Homogeneous Einstein metrics and butterflies

Abstract

In 2012, M. M. Graev associated to a compact homogeneous space G/H a nerve ${\text {X}}_{G/H}$, whose non-contractibility implies the existence of a G-invariant Einstein metric on G/H. The nerve ${\text {X}}_{G/H}$ is a compact, semi-algebraic set, defined purely Lie theoretically by intermediate subgroups. In this paper we present a detailed description of the work of Graev and the curvature estimates given by Böhm in 2004.

Appendix

In this section we will make a brief digression, first into semi-algebraic geometry. We also provide helpful background on the isotropy representation of a compact homogeneous space G/H.

1.1 Semi-algebraic sets

We refer the reader to Benedetti and Risler for more details [7]. A set $X\subset {{{\mathbb {R}}}}^m$ is called semi-algebraic if X is defined by finitely many polynomial equations and inequalities. The inequalities are allowed to be both strict and non-strict.

Given two semi-algebraic sets $X\subset {{{\mathbb {R}}}}^m$ and $Y\subset {{{\mathbb {R}}}}^n$, we say a map $f:X\rightarrow Y$ is semi-algebraic if the graph of f is a semi-algebraic subset of ${{{\mathbb {R}}}}^{m+n}$: see [7, Definition 2.3.2]. A semi-algebraic map is not necessarily continuous, see [7, Example 2.7.3], though some authors include a requirement of continuity in their definition of a semi-algebraic map: see [46, 29, Theorem 7.6], [31, I, Proposition 3.13]. The paradigm for semi-algebraic maps are polynomials.

A fundamental result in semi-algebraic geometry is the following:

Theorem 7.1

(Tarski-Seidenberg) The image of a semi-algebraic subset of ${{{\mathbb {R}}}}^n\times {{{\mathbb {R}}}}^m$ under the projection $\pi :{{{\mathbb {R}}}}^n \times {{{\mathbb {R}}}}^m \rightarrow {{{\mathbb {R}}}}^n$ such that $(x,y)\mapsto x$ is a semi-algebraic set.

The theorem implies that intersections, unions, and complements of semi-algebraic sets are semi-algebraic. Moreover, it is possible to use first-order formulae (in the language of ordered fields with parameters in ${{{\mathbb {R}}}}$) to obtain new semi-algebraic sets from known ones. For instance, given $X_1\subset {{{\mathbb {R}}}}^n,X_2 \subset {{{\mathbb {R}}}}^m$ semi-algebraic sets and a semi-algebraic map $F:{{{\mathbb {R}}}}^n\times {{{\mathbb {R}}}}^m \rightarrow {{{\mathbb {R}}}}^k$, the set

$$\begin{aligned} \{ x_1 \in X_1\mid \exists x_2 \in X_2 \text { s.t. } F(x_1,x_2) \le 0\} \end{aligned}$$

is semi-algebraic. As another example, given endomorphisms A, B of ${{{\mathbb {R}}}}^n$, the condition $\ker (A) \subsetneq \ker (B)$ is a semi-algebraic condition because it is equivalent to

$$\begin{aligned} \{ \forall x \in {{{\mathbb {R}}}}^n \mid A(x)=0 \Rightarrow B(x)=0\} \wedge \{ \exists y \in {{{\mathbb {R}}}}^n \mid B(y)=0 \wedge A(y) \ne 0\}. \end{aligned}$$

Connected components of semi-algebraic sets are semi-algebraic, and the closure, interior, or boundary of a semi-algebraic set is semi-algebraic.

The key property of compact semi-algebraic sets used in Sects. 3.1 and 3.2 is that such sets are absolute neighborhood retracts. The following result states a useful generalization of that.

Lemma 7.2

[39] Let $X,Y \subset {{{\mathbb {R}}}}^N$ be compact, semi-algebraic sets with $X \subset Y$ and $X \ne \emptyset$. Suppose that for some $\delta _0>0$, for all $\delta \in (0,\delta _0)$ there exists a continuous map

$$\begin{aligned} H_\delta :[0,1]\times Y \rightarrow Y \end{aligned}$$

with the following properties: $H_\delta (0,y)=y$ for all $y \in Y$, $H_\delta (t,x)=x$ for all $x \in X$ and all $t \in [0,1]$, and

$$\begin{aligned} d(H_\delta (1,y),X)< \delta \text { for all } y \in Y. \end{aligned}$$

Then X is a strong deformation retract of Y.

Proof

It follows from Theorem 1 of [30], see also [31, III, Theorem 1.1], that there exists an open semi-algebraic neighborhood U of X in Y and a semi-algebraic, continuous map $G:[0,1]\times {\overline{U}}\rightarrow {\overline{U}}$, such that the restriction $G\vert _{[0,1]\times U}$ yields a strong deformation retraction from U to X.

By compactness of X, there exists $\delta _0>0$ such that for all $0<\delta \le \delta _0$, we have

$$\begin{aligned} T_\delta (X):=\{y \in Y \mid d(y,X)<\delta \} \subset U. \end{aligned}$$

Hence we can define

$$\begin{aligned} H:[0,1]\times Y \rightarrow Y \text { s.t. } (t,y) \mapsto \left\{ \begin{array}{ll} H_{\delta }(2t,y) &{} \text { for } 0\le t \le \tfrac{1}{2} \\ G(2t-1,H_\delta (1,y)) &{}\text { for } \tfrac{1}{2}\le t \le 1. \end{array} \right. \end{aligned}$$

The function H is continuous since for $t=\tfrac{1}{2}$ we have $G(2t-1,H_\delta (1,y))=H_{\delta }(2t,y)$. Moreover $H(t,x)=x$ for all $t \in [0,1]$ and all $x \in X$ and $H(1,Y)=X$. This shows the claim. $\square$

1.2 Lie-theory I

In this section we provide details concerning the isotropy representation of a compact homogeneous space G/H and the normalizer $N_G(H)$ of H in G. We assume, as we have throughout the paper, that G and H are connected.

Definition 7.3

For any compact homogeneous space G/H, we denote by ${{{\mathfrak {m}}}}_0$ the subspace of ${{{\mathfrak {m}}}}$ on which ${\text {Ad}}(H)$ acts trivially.

Notice that if $\text {rk} \,G=\text {rk} \,H$, then ${{{\mathfrak {m}}}}_0 = \{0\}$, while if $H=\{e\}$, ${{{\mathfrak {m}}}}_0={{{\mathfrak {g}}}}$. Recall also that

$$\begin{aligned} \exp ({\text {ad}}(X))={\text {Ad}}(\exp _G(X)) \end{aligned}$$

(7.1)

for all $X \in {{{\mathfrak {g}}}}$, where $\exp _G:{{{\mathfrak {g}}}}\rightarrow G$ denotes the exponential of G and $\exp (D)=\sum _{k=0}^\infty \tfrac{D^k}{k!}$ for any endomorphism $D:{{{\mathfrak {g}}}}\rightarrow {{{\mathfrak {g}}}}$ [44, II].

Lemma 7.4

A subspace $\tilde{{{\mathfrak {m}}}}$ of ${{{\mathfrak {m}}}}$ is ${\text {Ad}}(H)$-invariant if and only if $[{{{\mathfrak {h}}}},\tilde{{{{\mathfrak {m}}}}}]\subset \tilde{{{{\mathfrak {m}}}}}$. Moreover, $[{{{\mathfrak {h}}}},{{{\mathfrak {m}}}}_0]=0$ and for any ${\text {Ad}}(H)$-invariant subspace ${\tilde{{{{\mathfrak {m}}}}}}$ of ${{{\mathfrak {m}}}}\ominus {{{\mathfrak {m}}}}_0$, we have $\{0\} \ne [{{{\mathfrak {h}}}},{\tilde{{{{\mathfrak {m}}}}}}]$.

Proof

If $\tilde{{{{\mathfrak {m}}}}}$ is ${{\text {Ad}}}(H)$-invariant, then differentiation yields $[{{{\mathfrak {h}}}},\tilde{{{{\mathfrak {m}}}}}]\subset \tilde{{{{\mathfrak {m}}}}}$. Conversely, suppose we know $[{{{\mathfrak {h}}}},\tilde{{{{\mathfrak {m}}}}}]\subset \tilde{{{{\mathfrak {m}}}}}$. Since H is connected, for any $Z \in {{{\mathfrak {h}}}}$, ${{\text {Ad}}}(\exp (Z))=\exp ({{\text {ad}}}(Z))$. Hence, $\tilde{{{{\mathfrak {m}}}}}$ is ${{\text {Ad}}}(H)$-invariant. This shows the first claim. The second and the third claim follow immediately. $\square$

Let ${{{\mathfrak {n}}}}({{{\mathfrak {h}}}})$ denote the Lie algebra of the normalizer $N_G(H)$ of H in G.

Lemma 7.5

Let G/H be a compact homogeneous space. Then ${{{\mathfrak {n}}}}({{{\mathfrak {h}}}})= {{{\mathfrak {h}}}}\oplus {{{\mathfrak {m}}}}_0$. Moreover, ${{{\mathfrak {m}}}}_0$ is a compact subalgebra.

