---
address: 'BICMR, Peking University, No. 5 Yiheyuan Road, Haidian District, Beijing 100871, China.'
author:
- Jun Yu
date: June 2018
title: Acceptable compact Lie groups
---

[**Mathematics Subject Classification (2010).**]{} 22C05, 22E15.

[**Keywords.**]{} Acceptable group, strongly controlling fusion.

Introduction
============

Acceptability of a group is defined by Michael Larsen in [@Larsen1]. The motivation of studying acceptability comes from its connection with multiplicity one question in the automorphic form theory ([@Arthur]). Recently it is also found its connection with Langlands correspondence through Lafforgue’s operators ([@Lafforgue]).

This paper contributes to the goal of classifying acceptable compact Lie groups. The major results in this paper is summarized with the following theorem.

\[T1\] Let $G$ be a connected compact semisimple Lie group. For $G$ to be acceptable it is necessary and sufficient that $G$ is isomorphic to a direct product of the following groups,

-   $\SU(n)$ ($n\geq 1$);

-   $\Sp(n)$ ($n\geq 1$);

-   $\SO(2n+1)$ ($n\geq 1$);

-   $\G_2$;

-   $\SO(4)$.

This paper contains two main aspects. In the first aspect, we study the notion of strongly controlling fusion (“SCF" for short) defined by Robert Griess ([@Griess]). We give new proofs for “SCF" pairs shown by Griess in [@Griess], and show new “SCF" pairs with our method. Our proof of “SCF" property for a pair $H\subset G$ is through studying the intersections $H\cap\Ad(g)H$ ($g\in G$), decomposing the double coset space $H\backslash G/H$, and calculating the centralizer subgroup $Z_{H}(g)=
H\cap Z_{G}(g)$ ($g\in G$). These are of independent interest. Using “SCF" property we show that groups in the items (1)-(4) of Theorem \[T1\] are acceptable. The connection between acceptability and “SCF" property is through the following proposition.

\[P1\] Let $H$ be a closed subgroup of a real Lie group $G$. Suppose $H$ strongly controls its fusion in $G$. If $G$ is acceptable (or strongly acceptable), then so is $H$.

In the second aspect, we show many groups are unacceptable. For doing this the following Lemma \[L:product2\] and Proposition \[P:centralizer2\] play important roles, though they are very simple. Lemma \[L:product2\] is clear to see, and Proposition \[P:centralizer2\] is also easy to show.

\[L:product2\] Let $Z\subset Z(G_1)\times Z(G_2)$ with $Z\cap Z(G_1)=Z\cap Z(G_2)=1$. Suppose $(G_1\times G_2)/Z$ is acceptable (or strongly acceptable), then so are $G_1$ and $G_2$.

\[P:centralizer2\] Let $G$ be a compact Lie group and $A\subset G$ be a closed abelian subgroup. If $G$ is strongly acceptable (or acceptable), then so is $Z_{G}(A)$.

Note that the effect of Proposition \[P:centralizer2\] is bi-sided. Besides showing many groups are unacceptable using it, we also use it to show $\SO(4)\cong\Sp(1)^{2}/\langle(-1,-1)\rangle$ is acceptable based on the acceptability of $\G_2$.

A critical part in the process of showing Theorem \[T1\] is the study of acceptability/unacceptability of rank $3$ compact simple Lie groups. Among these it is well known that $\SU(4)$, $\SO(7)$, $\Sp(3)$ are acceptable, and Larsen showed that $\PSU(4)$ is unacceptable in [@Larsen1]. The remaining ones are $\SU(4)/\langle-I\rangle\cong\SO(6)$, $\Spin(7)$, $\PSp(3)$. The unacceptability of $\SO(6)$ is first shown by Matthew Weidner in [@Weidner] (cf. Example \[E:SU4\]). The unacceptability of $\Spin(7)$ is due to Gaetan Chenevier and Wee Teck Gan (cf. Theorem \[T:unacceptable-ChenevierGan\]). We show that $\PSp(3)$ is unacceptable (cf. Example \[E:Sp3\]). Curiously, the unacceptability of all of the groups $\SU(4)/\langle-I\rangle\cong\SO(6)$, $\Spin(7)$, $\PSp(3)$ are shown by constructing element-conjugate but not globally conjugate homomorphisms from $(C_4)^{2}$ to them.

Another crucial step in showing Theorem \[T1\] is to show that the group $$\Sp(1)^{3}/\langle
(1,-1,-1), (-1,1,-1)\rangle$$ and groups of the form $$\Sp(1)^{m}/\langle(-1,\dots,-1)\rangle$$ ($m\geq 3$) are unacceptable (cf. Theorem \[T:3A1\] and Example \[E:Sp3-2\]).

[*Notation and conventions.*]{} Let $G$ be a group. For $x,y\in G$, we write $x\sim y$ if $y=
\Ad(g)x=gxg^{-1}$ for some $g\in G$, and we say $x$ and $y$ are conjugate (or $G$-conjugate). For a subgroup $G_1\subset G$ (or an over-group $G_{2}\supset G$), we write $x\sim_{G_1}y$ if $y=\Ad(g)x=
gxg^{-1}$ for some $g\in G_1$ (or $g\in G_2$).

Let $G$ be a connected compact Lie group. Write $Z(G)$ for the center of $G$, and $G_{der}=[G,G]$ for the derived subgroup of $G$.

In this paper we consider only real Lie group with finitely many connected components. This includes particularly all real algebraic groups.

[*Acknowledgement.*]{} I would like to thank Xinwen Zhu for bringing up this question to my attention, and to thank Gaetan Chenevier, Wee Teck Gan, Matthew Weidner for helpful communications. Part of this paper was written when the author visited Paris Diderot University (Paris 7) in May 2018. I would like to thank Huayi Chen for the invitation.

Acceptability conditions
========================

Let $G$ be a real Lie group, and $\Gamma$ be a group. Two homomorphisms $\phi_{1},\phi_{2}:\Gamma
\rightarrow G$ are called element-conjugate if $\phi_{1}(x)\sim\phi_{2}(x)$ for any $x\in\Gamma$. They are called globally conjugate if there exists $g\in G$ such that $\phi_{2}(x)=\Ad(g)(\phi_{1}(x))$ for all $x\in
\Gamma$. They are called conjugate in image if $\phi(2)(\Gamma)=\Ad(g)(\phi_{1}(\Gamma))$ for some $g\in G$.

Following [@Larsen1], we define acceptability and unacceptability as follows. The only new addition is defining a kind of “strongly acceptability" by allowing homomorphisms from any compact group.

\[D:acceptable1\] A real Lie group $G$ with finitely many connected components is called acceptable if for any finite group $\Gamma$ and every pair of element-conjugate homomorphisms $\phi_1,\phi_2:\Gamma\rightarrow G$, $\phi_1$ and $\phi_2$ are globally conjugate.

A real Lie group $G$ with finitely many connected components is called strongly acceptable if for any compact group $\Gamma$ and every pair of element-conjugate homomorphisms $$\phi_1,\phi_2:\Gamma\rightarrow G,$$ $\phi_1$ and $\phi_2$ are globally conjugate.

\[D:acceptable2\] A real Lie group $G$ with finitely many connected components is called unacceptable if there exists a finite group $\Gamma$ and two element-conjugate homomorphisms $\phi_1,\phi_2:\Gamma\rightarrow G$, $\phi_1$ and $\phi_2$ are not globally conjugate.

A real Lie group $G$ with finitely many connected components is called strongly unacceptable if there is a compact group $\Gamma$ and two element-conjugate homomorphisms $\phi_1,\phi_2:\Gamma\rightarrow G$, $\phi_1$ and $\phi_2$ are not conjugate in image.

Maximal compact subgroups and “SCF” subgroups
---------------------------------------------

[*Maximal compact subgroup.*]{} Any Lie group $G$ with finitely many connected components has a maximal compact subgroup. A maximal compact subgroup $K$ has the following property ([@Borel Page 124, Theorem 1.2 and Corollary 1.3]):

-   if $K'$ is another maximal compact subgroup, then $K'=\Ad(g)K$ for some $g\in G^{0}$.

-   for any compact group $\Gamma$ and homomorphism $\phi:\Gamma\rightarrow G$, there exists $g\in G$ such that $\phi(\Gamma)\subset\Ad(g)K$.

-   for any compact group $\Gamma$ and homomorphisms $\phi_1,\phi_2:\Gamma\rightarrow K$, if there exists $g\in G$ such that $\phi_{2}(x)=\Ad(g)(\phi_{1}(x))$ for any $x\in\Gamma$, then there exists $k\in K$ such that $\phi_{2}(x)=\Ad(k)(\phi_{1}(x))$ for any $x\in\Gamma$.

By conditions (2) and (3) in the above, $G$ is acceptable (or strongly acceptable) if and only if $K$ is. Thus, it suffices to study acceptability of compact Lie groups.

[*SCF pairs.*]{} The notion of strongly controlling fusion is invented by Griess ([@Griess]). Considering only homomorphisms from a compact group, we give the following definition.

\[D:SCF\] Let $H$ be a closed subgroup of a real Lie group $G$. We say $H$ strongly controls its fusion in $G$, if for any compact group $\Gamma$ and homomorphisms $\phi_1,\phi_2:\Gamma\rightarrow K$, whenever there exists $g\in G$ such that $\phi_{2}(x)=\Ad(g)(\phi_{1}(x))$ for any $x\in\Gamma$, then there exists $h\in H$ such that $\phi_{2}(x)=\Ad(h)(\phi_{1}(x))$ for any $x\in\Gamma$.

For short, in the below we say $H$ is an “SCF" subgroup of $G$ if $H$ strongly controls its fusion in $G$, and we call $H\subset G$ an “SCF” pair. The condition (3) for a maximal compact subgroup just means a maximal compact subgroup $K$ is an “SCF" subgroup of $G$. Generally we have the following statement.

\[P:SCF\] Let $H$ be a closed subgroup of a real Lie group $G$. Suppose $H$ strongly controls its fusion in $G$. If $G$ is acceptable (or strongly acceptable), then so is $H$.

We show for strongly acceptability. The proof for acceptability is similar. Suppose $G$ is strongly acceptable. Let $\Gamma$ be a compact group and $\phi_1,\phi_2:\Gamma\rightarrow H$ be two element-conjugate homomorphisms. Considering $\phi_1,\phi_2$ as homomorphism to $G$, by the strongly acceptability of $G$ there exists $g\in G$ such that $\phi_{2}(x)=\Ad(g)(\phi_{1}(x))$ for any $x\in\Gamma$. As $H$ is an “SCF" subgroup of $G$, there exists $h\in H$ such that $\phi_{2}(x)=\Ad(h)(\phi_{1}(x))$ for any $x\in\Gamma$. This means that $\phi_1$ and $\phi_2$ are globally conjugate. Thus, $H$ is strongly acceptable.

First reduction
---------------

\[L:acceptable1\] For a compact Lie group $G$ to be strongly acceptable, if and only if for any compact Lie group $H$ and every pair of element-conjugate homomorphisms $\phi_1,\phi_2:H\rightarrow G$, $\phi_1$ and $\phi_2$ are globally conjugate.

The “only if" part is trivial. For the “if" part, suppose the statement above for compact Lie groups $H$ holds. Write $G^{\#}$ for the set of conjugacy classes in $G$. We know that $G^{\#}$ has a topology of a compact space (actually it is the union of finitely many compact orbifolds), and the natural map $\pi: G
\rightarrow G^{\#}$ is continuous. For an element $g\in G$, write $[g]=\pi(g)\in G^{\#}$ for its conjugacy class. The map $\pi$ has a further property: for any sequence $\{g_{n}\}_{n\geq 1}\subset G$, $$\label{Eq1}\lim_{n\to\infty}g_{n}=1\Leftrightarrow\lim_{n\to\infty}[g_{n}]=[1].$$

Let $\Gamma$ be a compact group and $\phi_1,\phi_2:\Gamma\rightarrow G$ be two element-conjugate homomorphisms. Then for any $x\in\Gamma$, $\phi_{1}(x)=1$ if and only if $\phi_{2}(x)=1$. Thus, $\ker\phi_1=\ker\phi_2$. Considering $\Gamma/\ker\phi_{1}$ instead, we may assume that $\phi_1,\phi_2$ are injections. Write $H=
\overline{\phi_{1}(\Gamma)}$ for the closure of the image of $\phi_1$, which is a closed subgroup of $G$. We also use $\phi_1$ to denote the injection $\Gamma\rightarrow H$. We first show that there is another injection $\psi': H\rightarrow G$ which is element-conjugate to the inclusion $\psi: H\hookrightarrow G$, such that $\phi_2=\psi'\circ\phi_1$. For any $h\in H$, choose a sequence $\{\gamma_{n}\}_{n\geq 1}\subset
\Gamma$ such that $$\lim_{n\to\infty}\phi_{1}(\gamma_{n})=h.$$ Then, $$\lim_{n,m\to\infty}\phi_{1}
(\gamma_{n}^{-1}\gamma_{m})=1.$$ By Property (\[Eq1\]), we get $$\lim_{n,m\to\infty}[\phi_{1}
(\gamma_{n}^{-1}\gamma_{m})]=1.$$ As $\phi_1$ and $\phi_2$ are element-conjugate, we get $$\lim_{n,m\to
\infty}[\phi_{2}(\gamma_{n}^{-1}\gamma_{m})]=1.$$ By Property (\[Eq1\]) again, we get $$\lim_{n,m\to
\infty}\phi_{2}(\gamma_{n}^{-1}\gamma_{m})=1.$$ Thus, $\{\phi_{2}(\gamma_{n})\}_{n\geq 1}$ is a Cauchy sequence in $G$. Define $$\psi'(h)=\lim_{n\to\infty}\phi_{2}(\gamma_{n}).$$ Using Property (\[Eq1\]), one can show that $\psi'(h)$ does not depend on the choice of the sequence $\{\gamma_{n}\}_{n\geq 1}$. One can show that the map $\psi'$ is an injective homomorphism and it satisfies all required properties.

