4.1 Cartan subalgebra

TLDR

To describe the case where the generalized $0$-weight space of an adjoint representation of a nilpotent subalgebra is exactly the subalgebra itself, Cartan subalgebra is defined. The main theorem show that each finite dimensional complex Lie algebra has a Cartan subalgebra and it is unique up to conjugate. In fact, a Cartan subalgebra is a generalized $0$-eigenspace of a element with the minimal dimension.

What is Cartan algebra?

Motivation. Suppose $g$ is a finite dimensional complex Lie algebra, and $h\subseteq g$ is a nilpotent subalgebra. Then with respect to $h\stackrel{\mathrm{ad}}{\curvearrowright }g$ one may write

$$g=L_0\oplus ({\oplus }_{{\lambda }_i\ne 0}{L}_{{\lambda }_i})$$

where $L_0$ is the generalized weight space associated to the $0$-weight of $h$ and $L_{{\lambda }_i}$ is the generalized weight space associated to the non-zero weight ${\lambda }_i$ of $h$. Obviously, $h\subseteq L_0$. Naturally, one may hope that $h=L_0$. In Observation 1, we can prove $h=L_0$ iff $h$ is Cartan.

Cartan subalgebra

$h\subseteq g$ is called a Cartan subalgebra if $h$ is nilpotent and

$$N(h):=\{x\in g:[x,h]\subseteq h\}$$

is equal to $h$.

Remark. For a Lie subalgebra $h$, $N(h)$ is called the normalizer of $h$. Hence, a Cartan subalgebra is a nilpotent self-normalizing subalgebra.

Some easy observations

Observation 1.

• $N(h)$ is a Lie subalgebra.
• $h$ is an ideal of $N(h)$ and $N(h)$ is the maximal Lie subalgebra in which $h$ is an ideal.
• $h=L_0$ in $(\ast )$ iff $h$ is a Cartan subalgebra.

\begin{proof} (i) and (ii) are easy. Now we prove (iii). Assume $h$ is a Cartan subalgebra. If $h⊊L_0$, then $L_0/h$ is an $h$-modulo whose only weight is the zero weight and hence has an non-trivial $0$-weight space $M/h$ where $M\subseteq L_0$. So $[M,h]\subseteq h$ and $N(h)\supseteq M⊋h$, contradiction. Conversely, if $h=L_0$, then $h$ is the generalized weight space associated to the $0$-weight and so $(\mathrm{ad}h{)}^{N}h=0$ for some $N$. Thus $h$ is nilpotent. Note that for any $x\in N(h)$, $[x,h]\subseteq h$ yields $(\mathrm{ad}h{)}^{m}x\subseteq (\mathrm{ad}h{)}^{m-1}h=0$ for big enough $m$. Thus $x\in L_0=h$ and $h$ is a Cartan subalgebra. \end{proof}

Observation 2.

• $[L_0,L_{{\lambda }_i}]\subseteq L_{{\lambda }_i}$.
• $[L_{{\lambda }_i},L_{{\lambda }_j}]\subseteq L_{{\lambda }_i+{\lambda }_j}$, if ${\lambda }_i+{\lambda }_j$ is not a weight, then $[L_{{\lambda }_i},L_{{\lambda }_j}]=0$.

\begin{proof} We need to verify that for any $y\in L_{{\lambda }_i}$ and $z\in L_{{\lambda }_j}$, $[y,z]\in L_{{\lambda }_i+{\lambda }_j}$, that is,

$$(\mathrm{ad}x-{\lambda }_i(x)-{\lambda }_j(x){)}^N[y,z]=0$$

for any $x\in h$. Since

$$\begin{array}{rl}& (\mathrm{ad}x-{\lambda }_i(x)-{\lambda }_j(x){)}^N[y,z]\\ =& \sum _{L=0}^N\left(\genfrac{}{}{0ex}{}{N}{L}\right)[(\mathrm{ad}x-{\lambda }_i(x){)}^{L}y,(\mathrm{ad}x-{\lambda }_j(x){)}^{N-L}z]\\ =& 0\end{array}$$

with big enough $N$, $[y,z]\in L_{{\lambda }_i+{\lambda }_j}$. \end{proof}

Remark. It is similar to graded ring.

Observation 3. When $h$ is a Cartan subalgebra, $g=h\oplus ({\oplus }_{{\lambda }_i\ne 0}{L}_{{\lambda }_i})$. Then for any $x\in h$, the generalized eigenspace of $\mathrm{ad}x$ with eigenvalue $0$ is

$$h\oplus ({\oplus }_{{\lambda }_i(x)=0}{L}_{{\lambda }_i}).$$

Thus when all ${\lambda }_i(x)$ are non-zero, the generalized $0$-weight space is the generalized eigenspace of $\mathrm{ad}x$.