Proof

Since $[{{{\mathfrak {h}}}},{{{\mathfrak {m}}}}_0]=0 \in {{{\mathfrak {h}}}}$, see above, we conclude ${{{\mathfrak {m}}}}_0 \subset {{{\mathfrak {n}}}}({{{\mathfrak {h}}}})$. Conversely, suppose that $\tilde{{{\mathfrak {m}}}}\subset {{{\mathfrak {n}}}}({{{\mathfrak {h}}}}) \cap {{{\mathfrak {m}}}}$. Then on the one hand, $[{{{\mathfrak {h}}}}, \tilde{{{\mathfrak {m}}}}]\subset {{{\mathfrak {h}}}}$ by definition of ${{{\mathfrak {n}}}}({{{\mathfrak {h}}}})$, while on the other hand, $[{{{\mathfrak {h}}}}, \tilde{{{\mathfrak {m}}}}] \subset {{{\mathfrak {m}}}}$ because $[{{{\mathfrak {h}}}},{{{\mathfrak {m}}}}]\subset {{{\mathfrak {m}}}}$. This shows $[{{{\mathfrak {h}}}}, \tilde{{{\mathfrak {m}}}}]=0$, and thus ${{{\mathfrak {m}}}}\cap {{{\mathfrak {n}}}}({{{\mathfrak {h}}}})\subset {{{\mathfrak {m}}}}_0$, proving our equality.

Since $[{{{\mathfrak {m}}}}_0,{{{\mathfrak {m}}}}_0] \subset {{{\mathfrak {n}}}}({{{\mathfrak {h}}}})={{{\mathfrak {h}}}}\oplus {{{\mathfrak {m}}}}_0$ and $[{{{\mathfrak {h}}}},{{{\mathfrak {m}}}}_0]=0$, then because Q is self-adjoint (4.3), we know that ${{{\mathfrak {m}}}}_0$ is a subalgebra of ${{{\mathfrak {n}}}}({{{\mathfrak {h}}}})$ and thus is a subalgebra of ${{{\mathfrak {g}}}}$. In order to show that the corresponding connected subgroup of G is compact we now use the special property of the biinvariant metric Q chosen in Sect. 2.1. We decompose the compact Lie algebra ${{{\mathfrak {n}}}}({{{\mathfrak {h}}}})={{{\mathfrak {h}}}}\oplus {{{\mathfrak {m}}}}_0$ into the Q-orthogonal sum of its semisimple part $({{{\mathfrak {h}}}}\oplus {{{\mathfrak {m}}}}_0)_s= {{{\mathfrak {h}}}}_s \oplus ({{{\mathfrak {m}}}}_0)_s$ and its center ${{{\mathfrak {z}}}}({{{\mathfrak {h}}}}\oplus {{{\mathfrak {m}}}}_0)$. Each of the Lie algebras ${{{\mathfrak {h}}}}_s$, $({{{\mathfrak {m}}}}_0)_s$ and ${{{\mathfrak {z}}}}({{{\mathfrak {h}}}}\oplus {{{\mathfrak {m}}}}_0)$ is compact, and so ${{{\mathfrak {h}}}}\cap {{{\mathfrak {z}}}}({{{\mathfrak {h}}}}\oplus {{{\mathfrak {m}}}}_0)$ is also compact. It remains to show that the Q-orthogonal complement of ${{{\mathfrak {h}}}}\cap {{{\mathfrak {z}}}}({{{\mathfrak {h}}}}\oplus {{{\mathfrak {m}}}}_0)$ in ${{{\mathfrak {z}}}}({{{\mathfrak {h}}}}\oplus {{{\mathfrak {m}}}}_0)$ is compact.

By the definition of Q, the compact, abelian subalgebras ${{{\mathfrak {h}}}}\cap {{{\mathfrak {z}}}}({{{\mathfrak {h}}}}\oplus {{{\mathfrak {m}}}}_0)$ and ${{{\mathfrak {z}}}}({{{\mathfrak {h}}}}\oplus {{{\mathfrak {m}}}}_0)$ can be viewed as subalgebras of ${{{\mathfrak {s}} {\mathfrak {o}}}}(6N)$. Notice we can choose a Q-orthonormal (standard) basis $({\hat{e}}_1,\ldots ,{\hat{e}}_{3N})$ for the standard maximal (diagonal) torus ${{{\mathfrak {t}}}}_{6N}$ of ${{{\mathfrak {s}} {\mathfrak {o}}}}(6N)$ so that every compact subtorus ${{{\mathfrak {t}}}}$ of ${{{\mathfrak {t}}}}_{6N}$ has a basis $(e_1,\ldots ,e_{\dim {{{{\mathfrak {t}}}}}})$, for which each $e_i$ is a rational linear combination of the above basis elements $({\hat{e}}_i)$ of ${{{\mathfrak {s}} {\mathfrak {o}}}}(6N)$. Then, the Q-orthogonal complement in ${{{\mathfrak {t}}}}_{6N}$ of the a subtorus ${{{\mathfrak {t}}}}$ must also have such a basis, and hence it corresponds to a compact subtorus in ${{{\mathfrak {s}} {\mathfrak {o}}}}(6N)$. As a consequence, the Q-orthogonal complement of ${{{\mathfrak {h}}}}\cap {{{\mathfrak {z}}}}({{{\mathfrak {h}}}}\oplus {{{\mathfrak {m}}}}_0)$ in ${{{\mathfrak {z}}}}({{{\mathfrak {h}}}}\oplus {{{\mathfrak {m}}}}_0)$ is the intersection of two compact subalgebras, thus a compact subalgebra. $\square$

Remark 7.6

We decompose ${{{\mathfrak {m}}}}$ into its ${\text {Ad}}(H)$-invariant isotypical summands (see [21, II, Proposition 6.9]):

$$\begin{aligned} {{{\mathfrak {m}}}}=\bigoplus _{i=1}^{\ell _{\text {iso} }}\,{{{\mathfrak {p}}}}_i . \end{aligned}$$

Each ${{{\mathfrak {p}}}}_i$ is a direct sum of ${\text {Ad}}(H)$-irreducible summands which are equivalent (as ${\text {Ad}}(H)$-representations), while, for $i \ne j$, irreducible summands in ${{{\mathfrak {p}}}}_i$ and ${{{\mathfrak {p}}}}_j$ are inequivalent. Note by Lemma 7.4, if ${{{\mathfrak {m}}}}_0$ is non-trivial, then ${{{\mathfrak {m}}}}_0$ is one of the isotypical summands of ${{{\mathfrak {m}}}}$.

Thus by Schur’s Lemma, for every $P_g \in \mathcal {M}^G$ we have

$$\begin{aligned} Q(P_g\,\cdot ,\,\cdot \,)= Q((P_g)_1\,\cdot ,\,\cdot \,)\vert _{{{{\mathfrak {p}}}}_1} \perp \cdots \perp Q((P_g)_{\ell _{\text {iso} }}\,\cdot ,\,\cdot \,) \vert _{{{{\mathfrak {p}}}}_{\ell _{\text {iso} }}} \end{aligned}$$

(7.2)

where, for each $1 \le i \le \ell _{\text {iso} }$, $(P_g)_i:{{{\mathfrak {p}}}}_i \rightarrow {{{\mathfrak {p}}}}_i$ is an ${\text {Ad}}(H)$-equivariant, self-adjoint, positive definite endomorphism. Consequently, every $P_g \in \mathcal {M}^G$ respects the decomposition ${{{\mathfrak {m}}}}=\oplus _{i=1}^{\ell _{\text {iso} }}\,{{{\mathfrak {p}}}}_i$. Notice this is also true for every $v \in \text {Sym} _{{{{\mathfrak {m}}}}}^{H}$.

For ${{{\mathfrak {k}}}}$ a subalgebra in $\mathcal {K}$, let ${{{\mathfrak {m}}}}_{{{{\mathfrak {k}}}}}:={{{\mathfrak {m}}}}\cap {{{\mathfrak {k}}}}$. For convenience, ${{{\mathfrak {m}}}}_{{{{{\mathfrak {h}}}}}}:= {{{\mathfrak {m}}}}$.

Lemma 7.7

Let ${{{\mathfrak {k}}}}$ be a subalgebra, ${{{\mathfrak {h}}}}\le {{{\mathfrak {k}}}}< {{{\mathfrak {g}}}}$, let K be the connected subgroup of G corresponding to ${{{\mathfrak {k}}}}$, and let $A \in \text {Sym} _{{{{\mathfrak {g}}}}}^{H}$ with ${{{\mathfrak {k}}}}\subset \ker (A)$. Then, the condition $[A,{\text {ad}}({{{\mathfrak {k}}}})]=0$ is equivalent to the ${\text {Ad}}(K)$-equivariance of A. Furthermore, $[A,{\text {ad}}({{{\mathfrak {k}}}})]=0$ if and only if $[A\vert _{{{{\mathfrak {k}}}}^\perp },{\text {ad}}({{{\mathfrak {m}}}}_{{{{\mathfrak {k}}}}})\vert _{{{{\mathfrak {k}}}}^\perp }]=0$ if and only if $[A\vert _{{{{{\mathfrak {m}}}}}},{\text {ad}}({{{\mathfrak {m}}}}_{{{{\mathfrak {k}}}}})\vert _{{{{{\mathfrak {m}}}}}}]=0$.

Proof

Let K be the connected subgroup of G with Lie algebra ${{{\mathfrak {k}}}}$. We know by differentiation that ${\text {Ad}}(K)$-equivariance of A implies $[A,{\text {ad}}({{{\mathfrak {k}}}})]=0$. To see the converse, let $A \in \text {Sym} _{{{{\mathfrak {g}}}}}^{H}$ and suppose that $[A,{\text {ad}}({{{\mathfrak {k}}}})]=0$. Using (7.1) shows that A is ${\text {Ad}}(K)$-equivariant.

Next, let $A \in \text {Sym} _{{{{\mathfrak {g}}}}}^{H}$ with ${{{\mathfrak {k}}}}\subset \ker (A)$. Since ${\text {Ad}}(K)$, ${\text {ad}}({{{\mathfrak {k}}}})$ respect the decomposition ${{{\mathfrak {g}}}}= {{{\mathfrak {k}}}}\oplus {{{\mathfrak {k}}}}^\perp$, it follows that $A\vert _{{{{\mathfrak {k}}}}^\perp }$ is ${\text {Ad}}(K)\vert _{{{{\mathfrak {k}}}}^\perp }$-equivariant if and only if $[A\vert _{{{{\mathfrak {k}}}}^\perp },{\text {ad}}({{{\mathfrak {k}}}})\vert _{{{{\mathfrak {k}}}}^\perp }]=0$.

