Conjugacy of Cartan subalgebras

For $x\in \mathfrak{g}$ let $n(x)$ be the multiplicity of $0$ as an eigenvalue of ${\mathrm{ad}}_x$, i.e. the dimension of its generalized kernel, and let $\rho ={\min }_{x\in \mathfrak{g}}n(x)$. Writing $\det (t\cdot 1-{\mathrm{ad}}_x)={\sum }_j{d}_j(x)t^j$, the coefficients $d_j$ are polynomial functions of $x$ and $n(x)=\min \{j:d_j(x)\mathrm{\ne }0\}$, so $\rho =\min \{j:d_j\equiv ̸0\}$ and the strongly regular locus ${\mathfrak{g}}^{\mathrm{s}\mathrm{r}}=\{x:n(x)=\rho \}$ equals $\{x:d_{\rho }(x)\mathrm{\ne }0\}$, the complement of the zero set of the nonzero polynomial $d_{\rho }$.

Let $\mathfrak{h}$ be a Cartan subalgebra. By [L3] there are finitely many nonzero weights $\lambda$ with ${\mathfrak{g}}_{\lambda }\mathrm{\ne }0$ and $\mathfrak{g}=\mathfrak{h}\oplus {⨁}_{\lambda \mathrm{\ne }0}{\mathfrak{g}}_{\lambda }$, and for $y\in \mathfrak{h}$ one has $\ker ({\mathrm{ad}}_y)=\mathfrak{h}\oplus {⨁}_{\lambda \mathrm{\ne }0,\lambda (y)=0}{\mathfrak{g}}_{\lambda }$. Hence the set ${\mathfrak{h}}_{\mathrm{r}\mathrm{e}\mathrm{g}}=\{y\in \mathfrak{h}:\ker ({\mathrm{ad}}_y)=\mathfrak{h}\}$ is the complement in $\mathfrak{h}$ of the finitely many proper subspaces $\ker (\lambda {\mathrm{\mid }}_{\mathfrak{h}})$, which cannot exhaust $\mathfrak{h}$: the product of their nonzero defining linear forms is a nonzero polynomial and cannot vanish on all of the complex vector space $\mathfrak{h}$ (induct on the number of coordinates, using that a nonzero one-variable polynomial has finitely many roots). For an empty family this product is $1$. Thus ${\mathfrak{h}}_{\mathrm{r}\mathrm{e}\mathrm{g}}\mathrm{\ne }\mathrm{\varnothing }$.

Define $x\sim y$ on ${\mathfrak{g}}^{\mathrm{s}\mathrm{r}}$ when some $a\in \mathrm{Ad}(G)$ satisfies $a(C_{\mathfrak{g}}(x))=C_{\mathfrak{g}}(y)$. This is an equivalence relation because $\mathrm{Ad}(G)$ is a group: reflexivity uses $a=1$, symmetry uses $a^{-1}$, and transitivity uses the product of the two group elements.

The complement of the zero set of a nonzero complex polynomial $P$ on a finite-dimensional complex vector space $V$ is path-connected and dense: density holds because a polynomial vanishing on a nonempty open set vanishes identically, and for $P(x)\mathrm{\ne }0\mathrm{\ne }P(y)$ the one-variable polynomial $t\mapsto P(x+t(y-x))$ has finitely many zeros, so the line through $x$ and $y$ with finitely many points removed is path-connected and avoids the zero set of $P$. Applying this to $V=\mathfrak{g}$ and $P=d_{\rho }$, the locus ${\mathfrak{g}}^{\mathrm{s}\mathrm{r}}$ of step 1.1 is nonempty, dense and path-connected, hence connected.