As $H$ is a compact Lie group and $\psi,\psi'$ are element-conjugate, by the assumption there exists $g\in G$ such that $\psi'(h)=\Ad(g)(\psi(h))$ for any $h\in H$. Then, $\phi_{2}(x)=\Ad(g)(\phi_{1}(x))$ for any $x\in\Gamma_1$. Thus, $\phi_1$ and $\phi_2$ are globally conjugate. Therefore, $G$ is strongly acceptable.

By Lemma \[L:acceptable1\], to check whether a Lie group $G$ with finitely many connected components is strongly acceptable, it suffices to consider element-conjugate homomorphisms from compact Lie groups to $G$.

It looks to us that strong acceptability should be a consequence of acceptability.

\[Q:acceptable1\] Is a Lie group $G$ acceptable if and only if it is strongly acceptable?

We call a compact Lie group $H$ a quasi-torus if $H^{0}$ is a torus. In the Proposition \[P:acceptable2\] below, we show that acceptability implies that element-conjugate homomorphisms from a a quasi-torus are globally conjugate.

\[L:centralizer1\] Let $G$ be a compact Lie group. Then there exist only finitely many conjugacy classes of subgroups $H$ of $G$ of the form $H=Z_{G}(A)$ with $A$ an abelian subgroup of $G$.

Prove by induction on $\dim G$. First, using Kac coordinate one shows that there are only finitely many conugacy classes of subgroups $H$ of $G$ of the form $H=Z_{G}(g)$ for some $g\in G$.

If $A\subset Z_{G}(G^{0})$, then $Z_{G}(A)\supset G^{0}$. As $G$ is a compact Lie group, $G/G^{0}$ is a finite group. Thus, there are only finitely many subgroups containing $G^{0}$.

If $A\not\subset Z_{G}(G^{0})$, take an $g\in A-Z_{G}(G^{0})$ and set $G'=Z_{G}(g)$. Then, $A\subset Z_{G}(A)\subset
G'$. Thus, $Z_{G}(A)=Z_{G'}(A)$. In the beginning we have shown that the conjugacy class of $G'=Z_{G}(g)$ has only finitely many possibilities. Since $g\not\in Z_{G}(G^{0})$, we have $G'\not\supset G^{0}$. Thus, $\dim G'<\dim G$. By induction subgroups $H$ of $G'$ of the form $H=Z_{G'}(A)$ and $A$ abelian have only finitely many possibilities. This finishes the proof.

\[P:acceptable2\] If a compact Lie group $G$ is acceptable, then for any quasi-torus $H$ and element-conjugate homomorphisms $\phi_{1},\phi_{2}:H\rightarrow G$, $\phi_1$ and $\phi_2$ are globally conjugate.

As $\phi_1$ and $\phi_2$ are element-conjugate, we have $\ker\phi_1=\ker\phi_2$. Considering $H/\ker\phi_1$ instead, we may assume that $\phi_1$ and $\phi_2$ are injections. Write $$H^{0}[n]=\{x\in H^{0}:x^{n}=1\}$$ for the group of $n$-torsion elements in $H^{0}$. For any $m\geq 1$, set $$H_{m}=Z_{G}(\phi_{1}(H^{0}[m])).$$ Then, $$H_1\supset H_2
\supset\cdots H_{m}\supset\cdots.$$ By Lemma \[L:centralizer1\], there exists $k\geq 1$ such that $H_{m}=H_{m+1}$ for any $m\geq k$. As $Z_{G}(H^{0})=\cap_{m\geq 1}Z_{G}(H^{0}[m])$, we get $Z_{G}(H^{0})=Z_{G}(H^{0}[m])$ for any $m\geq k$.

There is an exact sequence $1\rightarrow H^{0}\rightarrow H\rightarrow H/H^{0}\rightarrow 1.$ Then, the group $H$ is determined by a homomorphism $\phi: H/H^{0}\rightarrow\Aut(H^{0})$ and a class $\alpha\in
H^{2}_{\phi}(H/H^{0},H^{0})$. Write $n=|H/H^{0}|$. Then, $n\alpha=0$. There is an exact sequence $$1\rightarrow
H^{0}[n]\rightarrow H^{0}\rightarrow H^{0}\rightarrow 1,$$ where the homomorphism $H^{0}\rightarrow H^{0}$ is $$\psi_{n}(x)=x^{n}, \forall x\in H^{0}.$$ Due to $\psi_{n}(\alpha)=n\alpha=0$, the class $\alpha$ comes from $H^{2}_{\phi}(H/H^{0},H^{0}[n])$. That means there is a finite subgroup $J'$ of $H$ with $J'\cap H^{0}=H^{0}[n]$ and $H=H^{0}J'$. Set $J=J'H^{0}[nk]$. For any $m\geq 1$, set $J_{m}=JH^{0}[nkm]$. Since $G$ is acceptable, there exists $g\in G$ such that $$\phi_{2}|_{J}=\Ad(g)\circ\phi_{1}|_{J}.$$ Set $\phi'_{2}=\Ad(g^{-1})\circ\phi_2$. Then, $\phi'_{2}|_{J}=\phi_{1}|_{J}$. For any $m\geq 1$, by the acceptableness of $G$ there exists $g_{m}\in G$ such that $$\phi_2'|_{J_{m}}=\Ad(g_{m})\circ\phi_{1}|_{J_{m}}.$$ Then, $$g_{m}\in Z_{G}(\phi_{1}(H^{0}[nk]))=Z_{G}(H^{0}).$$ Thus, $\phi'_{2}|_{H^{0}}=\phi_{1}|_{H^{0}}$. By $H=H^{0}J$, we get $\phi'_{2}=\phi_{1}$. Hence, $\phi_1$ and $\phi_2$ are globally conjugate.

Further reduction
-----------------

The following lemma is trivial.

\[L:product\] Let $G_1$ and $G_2$ are compact Lie groups, then $G_1\times G_2$ is acceptable (or strongly acceptable) if and only if both of $G_1$ and $G_2$ are.

The following statement follows from [@Larsen1 Prop. 1.4].

\[L:Larsen\] Let $G$ be a connected compact Lie group. Then $G$ is acceptable (or strongly acceptable) if and only if its derived subgroup $G_{der}$ is.

Now we show Proposition \[P:centralizer2\].

We first give the proof for the strongly acceptability. Suppose $G$ is strongly acceptable. Write $H=
Z_{G}(A)$. Let $\Gamma$ be a group, and $\phi_{1},\phi_{2}:\Gamma\rightarrow H$ be two element-conjugate homomorphisms. Put $\Gamma'=\Gamma\times A$. Define $\phi'_{i}: \Gamma'\rightarrow G$ ($i=1,2$) by $$\phi'_{i}(\gamma,a)=\phi_{i}(\gamma)a,\ \forall\gamma\in\Gamma,a\in A.$$ Apparently, $\phi'_{1}$ and $\phi'_{2}$ are element-conjugate homomorphisms. Since $G$ is strongly acceptable, there exists $g\in G$ such that $$\phi'_{2}(\gamma,a)=\Ad(g)(\phi'_{1}(\gamma,a)),\ \forall(\gamma,a)\in\Gamma'.$$ Applying to $\gamma=1$ and $a\in A$, we get $g\in Z_{G}(A)=H$. Applying to $a=1$ and $\gamma\in\Gamma$, we get $\phi_{2}(\gamma)=\Ad(g)(\phi_{1}(\gamma))$ for any $\gamma\in\Gamma$. The just means, $\phi_1$ and $\phi_2$ are globally conjugate. Thus, $H$ is strongly acceptable.

For the acceptability. We could take $m>>1$ such that $Z_{G}(A(m))=Z_{G}(A)$, where $$A(m)=\{x\in A:
x^{m}=1\}.$$ Using $A(m)$ instead of $A$ in the above argument, the proof is the same.

Examples of “SCF" subgroups
===========================

In [@Griess], Griess showed several “SCF” pairs. The following proposition follows from results shown in [@Griess].

\[T:Griess\] The following pairs are “SCF” pairs.

-   $\O(n)\subset\U(n)$ and $\Sp(n)\subset\U(2n)$ (cf. [@Griess Theorem 2.3]);

-   $\G_2\subset\O(7)$ and $\G_2\subset\Pin(7)$ (cf. [@Griess Theorem 1.1]);

-   $\F_4\subset\E_6$ (cf. [@Griess Theorem 1.5]).

In this section, we give a new proof for Theorem \[T:Griess\]. Our method is to to study the intersections $H\cap\Ad(g)H$ ($g\in G$ and the double coset decomposition $H\backslash G/H$. This consideration has independent interest.

\[L:SCF2\] Let $G$ be a compact Lie group and $H\subset G$ be a closed subgroup. Then $H$ is an “SCF" subgroup of $G$ if and only if $$\label{Eq2}g\in Z_{G}(H\cap\Ad(g)H)H$$ for any $g\in G$.

Necessarity. Suppose $H$ is an “SCF" subgroup of $G$. For any $g\in G$, set $$\Gamma=H\cap\Ad(g)H.$$ Let $\phi_1:\Gamma\rightarrow H$ be the inclusion map, and $\phi_2: \Gamma\rightarrow H$ be $\phi_2=\Ad(g^{-1})
\circ\phi_{1}$. Then, $\phi_1,\phi_2:\Gamma\rightarrow H$ and $\phi_2=\Ad(g^{-1})\circ\phi_1$. As $H$ is an “SCF" subgroup of $G$, there exists $h\in H$ such that $\phi_2=\Ad(h^{-1})\circ\phi_{1}$. Then, $\phi_1=\Ad(gh^{-1})\circ\phi_1$. Thus, $gh^{-1}\in Z_{G}(\Gamma)$, and $$g=(gh^{-1})h\in Z_{G}(H\cap\Ad(g)H)H.$$

Sufficiency. Suppose $g\in Z_{G}(H\cap\Ad(g)H)H$ for any $g\in G$. Let $\Gamma$ be a compact group and $\phi_1,\phi_2:\Gamma\rightarrow H$ be two homomorphisms with $\phi_1=\Ad(g)\circ\phi_2$ for some $g\in G$. Then, $\Im\phi_1\subset H$ and $$\Im\phi_1=\Im(\Ad(g)\circ\phi_2)\subset\Ad(g)H.$$ Hence, $\Im\phi_1\subset
H\cap\Ad(g)H$. By the assumption, we then have $$g\in Z_{G}(H\cap\Ad(g)H)H\subset Z_{G}(\Im\phi_1)H.$$ Thus, $g=g'h$ for some $g'\in Z_{G}(\Im\phi_1)$ and $h\in H$. Hence, $$\phi_2= \Ad(g^{-1})\circ\phi_1=
\Ad(h^{-1})\circ\Ad(g'^{-1})\circ\phi_1=\Ad(h^{-1})\circ\phi_1.$$ Therefore, $H$ is an “SCF" subgroup of $G$.

\[R:SCF2\] It is easy to see that if the condition (\[Eq2\]) in Lemma \[L:SCF2\] holds for an element $g\in G$, then it holds for all elements in the double coset $HgH$. By this, to verify the “SCF” condition of the pair $H\subset G$ it suffices to check the condition (\[Eq2\]) for a set of representatives of the double coset space $H\backslash G/H$.

Symmetric pairs
---------------

Let $G$ be a connected compact Lie group. Endow $G$ with a biinvariant Riemannian metric. Particularly it gives a $G$ conjugation invariant positive-definite inner product on ${\mathfrak{g}}_0$. Write ${\mathfrak{p}}_0$ for the orthogonal complement of ${\mathfrak{h}}_0$ in ${\mathfrak{g}}_0$.

\[L:symmetric1\] We have $G=H^{0}\exp({\mathfrak{p}}_0)$.

For any $g\in G$. Choose a minimal length geodesic $\gamma:[0,1]\rightarrow G$ connecting $H^{0}$ and $H^{0}g$. By the biinvariance of the metric, we may assume that $\gamma(0)=1\in H^0$. Then, there exists $X\in{\mathfrak{g}}_0$ such that$\gamma(t)=\exp(tX)$ ($\forall t\in[0,1]$). Due to $\gamma$ is of minimal length, we have $X\perp
{\mathfrak{h}}_0$. That means, $X\in{\mathfrak{p}}_0$. Write $\gamma(1)=xg$ for some $x\in H^{0}$. Then, $$g=x^{-1}\gamma(1)=x^{-1}\exp(X)\in H^{0}\exp({\mathfrak{p}}_0).$$ This shows the conclusion.

In case $H\subset G$ is a symmetric pair, Lemma \[L:symmetric1\] could be further simplified. Let $G$ be a connected compact Lie group and $\theta\in\Aut(G)$ be an involutive automorphism. Write $H=G^{\theta}$. Choose a maximal abelian subspace ${\mathfrak{a}}_0$ of ${\mathfrak{p}}_0$, and set $A=\exp({\mathfrak{a}}_0)$.

\[L:symmetric2\] We have $G=H^{0}AH^{0}$.

By [@Knapp Proposition 7.29], we have $${\mathfrak{p}}_0=\Ad(H^{0}){\mathfrak{a}}_0=\{\Ad(k)X:k\in H^{0},X\in{\mathfrak{a}}_0\}.$$ Then, by Lemma \[L:symmetric1\] we get $$G\!=\!H^{0}\exp({\mathfrak{p}}_0)\!=\!H^{0}\exp(\Ad(H^{0}){\mathfrak{a}}_0)\!=\!
H^{0}\exp({\mathfrak{a}}_0)H^{0}\!=\!H^{0}AH^{0}.$$

\[L:symmetric3\] For any $g\in\exp({\mathfrak{p}}_0)$, $H\cap\Ad(g)H=H^{g^{2}}$.

