Iwasawa Decomposition for Lie Superalgebras

Let $(\mathfrak{g},\mathfrak{k})$ be a symmetric pair coming from an involution $\theta$ of $\mathfrak{g}$, where $\mathfrak{g}$ is a reductive Lie algebra over an algebraically closed field of characteristic zero. Then we have the well-known Iwasawa decomposition of $\mathfrak{g}$ given by $\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{a}\oplus\mathfrak{n}$, which plays an important role in the study of symmetric spaces. Here $\mathfrak{a}$ is a maximal toral subalgebra of $\mathfrak{p}$, where $\mathfrak{p}$ is the $(-1)$-eigenspace of $\theta$, and $\mathfrak{n}$ is the sum of positive weight spaces for the adjoint action of $\mathfrak{a}$ on $\mathfrak{g}$, for some choice of positivity.

A close analogue of this situation for Lie superalgebras is to consider a supersymmetric pair $(\mathfrak{g},\mathfrak{k})$ coming from an involution $\theta$ of $\mathfrak{g}$, where $\mathfrak{g}$ is a basic simple Lie superalgebra, i.e. $\mathfrak{g}$ is simple, admits a nondegenerate invariant form, and $\mathfrak{g}_{\overline{0}}$ is reductive. However it is well known that even if $\theta$ preserves an invariant form on $\mathfrak{g}$, there need not be an Iwasawa decomposition in this setting. We seek to clarify the situation by proving that if $\theta$ does not admit an Iwasawa decomposition then $\delta\circ\theta$ does, where $\delta(x)=(-1)^{\overline{x}}x$. Note that $\delta\circ\theta|_{\mathfrak{g}_{\overline{0}}}=\theta|_{\mathfrak{g}_{\overline{0}}}$, so these involutions are closely related to one another. The theorem fails if we do not assume that $\theta$ preserves a nondegenerate form – see Remark 5.6 for an example.

An important consequence of the Iwasawa decomposition is the existence of a Borel subalgebra of $\mathfrak{g}$ complementary to $\mathfrak{k}$ – in particular one can find a Borel subalgebra containing $\mathfrak{a}\oplus\mathfrak{n}$. Thus a corresponding global symmetric space $G/K$ will be a spherical variety. Algebraic symmetric spaces give rise to a beautiful and well-understood source of spherical varieties. Many of the features and structures enjoyed by symmetric varieties have been generalized to spherical varieties such as the little Weyl group ([Kno90] and [Kno94]), wonderful compactifications ([DCP83]), and (restricted) root systems ([Bri90] and [Kno96]). The author has begun a study of spherical supervarieties and their properties in [She21] and [She20], and this paper shows that many symmetric supervarieties are spherical using the existence of an Iwasawa decomposition.

Another important use of the Iwasawa decomposition is in the study of invariant differential operators on the symmetric space $G/K$. One uses the decomposition to obtain a natural projection $\mathcal{U}\mathfrak{g}/\mathcal{U}\mathfrak{g}\mathfrak{k}\to S(\mathfrak{a})$, giving rise to the Harish-Chandra homomorphism associated to this pair. In [All12] a characterization of the image of the Harish-Chandra homomorphism was given for supersymmetric pairs which admit an Iwasawa decomposition. See also [SS16] and [SSS20] for work on the Capelli eigenvalue problem on symmetric supervarieties.

Restricted root systems coming from supersymmetric pairs were used in [SV04] to construct new families of Calogero-Moser-Sutherland systems which are completely integrable. In at the end of Section 6, we explain a relationship between the deformed root systems used in [SV04] and the restricted root systems obtained from supersymmetric pairs.

We now explain what will be shown. Let $V$ be a vector space with a symmetric bilinear form, and $R\subseteq V\setminus\{0\}$ a finite irreducible generalized reflection root system (GRRS) (see Definition 2 for full definitions). GRRSs were defined in [GS17]. Finite GRRSs are a very mild generalization of generalized root systems (GRSs) as defined in [Ser96], and they are more suitable for our purposes. An irreducible GRRS should be viewed as the root system of a basic (almost) simple Lie superalgebra $\mathfrak{g}$.