Since $A \in \text {Sym} _{{{{\mathfrak {g}}}}}^{H}$ with ${{{\mathfrak {k}}}}\subset \ker (A)$ is ${\text {Ad}}(K)$-equivariant if and only if $A\vert _{{{{\mathfrak {k}}}}^\perp }$ is ${\text {Ad}}(K)\vert _{{{{\mathfrak {k}}}}^\perp }$-equivariant we deduce that $[A,{\text {ad}}({{{\mathfrak {k}}}})]=0$ is equivalent to $[A\vert _{{{{\mathfrak {k}}}}^\perp },{\text {ad}}({{{\mathfrak {k}}}})\vert _{{{{\mathfrak {k}}}}^\perp }]=0$.

Since for $A \in \text {Sym} _{{{{\mathfrak {g}}}}}^{H}$ with ${{{\mathfrak {k}}}}\subset \ker (A)$ A preserves ${{{\mathfrak {m}}}}$ we deduce from $[A,{\text {ad}}({{{\mathfrak {k}}}})]=0$ that $[A\vert _{{{{{\mathfrak {m}}}}}},{\text {ad}}({{{\mathfrak {k}}}})\vert _{{{{{\mathfrak {m}}}}}}]=0$, which implies $[A\vert _{{{{{\mathfrak {m}}}}}},{\text {ad}}({{{\mathfrak {m}}}}_{{{{\mathfrak {k}}}}})\vert _{{{{{\mathfrak {m}}}}}}]=0$ since ${{{\mathfrak {m}}}}_{{{{\mathfrak {k}}}}}\subset {{{\mathfrak {k}}}}$.

Suppose now $[A\vert _{{{{{\mathfrak {m}}}}}},{\text {ad}}({{{\mathfrak {m}}}}_{{{{\mathfrak {k}}}}})\vert _{{{{{\mathfrak {m}}}}}}]=0$. All elements of $\text {Sym} _{{{{\mathfrak {g}}}}}^{H}$ are ${\text {Ad}}(H)$-equivariant, by definition, thus $[A\vert _{{{{{\mathfrak {m}}}}}},{\text {ad}}({{{\mathfrak {h}}}})\vert _{{{{{\mathfrak {m}}}}}}]=0$. Using ${{{\mathfrak {k}}}}={{{\mathfrak {h}}}}\oplus {{{\mathfrak {m}}}}_{{{{\mathfrak {k}}}}}$ we conclude $[A\vert _{{{{{\mathfrak {m}}}}}},{\text {ad}}({{{\mathfrak {k}}}})\vert _{{{{{\mathfrak {m}}}}}}]=0$. Since ${\text {ad}}({{{\mathfrak {k}}}})$ preserves the decomposition ${{{\mathfrak {g}}}}={{{\mathfrak {k}}}}\oplus {{{\mathfrak {k}}}}^\perp$ it follows that ${\text {ad}}({{{\mathfrak {k}}}})\vert _{{{{{\mathfrak {m}}}}}}$ preserves the decomposition ${{{\mathfrak {m}}}}={{{\mathfrak {m}}}}_{{{{\mathfrak {k}}}}} \oplus {{{\mathfrak {k}}}}^\perp$. We deduce $[A\vert _{{{{\mathfrak {k}}}}^\perp },{\text {ad}}({{{\mathfrak {k}}}})\vert _{{{{\mathfrak {k}}}}^\perp }]=0$. Using that for $A \in \text {Sym} _{{{{\mathfrak {g}}}}}^{H}$ with ${{{\mathfrak {k}}}}\subset \ker (A)$, this is equivalent to $[A,{\text {ad}}({{{\mathfrak {k}}}})]=0$, proving the claim. $\square$

Any such A induces a submersion metric on G/H with respect to $K/H \rightarrow G/H \rightarrow G/K$. Notice however, that K need not be a compact subgroup of G.

Corollary 7.8

Let $A \in \text {Sym} _{{{{\mathfrak {g}}}}}^{H}$ ($v \in {{\text {S}}}$) and let ${{{\mathfrak {m}}}}_{I_1},\dots , {{{\mathfrak {m}}}}_{I_\ell } \subset {{{\mathfrak {m}}}}$ denote the eigenspaces of $A\vert _{{{{\mathfrak {m}}}}}$ corresponding to distinct eigenvalues in increasing order.

• (i) Then $[{{{\mathfrak {m}}}}_{I_1},{{{\mathfrak {m}}}}_{I_1}]\perp {{{\mathfrak {m}}}}_{I_p}$ (i.e. $[I_1 I_1 I_p]=0$) for all $1<p \le \ell$ if and only if ${{{\mathfrak {k}}}}= {{{\mathfrak {h}}}}\oplus {{{\mathfrak {m}}}}_{I_1}$ is a subalgebra.

• (ii) Furthermore, when ${{{\mathfrak {k}}}}= {{{\mathfrak {h}}}}\oplus {{{\mathfrak {m}}}}_{I_1}$ is a subalgebra, $A \in {\text {D}}({{{\mathfrak {k}}}})$ if and only if ${{{\mathfrak {k}}}}\le \ker (A)$ and for all $1 \le p \ne q \le \ell$, we have $[{{{\mathfrak {m}}}}_{I_1},{{{\mathfrak {m}}}}_{I_p}]\perp {{{\mathfrak {m}}}}_{I_q}$ (i.e. $[I_1 I_p I_q]=0$).

Proof

(i) If for all $1 < p$ we have $[{{{\mathfrak {m}}}}_{I_1},{{{\mathfrak {m}}}}_{I_1}]\perp {{{\mathfrak {m}}}}_{I_p}$, then since $[{{{\mathfrak {h}}}},{{{\mathfrak {m}}}}_{I_1}] \subset {{{\mathfrak {m}}}}_{I_1}$ we have that ${{{\mathfrak {k}}}}= {{{\mathfrak {h}}}}\oplus {{{\mathfrak {m}}}}_{I_1}$ is a subalgebra of ${{{\mathfrak {g}}}}$. Conversely, if ${{{\mathfrak {k}}}}= {{{\mathfrak {h}}}}\oplus {{{\mathfrak {m}}}}_{I_1}\in \mathcal {K}$ is a subalgebra, then we know $[{{{\mathfrak {m}}}}_{I_1},{{{\mathfrak {m}}}}_{I_1}] \subset {{{\mathfrak {k}}}}\perp {{{\mathfrak {m}}}}_{I_p}$ for all $1<p$.

(ii) Suppose ${{{\mathfrak {k}}}}= {{{\mathfrak {h}}}}\oplus {{{\mathfrak {m}}}}_{I_1}\in \mathcal {K}$ is a subalgebra, and $A \in {\text {D}}({{{\mathfrak {k}}}})$ or equivalently, $v \in {\text {D}}({{{\mathfrak {k}}}})^{{\text {S}}}$. Since $A \in {\text {D}}({{{\mathfrak {k}}}})$ we know $[A,{\text {ad}}({{{\mathfrak {k}}}})]=0$. Thus, by Lemma 7.7 we deduce $[A\vert _{{{{\mathfrak {k}}}}^\perp },{\text {ad}}({{{\mathfrak {m}}}}_{{{{\mathfrak {k}}}}})\vert _{{{{\mathfrak {k}}}}^\perp }]=0$. This shows that ${\text {ad}}({{{\mathfrak {m}}}}_{{{{\mathfrak {k}}}}})$ respects the eigenspaces of $A\vert _{{{{{\mathfrak {m}}}}}}$. That is, $[{{{\mathfrak {m}}}}_{I_1}, {{{\mathfrak {m}}}}_{I_p}] \subset {{{\mathfrak {m}}}}_{I_p} \perp {{{\mathfrak {m}}}}_{I_q}$ for $p \ne q$. Conversely, suppose ${{{\mathfrak {k}}}}= {{{\mathfrak {h}}}}\oplus {{{\mathfrak {m}}}}_{I_1} \le \ker (A)$ and for all $p \ne q$, we have $[{{{\mathfrak {m}}}}_{I_1},{{{\mathfrak {m}}}}_{I_p}]\perp {{{\mathfrak {m}}}}_{I_q}$. Thus ${\text {ad}}({{{\mathfrak {k}}}})\vert _{{{{\mathfrak {k}}}}^\perp }$ preserves the eigenspaces of $A\vert _{{{{\mathfrak {k}}}}^\perp }$. By Lemma 7.7, to see that $A \in {\text {D}}({{{\mathfrak {k}}}})$ we need that $[A\vert _{{{{\mathfrak {k}}}}^\perp },{\text {ad}}({{{\mathfrak {k}}}})\vert _{{{{\mathfrak {k}}}}^\perp }]=0$ which follows from $[A|_{{{{\mathfrak {m}}}}_{I_p}},{\text {ad}}({{{\mathfrak {m}}}}_{I_1})|_{{{{\mathfrak {m}}}}_{I_p}}] =0$ for every $p>1$. Let $X_1 \in {{{\mathfrak {m}}}}_{I_i}$ and $X_p \in {{{\mathfrak {m}}}}_{I_p}$, $p>1$. Then

$$\begin{aligned} {[}A|_{{{{\mathfrak {m}}}}_{I_p}},{\text {ad}}(X_1)|_{{{{\mathfrak {m}}}}_{I_p}}](X_p) = A[X_1,X_p] - [X_1,A X_p]=0, \end{aligned}$$

since by hypothesis, $[{{{\mathfrak {m}}}}_{I_1},{{{\mathfrak {m}}}}_{I_p}] \subset {{{\mathfrak {m}}}}_{I_p}$. Thus our equality holds. $\square$

