2.1 Reductive groups and symmetric spaces

Let $G$ be a connected reductive real Lie group, and let $\mathfrak g$ be its Lie algebra. Let $\theta \in \mathrm {Aut}(G)$ be a Cartan involution. Then $\theta$ acts as an automorphism of $\mathfrak g$. Let $K \subset G$ be the fixed point set of $\theta$. Then $K$ is a compact connected subgroup of $G$, which is a maximal compact subgroup. If $\mathfrak k \subset \mathfrak g$ is the Lie algebra of $K$, then $\mathfrak k$ is the fixed point set of $\theta$ in $\mathfrak g$. Let $\mathfrak p \subset \mathfrak g$ be the eigenspace of $\theta$ corresponding to the eigenvalue $-1$, so that we have the Cartan decomposition

(2.1)\begin{equation} \mathfrak g= \mathfrak p \oplus \mathfrak k. \end{equation}

Put

(2.2)\begin{equation} m=\dim \mathfrak p, \quad n=\dim \mathfrak k, \end{equation}

so that

(2.3)\begin{equation} \dim \mathfrak g=m+n. \end{equation}

Let $B$ be a $G$ and $\theta$ invariant bilinear symmetric nondegenerate form on $\mathfrak g$. Then (2.1) is a $B$-orthogonal splitting. We assume that $B$ is positive on $\mathfrak p$ and negative on $\mathfrak k$. Let $\langle \,\,\rangle =-B(\cdot,\theta \cdot )$ be the corresponding scalar product on $\mathfrak g$. Let $B^{*}$ be the bilinear symmetric form on $\mathfrak g^{*}= \mathfrak p^{*} \oplus \mathfrak k^{*}$ which is dual to $B$.

Let $\omega ^{\mathfrak g}$ be the canonical left-invariant $1$-form on $G$ with values in $\mathfrak g$. By (2.1), $\omega ^{\mathfrak g}$ splits as

(2.4)\begin{equation} \omega^{\mathfrak g}=\omega^{\mathfrak p} + \omega^{\mathfrak k}. \end{equation}

Let $X=G/K$ be the corresponding symmetric space. Then $p:G\to X=G/K$ is a $K$-principal bundle, and $\omega ^{\mathfrak k}$ is a connection form. In addition, the tangent bundle $TX$ is given by

(2.5)\begin{equation} TX =G\times_{K}\mathfrak p. \end{equation}

Then $TX$ is equipped with the scalar product $\langle \,\,\rangle$ induced by $B$, so that $X$ is a Riemannian manifold. The connection $\nabla ^{TX}$ on $TX$ which is induced by $\omega ^{\mathfrak k}$ is the Levi-Civita connection of $TX$, and its curvature is parallel and nonpositive. In addition, $G$ acts isometrically on the left on $X$, and $\theta$ acts as an isometry of $X$.

By [Reference KnappKna86, Proposition 1.2], any element $\gamma \in G$ factorizes uniquely in the form

(2.6)\begin{equation} \gamma=e^{a}k^{-1}, \quad a\in \mathfrak p, k\in K. \end{equation}

If $\gamma,g\in G$, set

(2.7)\begin{equation} C(g)\gamma=g\gamma g^{-1}. \end{equation}

Then $C(g)$ is an automorphism of $G$. Its derivative at the identity is the adjoint representation $g\in G\to \mathrm {Ad}(g)\in \mathrm {Aut}(\mathfrak g)$. The derivative of this last map is given by $a\in \mathfrak g\to \mathrm {ad}(a)\in \mathrm {End}(\mathfrak g)$, with $\mathrm {ad}(a)b=[a,b]$. If $\gamma \in G$, the fixed point set of $C(\gamma )$ is the centralizer $Z(\gamma ) \subset G$, whose Lie algebra $\mathfrak z(\gamma )$ is given by

(2.8)\begin{equation} \mathfrak z(\gamma)=\ker (1-\mathrm{Ad}(\gamma)). \end{equation}

If $f\in \mathfrak g$, let $Z(f) \subset G$ be the stabilizer of $f$. Its Lie algebra $\mathfrak z(f) \subset \mathfrak g$ is given by

(2.9)\begin{equation} \mathfrak z(f)=\ker \mathrm{ad}(f). \end{equation}

In the following, if $M$ is a Lie group, we denote by $M^{0}$ the connected component of the identity.

2.2 Semisimple elements and their displacement function

Let $d$ be the Riemannian distance on $X$. By [Reference Ballmann, Gromov and SchroederBGS85, § 6.1], $d$ is a convex function on $X\times X$. If $\gamma \in G$, let $d_{\gamma }$ be the corresponding displacement function on $X$, i.e.

(2.10)\begin{equation} d_{\gamma}(x)=d(x,\gamma x). \end{equation}

If $g\in G$, then

(2.11)\begin{equation} d_{C(g)\gamma}(gx)=d_{\gamma}(x). \end{equation}

Moreover,

(2.12)\begin{equation} d_{\theta(\gamma)}(\theta x)=d_{\gamma}(x). \end{equation}

Set

(2.13)\begin{equation} m_{\gamma}=\inf d_{\gamma}. \end{equation}

Let $X(\gamma ) \subset X$ be the closed subset where $d_{\gamma }$ reaches its minimum. By [Reference Ballmann, Gromov and SchroederBGS85, p. 78 and § 1.2], $X(\gamma )$ is a closed convex subset, $d_{\gamma }$ is smooth on $X{\setminus} X(\gamma )$ and has no critical points on $X{\setminus} X(\gamma )$. In addition, by (2.11) and (2.12),

(2.14)\begin{equation} \begin{array}{l} X(C(g)\gamma)=gX(\gamma), \\ m_{C(g)\gamma}=m_{\gamma}, \end{array} \quad \begin{array}{l} X(\theta\gamma)=\theta X(\gamma),\\ m_{\theta(\gamma)}=m_{\gamma}. \end{array} \end{equation}

By [Reference EberleinEbe96, Definition 2.19.21], $\gamma$ is said to be semisimple if $X(\gamma )$ is nonempty. If $\gamma$ is semisimple, then $C(g)\gamma$ and $\theta (\gamma )$ are semisimple. In addition, $\gamma$ is said to be elliptic if it is semisimple and $m_{\gamma }=0$. Elliptic elements are exactly the group elements that are conjugate to elements of $K$. Finally, $\gamma$ is said to be hyperbolic if it is conjugate to $e^{a}, a\in \mathfrak p$.

By [Reference KostantKos73, Proposition 2.1], [Reference Ballmann, Gromov and SchroederBGS85, Theorems 2.19.23 and 2.19.24], $\gamma \in G$ is semisimple if and only if it factorizes as $\gamma =he=eh$, with commuting hyperbolic $h$ and elliptic $e$. In addition, $e$ and $h$ are uniquely determined by $\gamma$, and

(2.15)\begin{equation} Z(\gamma)=Z(h)\cap Z(e). \end{equation}

Set

(2.16)\begin{equation} x_{0}=p1. \end{equation}

Theorem 2.1 Let $\gamma \in G$ be semisimple. If $g\in G,x=p g\in X$, then $x\in X(\gamma )$ if and only if there exist $a\in \mathfrak p$, $k\in K$ such that $\mathrm {Ad}(k)a=a$, and

(2.17)\begin{equation} \gamma=C(g)(e^{a}k^{-1}).\end{equation}

In addition, $C(g)e^{a}\in G, \ C(g)k\in G$ are uniquely determined by $\gamma$. If $g_{t}=ge^{ta}$, then $t\in [0,1]\to y_{t}=pg_{t}$ is the unique geodesic connecting $x$ and $\gamma x$. Moreover,

(2.18)\begin{equation} m_{\gamma}=\vert a\vert.\end{equation}

If $\gamma \in G$ is semisimple, then $x_{0}\in X(\gamma )$ if and only if there exist $a\in \mathfrak p,\ k\in K$ such that

(2.19)\begin{equation} \gamma=e^{a}k^{-1}, \quad a\in \mathfrak p, \mathrm{Ad}(k)a=a. \end{equation}

In addition, $a$ and $k$ are uniquely determined by (2.19).

Let $\gamma \in G$ be a semisimple element written as in (2.19). By [Reference BismutBis11, Proposition 3.2.8, (3.3.4), and (3.3.6)],

(2.20)\begin{equation} Z(e^{a})=Z(a), \quad Z(\gamma)=Z(a)\cap Z(k),\quad \mathfrak z(\gamma)=\mathfrak z(a)\cap \mathfrak z(k). \end{equation}

By (2.19), $a\in \mathfrak z(\gamma )$, and by (2.20), $\mathfrak z(\gamma ) \subset \mathfrak z(a)$, so that $a$ is an element of the center of $\mathfrak z(\gamma )$.

Clearly,

(2.21)\begin{equation} \theta(\gamma)=e^{-a}k^{-1}. \end{equation}

Therefore, $\theta (\gamma )\in Z(\gamma )$, so that the above centralizers and Lie algebras are preserved by $\theta$. Set

(2.22)\begin{equation} K(\gamma)=Z(\gamma)\cap K. \end{equation}

By [Reference BismutBis11, Theorem 3.3.1], we have the identity

(2.23)\begin{equation} K^{0}(\gamma)=Z^{0}(\gamma)\cap K, \end{equation}

and $K^{0}(\gamma )$ is a maximal compact subgroup of $Z^{0}(\gamma )$.

Put

(2.24)\begin{equation} \mathfrak p(\gamma)= \mathfrak p\cap \mathfrak z(\gamma), \quad \mathfrak k(\gamma)=\mathfrak k\cap \mathfrak z(\gamma). \end{equation}

Then $\mathfrak k(\gamma )$ is the Lie algebra of $K(\gamma )$. We use similar notation for the Lie algebras $\mathfrak z(k)$ and $\mathfrak z(a)$. We have the Cartan decompositions of Lie algebras,

(2.25)\begin{equation} \mathfrak z(\gamma)= \mathfrak p(\gamma) \oplus \mathfrak k(\gamma), \quad \mathfrak z(k)= \mathfrak p(k) \oplus \mathfrak k(k),\quad \mathfrak z(a)= \mathfrak p(a) \oplus \mathfrak k(a). \end{equation}

Then $B$ restricts to a nondegenerate form on $\mathfrak z(\gamma )$, $\mathfrak z(k)$, and $\mathfrak z(a)$, so that $Z(\gamma )$, $Z(k)$, and $Z(a)$ are possibly nonconnected reductive subgroups of $G$. By [Reference BismutBis11, Theorem 3.3.1], we have the identification of finite groups,

(2.26)\begin{equation} Z^{0}(\gamma){\setminus} Z(\gamma)=K^{0}(\gamma){\setminus} K(\gamma). \end{equation}

Let $\mathfrak z^{\perp }(\gamma )$ and $\mathfrak z^{\perp }(a)$ be the orthogonal spaces to $\mathfrak z(\gamma )$ and $\mathfrak z(a)$ in $\mathfrak g$ with respect to $B$. We have splittings

(2.27)\begin{equation} \mathfrak z^{\perp}(\gamma)= \mathfrak p^{\perp}(\gamma) \oplus \mathfrak k^{\perp}(\gamma), \quad \mathfrak z^{\perp}(a)= \mathfrak p^{\perp}(a) \oplus \mathfrak k^{\perp}(a). \end{equation}

Let $\mathfrak z^{\perp }_{a}(\gamma )$ denote the orthogonal to $\mathfrak z(\gamma )$ in $\mathfrak z(a)$. We still have a splitting

(2.28)\begin{equation} \mathfrak z^{\perp}_{a}(\gamma)= \mathfrak p^{\perp}_{a}(\gamma) \oplus \mathfrak k^{\perp}_{a}(\gamma). \end{equation}

Now we recall a result established in [Reference BismutBis11, Theorem 3.3.1].

Theorem 2.2 The set $X(\gamma )$ is preserved by $\theta$. Moreover,

(2.29)\begin{equation} X(\gamma)=X(e^{a})\cap X(k).\end{equation}

In addition, $X(\gamma )$ is a totally geodesic submanifold of $X$. In the geodesic coordinate system centered at $x_{0}=p1$, then

(2.30)\begin{equation} X(\gamma)= \mathfrak p(\gamma). \end{equation}

The actions of $Z^{0} (\gamma ), Z(\gamma )$ on $X(\gamma )$ are transitive. More precisely the maps $g\in Z^{0}(\gamma ) \to p g\in X ,\ g\in Z(\gamma )\to pg\in X$ induce the identification of $Z^{0}(\gamma )$-manifolds,

(2.31)\begin{equation} X(\gamma) = Z^{0}(\gamma)/K^{0}(\gamma)=Z(\gamma)/K(\gamma). \end{equation}

We now establish a simple important consequence of Theorem 2.2.