By definition $H=G^{\theta}$. Thus, $x\in H\cap\Ad(g)H$ if and only if $\theta(x)=x$ and $\theta(g)^{-1}
\theta(x)\theta(g)=\theta(g^{-1}xg)=g^{-1}xg$. This is also equivalent to: $x\in H$ and $x\in G^{\theta(g)g^{-1}}$. Due to $g\in\exp({\mathfrak{p}}_0)$, we have $\theta(g)=g^{-1}$. Therefore, we get $H\cap\Ad(g)H=H^{g^{2}}$.

Combining Lemma \[L:SCF2\], Lemma \[L:symmetric2\] and Lemma \[L:symmetric3\], we get the following statement.

\[L:symmetric4\] Assume that $\theta$ is an involutive automorphism of $G$, and $H=G^{\theta}$. Then, $H$ is an “SCF" subgroup of $G$ if and only if $g\in Z_{G}(H^{g^{2}})H$ for any $g\in A$.

\[L:symmetric5\] Let $(G,H)$ be one of the following pairs: $$(\U(n),\!\O(n)),\ (\U(2n),\!\Sp(n)),\ (\E_6,\!\F_4).$$ For any $g\in A-Z_{G}$, there exists $X\in{\mathfrak{a}}_0$ such that $g^{2}=\exp(2X)$ and $G^{g^{2}}=\Stab_{G}(X)$.

We know that the set of nonzero $A$ weights appearing in ${\mathfrak{g}}$ forms a restricted root system (cf. [@Knapp Page 370]). The centralizer $G^{g}=Z_{G}(g)$ ($g\in A$) and the stabilizer $\Stab_{G}(X)$ ($X\in{\mathfrak{a}}_0$) could be calculated from the evaluation of restricted roots on $g$ (or on $X$). The restricted root system for the pairs $(\U(n),\O(n))$, $(\U(2n),\Sp(n))$, $(\E_6,\F_4)$ are of types $\A_{n-1}$, $\A_{n-1}$, $\A_{2}$ respectively. One could show the conclusion by a case by case verification.

Let $(G,H)$ be one of the following pairs: $(\U(n),\O(n))$, $(\U(2n),\Sp(n))$, $(\E_6,\F_4)$. For any $g\in Z_{G}$, we have $$g\in Z_{G}(H^{g^2})\subset Z_{G}(H^{g^{2}})H.$$ For any $g\in A-Z_{G}$, by Lemma \[L:symmetric5\] there exists $X\in{\mathfrak{a}}_0$ such that $$g^{2}=\exp(2X)\textrm{ and }G^{g^{2}}=
\Stab_{G}(X).$$ Write $z=\exp(X)$ and $k=\exp(-X)g$. By $G^{g^{2}}=\Stab_{G}(X)$, we get $z\in Z_{G}
(H^{g^{2}})$. By $g^{2}=\exp(2X)$, we get $k^2=1$. Then, $\theta(k)=k^{-1}=k$. Thus, $k\in H$. Therefore, $g=zk\in Z_{G}(H^{g^{2}})H$. By Lemma \[L:SCF2\], the conclusion follows.

\[P:SO(2n+1)\] For any $n\geq 1$, $\SO(2n+1)$ is an “SCF" subgroup of $\U(2n+1)$.

Due to $\O(2n+1)=\SO(2n+1)\times\{\pm{I}\}$, $\SO(2n+1)$ is an “SCF” subgroup of $\O(2n+1)$. Then, the conclusion follows from Theorem \[T:Griess\](1).

$\G_2$
------

Let $(G,H)=(\SO(7),\G_2)$. Write $V={\mathfrak{p}}_0$. It is clear that $V$ is an irreducible real representation of $\G_2$ with $V\otimes_{\mathbb{R}}\mathbb{C}$ a nontrivial irreducible representation of $\G_2$ with minimal dimension.

For any $X\in V$, write $G_{X}=\Stab_{\SO(7)}(X)$ and $$\overline{G}_{X}=\{g\in\SO(7), g\cdot X=\pm{X}.\}$$

\[L:G2\] For any $0\neq X\in V$, we have $HG_{X}=G$, $\Stab_{H}(X)=H\cap G_{X}\cong\SU(3)$, and $\Stab_{H}(X)$ is an index $2$ subgroup of $H\cap\overline{G}_{X}$.

Write $M$ for the unit sphere in $V$. Then, $M\cong S^{6}$, and $G$ acts on $M$ transitively. Note that a proper closed subgroup of $H$ has dimension at most $8$. Thus, $\dim H\cdot X\geq 14-8=6=\dim M$ for any $X\in M$. By the connectedness of $M$. We see that $H$ acts on $M$ transitively. Then, for any $X\in M$, we have $HG_{X}=G$ and $\dim\Stab_{H}(X)=8$. Due to $\pi_1(H)=\pi_{0}(H)=1$, applying the spectral homotopy exact sequence to the fibration $$\Stab_{H}\mapsto H\mapsto M$$ we get $\pi_{0}(\Stab_{H}(X))=\pi_{1}(M)=1$. Thus, $\Stab_{H}(X)=8$. Therefore, $\Stab_{H}(X)=H\cap G_{X}\cong\SU(3)$.

Similarly, considering the transitive action of $H$ on $\mathbb{P}(M)=M/\pm{1}$, we have a fibration $$H\cap\overline{G}_{X}\mapsto H\mapsto \mathbb{P}(M).$$ Due to $\pi_1(H)=\pi_{0}(H)=1$, applying the spectral homotopy exact sequence we get $\pi_{0}(H\cap\overline{G}_{X})=\pi_{1}(\mathbb{P}(M))=C_2$. Thus, $\Stab_{H}(X)$ is an index $2$ subgroup of $H\cap\overline{G}_{X}$.

\[L:G2-1\] For any $1\neq g\in\exp(V)$, if $g^2\neq 1$, then $Z_{H}(g)\cong\SU(3)$; if $g^2=1$, then $Z_{H}(g)^{0}\cong
\SU(3)$ and the index $[Z_{H}(g),Z_{H}(g)^{0}]=2$.

For $1\neq g=\exp(X)\in\exp(V)$ ($X\in V$), we have $$\dim Z_{H}(g)\!=\!\dim H\!-\!\dim \Ad(H)g\geq 14-
\dim\exp(V)=\!7.$$ By considering the possible connected subgroups of $H\cong\G_2$, we get $$Z_{H}(g)^{0}
\cong\SU(3).$$ We may assume that $Z_{H}(g)^{0}$ equals to $K\subset H\subset G$, where $\SU(3)\cong K
\hookrightarrow G=\SO(7)$ through the standard embedding $$A+iB\mapsto\left(\begin{array}{ccc}A&B&0\\-B&A&0
\\0&0&1\end{array}\right).$$ The above map also give an inclusion of $\U(3)$ in $\SO(7)$. Write $K'$ for the image of $\U(3)$ in $\SO(7)$, and write $Z=Z(K')$. Then, $Z\cong\U(1)$ and $H\cap Z\subset K$. Using the form of embedding of $\U(3)$ in $\SO(7)$ above, one sees that $Z_{G}(K)=Z$. From $Z_{G}(g)^{0}=K$, we get $g\in Z_{G}(K)=Z$.

For any $1\neq g=\exp(X)\in Z$ ($X\in V$), if $g^2\neq 1$, then $G^{g}=K'=KZ\subset G_{X}$ and $$Z_{H}(g)=
H\cap G^{g}=H\cap KZ=K(H\cap Z)=K\cong\SU(3).$$ This shows $Z_{H}(g)=\Stab_{H}(X)$ by Lemma \[L:G2\]. If $g^2=1$, then $G^{g}=\overline{G}_{X}\cong\O(6)$ and $$Z_{H}(g)=H\cap G^{g}=H\cap\overline{G}_{X}.$$ By Lemma \[L:G2\] again, we get $Z_{H}(g)^{0}\cong\SU(3)$ and the index $$[Z_{H}(g),Z_{H}(g)^{0}]=2.$$

\[L:G2-2\] For any $g\in\exp(V)$, $H\cap\Ad(g)H=H^{g^{3}}$.

Consider $H=\G_2$ as a subgroup of $G'=\PSO(8)$. There is an order three element $\theta\in\Aut(G')$ such that $H=G^{\theta}$. Analogous to the proof of Lemma \[L:symmetric3\], one can show that $$H\cap\Ad(g)H=Z_{H}(\theta(g)^{-1}g).$$

Write $\mathfrak{q}_0$ for the orthogonal complement of ${\mathfrak{h}}_0$ in ${\mathfrak{g}}'_0$. Then, $\theta$ acts on $\mathfrak{q}_0$ as a linear transformation with $\theta^2+\theta+1=0$. For any $X\in V$, we show that $[\theta(X),X]=0$. Write $Y=[\theta(X),X]$. Write $\sigma\in\Aut(G')$ for an element with $G'^{\sigma}=G$. Then, $\sigma^2=1$ and $\sigma\theta\sigma^{-1}=\theta^{-1}$. Then, $$\theta(Y)=[\theta^{2}(X),\theta(X)]
=[-\theta(X)-X,\theta(X)]=[\theta(X),X]=Y$$ and $$\!\sigma(Y)\!=\![\sigma\theta(X),\!\sigma(X)]\!=\!
[\theta^{-1}\!\sigma(X),\!X]\!=\![\theta^{2}\!(X),\!X]\!=\![-\theta(X)\!-\!X,\!X]\!=\!-Y.$$ Since ${\mathfrak{g}}^{\theta}={\mathfrak{h}}\subset{\mathfrak{g}}^{\sigma}={\mathfrak{g}}$, we get $Y=\sigma(Y)=-Y$. Thus, $Y=0$.

Write $g=\exp(X)$ for some $X\in{\mathfrak{p}}_0$. Write $X'=X-\theta(X)$. Then, $\theta(X')=\theta(X)-\theta^{2}(X)
=X+2\theta(X)$. Thus, $2X'+\theta(X')=3X$. By this, $$Z_{H}(\theta(g)^{-1}g)=Z_{H}(\exp(X'))\subset
Z_{H}(\exp(3X))=Z_{H}(g^{3}).$$ Thus, $Z_{H}(g)\subset Z_{H}(\theta(g)^{-1}g)\subset Z_{H}(g^{3})$.

When $g^6\neq 1$, by Lemma \[L:G2-1\], we get $Z_{H}(g)=Z_{H}(g^3)\cong\SU(3)$. Thus, $Z_{H}(\theta(g)^{-1}g)=Z_{H}(g^{3})$. When $g^2=1$, then $g=g^3$ and $Z_{H}(g)=Z_{H}(g^3)$. Thus, $Z_{H}(\theta(g)^{-1}g)=Z_{H}(g^{3})$. When $g^3=1$, we have $g\in Z\cap K\subset H$. Thus, $\theta(g)
=g$ and $Z_{H}(\theta(g)^{-1}g)=Z_{H}(g^{3})=H$. When $o(g)=6$, we have $g^2\in H$, $\theta(g)^{-1}g
=\theta(g^3)^{-1}g^{3}$, and $$Z_{H}(\theta(g)^{-1}g)=Z_{H}(\theta(g^3)^{-1}g^3)=Z_{H}(g^3).$$

It suffices to show $(G,H)=(\SO(7),\G_2)$ is an “SCF” pair. By Lemma \[L:SCF2\] and Lemma \[L:symmetric1\], we just need to show: for any $g\in\exp(V)$, $$g\in Z_{G}(H\cap
\Ad(g)H)H.$$ By Lemma \[L:G2-2\], we have $H\cap\Ad(g)H=H^{g^{3}}$. Replacing $g$ by an $H$ conjugate element if necessary, we may assume that $g\in Z$ for $Z=Z(K')$ as in the proof of Lemma \[L:G2-1\].

When $g^6\neq 1$, we have $Z_{H}(g)=Z_{H}(g^{3})$ by Lemma \[L:G2\]. Thus, $$g\in Z_{G}(H^{3})\subset
Z_{G}(H\cap\Ad(g)H)H.$$ When $g^6=1$, we have $g^2\in Z\cap K\subset H$ and $g^3\in Z_{G}(H\cap\Ad(g)H)$. Thus, $$g=g^{3}g^{-2}\in Z_{G}(H\cap\Ad(g)H)H.$$

Pinned automorphisms
--------------------

[*The pair $\O(2n+1)\subset\SO(2n+2)$.*]{} Let $n\geq 1$, $G=\SO(2n+2)$ and $\theta=
\Ad(I_{2n+1,1})$. Then, $\theta^2=1$ and $G^{\theta}=\O(2n+1)$. Write $H=G^{\theta}$. Write $$g_{\theta}=\left(\begin{array}{ccc}I_{2n}&0&0\\0&\cos\theta&\sin\theta\\0&-\sin\theta&\cos\theta
\\\end{array}\right)$$ ($\theta\in\mathbb{R}$). By Lemma \[L:symmetric2\], we have $G=HAH$ for $$A=\{g_{\theta}:\theta\in[0,2\pi)\}.$$ We know that the condition $(\ref{Eq2})$ in Lemma \[L:SCF2\] fails for an element $g\in A$ if and only if it fails for all elements in the double coset $HgH$.

\[P:O(2n+1)-1\] For $g=g_{\theta}\in A$ ($\theta\in[0,2\pi)$), the condition $(\ref{Eq2})$ in Lemma \[L:SCF2\] fails if and only if $\theta=\frac{\pi}{2}$ or $\frac{3\pi}{2}$.