Here we provide a brief overview of representation theory of compact Lie groups K. Let V be a vector space over ${{{\mathbb {C}}}}$. We call a representation $\rho _{{{{{\mathbb {C}}}}}}:K \rightarrow {\text {U}}(V) \subset \textrm{Gl}(V,{{{\mathbb {C}}}})$ an irreducible complex representation if there exists no proper, non-trivial ${{{\mathbb {C}}}}$-vector subspace ${\tilde{V}}$ of V which is $\rho _{{{{{\mathbb {C}}}}}}(K)$-invariant. The realization of an irreducible complex representation $\rho _{{{{{\mathbb {C}}}}}}:K \rightarrow {\text {U}}(V,{{{\mathbb {C}}}})$ is given by $\rho _{{{{{\mathbb {R}}}}}}:K\rightarrow {\text {SO}}(W) \subset \text {Gl} (W,{{{\mathbb {R}}}})\,;\,\, k \mapsto \rho _{{{{{\mathbb {C}}}}}}(k)$, where W is the ${{{\mathbb {R}}}}$-vector space induced by V; that is, $W=V$ as sets, but $\dim _{{{{{\mathbb {R}}}}}} W =2 \dim _{{{{{\mathbb {C}}}}}} V$. Such representations are called real representations. Notice that the realization of an irreducible complex representation may not be irreducible.

An irreducible real representation $\rho _{{{{{\mathbb {R}}}}}}:K \rightarrow {\text {SO}}(W)$ is called of real type, complex type or quaternionic type, respectively, if the group of intertwining operators

$$\begin{aligned} G(W,K):=\{ A \in \text {End} (W,{{{\mathbb {R}}}}) \mid A \cdot \rho _{{{{{\mathbb {R}}}}}}(k) = \rho _{{{{{\mathbb {R}}}}}}(k) \cdot A \text { for all } k\in K\} \end{aligned}$$

is isomorphic to ${{{\mathbb {R}}}}$, ${{{\mathbb {C}}}}$ or ${{{\mathbb {H}}}}$, respectively: see [21, II, 6.2, II, Theorem 6.7].

Lemma 7.9

Let $\rho _{{{{{\mathbb {C}}}}}}:{\text {SU}}(n)\rightarrow {\text {U}}(V,{{{\mathbb {C}}}})$ be an irreducible complex representation, $n \ge 2$. Suppose that $\rho _{{{{{\mathbb {C}}}}}}$ can be extended to a representation of ${\text {U}}(n)$ such that the center of ${\text {U}}(n)$ acts non-trivially on V by multiples of the identity. Then the realization $\rho _{{{{{\mathbb {R}}}}}}:{\text {U}}(n)\rightarrow {\text {SO}}(W)$, $W=V$, is an irreducible real representation.

Proof

For a contradiction, suppose that ${\tilde{W}}$ is a non-trivial proper real subspace of W which is $\rho _{{{{{\mathbb {R}}}}}}({\text {U}}(n))$-invariant. By assumption there exists some $\alpha \in {{{\mathbb {R}}}}$ such that for every $e^{i\varphi } \cdot I_n$ in the center of ${\text {U}}(n)$, we have $\rho _{{{{{\mathbb {C}}}}}}(e^{i\varphi } \cdot I_n)=e^{i\varphi \alpha }\cdot I_V$. Thus whenever ${\tilde{w}} \in {\tilde{W}}$, then $i\cdot {\tilde{w}} \in {\tilde{W}}$ as well. Consequently ${\tilde{W}}$ is an ${\text {SU}}(n)$-invariant complex subspace of $W=V$, contradicting our hypothesis. Consequently W is an irreducible real representation. $\square$

1.3 Lie-theory II

In this section we describe the Casimir operator of the isotropy representation of a compact homogeneous space G/H and its relation to the Killing form and the structure constants.

Let $(Z_m)_{1\le m \le \dim H}$ denote a Q-orthonormal basis of ${{{\mathfrak {h}}}}$. Let ${{{\mathfrak {m}}}}_i$ be an ${\text {Ad}}(H)$-irreducible summand of ${{{\mathfrak {m}}}}$. The Casimir operator on ${{{\mathfrak {m}}}}_i$ is given by

$$\begin{aligned} C_{{{{{\mathfrak {m}}}}}_i,Q\vert _{{{{{\mathfrak {h}}}}}}}:=-\sum _{m}{{\text {ad}}}(Z_m)\circ {{\text {ad}}}(Z_m):{{{\mathfrak {m}}}}_i \rightarrow {{{\mathfrak {m}}}}_i \end{aligned}$$

and it satisfies

$$\begin{aligned} C_{{{{{\mathfrak {m}}}}}_i,Q\vert _{{{{{\mathfrak {h}}}}}}}=c_i\cdot {{\text {Id}}}_{{{{{\mathfrak {m}}}}}_i}. \end{aligned}$$

A short computation using (4.3) shows $c_i\ge 0$. Clearly, $c_i=0$ if and only if ${{{\mathfrak {m}}}}_i \subset {{{\mathfrak {m}}}}_0$: see Lemma 7.5.

Recall, the definitions of $d_i$, $b_i$ and [ijk] can be found in Sect. 4.1 and that these numbers depend on the decomposition $f=\oplus _{i=1}^\ell {{{\mathfrak {m}}}}_i$ of ${{{\mathfrak {m}}}}$ choosen.

Lemma 7.10

[77] Let G/H be a compact homogeneous space and let f be a decomposition of ${{{\mathfrak {m}}}}$. Then for all $1 \le i \le \ell$,

$$\begin{aligned} d_ib_i-\tfrac{1}{2}\sum _{j,k=1}^\ell [ijk] =2d_ic_i+\tfrac{1}{2}\sum _{j,k=1}^\ell [ijk] \ge 0. \end{aligned}$$

Moreover, the expression is zero if and only if ${{{\mathfrak {m}}}}_i \subset {{{\mathfrak {z}}}}({{{\mathfrak {g}}}})$.

Proof

The equality is proved in [77, La.1.5].

Suppose that $i=1$ and the left-hand side is zero. Then $c_1=0$, thus $[{{{\mathfrak {h}}}},{{{\mathfrak {m}}}}_1]=0$ and ${{{\mathfrak {m}}}}_1 \subset {{{\mathfrak {m}}}}_0$ by Lemma 7.4. Since $[11j]=0$ for all $1\le j\le \ell$, the module ${{{\mathfrak {m}}}}_1$ is an abelian subalgebra of ${{{\mathfrak {m}}}}_0$. Moreover, since for all $2 \le j,k\le \ell$ we have $[1jk]=0$, the subspace ${{{\mathfrak {g}}}}':={{{\mathfrak {h}}}}\oplus (\bigoplus _{j=2}^\ell {{{\mathfrak {m}}}}_j)$ is a subalgebra of ${{{\mathfrak {g}}}}$ commuting with ${{{\mathfrak {m}}}}_1$: see Lemma 5.21. It follows that ${{{\mathfrak {m}}}}_1 \subset {{{\mathfrak {z}}}}({{{\mathfrak {g}}}})$.

The converse direction is clear. $\square$

Lemma 7.11

Let ${{{\mathfrak {k}}}}\in \mathcal {K}$ and let f be a decomposition of ${{{\mathfrak {m}}}}$ such that ${{{\mathfrak {k}}}}$ is f-adapted, and let $I_1=I^{{{{\mathfrak {k}}}}}_1$. Then

$$\begin{aligned} a_{{{{\mathfrak {k}}}}}:=\sum _{j\in I_1}d_jc_j+\tfrac{1}{4}[I_1I_1I_1] \ge 0. \end{aligned}$$

Furthermore, ${{{\mathfrak {k}}}}$ is toral if and only if $a_{{{{\mathfrak {k}}}}} =0$.

Proof

Suppose that ${{{\mathfrak {k}}}}={{{\mathfrak {h}}}}\oplus {{{\mathfrak {a}}}}$ is a toral subalgebra of ${{{\mathfrak {g}}}}$, that is, ${{{\mathfrak {a}}}}\subset {{{\mathfrak {m}}}}_0$. Then clearly $a_{{{{\mathfrak {k}}}}}=0$. Conversely, when $a_{{{{\mathfrak {k}}}}}=0$, then $\sum _{j\in I_1}d_jc_j=0$, so we know ${{{\mathfrak {m}}}}_{I_1}:={{{\mathfrak {m}}}}_{{{{\mathfrak {k}}}}}\subset {{{\mathfrak {m}}}}_0$. Because ${{{\mathfrak {k}}}}$ is a subalgebra we conclude $[{{{\mathfrak {m}}}}_{I_1},{{{\mathfrak {m}}}}_{I_1}]\subset {{{\mathfrak {m}}}}_{I_1}$. Since $[I_1I_1I_1]=0$, the claim follows. $\square$

Corollary 7.12

For every a compact homogeneous space G/H, there exists a constant $n_{G/H}>0$ such that the following holds: given any non-toral subalgebra ${{{\mathfrak {k}}}}$, with f a decomposition of ${{{\mathfrak {m}}}}$ such that ${{{\mathfrak {k}}}}$ is f-adapted, and letting $I_1=I^{{{{\mathfrak {k}}}}}_1$, then

$$\begin{aligned} a_{{{{\mathfrak {k}}}}}=\sum _{j\in I_1}d_j c_j+\tfrac{1}{4}[I_1I_1I_1] \ge n_{G/H}. \end{aligned}$$