For $y\in {\mathfrak{h}}_{\mathrm{r}\mathrm{e}\mathrm{g}}$ the orbit $\mathrm{Ad}(G)y$ has $\mathfrak{g}$ as the direct sum $\ker ({\mathrm{ad}}_y)\oplus \mathrm{im}({\mathrm{ad}}_y)=\mathfrak{h}\oplus [\mathfrak{g},y]$, because ${\mathrm{ad}}_y$ is semisimple. Consider the smooth map $\sigma :G\times {\mathfrak{h}}_{\mathrm{r}\mathrm{e}\mathrm{g}}\to \mathfrak{g}$, $\sigma (a,z)=\mathrm{Ad}(a)z$, for the group $G$ and its immersed image $\mathrm{Ad}(G)$ of [L6], [L7]. Its differential at $(1,y)$ is $(X,v)\mapsto [X,y]+v$ for $X\in \mathfrak{g}$ and $v\in \mathfrak{h}$, by $d{\mathrm{Ad}}_e=\mathrm{ad}$ in [L7] and the bilinear evaluation map; its image is $[\mathfrak{g},y]+\mathfrak{h}=\mathfrak{g}$. Hence by [L8] the image of $\sigma$ contains a neighbourhood of $y$, and by [L7] it equals the set $U_{\mathfrak{h}}:=\mathrm{Ad}(G)\cdot {\mathfrak{h}}_{\mathrm{r}\mathrm{e}\mathrm{g}}$, which is therefore open in $\mathfrak{g}$ and nonempty.

Every element of $U_{\mathfrak{h}}$ is semisimple, and has $n$-value $\dim \mathfrak{h}$: automorphisms preserve the adjoint action, so ${\mathrm{ad}}_{\mathrm{Ad}(a)z}=\mathrm{Ad}(a){\mathrm{ad}}_z\mathrm{Ad}(a{)}^{-1}$ has the same generalized nullity as ${\mathrm{ad}}_z$, and $n(z)=\dim \ker ({\mathrm{ad}}_z)=\dim \mathfrak{h}$ for $z\in {\mathfrak{h}}_{\mathrm{r}\mathrm{e}\mathrm{g}}$ by step 1.2; semisimplicity is preserved because the operator is conjugate to a semisimple one. Since $U_{\mathfrak{h}}$ is nonempty open and ${\mathfrak{g}}^{\mathrm{s}\mathrm{r}}$ is dense by step 2.1, $U_{\mathfrak{h}}$ meets ${\mathfrak{g}}^{\mathrm{s}\mathrm{r}}$; at such a point $n=\rho$, so $\rho =\dim \mathfrak{h}$. Consequently $U_{\mathfrak{h}}\subseteq {\mathfrak{g}}^{\mathrm{s}\mathrm{r}}$, and this holds for every Cartan subalgebra.

Let $x\in {\mathfrak{g}}^{\mathrm{s}\mathrm{r}}$ and put $A={\mathrm{ad}}_x=S+N$, where $S={\mathrm{ad}}_{x_s}$ and $N={\mathrm{ad}}_{x_n}$ are the commuting semisimple and nilpotent parts from [L2]. Decompose $\mathfrak{g}={⨁}_{\lambda }{V}_{\lambda }$ into the eigenspaces of $S$. Commutation makes each $V_{\lambda }$ invariant under $N$. On $V_0$, $A=N$ is nilpotent. On $V_{\lambda }$ for $\lambda \mathrm{\ne }0$, $A=\lambda I+N$ is invertible, with inverse ${\lambda }^{-1}{\sum }_{j=0}^{m-1}(-N\mathrm{/}\lambda {)}^j$ if $N^m=0$. Thus the generalized zero-eigenspace of $A$ is exactly $V_0=\ker S$, proving $n(x_s)=n(x)=\rho$. The line $\mathbb{C}{x}_s$ is toral. Choose a toral subalgebra $\mathfrak{h}$ containing it of largest possible dimension; dimensions are bounded by $\dim \mathfrak{g}$, so such a subalgebra exists and is maximal toral. By [L1] it is Cartan. Then $\mathfrak{h}\subseteq C_{\mathfrak{g}}(x_s)$, and both have dimension $\rho$, the former by step 3.1 and the latter by the equality just proved.

Consequently $\mathfrak{l}=C_{\mathfrak{g}}(x_s)=\mathfrak{h}$ is a Cartan subalgebra. This conclusion uses the dimension equality in step 4.1 and the maximal-toral characterization; it does not infer equality merely from a lower bound on generalized nullity.

Every $x\in {\mathfrak{g}}^{\mathrm{s}\mathrm{r}}$ is semisimple and $C_{\mathfrak{g}}(x)$ is a Cartan subalgebra: by step 5.1 applied to $x_s$ we get that $\mathfrak{l}=C_{\mathfrak{g}}(x_s)$ is a Cartan subalgebra, and by [L1] it is maximal toral, hence consists of semisimple elements; since $[x,x_s]=0$, we have $x\in \mathfrak{l}$, so $x$ is semisimple. Now $\mathfrak{l}$ is abelian and contains $x$, so $\mathfrak{l}\subseteq C_{\mathfrak{g}}(x)$. Since $x$ is semisimple, $\dim C_{\mathfrak{g}}(x)=n(x)=\rho =\dim \mathfrak{l}$, whence $C_{\mathfrak{g}}(x)=\mathfrak{l}$ is Cartan.