When $\theta\in[0,2\pi)-\{0,\frac{\pi}{2},\pi,\frac{3\pi}{2}\}$, one could verify that $G^{g^2}=G^{g}$. Thus, $$g\in Z_{G}(H^{g})=Z_{G}(H^{g^2})\subset Z_{G}(H^{g^2})H.$$ Hence, the condition $(\ref{Eq2})$ in Lemma \[L:SCF2\] holds. When $\theta=0$ or $\pi$, $$g\in H\subset Z_{G}(H^{g^2})H.$$ Hence, the condition $(\ref{Eq2})$ in Lemma \[L:SCF2\] holds.

When $\theta=\frac{\pi}{2}$ or $\frac{3\pi}{2}$, $g^2=g_{\pi}=\diag\{I_{2n},-1,-1\}$. In this case $H^{g^{2}}=S(\O(2n)\!\times\!\O(1)\!\times\!\O(1))$. Thus, $Z_{G}(H^{g^2})\subset H$ and $Z_{G}(H^{g^2})H=H$. However, $g\not\in H$. Hence, the condition $(\ref{Eq2})$ in Lemma \[L:SCF2\] fails.

Different with $\O(2n+1)\subset\SO(2n+2)$, the pair $\SO(2n+1)\subset\SO(2n+2)$ is an “SCF” pair.

\[P:O(2n+1)-2\] For any $n\geq 1$, $\SO(2n+1)$ is and “SCF” subgroup of $\SO(2n+2)$.

Following the notation as in the proof of Proposition \[P:O(2n+1)-1\]. We have $H=\O(2n+1)$, and $H^{0}=\SO(2n+1)$.

When $\theta\in[0,2\pi)-\{0,\frac{\pi}{2},\pi,\frac{3\pi}{2}\}$, we have $$H\cap\Ad(g_{\theta})H
=H^{g_{\theta}^2}=S(\O(2n)\!\times\!\O(1)\!\times\!\O(1)).$$ Then, $$H^{0}\cap\Ad(g_{\theta})H=
S(\O(2n)\!\times\!\O(1)).$$ By this, $$H^{0}\cap\Ad(g_{\theta})H^0=\SO(2n)\textrm{ or }
S(\O(2n)\!\times\!\O(1)).$$ By calculation one sees that $\diag\{I_{2n-1},-1,-1,1\}\not\in
\Ad(g_{\theta})H^0$. Thus, $$H^{0}\cap\Ad(g_{\theta})H^0=\SO(2n).$$ Hence, $$g_{\theta}\in
Z_{G}(H^{0}\cap\Ad(g_{\theta})H^0)\subset Z_{G}(H^{0}\cap\Ad(g_{\theta})H^0)H^{0}.$$

When $\theta=\frac{\pi}{2}$ or $\frac{3\pi}{2}$, by direct calculation one sees that $H^{0}
\cap\Ad(g_{\theta})H^0=\SO(2n)$. Thus, $$g_{\theta}\in Z_{G}(H^{0}\cap\Ad(g_{\theta})H^0)
\subset Z_{G}(H^{0}\cap\Ad(g_{\theta})H^0)H^{0}.$$

When $\theta=0$ or $\pi$, $$g_{\theta}\in Z_{G}H^0\subset Z_{G}(H^{0}\cap\Ad(g_{\theta})H^0)H^{0}.$$

[*The pair $\Spin(2n+1)\subset\Spin(2n+2)$.*]{} Let $n\geq 1$, $G=\Spin(2n+2)$ and $\theta
=\Ad(e_{2n+2})$. Then, $\theta^2=1$. Write $H=G^{\theta}$. Then, $H=\Spin(2n+1)$. Write $$g_{\theta}
=\cos\theta+\sin\theta e_{2n+1}e_{2n+2}$$ ($\theta\in\mathbb{R}$). By Lemma \[L:symmetric2\], we have $G=HAH$ for $$A=\{g_{\theta}:\theta\in[0,2\pi)\}.$$

\[L:spin(2n+1)\] For any $n\geq 1$, $\Spin(2n+1)$ is and “SCF” subgroup of $\Spin(2n+2)$.

By Lemma \[L:symmetric4\], it suffices to show that $g\in Z_{G}(H^{g^2})H$ for any $g=
g_{\theta}$ ($\theta\in[0,2\pi)$).

When $\theta\in[0,2\pi)-\{0,\frac{\pi}{2},\pi,\frac{3\pi}{2}\}$, we have $g^2=g_{2\theta}$ and $H^{g^{2}}=\Spin(2n)$. Thus, $$g\in Z_{G}(H^{g^2})\subset Z_{G}(H^{g^2})H.$$

When $\theta=\frac{\pi}{2}$ or $\frac{3\pi}{2}$, we have $e_1\dots e_{2n+2}\in Z_{G}\subset
Z_{G}(H^{g^2})$ and $(e_1\dots e_{2n+2})^{-1}g=\pm{e_1\dots e_{2n}}\in H$. Thus, $$g=(e_1\dots e_{2n+2})((e_1\dots e_{2n+2})^{-1}g)\in Z_{G}(H^{g^2})H.$$

When $\theta=0$ or $\pi$, $$g=g_{\theta}\in Z_{G}\subset Z_{G}(H^{g^2})H.$$

[*Pinned automorphisms.*]{}

\[T:pinned\] Let $G$ be a connected and simply-connected compact simple Lie group and $\theta$ be a (nontrivial) pinned automorphism of $G$. Let $\overline{G}=G\rtimes\langle\overline{\theta}\rangle$, where $o(\overline{\theta})=o(\theta)$ and $\Ad(\overline{\theta})|_{G}=\theta$. Let $H=G^{\theta}$. Then $H$ is an “SCF" subgroup of $\overline{G}$ (and of $G$).

For any $g\in G$ and any $k\in\mathbb{Z}$, write $g'=\overline{\theta}^{k}g$. Then, $\Ad(g')H=\Ad(g)H$, and $$Z_{\overline{G}}(H\cap\Ad(g')H)H=\langle\overline{\theta}\rangle Z_{G}(H\cap\Ad(g)H)H.$$ Thus, it suffices to show $H$ is an “SCF" subgroup of $G$. The possible pairs $(G,G^{\theta})$ are as follows,

1.  $\SO(2n+1)\subset\SU(2n+1)$;

2.  $\Sp(n)\subset\SU(2n)$;

3.  $\Spin(2n+1)\subset\Spin(2n+2)$;

4.  $\G_2\subset\Spin(8)$;

5.  $\F_4\subset\E_6$.

All of these follow results shown above. Particularly for $\G_2\subset\Spin(8)$, we know that $\G_2$ is an “SCF" subgroup of $\Spin(7)$ (Theorem \[T:Griess\](2)), and $\Spin(7)$ is an “SCF" subgroup of $\Spin(8)$ (Lemma \[L:spin(2n+1)\]). Thus, $\G_2$ is an “SCF" subgroup of $\Spin(8)$.

$\Spin(7)$
----------

The following lemma is a generalization of Lemma \[L:symmetric1\].

\[L:double\] Let $G$ be a connected compact Lie group endowed with a biinvariant Riemannian metric. Let $H_1$, $H_2$ be two closed subgroups of $G$. Write ${\mathfrak{p}}_0$ for the orthogonal complement of ${\mathfrak{h}}_{1,0}+{\mathfrak{h}}_{2,0}$ in ${\mathfrak{g}}_0$. Then, $G=H_{1}^{0}\exp({\mathfrak{p}}_0)H_{2}^{0}$.

For any $g\in G$. Choose a minimal length geodesic $\gamma:[0,1]\rightarrow G$ connecting $H_{1}^{0}$ and $H_{2}^{0}g$. By the biinvariance of the metric, we may assume that $\gamma(0)=1\in H^0$ and $\gamma(1)=g$. Then, there exists $X\in{\mathfrak{g}}_0$ such that$\gamma(t)=\exp(tX)$ ($\forall t\in[0,1]$). Due to $\gamma$ is of minimal length, the tangent vector of $\gamma$ at $t=0$ is orthogonal to $T_{e}(H_1^{0})$, and the tangent vector of $\gamma$ at $t=1$ is orthogonal to $T_{g}(H_2^{0}g)$. By the biinvariance of the Riemannina metric, we get $X\perp{\mathfrak{h}}_{1,0}$ and $X\perp{\mathfrak{h}}_{2,0}$. That means, $X\in{\mathfrak{p}}_0$. Then, $$g=\gamma(1)=\exp(X)\in
H_{1}^{0}\exp({\mathfrak{p}}_0)H_{2}^{0}.$$ This shows the conclusion.

Write $G=\Spin(8)$. Let $\theta,\sigma$ be diagram automorphisms of $G$ with $G^{\theta}=\G_2\subset
G^{\sigma}=\Spin(7)$. Then, $o(\theta)=3$, $o(\sigma)=2$, and $\sigma\theta\sigma^{-1}=\theta^{-1}$. Put $$B=\langle\theta,\sigma\rangle\subset\Aut(G),$$ and $$\overline{G}=G\rtimes B.$$ Let $H=G^{\sigma}$. For $\theta\in\mathbb{R}$, write $g_{\theta}=\cos\theta+\sin\theta e_{7}e_{8}$.

\[P:spin7-1\] The pair $H\subset\overline{G}$ is an “SCF" pair.

Since $\sigma$ commutes with $H$, it suffices to show that $$g\in Z_{G}(H\cap\Ad(g)H)H$$ for elements $g\in G\cup G\theta\cup G\theta^{2}$.

Firstly we consider elements $g\in G\theta\cup G\theta^{2}$. For an element $X\in{\mathfrak{g}}_0$, $X\in{\mathfrak{h}}_0
\cap\Ad(\theta){\mathfrak{h}}_0$ if and only if $\sigma(X)=X$ and $\sigma(\theta(X))=\theta(X)$. Using $\theta^{-1}\sigma\theta=\theta\sigma$ and ${\mathfrak{g}}_0^{\theta}\subset{\mathfrak{g}}_0^{\sigma}$, the above is equivalent to $\theta(X)=X$. Thus, ${\mathfrak{h}}_0\cap\theta({\mathfrak{h}}_0)={\mathfrak{g}}_0^{\theta}$. Counting dimension, we get ${\mathfrak{h}}_0+\theta({\mathfrak{h}}_0)={\mathfrak{g}}_0$. By Lemma \[L:double\], we get $G=H\theta(H)$. Then, $G\theta=H\theta H$. Similarly one shows $G\theta^2=H\theta^{2}H$. Thus, $H\backslash G\theta/H$ (or $H\backslash G\theta^{2}/H$) is a single double coset. Hence, it suffices to verify the condition (\[Eq2\]) in Lemma \[L:SCF2\] for $g=\theta$ or $\theta^2$. In this case, $H\cap\Ad(g)(H)=
G^{\theta}$, and $$g\in Z_{G}(H\cap\Ad(g)H)\subset Z_{G}(H\cap\Ad(g)H)H.$$ Therefore, the condition $(\ref{Eq2})$ in Lemma \[L:SCF2\] for elements $g\in G\theta\cup G\theta^{2}$.

Secondly we consider elements $g\in G$. By Lemma \[L:spin(2n+1)\], we get $g\in Z_{G}(H\cap\Ad(g)H)H$ for elements $g\in G$.

We have $$G^{\sigma}\subset\langle G,\sigma\theta\rangle/\langle\theta(-1)\rangle=\Ad(\theta)(\langle G,
\sigma\rangle/\langle-1\rangle).$$ Since $\langle G,\sigma\rangle/\langle-1\rangle\cong\O(8)$, we get $$\langle G,\sigma\theta\rangle/\langle\theta(-1)\rangle\cong\O(8).$$ From this we get an inclusion $\Spin(7)\hookrightarrow\O(8)$, which is just the spinor module of $\Spin(7)$. Write $$G'=\langle G,
\sigma\theta\rangle/\langle\theta(-1)\rangle,$$ and consider the pair $H=G^{\sigma}\subset G'$. We still write $g_{\theta}=\cos\theta+\sin\theta e_{7}e_{8}$, and write $[g_{\theta}]$ for its projection in $G'$. Write $A'=\{[g_{\theta}]:\theta\in[0,2\pi)\}$.

\[P:spin7-2\] For $g\in G'$, the condition $(\ref{Eq2})$ in Lemma \[L:SCF2\] fails if and only if $$g\in
H[g_{\frac{k\pi}{4}}]H$$ ($k=1,3,5,7$).

Consider elements $g\in G'^{0}\sigma\theta$ and $g\in G'^{0}$ separately.