Now let $\theta$ be an automorphism of $V$ preserving both the form and $R$. This automorphism may come from a semisimple automorphism of $\mathfrak{g}$, and if $\theta$ comes from an involution of $\mathfrak{g}$ then it will be of order 2. Write $S\subseteq R$ for those roots fixed by $\theta$. A root $\alpha\in R$ is odd if the corresponding root space in $\mathfrak{g}$ is odd (for the definition of odd roots in terms of GRRSs, see Definition 2). The following theorem is the main technical result upon which all other results are based. {Theorem} Let $T\subseteq S$ be the smallest subset of $S$ containing all odd roots of $S$ and such that if $\alpha\in T,\beta\in S$ with $(\alpha,\beta)\neq 0$, then $\beta\in T$. Then we have one of the following possibilities for $T$:

• •

  $T=\emptyset$;

• •

  $T=\{\pm\alpha\}$ for an isotropic root $\alpha$;

• •

  $T\subseteq\operatorname{span}(T)$ is a finite irreducible GRRS containing at least one odd root.

Now either let $\mathfrak{g}$ be a basic simple Lie superalgebra not equal to $\mathfrak{p}\mathfrak{s}\mathfrak{l}(2|2)$ or let $\mathfrak{g}$ be $\mathfrak{g}\mathfrak{l}(m|n)$. Recall that being basic means there is an even invariant nondegenerate form on $\mathfrak{g}$. Let $\theta\in\operatorname{Aut}(\mathfrak{g})$ be a semisimple automorphism preserving such a form. Let $\mathfrak{h}$ be a $\theta$-stable Cartan subalgebra of $\mathfrak{g}_{\overline{0}}$. Then $\theta$ induces an automorphism of the GRRS $R\subseteq\mathfrak{h}^{*}$ corresponding to the choice of $\mathfrak{h}$. Write $\mathfrak{a}\subseteq\mathfrak{h}$ for the sum of the eigenspaces of $\theta$ on $\mathfrak{h}$ with eigenvalue not equal to one. If we write $S$ for the roots fixed by $\theta$, then the centralizer of $\mathfrak{a}$ is given by $\mathfrak{c}(\mathfrak{a})=\mathfrak{h}+\bigoplus\limits_{\alpha\in S}\mathfrak{g}_{\alpha}$. Using Theorem 1 we obtain:

The Lie superalgebra $\mathfrak{c}(\mathfrak{a})$ is an extension of an abelian Lie superalgebra by the product of ideals $\mathfrak{a}\times\tilde{\mathfrak{l}}\times\mathfrak{l}$, where $\mathfrak{l}$ is an even semisimple Lie algebra and $\tilde{\mathfrak{l}}$ is isomorphic to either a basic simple Lie superalgebra, $\mathfrak{s}\mathfrak{l}(n|n)$ for some $n\geq 1$, or is trivial.

Note that if $\mathfrak{g}$ is Kac-Moody, then $\mathfrak{c}(\mathfrak{a})$ will also be Kac-Moody, see Theorem 4. We emphasize that the nontrivial statement in Theorem 1 is that the centralizer has only one simple superalgebra appearing which is not purely even. This need not be true for centralizers of an arbitrary torus in $\mathfrak{g}$ – in particular it is false for many Levi subalgebras.