Theorem 2.3 Let $\gamma$ be a semisimple element of $G$ as in (2.19). Let $\gamma '$ be another semisimple element of $G$ such that

(2.32)\begin{equation} \gamma'=e^{a'}k^{\prime -1}, \quad a'\in \mathfrak p, \quad \mathrm{Ad}(k')a'=a'. \end{equation}

Then there exists $g\in G$ such that $\gamma '=C(g)\gamma$ if and only if there exists $k''\in K$ such that $C(k'')\gamma =\gamma '$, in which case

(2.33)\begin{equation} a'=\mathrm{Ad}(k^{\prime \prime })a, \quad k'=C(k^{\prime \prime })k. \end{equation}

Proof. Assume that $\gamma '=C(g)\gamma$. By (2.14), we obtain

(2.34)\begin{equation} X(\gamma')=gX(\gamma). \end{equation}

By Theorem 2.1, $x_{0}\in X(\gamma )\cap X(\gamma ')$. By (2.34), $gx_{0}\in X(\gamma ')$. By Theorem 2.2, there exists $h\in Z(\gamma ')$ such that

(2.35)\begin{equation} hgx_{0}=x_{0}, \end{equation}

which is equivalent to

(2.36)\begin{equation} k''=hg\in K. \end{equation}

As $h\in Z(\gamma ')$, we conclude that $C(k'')\gamma =\gamma '$. Using the uniqueness of decomposition in (2.32) established in Theorem 2.1, (2.33) follows. The proof of our theorem is complete.

2.3 Enveloping algebra and the Casimir

We identify $\mathfrak g$ with the Lie algebra of left-invariant vector fields on $G$. Let $U(\mathfrak g)$ be the enveloping algebra of $\mathfrak g$. Then $U(\mathfrak g)$ can be identified with the algebra of left-invariant differential operators on $G$. Let $ Z(\mathfrak g) \subset U(\mathfrak g)$ denote the center of $U(\mathfrak g)$.

If $E$ is a finite-dimensional real or complex vector space, and if $\rho ^{E}: \mathfrak g\to \mathrm {End}(E)$ is a morphism of Lie algebras, the map $\rho ^{E}$ extends to a morphism $U(\mathfrak g)\to \mathrm {End}(E)$.

Among the elements of $Z (\mathfrak g)$, there is the Casimir element $C^{\mathfrak g}$. If $e_{1},\ldots,e_{m+n}$ is a basis of $\mathfrak g$, and if $e^{*}_{1},\ldots,e_{m+n}^{*}$ is the dual basis of $\mathfrak g$ with respect to $B$, thenFootnote ⁵

(2.37)\begin{equation} C^{\mathfrak g}=-\sum_{i=1}^{m+n}e_{i}^{*}e_{i}. \end{equation}

If we consider instead the Lie algebra $(\mathfrak k, B\vert _{\mathfrak k})$, $C^{\mathfrak k}\in Z(\mathfrak k)$ denotes the associated Casimir element.

If $e_{1},\ldots, e_{m}$ is a basis of $\mathfrak p$, and if $e^{*}_{1},\ldots,e^{*}_{m}$ is the dual basis of $\mathfrak p$ with respect to $B\vert _{\mathfrak p}$, set

(2.38)\begin{equation} C^{\mathfrak p}=-\sum_{i=1}^{m}e^{*}_{i}e_{i}. \end{equation}

Then $C^{\mathfrak p}\in U(\mathfrak g)$. Using (2.37) and (2.38), we obtain

(2.39)\begin{equation} C^{\mathfrak g}=C^{\mathfrak p}+C^{\mathfrak k}. \end{equation}

In addition, $C^{\mathfrak p}$ and $C^{\mathfrak k}$ commute.

If $\rho ^{E}: \mathfrak g\to \mathrm {End}(E)$ is taken as previously, put

(2.40)\begin{equation} C^{\mathfrak g,E}=\rho^{E}(C^{\mathfrak g}). \end{equation}

Under the above conditions, we can define $C^{\mathfrak p, E}, C^{\mathfrak k,E}$.

As $\mathfrak g$ is itself a representation of $\mathfrak g$, $C^{\mathfrak g, \mathfrak g}$ is the action of $C^{\mathfrak g}$ on $\mathfrak g$. Since $\mathfrak k$ acts on $\mathfrak p, \mathfrak k$, $C^{\mathfrak k, \mathfrak p}, C^{\mathfrak k, \mathfrak k}$ are also well-defined.

Proposition 2.4 The following identity holds:

(2.41)\begin{equation} \mathrm{Tr}[C^{\mathfrak g, \mathfrak g}]=3\mathrm{Tr} [C^{\mathfrak k, \mathfrak p}]+\mathrm{Tr} [C^{\mathfrak k, \mathfrak k}]. \end{equation}

Proof. By (2.39), we obtain

(2.42)\begin{equation} \mathrm{Tr}[C^{\mathfrak g,\mathfrak g}]=\mathrm{Tr} [C^{\mathfrak p,\mathfrak g}]+\mathrm{Tr}[C^{\mathfrak k,\mathfrak g}]. \end{equation}

Let $e_{1},\ldots,e_{m}$ be an orthonormal basis of $\mathfrak p$, and let $e_{m+1},\ldots,e_{n}$ be an orthonormal basis of $\mathfrak k$. Then

(2.43)\begin{equation} \begin{aligned} & \mathrm{Tr}[C^{\mathfrak p, \mathfrak g}]=- \sum_{\substack{1\le i\le m\\ 1\le j\le m+n}} \vert [e_{i},e_{j}]\vert^{2},\\ & \mathrm{Tr}[C^{\mathfrak k, \mathfrak p}]= -\sum_{\substack{1\le i\le m\\ m+1\le j\le m+n}}\vert [e_{i},e_{j}]\vert^{2} = -\sum_{1\le i,j\le m}\vert [e_{i},e_{j}]\vert^{2}. \end{aligned} \end{equation}

By (2.43), we deduce that

(2.44)\begin{equation} \mathrm{Tr}[C^{\mathfrak p, \mathfrak g}]=2\mathrm{Tr} [C^{\mathfrak k, \mathfrak p}]. \end{equation}

In addition,

(2.45)\begin{equation} \mathrm{Tr}[C^{\mathfrak k,\mathfrak g}]=\mathrm{Tr} [C^{\mathfrak k, \mathfrak p}]+\mathrm{Tr} [C^{\mathfrak k, \mathfrak k}]. \end{equation}

By (2.42), (2.44), and (2.45), we obtain (2.41). The proof of our proposition is completed.

Let $\mathfrak h \subset \mathfrak g$ be a $\theta$-stable Cartan subalgebra. Then we have the splitting $\mathfrak h= \mathfrak h_{\mathfrak p} \oplus \mathfrak h_{\mathfrak k}$.Footnote ⁶ Let $R \subset \mathfrak h_{\mathbf {C}}^{*}$ be the corresponding root system. Let $R_{+}$ be a positive root system. Set $R_{-}=-R_{+}$. Then $R$ is the disjoint union of $R_{+}$ and $R_{-}$. Let $\rho ^{\mathfrak g}\in \mathfrak h^{*}_{\mathbf {C}}$ be the half sum of the positive roots. Here $\rho ^{\mathfrak g}\in \mathfrak h_{\mathfrak p}^{*} \oplus i \mathfrak h_{\mathfrak k}^{*}$. By Kostant's strange formula [Reference KostantKos76], we have the identity

(2.46)\begin{equation} B^{*}(\rho^{\mathfrak g},\rho^{\mathfrak g})=-\tfrac{1}{24}\mathrm{Tr} [C^{\mathfrak g, \mathfrak g}]. \end{equation}

By (2.41) and (2.46), we obtain

(2.47)\begin{equation} B^{*}(\rho^{\mathfrak g}, \rho^{\mathfrak g})=-\tfrac{1}{8}\mathrm{Tr} [C^{\mathfrak k, \mathfrak p}]-\tfrac{1}{24}\mathrm{Tr} [C^{\mathfrak k, \mathfrak k}]. \end{equation}

Using the notation in [Reference BismutBis11, (2.6.11)],Footnote ⁷ by (2.47), we obtain

(2.48)\begin{equation} B^{*}(\rho^{\mathfrak g},\rho^{\mathfrak g})=-\tfrac{1}{4}B^{*}(\kappa^{\mathfrak g}, \kappa^{\mathfrak g}). \end{equation}

2.4 The elliptic operator $C^{\mathfrak g,X}$

Let $E$ be a finite-dimensional Hermitian vector space, and let $\rho ^{E}:K\to U(E)$ denote a unitary representation of $K$. The Casimir $C^{\mathfrak k,E}$ is a self-adjoint nonpositive endomorphism of $E$. If $\rho ^{E}$ is irreducible, then $C^{\mathfrak k,E}$ is a scalar.

Let $F$ be the vector bundle on $X$,

(2.49)\begin{equation} F=G\times_{K}E. \end{equation}

Then $F$ is a Hermitian vector bundle on $X$, which is equipped with a canonical connection $\nabla ^{F}$. In addition, $C^{\mathfrak k,E}$ descend to a parallel section $C^{\mathfrak k,F}$ of $\mathrm {End}(F)$. Moreover, $G$ acts on $C^{ \infty }(X,F)$, so that if $g\in G, s\in C^{\infty }(X,F)$, if $g_{*}$ denotes the lift of the action of $g$ to $F$,

(2.50)\begin{equation} gs(x)=g_{*}s(g^{-1}x). \end{equation}

The Casimir operator $C^{\mathfrak g}$ descends to a second-order elliptic operator $C^{\mathfrak g,X}$ acting on $C^{ \infty }(X,F)$, which commutes with $G$. Let $\Delta ^{X}$ denote the classical Bochner Laplacian acting on $C^{\infty }(X,F)$. By [Reference BismutBis11, (2.13.2)], the splitting (2.39) of $C^{\mathfrak g}$ descends to the splitting of $C^{\mathfrak g,X}$,

(2.51)\begin{equation} C^{\mathfrak g,X}=-\Delta^{X}+C^{\mathfrak k,F}. \end{equation}

2.5 Orbital integrals and the Casimir

Let $\mu \in \mathcal {S}^{\mathrm {even}}(\mathbf {R})$. Let $\widehat {\mu } \in \mathcal {S}^{\mathrm {even}}(\mathbf {R})$ be its Fourier transform, i.e.

(2.52)\begin{equation} \widehat{\mu}(y)=\int_{\mathbf{R}}e^{-2i\pi yx}\mu(x) \, d x. \end{equation}

We assume that there exists $C>0$ such that for any $k\in \mathbf {N}$, there is $c_{k}>0$ such that

(2.53)\begin{equation} \vert \widehat{\mu}^{(k)}(y)\vert\le c_{k}\exp(-Cy^{2}). \end{equation}

This condition is verified if $\widehat {\mu }$ has compact support. For $t>0$, it is also verified by the Gaussian function $e^{-tx^{2}}$.

If $A\in \mathbf {R}$, the operator $\mu (\sqrt {C^{\mathfrak g,X}+A}\,)$ is self-adjoint with a smooth kernel $\mu (\sqrt {C^{\mathfrak g,X}+A}\,) (x,x')$ with respect to the Riemannian volume $d x'$ on $X$. As explained in [Reference BismutBis11, § 6.2], using finite propagation speed for the wave equation, condition (2.53) implies that there exist $C>0,c>0$ such that if $x,x'\in X$, then

(2.54)\begin{equation} \vert \mu(\sqrt{C^{\mathfrak g,X}+A}\,)(x,x')\vert \le Ce^{-cd^{2}(x,x')}. \end{equation}

If $\widehat {\mu }$ has compact support, then $\mu (\sqrt {C^{\mathfrak g,X}+A}\,)(x,x')$ vanishes when $d(x,x')$ is large enough.

As explained in [Reference BismutBis11, § 6.2], the above condition guarantees that if $\gamma \in G$ is semisimple, the orbital integral $\mathrm {Tr}^{[\gamma ]}[\mu (\sqrt {C^{\mathfrak g,X}+A}\,)]$ is well-defined. Let us give more details on our conventions.

