2.1 Existence of (generalized) weight space

TLDR

Just like linear transformation classified by Jordan canonical form, we use “(generalized) eigenspace”, which is called (generalized) weight space, to analyze representations of Lie algebra. However, for a set of matrices, it is not easy to find a common eigenspace. Luckily, with additional conditions, named solvable and nilpotent, we have some good results.

Two important conclusions will be showed:

• every solvable Lie algebra has a weight space;
• for any representation $(\rho ,V)$ of nilpotent Lie algebra, $V$ can be decomposed as direct sum of generalized weight spaces.

Solvable Lie algebra

Eigenspace contained in an ideal is $g$-invariant

Naive try. Suppose $\rho :g\to \mathrm{End}(V)$, $h\subseteq g$ is a subalgebra. Suppose $\rho {|}_h$ has a weight $\lambda$ and $V_{\lambda }$ is the according weight space. Hope that $V_{\lambda }$ to be $g$-invariant. That is, for any $Y\in g$, $\rho (Y)v=\tilde{\lambda }(Y)v$ and $\tilde{\lambda }{|}_h=\lambda$.

For any $x\in h$ and $y\in g$, by above assumptions we have

$$\begin{array}{rl}\rho (X)\rho (Y)v& =\rho (Y)\rho (X)v+\rho ([Y,X])v=\lambda (X)\rho (Y)v+\rho ([Y,X])v,\\ \rho (X)\rho (Y)v& =\lambda (X)\rho (Y)v,\end{array}$$

so $\rho ([Y,X])$ should be zero. If $h$ is an ideal, then $\rho ([Y,X])=\lambda ([Y,X])$ should be zero.

Lemma

Suppose $g$ is a finite dimensional complex Lie algebra and $h\subseteq g$ is an ideal. Let $\lambda$ be a weight of $\rho {|}_h$ and $V_{\lambda }$ be the corresponding weight space, where $(\rho ,V)$ is a finite dimensional complex representation. Then $V_{\lambda }$ is $g$-invariant.

TL; DR: 找到一个$h\oplus \mathbb{C}X$下的不变子空间，考虑$\rho ([X,Y])$限制在子空间上的矩阵：它的迹为$k\cdot \lambda ([X,Y])=0$.

\begin{proof}To show $V_{\lambda }$ is $g$-invariant, by the above argument it remains to show $\lambda ([h,g])=0$. For any $X\in g$, $Y\in h$ and $v\in V_{\lambda }$, we can verify

$$W:=\mathrm{span}\{v,\rho (X)v,\cdots ,{\rho }^{n-1}(X)v\}$$

is $h\oplus \mathbb{C}X$-invariant.

Note that for any $Z\in h$, $\rho (Z)$ acts on this space by the matrix

$$\left[\begin{array}{ccc}\lambda (Z)& & \ast \\ & \ddots & \\ & & \lambda (Z)\end{array}\right].$$

Since $h$ is an ideal, $[X,Y]\in h$. Then the trace of $\rho ([X,Y])$ is $k\cdot \lambda ([X,Y])$. Also $\rho ([X,Y])=[\rho (X),\rho (Y)]$ is trace zero. Thus $\lambda ([X,Y])=0$. \end{proof}

Main result

Proposition

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

\begin{proof}By induction, we only need to prove that if $\rho {|}_{[g,g]}$ has a weight, then $\rho$ has a weight. Suppose $\rho {|}_{[g,g]}$ has a weight $\mu :[g,g]\to \mathbb{C}$ and the corresponding weight space $U_{\mu }$. For any $Y\in g\backslash [g,g]$, by lemma $U_{\mu }$ is $\rho (Y)$-invariant. Choose eigenvalue $c_Y$ and let $U_{\mu }^{\prime }\subset U_{\mu }$ be the corresponding eigenspace. Now $U_{\mu }^{\prime }$ is a weight space of $[g,g]\oplus \mathbb{C}Y$. Then repeat the procedure, by $\dim (g/[g,g])$ is finite we know $\rho$ has a weight. \end{proof}

