可解李代数表示的分解

本篇笔记中将会展示有限维复数空间可解李代数的表示的分解性

引言

为什么我们需要一个 bracket structure ？ 当我们试图去理解非交换性给我带来的变化的时候，我们会很自然的去考虑$XY-YX$.

Question. what can we do with a general f.d. complex rep of a f.d. Lie algebra over $\mathbb C$?

Recall. weight generalized weight

Question.

• when would weight exist?
• when would a rep be written as the direc sum of generalized weight space?

As a quick answer, Q1 will be true if $\mathcal g$ is sovable, Q2 will be true if $g$ is nilpotent.

旅途开始

$g$ f.d. Lie algebra, $V$ f.d. complex space $\rho: g \to \mathrm{End}(V)$ Lie algebra homo For $X \in g$, denote $V_c$ is a $c$-eigenspace of $\rho(X)$, we want to find $Y\in g$ s.t. $V_c$ to be $\rho(Y)$ invarient. \(\begin{eqnarray} \forall v \in V_c, \rho(X)\rho(Y)v & = & \rho(Y)\rho(X)v +[\rho(X), \rho(Y)]v\\ & = & c\rho(Y)v +[\rho(X), \rho(Y)]v \end{eqnarray}\) To make $\rho(X)$ and $\rho(Y)$ commutative, we hope the last part to be zero.

In a slightly more general setting \(\rho:g \to \mathrm{End}(V)\) then we hope to find a Lie subalgebra $h \subset g$, s.t. $\rho|h$ has a weight $\lambda$ and $V\lambda$ the according weight space \(\rho(Y)v = \lambda(Y)v, \forall Y \in h,\forall v \in V_\lambda\)

对于这个等式，我们似乎需要增加一些条件，比如说$[h,g] \subset h$

Lemma. $g$ a f.d. complex Lie algebra $h \subset g$ is an ideal $(\rho, V)$ a f.d. complex rep of $g$. Let $\lambda$ be a weight of $\rho|h$ and $V\lambda$ the according weight space of $\rho|h$, then $V\lambda$ is $g$-invariant.

proof. For $v \in V_\lambda$ and $X \in g, Y \in h$, we need to show $\rho(Y)\rho(X)v = \lambda(Y) \rho(X)v$.

In the other hand, we have \(\begin{eqnarray} \rho(Y)\rho(X)v & = & \rho(X)\rho(Y)v + [\rho(Y),\rho(X)]v\\ & = & \lambda(Y)\rho(X)v + \rho([Y,X])v\\ & = & \lambda(Y)\rho(X)v + \lambda([Y,X])v \end{eqnarray}\) It suffices to verify $\lambda([h,g])=0$,

Consider the sequence generated by $\rho(X)$ act on $v \in V_\lambda$. \(v \to \rho(X)v \to \rho(X)^2v \to \cdots \to \rho(X)^nv\) where $n$ is the minimal poly of $\rho(X)$ w.r.t. $v$ For $Y \in h$, we act $Y$ on it, then we have $$ \begin{eqnarray} v & \to & \lambda(Y)v
\rho(X)v & \to & \lambda(Y)\rho(X)v + \lambda([Y,X])v
&\vdots &\

\end{eqnarray} \(Using the exercise below. we observe that $W = \mathrm{span}\{v, \rho(X)v,\dots,\rho(X)^{n-1}\}$ is **$h \oplus \mathbb C X$- invariant** and $\rho(Y)$ acts on this space by the matrix\) \begin{pmatrix} \lambda(Y) & * & *
& \ddots & *
& &\lambda(Y) \end{pmatrix} \(Then, we calculate the trace of $\rho([Y,X])|_W$, but\) \rho([Y,X])|_W = \rho(Y)|_W\rho(X)|_W-\rho(X)|_W\rho(Y)|_W $$ so, we have $n \cdot \lambda([Y,X]) = 0$ implies $\lambda([Y,X]) = 0$. Tha is all.