Let $\gamma \in G$ be taken as in (2.19). Let $N_{X(\gamma )/X}$ be the orthogonal bundle to $TX(\gamma )$ in $TX$. By [Reference BismutBis11, (3.4.1)], we have the identity

(2.55)\begin{equation} N_{X(\gamma)/X}=Z^{0}(\gamma)\times_{K^{0}(\gamma)} \mathfrak p^{\perp}(\gamma). \end{equation}

Let $\mathcal {N}_{X(\gamma )/X}$ be the total space of $N_{X(\gamma )/X}$. By [Reference BismutBis11, Theorem 3.4.1], the normal geodesic coordinate system based at $X(\gamma )$ gives a smooth identification of $\mathcal {N}_{X(\gamma )/X}$ with $X$. Let $dx,dy,df$ be the Riemannian volumes on $X,X(\gamma ), N_{X(\gamma )/X}$. Then $dydf$ is a volume on $\mathcal {N}_{X(\gamma )/X}$. Let $r(f)$ denote the corresponding Jacobian, so that we have the identity of volumes on $X$,

(2.56)\begin{equation} dx=r(f)dydf. \end{equation}

By [Reference BismutBis11, (3.4.36)], there are constants $C>0,C'>0$ such that

(2.57)\begin{equation} r(f)\le Ce^{C'\vert f\vert}. \end{equation}

By [Reference BismutBis11, Theorem 3.4.1], there exists $C_{\gamma }>0$ such that for $f\in \mathfrak p^{\perp }(\gamma ),\ \vert f\vert \ge 1$,

(2.58)\begin{equation} d_{\gamma}(e^{f}x_{0})\ge \vert a\vert+C_{\gamma}\vert f\vert. \end{equation}

As explained in [Reference BismutBis11, (4.2.6)], by (2.54) and (2.58), there exist $C_{\gamma }>0$ and $c_{\gamma }>0$ such that if $f\in \mathfrak p^{\perp }(\gamma )$, then

(2.59)\begin{equation} \vert \mu(\sqrt{C^{\mathfrak g,X}+A}\,)(\gamma^{-1}e^{f}x_{0},e^{f}x_{0} )\vert\le C_{\gamma}\exp(-c_{\gamma}\vert f\vert^{2}). \end{equation}

We denote by $\gamma _{*}$ the action of $\gamma$ on $F$. More precisely, if $x\in X$, $\gamma _{*}$ maps $F_{x}$ into $F_{\gamma x}$.

In [Reference BismutBis11, Definition 4.2.2], the orbital integral

\[ \mathrm{Tr}^{[\gamma]} [\mu(\sqrt{C^{\mathfrak g,X}+A}\,)] \]

is defined by the formula

(2.60)\begin{align} \mathrm{Tr}^{[\gamma]} [\mu(\sqrt{C^{\mathfrak g,X}+A}\,)] =\int_{\mathfrak p^{\perp}(\gamma)}\mathrm{Tr}[\gamma_{*} \mu(\sqrt{C^{\mathfrak g,X}+A}\,)(\gamma^{-1}e^{f}x_{0}, e^{f}x_{0}) ]r(f) \, df. \end{align}

Equations (2.57) and (2.59) guarantee that the integral in (2.60) converges.

Let $dk$ be the Haar measure on $K$ such that $\mathrm {Vol}(K)=1$. Then $dg=dxdk$ is a Haar measure on $G$. Let $dy$ be the Riemannian volume on $X(\gamma )$. Let $dk^{\prime 0}$ be the Haar measure on $K^{0}(\gamma )$ such that $\mathrm {Vol}(K^{0}(\gamma ))=1$. Then $dz^{0}=dydk^{\prime 0}$ is a Haar measure on $Z^{0}(\gamma )$. Let $dv^{0}$ be the volume on $Z^{0}(\gamma ){\setminus} G$ such that $dg=dz^{0}dv^{0}$.

As explained in [Reference BismutBis11, § 4.2], the smooth kernel

\[ \mu(\sqrt{C^{\mathfrak g,X}+A}\,)(x,x') \]

lifts to a smooth function on $G$ with values in $\mathrm {End}(E)$, denoted by

\[ \mu^{E}(\sqrt{C^{\mathfrak g,X}+A}\,)(g), \]

and by [Reference BismutBis11, (4.2.11)], we have the identity

(2.61)\begin{align} \mathrm{Tr}^{[\gamma]} [\mu(\sqrt{C^{\mathfrak g,X}+A}\,)] =\int_{Z^{0}(\gamma){\setminus} G}\mathrm{Tr}^{E} [\mu(\sqrt{C^{\mathfrak g,X}+A}\,)((v^{0})^{-1}\gamma v^{0})] \, dv^{0}. \end{align}

This definition of orbital integrals coincides with the definition given by Selberg [Reference SelbergSel56, p. 66].

2.6 The function $\mathcal {J}_{\gamma }$

We use the assumptions in § 2.2 and the corresponding notation. In particular, $\gamma \in G$ is a semisimple element as in (2.19).

Then $\mathrm {Ad}(\gamma )$ preserves $\mathfrak z(a), \mathfrak z^{\perp }(a)$. In addition, $\mathrm {Ad}(k^{-1})$ preserves $\mathfrak z_{a}^{\perp }(\gamma )$. If ${Y_{0}^{\mathfrak k}}\in \mathfrak k(\gamma )$, $\mathrm {ad}({Y_{0}^{\mathfrak k}})$ preserves $\mathfrak z_{a}^{\perp }(\gamma )$. The splitting (2.28) is preserved by $\mathrm {Ad}(k^{-1})$ and $\mathrm {ad}({Y_{0}^{\mathfrak k}})$.

If $x\in \mathbf {R}$, put

(2.62)\begin{equation} \widehat{A}(x)=\frac{x/2}{\sinh(x/2)}. \end{equation}

If ${Y_{0}^{\mathfrak k}}\in \mathfrak k(\gamma )$, $\mathrm {ad}({Y_{0}^{\mathfrak k}})$ acts as an antisymmetric endomorphism of $\mathfrak p(\gamma ), \mathfrak k(\gamma )$, so that its eigenvalues are either $0$, or they come by nonzero conjugate imaginary pairs. If ${Y_{0}^{\mathfrak k}}\in i \mathfrak k(\gamma )$, putFootnote ⁸

(2.63)\begin{equation} \begin{aligned} & \widehat{A}(\mathrm{ad}({Y_{0}^{\mathfrak k}})\vert _{ \mathfrak p(\gamma)})= [\det(\widehat{A}(\mathrm{ad}({Y_{0}^{\mathfrak k}})\vert_{\mathfrak p(\gamma)}) ) ]^{1/2},\\ & \widehat{A}(\mathrm{ad}({Y_{0}^{\mathfrak k}})\vert _{ \mathfrak k(\gamma)})=[\det(\widehat{A}(\mathrm{ad}({Y_{0}^{\mathfrak k}})\vert _{ \mathfrak k(\gamma)}))]^{1/2}. \end{aligned} \end{equation}

The square root in (2.63) is the positive square root of a positive real number.

We follow [Reference BismutBis11, Theorem 5.5.1], while slightly changing the notation.

Definition 2.5 If ${Y_{0}^{\mathfrak k}}\in i \mathfrak k(\gamma )$, put

(2.64)\begin{equation} \mathcal{L}_{\gamma}({Y_{0}^{\mathfrak k}})=\frac{\det(1- \mathrm{Ad}(k^{-1}e^{-{Y_{0}^{\mathfrak k}}}))\vert_{\mathfrak k^{\perp}_{a}(\gamma)}}{\det(1- \mathrm{Ad}(k^{-1}e^{-{Y_{0}^{\mathfrak k}}}))\vert_{\mathfrak p_{a}^{\perp}(\gamma)}}. \end{equation}

Set

(2.65)\begin{equation} \mathcal{M}_{\gamma}({Y_{0}^{\mathfrak k}}) =\bigg[\frac{1}{\det(1-\mathrm{Ad}(k^{-1}))\vert_{ \mathfrak z^{\perp}_{a}(\gamma)} }\mathcal{L}_{\gamma}({Y_{0}^{\mathfrak k}})\bigg]^{1/2}. \end{equation}

The fact that the square root in (2.65) is unambiguously defined is established in [Reference BismutBis11, § 5.5]. Let us explain the details. First we make ${Y_{0}^{\mathfrak k}}=0$. Then

(2.66)\begin{align} \frac{1}{\det(1-\mathrm{Ad}(k^{-1}))\vert_{ \mathfrak z^{\perp}_{a}(\gamma)} }\frac{\det(1-\mathrm{Ad}(k^{-1}))\vert_{\mathfrak k^{\perp}_{a}(\gamma)}}{\det(1- \mathrm{Ad}(k^{-1}))\vert_{\mathfrak p_{a}^{\perp}(\gamma)}} =\bigg[\frac{1}{\det(1-\mathrm{Ad}(k^{-1}))\vert_{\mathfrak p_{a}^{\perp}(\gamma)}} \bigg]^{2}. \end{align}

The conventions in [Reference BismutBis11] say that the square root of (2.66) is the obvious positive square root, i.e.

(2.67)\begin{equation} \mathcal{M}_{\gamma}(0) =\frac{1}{\det(1-\mathrm{Ad}(k^{-1}))\vert_{\mathfrak p_{a}^{\perp}(\gamma)}}. \end{equation}

Using analyticity in the variable ${Y_{0}^{\mathfrak k}}\in i \mathfrak k(\gamma )$, the choice of the square root in (2.67) determines a choice of the square root in (2.65). This point is discussed at length in § 4. No choice of a Cartan subalgebra or of a positive root system is needed at this stage.

Definition 2.6 Let $\mathcal {J}_{\gamma }({Y_{0}^{\mathfrak k}})$ be the smooth function of ${Y_{0}^{\mathfrak k}}\in i\mathfrak k(\gamma )$,

(2.68)\begin{equation} \mathcal{J}_{\gamma}(Y_{0}^{ \mathfrak k})=\frac{1}{\vert \det(1-\mathrm{Ad}(\gamma))\vert_{ \mathfrak z^{\perp}(a)}\vert^{1/2}} \frac{\widehat{A}(\mathrm{ad}({Y_{0}^{\mathfrak k}})\vert _{ \mathfrak p(\gamma)})}{\widehat{A}(\mathrm{ad}({Y_{0}^{\mathfrak k}}) \vert_{\mathfrak k(\gamma)})}\mathcal{M}_{\gamma}({Y_{0}^{\mathfrak k}}). \end{equation}

With the conventions in [Reference BismutBis11, Chapter 5], where instead a function $J_{\gamma }({Y_{0}^{\mathfrak k}})$ is defined on $\mathfrak k(\gamma )$, we have

(2.69)\begin{equation} J_{\gamma}({Y_{0}^{\mathfrak k}})=\mathcal{J}_{\gamma}(i{Y_{0}^{\mathfrak k}}). \end{equation}

By [Reference BismutBis11, (5.5.11)] or by (2.68), if ${Y_{0}^{\mathfrak k}}\in i\mathfrak k$, then

(2.70)\begin{equation} \mathcal{J}_{1}({Y_{0}^{\mathfrak k}})=\frac{\widehat{A}(\mathrm{ad}({Y_{0}^{\mathfrak k}})\vert _{ \mathfrak p})}{\widehat{A}(\mathrm{ad}({Y_{0}^{\mathfrak k}}) \vert_{\mathfrak k})}. \end{equation}

2.7 Some properties of the function $\mathcal {J}_{\gamma }$

Let $\rho ^{E^{*}}: K\to U(E^{*})$ denote the representation of $K$ which is dual to the representation $\rho ^{E}$.

Proposition 2.7 If ${Y_{0}^{\mathfrak k}}\in i \mathfrak k(\gamma )$, then

(2.71)\begin{equation} \begin{aligned} & \mathcal{J}_{\gamma^{-1}}({Y_{0}^{\mathfrak k}})=\mathcal{J}_{\gamma}(-{Y_{0}^{\mathfrak k}}), \quad \mathrm{Tr}^{E^{*}} [\rho^{E^{*}}(ke^{-{Y_{0}^{\mathfrak k}}}) ]= \mathrm{Tr}^{E}[\rho^{E}(k^{-1}e^{{Y_{0}^{\mathfrak k}}})],\\ & \overline{\mathcal{J}_{\gamma}({Y_{0}^{\mathfrak k}})}=\mathcal{J}_{\gamma}(-{Y_{0}^{\mathfrak k}}), \quad \overline{\mathrm{Tr}^{E}[\rho^{E} ( k^{-1}e^{-{Y_{0}^{\mathfrak k}}} ) ]}=\mathrm{Tr}^{E^{*}}[\rho^{E^{*}}(k^{-1}e^{{Y_{0}^{\mathfrak k}}})]. \end{aligned} \end{equation}

Proof. If $f\in \mathrm {End}(\mathfrak g)$, let $\widetilde f\in \mathrm {End}(\mathfrak g)$ denote the adjoint of $f$ with respect to $B$. We have the identity

(2.72)\begin{equation} \mathrm{Ad}(\gamma^{-1})=\widetilde {\mathrm{Ad}(\gamma)}. \end{equation}

By (2.72), we deduce that

(2.73)\begin{equation} \det(1-\mathrm{Ad}(\gamma^{-1})) \vert_{\mathfrak z^{\perp}(a)}=\det(1-\mathrm{Ad}(\gamma))\vert_{\mathfrak z^{\perp}(a)}. \end{equation}

A similar argument shows that

(2.74)\begin{equation} \begin{aligned} & \det(1-\mathrm{Ad}(ke^{-{Y_{0}^{\mathfrak k}}}))\vert_{\mathfrak p_{a}^{\perp}(\gamma)} =\det(1-\mathrm{Ad}(k^{-1}e^{{Y_{0}^{\mathfrak k}}}))\vert_{\mathfrak p_{a}^{\perp}(\gamma)},\\ & \det(1-\mathrm{Ad}(ke^{-{Y_{0}^{\mathfrak k}}}))\vert_{\mathfrak k_{a}^{\perp}(\gamma)} =\det(1-\mathrm{Ad}(k^{-1}e^{{Y_{0}^{\mathfrak k}}}))\vert_{\mathfrak k_{a}^{\perp}(\gamma)}. \end{aligned} \end{equation}

By (2.64), (2.65), (2.68), (2.73), and (2.74), we obtain the first identity in (2.71). The second identity in (2.71) is trivial.