Nilpotent Lie algebra

Proposition

Let $g$ be a finite dimensional nilpotent Lie algebra over $\mathbb{C}$ and $(\rho ,V)$ be a finite dimensional complex representation. Then:

• for $Y\in g$, the generalized eigenspaces of $\rho (Y)$ are $g$-invariant;
• $V$ is the direct sum of generalized weight spaces.

\begin{proof}i) Let $c$ be an eigenvalue of $\rho (Y)$ and $V_c^{\prime }$ the according generalized weight space. It suffices to show for any $x\in g$ and any $v\in V_c^{\prime }$, $\rho (X)v\in V_c^{\prime }$, that is,

$$(\rho (Y)-c{)}^N\rho (X)v=0$$

for some $N\in \mathbb{N}$. Note that

$$\begin{array}{rl}& (\rho (Y)-c{)}^N\rho (X)v\\ =& \rho (X)(\rho (Y)-c{)}^{N}v+\sum _{i=1}^n(\genfrac{}{}{0ex}{}{n}{i})(\mathrm{ad}(\rho (Y)-c){)}^{N-i}\rho (X)(\rho (Y)-c{)}^{i}v\\ =& \rho (X)(\rho (Y)-c{)}^{N}v+\sum _{i=1}^n(\genfrac{}{}{0ex}{}{n}{i}\left)\right(\mathrm{ad}\rho (Y{)}^{N-i}\rho (X))(\rho (Y)-c{)}^{i}v,\end{array}$$

and $\rho (g)$ is also nilpotent. So for a big enough $N$, $\mathrm{RHS}=0$.

ii) Choose a basis $\{Y_1,\cdots ,Y_n\}$ of $g$ where $\dim g=n$. Write $V=\oplus E_{i_1}$ where each $E_{i_1}$ is a generalized eigenspace of $\rho (Y_1)$. Each $E_{i_1}$ is $\rho (Y_2)$-invariant and hence can be decomposed as the direct sum of generalized eigenspaces of $\rho (Y_2){|}_{E_{i_1}^{\prime }}$, written as $E_{i_1}=\oplus E_{i_1{i}_2}$. Continue the process we get $V=\oplus E_{i_1\cdots i_n}$ where

$$(\rho (Y_j)-c_{j,i_1\cdots i_n}I{)}^{n_{i_1\cdots i_n}}{|}_{E_{i_1\cdots i_n}}=0.$$

Define

\begin{aligned} \lambda_{i_1\cdots i_n}:g\to&\mathbb C\\\sum a_jY_j\mapsto&\sum a_jc_{j,i_1\cdots i_n}\end{aligned}.$$ Claim that $E_{i_1\cdots i_n}$ is the generalized weight space associated to $\lambda_{i_1\cdots i_n}$. $g$ is nilpotent yields $g$ is solvable and then there is a basis such that $\rho(Y)|_{E_{i_1\cdots i_n}}$ is an upper triangular matrix for all $Y\in g$. Then

\begin{aligned} Y_j&\to\left[ \begin{matrix} c_{j,i_1\cdots i_n} & & * \ & \ddots & \ & & c_{j,i_1\cdots i_n} \end{matrix} \right]\ Y=\sum a_j Y_j&\to\left[ \begin{matrix} \sum a_jc_{j,i_1\cdots i_n} & & * \ & \ddots & \ & & \sum a_jc_{j,i_1\cdots i_n} \end{matrix} \right] \end{aligned}

So $\rho(\sum a_j Y_j)v_1=(\sum a_jc_j)v_1$, whence $\lambda_{i_1\cdots i_n}$ is a weight. And $[\rho(Y)-\lambda_{i_1\cdots i_n}(Y)]^m=0$ on $E_{i_1\cdots i_n}$ for some $m$. Therefore $E_{i_1\cdots i_n}$ is the generalized $\lambda_{i_1\cdots i_n}$-weight space. `\end{proof}`