In the case when $\theta$ is of order $2$, we can construct $\mathfrak{h}$ so that $\mathfrak{a}$ is a maximal toral subspace of $\mathfrak{p}$, the $(-1)$-eigenspace of $\theta$ acting on $\mathfrak{g}$. Classically it is known that $\mathfrak{c}(\mathfrak{a})_{\overline{0}}\subseteq\mathfrak{a}+\mathfrak{k}$. However it is possible that $\mathfrak{c}(\mathfrak{a})_{\overline{1}}\cap\mathfrak{p}\neq 0$, in which case the Iwasawa decomposition doesn’t hold. However Theorem 1 implies that if $\mathfrak{c}(\mathfrak{a})_{\overline{1}}\cap\mathfrak{p}\neq 0$, then $\mathfrak{c}(\mathfrak{a})_{\overline{1}}\subseteq\mathfrak{p}$. Therefore if we look at $\delta\circ\theta$ instead, where $\delta(x)=(-1)^{\overline{x}}x$ is the canonical grading automorphism, then for this automorphism we have $\mathfrak{c}(\mathfrak{a})_{\overline{1}}\subseteq\mathfrak{k}$, and thus the Iwasawa decomposition will hold. We state this as the following result (where the case of $\mathfrak{p}\mathfrak{s}\mathfrak{l}(2|2)$ is checked separately).

If $\theta$ is an involution on a basic simple superalgebra or $\mathfrak{g}\mathfrak{l}(m|n)$ which preserves the nondegenerate invariant form, then either $\theta$ or $\delta\circ\theta$ admits an Iwasawa decomposition. In particular, either the fixed points of $\theta$ or the fixed points of $\delta\circ\theta$ have a complementary Borel subalgebra.

Observe that it is possible for both $\theta$ and $\delta\circ\theta$ to admit Iwasawa decompositions; indeed, in many cases these involutions are conjugate to one another, for example any involution of $\mathfrak{a}\mathfrak{b}(1|3)$ satisfies this.

Finally, let us summarize the contents of the paper. In Section 2 we recall the definition of finite GRRSs, state the classification of finite irreducible GRRSs, and prove a few facts we will need later on about them. In Section 3 we introduce automorphisms of GRRSs and prove Theorem 1. Section 4 applies the results from Section 3 to prove Theorem 1. Section 5 proves Theorem 1 and describes supersymmetric pairs that fit into our framework. In Section 6 we study restricted root systems that arise from supersymmetric pairs, discuss their general properties, and relate them to the work of Sergeev and Veselov in [SV04]. Finally, the appendix classifies all supersymmetric pairs of $\mathfrak{a}\mathfrak{g}(1|2)$ and $\mathfrak{a}\mathfrak{b}(1|3)$.

We work over an algebraically closed field $\Bbbk$ of characteristic zero. In [Ser96] the notion of a generalized root system (GRS) was introduced, and GRSs were completely classified. In [GS17], this notion was generalized to that of a generalized reflection root system (GRRS) that was designed to encompass root systems of affine Lie superalgebras. Finite GRRSs come from root systems of certain (almost) simple Lie superalgebras, and we have found they are a natural object to consider for our problem.

The proofs of properties of GRSs stated in [Ser96] carry over almost entirely to finite GRRSs. We will restate some of these results without proof with this understanding.

Let $V$ be a finite-dimensional $\Bbbk$-vector space equipped with a symmetric bilinear form $(\cdot,\cdot)$ (not necessarily nondegenerate). A finite generalized reflection root system (GRRS) is a nonempty finite set $R\subseteq V\setminus\{0\}$ satisfying the following axioms:

1. 1.

   $\operatorname{span}(R)=V$;

2. 2.

   for $\alpha\in R$, $(\alpha,-)\neq 0$ as an element of $V^{*}$.

3. 3.

   for $\alpha,\beta\in R$ with $(\alpha,\alpha)\neq 0$ we have $k_{\alpha,\beta}:=\frac{2(\alpha,\beta)}{(\alpha,\alpha)}\in\mathbb{Z}$ and $r_{\alpha}(\beta):=\beta-k_{\alpha,\beta}\alpha\in R$;

4. 4.

   for $\alpha\in R$ such that $(\alpha,\alpha)=0$ there exists a bijection $r_{\alpha}:R\to R$ such that $r_{\alpha}(\beta)=\beta$ if $(\alpha,\beta)=0$, and $r_{\alpha}(\beta)=\beta\pm\alpha$ if $(\alpha,\beta)\neq 0$;

5. 5.