If ${Y_{0}^{\mathfrak k}}\in i\mathfrak k(\gamma )$,

(2.75)\begin{equation} \begin{aligned} & \overline{\det(1-\mathrm{Ad}(k^{-1}e^{-{Y_{0}^{\mathfrak k}}}))\vert_{\mathfrak p_{a}^{\perp}(\gamma)}}=\det(1-\mathrm{Ad}(k^{-1}e^{{Y_{0}^{\mathfrak k}}}))\vert_{\mathfrak p_{a}^{\perp}(\gamma)},\\ & \overline{\det(1-\mathrm{Ad}(k^{-1}e^{-{Y_{0}^{\mathfrak k}}}))\vert_{\mathfrak k_{a}^{\perp}(\gamma)}}=\det(1-\mathrm{Ad}(k^{-1}e^{{Y_{0}^{\mathfrak k}}}))\vert_{\mathfrak k_{a}^{\perp}(\gamma)}. \end{aligned} \end{equation}

By (2.75), we obtain the third identity in (2.71). As ${Y_{0}^{\mathfrak k}}\in i\mathfrak k(\gamma )$, the fourth identity is trivial. The proof of our proposition is complete.

2.8 A geometric formula for the orbital integrals associated with the Casimir

Note that $i \mathfrak k(\gamma )$ is naturally an Euclidean vector space. If ${Y_{0}^{\mathfrak k}}\in i\mathfrak k(\gamma )$, we denote by $\vert {Y_{0}^{\mathfrak k}}\vert$ its Euclidean norm. More precisely, if ${Y_{0}^{\mathfrak k}}\in i\mathfrak k(\gamma )$, then

(2.76)\begin{equation} \vert {Y_{0}^{\mathfrak k}}\vert^{2}=B({Y_{0}^{\mathfrak k}},{Y_{0}^{\mathfrak k}}). \end{equation}

By [Reference BismutBis11, (6.1.1)], there exist $c>0,C>0$ such that if ${Y_{0}^{\mathfrak k}} \in i \mathfrak k(\gamma )$,

(2.77)\begin{equation} \vert \mathcal{J}_{\gamma}({Y_{0}^{\mathfrak k}})\vert\le c \exp(C\vert {Y_{0}^{\mathfrak k}}\vert). \end{equation}

In the following, $\int _{i \mathfrak k(\gamma )}$ denotes integration on the real vector space $i \mathfrak k(\gamma )$.

Let $d{Y_{0}^{\mathfrak k}}$ be the Euclidean volume on $i \mathfrak k(\gamma )$. Set $p=\dim \mathfrak p(\gamma ), q=\dim \mathfrak k(\gamma )$. Now we state the result obtained in [Reference BismutBis11, Theorem 6.1.1]. Our reformulation takes (2.48) into account.

Theorem 2.8 For $t>0$, the following identity holds:

(2.78)\begin{align} \mathrm{Tr}^{[\gamma]}[\exp(-tC^{\mathfrak g,X}/2) ]&=\exp(-tB^{*}(\rho^{\mathfrak g},\rho^{\mathfrak g})/2) \frac{\exp(-\vert a\vert^{2}/2t)}{(2\pi t)^{p/2}}\nonumber\\ &\quad \times \int_{i\mathfrak k(\gamma)} \mathcal{J}_{\gamma}(Y_{0}^{ \mathfrak k})\mathrm{Tr}^{E} [\rho^{E}(k^{-1}e^{-{Y_{0}^{\mathfrak k}}})] \exp(-\vert Y_{0}^{ \mathfrak k}\vert^{2}/2t)\frac{dY_{0}^{ \mathfrak k}}{ (2\pi t)^{q/2}}. \end{align}

Let $B\vert _{\mathfrak z(\gamma )}$ be the restriction of $B$ to $\mathfrak z(\gamma )$, and let $B^{*}\vert _{\mathfrak z(\gamma )}$ be the corresponding quadratic form on $\mathfrak z^{*}(\gamma )$. Let $\Delta ^{\mathfrak z(\gamma )}$ denote the associated generalized LaplacianFootnote ⁹ on $\mathfrak z(\gamma )$. We can extend $\Delta ^{\mathfrak z(\gamma )}$ to an operator acting via constant holomorphic vector fields on $\mathfrak z(\gamma )_{\mathbf {C}}$.

Put

(2.79)\begin{equation} \mathfrak z_{i}(\gamma)=\mathfrak p(\gamma) \oplus i \mathfrak k(\gamma). \end{equation}

Then $B\vert _{\mathfrak z_{i}(\gamma )}$ is a scalar product on $\mathfrak z_{i}(\gamma )$. The generalized Laplacian $\Delta ^{\mathfrak z(\gamma )}$ restricts on $\mathfrak z_{i}(\gamma )$ to the standard Euclidean Laplacian of $\mathfrak z_{i}(\gamma )$.

We take $\mu \in \mathcal {S}^{\mathrm {even}}(\mathbf {R})$ as in § 2.5. If $f\in \mathfrak z_{i}(\gamma )$, let

\[ \mu ( \sqrt{-\Delta^{ \mathfrak z(\gamma)}+B^{*}(\rho^{\mathfrak g},\rho^{\mathfrak g})+A}\,)(f) \]

be the smooth convolution kernel for $\mu ( \sqrt {-\Delta ^{ \mathfrak z(\gamma )}+B^{*}(\rho ^{\mathfrak g},\rho ^{\mathfrak g})+A}\, )$ on $\mathfrak z_{i}(\gamma )$ with respect to the volume associated with the scalar product of $\mathfrak z_{i}(\gamma )$. Using (2.53) and finite propagation speed for the wave equation, there exist $C>0$ and $c>0$ such that if $f\in \mathfrak z_{i}(\gamma )$, then

(2.80)\begin{equation} \vert \mu ( \sqrt{-\Delta^{ \mathfrak z(\gamma)}+B^{*}(\rho^{\mathfrak g},\rho^{\mathfrak g})+A}\, ) (f)\vert\le Ce^{-c\vert f\vert^{2}}. \end{equation}

Let $\delta _{a}$ be the Dirac mass at $a\in \mathfrak p(\gamma )$. Then

\[ \mathcal{J}_{\gamma}({Y_{0}^{\mathfrak k}})\mathrm{Tr}^{E}[\rho^{E}(k^{-1}e^{-{Y_{0}^{\mathfrak k}}} )]\delta_{a} \]

is a distribution on $\mathfrak z_{i}(\gamma )$, to which the smooth convolution kernel

\[ \mu ( \sqrt{-\Delta^{ \mathfrak z(\gamma)}+B^{*}(\rho^{\mathfrak g},\rho^{\mathfrak g})+A}\, ) \]

can be applied. By definition,

(2.81)\begin{align} &\mu(\sqrt{-\Delta^{ \mathfrak z(\gamma)} +B^{*}(\rho^{\mathfrak g},\rho^{\mathfrak g})+A}\, ) [\mathcal{J}_{\gamma}({Y_{0}^{\mathfrak k}})\mathrm{Tr}^{E}[\rho^{E}(k^{-1}e^{-{Y_{0}^{\mathfrak k}}} )]\delta_{a}](0)\nonumber\\ &\quad =\int_{i \mathfrak k(\gamma)}\mu(\sqrt{-\Delta^{ \mathfrak z(\gamma)} +B^{*}(\rho^{\mathfrak g},\rho^{\mathfrak g})+A}\, ) (-{Y_{0}^{\mathfrak k}},-a) \mathcal{J}_{\gamma}({Y_{0}^{\mathfrak k}})\mathrm{Tr}^{E}[\rho^{E}(k^{-1}e^{-{Y_{0}^{\mathfrak k}}} )] \,d{Y_{0}^{\mathfrak k}}. \end{align}

In the right-hand side of (2.81), $(-{Y_{0}^{\mathfrak k}},-a)$ can also be replaced by $({Y_{0}^{\mathfrak k}},a)$.

In [Reference BismutBis11, Theorem 6.2.2], the following extension of Theorem 2.8 was established.

Theorem 2.9 The following identity holds:

(2.82)\begin{align} \mathrm{Tr}{}^{[\gamma]}[\mu(\sqrt{C^{\mathfrak g,X}+A}\,)] =\mu(\sqrt{-\Delta^{ \mathfrak z(\gamma)} +B^{*}(\rho^{\mathfrak g},\rho^{\mathfrak g})+A}\, ) [\mathcal{J}_{\gamma}({Y_{0}^{\mathfrak k}})\mathrm{Tr}^{E}[\rho^{E}(k^{-1}e^{-{Y_{0}^{\mathfrak k}}} )]\delta_{a}](0). \end{align}

3.1 Linear algebra

Let $V$ be a finite-dimensional real vector space. The symmetric algebras $S{{}^{\cdot }} (V), S{{}^{\cdot }} (V^{*})$ are the algebras of polynomials on $V^{*},V$. If $v\in V$, $v$ acts as a derivation of $S{{}^{\cdot }}(V^{*})$. More generally, $S{{}^{\cdot }}(V)$ acts on $S{{}^{\cdot }}(V^{*})$, and this action identifies $S{{}^{\cdot }}(V)$ with the algebra $D{{}^{\cdot }}(V)$ of real partial differential operators on $V$ with constant coefficients. In particular, if $B^{*}\in S^{2}(V)$ is a bilinear symmetric form on $V^{*}$, the associated element in $D{{}^{\cdot }}(V)$ is the corresponding Laplacian $\Delta ^{V}$. If $B^{*}$ is positive, $\Delta ^{V}$ is just a classical Laplacian. If $B^{*}$ is negative, then $\Delta ^{V}$ is the negative of a classical Laplacian on $V$.

Let $V_{\mathbf {C}}=V \otimes _{\mathbf {R}}\mathbf {C}$ be the complexification of $V$, a complex vector space. Its complex dual is given by $V^{*}_{\mathbf {C}}=V^{*} \otimes _{\mathbf {R}}\mathbf {C}$. The algebras $S{{}^{\cdot }}(V_{\mathbf {C}})$ and $S{{}^{\cdot }}(V^{*}_{\mathbf {C}})$ are the algebras of complex polynomials on $V^{*},V$. Note that

(3.1)\begin{equation} S{{}^{\cdot}}(V_{\mathbf{C}})=S{{}^{\cdot}}(V) \otimes _{\mathbf{R}}\mathbf{C}, \quad S{{}^{\cdot}}(V^{*}_{\mathbf{C}})=S{{}^{\cdot}}(V^{*}) \otimes _{\mathbf{R}}\mathbf{C}. \end{equation}

Put

(3.2)\begin{equation} D{{}^{\cdot}}(V_{\mathbf{C}})=D(V) \otimes _{\mathbf{R}}\mathbf{C}, \quad D{{}^{\cdot}}(V^{*}_{\mathbf{C}})=D{{}^{\cdot}}(V^{*}) \otimes _{\mathbf{R}}\mathbf{C}. \end{equation}

Then $D{{}^{\cdot }}(V_{\mathbf {C}})$ and $D{{}^{\cdot }}(V^{*}_{\mathbf {C}})$ are the complexifications of $D{{}^{\cdot }}(V)$ and $D{{}^{\cdot }}(V^{*})$, and also the spaces of complex holomorphic differential operators with constant coefficients on $V_{\mathbf {C}}$ and $V^{*}_{\mathbf {C}}$.

In particular, if $B^{*}\in S^{2}(V)$, $\Delta ^{V}$ is now viewed as a holomorphic operator on $V_{\mathbf {C}}$, that coincides with the corresponding Laplacian $\Delta ^{V}$ on $V$, and with $-\Delta ^{V}$ on $iV \simeq V$.Footnote ¹⁰

In addition, $S{{}^{\cdot }}(V^{*}) \subset C^{ \infty }(V,\mathbf {R})$, and the action of $D{{}^{\cdot }}(V)$ extends to $C^{ \infty }(V,\mathbf {R})$.