Proof

For a contradiction, suppose that there exists a sequence of non-toral subalgebras $({{{\mathfrak {k}}}}_\alpha )_{\alpha \in {{{\mathbb {N}}}}}$ and a sequence $(f_\alpha )_{\alpha \in {{{\mathbb {N}}}}}$ of decompositions of ${{{\mathfrak {m}}}}$, such that for each $\alpha \in {{{\mathbb {N}}}}$, ${{{\mathfrak {k}}}}_\alpha$ is $f_\alpha$-adapted, with $a_{{{{\mathfrak {k}}}}_\alpha } \rightarrow 0$ as $\alpha \rightarrow \infty$. Because the Casimir operator $C\vert _{{{{\mathfrak {m}}}}}:{{{\mathfrak {m}}}}\rightarrow {{{\mathfrak {m}}}}$ is non-negative with kernel ${{{\mathfrak {m}}}}_0$, for $\alpha$ large we must have ${{{\mathfrak {m}}}}_{{{{{\mathfrak {k}}}}}_\alpha }\subset {{{\mathfrak {m}}}}_0$. But since ${{{\mathfrak {m}}}}_0$ and ${{{\mathfrak {k}}}}_\alpha$ are subalgebras of ${{{\mathfrak {g}}}}$, so is ${{{\mathfrak {k}}}}_\alpha ^0:={{{\mathfrak {m}}}}_{{{{{\mathfrak {k}}}}}_\alpha }={{{\mathfrak {m}}}}\cap {{{\mathfrak {k}}}}_{\alpha }$. Since each ${{{\mathfrak {k}}}}_\alpha$ is non-toral by assumption, ${{{\mathfrak {k}}}}_\alpha ^0$ is a non-abelian subalgebra of ${{{\mathfrak {m}}}}_0$ and consequently, ${{{\mathfrak {k}}}}_\alpha$ has a nonvanishing semisimple part $({{{\mathfrak {k}}}}_\alpha ^0)^{ss}$. Passing to a subsequence, we may assume that these semisimple subalgebras converge to a limit subalgebra ${{{\mathfrak {k}}}}_\infty ^0$ of ${{{\mathfrak {m}}}}_0 \subset {{{\mathfrak {g}}}}$, and also that $f_\alpha \rightarrow f_\infty$ as $\alpha \rightarrow \infty$. Up to conjugation, ${{{\mathfrak {g}}}}$ has only finitely many semisimple subalgebras, see [17, Corollary 4.5]; thus ${{{\mathfrak {k}}}}_\infty ^0$ must be semisimple. Hence $[I_1I_1I_1]_{f_\infty }>0$, a contradiction. $\square$

Corollary 7.13

Let G/H be a compact homogeneous space with finite fundamental group and let f be a decomposition of ${{{\mathfrak {m}}}}$ into ${\text {Ad}}(H)$-irreducible summands. For each $1\le i \le \ell$,

$$\begin{aligned} d_ib_i-\tfrac{1}{2}\sum _{j,k=1}^\ell [ijk]>0, \end{aligned}$$

and consequently

$$\begin{aligned} b_{G/H}:=\sum _{i=1}^{\ell }d_ib_i> 0. \end{aligned}$$

(7.3)

Proof

By Lemma 7.10 we know if $2d_ic_i+\frac{1}{2}\sum _{j,k=1}^\ell [ijk]=0$, then ${{{\mathfrak {m}}}}_i$ is an abelian subalgebra in ${{{\mathfrak {g}}}}$ and ${{{\mathfrak {m}}}}_i$ commutes with the subalgebra ${{{\mathfrak {h}}}}\oplus (\bigoplus _{j=1,j\ne i}^\ell {{{\mathfrak {m}}}}_j)$. This would imply that G/H has infinite fundamental group, see Lemma 5.21, contradicting our hypothesis. This proves the claim. $\square$

1.4 Lie theory III: A Łojasiewicz inequality for structure constants

We recall the Łojasiewicz inequality in semi-algebraic geometry (cf. [7, Proposition 2.3.11]): Let $K \subset {{{\mathbb {R}}}}^n$ be a compact semi-algebraic set and let $f,g:K \rightarrow {{{\mathbb {R}}}}$ be continuous and semi-algebraic. Suppose that for all $x \in K$ we have: $f(x)=0 \Rightarrow g(x)=0$. Thus $f^{-1}(0) \subset g^{-1}(0)$. Then there exists $N\in {{{\mathbb {N}}}}$ and $C=C(f,g,K)\ge 0$ such that for all $x \in K$ we have

$$\begin{aligned} \vert g(x)\vert ^N \le C\cdot \vert f(x) \vert . \end{aligned}$$

The Łojasiewicz inequality stated in Proposition 7.14 provides the key estimate for the proof of Theorem 4.10. Note that standard estimation methods cannot be applied to the above situation.

Let $I^{\textsf {c}}=\{1,\ldots ,\dim {{{\mathfrak {g}}}}\}\backslash I$.

Proposition 7.14

Let G be a compact, connected Lie group endowed with a biinvariant metric Q. Let $I\subset \{1,\ldots ,\dim \,{{{\mathfrak {g}}}}\}$ with $1\in I$. Let $Z_{\vert I\vert }$ denote the set of all Q-orthonormal bases $b:=(e_1,\ldots ,e_{\dim {{{{\mathfrak {g}}}}}})$ of $({{{\mathfrak {g}}}},Q)$ such that

$$\begin{aligned} {{{\mathfrak {t}}}}:=\oplus _{i\in I} \langle e_i \rangle \end{aligned}$$

is an abelian subalgebra. Then there exists a constant $C=C(G,Q)>0$ and an open neighborhood $U_{\vert I\vert }$ of $Z_{\vert I\vert }$ in the space of all Q-orthonormal bases, such that for all $\tilde{b}:=({\tilde{e}}_1,\ldots ,{\tilde{e}}_{\dim {{{{\mathfrak {g}}}}}}) \in U_{\vert I\vert }$ we have

$$\begin{aligned} \sum _{j,k \in I}Q([{\tilde{e}}_1,{\tilde{e}}_j],{\tilde{e}}_k)^2 \le C \cdot \!\!\!\!\sum _{j\in I,k \in I^{\textsf {c}}} Q([{\tilde{e}}_1,{\tilde{e}}_j],\tilde{e}_k)^2. \end{aligned}$$

(7.4)

Proof

First note that if $I^{\textsf {c}}=\emptyset$, then ${{{\mathfrak {g}}}}$ is abelian and the above claim is trivially true. Hence we may assume that $I^{\textsf {c}}\ne \emptyset$ and that ${{{\mathfrak {g}}}}$ is not abelian. We prove the result by induction on $\vert I\vert$. It is clear that for $\vert I\vert =1,2$ the above claim is true, since the left hand side of (7.4) is zero by ${\text {Ad}}$-invariance of Q. Thus we may assume that $\vert I\vert \ge 3$.

We first show that for an arbitrary choice of Q-orthonormal basis $b \in Z_{\vert I\vert }$ and an arbitrary choice of sequence $(b_\alpha )_{\alpha \in {{{\mathbb {N}}}}}:=(e_1(\alpha ),\ldots ,e_{\dim {{{{\mathfrak {g}}}}}}(\alpha ))_{\alpha \in {{{\mathbb {N}}}}}$ of Q-orthonormal bases for ${{{\mathfrak {g}}}}$ with $\lim _{\alpha \rightarrow \infty }b_\alpha =b$, such a constant $C>0$ exists (a priori, C may depend on b and the sequence $(b_\alpha )$).

For a contradiction, suppose that no such constant $C>0$ exists. In what follows we will pass to a subsequence whenever convenient, without explicitly mentioning it. For $1 \le j,k \le \dim {{{\mathfrak {g}}}}$, we set

$$\begin{aligned} {[}1jk]_\alpha :=Q([e_1(\alpha ),e_j(\alpha )], e_k(\alpha ))^2 . \end{aligned}$$

We set $I':= I\backslash \{1\}$ and get

$$\begin{aligned} \sum _{j,k \in I'} [1jk]_\alpha> & {} g(\alpha ) \cdot \sum _{j\in I',k \in I^{\textsf {c}}} [1jk]_\alpha \end{aligned}$$

(7.5)

where $g:{{{\mathbb {R}}}}\rightarrow {{{\mathbb {R}}}}$ with $\lim _{\alpha \rightarrow \infty }g(\alpha )=+\infty$.

Suppose that for each $\alpha \in {{{\mathbb {N}}}}$, $e_1(\alpha )\in {{{\mathfrak {t}}}}(\alpha )$, where ${{{\mathfrak {t}}}}(\alpha )$ denotes a maximal abelian subalgebra of ${{{\mathfrak {g}}}}$. Since maximal abelian subalgebras of ${{{\mathfrak {g}}}}$ are conjugate, there exists a maximal torus ${{{\mathfrak {t}}}}(\infty )$ and a sequence $(g_\alpha )_{\alpha \in {{{\mathbb {N}}}}}$ of group elements in G with $\textrm{Ad}(g_\alpha )({{{\mathfrak {t}}}}(\alpha ))={{{\mathfrak {t}}}}(\infty )$. We obtain a sequence $(b_\alpha )$ of Q-orthonormal bases of ${{{\mathfrak {g}}}}$ converging to b, with $e_1(\alpha )\in {{{\mathfrak {t}}}}(\infty )$ for all $\alpha \in {{{\mathbb {N}}}}$, which we again denote by $b_\alpha =(e_1(\alpha ),\ldots , e_{\dim {{{{\mathfrak {g}}}}}}(\alpha ))$. Since all terms in (7.5) are $\textrm{Ad}(G)$-invariant, Eq. (7.5) holds for each $b_\alpha$ in this sequence as well.