Firstly let $g\in G'^{0}\sigma\theta$. By the proof of Proposition \[P:spin7-1\], $H\backslash G'^{0}
\sigma\theta/H$ is a single double coset. Thus, it suffices to consider $g=\sigma\theta$. In this case, $H\cap\Ad(g)(H)=G^{\theta}$, and $$g\in Z_{G}(H\cap\Ad(g)H)\subset Z_{G}(H\cap\Ad(g)H)H.$$

Secondly let $g\in G'^{0}$. By Lemma \[L:symmetric2\] one shows that $G'^{0}=HA'H$. Then, it suffices to consider $g=[g_{\theta}]$ ($\theta\in[0,2\pi)$). When $\theta=0$ or $\pi$, $$g=
[g_{\theta}]\in Z_{G'}\subset Z_{G'}(H^{0}\cap\Ad(g_{\theta})H^0)H^{0}.$$ When $\theta=\frac{\pi}{2}$ or $\frac{3\pi}{2}$, we have $$g=[g_{\theta}]=[\theta(-1)g_{\theta}]=[\pm{e_1\dots e_{6}}]\in H
\subset Z_{G'}(H^{g^2})H.$$

Now assume $\theta\in[0,2\pi)-\{0,\frac{\pi}{2},\pi,\frac{3\pi}{2}\}$. In general, for $x\in H\subset G$, $[x]\in\Ad(g)H$ if and only if $g_{\theta}^{-1}xg_{\theta}\in H\cup H\theta(-1)$. This is also equivalent to $$\sigma(g_{\theta}^{-1}xg_{\theta})=g_{\theta}^{-1}xg_{\theta}$$ or $$\sigma(g_{\theta}^{-1}x
g_{\theta}\theta(-1))=g_{\theta}^{-1}xg_{\theta}\theta(-1).$$ Due to $(\sigma\theta(-1))\theta(-1)=-1$, the above is equivalent to $$[g_{2\theta},x]=\pm{1}.$$ The last condition just means the projections of $g_{2\theta}$ and $x$ in $G/\langle-1\rangle\cong\SO(8)$ commute. When $\frac{4\theta}{\pi}\not\in
\mathbb{Z}$, $\SO(7)^{g_{2\theta}}=\SO(6)$ is connected. Thus, $$H\cap\Ad(g)H=\Spin(6).$$ Hence, $$g=[g_{\theta}]\in Z_{G'}(H\cap\Ad(g)H)\subset Z_{G'}(H\cap\Ad(g)H)H.$$ When $\frac{4\theta}{\pi}
\in\mathbb{Z}$ (and $\theta\in[0,2\pi)-\{0,\frac{\pi}{2},\pi,\frac{3\pi}{2}\}$), we have $\theta=
\frac{k\pi}{4}$ with $k=1,3,5,7$. In this case $[g_{2\theta},x]=\pm{1}$ if and only if $x\in\Pin(6)
\subset\Spin(7)$. Thus, $$H\cap\Ad(g)H=\Pin(6).$$ One calculates that $Z_{G'}(\Pin(6))=Z_{G'}$. It is clear that $[g_{\frac{k\pi}{4}}]\not\in Z_{G'}H$ ($k=1,3,5,7$). Thus, $$g\not\in Z_{G'}(H\cap
\Ad(g)H)H$$ for $g=[g_{\frac{k\pi}{4}}]$ ($k=1,3,5,7$).

Remark: after posing a previous version of this paper on the arXiv, Gaetan Chenevier and Wee Teck Gan showed me that ¡°$\Spin(7)$ is unacceptable¡±. This led me to check statements and proofs in that version. Proposition \[P:O(2n+1)-1\] and Proposition \[P:spin7-2\] are the corrected statements.

Acceptable/unacceptable compact Lie groups
==========================================

Examples of acceptable compact Lie groups
-----------------------------------------

The acceptability of groups in items (1) and (2) of the following theorem are well known (cf. [@Larsen1],[@Larsen2],[@Griess]). Here we show their strong acceptability from the stronger “SCF” property. Acceptability of the group $\SO(4)$ is new.

\[T:acceptable1\] Groups isomorphic to any one in the following list are strongly acceptable,

-   $\U(n)$, $\SU(n)$, $\Sp(n)$, $\O(n)$, $\SO(2n+1)$.

-   $\G_2$;

-   $\SO(4)$.

For groups in item (1), strongly acceptability of $\U(n)$ and $\SU(n)$ follow from the character theory of representations of compact Lie groups. By Theorem \[T:Griess\] (1) and Proposition \[P:SCF\], $\Sp(n)$ and $\O(n)$ are strongly acceptable as $\U(n)$ is. By Propositions \[P:SO(2n+1)\] and \[P:SCF\], so is $\SO(2n+1)$.

By Theorem \[T:Griess\] (2) and Proposition \[P:SCF\], $\G_2$ is strongly acceptable as $\SO(7)$ is.

Take an involution $\theta\in\G_2$. Then, $$\G_{2}^{\theta}\cong\Sp(1)^{2}/\langle(-1,-1)\rangle$$ (cf. [@Huang-Yu Table2]). By Proposition \[P:centralizer2\], $\SO(4)\cong\Sp(1)^{2}/\langle(-1,-1)$ is strongly acceptable as $\G_2$ is.

Examples of unacceptable compact Lie groups
-------------------------------------------

In the following, we construct a concrete example of two element-conjugate homomorphisms from $(C_4)^{2}$ which are not globally conjugate. This construction not only shows that $\SU(4)/\langle
-I\rangle$ (and also $\SO(6)$) is unacceptable, but also is the building block in showing many connected compact semisimple Lie groups are unacceptable in Theorem \[T:unacceptable1\] and Theorem \[T:unacceptable3\].

\[E:SU4\] Let $G=\SU(4)/\langle-I\rangle$, and $\Gamma=(C_4)^{2}$ with two generators $\gamma_1,\gamma_2$. Define $\phi:\Gamma\rightarrow\SU(4)$ by $$\phi(\gamma_1)=\diag\{1,1,\mathbf{i},-\mathbf{i}\},
\quad\phi(\gamma_2)=\diag\{1,\mathbf{i},1,-\mathbf{i}\}.$$ Define $\phi':\Gamma\rightarrow\SU(4)$ by $$\phi'(\gamma)=\overline{\phi(\gamma)},\ \forall\gamma\in\Gamma.$$ Write $\pi:\SU(4)\rightarrow
\SU(4)/\langle-I\rangle$ for the projection. Set $\rho=\pi\circ\phi$, and $\rho'=\pi\circ\phi'$. In the below, we show that $\rho$ and $\rho'$ are element-conjugate, but not globally conjugate. Thus, $\SU(4)/\langle-I\rangle$ is unacceptable.

Write $\gamma=\gamma_1^{a}\gamma_2^{b}$ ($a,b\in\{1,2,3,4\}$) for a general element in $\Gamma$. Then, $$\gamma(\gamma)=\diag\{1,\mathbf{i}^{b},\mathbf{i}^{a},(-\mathbf{i})^{a+b}\}.$$ When $(a,b)
\equiv(0,1),(1,0),(0,3),(3,0),(1,3),(3,1)$, $$\phi'(\gamma)=\overline{\phi(\gamma)}\sim\phi(\gamma);$$ when $(a,b)\equiv(2,1),(1,2),(2,3),(3,2),(1,1),(3,3)$, $$\phi'(\gamma)=\overline{\phi(\gamma)}\sim
-\phi(\gamma);$$ when $a$ and $b$ are both even, $$\phi'(\gamma)=\overline{\phi(\gamma)}=\phi(\gamma).$$ In any case we have $\rho'(\gamma)\sim\rho(\gamma)$. Thus, $\rho$ and $\rho'$ are element-conjugate.

Suppose $\rho$ and $\rho'$ are globally conjugate. Then, there exists $g\in\SU(4)$ such that $$\phi'(\gamma)
=\pm{g\phi(\gamma)g^{-1}},\ \forall\gamma\in\Gamma.$$ Write $$\phi'(\gamma_1)=t_{1}g\phi(\gamma_1)g^{-1}
\textrm{ and }\phi'(\gamma_2)=t_{2}g\phi(\gamma_2)g^{-1},$$ where $t_1,t_2\in\{\pm{1}\}$. Then, $$\phi'(\gamma_1^{2})=g\phi(\gamma_1^{2})g^{-1}\textrm{ and }\phi'(\gamma_2^{2})=g\phi(\gamma_2^{2})g^{-1}.$$ Looking at the forms of $\phi'(\gamma_{i}^{2})=\phi(\gamma_{i}^{2})$ ($i=1,2$), we see that $g$ is a diagonal matrix. Then, $g$ commutes with $\phi(\gamma_1)$ and $\phi(\gamma_2)$. However, $\phi'(\gamma_1)\neq
\pm{\phi(\gamma_1)}$. This is a contradiction. Therefore, $\rho$ and $\rho'$ are not globally conjugate.

Due to $\SO(6)\cong\SU(4)/\langle-I\rangle$, $\SO(6)$ is also unacceptable.

Remark: the unacceptability of $\SO(6)$ is first shown by Matthew Weidner. The above counter-example for $\SU(4)/\langle-I\rangle$ is translated from a counter-example for $\SO(6)$ in [@Weidner].

\[E:Sp3-2\] For any any $m\geq 3$, let $G=\Sp(1)^{m}/\langle(-1,\dots,-1)\rangle$. Write $\Gamma=(C_4)^{2}$ with two generators $\gamma_1,\gamma_2$. Let $\epsilon=\pm{1}$. Define $\phi:\Gamma\rightarrow
\Sp(1)^{m}$ by $$\phi(\gamma_1)=(1,\dots,1,\mathbf{i},\mathbf{i}),\quad\phi(\gamma_2)=(\mathbf{i},
\dots,\mathbf{i},1,\mathbf{i}).$$ Define $\phi':\Gamma\rightarrow\Sp(1)^{3}$ by $$\phi'(\gamma_1)
=\phi(\gamma_1)=(1,\dots,1,\mathbf{i},\mathbf{i})$$ and $$\phi'(\gamma_2)=(\epsilon\mathbf{i},
\dots,\epsilon\mathbf{i},1,-\mathbf{i}).$$ Write $$\pi:\Sp(1)^{m}\rightarrow G=\Sp(1)^{m}/\langle
(-1,\dots,-1)\rangle$$ for the projection. Set $$\rho=\pi\circ\phi,\quad\rho'=\pi\circ\phi'.$$ In the below, we show that $\rho$ and $\rho'$ are element-conjugate, but not globally conjugate. Thus, $\Sp(1)^{m}/\langle(-1,\dots,-1)\rangle$ is unacceptable.

Write $\gamma=\gamma_1^{a}\gamma_2^{b}$ ($a,b\in\{1,2,3,4\}$) for a general element in $\Gamma$. Then, $$\phi(\gamma)=(\mathbf{i}^{b},\dots,\mathbf{i}^{b},\mathbf{i}^{a},\mathbf{i}^{a+b})$$ and $$\phi'(\gamma)=(\epsilon^{b}\mathbf{i}^{b},\dots,\epsilon^{b}\mathbf{i}^{b},\mathbf{i}^{a},
(-1)^{b}\mathbf{i}^{a+b}).$$ When $b$ is even, we have $\phi'(\gamma)=\phi(\gamma)$; when $b$ is odd and $a$ is even, we have $\phi'(\gamma)\sim\phi(\gamma)$; when $b$ and $a$ are both odd, we have $\phi'(\gamma)\sim(-1,\dots,-1)\phi(\gamma)$. In any case we have $\rho'(\gamma)\sim
\rho(\gamma)$. Thus, $\rho$ and $\rho'$ are element-conjugate.

Write $$z=(-1,\dots,-1)\in\Sp(1)^{m}.$$ Suppose $\rho$ and $\rho'$ are globally conjugate. Then, there exists $g\in\Sp(1)^{m}$ such that $$\phi'(\gamma)=z^{t}g\phi(\gamma)g^{-1},\ \forall\gamma
\in\Gamma$$ ($t=0$ or $1$, depending on $\gamma$). Write $$\phi'(\gamma_1)=z^{t_{1}}g
\phi(\gamma_1)g^{-1}\textrm{ and }\phi'(\gamma_2)=z^{t_{2}}g\phi(\gamma_2)g^{-1},$$ where $t_1,
t_2\in\{0,1\}$. Write $$g=\diag\{g_1,g_2,g_3\}$$ ($g_{i}\in\Sp(1)$). From $\phi'(\gamma_1)=
z^{t_{1}}g\phi(\gamma_1)g^{-1}$ and $\phi'(\gamma_2)=z^{t_{2}}g\phi(\gamma_2)g^{-1}$, and the precise forms of $\phi(\gamma_{j})$, $\phi'(\gamma_{j})$ ($j=1,2$), one sees that $$g_{j}
\mathbf{i}g_{j}^{-1}=\eta_{j}\mathbf{i}$$ for some $\eta_{j}=\pm{1}$ ($j\in\!\{1,2,3\}$). More precisely, from $\phi'(\gamma_1)=z^{t_{1}}g\phi(\gamma_1)g^{-1}$ where $$\phi'(\gamma_1)=
\phi(\gamma_1)=(1,\dots,1,\mathbf{i},\mathbf{i})$$ with the first components both equal to $1$, we get $t_1=0$ and $\eta_2=\eta_3=1$. From $\phi'(\gamma_2)=z^{t_{2}}g\phi(\gamma_2)g^{-1}$ where $$\phi(\gamma_2)=(\mathbf{i},\dots,\mathbf{i},1,\mathbf{i})\textrm{ and }\phi'(\gamma_2)
=(\epsilon\mathbf{i},\dots,\epsilon\mathbf{i},1,-\mathbf{i})$$ with the second components both equal to $1$, we get $t_2=0$ and $\eta_1=\epsilon$, $\eta_3=-1$. There is a contraction with $1=\eta_3=-1$. Therefore, $\rho$ and $\rho'$ are not globally conjugate.

\[E:Sp3\] Let $G=\PSp(3)=\Sp(3)/\langle-I\rangle$. Let $$A=\langle\diag\{(-1,1,1)\},\diag\{(1,-1,1)\}\rangle.$$ Then, $$Z_{G}(A)=\Sp(1)^{3}/\langle(-1,-1,-1)\rangle.$$ From Example \[E:Sp3-2\], $\Sp(1)^{3}/\langle(-1,-1,-1)\rangle$ is unacceptable. By Proposition \[P:centralizer2\], $\PSp(3)$ is also unacceptable.

The following theorem \[T:unacceptable-ChenevierGan\] is due to Gaetan Chenevier and Wee Teck Gan ([@Chenevier-Gan]). We have $$\Spin(7)^{e_1e_2e_3e_4}\cong\Sp(1)^{3}/\langle(-1,-1,-1)\rangle.$$ By Proposition \[P:centralizer2\] the unacceptability of $\Spin(7)$ also follows from the unacceptability of $\Sp(1)^{3}/\langle(-1,-1,-1)\rangle$.

\[T:unacceptable-ChenevierGan\] The group $\Spin(7)$ is unacceptable.