Let $S[[V^{*}]]$ be the algebra of formal power series $s=\sum _{i=0}^{+ \infty }s^{i}, s^{i}\in S^{i}(V^{*})$. Then $S[[V^{*}]]$ can be identified with the algebra $D[[V^{*}]]$ of differential operators of infinite order with constant coefficients on $V^{*}$. In particular, $S[[V^{*}]]$ acts on $S{{}^{\cdot }}(V)$.

3.2 The Cartan subalgebras of $\mathfrak g$

By [Reference WallachWal88, § 0.2], a Lie subalgebra $\mathfrak h \subset \mathfrak g$ is said to be a Cartan subalgebra if $\mathfrak h$ is maximal among the abelian subalgebras of $\mathfrak g$ whose elements act as semisimple endomorphisms of $\mathfrak g$. Cartan subalgebras are known to exist and have the same dimension $r$, which is called the complex rank of $G$. By [Reference KnappKna02, Proposition 6.64], there is a finite family of nonconjugate Cartan subalgebras in $\mathfrak g$. By [Reference WallachWal88, Lemma 2.3.3], in every conjugacy class of Cartan subalgebras, there is a unique $\theta$-stable Cartan algebra, up to conjugation by $K$. Therefore, there is a finite family of nonconjugate $\theta$-stable Cartan subalgebras, up to conjugation by $K$. By [Reference KnappKna02, Theorem 2.15], the Cartan subalgebras of $\mathfrak g_{\mathbf {C}}$ are unique up to automorphisms induced by the adjoint group $\mathrm {Ad}(\mathfrak g_{\mathbf {C}})$.

Let $\mathfrak h \subset \mathfrak g$ be a $\theta$-stable Cartan subalgebra. To the Cartan splitting of $\mathfrak g$ in (2.1) corresponds the splitting

(3.3)\begin{equation} \mathfrak h= \mathfrak h_{\mathfrak p} \oplus \mathfrak h_{\mathfrak k}. \end{equation}

In particular, the restriction $B\vert _{\mathfrak h}$ of $B$ to $\mathfrak h$ is nondegenerate. This is also the case if $\mathfrak h$ is any Cartan subalgebra.

Up to conjugation by $K$, there is a unique $\theta$-stable Cartan subalgebra $\mathfrak h$, which is called fundamental, such that $\mathfrak h_{\mathfrak k}$ is a Cartan subalgebra of $\mathfrak k$. Let us give more details on its construction [Reference KnappKna86, pp. 129 and 131]. Let $\mathfrak t \subset \mathfrak k$ be a Cartan subalgebra of $\mathfrak k$. Let $\mathfrak z(\mathfrak t) \subset \mathfrak g$ be the centralizer of $\mathfrak t$, i.e.

(3.4)\begin{equation} \mathfrak z(\mathfrak t)=\{f\in \mathfrak g, [\mathfrak t,f]=0\}. \end{equation}

Then $\mathfrak h= \mathfrak z(\mathfrak t)$ is a $\theta$-stable fundamental Cartan subalgebra of $\mathfrak g$, and $\mathfrak h_{\mathfrak k}= \mathfrak t$.

An element $f\in \mathfrak g$ is said to be semisimple if $\mathrm {ad}(f)\in \mathrm {End}(\mathfrak g)$ is semisimple. If $\mathfrak h$ is a Cartan subalgebra, elements of $\mathfrak h$ are semisimple. Any semisimple element of $\mathfrak g$ lies in a Cartan subalgebra.

If $\mathfrak h \subset \mathfrak g$ is a Cartan subalgebra, let $\mathfrak h^{\perp }$ be the orthogonal to $\mathfrak h$ in $\mathfrak g$. We have the $B$-orthogonal splitting,

(3.5)\begin{equation} \mathfrak g= \mathfrak h \oplus \mathfrak h^{\perp}, \end{equation}

and $B$ is also nondegenerate on $\mathfrak h^{\perp }$. If $\mathfrak h$ is $\theta$-stable, then $\mathfrak h^{\perp }$ is also $\theta$-stable, and so it splits as

(3.6)\begin{equation} \mathfrak h^{\perp}=\mathfrak h^{\perp}_{\mathfrak p} \oplus \mathfrak h^{\perp}_{\mathfrak k}. \end{equation}

Let $\mathfrak u= i \mathfrak p \oplus \mathfrak k$ be the compact form of $\mathfrak g$. Then $\mathfrak h_{\mathfrak u}=i \mathfrak h_{\mathfrak p}\oplus \mathfrak h_{\mathfrak k}$ is a Cartan subalgebra of $\mathfrak u$. If $\mathfrak h$ is $\theta$-stable, then $\mathfrak h_{\mathfrak u}$ is also $\theta$-stable.

An element $f\in \mathfrak g$ is said to be regular if $\mathfrak z(f)$ is a Cartan subalgebra. Regular elements in $\mathfrak g$ are semisimple.

If $\mathfrak h$ is a Cartan subalgebra, if $f\in \mathfrak h$, $\mathrm {ad}(f)$ acts as an endomorphism of $\mathfrak g/\mathfrak h$. Then $f\in \mathfrak h$ is regular if and only if $\det \mathrm {ad}(f)\vert _{\mathfrak g/\mathfrak h}\neq 0$.

3.3 A root system and the Weyl group

Let $\mathfrak h$ be a $\theta$-stable Cartan subalgebra.

Let $R \subset \mathfrak h^{*}_{\mathbf {C}}$ be the root system associated with $\mathfrak h, \mathfrak g$ [Reference KnappKna02, § II.4]. If $\alpha \in R$, then $-\alpha \in R,\overline {\alpha }\in R$. If $\alpha \in R$, let $\mathfrak g_{\alpha } \subset \mathfrak g_{\mathbf {C}}$ be the weight space associated with $\alpha$, which is of dimension one. Then we have the splitting

(3.7)\begin{equation} \mathfrak g_{\mathbf{C}}= \mathfrak h_{\mathbf{C}} \oplus \bigoplus_{\alpha\in R} \mathfrak g_{\alpha}. \end{equation}

If $\alpha \in R$, then

(3.8)\begin{equation} \mathfrak g_{\overline{\alpha}}=\overline{\mathfrak g}_{\alpha}. \end{equation}

If $f\in \mathfrak h$, $\mathrm {ad}(f)\in \mathrm {End}(\mathfrak g)$ is antisymmetric with respect to $B$, so that the $\mathfrak g_{\alpha }\vert _{\alpha \in R}$ are $B$-orthogonal to $\mathfrak h_{\mathbf {C} }$. If $\alpha,\beta \in R$, then $\mathfrak g_{\alpha }, \mathfrak g_{\beta }$ are $B$-orthogonal except when $\beta =-\alpha$, and the pairing between $\mathfrak g_{\alpha }, \mathfrak g_{-\alpha }$ is nondegenerate, so that if $\alpha \in R$, the form $B$ induces the identification

(3.9)\begin{equation} \mathfrak g_{-\alpha} \simeq \mathfrak g_{\alpha}^{*}. \end{equation}

Also

(3.10)\begin{equation} \mathfrak h^{\perp}_{\mathbf{C}}=\bigoplus_{\alpha\in R}\mathfrak g_{\alpha}. \end{equation}

If $\alpha \in R$, $\alpha$ takes real values on $\mathfrak h_{\mathfrak p}$, and imaginary values on $\mathfrak h_{\mathfrak k}$, i.e. $\alpha \in \mathfrak h^{*}_{\mathfrak p} \oplus i \mathfrak h^{*}_{\mathfrak k}$. In addition, $\theta$ preserves the splitting (3.5) of $\mathfrak g$, and it maps $R$ into itself. More precisely, if $\alpha \in R$,

(3.11)\begin{equation} \theta \alpha=-\overline{\alpha}, \quad \mathfrak g_{\theta\alpha}=\theta \mathfrak g_{\alpha}. \end{equation}

Let $W(\mathfrak h_{\mathbf {C}}:\mathfrak g_{\mathbf {C}}) \subset \mathrm {Aut}(\mathfrak h_{\mathbf {C}})$ be the algebraic Weyl group [Reference KnappKna86, p. 131]. Then $R \subset i \mathfrak h^{*}_{\mathfrak u}$, and $W(\mathfrak h_{\mathbf {C}}:\mathfrak g_{\mathbf {C}}) \subset \mathrm {Aut}(\mathfrak h_{\mathfrak u})$, i.e. $W(\mathfrak h_{\mathbf {C}}:\mathfrak g_{\mathbf {C}})$ preserves the real vector space $\mathfrak h_{\mathfrak u}$. In addition, $\theta$ acts as an automorphism of the Lie algebras $\mathfrak g,\mathfrak u$, and $W(\mathfrak h_{\mathbf {C}}: \mathfrak g_{\mathbf {C}})$ is preserved by conjugation by $\theta$.

In general, $W(\mathfrak h_{\mathbf {C}}: \mathfrak g_{\mathbf {C}})$ does not preserve the real vector space $\mathfrak h$. Recall that if $h\in \mathfrak h_{\mathbf {C}}$, we can define its complex conjugate $\overline {h}\in \mathfrak h_{\mathbf {C}}$. If $u\in \mathrm {End}(\mathfrak h_{\mathbf {C}})$, its complex conjugate $\overline {u}\in \mathrm {End}(\mathfrak h_{\mathbf {C}})$ is such that if $h\in \mathfrak h_{\mathbf {C}}$, then

(3.12)\begin{equation} \overline{u}(h)=\overline{u(\overline{h})}. \end{equation}

Proposition 3.1 If $w\in W(\mathfrak h_{\mathbf {C}}: \mathfrak g_{\mathbf {C}})$, then

(3.13)\begin{equation} \overline{w}=\theta w\theta^{-1}. \end{equation}

In particular, the group $W(\mathfrak h_{\mathbf {C}}:\mathfrak g_{\mathbf {C}})$ is preserved by complex conjugation.

Proof. The group $W(\mathfrak h_{\mathbf {C}}:\mathfrak g_{\mathbf {C}})$ is generated by the symmetries $s_{\alpha }, \alpha \in R$ with respect to the vanishing locus of the $\alpha \in R$. By (3.11), we deduce that if $\alpha \in R$,

(3.14)\begin{equation} \overline{s_{\alpha}}= \theta s_{\alpha}\theta^{-1}, \end{equation}

from which we obtain (3.13). As $W(\mathfrak h_{\mathbf {C}}: \mathfrak g_{\mathbf {C}})$ is stable by conjugation by $\theta$, the group $W(\mathfrak h_{\mathbf {C}}:\mathfrak g_{\mathbf {C}})$ is preserved by complex conjugation.

Another proof is as follows. Observe that there is a canonical identification of complex vector spaces $\varphi : \mathfrak h_{\mathbf {C}} \simeq \mathfrak h_{\mathfrak u,\mathbf {C}}$, but the complex conjugations on $\mathfrak h_{\mathbf {C}}$ and on $\mathfrak h_{\mathfrak u,\mathbf {C}}$ are not the same. More precisely, if $h\in \mathfrak h_{\mathbf {C}}$,

(3.15)\begin{equation} \overline{\varphi h}=\varphi\theta\overline{h}=\theta\varphi\overline{h}. \end{equation}

If $w\in W(\mathfrak h_{\mathbf {C}}:\mathfrak g_{\mathbf {C}})$, then

(3.16)\begin{equation} w\vert_{\mathfrak h_{\mathbf{C}}}=\varphi^{-1}w\vert_{\mathfrak h_{\mathfrak u,\mathbf{C}}}\varphi. \end{equation}

Recall that $w$ is a real automorphism of the real vector space $\mathfrak h_{\mathfrak u}$. By (3.15) and (3.16), we obtain

(3.17)\begin{equation} \overline{w}\vert_{\mathfrak h_{\mathbf{C}}}=\overline{\varphi}^{-1}w\vert_{\mathfrak h_{\mathfrak u,\mathbf{C}}}\overline{\varphi}. \end{equation}

By (3.15)–(3.17), we obtain (3.13). The proof of our proposition is complete.