   $R=-R$.

We call the elements of $R$ roots. For the rest of this paper we will call a finite GRRS $R$ just a GRRS with the understanding that it is finite. We will not consider infinite GRRSs.

Let $R\subseteq V$ be a GRRS and suppose $S\subseteq R$ is a subset of $R$ such that

• •

  $S=-S$;

• •

  for each $\alpha\in S$ there exists $\beta\in S$ such that $(\alpha,\beta)\neq 0$;

• •

  for each $\alpha\in S$, $r_{\alpha}(S)=S$.

Then $S\subseteq\operatorname{span}(S)$ is a GRRS.

If $R$ is a GRRS we define the subset of real (non-isotropic) and imaginary (isotropic) roots as

Further, we call $\alpha\in R$ odd if $\alpha\in R_{im}$ or $2\alpha\in R_{re}$. Otherwise we say a root is even.

By Chapter VI of [Bou02], $R_{re}\subseteq\operatorname{span}(R_{re})=U$ will be a (potentially non-reduced) root system in the usual sense, and in particular the form is nondegenerate when restricted to $U$. Thus we can decompose $U$ as $U=V_{1}\oplus\cdots\oplus V_{k}$, where $R_{re}^{i}:=R_{re}\cap V_{i}\subseteq V_{i}$ is irreducible and $R_{re}=\coprod\limits_{i}R_{re}^{i}$. Let $W_{i}$ denote the Weyl group of $R_{re}^{i}$, and let $W=W_{1}\times\cdots\times W_{k}$, the Weyl group of $R_{re}\subseteq U$. Then $W$ acts naturally on $V$ and preserves $R$ and the form $(-,-)$. Finally let $V_{0}$ be the orthogonal complement to $U$ in $V$ so that

where $R_{re}\cap V_{0}=\emptyset$. We write $p_{i}:V\to V_{i}$ $i=0,1,\dots,k$ for the projection maps. Note that $(-,-)$ may be degenerate when restricted to $V_{0}$.

A GRRS $R$ is reducible if we can write $R=R^{\prime}\coprod R^{\prime\prime}$, where $R^{\prime}$ and $R^{\prime\prime}$ are nonempty and orthogonal to one another. In this case each of $R^{\prime}$ and $R^{\prime\prime}$ will form GRRSs in the respective subspaces they span. A GRRS $R$ is irreducible if it is not reducible. Every GRRS can be decomposed into a finite direct sum of irreducible GRRSs.

[Prop. 2.6, [Ser96]] For an irreducible GRRS $R$, either $\operatorname{dim}V_{0}=1$ and $k\leq 2$, or $\operatorname{dim}V_{0}=0$ and $k\leq 3$. If $V_{0}\neq 0$, then $p_{0}(R_{im})=\{\pm v\}$ for some nonzero vector $v\in V_{0}$.

For the irreducible root system $R^{i}_{re}\subseteq V_{i}$, we write

for the weight lattice of $V_{i}$.

A $W_{i}$-orbit $X\subseteq P_{i}$ is small if $x-y\in R_{re}^{i}$ for any $x,y\in X$, where $x\neq\pm y$.

[Prop. 3.5 of [Ser96]] Let $R$ be a GRRS. Then $p_{i}(R_{im})$ is a subset of $P_{i}\setminus\{0\}$, and is the union of small $W_{i}$-orbits. In particular $(p_{i}(\alpha),p_{i}(\alpha))\neq 0$ for all $\alpha\in R_{im}$ and $i>0$.

Let $R$ be a GRRS. Then $R_{im}$ is $W$-invariant, and thus we may break it up into its orbits

We call the orbits the imaginary components of $R$.

Let $R$ be an irreducible GRRS. If $\alpha,\beta$ are isotropic roots that lie in the same imaginary component of $R$, and $p_{i}(\alpha)=\pm p_{i}(\beta)$ for all $i$, then either $\alpha=\pm\beta$ or $\alpha\pm\beta=2p_{\ell}(\alpha)\in R_{re}^{\ell}$ for some $\ell\in\{1,\dots,k\}$.

