Correction to: Classification of generalized Wallach spaces

1 Correction to: Geom Dedicata (2016) 111:193–212 https://doi.org/10.1007/s10711-015-0119-z

The paper [2] is devoted to the classification of generalized Wallach spaces. A generalized Wallach space is a homogeneous spaces G/H of a connected compact semisimple Lie group G (H is a compact subgroup of G), such that there is a $\langle \cdot , \cdot \rangle$-orthogonal and ${\text {Ad}}(H)$-invariant decomposition ${\mathfrak {g}}={\mathfrak {h}}\oplus {\mathfrak {p}}_1\oplus {\mathfrak {p}}_2\oplus {\mathfrak {p}}_3$, where ${\mathfrak {g}}$ and ${\mathfrak {h}}$ are Lie algebras of G and H respectively, $\langle \cdot , \cdot \rangle$ is the minus Killing form of ${\mathfrak {g}}$, the modules ${\mathfrak {p}}_i$ are ${\text {Ad}}(H)$-irreducible, $[{\mathfrak {p}}_i,{\mathfrak {p}}_i]\subset {\mathfrak {h}}$ for $i=1,2,3$. Here we will make a correction to the obtained classification. In what follows, we use the notation from [2].

The main result of the paper [2] (Theorem 1) should be stated as follows (in fact, we just add the item (4)):

Theorem 1

Let G/H be a connected and simply connected compact homogeneous space. Then G/H is a generalized Wallach space if and only if it is of one of the following types:

• (1) G/H is a direct product of three irreducible symmetric spaces of compact type $(A=a_1=a_2=a_3=0$ in this case);

• (2) The group G is simple and the pair $({\mathfrak {g}}, {\mathfrak {h}})$ is one of the pairs in Table 1 of [2] (the embedding of ${\mathfrak {h}}$ to ${\mathfrak {g}}$ is determined by the following requirement: the corresponding pairs $({\mathfrak {g}}, {\mathfrak {k}}_i)$ and $({\mathfrak {k}}_i,{\mathfrak {h}})$, $i=1,2,3$, in Table 2 of [2] are symmetric);

• (3) $G=F\times F\times F \times F$ and $H={\text {diag}}(F)\subset G$ for some connected simply connected compact simple Lie group F, with the following description on the Lie algebra level:

  $$\begin{aligned} ({\mathfrak {g}}, {\mathfrak {h}})= \bigl ({\mathfrak {f}}\oplus {\mathfrak {f}}\oplus {\mathfrak {f}}\oplus {\mathfrak {f}},\,{\text {diag}}({\mathfrak {f}})=\{(X,X,X,X)\,|\, X\in f\}\bigr ), \end{aligned}$$

  where ${\mathfrak {f}}$ is the Lie algebra of F, and (up to permutation) ${\mathfrak {p}}_1=\{(X,X,-X,-X)\,|\, X\in f\}$, ${\mathfrak {p}}_2\!=\!\{(X,-X,X,-X)\,|\, X\in f\}$, ${\mathfrak {p}}_3\!=\!\{(X,-X,-X,X)\,|\, X\in f\}$ ($a_1\!=\!a_2\!=\!a_3\!=\!1/4$ in this case).

• (4) $H={{\,\mathrm{diag}\,}}(K)\subset K\times K\subset F\times F=G$, where $({\mathfrak {f}},{\mathfrak {k}})$ is a compact irreducible symmetric pair with simple ${\mathfrak {f}}$ and with simple or one-dimensional ${\mathfrak {k}}$, (up to permutation) ${\mathfrak {p}}_1=\{(X,X)\,|\, X\in {\mathfrak {q}}\}$, ${\mathfrak {p}}_2=\{(X,-X)\,|\, X\in {\mathfrak {q}}\}$, ${\mathfrak {p}}_3=\{(Y,-Y)\,|\, Y\in {\mathfrak {k}}\}$, and ${\mathfrak {q}}$ is the orthogonal complement to ${\mathfrak {k}}$ in ${\mathfrak {f}}$ with respect to the Killing form of the Lie algebra ${\mathfrak {f}}$.