Now let $$G=\Sp(1)^{3}/\langle(1,-1,-1),(-1,1,-1)\rangle.$$ Write $\overline{G}=\SO(3)^{3}$. There is a natural projection $$\pi':\Sp(1)\rightarrow\SO(3).$$ Write $$\pi: G\rightarrow\overline{G}=\SO(3)^{3}$$ for the projection induced by $\pi'$. In $\SO(3)$, put $$S=\left(\begin{array}{ccc}0&1&0\\-1&0&0\\0&0&
1\\\end{array}\right)$$ and $$T=\left(\begin{array}{ccc}1&0&0\\0&-1&0\\0&0&-1\\\end{array}\right).$$ We know that all elements of order $2$ in $\SO(3)$ are conjugate to $T$. Paritularly, $S^2\sim T$.

\[L:3A1-1\] For an element $x\in\Sp(1)$ to satisfy $x\sim-x$ it is necessary and sufficient that $\pi'(x)\sim T$.

For any $x\in\Sp(1)$, $x\sim-x$ if and only if $x\sim\mathbf{i}$. The latter is also equivalent to $\pi'(x)
\sim T$.

For a finite subgroup $\overline{\Gamma}$ of $\overline{G}$, define $\overline{\Gamma}'$ as the subgroup generated by all elements $\overline{g}=(\overline{g}_1,\overline{g}_2,\overline{g}_3)\in\overline{\Gamma}$ with $\overline{g}_{j}\not\sim T$ for each $j=1,2,3$, and all elements $\overline{g}^{2}$ ($\overline{g}
\in\overline{\Gamma}$). Then, $\overline{\Gamma}/\overline{\Gamma}'$ is an elementary abelian $2$-group. Define $$Y_{\overline{\Gamma}}=\Hom(\overline{\Gamma}/\overline{\Gamma}',Z_{G}).$$ Note that $Z_{G}\cong
\{\pm{1}\}$. Define $$X_{\overline{\Gamma}}=Z_{\overline{G}}(\overline{\Gamma})/\pi(Z_{G}(\pi^{-1}
(\overline{\Gamma}))).$$

\[L:3A1-2\] There is a natural injective homomorphism $$\phi=\phi_{\overline{\Gamma}}:X_{\overline{\Gamma}}
\rightarrow Y_{\overline{\Gamma}}.$$

For any $\overline{g}=\pi(g)\in\overline{G}$ ($g\in G$), $\overline{g}\in Z_{\overline{G}}
(\overline{\Gamma})$ if and only if there is a map $\chi=\chi_{g}:\overline{\Gamma}\rightarrow 
Z_{G}$ such that $$gxg^{-1}=\chi(\overline{x})x$$ for all $x\in G$ with $\pi(x)=\overline{x}
\in\overline{\Gamma}$. It is easy to show that the map $\chi_{g}:\overline{\Gamma}\rightarrow 
Z_{G}$ is a homomorphism.

By Lemma \[L:3A1-1\], $\chi_{g}$ is trivial on elements $\overline{g}=(\overline{g}_1,\overline{g}_2,
\overline{g}_3)\in\overline{\Gamma}$with $\overline{g}_{j}\not\sim T$ for each $j=1,2,3$. Since $\chi_{g}$ is a homomorphism and $Z_{G}$ is of order $2$, $\chi_{g}$ is trivial on elements $\overline{g}^{2}$ ($\overline{g}\in\overline{\Gamma}$). On the other hand, $\chi_{g}$ is induced from the conjugation action of $g$ on $\pi^{-1}(\overline{\Gamma})$. Thus, $\chi_{g}$ is trivial if and only if $g\in Z_{G}(\pi^{-1}(\overline{\Gamma}))$, which is equivalent to $\overline{g}\in
\pi(Z_{G}(\pi^{-1}(\overline{\Gamma})))$. All in all, we get an injective map $\phi: X_{\overline{\Gamma}}
\rightarrow Y_{\overline{\Gamma}}$ by defining $$\phi([\overline{g}])=\chi_{g},$$ which is clearly a homomorphism.

\[L:3A1-3\] For $G$ to be unacceptable it is necessary and sufficient that there is a finite subgroup $\overline{\Gamma}$ of $\overline{G}$ with $\phi_{\overline{\Gamma}}$ not an isomorphism.

Sufficiency. Suppose $\phi_{\overline{\Gamma}}$ is not an isomorphism for some finite subgroup $\overline{\Gamma}$ of $\overline{G}$. Then it is not surjective. Hence, there is an element $\chi\in Y_{\overline{\Gamma}}$ such that $$\chi\neq\chi_{g}$$ for any $g\in\pi^{-1}(Z_{\overline{G}}(\overline{\Gamma}))$. Put $$\Gamma=
\pi^{-1}(\overline{\Gamma}).$$ Let $\rho:\Gamma\rightarrow G$ be the inclusion map. Define $\rho':\Gamma\rightarrow G$ by $$\phi'(x)=\chi(\pi(x))x.$$ By our definition of the subgroup $\overline{\Gamma}'\subset\overline{\Gamma}$ and the set $Y_{\overline{\Gamma}}$, $\phi$ and $\phi'$ are element-conjugate homomorphisms. Suppose $\phi'$ is conjugate to $\phi$. Then there exists $g\in G$ such that $\phi'(x)=g\phi(x)g^{-1}$ for all $x\in\Gamma$. Projecting to $\overline{G}$, one sees that $g\in\pi^{-1}(Z_{\overline{G}}(\overline{\Gamma}))$ and $$g\phi(x)g^{-1}=\chi_{g}(\pi(x))x$$ for any $x\in\Gamma$. Thus, $$\chi(\pi(x))x=\phi'(x)=
g\phi(x)g^{-1}=\chi_{g}(\pi(x))x$$ for all $x\in\Gamma$. That just means $\chi=\chi_{g}$, which is in contraction with the condition $\chi\neq\chi_{g}$ ($g\in\pi^{-1}(Z_{\overline{G}}
(\overline{\Gamma}))$. Hence, $\phi$ and $\phi'$ are element-conjugate, but not globally conjugate. Therefore, $G$ is unacceptable.

Necessarity. Let $\phi,\phi':\Gamma\rightarrow G$ be two element-conjugate but not globally conjugate homomorphisms from a finite group $\Gamma$. Then it is clear that $\ker\phi=\ker\phi'$. Considering $\Gamma/\ker\phi$ instead, we may that $\phi$ and $\phi'$ are injective. Moreover, we many assume that $\Gamma\subset G$ and $\phi$ is the inclusion map from $\Gamma$ to it. Since $\overline{G}\cong\SO(3)^{3}$ is acceptable, there exists $g\in G$ such that $\pi\circ\phi=
\Ad(\pi(g))\circ\pi\circ\phi'=\pi\circ\Ad(g)\circ\phi'$. Considering $\Ad(g)\circ\phi'$ instead, we may further assume that $\pi\circ\phi=\pi\circ\phi'$. Then, there exists a homomorphism $\chi:\Gamma\rightarrow Z_{G}$ such that $$\phi'(x)=\chi(x)x$$ for all $x\in\Gamma$. Write $\overline{\Gamma}=\pi(\Gamma)$. One can show that: $\phi$ and $\phi'$ being element-conjugate is equivalent to $$\chi\in Y_{\overline{\Gamma}}$$; $\phi$ and $\phi'$ being not globally conjugate is equivalent to $$\chi\not\in\phi(X_{\overline{\Gamma}}).$$ Thus, $\phi$ is not surjective, hence not an isomorphism.

Recall that we define matrices $S,T\in\SO(3)$ ahead of Lemma \[L:3A1-1\].

\[L:3A1-4\] Let $\overline{\Gamma}=\langle(T,S,S),(S,T,S),(S,S,T),(S^2,S^2,S^2)\rangle$. Then, $\phi_{\overline{\Gamma}}$ is not an isomorphism.

The image of projection of $\overline{\Gamma}$ to each component of $\overline{G}=\SO(3)^{3}$ is equal to $\langle S,T\rangle$, which has centralizer in $\SO(3)$ equal to $\langle S^2\rangle$. Thus, $$Z_{\overline{G}}(\overline{\Gamma})=\langle(S^2,1,1),(1,S^2,1),(1,1,S^2)\rangle\cong\{\pm{1}\}^{3}.$$ By calculation one shows that $\pi(Z_{G}(\pi^{-1}(\Gamma)))$ ($\subset Z_{\overline{G}}
(\overline{\Gamma})$) is the trivial group. Thus, $$X_{\overline{\Gamma}}\cong\{\pm{1}\}^{3}.$$

Due to $S^2\sim T$, for any element $1\neq x=(x_1,x_2,x_3)\in\overline{\Gamma}$ ($x_1,x_2,x_3\in
\SO(3)$), at least one of $x_1,x_2,x_3$ is conjugate to $T$. Then one shows $$\overline{\Gamma}'=
\langle(1,S^2,S^2),(S^2,1,S^2)\rangle.$$ Thus, $$Y_{\overline{\Gamma}}\cong
\overline{\Gamma}/\overline{\Gamma}\cong\{\pm{1}\}^{4}.$$ As the order of $Y_{\overline{\Gamma}}$ is larger than the order of $X_{\overline{\Gamma}}$, $\phi_{\overline{\Gamma}}$ is not an isomorphism.

By Lemma \[L:3A1-3\] and Lemma \[L:3A1-4\], we get the following statement.

\[T:3A1\] The group $$\Sp(1)^{3}/\langle(1,-1,-1),(-1,1,-1)\rangle$$ is unacceptable.

\[E:3A1\] Write $\eta=\frac{1+\mathbf{i}}{\sqrt{2}}$. Precisely, the group $\Gamma=\pi^{-1}(\overline{\Gamma})$ in Lemma \[L:3A1-4\] is generated by $$(\mathbf{j},\eta,\eta),(\eta,\mathbf{j},\eta),(\eta,\eta,
\mathbf{j}),(\mathbf{i},\mathbf{i},\mathbf{i}).$$

Write $$\gamma_1=(\mathbf{j},\eta,\eta),\gamma_2 =(\eta,\mathbf{j},\eta),\gamma_3=(\eta,\eta,\mathbf{j}),
\gamma_0=(\mathbf{i},\mathbf{i},\mathbf{i})$$ and $$z_1=(\mathbf{i},1,1),z_2=(1,\mathbf{i},1),z_3=
(1,1,\mathbf{i}),z_0=(-1,-1,-1).$$ Put $$\Gamma'=\langle z_0,z_1z_2,z_1z_3\rangle,$$ $$\Gamma_{0}=
\langle z_0,\gamma_1,\gamma_2,\gamma_3\rangle.$$ Then, $$\pi^{-1}(\overline{\Gamma}')=\Gamma'.$$ We take $\phi:\Gamma\rightarrow G$ be the inclusion, and $\phi':\Gamma\rightarrow G$ be defined by $\phi'|_{\Gamma_0}=\id$ and $\phi'(\gamma_0)=z_{0}\gamma_0$.

Unacceptable compact Lie groups
-------------------------------

In the following theorem, we show all other connected compact simple Lie groups except those in Theorem \[T:acceptable1\] are unacceptable. Unacceptability of many of these groups have been shown by Larsen ([@Larsen1],[@Larsen2]). What we do here is to make the list as complete as possible.

\[T:unacceptable1\] Groups isomorphic to any one in the following list are unacceptable,

-   $\SU(n)/\mu_{m}$ ($m|n$, $m\geq 2$, $(n,m)\neq(2,2)$);

-   $\PSp(n)$ ($n\geq 4$);

-   $\SO(2n)$ ($n\geq 3$);

-   $\PSO(2n)$ ($n\geq 3$);

-   $\Spin(n)$ ($n\geq 8$);

-   $\HSpin(4n)$ ($n\geq 2$);

-   $\F_4$, $\E_6$, $\E_7$, $\E_8$;

-   $\E_{6}^{ad}$, $\E_{7}^{ad}$;

For a group $G=\SU(n)/\mu_{m}$ in item (1), we have $m>1$ and $m$ is odd, or $m$ is even and $n\geq 4$. When $m>1$ and $m$ is odd, take an odd prime $p|m$. Write $$A_{p}=\left(\begin{array}{ccccc}
0_{\frac{n}{p}}&I_{\frac{n}{p}}&0_{\frac{n}{p}}&\ldots&0_{\frac{n}{p}}\\0_{\frac{n}{p}}&0_{\frac{n}{p}}
&I_{\frac{n}{p}}&\ldots&0_{\frac{n}{p}}\\0_{\frac{n}{p}}&0_{\frac{n}{p}}&0_{\frac{n}{p}}&\ldots&
0_{\frac{n}{p}}\\\vdots&\vdots&\vdots&\ddots&\vdots\\1_{\frac{n}{p}}&0_{\frac{n}{p}}&0_{\frac{n}{p}}&
\ldots&0_{\frac{n}{p}}\\\end{array}\right)$$ and $$B_{p}=\diag\{I_{\frac{n}{p}},e^{\frac{2\pi i}{p}}
I_{\frac{n}{p}},\dots,e^{\frac{2(p-1)\pi i}{p}}I_{\frac{n}{p}}\}.$$ Set $\Gamma=(C_{p})^{2}$ with two generators $\gamma_1$ and $\gamma_2$. Define homomorphisms $\rho,\rho':\Gamma\rightarrow G$ by $$\rho(\gamma_1)=\rho'(\gamma_1)=[A_{p}],\ \rho(\gamma_2)=[B_{p}],\ \rho'(\gamma_2)=[B_{p}^{2}].$$ Then, one can verify that $\rho$ and $\rho'$ are element-conjugate, but not globally conjugate. Thus, $G=
\SU(n)/\mu_{m}$ ($m|n$, odd $m\geq 3$) is unacceptable.