Main result

Definitions

regular

For $x\in g$, put $L_{0,x}=$ the generalized eigenspace of $\mathrm{ad}x$ with eigenvalue $0$. Then $x\in g$ is called regular if $\dim L_{0,x}={\min }_{y\in g}\dim L_{0,y}$.

derivation

A linear map $\delta :g\to g$ is called a derivation if $\delta [y,z]=[\delta y,z]+[y,\delta z]$. For any $x\in g$, define $\exp (\mathrm{ad}x)={\sum }_{n=0}^{\infty }(\mathrm{ad}x{)}^n/n!$. If $\mathrm{ad}x$ is nilpotent, it is a finite sum. If $\delta$ is nilpotent derivation, then

$$\exp (\delta ):=\sum _{i=0}^{\infty }\frac{{\delta }^n}{n!}$$

is a finite sum and defines an element in $\mathrm{Aut}(g)$, which is verified in the following remark. Also define

$$\mathrm{Inn}(g)=⟨\exp (\mathrm{ad}x):\mathrm{ad}x\text{ is nilpotent }⟩.$$

Remark. Note that $\exp (\delta )[y,z]=[\exp (\delta )y,\exp (\delta )z]$, and $\exp (\delta )\exp (-\delta )=\mathrm{id}$. Thus $\exp (\delta )$ is an automorphism.

Main theorem: existence and uniqueness of Cartan subalgebra

Theorem

Suppose $g$ is a finite dimensional complex Lie algebra.

• If $x\in g$ is regular, then $L_{0,x}$ is the Cartan subalgebra.
• If $h$ is a Cartan subalgebra, then there exists a regular element $x$ such that $h=L_{0,x}$.
• If $h_1,h_2$ are two Cartan subalgebras, then there is a $\sigma \in \mathrm{Inn}(g)$ such that $\sigma (h_1)=h_2$.

\begin{proof}i) Suppose $x$ is regular. By the argument of Observation 2, $L_{0,x}$ is a subalgebra. Since $x\in L_{0,x}$, for any $z\in N(L_{0,x})$, $[z,x]\in L_{0,x}$ yields $z\in L_{0,x}$. Thus $N(L_{0,x})=L_{0,x}$. It remains to check $L_{0,x}$ is nilpotent, so we only need to show $\mathrm{ad}y{|}_{L_{0,x}}$ is nilpotent for all $y\in L_{0,x}$.

Claim that $x$ is regular iff $a_L(x)\ne 0$, where $L={\min }_{y\in g}\dim L_{0,y}$ and $a_i(x)$ is the coefficient of the characteristic polynomial of degree $i$. For any $y\in g$,

$$\det (t{\mathrm{Id}}_g-\mathrm{ad}y)=t^n+a_{n-1}(y)t^{n-1}+\cdots +a_1(y)t+a_0(y)$$

where $n=\dim g$ and $a_0(y),\cdots ,a_{n-1}(y)$ are polynomials functions on $g$. Let $L^{\prime }$ be the minimal $i$ such that $a_i(y)\not\equiv 0$. Then

$$\det (t{\mathrm{Id}}_g-\mathrm{ad}y)=t^{L^{\prime }}(a_{L^{\prime }}(y)+a_{L^{\prime }+1}(y)t+\cdots +a_{n-1}{t}^{n-1-L^{\prime }}+t^{n-L^{\prime }}),$$

and $\dim L_{0,y}=L$. So we prove the claim.

In the other word, we need to show that if $a_L(x)\ne 0$, then the characteristic polynomial of $\mathrm{ad}y{|}_{L_{0,x}}$ is $t^L$ for $y\in L_{0,x}$. Note that

$$\begin{array}{rlr}\det (t{\mathrm{Id}}_g-\mathrm{ad}y)=& \det (t{\mathrm{Id}}_{L_{0,x}}-\mathrm{ad}y{|}_{L_{0,x}})\cdot \det (t{\mathrm{Id}}_{g/L_{0,x}}-\mathrm{ad}y{|}_{g/L_{0,x}})& \\ :=& {\chi }_1(t)\cdot {\chi }_2(t),& \end{array}$$

For a special case of $y=x$, ${\chi }_1(t)=t^L$ and ${\chi }_2(t)=d_0(y)+d_1(y)t+\cdots +d_{n-L}{t}^{n-L}$. Then $d_0(y)$ is non-zero polynomial of $y$. Claim that ${\chi }_1(t)=t^L$. Otherwise, ${\chi }_1(t)=t^m(t^{L-m}+\cdots +c_m(y))$ and $c_m$ is a nonzero polynomial with $m<L$. Then $\det (t{\mathrm{Id}}_g-\mathrm{ad}y)=t^m(d_0{c}_m+\cdots +t^{n-m})$. So $\min \dim L_{0,y}=m<L$ with $y\in L_{0,x}$, contradiction.

ii) If $h$ is a Cartan subalgebra, then $g=h\oplus ({\oplus }_{{\lambda }_i\ne 0}{L}_{{\lambda }_i})$. One need to find an element $z$ such that $h=L_{0,z}$.