3.4 Real roots and imaginary roots

Let $R^{\mathrm {re}} \subset R$ be the roots $\alpha \in R$ such that $\theta \alpha =-\alpha$, let $R^{\mathrm {im}}$ be the roots $\alpha \in R$ such that $\theta \alpha =\alpha$. These are the real roots and the imaginary roots, respectively. Imaginary roots vanish on $\mathfrak h_{\mathfrak p}$, real roots vanish on $\mathfrak h_{\mathfrak k}$. By [Reference KnappKna86, p. 349], the set of complex roots $R^{\mathrm {c}} \subset R$ is defined to be

(3.18)\begin{equation} R^{\mathrm{c}}=R{\setminus}(R^{\mathrm{re}}\cup R^{\mathrm{im}}). \end{equation}

Proof. If $b\in \mathfrak h,f\in \mathfrak g_{\alpha }$, then

(3.19)\begin{equation} [b,f]=\langle \alpha,b\rangle f. \end{equation}

Using (3.11) and taking the conjugate of (3.19), we obtain

(3.20)\begin{equation} [b,\overline{f}]=\langle -\theta \alpha,b\rangle\overline{f}. \end{equation}

By (3.20), we get the first part of our proposition. By composing this isomorphism with $\theta$, we obtain the second part. If $\alpha \in R^{\mathrm {re}}$, then $-\theta \alpha =\alpha$, so that $\mathfrak g_{\mathfrak \alpha }$ is real. The proof of our proposition is complete.

Definition 3.3 Put

(3.21)\begin{equation} \mathfrak i=\ker \mathrm{ad}(\mathfrak h_{\mathfrak p})\cap \mathfrak h^{\perp}, \quad \mathfrak r=\ker \mathrm{ad}(\mathfrak h_{\mathfrak k})\cap \mathfrak h^{\perp}. \end{equation}

Then $\theta$ acts on $\mathfrak i, \mathfrak r$, so that we have the splittings,

(3.22)\begin{equation} \mathfrak i= \mathfrak i_{\mathfrak p} \oplus \mathfrak i_{\mathfrak k}, \quad \mathfrak r= \mathfrak r_{\mathfrak p} \oplus \mathfrak r_{\mathfrak k}. \end{equation}

Proposition 3.4 The vector spaces $\mathfrak i$ and $\mathfrak r$ are orthogonal in $\mathfrak h^{\perp }$. Moreover,

(3.23)\begin{equation} \mathfrak i_{\mathbf{C}}=\bigoplus_{\alpha\in R^{\mathrm{im}}}\mathfrak g_{\alpha}, \quad \mathfrak r_{\mathbf{C}}= \bigoplus_{\alpha\in R^{\mathrm{re}}}\mathfrak g_{\alpha}. \end{equation}

If $\alpha \in R^{\mathrm {im}}$, then either $\mathfrak g_{\alpha } \subset \mathfrak p_{\mathbf {C}}$, or $\mathfrak g_{\alpha } \subset \mathfrak k_{\mathbf {C}}$.

Proof. If $f\in \mathfrak h^{\perp }$ and $f\in \mathfrak i \cap \mathfrak r$, then $f$ commutes with $\mathfrak h$. As $\mathfrak h$ is a Cartan subalgebra, $f=0$, so that $\mathfrak i\cap \mathfrak r=0$. If $\alpha \in R{\setminus} R^{\mathrm {im}}$, $\alpha$ does not vanish identically on $\mathfrak h_{\mathfrak p}$, and its vanishing locus in $\mathfrak h_{\mathfrak p}$ is a hyperplane. Thus, one can find $b_{\mathfrak p}\in \mathfrak h_{\mathfrak p}{\setminus} 0$ such that for any $\alpha \in R{\setminus} R^{\mathrm {im}}$, $\langle \alpha,b_{\mathfrak p}\rangle \neq 0$. Then

(3.24)\begin{equation} \mathfrak i=\ker \mathrm{ad}(b_{ \mathfrak p})\cap \mathfrak h^{\perp}. \end{equation}

As $\mathfrak i\cap \mathfrak r=0$, $\mathrm {ad}(b_{\mathfrak p})$ acts as an invertible morphism of $\mathfrak r$. Therefore, any element of $\mathfrak r$ lies in the image of $\mathrm {ad}(b_{\mathfrak p})$. As $\mathrm {ad}(b_{\mathfrak p})$ is symmetric in the classical sense, $\mathfrak i$ and $\mathfrak r$ are orthogonal. Equation (3.23) is elementary. If $\alpha \in R^{\mathrm {im}}$, the action of $\mathfrak h$ on $\mathfrak g_{\alpha }$ factors through $\mathfrak h_{\mathfrak k}$. In addition, $\mathrm {ad}(\mathfrak h_{\mathfrak k})$ preserves the splitting $\mathfrak g = \mathfrak p \oplus \mathfrak k$. Therefore, if $\alpha \in R^{\mathrm {im}}$, either $\mathfrak g_{\alpha } \subset \mathfrak p_{\mathbf {C}}$, or $\mathfrak g_{\alpha }\subset \mathfrak k_{\mathbf {C}}$. The proof of our proposition is complete.

Definition 3.5 Put

(3.25)\begin{equation} R^{\mathrm{im}}_{\mathfrak p}=\{\alpha\in R^{\mathrm{im}}, \mathfrak g_{\alpha}\subset \mathfrak p_{\mathbf{C}}\}, \quad R^{\mathrm{im}}_{\mathfrak k}=\{\alpha\in R^{\mathrm{im}},\mathfrak g_{\alpha}\subset \mathfrak k_{\mathbf{C}}\}. \end{equation}

By Proposition 3.4, we obtain

(3.26)\begin{equation} R^{\mathrm{im}}=R^{\mathrm{im}}_{\mathfrak p}\cup R^{\mathrm{im}}_{\mathfrak k}. \end{equation}

Let $\mathfrak c$ denote the orthogonal to $\mathfrak i \oplus \mathfrak r$ in $\mathfrak h^{\perp }$. Again $\mathfrak c$ splits as

(3.27)\begin{equation} \mathfrak c=\mathfrak c_{\mathfrak p} \oplus \mathfrak c_{\mathfrak k}. \end{equation}

Moreover, we have the orthogonal splitting

(3.28)\begin{equation} \mathfrak h^{\perp}=\mathfrak i \oplus \mathfrak r \oplus \mathfrak c. \end{equation}

Proposition 3.6 The following identity holds:

(3.29)\begin{equation} \mathfrak c_{\mathbf{C}}=\bigoplus _{\alpha\in R^{\mathrm{c}}}\mathfrak g_{\alpha}. \end{equation}

Now we give a result taken from [Reference WallachWal88, Lemma 2.3.5].

Proof. If $\alpha \in R$, then $\alpha \in R^{\mathrm {re}}$ if and only if when $f\in \mathfrak g_{\alpha }$,

(3.30)\begin{equation} [\mathfrak h_{\mathfrak k},f]=0. \end{equation}

If $\mathfrak h$ is fundamental, by (3.4), then $f\in \mathfrak h_{\mathbf {C}}$, so that $f=0$, which proves there are no real roots. Put

(3.31)\begin{equation} \mathfrak z(\mathfrak h_{\mathfrak k})=\{f\in \mathfrak g, [\mathfrak h_{\mathfrak k},f]=0\}. \end{equation}

Then $\mathfrak z(\mathfrak h_{\mathfrak k})$ is a Lie subalgebra of $\mathfrak g$ such that $\mathfrak h \subset \mathfrak z(\mathfrak h_{\mathfrak k})$. If $\mathfrak h$ is not fundamental, $\mathfrak z(\mathfrak h_{\mathfrak k})$ is strictly larger than $\mathfrak h$, and there is a real root. The proof of our proposition is complete.

If $\mathfrak h$ is a Cartan subalgebra, the associated Cartan subgroup $H \subset G$ is the stabilizer of $\mathfrak h$. Then $H$ is a Lie subgroup of $G$, with Lie algebra $\mathfrak h$.

Proposition 3.8 The vector spaces $\mathfrak i_{\mathfrak p}, \mathfrak i_{\mathfrak k}, \mathfrak c_{\mathfrak p}, \mathfrak c_{\mathfrak k}$ have even dimension. In addition, $H\cap K$ preserves these vector spaces, and the corresponding determinants are equal to 1. Also $\mathfrak r_{\mathfrak p}, \mathfrak r_{\mathfrak k}$ (respectively, $\mathfrak c_{\mathfrak p}, \mathfrak c_{\mathfrak k}$) have the same dimension, and the actions of $H\cap K$ on these two vector spaces are equivalent. In particular, we have the identity

(3.32)\begin{equation} \dim \mathfrak p-\dim \mathfrak h_{\mathfrak p}= \dim \mathfrak i_{\mathfrak p}+\tfrac{1}{2}\dim \mathfrak r+\tfrac{1}{2}\dim \mathfrak c. \end{equation}

Proof. If $\alpha \in R{\setminus} R^{\mathrm {re}}$, the vanishing locus of $\alpha$ in $\mathfrak h_{\mathfrak k}$ is a hyperplane. Therefore, we can find $f_{\mathfrak k}\in \mathfrak h_{\mathfrak k}{\setminus} 0$ such for any $\alpha \in R{\setminus} R^{\mathrm {re}}$, $\langle \alpha,f_{\mathfrak k}\rangle \neq 0$, which just says that $\mathrm {ad}(f_{\mathfrak k})$ acts as an invertible endomorphism of $\mathfrak i, \mathfrak c$. This endomorphism preserves their $\mathfrak p$ and $\mathfrak k$ components, and it is classically antisymmetric. This is only possible if these vector spaces are even-dimensional. If $k\in H\cap K$, $\mathrm {Ad}(k^{-1})$ preserves these vector spaces and commutes with $\mathrm {ad}(f_{\mathfrak k})$. Therefore, the eigenspaces associated with the eigenvalue $-1$ are preserved by $\mathrm {ad}(f_{\mathfrak k})$, so they are even-dimensional. This forces the determinant of $\mathrm {Ad}(k^{-1})$ to be equal to 1 on each of these vector spaces.

We choose $b_{\mathfrak p}\in \mathfrak h_{\mathfrak p}{\setminus} 0$ such that for any $\alpha \in R{\setminus} R^{\mathrm {im}}$, $\langle \alpha,b_{\mathfrak p}\rangle \neq 0$. Therefore, $\mathrm {ad}(b_{\mathfrak p})$ acts as an automorphism of $\mathfrak r, \mathfrak c$ that exchanges the corresponding $\mathfrak p$ and $\mathfrak k$ parts, and commutes with $\mathrm {Ad}(k^{-1})$.

By (3.28), we obtain

(3.33)\begin{equation} \mathfrak h_{\mathfrak p}^{\perp}= \mathfrak i_{\mathfrak p} \oplus \mathfrak r_{\mathfrak p} \oplus \mathfrak c_{\mathfrak p}. \end{equation}

Using the results we already established and (3.33), we obtain (3.32). The proof of our proposition is complete.

3.5 A positive root system

Let $\mathfrak h$ be a $\theta$-stable Cartan subalgebra, and let $R$ denote the corresponding root system. Let $R_{+} \subset R$ be a positive root system. Set

(3.34)\begin{equation} R_{+}^{\mathrm{re}}=R_{+}\cap R^{\mathrm{re}}, \quad R_{+}^{\mathrm{im}}=R_{+}\cap R^{\mathrm{im}}, \quad R^{\mathrm{c}}_{+}=R_{+}\cap R^{\mathrm{c}}, \end{equation}

so that

(3.35)\begin{equation} R_{+}=R^{\mathrm{re}}_{+}\cup R_{+}^{\mathrm{im}}\cup R_{+}^{\mathrm{c}}. \end{equation}

In the whole paper, we choose $R_{+}$ such that $-\theta$ preserves $R_{+}{\setminus} R^{\mathrm {im}}_{+}$. Equivalently, we assume that if $\alpha \in R_{+}{\setminus} R^{\mathrm {im}}_{+}$, then $\overline {\alpha }\in R_{+}$.

Let us explain how to do this. If $\alpha \in R{\setminus} R^{\mathrm {im}}$, the vanishing locus of $\alpha$ in $\mathfrak h_{\mathfrak p}$ is a hyperplane, and so there is $b_{\mathfrak p}\in \mathfrak h_{\mathfrak p}, \vert b_{\mathfrak p}\vert =1$ such that for any $\alpha \in R{\setminus} R^{\mathrm {im}}$, $\langle \alpha,b_{\mathfrak p}\rangle \neq 0$. The same argument shows that there is $b_{\mathfrak k}\in h_{\mathfrak k}, \vert b_{\mathfrak k}\vert =1$ such that for $\alpha \in R^{\mathrm {im}}$, $\langle \alpha,b_{\mathfrak k}\rangle \neq 0$. For $\epsilon >0$, $b_{\pm }=\pm b_{\mathfrak p}+ i\epsilon b_{\mathfrak k}\in \mathfrak h_{\mathfrak p} \oplus i \mathfrak h_{\mathfrak k}$, and $\theta$ interchanges $b_{+}$ and $b_{-}$. Also for $\epsilon >0$ small enough, for $\alpha \in R$, the real numbers $\langle \alpha,b_{\pm }\rangle$ do not vanish, and if $\alpha \in R{\setminus} R^{\mathrm {im}}$, they have opposite signs. Put

(3.36)\begin{equation} R_{+}=\{\alpha\in R,\langle \alpha,b_{+}\rangle>0\}. \end{equation}

Then $R_{+}$ is a positive root system such that $-\theta$ preserves $R_{+}{\setminus} R_{+}^{\mathrm {im}}$.