Thm. 5.10 of [Ser96] classified irreducible GRSs. However from an analysis of the proof one sees that it also classifies GRRSs, and only one extra family of GRRSs arises that are not already GRSs, and this is the family $\tilde{A}(n,n)$. This is verified in [GS17] as well. In terms of Lie superalgebras, $\tilde{A}(n-1,n-1)$ is the root system of $\mathfrak{p}\mathfrak{g}\mathfrak{l}(n|n)=\mathfrak{g}\mathfrak{l}(n|n)/\Bbbk I_{n|n}$. To be precise, if we write $\mathfrak{h}\subseteq\mathfrak{g}\mathfrak{l}(n|n)$ for the subalgebra of diagonal matrices, then $\mathfrak{h}^{*}$ has a nondegenerate inner product from the supertrace form. If we take the subspace of $\mathfrak{h}^{*}$ spanned by roots of $\mathfrak{g}\mathfrak{l}(n|n)$ and restrict the form to it, we get the GRRS $\tilde{A}(n-1,n-1)$.

In the following theorem we give the classification of irreducible GRRSs. In each case we will describe $R_{re}$ and $R_{im}$. We will write $W$ for the Weyl group of $R_{re}$ in each case, and $\omega_{i}^{(j)}\in V_{j}$ for the $i$th fundamental weight of $R_{re}^{j}$; for instance if $R_{re}^{(2)}=A_{n}$, then $\omega_{1}^{(2)}\in V_{2}$ denotes the first fundamental weight of the root system $A_{n}$, i.e. the dominant weight corresponding to the standard representation. In the case that $V_{0}\neq 0$, we write $v\in V_{0}$ for the element describe in Proposition 2.2.

The irreducible GRRSs with $R_{im}\neq 0$ are as follows.

1. (0)

   $\tilde{A}(n,n)$, $n\geq 1$: $R_{re}=A_{n}\sqcup A_{n}$, $R_{im}=(W\omega_{1}+v)\sqcup(W\omega_{n}-v)$;

2. 1.

   $A(0,n)$, $n\geq 1$: $R_{re}=A_{n}$, $R_{im}=(W\omega_{1}+v)\sqcup(W\omega_{n}-v)$;

3. 2.

   $C(0,n)$, $n\geq 2$: $R_{re}=C_{n}$, $R_{im}=(W\omega_{1}+v)\sqcup(W\omega_{1}-v)$;

4. 3.

   $A(m,n)$, $m\neq n,m\geq 1$: $R_{re}^{1}=A_{m}$, $R_{re}^{2}=A_{n}$, $R_{im}=(W(\omega_{1}^{(1)}+\omega_{n}^{(2)})+v)\sqcup(W(\omega_{m}^{(1)}+\omega_{1}^{(2)})-v)$;

5. 4.

   $A(n,n)$, $n\geq 2$: $R_{re}^{1}=A_{n}$, $R_{re}^{2}=A_{n}$, $R_{im}=W(\omega_{1}^{(1)}+\omega_{n}^{(2)})\sqcup W(\omega_{n}^{(1)}+\omega_{1}^{(2)})$;

6. 5.

   $B(m,n)$, $m,n\geq 1$: $R_{re}^{1}=B_{m}$, $R_{re}^{2}=BC_{n}$, $R_{im}=W(\omega_{1}^{(1)}+\omega_{1}^{(2)})$;

7. 6.

   $G(1,2)$: $R_{re}^{1}=BC_{1}$, $R_{re}^{2}=G_{2}$, $R_{im}=W(\omega_{1}^{(1)}+\omega_{1}^{(2)})$;

8. 7.

   $D(m,n)$, $m>2$, $n\geq 1$: $R_{re}^{1}=D_{m}$, $R_{re}^{2}=C_{n}$, $R_{im}=W(\omega_{1}^{(1)}+\omega_{1}^{(2)})$;

9. 8.