The same item (4) should be added in the statement of Theorem 3 in [2]. The reason for the above correction is the fact that Corollary 2 in [2] is not correct in general, but it is true under some additional assumptions. The correct version of this corollary is as follows.

Corollary 1

If $p \ge 2$ and at least one of the modules ${\mathfrak {p}}_i$, $i=1,2,3$, is situated in some simple ideal ${\mathfrak {g}}_j$ of the Lie algebra ${\mathfrak {g}}$, then $A=0$, consequently, G/H locally is a direct product of three irreducible symmetric spaces of compact type.

Proof

Without loss of generality we may suppose that ${\mathfrak {p}}_1\subset {\mathfrak {g}}_1$, then $[{\mathfrak {p}}_1,{\mathfrak {p}}_2]\subset {\mathfrak {p}}_3 \cap {\mathfrak {g}}_1$ and $[{\mathfrak {p}}_1,{\mathfrak {p}}_3]\subset {\mathfrak {p}}_2 \cap {\mathfrak {g}}_1$. If $[{\mathfrak {p}}_1,{\mathfrak {p}}_2]=0$ or $[{\mathfrak {p}}_1,{\mathfrak {p}}_3]=0$, we get $A=0$. Otherwise, ${\mathfrak {p}}_2,{\mathfrak {p}}_3 \subset {\mathfrak {g}}_1$ (note that all the modules ${\mathfrak {p}}_2$, ${\mathfrak {p}}_3$, ${\mathfrak {p}}_2 \cap {\mathfrak {g}}_1$, and ${\mathfrak {p}}_3 \cap {\mathfrak {g}}_1$ are ${{\,\mathrm{Ad}\,}}(H)$-irreducible), which implies $p=1$. $\square$

This result should be completed with the following proposition (that provides the case (4) for Theorems 1 and 3 in [2]).

Proposition 1

If $p \ge 2$ and no one module ${\mathfrak {p}}_i$, $i=1,2,3$, is in some simple ideal ${\mathfrak {g}}_j$ of ${\mathfrak {g}}$, then $p=2$ and $({\mathfrak {g}}, {\mathfrak {h}})=({\mathfrak {f}}\oplus {\mathfrak {f}}, {{\,\mathrm{diag}\,}}({\mathfrak {k}}))$, where $({\mathfrak {f}},{\mathfrak {k}})$ is a compact irreducible symmetric pair with simple ${\mathfrak {f}}$ and with simple or one-dimensional ${\mathfrak {k}}$. Moreover, up to permutation of indices, we have ${\mathfrak {p}}_1=\{(X,X)\,|\, X\in {\mathfrak {q}}\}$, ${\mathfrak {p}}_2=\{(X,-X)\,|\, X\in {\mathfrak {q}}\}$, ${\mathfrak {p}}_3=\{(Y,-Y)\,|\, Y\in {\mathfrak {k}}\}$, where ${\mathfrak {q}}$ is the orthogonal complement to ${\mathfrak {k}}$ in ${\mathfrak {f}}$ with respect to the Killing form of the Lie algebra ${\mathfrak {f}}$.

