, we have \((\ad_x)^{l+i}(y) \in (\ad_x)^l(K)\) for any \(x,y\in L\), vanishing for large enough \(l\). Hence \(L\) is \(ad\)-nilpotent and by Engel nilpotent.