Each class of the relation of step 1.3 is open in ${\mathfrak{g}}^{\mathrm{s}\mathrm{r}}$. Let $x\in {\mathfrak{g}}^{\mathrm{s}\mathrm{r}}$ and put ${\mathfrak{h}}_x=C_{\mathfrak{g}}(x)$. By step 6.1 the subalgebra ${\mathfrak{h}}_x$ is a Cartan subalgebra and $x$ is semisimple, so $\ker ({\mathrm{ad}}_x)={\mathfrak{h}}_x$; hence $x$ is a regular element of ${\mathfrak{h}}_x$ in the sense of step 1.2, and ${\mathfrak{h}}_x$ satisfies the hypotheses of step 2.2. Therefore $U_{{\mathfrak{h}}_x}:=\mathrm{Ad}(G)\cdot ({\mathfrak{h}}_x{)}_{\mathrm{r}\mathrm{e}\mathrm{g}}$ is open in $\mathfrak{g}$ by step 2.2. Moreover $U_{{\mathfrak{h}}_x}$ is exactly the class of $x$: every $\mathrm{Ad}(a)z$ with $z\in ({\mathfrak{h}}_x{)}_{\mathrm{r}\mathrm{e}\mathrm{g}}$ has centralizer $\mathrm{Ad}(a)C_{\mathfrak{g}}(z)=\mathrm{Ad}(a){\mathfrak{h}}_x$, which is conjugate to ${\mathfrak{h}}_x=C_{\mathfrak{g}}(x)$; conversely if $x^{\mathrm{\prime }}\in {\mathfrak{g}}^{\mathrm{s}\mathrm{r}}$ has $C_{\mathfrak{g}}(x^{\mathrm{\prime }})=\mathrm{Ad}(a){\mathfrak{h}}_x$, then $z:=\mathrm{Ad}(a^{-1})x^{\mathrm{\prime }}$ has centralizer ${\mathfrak{h}}_x$, so $z\in ({\mathfrak{h}}_x{)}_{\mathrm{r}\mathrm{e}\mathrm{g}}$ and $x^{\mathrm{\prime }}=\mathrm{Ad}(a)z\in U_{{\mathfrak{h}}_x}$. As classes of an equivalence relation are pairwise disjoint and ${\mathfrak{g}}^{\mathrm{s}\mathrm{r}}\mathrm{\ne }\mathrm{\varnothing }$ by step 2.1, every class is a nonempty open subset of ${\mathfrak{g}}^{\mathrm{s}\mathrm{r}}$.

The classes of step 1.3 are pairwise disjoint nonempty open subsets of the connected set ${\mathfrak{g}}^{\mathrm{s}\mathrm{r}}$ of step 2.1, so there is exactly one class by step 7.1. Hence $C_{\mathfrak{g}}(x_1)$ and $C_{\mathfrak{g}}(x_2)$ are conjugate for all $x_1,x_2\in {\mathfrak{g}}^{\mathrm{s}\mathrm{r}}$; by step 6.1 they are Cartan subalgebras, and every Cartan subalgebra $\mathfrak{h}$ arises in this way, since for $y\in {\mathfrak{h}}_{\mathrm{r}\mathrm{e}\mathrm{g}}$ (nonempty by step 1.2) step 3.1 gives $n(y)=\dim \mathfrak{h}=\rho$, so $y\in {\mathfrak{g}}^{\mathrm{s}\mathrm{r}}$ and $C_{\mathfrak{g}}(y)=\ker ({\mathrm{ad}}_y)=\mathfrak{h}$. Therefore ${\mathfrak{h}}_1$ and ${\mathfrak{h}}_2$ are conjugate by an element of $\mathrm{Ad}(G)$, an inner automorphism in the connected adjoint group, and conjugate subalgebras have the same dimension. If $\mathfrak{g}=0$ both Cartan subalgebras are zero and the identity conjugates them. The Axiom of Choice supplies the hypotheses of [L1] and [L2] and, via [A1], the countable-choice hypotheses in [L6] and [L7].