Lie Algebras
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1 Nilpotent and Solvable Lie Algebras
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Proof. Let \(\mh \) be a codimension \(1\) ideal. By induction, there is a nontrivial subspace such that \(\mh \) acts trivially. Since \(\mh \) contains \([\mg ,\mg ]\), \(\ker \mh \)
is an invariant subspace since \(hgv= [h,g]v + ghv = 0\) if \(h \in \mh , g \in \mg , v \in \ker \mh \). Any \(a \notin h\) has an element it kills \(v \in \ker \mh \) since it acts nilpotently. Since \(ka + \mh = \mg
\), we see \(\mg v = 0\). □
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Proof. Since the lower central series eventually vanishes, \(\ad a\) is nilpotent for a nilpotent Lie algebra. Conversely, by Engel’s theorem, if \(\ad a\) is nilpotent, then \(\mg /Z(\mg
)\) is nilpotent, so \(\mg \) is too. □
We can classify \(2\)-step nilpotent Lie algebras \(\mg \). They are completely determined by the bracket \(\mg /Z(\mg ) \to Z(\mg )\). This is an alternating form, and is nondegenerate because \(Z(\mg )\) is the center.
The idea in Engel’s Theorem can be upgraded to nonzero weights when the characteristic is sufficiently large.
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Lemma 1.7 (Lie’s Lemma). If \(\mh \subset \mg \) is an
ideal, \(V^{\mh }_{\lambda }\) a weight space of \(\mh \) in a f.d representation \(V\) of \(\mg \), with \(\ch k > V\), then it is \(\mg \)-invariant.
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Proof. For \(v \in V^{mh}_{\lambda }, g \in \mg , h \in \mh \) we need to show that \(hgv = \lambda (h)gv\). But it is \([h,g]v + ghv = [h,g] v + \lambda (h)gv\), so it
is equivalent to show \([h,g]v=0\).
Define \(W_m\) as the span of \(g^iv\) for \(i \leq m\). By induction, we can see that \(W_m\) is \(\mh \)-invariant and \(hW_m/W_{m-1} = \lambda (h)W_m/W_{m-1}\), since \(hg^iv = \sum g^j[h,g]g^{i-1-j}v + g^ihv\).
For sufficiently large \(m\), \(W_m\) stabilizes, so is \(g\)-invariant. Then \(\tr ([g,h]) = 0\), but it is upper triangular with eigenvalue \(\lambda ([g,h])\). By our assumption on characteristic, we are done. □
Now we can copy the proof of Engel’s theorem to get a result about solvable Lie algebras.
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Theorem 1.8 (Lie’s Theorem). If \(V\) is a finite
dimensional representation over \(\mg \), a solvable Lie algebra, and \(\ch k> \dim V\) and \(k\) is algebraically closed, then there is a nonzero weight space.
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Proof. By induction, a codimension \(1\) ideal of \(\mg \) has a nonzero weight space \(W\). By Lie’s lemma, it is an invariant subspace. Any \(a \notin h\) has an eigenvector \(v\)
in \(W\) since \(k\) is algebraically closed. Since \(ka + \mh = \mg \), we see \(\mg \) acts by scalars on \(v\). □
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Proof. By extending scalars, we can assume \(k\) is algebraically closed. Then \(\ad \mg \) acts by upper triangular matricies by Lie’s theorem, so \(\ad [\mg ,\mg ]\) acts by strictly
upper triangular matricies, so by Engel’s theorem, we are done. □
Now we will study generalized weight space decompositions.
Let \(k\) be algebraically closed. Via the Jordan decomposition any operator \(A\) on a finite dimensional vector space \(V\) decomposes as \(A_s + A_n\) where \(A_s\) is semisimple, \(A_n\) is nilpotent, and they commute.
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Lemma 1.11. If \(A,B\) commute, they preserve eachother’s generalized eigenspaces, and their Jordan
decompositions commute with eachother.
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Proof. For the first statement, the support of the \(k[A,B]\)-module \(V\) is a finite set of points so it splits as a sum of modules supported at a point, which corresponds to finding
simultaneous generalized eigenspaces of \(A,B\). The semi-simple parts act diagonally on each component of the sum, so \(A,A_s\) commute with \(B,B_s\), so we are done. □
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Proof. Suppose \(A = A_s+A_n = A'_s+A'_n\), where the former is the usual decomposition. Then since \(A'_s\) commutes with \(A\), it commutes with \(A_s\), and
hence \(A_n\). Since \(A'_n = A-A'_s\), it also commutes with everything else. Then \(A_s-A'_s=A_n-A'_n\) is a semisimple and nilpotent so it is \(0\). □
We would like a theory of generalized eigenspaces that works for representations of nilpotent Lie algebra, rather than just for endomorphisms.
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Lemma 1.14. In an associative unital algebra \(U\) over \(k\), if \(a,b \in U\) and \(\lambda
,\alpha \in k\), then
\[(a-\alpha -\lambda )^{N}b = \sum _0^N {N \choose j} (\ad a-\alpha )^jb(a-\lambda )^{N-j}\]
To avoid writing a more precise assumption, from now on assume \(\ch k = 0\) and \(k\) is algebraically closed. Some statements will hold without the algebraically closed assumption via extension of scalars.
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Proof. The very last statement is obtained by having both sides of the equation in the previous lemma act on \(v\in V\), with \(U\) the universal enveloping algebra \(U(\mg )\), \(a
\in \mh ,b \in \mg \).
To obtain the decomposition, first note that for a \(1\)-dimensional Lie algebra it is just the generalized eigenspace decomposition. If there is some \(a \in \mh \) with more than one eigenvalue, first note that \(\mh = \mh
_0^a\) because \(a\) acts nilpotently. Then by the last statement of the theorem, \(\mh \) preserves the generalized eigenspaces of \(a\), so by induction on \(\dim V\) we are done.
In the case that everything has one eigenvalue, Lie’s theorem shows that \(\mh \) acts by upper triangular matricies where the diagonals are constant, so it is a weight space. □
For a Cartan subalgebra, we drop the superscript \(\mh \) in the notation.