Step 1: Let $V_1(\alpha ):=\langle e_2(\alpha ),\ldots , e_{\vert I\vert }(\alpha )\rangle _{{{{{\mathbb {R}}}}}}$, and let $V_1(\alpha )^\perp$ denote the Q-orthogonal complement of $V_1(\alpha )$ in ${{{\mathfrak {g}}}}$. Let $j\in I'$. Since $e_1(\alpha )\in {{{\mathfrak {t}}}}(\infty )$ we have

$$\begin{aligned}{}[e_1(\alpha ),e_j(\alpha )]= & {} \text {pr} _{V_1(\alpha )} ([e_1(\alpha ),e_j(\alpha )]) + \text {pr} _{V_1(\alpha )^\perp }([e_1(\alpha ),e_j(\alpha )])\\= & {} \sum _{k\in I'\backslash \{j\}}s_{jk}(\alpha ) \cdot e_k(\alpha )+ \textrm{pr}_{V_1(\alpha )^\perp }([e_1(\alpha ),e_j(\alpha )]) \,\,\,\in \,\,\,{{{\mathfrak {t}}}}(\infty )^\perp . \end{aligned}$$

As mentioned above, we prove our proposition by induction on $\vert I\vert$. Suppose that, after renumbering, $I=\{1,\ldots ,\vert I\vert \}$, and that (after passing to a subsequence)

$$\begin{aligned} \vert s_{23}(\alpha )\vert \ge \vert s_{jk}(\alpha )\vert \end{aligned}$$

for all $j,k\in I'$ and all $\alpha \in {{{\mathbb {N}}}}$. By (7.5) we know that $\vert s_{23}(\alpha )\vert >0$ for all sufficiently large $\alpha$. Thus, for all $j,k\in I'$

$$\begin{aligned} \lim _{\alpha \rightarrow \infty }\tfrac{\vert s_{jk}(\alpha ) \vert }{\vert s_{23}(\alpha )\vert }\in [0,1]. \end{aligned}$$

(7.6)

Note that $\lim _{\alpha \rightarrow \infty } s_{23}(\alpha )=0$, since ${{{\mathfrak {t}}}}=\oplus _{i=1}^{\vert I \vert }\langle e_i \rangle$ is abelian.

By (7.5) we claim that for each $j\in I'$ there exists some $k\in I'$ such that the limit in (7.6) is positive. For $\vert I\vert =3$ this is trivially true. To see the claim, suppose that there exists $j_0 \in I'\backslash \{2,3\}$ such that this limit is zero for all $k\in I'$. It follows that there exists a function ${\tilde{g}}:{{{\mathbb {R}}}}\rightarrow {{{\mathbb {R}}}}$ with ${\displaystyle \lim _{\alpha \rightarrow \infty } {\tilde{g}}(\alpha )=+\infty }$ and

$$\begin{aligned} \tfrac{1}{ \vert I\vert } \cdot [123]_\alpha = \tfrac{1}{ \vert I\vert }\cdot \vert s_{23}(\alpha ) \vert ^2 \ge \vert s_{j_0k}(\alpha )\vert ^2 \cdot {\tilde{g}}(\alpha ) =[1j_0k]_\alpha \cdot {\tilde{g}}(\alpha ) \end{aligned}$$

(7.7)

for all $k \in I'$ and all $\alpha \in {{{\mathbb {N}}}}$. Let $I_{j_0}:= I \backslash \{j_0\}$ and $I'_{j_0}:= I'\backslash \{j_0\}$. Then

$$\begin{aligned} \sum _{j,k \in I'} [1jk]_\alpha =\sum _{j,k \in I'_{j_0}} [1jk]_\alpha +2 \sum _{k \in I'_{j_0}}[1j_0k]_\alpha , \end{aligned}$$

and

$$\begin{aligned} \sum _{j\in I', k\in I^{\textsf {c}}} [1jk]_\alpha= & {} \sum _{j \in I_{j_0}', k \in I^{\textsf {c}}} [1jk]_\alpha + \sum _{k\in I^{\textsf {c}}} [1j_0k]_\alpha \\= & {} \sum _{j \in I_{j_0}', k \in (I_{j_0})^\textsf {c}} [1jk]_\alpha + \sum _{k\in I^{\textsf {c}}} [1j_0k]_\alpha - \sum _{j \in I'_{j_0}} [1jj_0]_\alpha . \end{aligned}$$

We deduce from (7.5)

$$\begin{aligned} \sum _{j,k \in I'_{j_0}} [1jk]_\alpha + (2 +g(\alpha )) \cdot \sum _{k \in I'_{j_0}} [1j_0k]_\alpha> & {} g(\alpha ) \cdot \sum _{j \in I'_{j_0}, k \in (I_{j_0})^\textsf {c}} [1jk]_\alpha . \end{aligned}$$

By (7.7) this yields

$$\begin{aligned} {\tilde{g}}(\alpha ) \cdot \sum _{j,k \in I'_{j_0}} [1jk]_\alpha + (2 + g(\alpha )) \cdot [123]_\alpha> & {} {\tilde{g}}(\alpha ) \cdot g(\alpha ) \cdot \sum _{j \in I'_{j_0}, k \in (I_{j_0})^\textsf {c}} [1jk]_\alpha . \end{aligned}$$

Since $\sum _{j,k \in I'_{j_0}} [1jk]_\alpha \ge [123]_\alpha$, using $j_0 \ne 2,3$, this yields

$$\begin{aligned} \sum _{j,k \in I'_{j_0}} [1jk]_\alpha> & {} \tfrac{{\tilde{g}}(\alpha ) \cdot g(\alpha ) }{{\tilde{g}}(\alpha )+2+g(\alpha )} \cdot \sum _{j \in I'_{j_0}, k \in (I_{j_0})^\textsf {c}} [1jk]_\alpha . \end{aligned}$$

This shows that for the torus ${\displaystyle {{{\mathfrak {t}}}}_{j_0}:=\oplus _{i\in I_{j_0}} \langle e_i \rangle }$ of dimension $\vert I\vert -1$ we obtain an inequality like (7.5), because as $\alpha \rightarrow \infty$,

$$\begin{aligned} \tfrac{{\tilde{g}}(\alpha ) \cdot g(\alpha ) }{\tilde{g}(\alpha )+2+g(\alpha )} \rightarrow +\infty . \end{aligned}$$

By the induction hypothesis, we obtain a contradiction. This proves for each $j\in I'$ there exists some $k\in I'$ such that the limit in (7.6) is positive.

Step 2: For every $j\in I'$ there exists some $k(j)\in I'\backslash \{j\}$ with

$$\begin{aligned} \vert s_{jk(j)}(\alpha )\vert \ge \vert s_{jk}(\alpha )\vert \end{aligned}$$

for all $k\in I'$ and $\alpha \in {{{\mathbb {N}}}}$. By Step 1 we have

$$\begin{aligned} \lim _{\alpha \rightarrow \infty } \tfrac{\vert s_{jk(j)}(\alpha )\vert }{\vert s_{23} (\alpha )\vert } >0\,, \end{aligned}$$

(7.8)

which yields $\vert s_{jk(j)}(\alpha )\vert >0$ for all sufficiently large $\alpha \in {{{\mathbb {N}}}}$. Thus we obtain, for all $j \in I'$,

$$\begin{aligned}{}[e_1(\alpha ),e_j(\alpha )]= s_{jk(j)}(\alpha )\cdot (E_j^1(\alpha )+X_j^1(\alpha ))\in {{{\mathfrak {t}}}}(\infty )^\perp , \end{aligned}$$

(7.9)

with

$$\begin{aligned} E_j^1(\alpha ):=\tfrac{ \textrm{pr}_{V_1(\alpha )}([e_1(\alpha ),e_j(\alpha )])}{ s_{jk(j)}(\alpha )} \quad \text { and } \quad X_j^1(\alpha ):=\tfrac{ \textrm{pr}_{V_1(\alpha )^\perp }([e_1(\alpha ),e_j(\alpha )])}{ s_{jk(j)}(\alpha )}. \end{aligned}$$

The left hand side of (7.5) is bounded from above by $C' \cdot (s_{23}(\alpha ))^2$ by the very definition of $s_{23}(\alpha )$, where $C'$ is independent of $\alpha$. Meanwhile, $\Vert X_j^1(\alpha )\Vert ^2 \cdot s_{jk(j)}(\alpha )^2$ contributes to the right hand side. Since by (7.8), the limit behavior of $s_{jk(j)}(\alpha )$ is like that of $s_{23}(\alpha )$, in order for the inequality in (7.5) to hold while $g(\alpha ) \rightarrow \infty$, we must have

$$\begin{aligned} \lim _{\alpha \rightarrow \infty } X_j^1(\alpha )= 0. \end{aligned}$$

(7.10)

Next, let ${{{\mathfrak {c}}}}(e_1)\le {{{\mathfrak {g}}}}$ denote the centralizer of $e_1$. Since $e_1 \in {{{\mathfrak {t}}}}(\infty )$ and $[e_1,{{{\mathfrak {t}}}}']=0$, where ${\displaystyle {{{\mathfrak {t}}}}':=\oplus _{i\in I'} \langle e_j\rangle \subset {{{\mathfrak {t}}}}}$ is abelian, we obtain

$$\begin{aligned} {{{\mathfrak {t}}}}(\infty ) \subset {{{\mathfrak {t}}}}(\infty ) \oplus {{{\mathfrak {t}}}}' \subset {{{\mathfrak {c}}}}(e_1). \end{aligned}$$

(7.11)

Using that $1 \le \Vert E_j^1(\alpha )\Vert \le C''$ by the definition of $s_{jk(j)}(\alpha )$ this implies

$$\begin{aligned} \underbrace{E_j(\infty )}_{\ne 0}:= \lim _{\alpha \rightarrow \infty } E_j^1(\alpha )=\lim _{\alpha \rightarrow \infty } (E_j^1(\alpha )+X_j^1(\alpha )) \in {{{\mathfrak {c}}}}(e_1)\cap {{{\mathfrak {t}}}}(\infty )^\perp \cap {{{\mathfrak {t}}}}'. \end{aligned}$$

(7.12)

This is seen as follows: By (7.9) we have $E_j^1(\alpha )+X_j^1(\alpha ) \in {{{\mathfrak {t}}}}(\infty )^\perp$. Because we know $\lim _{\alpha \rightarrow \infty }X_j^1(\alpha )=0$ this implies $E_j(\infty ) \in {{{\mathfrak {t}}}}(\infty )^\perp$. Moreover, we have $E_j^1(\alpha )\in V_1(\alpha )$ by definition. Since $V_1(\alpha )\rightarrow {{{\mathfrak {t}}}}'$ as $\alpha \rightarrow \infty$, we deduce $E_j(\infty )\in {{{\mathfrak {t}}}}'$, so it follows that $E_j(\infty ) \in {{{\mathfrak {c}}}}(e_1)$ by (7.11).