When $m$ is even and $n\geq 4$, write $n=4k$ or $4k+2$ ($k\geq 1$). Suppose $G$ is acceptable. When $n=4k$. Put $$A=\{[\diag\{\lambda_1I_{4},\dots,\lambda_{k}I_{4}\}]:|\lambda_{i}|=1,\prod_{1\leq i\leq k}\lambda_{i}
=1\}.$$ Then, $$Z_{G}(A)=S(\U(4)^{k})/\mu_{m}.$$ By Proposition \[P:centralizer2\], $S(\U(4)^{k})/\mu_{m}$ is acceptable. Take $\Gamma=(C_4)^{2}$, using the homomorphisms $\phi,\phi':\Gamma\rightarrow\SU(4)$ as in Example \[E:SU4\]. Set $$\rho(\gamma)=[(\phi(\gamma),\dots,\phi(\gamma))]$$ and $$\rho'(\gamma)=
[(\phi'(\gamma),\dots,\phi'(\gamma))].$$ Then, $\rho,\rho':\Gamma\rightarrow S(\U(4)^{k})/\mu_{m}$ are element-conjugate, but not globally conjugate. This in contradiction with $S(\U(4)^{k})/\mu_{m}$ is acceptable.

When $n=4k+2$. Put $$A=\{[\diag\{\lambda_1I_{4},\dots,\lambda_{k}I_{4},I_{2}\}]:|\lambda_{i}|=1,
\prod_{1\leq i\leq k}\lambda_{i}=1\}.$$ Then, $$Z_{G}(A)=S(\U(4)^{k}\times\U(2))/\mu_{m}.$$ By Proposition \[P:centralizer2\], $S(\U(4)^{k}\times\U(2))/\mu_{m}$ is acceptable. Let $\Gamma=\langle\gamma_0,
\gamma_1,\gamma_2\rangle$ be defined by $$\gamma_{1}^{4}=\gamma_{2}^{4}=\gamma_0^{2}=[\gamma_0,\gamma_1]
=[\gamma_0,\gamma_1]=1$$ and $[\gamma_1,\gamma_2]=\gamma_0$. Then, there is an exact sequence $$1\rightarrow C_2\rightarrow\Gamma\rightarrow(C_4)^{2}\rightarrow 1.$$ From Example \[E:SU4\], we have homomorphisms $\phi,\phi':\Gamma\rightarrow\SU(4)$ by composing the homomorphisms there with the projection $\Gamma\rightarrow(C_4)^{2}$. Define $\psi:\Gamma\rightarrow\SU(2)$ by $$\psi(\gamma_1)
=\diag\{\mathbf{i},-\mathbf{i}\}\textrm{ and }\psi(\gamma_2)=\left(\begin{array}{cc}0&1\\-1&0\\\end{array}
\right).$$ Set $$\rho(\gamma)=[(\phi(\gamma),\dots,\phi(\gamma),\psi(\gamma))]$$ and $$\rho'(\gamma)=
[(\phi'(\gamma),\dots,\phi'(\gamma),\psi(\gamma))].$$ Then, $\rho,\rho':\Gamma\rightarrow S(\U(4)^{k}
\times\U(2))/\mu_{m}$ are element-conjugate, but not globally conjugate. This in contradiction with $S(\U(4)^{k}\times\U(2))/\mu_{m}$ is acceptable. Thus, $G=\SU(n)/\mu_{m}$ ($m|n$, $m$ even, $n\geq 4$) is unacceptable.

In item (2), suppose $G=\PSp(n)$ ($n\geq 4$) is acceptable. Write $n=4k+l$ ($k\geq 1$, $l\in\{0,1,2,3\}$). Take $$A\!=\!\{\diag\{\!(\!a_1\!+\!b_1\!\mathbf{i})\!I_{4},\!\dots,\!(\!a_{k}\!+\!b_{k}\!\mathbf{i}\!)
\!I_{4},\!t_1,\!\dots,\!t_{l}\}:\!a_{i},\!b_{i}\!\in\mathbb{R},\!|a_{i}\!+\!b_{i}\mathbf{i}\!|\!=\!1,
\!t_{j}\!=\!\pm{1}\}.$$ Then, $$Z_{G}(A)=(\U(4)^{k}\times\Sp(1)^{l})/\langle-I\rangle.$$ By Proposition \[P:centralizer2\], $(\U(4)^{k}\times\Sp(1)^{l})/\langle-I\rangle$ is acceptable. As in the last paragraph, we have $\Gamma=\langle\gamma_0,\gamma_1,\gamma_2\rangle$ be defined by $$\gamma_{1}^{4}=
\gamma_{2}^{4}=\gamma_0^{2}=[\gamma_0,\gamma_1]=[\gamma_0,\gamma_1]=1$$ and $[\gamma_1,\gamma_2]=
\gamma_0$, and homomorphisms $\phi,\phi':\Gamma\rightarrow\SU(4)$. Define $\psi:\Gamma\rightarrow
\Sp(1)$ by $$\psi(\gamma_1)=\mathbf{i}\textrm{ and }\psi(\gamma_2)=\mathbf{j}.$$ Set $$\rho(\gamma)= [(\phi(\gamma),\dots,\phi(\gamma),\psi(\gamma),\dots,\psi(\gamma))]$$ and $$\rho'(\gamma)=
[(\phi'(\gamma),\dots,\phi'(\gamma),\psi(\gamma),\dots,\psi(\gamma))].$$ Then, $\rho,\rho':\Gamma
\rightarrow(\U(4)^{k}\times\Sp(1)^{l})/\langle-I\rangle$ are element-conjugate, but not globally conjugate. This is a contradiction. Thus, $G=\PSp(n)$ ($n\geq 4$) is unacceptable.

In item (3), take a maximal torus $T$ of $\SO(2n-6)\subset\SO(2n)$. Then, $$Z_{\SO(2n)}(T)=T\times
\SO(6).$$ Suppose $\SO(2n)$ is acceptable. By Proposition \[P:centralizer2\] and Lemma \[L:product\], $\SO(6)$ is also acceptable. This is in contradiction with Example \[E:SU4\]. Thus, $G=\SO(2n)$ ($n\geq 3$) is unacceptable.

In item (4), when $n\geq 4$, take a maximal torus $T$ of $\SO(2n-6)\subset\PSO(2n)$. Then, $$Z_{\PSO(2n)}(T)
=(T\times\SO(6))/\langle(-I,-I)\rangle.$$ Suppose $\PSO(2n)$ ($n\geq 4$) is acceptable. By Proposition \[P:centralizer2\] and Lemma \[L:product2\], so is $\SO(6)$. This is in contradiction with Example \[E:SU4\]. When $n=3$, $\PSO(6)\cong\PSU(4)$ is unacceptable by Theorem \[T:unacceptable1\](1). Thus, $G=\PSO(2n)$ ($n\geq 3$) is unacceptable.

In item (5), take a maximal torus $T$ of $\Spin(n-8)\subset\Spin(2n)$, and choose $c'=e_1e_2\cdots e_{8}
\in\Spin(8)\subset Z_{\Spin(n)}(T)$. Put $A=T\times\langle c'\rangle$. Then, $$Z_{\Spin(n)}(A)=(T\times
\Spin(8)/\langle(-1,-1)\rangle.$$ Suppose $\Spin(n)$ ($n\geq 8$) is acceptable. By Proposition \[P:centralizer2\] and Lemma \[L:product2\], so is $\Spin(8)$. This is in contradiction with [@Larsen2 Proposition 2.5]. Thus, $\Spin(n)$ ($n\geq 8$) is unacceptable.

In item (6), take a maximal torus $T$ of $\Spin(4n-8)\subset\HSpin(4n)$. Then, $$Z_{\HSpin(4n)}(T)=
(T\times\Spin(8))/\langle(-1,-1),(c',c'')\rangle.$$ Suppose $\HSpin(4n)$ is acceptable. By Proposition \[P:centralizer2\] and Lemma \[L:product2\], so is $\Spin(8)$ (when $n>2$) or $\HSpin(8)$ (when $n=2$). By [@Larsen2 Proposition 2.5], $\Spin(8)$ is unacceptable. Due to $\HSpin(8)\cong\SO(8)$, $\HSpin(8)$ is unacceptable. This is a contradiction. Thus, $\HSpin(4n)$ ($n\geq 3$) is strongly unacceptable.

Groups in items (7) and (8) are treated in [@Larsen2 Theorem 3.4]. Alternatively, we could use the fact that each of $G=\E_6$, $\E_7$, $\E_8$, $\E_6^{ad}$, $\E_7^{\ad}$ has a Levi subgroup with derived subgroup isomorphic to $\Spin(8)$, which means there is a torus $T\subset G$ such that $Z_{G}(T)$ is connected and $(Z_{G}(T))_{der}\cong\Spin(8)$; and $G=\F_4$ possess a Klein four subgroup $A$ such that $Z_{G}(A)\cong\Spin(8)$. By Proposition \[P:centralizer2\] and Lemma \[L:product2\], if any of the groups in items (7) and (8) is acceptable, then so is $\Spin(8)$. This is in contradiction with [@Larsen2 Proposition 2.5]. Thus, the groups in items (7) and (8) are all unacceptable.

In the following theorem, we consider disconnected groups with a simple Lie algebra.

\[T:unacceptable2\] Groups isomorphic to one in the following list are unacceptable,

-   $\SU(2n)\rtimes\langle\tau\rangle$ ($n\geq 3$, $\tau^{2}=1$, $\Ad(\tau)X=\overline{X}$);

-   $\Pin(n)$ ($n\geq 9$),

-   $\PO(4n)$ ($n\geq 2$);

-   $\Spin(8)\rtimes\langle\tau\rangle$ ($\tau^3=1$, $\Spin(8)^{\tau}=\G_2$);

-   $\PSO(8)\rtimes\langle\tau\rangle$ ($\tau^3=1$, $\Spin^{\tau}=\G_2$);

-   $\E_{6}\rtimes\langle\tau\rangle$ ($\tau^2=1$, $\E_{6}^{\tau}=\F_4$);

-   $\E_{6}^{ad}\rtimes\langle\tau\rangle$ ($\tau^2=1$, $(\E_{6}^{ad})^{\tau}=\F_4$).

Write $G$ for the group in consideration in each item.

In item (1), we have $$Z_{G}(\tau)=\SO(2n)\times\langle\tau\rangle.$$ Suppose $G$ is acceptable. By Proposition \[P:centralizer2\] and Lemma \[L:product\], so is $\SO(2n)$. This is in contradiction with Theorem \[T:unacceptable1\](3). Thus, $\SU(2n)\rtimes\langle\tau\rangle$ ($n\geq 3$ is strongly unacceptable.

In item (2), we have $$Z_{\Pin(n)}(e_1)=\Spin(n-1)\cdot\langle e_1\rangle.$$ Suppose $\Pin(n)$ is acceptable. By Proposition \[P:centralizer2\] and Lemma \[L:product2\], so is $\Spin(n-1)$. This is in contradiction with Theorem \[T:unacceptable1\](5). Thus, $\Pin(n)$ ($n\geq 9$) is unacceptable.

In item (3), choose $$A=\{[\left(\begin{array}{cc}aI_{2n}&bI_{2n}\\-bI_{2n}&aI_{2n}\\\end{array}\right)]:a,b
\in\mathbb{R},|a+b\mathbf{i}|=1\}.$$ Then, $$Z_{\PO(4n)}(A)=\U(2n)/\langle-I\rangle.$$ Suppose $\PO(4n)$ is acceptable. By Proposition \[P:centralizer2\] and Lemma \[L:product2\], so is $\U(2n)/\langle-I\rangle$. This is in contradiction with Theorem \[T:unacceptable1\](1). Thus, $\PO(4n)$ ($n\geq 2$) is strongly unacceptable.

In items (4) and (5), there exists an element $\tau'\in G^{0}\tau$ with $o(\tau')=3$ such that $$Z_{G}(\tau')
\cong\PSU(3)\times\langle\tau'\rangle.$$ Suppose $G$ is acceptable. By Proposition \[P:centralizer2\] and Lemma \[L:product2\], so is $\PSU(3)$. This is in contradiction with Theorem \[T:unacceptable1\](1). Thus, $G$ is unacceptable.

In items (6) and (7), there exists an element $\tau'\in G^{0}\tau$ with $o(\tau')=2$ such that $$Z_{G}(\tau')
\cong\PSp(4)\times\langle\tau'\rangle.$$ Suppose $G$ is acceptable. By Proposition \[P:centralizer2\] and Lemma \[L:product2\], so is $\PSp(4)$. This is in contradiction with Theorem \[T:unacceptable1\](2). Thus, $G$ is unacceptable.

We call a compact Lie group $G$ [*nearly simple*]{} if $({\mathfrak{g}}_{0})_{der}=[{\mathfrak{g}}_0,{\mathfrak{g}}_0]$ is a compact simple Lie algebra. In the following, we ask the acceptability of some interesting nearly simple compact Lie groups which we haven’t known of its acceptability/unacceptability yet.

\[Q:acceptable2\] Are the following groups acceptable or unacceptable?

-   $\SU(4)\!\rtimes\!\langle\tau\rangle$ ($\tau^{2}=1$, $\Ad(\tau)X=\overline{X}$);

-   $\SU(\!2n\!+\!1)\!\rtimes\!\langle\tau\rangle$ ($n\geq 2$, $\tau^{2}=1$, $\Ad(\tau)X=\overline{X}$);

-   $\U(n)\!\rtimes\!\langle\tau\rangle$ ($n\geq 2$, $\tau^{2}=1$, $\Ad(\tau)X=\overline{X}$);

-   $\PO(4)$;

-   $\PO(4n+2)$ ($n\geq 1$).