Note that $-\theta$ acts without fixed points on $R^{\mathrm {c}}_{+}$, so that $\vert R^{\mathrm {c}}_{+} \vert$ is even.

Definition 3.9 Put

(3.37)\begin{equation} \mathfrak c_{+,\mathbf{C}}=\bigoplus_{\alpha\in R^{\mathrm{c}}_{+}} \mathfrak g_{\alpha}, \quad \mathfrak c_{-,\mathbf{C}}=\bigoplus_{\alpha\in -R^{\mathrm{c}}_{+}} \mathfrak g_{\alpha}. \end{equation}

Proposition 3.10 The vector spaces $\mathfrak c_{+,\mathbf {C}}$ and $\mathfrak c_{-,\mathbf {C}}$ are the complexifications of real Lie subalgebras $\mathfrak c_{+}$ and $\mathfrak c_{-}$ of $\mathfrak g$, which have the same even dimension, and are such that

(3.38)\begin{equation} \mathfrak c= \mathfrak c_{+} \oplus \mathfrak c_{-}, \quad \mathfrak c_{-}=\theta \mathfrak c_{+}. \end{equation}

In addition, $B$ vanishes on $\mathfrak c_{+}, \mathfrak c_{-}$ and induces the identification,

(3.39)\begin{equation} \mathfrak c_{-} \simeq \mathfrak c_{+}^{*}. \end{equation}

The projections on $\mathfrak p$ and $\mathfrak k$ map $\mathfrak c_{\pm }$ into $\mathfrak c_{\mathfrak p}$ and $\mathfrak c_{\mathfrak k}$ isomorphically. Finally, the actions of $H\cap K$ on $\mathfrak c_{+}$, $\mathfrak c_{-}$, $\mathfrak c_{\mathfrak p}$, and $\mathfrak c_{\mathfrak k}$ are equivalent.

If $f\in \mathfrak h$, then

(3.40)\begin{equation} \det\mathrm{ad}(f)_{\mathfrak h^{\perp}}=\prod_{\alpha\in R}\langle \alpha,f\rangle. \end{equation}

Definition 3.11 Let $\pi ^{\mathfrak h, \mathfrak g}\in S{{}^{\cdot }}(\mathfrak h^{*}_{\mathbf {C}})$ be such that if $h\in \mathfrak h_{\mathbf {C}}$, then

(3.41)\begin{equation} \pi^{\mathfrak h, \mathfrak g}(h)=\prod_{\alpha\in R_{+}}\langle \alpha,h\rangle. \end{equation}

By (3.40), if $f\in \mathfrak h$,

(3.42)\begin{equation} \det \mathrm{ad}(f)_{\vert_{\mathfrak h^{\perp}}}=\pi^{\mathfrak h,\mathfrak g}(f) \pi^{\mathfrak h, \mathfrak g}(-f). \end{equation}

In addition, $f\in \mathfrak h$ is regular if and only if $\pi ^{\mathfrak h, \mathfrak g}(f)\neq 0$.

3.6 The case when $\mathfrak h$ is fundamental and the root system of $(\mathfrak h_{\mathfrak k},\mathfrak k)$

In this subsection, we assume that $\mathfrak h$ is a $\theta$-stable fundamental Cartan subalgebra of $\mathfrak g$. By Proposition 3.7 and by (3.18), we obtain

(3.43)\begin{equation} R^{\mathrm{c}}=R{\setminus} R^{\mathrm{im}}. \end{equation}

By Proposition 3.4, $\mathfrak r=0$. By (3.28), we have the orthogonal splitting,

(3.44)\begin{equation} \mathfrak h^{\perp}_{\mathfrak p}=\mathfrak i_{\mathfrak p} \oplus \mathfrak c_{\mathfrak p}, \quad \mathfrak h^{\perp}_{\mathfrak k}=\mathfrak i_{\mathfrak k} \oplus \mathfrak c_{\mathfrak k}. \end{equation}

In addition, $\mathfrak i_{\mathfrak p}, \mathfrak i_{\mathfrak k}$ have even dimension, $\mathfrak c_{\mathfrak p}, \mathfrak c_{\mathfrak k}$ have the same even dimension, and the action of $H\cap K$ on these last two vector spaces are conjugate.

The roots in $R$ do not vanish identically on $\mathfrak h_{\mathfrak k}$. We now reinforce the choice of positive roots made in § 3.5. We may and we will assume that $b_{\mathfrak k}\in \mathfrak h_{\mathfrak k}, \vert b_{\mathfrak k}\vert =1$ has been chosen so that if $\alpha \in R$, $\langle \alpha,b_{\mathfrak k}\rangle\neq 0$.

As we saw in § 3.5, $-\theta$ acts without fixed points on $R^{\mathrm {c}}_{+}$. In addition, if $\alpha \in R^{\mathrm {c}}_{+}$, $-\theta \alpha \vert _{\mathfrak h_{\mathfrak k}}=-\alpha \vert _{\mathfrak h_{\mathfrak k}}$, so that the nonzero real numbers $\langle \alpha, ib_{\mathfrak k}\rangle$ and $\langle -\theta \alpha,ib_{\mathfrak k}\rangle$ have opposite signs.

Set

(3.45)\begin{equation} R^{\mathrm{im}}_{\mathfrak k,+}=R^{\mathrm{im}}_{\mathfrak k}\cap R_{+}, \quad R^{\mathrm{c}}_{++}=\{\alpha\in R^{\mathrm{c}}_{+}, \langle \alpha,ib_{\mathfrak k}\rangle>0 \}. \end{equation}

Definition 3.13 Let $R(\mathfrak h_{\mathfrak k}, \mathfrak k)$ be the root system associated with the pair $(\mathfrak h_{\mathfrak k}, \mathfrak k)$. If $R_{+}(\mathfrak h_{\mathfrak k}, \mathfrak k)$ is a positive root system for $(\mathfrak h_{\mathfrak k}, \mathfrak k)$, if $h_{\mathfrak k}\in \mathfrak h_{\mathfrak k,\mathbf {C}}$, put

(3.46)\begin{equation} \pi^{\mathfrak h_{\mathfrak k}, \mathfrak k}(h_{\mathfrak k})=\prod_{\beta\in R_{+}(\mathfrak h_{\mathfrak k}, \mathfrak k)}\langle \beta, h_{\mathfrak k}\rangle. \end{equation}

Then $[\pi ^{\mathfrak h_{\mathfrak k}, \mathfrak k}]^{2}(h_{\mathfrak k})$ does not depend on the choice of $R_{+}(\mathfrak h_{\mathfrak k}, \mathfrak k)$. The arguments above (3.45) show that if $h_{\mathfrak k}\in \mathfrak h_{\mathfrak k}$, then

(3.47)\begin{equation} \prod_{\alpha\in R_{+}^{\mathrm{c}}}\langle \alpha,h_{\mathfrak k}\rangle\ge 0. \end{equation}

Proposition 3.14 The map $\alpha \in R_{\mathfrak k}^{\mathrm {im}}\cup R^{\mathrm {c}}_{+} \to \alpha \vert _{\mathfrak h_{\mathfrak k}}$ is injective, and gives the identification

(3.48)\begin{equation} R(\mathfrak h_{\mathfrak k}, \mathfrak k)=R^{\mathrm{im}}_{\mathfrak k} \cup R^{\mathrm{c}}_{+}. \end{equation}

A positive root system $R_{+}(\mathfrak h_{\mathfrak k}, \mathfrak k)$ for $(\mathfrak h_{\mathfrak k}, \mathfrak k)$ is given by

(3.49)\begin{equation} R_{+}(\mathfrak h_{\mathfrak k}, \mathfrak k)=R_{\mathfrak k,+}^{\mathrm{im}} \cup R_{++}^{\mathrm{c}}. \end{equation}

If $h_{\mathfrak k}\in \mathfrak h_{\mathfrak k,\mathbf {C}}$, then

(3.50)\begin{equation} [\pi^{h_{\mathfrak k}, \mathfrak k}(h_{\mathfrak k})]^{2}=(-1)^{({1}/{2})\vert R_{+}^{\mathrm{c}}\vert}\biggl[\prod_{\alpha\in R^{\mathrm{im}}_{\mathfrak k,+}}\langle \alpha,h_{\mathfrak k}\rangle\biggr]^{2} \prod_{\alpha\in R_{+}^{\mathrm{c}}}\langle \alpha,h_{\mathfrak k}\rangle. \end{equation}

Proof. By (3.28), we obtain

(3.51)\begin{equation} \mathfrak h_{\mathfrak k}^{\perp}= \mathfrak i_{\mathfrak k} \oplus \mathfrak c_{\mathfrak k}, \end{equation}

and the above splitting is preserved by $\mathfrak h_{\mathfrak k}$. The weights for this action on $\mathfrak i_{\mathfrak k}$ are given by $R^{\mathrm {im}}_{\mathfrak k}$. By Proposition 3.10, $\mathfrak c_{\mathfrak k}$ and $\mathfrak c_{+}$ are equivalent under the action of $\mathfrak h_{\mathfrak k}$. By the first equation in (3.37), the weights for the action of $\mathfrak h_{\mathfrak k}$ on $\mathfrak c_{+}$ are given by the restriction of $R_{+}^{\mathrm {c}}$ to $\mathfrak h_{\mathfrak k}$. As the weights for the action of $\mathfrak h_{\mathfrak k}$ on $\mathfrak c_{\mathfrak k}$ are nonzero and of multiplicity $1$, the map $\alpha \in R_{\mathfrak k}^{\mathrm {im}}\cup R_{+}^{\mathrm {c}} \to \alpha \vert _{\mathfrak h_{\mathfrak k}}$ gives the identification in (3.48). By (3.48), we obtain (3.49). Using (3.46) and the above results, we obtain (3.50). The proof of our proposition is complete.

3.7 Cartan subgroups and regular elements

Assume that $\mathfrak h$ is $\theta$-stable. Then $\theta$ restricts to an involution of $H$, and (3.3) is the corresponding Cartan splitting of $\mathfrak h$. In addition, $B$ restricts to a $H$ and $\theta$ invariant symmetric nondegenerate bilinear form $B\vert _{\mathfrak h}$ on $\mathfrak h$, so that $H$ is a reductive subgroup of $G$.

We still assume $\mathfrak h$ to be $\theta$-stable. Let $Z_{G}(H) \subset G$ be the centralizer of $H$, and let $N_{G}(H) \subset G$ be its normalizer. Then $Z_{G}(H)$ is included in $H$, it is just the center $Z(H)$ of $H$. As in [Reference KnappKna86, p. 131], the analytic Weyl group $W(H:G)$ is defined as the quotient

(3.52)\begin{equation} W(H:G)=N_{G}(H)/Z_{G}(H). \end{equation}

Put

(3.53)\begin{equation} Z_{K}(H)=Z_{G}(H) \cap K, \quad N_{K}(H)=N_{G}(H)\cap K. \end{equation}

Then $N_{K}(H)/Z_{K}(H)$ embeds in $W(H:G)$. By [Reference KnappKna86, p. 131], this embedding is an isomorphism, i.e.

(3.54)\begin{equation} W(H:G)=N_{K}(H)/Z_{K}(H). \end{equation}

By [Reference KnappKna86, (5.6)], $W(H:G) \subset W(\mathfrak h_{\mathbf {C}}:\mathfrak g_{\mathbf {C}})$.

By [Reference KnappKna86, p. 130], an element $\gamma \in G$ is said to be regular if $\mathfrak z(\gamma )$ is a Cartan subalgebra. If $H$ is the corresponding Cartan subgroup, then $\gamma \in H$. By [Reference KnappKna86, Theorem 5.22], the set $G^{\mathrm {reg}} \subset G$ of regular elements is open and conjugation invariant. More precisely, if $H_{1},\ldots,H_{\ell }$ denotes the finite family of nonconjugate Cartan subgroups, by [Reference KnappKna86, Theorem 5.22], $G^{\mathrm {reg}}$ splits as the disjoint union of open sets

(3.55)\begin{gather} G^{\mathrm{reg}}=\bigcup_{i=1}^{\ell}G^{\mathrm{reg}}_{H_{i}}, \end{gather}

where $G^{\mathrm {reg}}_{H_{i}}$ denote the open set of elements of $G^{\mathrm {reg}}$ that are conjugate to an element of $H_{i}$.