An important consequence is that

$$\begin{aligned} {{{\mathfrak {t}}}}(\infty ) \subsetneq {{{\mathfrak {t}}}}(\infty ) \oplus {{{\mathfrak {t}}}}' \subset {{{\mathfrak {c}}}}(e_1). \end{aligned}$$

(7.13)

In fact we have that

$$\begin{aligned} e_2,\ldots ,e_{\vert I\vert }, E_2(\infty ),\ldots ,E_{\vert I\vert }(\infty ) \in {{{\mathfrak {c}}}}(e_1). \end{aligned}$$

Step 3: We show by induction on p, for $1 \le p \le \dim {{{\mathfrak {t}}}}(\infty )$, that ${{{\mathfrak {t}}}}(\infty )\subsetneq {{{\mathfrak {c}}}}({{{\mathfrak {t}}}}(\infty ))$, which is clearly impossible. This will prove our contradiction for the specific choice of Q-orthonormal basis $b \in Z_{\vert I\vert }$ and sequence $(b_\alpha )_{\alpha \in {{{\mathbb {N}}}}}:=(e_1(\alpha ),\ldots ,e_{\dim {{{{\mathfrak {g}}}}}}(\alpha ))_{\alpha \in {{{\mathbb {N}}}}}$ of Q-orthonormal bases for ${{{\mathfrak {g}}}}$ with $\lim _{\alpha \rightarrow \infty }b_\alpha =b$. Our induction hypotheses for each p, denoted by $(A1)_p$,...,$(A4)_p$, are defined here:

Assumption $(A1)_p$: There exist a Q-orthonormal frame $(e_1^1:=e_1,\ldots ,e_1^p)$ in ${{{\mathfrak {t}}}}(\infty )$, positive numbers $d_1(\alpha ),\ldots ,d_p(\alpha )>0$, and $r_1^{p+1}(\alpha )\in {{{\mathfrak {t}}}}(\infty )$, $r_1^{p+1}(\alpha ) \perp e_1^1,\ldots ,e_1^p$, such that

$$\begin{aligned} e_1(\alpha )=d_1(\alpha )\cdot e_1^1+\cdots +d_p(\alpha )\cdot e_1^p +r_1^{p+1}(\alpha ). \end{aligned}$$

(7.14)

Before we define $(A2)_p$, we set ${{{\mathfrak {c}}}}(0):={{{\mathfrak {g}}}}$ and

$$\begin{aligned} {{{\mathfrak {c}}}}(p):= \bigcap _{q=1}^p {{{\mathfrak {c}}}}(e_1^q). \end{aligned}$$

For $j \in I'$ let

$$\begin{aligned} e_j^1=e_j \quad \text { and } \quad e_j^{p}(\alpha )= \textrm{pr}_{{{{\mathfrak {c}}}}(p-1)}(e_j(\alpha )). \end{aligned}$$

Let

$$\begin{aligned} V_p(\alpha ):=\langle e_j^p(\alpha ),\ldots , e_{\vert I\vert }^p(\alpha )\rangle _{{{{{\mathbb {R}}}}}} \end{aligned}$$

and let $V_p(\alpha )^\perp$ denote the Q-orthogonal complement of $V_p(\alpha )$ in ${{{\mathfrak {c}}}}(p-1)$. Let

$$\begin{aligned} E_j^p(\alpha ):=\tfrac{ \textrm{pr}_{V_p(\alpha )}([e_1(\alpha ),e_j^p(\alpha )])}{ s_{jk(j)}(\alpha )} \quad \text { and } \quad X_j^p(\alpha ):=\tfrac{ \textrm{pr}_{V_p(\alpha )^\perp }([e_1(\alpha ),e_j^p(\alpha )])}{ s_{jk(j)}(\alpha )}, \end{aligned}$$

recalling, as we showed in Step 2, that $s_{jk(j)}(\alpha )>0$ for all sufficiently large $\alpha \in {{{\mathbb {N}}}}$.

Assumption $(A2)_p$: We have

$$\begin{aligned} {{{\mathfrak {t}}}}(\infty )\subsetneq {{{\mathfrak {t}}}}(\infty ) \oplus {{{\mathfrak {t}}}}' \subset {{{\mathfrak {c}}}}(p). \end{aligned}$$

(7.15)

Notice that because $e_1(\alpha )\in {{{\mathfrak {t}}}}(\infty )\subset {{{\mathfrak {c}}}}(p)\subset {{{\mathfrak {c}}}}(p-1)$, we have

$$\begin{aligned} \text {pr} _{{{{\mathfrak {c}}}}(p-1)}([e_1(\alpha ),e_j(\alpha )])= [e_1(\alpha ),e_j^{p}(\alpha )]=s_{jk(j)}\cdot ( E^{p}_j(\alpha )+ X^{p}_j(\alpha )) \in {{{\mathfrak {c}}}}(p-1) \cap {{{\mathfrak {t}}}}(\infty )^\perp . \end{aligned}$$

Assumption $(A3)_p$: For every $j \in I'$ we know ${\displaystyle \lim _{\alpha \rightarrow \infty }X_j^p(\alpha )=0}$ and

$$\begin{aligned} \lim _{\alpha \rightarrow \infty } (E^{p}_j(\alpha )+X_j^p(\alpha ))=E_j(\infty ) \in ({{{\mathfrak {c}}}}(p)\cap {{{\mathfrak {t}}}}'\cap {{{\mathfrak {t}}}}(\infty )^\perp )\backslash \{0\}. \end{aligned}$$

(7.16)

Assumption $(A4)_p$: For every $j \in I'$, we have

$$\begin{aligned} \lim _{\alpha \rightarrow \infty } e_j^{p}(\alpha )=e_j \in {{{\mathfrak {c}}}}(p) \cap {{{\mathfrak {t}}}}'. \end{aligned}$$

(7.17)

The base case $p=1$ of our induction was essentially established in Step 2: By definition we have ${\displaystyle \lim _{\alpha \rightarrow \infty }e_1(\alpha )=e_1}$, that is, we can write $e_1(\alpha )=d_1(\alpha )\cdot e_1 + r_1^2(\alpha )$ with $d_1(\alpha )>0$, $r_1^2(\alpha )\in {{{\mathfrak {t}}}}(\infty )$ and $e_1 \perp r_1^2(\alpha )$. Thus $(A1)_1$ holds. Next, $(A2)_1$ is established in (7.13). Equations (7.10) and (7.12) give $(A3)_1$. And then, because $e_j^1(\alpha )=e_j(\alpha ) \rightarrow e_j \in {{{\mathfrak {t}}}}' \subset {{{\mathfrak {c}}}}(e_1)$, $(A4)_1$ is also clear.

We now prove the induction step $p \rightarrow p+1\le \dim {{{\mathfrak {t}}}}(\infty )$: By the induction hypothesis $(A2)_p$ we have ${{{\mathfrak {t}}}}(\infty ) \subset {{{\mathfrak {c}}}}(p)$, thus $[{{{\mathfrak {t}}}}_\infty ,{{{\mathfrak {c}}}}(p)^\perp ]\subset {{{\mathfrak {c}}}}(p)^\perp$. For $j \in I'$ we define $e_j^{p+1}(\alpha ):=\textrm{pr}_{{{{{\mathfrak {c}}}}}(p)}(e_j(\alpha ))$. Then by $(A1)_p$ we have

$$\begin{aligned} \text {pr} _{{{{{\mathfrak {c}}}}}(p)}( [e_1(\alpha ),e_j(\alpha ) ] ) = [e_1(\alpha ),e_j^{p+1}(\alpha ) ] = [ r_1^{p+1}(\alpha ),e_j^{p+1}(\alpha ) ] \in {{{\mathfrak {c}}}}(p) \cap {{{\mathfrak {t}}}}(\infty )^\perp . \end{aligned}$$

Using $(A2)_p$, the equation after (7.15), and ${{{\mathfrak {c}}}}(p-1) \supset {{{\mathfrak {c}}}}(p)$, we see

$$\begin{aligned} {[}e_1(\alpha ),e_j^{p+1}(\alpha )]= & {} s_{jk(j)}(\alpha )\cdot \underbrace{ \text {pr} _{{{{{\mathfrak {c}}}}}(p)} (E_j^p(\alpha )+X_j^p(\alpha ))}_{=:\tilde{E}^{p+1}_j(\alpha )} \in {{{\mathfrak {c}}}}(p)\cap {{{\mathfrak {t}}}}(\infty )^\perp . \end{aligned}$$

(7.18)

By $(A3)_p$ we know

$$\begin{aligned} \lim _{\alpha \rightarrow \infty }X_j^p(\alpha )=0 \text { and } \lim _{\alpha \rightarrow \infty }\tilde{E}^{p+1}_j(\alpha )=E_j(\infty ) \in ({{{\mathfrak {c}}}}(p)\cap {{{\mathfrak {t}}}}' \cap {{{\mathfrak {t}}}}(\infty )^\perp )\backslash \{0\}. \end{aligned}$$

(7.19)

Since $s_{jk(j)}(\alpha )>0$ we must have $r_1^{p+1}(\alpha )\ne 0$ for all sufficiently large $\alpha \in {{{\mathbb {N}}}}$. Set

$$\begin{aligned} e_1^{p+1}(\alpha ):=\tfrac{ r_1^{p+1}(\alpha ) }{\Vert r_1^{p+1}(\alpha )\Vert }. \end{aligned}$$

Then, by passing to a further subsequence, we may assume ${\displaystyle \lim _{\alpha \rightarrow \infty } e_1^{p+1}(\alpha )=e_1^{p+1} \in {{{\mathfrak {t}}}}(\infty )}$, with $e_1^{p+1}\perp e_1,\ldots ,e_1^p$ and $\Vert e_1^{p+1} \Vert =1$. This shows $(A1)_{p+1}$.