Proof

Recall that $\varphi _i({\mathfrak {h}})$ is the $\langle \cdot , \cdot \rangle$-orthogonal projection of ${\mathfrak {h}}$ to ${\mathfrak {g}}_i$. Let ${\mathfrak {q}}_i$ be the $\langle \cdot , \cdot \rangle$-orthogonal complement to $\varphi _i({\mathfrak {h}})$ in ${\mathfrak {g}}_i$, $1\le i \le p$. It is clear that ${\mathfrak {q}}_1 \oplus {\mathfrak {q}}_2 \oplus \cdots \oplus {\mathfrak {q}}_p\subset {\mathfrak {p}}$. Obviously, we have $p\le 3$. If $p=3$, then all ${\mathfrak {q}}_i$, $i=1,2,3$, are ${{\,\mathrm{Ad}\,}}(H)$-irreducible and $\varphi _1({\mathfrak {h}})\oplus \varphi _2({\mathfrak {h}})\oplus \varphi _2({\mathfrak {h}})\subset {\mathfrak {h}}$. Since $[\varphi _i({\mathfrak {h}}),{\mathfrak {q}}_i]\ne 0$ and $[\varphi _i({\mathfrak {h}}),{\mathfrak {q}}_j]=0$ for $i\ne j$, the ${{\,\mathrm{Ad}\,}}(H)$-modules ${\mathfrak {q}}_i$, $i=1,2,3$, are pairwise non-isomorphic, hence, they coincides with the corresponding modules ${\mathfrak {p}}_i$, $i=1,2,3$. By the above corollary we have $A=0$ in this case.

If $p=2$, then there are some isomorphic ${{\,\mathrm{Ad}\,}}(H)$-irreducible submodules ${\mathfrak {q}}_1^\prime \subset {\mathfrak {q}}_1$ and ${\mathfrak {q}}_2^\prime \subset {\mathfrak {q}}_2$. Therefore, by the above arguments, $\varphi _1({\mathfrak {h}})\not \subset {\mathfrak {h}}$ and $\varphi _2({\mathfrak {h}})\not \subset {\mathfrak {h}}$ (otherwise, ${\mathfrak {q}}_1^\prime$ is not isomorphic to ${\mathfrak {q}}_2^\prime$). Hence, ${\mathfrak {h}} \subsetneq \varphi _1({\mathfrak {h}})\oplus \varphi _2({\mathfrak {h}})$, ${\mathfrak {q}}_1^\prime ={\mathfrak {q}}_1$, and ${\mathfrak {q}}_2^\prime = {\mathfrak {q}}_2$. Without loss of generality, we may assume that ${\mathfrak {p}}_1 \oplus {\mathfrak {p}}_2={\mathfrak {q}}_1 \oplus {\mathfrak {q}}_2$, and ${\mathfrak {h}}\oplus {\mathfrak {p}}_3=\varphi _1({\mathfrak {h}})\oplus \varphi _2({\mathfrak {h}})$. Therefore, $\bigl (\varphi _1({\mathfrak {h}})\oplus \varphi _2({\mathfrak {h}}),{\mathfrak {h}}\bigr )$ is a compact irreducible symmetric pair, which has the form $\bigl ({\mathfrak {k}} \oplus {\mathfrak {k}}, {{\,\mathrm{diag}\,}}({\mathfrak {k}})\bigr )$, where ${\mathfrak {k}}$ is a compact simple Lie algebra or ${\mathbb {R}}$ [1, Theorem 7.81]. Hence, $\varphi _1$ and $\varphi _2$ determine Lie algebra isomorphisms between ${\mathfrak {h}}$ and $\varphi _i({\mathfrak {h}})$, $i=1,2$. Let us consider $\theta : \varphi _1({\mathfrak {h}}) \mapsto \varphi _2({\mathfrak {h}})$, such that $\theta =\varphi _2\circ \varphi _1^{-1}$. It is clear that ${\mathfrak {h}}=\{(Y,\theta (Y))\,|\, Y\in \varphi _1({\mathfrak {h}}) \}$.