在这个证明中，其实我们稍微使用了一个看起来没有那么显然的等式，即下面这个等式。

Exercise. For any $A, B \in \mathrm{End}(V)$ \(AB^n=B^nA + \sum_{i=1}^n \binom{n}{i} B^{n-i}[\cdots[A,B]\cdots,B] \text{ (i times insert sleeve)}\)

Prop. $g$ a f.d. complex solvable Lie algebra $(\rho,V)$ a f.d. complex rep of $g$. Then, $\rho$ has weights (i.e. for some $1$-dim rep $\lambda$ of $g$, $V_\lambda \not = 0$)

proof. Obs: If $\rho|{[g,g]}$ has a weight $\mu:[g,g] \to \mathbb C$, then for some $1$-dimensional rep $\lambda: g \to \mathbb C$ that extends $\mu$ with $V\lambda \not = 0$. Actually, let $U_\mu$ be the $\mu$-weight space of $\rho|{[g,g]}$. For any $Y \in g -[g,g]$, $U\mu$ is $\rho(Y)$-invariant by the previous lemma and has nonzero eigenspace w.r.t. the action of $\rho(Y) \subset U_\mu$. （任何一个线性映射都有一个非平凡的不变子空间） Choose an eigenvalue $c_Y$ and let $U_Y$ be the accorfing $c_Y$-eigensubspace of $U_]mu$, then $\mathbb C Y \oplus [g,g]$ acts on $U_Y$ via the linear functional $\mu’: \mathbb C Y \oplus [g,g] \to \mathbb C$ with $\mu’|_h = \mu$ and $\mu’(Y) = c_Y$. Similarly, we do it for $Y_2 \in g-(\mathbb C Y \oplus [g,g])$, and so on. Since $g$ is finite dimension, then we will finally stop and get a extension $\lambda$ of $\mu$ on $g$. Then, we prove the Prop using the observation. Since $g$ is sovable, then we have \(g \supsetneq g^{(1)}=[g,g] \supsetneq g^{(2)} = [g^{(1)}, g^{(1)}]\supsetneq \cdots \supsetneq [g^{(n)},g^{(n)}] = 0\) Since $0$ must have such a weight, we repeatly apply the observation to lift $[g^{(k)},g^{(k)}]$ to $[g^{(k-1)},g^{(k-1)}]$. Finally, we get the weight on $g$, that is what we need.

Cor. 1 Let $g$ be a f.d. sovable Lie algebra over $\mathbb C$ and $(\rho, V)$ a f.d. complex representation. Then one can choose a basis ${v_1,\dots,v_n}$ of $V$ and $\rho(g) \subset {\text{upper triangular matrices}}$ w.r.t. the identification $\mathrm{End}(V) \simeq M_{n \times n}(\mathbb C)$.

Cor. 2 A $n$-dim Lie algebra is solvable iff it admits a sequence of ideals $0=I_0\subsetneq I_1 \subsetneq \cdots \subsetneq I_{n-1} \subsetneq I_n = g$ with $\dim I_k = k$.

Lemma. let $g$ be a f.d. nilpotent algebra over $\mathbb C$, $(\rho, V)$ a f.d. complex representation. 1) for $Y \in g$, the generalized eigenspaces of $\rho(Y)$ are $g$-invarianct 2) $V$ is the direct sum of generalized weight spaces.

proof. For the proof of 1), let $c$ be and eighenvalue of $\rho(Y)$ of $\rho(Y)$ and $V_c’$ the according generalized $c$-eigenspace. \(V_c'=\{v \in V:(\rho(Y)-c)^nv =0\text{ for some }n\}\) actully, there is a common $n$ working for all $v \in V_c$. We need to show $\forall x \in g, v\in V_c’$, we have $\rho(X)v = V_c’$. In other word, we need to show \(\left(\rho(Y)-c\right)^N\rho(X)v = 0\) for some $N \in \mathbb N$. Observe that \(\left(\rho(Y)-c\right)^N\rho(X)v =\rho(X)(\rho(Y)-c)^N - \sum_{i=1}^N \binom{N}{i} \left(\mathrm{ad}{(\rho(Y)-c)}^{(N-i)}\rho(X)\right)(\rho(Y)-c)^i v\) suppose $m$-apies $g^{(m)} = [g,[g,[\cdots[g,g]]]] = 0$, then for $N \ge m - 1 + n$, the right hand side will be zero, since \(\mathrm{ad}{(\rho(Y)-c)}^{(N-i)}\rho(X) = \mathrm{ad}{(\rho(Y))}^{(N-i)}\rho(X)=\rho(\mathrm{ad}{(Y)}^{(N-i)}X)\) We finish the proof of 1).