The induction hypothesis $(A4)_p$ guarantees for all $j \in I'$,

$$\begin{aligned} \lim _{\alpha \rightarrow \infty } e_j^{p+1}(\alpha )=e_j \in {{{\mathfrak {c}}}}(p) \cap {{{\mathfrak {t}}}}' . \end{aligned}$$

(7.20)

Thus for all sufficiently large $\alpha \in {{{\mathbb {N}}}}$,

$$\begin{aligned} V_{p+1}(\alpha ):=\langle e_2^{p+1}(\alpha ),\ldots , e_{\vert I\vert }^{p+1}(\alpha )\rangle _{{{{{\mathbb {R}}}}}} \end{aligned}$$

is a subspace of ${{{\mathfrak {c}}}}(p)$ of dimension $\vert I'\vert$, converging to ${{{\mathfrak {t}}}}'$ as $\alpha \rightarrow \infty$. Using (7.14), the definition of ${{{\mathfrak {c}}}}(p)$, and $e_1^{p+1}(\alpha )$ we deduce from (7.18) for all $j \in I'$

$$\begin{aligned} {[}e_1^{p+1}(\alpha ),e_j^{p+1}(\alpha )]= \tfrac{s_{jk(j)}(\alpha )}{\Vert r_1^{p+1}(\alpha ) \Vert }\cdot ( E^{p+1}_j(\alpha )+ X^{p+1}_j(\alpha )) \in {{{\mathfrak {c}}}}(p) \cap {{{\mathfrak {t}}}}(\infty )^\perp , \end{aligned}$$

(7.21)

where ${\displaystyle E^{p+1}_j(\alpha ):= \text {pr} _{V_{p+1}(\alpha ) }(\tilde{E}_j^{p+1}(\alpha ))\in {{{\mathfrak {c}}}}(p)}$ and ${\displaystyle X^{p+1}_j(\alpha ):= \textrm{pr}_{V_{p+1}(\alpha )^\perp }(\tilde{E}_j^{p+1}(\alpha ))\in {{{\mathfrak {c}}}}(p)}$.

Notice, by Eq. (7.21),

$$\begin{aligned} {[}e_1(\alpha ),e_j^{p+1}(\alpha )]=s_{jk(j)}\cdot ( E^{p+1}_j(\alpha )+ X^{p+1}_j(\alpha )) \in {{{\mathfrak {c}}}}(p) \cap {{{\mathfrak {t}}}}(\infty )^\perp . \end{aligned}$$

Since by (7.19) ${\displaystyle \lim _{\alpha \rightarrow \infty }\tilde{E}^{p+1}_j(\alpha )=E_j(\infty )\in {{{\mathfrak {t}}}}'}$ and $V_{p+1}(\alpha )\rightarrow {{{\mathfrak {t}}}}'$ as $\alpha \rightarrow \infty$, we get

$$\begin{aligned} \lim _{\alpha \rightarrow \infty } X^{p+1}_j(\alpha )=0 \quad \text { and } \quad \lim _{\alpha \rightarrow \infty } E^{p+1}_j(\alpha )=E_j(\infty ). \end{aligned}$$

This shows the first part of $(A3)_{p+1}$. Next, we claim ${\displaystyle \lim _{\alpha \rightarrow \infty }\tfrac{s_{jk(j)}(\alpha )}{\Vert r_1^{p+1}(\alpha ) \Vert } =0}$. To see this, notice that $\lim _{\alpha \rightarrow \infty }e_1^{p+1}(\alpha )=e_1^{p+1}\in {{{\mathfrak {t}}}}(\infty )$, that $\lim _{\alpha \rightarrow \infty } e_j^{p+1}(\alpha ) =e_j \in {{{\mathfrak {t}}}}'$ by (7.20) and that $\lim _{\alpha \rightarrow \infty } (E_j^{p+1}(\alpha )+X_j^{p+1}(\alpha ))=E_j(\infty ) \in {{{\mathfrak {t}}}}'$ for all $j \in I'$. Now if ${\displaystyle \tfrac{s_{jk(j)}(\alpha )}{\Vert r_1^{p+1}(\alpha ) \Vert } }$ had a uniform positive lower bound, then by passing to a subsequence if necessary, then we deduce $Q([e_1^{p+1},e_j],E_j(\infty ))>0$ by (7.21), because $E_j(\infty )\ne 0$. But since $e_j,E_j(\infty )\in {{{\mathfrak {t}}}}'$ and since ${{{\mathfrak {t}}}}'$ is abelian, the left-hand side equals zero. This shows the above claim.

By (7.21) we now deduce $[e_1^{p+1},e_j]=0$ for all $j \in I'$, that is,

$$\begin{aligned} {{{\mathfrak {t}}}}(\infty ) \subset {{{\mathfrak {t}}}}(\infty ) \oplus {{{\mathfrak {t}}}}' \subset {{{\mathfrak {c}}}}(e_1^{p+1}). \end{aligned}$$

Using this and $E_j(\infty )\in {{{\mathfrak {c}}}}(p)\cap {{{\mathfrak {t}}}}'$ we deduce

$$\begin{aligned} E_j(\infty ) \in {{{\mathfrak {c}}}}(p+1) \cap {{{\mathfrak {t}}}}(\infty )^\perp \cap {{{\mathfrak {t}}}}', \end{aligned}$$

where ${{{\mathfrak {c}}}}(p+1)={{{\mathfrak {c}}}}(p)\cap {{{\mathfrak {c}}}}(e_1^{p+1})$. By (7.20), this yields $(A2)_{p+1}$:

$$\begin{aligned} {{{\mathfrak {t}}}}(\infty )\subsetneq {{{\mathfrak {t}}}}(\infty ) \oplus {{{\mathfrak {t}}}}' \subset {{{\mathfrak {c}}}}(p+1). \end{aligned}$$

Notice that this implies

$$\begin{aligned} e_2,\ldots ,e_{\vert I\vert }, E_2(\infty ),\ldots ,E_{\vert I\vert }(\infty ) \in {{{\mathfrak {c}}}}(p+1), \end{aligned}$$

implying both the second part of $(A3)_{p+1}$ and also $(A4)_{p+1}$.

So far we have proved that for a given $b\in Z_{\vert I\vert }$ and a given sequence $(b_\alpha )_{\alpha \in {{{\mathbb {N}}}}}$ converging to b, such a constant C exists for all sufficiently large $\alpha$. We next show that C can be chosen independently of both $b\in Z_{\vert I\vert }$ and the sequence $(b_\alpha )_{\alpha \in {{{\mathbb {N}}}}}$.

To this end, suppose that there exists a sequence $(b_\alpha )_{\alpha \in {{{\mathbb {N}}}}}$ in $Z_{\vert I\vert }$ and corresponding sequences $(b_{\alpha ,\beta })_{\beta \in {{{\mathbb {N}}}}}$ with $\lim _{\beta \rightarrow \infty }b_{\alpha ,\beta }=b_\alpha$ for all $\alpha \in {{{\mathbb {N}}}}$, such that for the corresponding optimal constants, as $\alpha \rightarrow \infty$, $C_\alpha \rightarrow +\infty$. Here by optimal we mean that the constants $C_\alpha -1$ would not work. Since the constants $C_\alpha$ are chosen to be optimal, for each $\alpha \in {{{\mathbb {N}}}}$ there exists $N(\alpha ) \in {{{\mathbb {N}}}}$, such that we may assume (after passing to a subsequence) that

$$\begin{aligned} \sum _{j,k \in I'} [1jk]_{\alpha ,\beta }\ge & {} (C_\alpha -1) \cdot \sum _{j\in I',k \in I^{\textsf {c}}} [1jk]_{\alpha ,\beta } \end{aligned}$$

(7.22)

for all $\beta \ge N(\alpha )$. By again passing to a subsequence, we may assume that this holds for all $\alpha ,\beta \ge 1$. Since $Z_{\vert I\vert }$ is compact, we may assume that $\lim _{\alpha \rightarrow \infty }b_\alpha =b$. Moreover, it is clear that we can construct a sequence $({\tilde{b}}_\alpha )_{\alpha \in {\textbf{N}} }$ converging to b and satisfying (7.22). Since $C_\alpha \rightarrow + \infty$ as $\alpha \rightarrow \infty$, we obtain a contradiction. This shows that C can be chosen independently.

To prove the theorem, suppose now that no such constant $C=C(G,Q)$ and no such open neighborhood $U_{\vert I\vert }$ of $Z_{\vert I\vert }$ exist. Then there exists a sequence of bases $(b_\alpha )_{\alpha \in {{{\mathbb {N}}}}}$ converging to $b \in Z_{\vert I\vert }$ for which (7.22) holds. Contradiction. $\square$