For a polynomial $f$ on ${\mathbb{C}}^n$, define

$$\begin{array}{rl}{U}_f& =\{v\in {\mathbb{C}}^n,f(v)\ne 0\}\\ V_f& =\{v\in {\mathbb{C}}^n,f(v)=0\}.\end{array}$$

Then if $h=L_{0,z}$, then $z\in U_p$ where $p={\prod }_i{\lambda }_i(x)$ where ${\lambda }_i(x)$ are generalized weights w.r.t $L_{{\lambda }_i}$.

Still suppose

$$\det (t{\mathrm{Id}}_g-\mathrm{ad}y)=t^n+a_{n-1}(y)t^{n-1}+\cdots +a_1(y)t+a_0(y)$$

and $g=\mathrm{span}\{v_1,\cdots ,v_n\}$. Then polynomials $a_i$ of $y$ are also polynomials of $v_1,\cdots ,v_n$. So we can define $U_{a_i}$ as above. Assume $\dim h=L$ and define

$$\begin{array}{rl}f:g& \to g\\ (x_0,x_1,\cdots ,x_r)& \mapsto \exp (\mathrm{ad}{x}_1)\cdots \exp (\mathrm{ad}{x}_r)x_0\end{array}$$

where $x_0\in h,x_i\in L_{{\lambda }_i}$ because $g=h\oplus ({\oplus }_{{\lambda }_i\ne 0}{L}_{{\lambda }_i})$. We hope that $f(U_p)\cap U_{a_L}\ne \varnothing$. If it is true, then some $(x_0,x_1,\cdots ,x_r)$ has its $f$-image in $U_{a_L}$. Notice for any $\sigma \in \mathrm{Aut}(g)$, there is $\sigma (\mathrm{ad}x){\sigma }^{-1}(y)=\sigma [x,{\sigma }^{-1}y]=[\sigma x,y]=\mathrm{ad}(\sigma x)(y)$ and $\mathrm{ad}{x}^{\sigma }=\mathrm{ad}(\sigma (x))$. Thus $\mathrm{ad}x$ and $\mathrm{ad}\sigma (\mathrm{x})$ have the same characteristic polynomial and so $\mathrm{Aut}(g)$ preserves regularity. Therefore, $x_0$ is regular and so $L_{0,x_0}=h$. So it is enough to show $f(U_p)\cap U_{a_L}\ne \varnothing$.

On $U:=\{v\in g:\det J(f)\ne 0\}$, $f$ is non-singular. If $\det J(f)\ne 0$, then $f(U)$ is open and non-empty. And ${\mathbb{C}}^n$ is irreducible w.r.t. the Zariski topology, which follows $f(U\cap U_p)\ne \varnothing$ and $f(U\cap U_p)\cap U_{a_L}\ne \varnothing$. Therefore, $f(U_p)\cap U_{a_L}\ne \varnothing$. So it remains to show $\det J(f)\not\equiv 0$.

Suppose $\{w_{ij}{\}}_j$ is a basis of $L_{{\lambda }_i}$ and $x\in g$ is written as $x={\sum }_{i,j}{\mu }_{ij}{w}_{ij}$. Then $\partial f/\partial {\mu }_{ij}=w_{ij}$ if $i=0$, and

$$\partial f/\partial {\mu }_{ij}{|}_v=\mathrm{ad}(w_{ij})v=-{\lambda }_i(v)w_{ij}+\text{a linear combination of }{b}_{i1},\cdots ,b_{i(j-1)}.$$

Thus $\det J(f){|}_v=\pm {\Pi }_i({\lambda }_i(v){)}^{d_i}$ where $d_i=\dim L_{{\lambda }_i}$ and so $J(f)\not\equiv 0$. Now we finish the proof.

iii) Suppose $h_1$ and $h_2$ are Cartan subalgebras. Then $g=h_1\oplus (\oplus L_{{\lambda }_i})=h_2\oplus (\oplus L_{{\nu }_j})$. Define $p={\Pi }_i{\lambda }_i(x)$ and $\tilde{p}={\Pi }_j{\nu }_j(x)$ and define $f,\tilde{f}$ similarly as ii). By the previous argument,

$$f(U_p)\cap \tilde{f}(U_{\tilde{p}})\cap U_{a_L}\ne \varnothing .$$

Take $z$ in this set. Then there is $x,x^{\prime }\in g$ s.t. $L_{0,x}=h_1,L_{0,x^{\prime }}=h_2$. Also there is $\delta \in \mathrm{Inn}(g)$ mapping $x$ to $x^{\prime }$ by the definition of $f$. Therefore $h_1$ and $h_2$ are conjugate. \end{proof}