Now, let us consider the $\langle \cdot , \cdot \rangle$-orthogonal projections $\pi _i:{\mathfrak {p}}_1 \rightarrow {\mathfrak {q}}_i$, $i=1,2$. We may assume that $\pi _1$ is a bijection (otherwise, we can take ${\mathfrak {p}}_2$ instead of ${\mathfrak {p}}_1$). Now, let us consider the ${{\,\mathrm{Ad}\,}}(H)$-equivariant linear map $\psi :=\pi _2\circ \pi _1^{-1}:{\mathfrak {q}}_1 \mapsto {\mathfrak {q}}_2$. We have ${\mathfrak {p}}_1=\{(X,\psi (X))\,|\, X\in {\mathfrak {q}}_1\}$. Since $[{\mathfrak {h}},{\mathfrak {p}}_1]\subset {\mathfrak {p}}_1$ and $[{\mathfrak {p}}_1,{\mathfrak {p}}_1]\subset {\mathfrak {h}}$, we get $[{\mathfrak {q}}_1, {\mathfrak {q}}_1]\subset \varphi _1({\mathfrak {h}})$, $\psi ([Y,X])=[\theta (Y),\psi (X)]$ and $\theta ([X,Z])=[\psi (X),\psi (Z)]$ for every $Y\in \varphi _1({\mathfrak {h}})$ and for every $X,Z \in {\mathfrak {q}}_1$. In particular, $({\mathfrak {q}}_1, \varphi _1({\mathfrak {h}}))$ is a compact irreducible symmetric pair with simple ${\mathfrak {q}}_1$ and and with simple or one-dimensional $\varphi _1({\mathfrak {h}})$.

If we extend the linear map $\psi$ from ${\mathfrak {q}}_1$ to ${\mathfrak {g}}_1$ setting $\psi (X):=\theta (X)$ for any $X\in \varphi _1({\mathfrak {h}})$, we obtain the isomorphism $\psi$ between ${\mathfrak {g}}_1$ and ${\mathfrak {g}}_2$. Indeed, $\psi ([X,Y])=[\psi (X),\psi (Y)]$ for every $X,Y \in {\mathfrak {g}}_1$, $\varphi _2({\mathfrak {h}})\subset \psi ({\mathfrak {g}}_1)$, and ${\mathfrak {g}}_1$ is simple. Therefore, $\psi ({\mathfrak {g}}_1)$ is a simple Lie subalgebra in ${\mathfrak {g}}_2$, and, moreover, $\psi ({\mathfrak {g}}_1)={\mathfrak {g}}_2$, since $\varphi _2({\mathfrak {h}})\subset \psi ({\mathfrak {g}}_1)$ and ${\mathfrak {p}}_2$ is ${{\,\mathrm{ad}\,}}({\mathfrak {h}})$-irreducible. Note that ${\mathfrak {p}}_2=\{(X,-\psi (X))\,|\, X\in {\mathfrak {q}}_1\}$ and ${\mathfrak {p}}_3=\{(Y,-\psi (Y))\,|\, Y\in \varphi _1({\mathfrak {h}})\}$. Therefore, we may consider ${\mathfrak {g}}_2$ as the copy ${\mathfrak {g}}_1$ under the isomorphism $\psi$. The proposition is proved. $\square$

The list of all generalized Wallach spaces of the type as in the Proposition 1 follows directly from the list of compact irreducible symmetric spaces, see e. g. [1, 7.102]. Using structure of symmetric spaces and the Casimir operators for the isotropy representations (see e. g. [1, Chapter 7]), one can easily compute the values A, $a_1$, $a_2$, and $a_3$ (see the formulas (5) and (6) in [2]) for the spaces in Proposition 1: $A=\frac{1}{4} \bigl (\dim ({\mathfrak {f}})- \dim ({\mathfrak {k}})\bigr )=\frac{1}{4} \dim ({\mathfrak {p}}_1)=\frac{1}{4} \dim ({\mathfrak {p}}_2)$, $a_1=\frac{A}{\dim ({\mathfrak {p}}_1)}=a_2=\frac{A}{\dim ({\mathfrak {p}}_2)}=1/4$, and $a_3=\frac{A}{\dim ({\mathfrak {p}}_3)}=\frac{\dim ({\mathfrak {f}})- \dim ({\mathfrak {k}})}{4\dim ({\mathfrak {k}})}\le 1/2$.