2 Properties of solvable and nilpotent Lie algebra

Solvable Lie Algebra

Existence of Weight Space

We show that every finite dimensional complex representation of a solvable Lie algebra has a weight space:

Proposition

For a finite dimensional complex solvable Lie algebra $g$ and its finite dimensional complex representation $(\rho ,g)$, this representation has a weight.

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Then we have the following corollary:

Corollary

Every irreducible representation of finite dimensional solvable Lie algebra is $1$-dimensional.

Question

Conversely, for a Lie algebra $g$, if every finite dimensional complex representation of $g$ has a weight, would it must be solvable?

Answer

The answer is yes. Consider the adjoint representation. Since $\mathrm{ad}:g\to \mathrm{End}(g)$ is an representation, using the same argument in Lie’s theorem, there exists a set of basis such that $\mathrm{ad}g$ are upper triangular matrices. Then $\mathrm{ad}([g,g])$ are nilpotent. By Engel’s theorem, $[g,g]$ is nilpotent and so $g$ is solvable by ^2be7fe.

Lie's theorem

Let $g$ be a finite dimensional solvable Lie algebra over $\mathbb{C}$ and $(\rho ,V)$ a finite dimensional representation. Then one can choose a basis $\{v_1,\cdots ,v_n\}$ of $V$ and $\rho (g)\subseteq \{\text{ upper triangular matrices }\}$ w.r.t. the identification $\mathrm{End}(V)\cong M_{n\times n}(\mathbb{C})$.

\begin{proof} Since $g$ is solvable, there is a weight $U\subseteq V$ and $\rho {|}_U$ is diagonalizable. Let $U=\mathrm{span}\{v_1,\cdots ,v_t\}$. Then consider $\tilde{\rho }:g\to V/U$, still $\tilde{\rho }$ has a weight $\widetilde{U_1}$. Let $U_1$ be the preimage of $\widetilde{U_1}$ and $U_1=\mathrm{span}\{v_1,\cdots ,v_t,v_{t+1},\cdots ,v_m\}$. Then repeat the procedure, that is consider ${\tilde{\rho }}^{\prime }:g\to V/U_1$ and its weight until we get a set of basis of $V$. It is easy to verify under this basis $\rho (g)\subseteq \{\text{ upper triangular matrices }\}$. \end{proof}

Finest Ideal Sequence

Proposition

A finite dimensional Lie algebra is solvable iff it admits a sequence of ideals

$$0=I_0⊊I_1⊊\cdots ⊊I_{n-1}⊊I_n=g\text{\hspace{1em}(*)}$$

with $\dim I_k=k$.

\begin{proof} i) If $g$ is solvable, then $g\supseteq g^{(1)}\supseteq \cdots \supseteq g^{(n)}=0$. Note that any subspace contains $[g,g]=g^{(1)}$ is an ideal, so there is a sequence of ideals

$$g^{(1)}⊊g^{(1)}\oplus \mathbb{C}{Y}_1⊊\cdots ⊊g^{(1)}\oplus (\mathbb{C}{Y}_1\oplus \cdots \oplus \mathbb{C}{Y}_k)⊊g$$

where $Y_i\in g\backslash g^{(1)}$. Similarly, use $Z_i\in g^{(1)}\backslash Y^{(2)}$ to construct subsequence of $(\ast )$ between $g^{(2)}$ and $g^{(1)}$. Repeat this process and get the sequence we want.

ii) Conversely, suppose $g$ admits a sequence $(\ast )$. Take a set of basis $\{e_1,\cdots ,e_n\}$ such that $I_k=\mathrm{span}\{e_1,\cdots ,e_k\}$. Then under this basis $\mathrm{ad}X$ is an upper triangular matrix for any $X\in g$, and so $\mathrm{ad}([g,g])$ are nilpotent. By Engel’s theorem, $[g,g]$ is nilpotent and so $g$ is solvable by ^2be7fe. \end{proof}

Unique Maximal Nilpotent Ideal of Solvable Lie Algebra

Corollary

Let $g$ be a finite dimensional complex Lie algebra. Then

• $g$ is solvable $⟺$ $[g,g]$ is nilpotent;
• when $g$ is solvable, $g$ has a unique maximal nilpotent ideal

$$n=\{x\in g:\mathrm{ad}X\text{ is nilpotent }\}.$$

\begin{proof} i) $[g,g]$ nilpotent $\Rightarrow$ $g$ solvable is trivial. Now assume $g$ is solvable. Then $\mathrm{ad}g$ can be represented by upper triangular matrices. So elements of $\mathrm{ad}[g,g]$ can be represented by strictly upper triangular matrices. By Engel’s theorem, $[g,g]$ is nilpotent.

ii) Since $g$ is solvable, we have the following diagram.

Thus $n$ is a nilpotent ideal. Let $n^{\prime }$ be a nilpotent ideal. Since $n^{\prime }$ is nilpotent, for each $Y\in n^{\prime }$, $\mathrm{ad}(Y)$ is nilpotent. Then $Y\in n$ yields $n^{\prime }\subseteq n$ and $n$ is the unique maximal nilpotent ideal. \end{proof}

Open Question

Could we “classify” all complex representation of a finite dimensional solvable Lie algebra?

Nilpotent Lie algebra

Decomposed as Direct Sum of Generalized Weight Space

Every representation of a nilpotent Lie algebra can be decomposed as a direct sum of generalized weight space. See here.

Question

If its each finite dimensional complex representation can be decomposed as a direct product sum of generalized weight spaces, would it must be nilpotent?

Answer

The answer is yes, see here.

Engel’s Theorem

Let $g$ be a finite dimensional complex Lie algebra. Then $g$ is nilpotent $⟺$ $\mathrm{ad}X$ is nilpotent for each $X\in g$.

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This theorem is proved here, and we get the following corollary.

Corollary

Let $g$ be a finite dimensional complex Lie algebra. Then $g$ is nilpotent iff it has a basis w.r.t. which the adjoint representation sends $g$ into the space spanned by strictly upper triangular matrices.

\begin{proof} i) Suppose any $\mathrm{ad}X$ is upper triangular matrix. By Cayley-Hamilton $\mathrm{ad}X$ is nilpotent. Then $g$ is nilpotent by Engel’s theorem.

ii) Suppose $g$ is nilpotent. Then $g$ is solvable and there is a basis s.t. $\mathrm{ad}X$ is upper triangular for all $X\in g$. By Engel’s theorem, $\mathrm{ad}X$ is nilpotent. By Cayley-Hamilton $\mathrm{ad}X$ is strictly upper triangular matrix. \end{proof}