   $AB(1,3)$: $R_{re}^{1}=A_{1}$, $R_{re}^{2}=B_{3}$, $R_{im}=W(\omega_{1}^{(1)}+\omega_{3}^{(2)})$;

10. 9.

    $D(2,n)$, $n\geq 1$: $R_{re}^{1}=A_{1}$, $R_{re}^{2}=A_{1}$, $R_{re}^{3}=C_{n}$, $R_{im}=W(\omega_{1}^{(1)}+\omega_{1}^{(2)}+\omega_{1}^{(3)})$;

11. 10.

    $D(2,1;a)$: $R_{re}^{1}=A_{1}$, $R_{re}^{2}=A_{1}$, $R_{re}^{3}=A_{1}$, $R_{im}=W(\omega_{1}^{(1)}+\omega_{1}^{(2)}+\omega_{1}^{(3)})$.

The only GRRS which is not a GRS (i.e. for which the inner product is degenerate) is $\tilde{A}(n,n)$.

Each inner product is determined up to proportionality, except for $D(2,1;a)$ where we get a family of distinct inner products parametrized by $a\in\mathbb{C}\setminus\{0,-1\}$ modulo an action of $S_{3}$. Further the inner products on two distinct real components of $D(2,1;a)$ agree if and only if $D(2,1;a)\cong D(2,1)$, which is when $a=1,-2$, or $-1/2$.

let $\alpha,\beta$ be linearly independent isotropic roots in an irreducible GRRS $R$. Then for some $i>0$, one of two things must occur:

1. 1.

   $p_{i}(\alpha)$ and $p_{i}(\beta)$ are orthogonal and either $p_{i}(\alpha)+p_{i}(\beta)\in R_{re}^{i}$ or $p_{i}(\alpha)-p_{i}(\beta)\in R_{re}^{i}$;

2. 2.

   $2p_{i}(\alpha)=\pm 2p_{i}(\beta)\in R_{re}^{i}$.

Let $R\subseteq V$ be an irreducible GRRS and $\theta$ an automorphism of $R$, meaning that $\theta:V\to V$ is a linear isomorphism preserving the bilinear form, with $\theta(R)=R$. Write $S\subseteq R$ for the roots fixed by $\theta$. By linearity, we have that $S=-S$, and if $\alpha,\beta\in S$ with $\alpha+\beta\in R$, then $\alpha+\beta\in S$. We now prove the main technical result of the paper.

Let $\alpha,\beta$ be linearly independent odd roots of $S$. Then there exists a real root $\gamma\in R_{re}$ with $\theta(\gamma)=\gamma$ (i.e. $\gamma\in S$) such that $(\gamma,\alpha)\neq 0$ and $(\gamma,\beta)\neq 0$.

If $S$ either contains at least 2 linearly independent odd roots or no odd roots at all, then $S\subseteq\operatorname{span}(S)$ is a GRRS.

Now let $T\subseteq S$ be the smallest subset of $S$ satisfying:

1. 1.

   all odd roots of $S$ lie in $T$;

2. 2.

   if $\alpha\in T$, $\beta\in S$ with $(\alpha,\beta)\neq 0$, then $\beta\in T$.

Then $T$ will be orthogonal to $T^{\prime}:=S\setminus T$, and $T^{\prime}$ will consist of only even roots.

$T^{\prime}\subseteq\operatorname{span}(T^{\prime})$ is a reduced root system. Further, we have the following possibilities for $T$:

1. 1.

   $T=\emptyset$.

2. 2.

   $T=\{\pm\alpha\}$ for an isotropic root $\alpha$.

3. 3.

   $T\subseteq\operatorname{span}(T)$ is an irreducible GRRS containing at least one odd root.

In all cases, $T$ is orthogonal to $T^{\prime}$ and we have both $S\cap\operatorname{span}(T)=T$ and $S\cap\operatorname{span}(T^{\prime})=T^{\prime}$.