Recall that the group $$\SU(3)\!\rtimes\!\langle\tau\rangle$$ (where $\tau^2=1$ and $\Ad(\tau)X=\overline{X}$ ($\forall X\in\SU(3)$)) is acceptable (cf. [@Larsen1 Lemma 2.7]). Set $$G=(\SU(4)/\langle\!-I\rangle)
\!\rtimes\!\langle\tau\rangle,$$ where $\tau^2=1$ and $\Ad(\tau)\![X]=[\overline{X}]$ ($\forall\!X\!\in\!
\SU(3)$). Then, $G\cong\O(6)$. Thus, $G$ is acceptable.

Now we study connected compact Lie groups which are not nearly simple. By Lemma \[L:Larsen\], it suffices to consider connected compact semisimple Lie groups. By Lemma \[L:product\], it suffices to consider such groups which are not the direct product of two non-trivial groups. We call such groups [non-decomposable]{}.

Suppose $G$ is a non-decomposable and non-simple connected compact semisimple Lie group. Then, $G$ is of the following form, $$\label{Eq-standard}G=(G_1\times\cdots G_{s})/Z$$ with each $G_{i}$ ($1\leq i\leq s$) a connected compact simple Lie group, where $Z\subset Z(G_1)\times\cdots\times Z(G_{s})$, $Z\cap Z(G_{i})=1$ ($1\leq i\leq s$), and the image of projection of $Z$ to each $Z(G_{i})$ is non-trivial.

The following lemma is easy to show.

\[L:centralizer3\]

-   For any integer $d\geq 2$ and positive integer $m$, there is a torus $T\subset\SU(dm)$ such that $$Z_{\SU(dm)}(T)=\SU(d)^{m}\cdot T.$$

-   For any $n\geq 1$, take $A=\{\diag\{t_1,\dots,t_{n}\}: t_{j}=\pm{1}\}\subset\Sp(n)$. Then, $$Z_{\Sp(n)}(A)=\Sp(1)^{n}.$$

-   Take $A=\langle e_1e_2e_3e_4\rangle\subset\Spin(7)$. Then, $$Z_{\Spin(7)}(A)\cong\Sp(1)^{3}/\langle
    (-1,-1,-1)\rangle.$$

-   For any $n\geq 2$, there is a torus $T\subset\SU(2n)$ such that $$Z_{\SU(2n)}(T)=(\SU(4)\times
    \SU(2)^{n-2})\cdot T.$$

-   For any $n\geq 4$, take $$T=\{(a+b\mathbf{i})I_{4}:a,b\in\mathbb{R},|a+b\mathbf{i}|=1\}\subset
    \Sp(4)$$ and $$A'=\{\diag\{t_1,\dots,t_{n-4}\}:t_{j}=\pm{1}\}\subset\Sp(n-4).$$ Put $A=T\times A'$. Then, $$Z_{\Sp(n)}(A)=(\SU(4)\times\Sp(1)^{n-4})\cdot T.$$

-   Take $A=\langle e_1e_2e_3e_4e_5e_6\rangle\subset\Spin(7)$. Then, $$Z_{\Spin(7)}(A)=\Spin(6)\cong
    \SU(4).$$

\[T:unacceptable3\] Suppose $G$ is a non-decomposable and non-simple connected compact Lie group of the form in (\[Eq-standard\]) and satisfies the conditions there. If $G$ is acceptable, then each $G_{i}$ is isomorphic to one of $\Sp(1)$.

By Lemma \[L:product2\], each $G_{i}$ is also acceptable. By the assumption on $G$, we have $Z(G_{i})\neq 1$. By Theorem \[T:acceptable1\] and Theorem \[T:unacceptable1\], we get $G_{i}\cong\SU(n)$, $\Sp(n)$ or $\Spin(7)$.

First we show that $Z$ is an elementary abelian 2-group. Suppose it is not this case. Then, we could find an element $z\in Z$ with order $d$ an odd prime or $d=4$. Let $G'$ be generated by the simple factors $G_{i}$ with $p_{i}(z)\in Z(G_{i})$ an element of order $d$. By Lemma \[L:product2\], $G'$ is also acceptable. Without loss of generality we assume that the projection $p_{i}(z)$ is of order $d$ if and only if $1\leq i
\leq t$, where $1\leq t\leq s$. Then, each $G_{i}\cong\SU(n_{i})$ with $d|n_{i}$ ($1\leq i\leq t$). Write $G'=(G_1\times\cdots G_{t})/Z'$. When $d$ is an odd prime, we have $z\in Z'$; when $d=4$, we have $z^2\in Z'$. By Lemma \[L:centralizer3\](1) we could take a torus $T$ of $G'$ such that $$Z_{G'}(T)=(\SU(d)^{m}/Z'')
\cdot T$$ where $Z''\subset Z(\SU(d)^{m})$, with $z\in Z''$ in case $d$ is an odd prime, and $z^{2}=
(-I,\dots,-I)\in Z''$ in case $d=4$. By Proposition \[P:centralizer2\] and Lemma \[L:product2\], $\SU(d)^{m}/Z''$ is also acceptable. Write $\omega_{d}=e^{\frac{2\pi i}{d}}$. In the case of $d$ is an odd prime, write $$z=(\omega_{d}^{t_1}I_{d},\dots,\omega_{d}^{t_{m}}I_{d}),$$ where $\gcd(t_{i},d)=1$ ($1\leq i\leq m$).

When $d=p$ is an odd prime, write $$A_{p}=\left(\begin{array}{ccccc}0&1&0&\ldots&0\\0&0&1&\ldots&0\\0&0&0&
\ldots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\1&0&0&\ldots&0\\\end{array}\right)$$ and $$B_{p}=
\diag\{1,\omega_{p},\dots,\omega_{p}^{p-1}\}.$$ Take $\Gamma=(C_{p})^{2}$ with two generators $\gamma_1,
\gamma_1$. Define $$\rho(\gamma_1)=\rho'(\gamma_1)=[(A_{p},\dots,A_{p})],$$ $$\rho(\gamma_2)=[(B_{p}^{t_1},
\dots,B_{p}^{t_{m}})],\ \rho'(\gamma_2)=[(B_{p}^{2t_1},\dots,B_{p}^{2t_{m}})].$$ Then, $\rho,\rho':\Gamma
\rightarrow\SU(d)^{m}/Z''$ are two homomorphisms which are element-conjugate, but not globally conjugate. This is a contradiction.

When $d=4$, write $$A=\diag\{1,1,i,-i\},\quad B=\diag\{1,i,1,-i\}.$$ Take $\Gamma=(C_{4})^{2}$ with two generators $\gamma_1,\gamma_2$. Define $$\rho(\gamma_1)=\rho'(\gamma_1)=[(A,\dots,A)],$$ $$\rho(\gamma_2)
=[(B,\dots,B)],\ \rho'(\gamma_2)=[(\overline{B},\dots,\overline{B})].$$ Then, $\rho,\rho':\Gamma
\rightarrow\SU(4)^{m}/Z''$ are two homomorphisms which are element-conjugate, but not globally conjugate. This is a contradiction.

Now assume that $Z$ is an elementary abelian 2-group. Suppose some simple factor $G_{i_0}$ of $G$ is not isomorphic to any of $\Sp(1)$, $\Sp(2)$, $\Sp(3)$. Choose $z\in Z$ such that the image of projection of $z$ to $Z(G_{i_0})$ is non-trivial, and $z$ contains the least number of non-trivial components among such central elements. Let $G'$ be generated by the simple factors $G_{i}$ of $G$ such that the image of projection of $z$ to $Z(G_i)$ is non-trivial. Then, $Z(G')\cap Z$ is generated by $z$. Thus, $G'$ is also non-decomposable and non-simple. By Lemma \[L:product2\], $G'$ is also acceptable. Without loss of generality, we may assume that $G=G'$. Note that, at least one simple factor $G_{i}$ of $G$ is isomorphic to one of $\SU(2n)$ ($n\geq 2$), $\Sp(n)$ ($n\geq 4$), $\Spin(7)$. By Lemma \[L:centralizer3\], there exists a closed abelian subgroup $A$ of $G$ such that $$Z_{G}(A)=
((\SU(4)\times\Sp(1)^{m})/Z')\cdot A,$$ where $m\geq 1$, $Z'\subset Z(\SU(4)\times\Sp(1)^{m}),$ and $Z'$ is generated by an element of the form $(-I,-1,\dots,-1)$. By Proposition\[P:centralizer2\] and Lemma \[L:product2\], $\SU(4)\times\Sp(1)^{m})/Z'$ is also acceptable. Let $\Gamma=
\langle\gamma_0,\gamma_1,\gamma_2\rangle$ be defined by $$\gamma_{1}^{4}=\gamma_{2}^{4}=\gamma_0^{2}
=[\gamma_0,\gamma_1]=[\gamma_0,\gamma_1]=1$$ and $[\gamma_1,\gamma_2]=\gamma_0$. Then, there is an exact sequence $$1\rightarrow C_2\rightarrow\Gamma\rightarrow(C_4)^{2}\rightarrow 1.$$ From Example \[E:SU4\], we have homomorphisms $\phi,\phi':\Gamma\rightarrow\SU(4)$ by composing the homomorphisms there with the projection $\Gamma\rightarrow(C_4)^{2}$. Define $$\rho(\gamma_1)=\rho'(\gamma_1)=
[(A,\mathbf{i},\dots,\mathbf{i})],$$ $$\rho(\gamma_2)=[(B,\mathbf{j},\dots,\mathbf{j})],\
\rho'(\gamma_2)=[(\overline{B},\mathbf{j},\dots,\mathbf{j})].$$ Then, $\rho,\rho': \Gamma\rightarrow
\SU(4)\times\Sp(1)^{m})/Z'$ are two homomorphisms which are element-conjugate, but not globally conjugate. This is a contradiction.

In the remaining each simple factor of $G$ is isomorphic to one of $\Sp(1)$, $\Sp(2)$, $\Sp(3)$, and $Z$ is an elementary abelian 2-group. We consider two separate cases: (1), $G$ has a simple factor isomorphic to $\Sp(3)$; (2), $G$ has no simple factor isomorphic to $\Sp(3)$, but contains a simple factor isomorphic to $\Sp(2)$. In case (1), analogous to the above argument in the last paragraph, it reduces to show the group $(\Sp(3)\times\Sp(1)^{m})/\langle(-I,-1,\dots,-1)\rangle$ is unacceptable. By Lemma \[L:centralizer3\](2), it further reduces to show $(\Sp(1)^{m+3})/\langle(-1,\dots,-1)
\rangle$ is unacceptable, which is done in Example \[E:Sp3-2\]. The treating for case (2) is similar.

We show Theorem \[T1\] now.

After Theorem \[T:acceptable1\], Example \[E:SU4\], Example \[E:Sp3\], Theorem \[T:unacceptable-ChenevierGan\], Theorem \[T:unacceptable1\] and Theorem \[T:unacceptable3\], it suffices to show: if some group $G=\Sp(1)^{m}/Z$ ($m\geq 2$, $Z\subset Z(\Sp(1)^{m})$) is in-decomposable, non-simple and acceptable, then $m=2$. Suppose $m\geq 3$. Write $Z_0=Z(\Sp(1)^{m})=\{\pm{1}\}^{m}$. Write $$I_0=\{1,\dots,m\}.$$ For any element $z\in Z$, define $I_{z}$ as the set of indices $i\in I_0$ such that the $i$-th component of $z$ is equal to $-1$. Let $|z|$ be the cardinality of $z$. Let $X$ be the subset of $Z$ consisting of elements $z\in Z$ with $|z|=2$.

First we show that $X$ generates $Z$. Suppose it is not this case. Choose an element $z\in Z-\langle X
\rangle$ with $|z|$ minimal. Then, $|I_{z}|=|z|\geq 3$; and for any element $z'\in Z_0$ with $I_{z'}
\subset I_{z}$, $z'\not\in Z$. Let $G'$ be generated by simple factors of $G$ with indices in $I_{z}$. Then, $$G'\cong\Sp(1)^{k}/\langle(-1,\dots,-1)\rangle$$ where $k=|z|\geq 3$. By Lemma \[L:product2\], $G'$ is also acceptable. This is in contradiction with the conclusion in Example \[E:Sp3-2\]. Since $G$ is in-decomposable and non-simple, $X$ generates $Z$ implies that $$\bigcup_{z\in X}I_{z}=I_{0}.$$

Secondly we show that $I_{z}\cap I_{z'}=\emptyset$ for any two distinct elements $z,z'\in X$. Suppose it is not this case. Let $G'$ be generated by simple factors of $G$ with indices in $I_{z}\cup I_{z'}$. Then, $$G'\cong\Sp(1)^{3}/\langle(1,-1,-1),(-1,1,-1)\rangle.$$ By Lemma \[L:product2\], $G'$ is also acceptable. This is in contradiction with Theorem \[T:3A1\].

Since $\bigcup_{z\in X}I_{z}=I_{0}$ and $I_{z}\cap I_{z'}=\emptyset$ for any two distinct $z,z'\in X$, thus $m$ is even and $G$ is isomorphic to the direct product of $\frac{m}{2}$-copies of $\Sp(1)^{2}/\langle(-1,-1)\rangle$. As we assume that $G$ is in-decomposable, we get $m=2$.

Note that we showed that the groups in items (1)-(5) of Theorem \[T1\] are all strongly acceptable.

Strongly unacceptability is not touched at all in this paper. One may think if the groups shown to be unacceptable are actually strongly unacceptable. In [@Larsen2 Proposition 2.6], it is shown that $\Spin(n)$ ($n\geq 35$) are strongly unacceptable. From [@Huang-Yu] we know that $\E_7^{ad}$ has two non-conjugate Klein four subgroup with all involutions in a conjugacy class (cf. [@Huang-Yu Table 4]). Thus, $\E_7^{ad}$ is strongly unacceptable.

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