If $\gamma \in H$, $\mathrm {Ad}(\gamma )$ acts on $\mathfrak g$ and fixes $\mathfrak h$. As $\mathrm {Ad}(\gamma )$ preserves $B$, it also acts on $\mathfrak h^{\perp }$, so that $1-\mathrm {Ad}(\gamma )$ acts on $\mathfrak h^{\perp }$. Then $\gamma$ is regular if and only this endomorphism is invertible, i.e. $\det ( 1-\mathrm {Ad}(\gamma ))\vert _{\mathfrak h^{\perp }}\neq 0$.

3.8 Cartan subgroups and semisimple elements

The following result is established in [Reference VaradarajanVar77, Part I, § 2.3, Theorem 4].

Let us give a direct classical proof of part of our proposition. Let $\mathfrak h$ be a $\theta$-stable Cartan subalgebra, and let $H$ be the corresponding Cartan subgroup. If $\gamma \in H$, then $\mathfrak h \subset \mathfrak z(\gamma )$. Moreover, $\gamma$ can be written uniquely in the form

(3.56)\begin{equation} \gamma=e^{a}k^{-1}, \quad a\in \mathfrak p, k\in H. \end{equation}

As $H$ is $\theta$-stable, $\gamma (\theta (\gamma ))^{-1}\in H$, i.e. $e^{2a}\in H$. By [Reference BismutBis11, Proposition 3.2.8], $Z(e^{2a})=Z(a)$, so that $\mathfrak h \subset Z(a)$. As $\mathfrak h$ is a Cartan subalgebra, $a\in \mathfrak h_{\mathfrak p}$. Therefore, (3.56) can be rewritten in the form,

(3.57)\begin{equation} \gamma=e^{a}k^{-1}, \quad a\in \mathfrak h_{\mathfrak p}, k\in H\cap K. \end{equation}

As $a\in \mathfrak h_{\mathfrak p}, k\in H$, then $\mathrm {Ad}(k)a=a$, which guarantees that $\gamma$ is semisimple in $G$.

Let $\mathfrak h \subset \mathfrak g$ be a Cartan subalgebra, and let $H \subset G$ be the associated Cartan subgroup. If $\gamma \in H$, then $\mathfrak h \subset \mathfrak z(\gamma )$, so that $\mathfrak h$ is a Cartan subalgebra of $\mathfrak z(\gamma )$. In particular, $G$ and $Z^{0}(\gamma )$ have the same complex rank.

3.9 Root systems and their characters

Let $\mathfrak h \subset \mathfrak g$ be a $\theta$-stable Cartan subalgebra. We use the notation of the previous subsections.

Take $\gamma \in H$. As we saw after Proposition 3.16, if $\gamma \in H$, we can write $\gamma$ uniquely in the form

(3.58)\begin{equation} \gamma=e^{a}k^{-1}, \quad a\in \mathfrak h_{\mathfrak p},\quad k\in H\cap K, \end{equation}

so that

(3.59)\begin{equation} \mathrm{Ad}(k)a=a. \end{equation}

Let $R(\gamma ),R(a)$ be the root systems associated with $(\mathfrak h, \mathfrak z(\gamma )), (\mathfrak h, \mathfrak z(a))$. We will denote with extra subscripts the corresponding real, imaginary, and complex roots. Note that

(3.60)\begin{equation} R^{\mathrm{im}} \subset R(a). \end{equation}

Theorem 3.18 If $\gamma \in H$, for any $\alpha \in R$, $\mathrm {Ad}(\gamma )$ preserves the $1$-dimensional complex line $\mathfrak g_{\alpha }$. For every $\alpha \in R$, there is a character $\xi _{\alpha }: H\to \mathbf {C}^{*}$ such that $\mathrm {Ad}(\gamma )$ acts on $\mathfrak g_{\alpha }$ by multiplication by $\xi _{\alpha }(\gamma )$. If $\alpha \in R$,

(3.61)\begin{equation} \xi_{\alpha}\xi_{-\alpha}=1. \end{equation}

If $\alpha \in R$, if $f\in \mathfrak h, k\in H\cap K$, then

(3.62)\begin{equation} \xi_{\alpha}(e^{f})=e^{\langle \alpha,f\rangle}, \quad \vert \xi_{\alpha}(k)\vert=1. \end{equation}

In particular, if $\gamma \in H$ is taken as in (3.58), then

(3.63)\begin{equation} \xi_{\alpha}(\gamma)=e^{\langle \alpha,a\rangle}\xi_{\alpha}(k^{-1}), \quad \xi_{-\theta\alpha}(\gamma)=\overline{\xi_{\alpha}(\gamma)}. \end{equation}

If $\alpha \in R^{\mathrm {re}}$, then $\xi _{\alpha }(\gamma )\in \mathbf {R}^{*}$, if $\alpha \in R^{\mathrm {im}}$, then $\vert \xi _{\alpha }(\gamma )\vert =1$. If $\alpha \in R^{\mathrm {re}}$, the restriction of $\xi _{\alpha }$ to $H\cap K$ takes its values in $\{-1,+1\}$.

In addition,

(3.64)\begin{equation} \det(1-\mathrm{Ad}(\gamma))\vert_{\mathfrak h^{\perp}}=\prod_{\alpha\in R}(1-\xi_{\alpha}(\gamma)), \end{equation}

and $\gamma$ is regular if and only if for any $\alpha \in R$, $\xi _{\alpha }(\gamma ) \neq 1$.

The following identities hold:

(3.65)\begin{gather} R(\gamma)=\{\alpha\in R, \xi_{\alpha}(\gamma)=1\},\quad R(a)=\{\alpha\in R, \langle\alpha,a\rangle=0\},\nonumber\\ R^{\mathrm{re}}(\gamma)=R(\gamma)\cap R^{\mathrm{re}}, \quad R^{\mathrm{im}}(\gamma)=R(\gamma)\cap R^{\mathrm{im}}, \quad R^{\mathrm{c}}(\gamma)=R(\gamma)\cap R^{\mathrm{c}}, \nonumber\\ R^{\mathrm{re}}(a)=R(a)\cap R^{\mathrm{re}}, \quad R^{\mathrm{im}}(a) = R^{\mathrm{im}}, \quad R^{\mathrm{c}}(a)=R(a)\cap R^{\mathrm{c}}. \end{gather}

In addition, $R_{+}(\gamma )=R(\gamma ) \cap R_{+}$ and $R_{+}(a)=R(a)\cap R_{+}$ are positive root systems for $(\mathfrak h, \mathfrak z(\gamma ))$ and $(\mathfrak h, \mathfrak z(a))$.

Proof. If $\gamma \in H$, then $\mathrm {Ad}(\gamma )$ fixes $\mathfrak h$, and so if $h\in \mathfrak h$, we have the commutation relation in $\mathrm {End}(\mathfrak g)$,

(3.66)\begin{equation} [\mathrm{Ad}(\gamma), \mathrm{ad}(h)]=0. \end{equation}

By (3.7) and (3.66), we deduce that for any $\alpha \in R$, $\mathrm {Ad}(\gamma )$ preserves $\mathfrak g_{\alpha }$. As $\mathfrak g_{\alpha }$ is a complex line, $H$ acts on $\mathfrak g_{\alpha }$ via a character $\xi _{\alpha }$.

As $\mathrm {Ad}(\gamma )$ preserves $B$, if $f,f'\in \mathfrak g_{\mathbf {C}}$, we obtain

(3.67)\begin{equation} B(\mathrm{Ad}(\gamma)f,f')=B(f,\mathrm{Ad}(\gamma)^{-1}f'). \end{equation}

Take $\alpha \in R$. By (3.67), if $f\in \mathfrak g_{\alpha }$ and $f'\in \mathfrak g_{-\alpha }$, then

(3.68)\begin{equation} \xi_{\alpha}(\gamma)B(f,f')=\xi_{-\alpha}^{-1}(\gamma)B(f,f'). \end{equation}

As we saw in § 3.3, if $\alpha \in R$, the pairing between $\mathfrak g_{\alpha }$ and $\mathfrak g_{-\alpha }$ via $B$ is nondegenerate. By (3.68), we obtain (3.61).

The first equation in (3.62) is trivial. As $\xi _{\alpha }$ restricts to a character of the compact group $H\cap K$, we obtain the second equation in (3.62). The first equation in (3.63) follows from the previous considerations. As $\theta (\gamma )=e^{-a}k^{-1}$, and because by (3.11) $\theta$ maps $\mathfrak g_{\alpha }$ into $\mathfrak g_{\theta \alpha }$, we obtain the second equation in (3.63). From this second equation, we deduce that if $\alpha \in R^{\mathrm {re}}$, then $\xi _{\alpha }(\gamma )$ is real, and if $\alpha \in R^{\mathrm {im}}$, then $\vert \xi _{\alpha }(\gamma )\vert =1$. If $\gamma \in H\cap K$ and $\alpha \in R^{\mathrm {re}}$, we know that $\xi _{\alpha }(\gamma )\in \mathbf {R}^{*}$ and $\vert \xi _{\alpha }(\gamma )\vert =1$, so that $\xi _{\alpha }(\gamma )=\pm 1$.

Equations (3.64) and (3.65) are trivial. By (3.64), $\gamma$ is regular if and only if for $\alpha \in R$, $\xi _{\alpha }(\gamma )\neq 1$.

Now we proceed as in [Reference Bismut and LabourieBL99, Theorem 1.38]. If $\kappa \subset \mathfrak h$ is a positive Weyl chamber for $(\mathfrak h, \mathfrak g)$, the forms in $R$ do not vanish on $\kappa$, so that $\kappa$ is included in a $\mathfrak z(\gamma )$ Weyl chamber. It follows that $R(\gamma )\cap R_{+}$ is a positive root system on $(\mathfrak h, \mathfrak z(\gamma ))$. The same argument is valid for $R(a)$. The proof of our theorem is complete.

3.10 Real roots, imaginary roots, and semisimple elements

We still take $\gamma$ as in § 3.9. When taking the intersection of $\mathfrak i$, $\mathfrak r$, and $\mathfrak c$ with $\mathfrak z(\gamma )$, $\mathfrak z(a)$, and $\mathfrak z(k)$, this will be indicated with a parenthesis containing the corresponding argument. The intersection with $\mathfrak z^{\perp }(\gamma )$, $\mathfrak z^{\perp }(a)$, and $\mathfrak z^{\perp }(k)$ are denoted with an extra $\perp$. These vector spaces also have a $\mathfrak p$ and a $\mathfrak k$ component.

By construction,

(3.69)\begin{equation} R^{\mathrm{im}}(\gamma)=R^{\mathrm{im}}(k). \end{equation}

As in (3.25) and (3.26), we obtain

(3.70)\begin{equation} R^{\mathrm{im}}(\gamma)=R^{\mathrm{im}}_{\mathfrak p}(\gamma)\cup R^{\mathrm{im}}_{\mathfrak k}(\gamma), \quad R^{\mathrm{im}}(k)=R^{\mathrm{im}}_{\mathfrak p}(k)\cup R^{\mathrm{im}}_{\mathfrak k}(k). \end{equation}

To make the notation simpler, in (3.70), we did not use instead the notation $\mathfrak p(\gamma )$, $\mathfrak k(\gamma )$, $\mathfrak p(k)$, and $\mathfrak k(k)$.

Proposition 3.19 The following identities hold:

(3.71)\begin{equation} \mathfrak i \subset \mathfrak z(a), \quad \mathfrak i(\gamma)= \mathfrak i(k),\quad\mathfrak i^{\perp}(k) \subset \mathfrak z_{a}^{\perp}(\gamma). \end{equation}

In addition,

(3.72)\begin{equation} \mathfrak i(k)_{\mathbf{C}}=\bigoplus_{\alpha\in R^{\mathrm{im}}(k)} \mathfrak g_{\alpha}, \quad \mathfrak i^{\perp}(k)_{\mathbf{C}}=\bigoplus _{\alpha\in R^{\mathrm{im}}{\setminus} R^{\mathrm{im}}(k)}\mathfrak g_{\alpha}. \end{equation}

Moreover,

(3.73)\begin{equation} \begin{aligned} &\mathfrak z(a)_{\mathbf{C}}=\mathfrak h_{\mathbf{C}} \oplus \bigoplus _{\alpha\in R(a)}\mathfrak g_{\alpha}, \\ &\mathfrak z^{\perp}(a) \subset \mathfrak r \oplus \mathfrak c, \\ &\mathfrak z^{\perp}(a)_{\mathbf{C}} =\bigoplus_{\alpha\in R{\setminus} R(a)} \mathfrak g_{\alpha}. \end{aligned} \end{equation}

If $\gamma$ is regular, then

(3.74)\begin{equation} \mathfrak i(k)=0. \end{equation}