Either $S\subseteq\operatorname{span}(S)$ is a GRRS or $S=T^{\prime}\sqcup\{\pm\alpha\}$ where $T^{\prime}\subseteq\operatorname{span}(T^{\prime})$ is an even reduced root system and $\alpha$ is an isotropic root orthogonal to $T^{\prime}$.

Suppose that $\mathfrak{g}$ is a Lie superalgebra such that:

1. 1.

   $\mathfrak{g}_{\overline{0}}$ is reductive and $\mathfrak{g}_{\overline{1}}$ is a semisimple $\mathfrak{g}_{\overline{0}}$-module;

2. 2.

   If $\mathfrak{h}\subseteq\mathfrak{g}_{\overline{0}}$ is a Cartan subalgebra (CSA) of $\mathfrak{g}_{\overline{0}}$, then it is self-centralizing in $\mathfrak{g}$.

3. 3.

   For any root $\alpha$ we have $\operatorname{dim}\mathfrak{g}_{\alpha}\leq 1$.

Then $\theta\in\operatorname{Aut}(\mathfrak{g})$ is semisimple if and only if $\theta|_{\mathfrak{g}_{\overline{0}}}$ is semisimple. In particular, $\theta$ is semisimple if and only if it preserves a Cartan subalgebra of $\mathfrak{g}_{\overline{0}}$.

Suppose that $\mathfrak{g}$ either is a basic simple Lie superalgebra not equal to $\mathfrak{p}\mathfrak{s}\mathfrak{l}(2|2)$ or is $\mathfrak{g}\mathfrak{l}(m|n)$ for some $m,n$ so that $\mathfrak{g}$ satisfies the hypothesis of Lemma 4. Let $\theta\in\operatorname{Aut}(\mathfrak{g})$ be a semisimple automorphism of $\mathfrak{g}$ which preserves a nondegenerate invariant form on $\mathfrak{g}$. We get an orthogonal decomposition $\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{p}$ of super vector spaces, where $\mathfrak{k}$ is the fixed subalgebra of $\theta$, and $\mathfrak{p}$ is the sum of the eigenspaces of $\theta$ with nonzero eigenvalues.

Now suppose $\mathfrak{h}\subseteq\mathfrak{g}_{\overline{0}}$ is a Cartan subalgebra which is $\theta$-invariant. Write $\mathfrak{h}=\mathfrak{t}\oplus\mathfrak{a}$, where $\mathfrak{t}=\mathfrak{k}\cap\mathfrak{h}$ and $\mathfrak{a}=\mathfrak{p}\cap\mathfrak{h}$. Then $\theta$ induces an automorphism of $\mathfrak{h}^{*}$ preserving the set of roots, $R$, and thus induces an automorphism of the GRRS $R\subseteq V=\operatorname{span}(R)$. In the case of $\mathfrak{g}\mathfrak{l}(m|n)$, $R\subseteq\operatorname{span}(R)$ will either be $A(m-1,n-1)$ if $m\neq n$ or $\tilde{A}(n-1,n-1)$ if $m=n\neq 1$, and this is the GRRS we consider. If $m=n=1$, we do not obtain a GRRS but the following will be easy to check in this case anyway.

We keep the notations as above for $S$, $T,$ and $T^{\prime}$. Write $\mathfrak{c}(\mathfrak{a})$ for the centralizer of $\mathfrak{a}$ in $\mathfrak{g}$. Notice that we have $\mathfrak{c}(\mathfrak{a})=\mathfrak{h}+\bigoplus\limits_{\alpha\in S}\mathfrak{g}_{\alpha}$

Define the following subalgebras of $\mathfrak{g}$:

• •

  $\mathfrak{l}$ the subalgebra of $\mathfrak{g}$ generated by $\{e_{\alpha}:\alpha\in T^{\prime}\}$;

• •

  $\tilde{\mathfrak{l}}$ the subalgebra of $\mathfrak{g}$ generated by $\{e_{\alpha}:\alpha\in T\}$.

Then $\mathfrak{l}$ is a semisimple Lie algebra, and $\tilde{\mathfrak{l}}$ either is isomorphic to a basic simple Lie superalgebra, isomorphic to $\mathfrak{s}\mathfrak{l}(n|n)$ for some $n\geq 1$, or is trivial. Further, the natural map

is an injective Lie algebra homomorphism, with image $\operatorname{Im}\iota$ an ideal of $\mathfrak{c}(\mathfrak{a})$, such that $\operatorname{Im}\iota+\mathfrak{t}=\mathfrak{c}(\mathfrak{a})$.

Suppose that $\mathfrak{g}$ is a finite dimensional Kac-Moody Lie superalgebra with indecomposable Cartan matrix (i.e. we remove the case $\mathfrak{g}=\mathfrak{p}\mathfrak{s}\mathfrak{l}(n|n)$). Then $\mathfrak{c}(\mathfrak{a})$ is a product of a reductive Lie algebra with a Kac-Moody Lie superalgebra with an indecomposable Cartan matrix.

Let us now assume that $\mathfrak{g}$ either is basic simple or is $\mathfrak{g}\mathfrak{l}(m|n)$ for some $m,n\in\mathbb{N}$, and that $\theta$ is an involution preserving a chosen nondegenerate invariant form on $\mathfrak{g}$. Then in our decomposition $\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{p}$ we have that $\mathfrak{p}$ is the $(-1)$-eigenspace of $\theta$. Recall that on a Lie superalgebra $\mathfrak{g}=\mathfrak{g}_{\overline{0}}\oplus\mathfrak{g}_{\overline{1}}$ there is a canonical involution $\delta\in\operatorname{Aut}(\mathfrak{g})$ defined by $\delta=\operatorname{id}_{\mathfrak{g}_{\overline{0}}}\oplus(-\operatorname{id}_{\mathfrak{g}_{\overline{1}}})$. This involution is central in $\operatorname{Aut}(\mathfrak{g})$.

If $\theta\neq\operatorname{id}_{\mathfrak{g}},\delta$, then $\mathfrak{p}_{\overline{0}}\neq 0$.

Since we have an involution on $\mathfrak{g}_{\overline{0}}$ preserving the nondegenerate form on it, by classical theory (see for instance Sec. 26 of [Tim11]) we may choose a maximal toral subalgebra $\mathfrak{a}\subseteq\mathfrak{p}_{\overline{0}}$ that can be extended to a $\theta$-invariant Cartan subalgebra of $\mathfrak{g}$, which we will call $\mathfrak{h}$. We obtain a decomposition $\mathfrak{h}=\mathfrak{t}\oplus\mathfrak{a}$, where $\mathfrak{t}$ is the fixed subspace of $\theta$. We again write $\mathfrak{c}(\mathfrak{a})$ for the centralizer of $\mathfrak{a}$ in $\mathfrak{g}$. Notice that $\mathfrak{a}$ is also a maximal toral subalgebra of the $(-1)$-eigenspace of the involution $\delta\circ\theta$.

We already described the structure of $\mathfrak{c}(\mathfrak{a})$ as an algebra in Proposition 4, and in particular we saw that $\mathfrak{c}(\mathfrak{a})_{\overline{1}}=\tilde{\mathfrak{l}}_{\overline{1}}$. Now $\theta$ restricts to an automorphism of $\mathfrak{c}(\mathfrak{a})$ preserving $\tilde{\mathfrak{l}}$, and by classical theory we have $\mathfrak{c}(\mathfrak{a})_{\overline{0}}\cap\mathfrak{p}=\mathfrak{a}$. Since $\mathfrak{a}\cap\tilde{\mathfrak{l}}=0$, by Lemma 5 either $\theta|_{\tilde{\mathfrak{l}}}=\operatorname{id}_{\tilde{\mathfrak{l}}}$ or $\theta|_{\tilde{\mathfrak{l}}}=\delta_{\tilde{\mathfrak{l}}}$.