# Characters for extended affine Lie algebras; a combinatorial approach

Saeid Azam  Department of Pure Mathematics
Faculty of Mathematics and Statistics
University of Isfahan
P.O.Box: 81746-73441
Isfahan, Iran, and
School of Mathematics, Institute for Research in Fundamental Sciences (IPM),
P.O. Box: 19395-5746<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

The behavior of objects associated with general extended affine Lie algebras
is typically distinct from their counterparts in affine Lie algebras. Our
research focuses on studying characters and Cartan automorphisms, which appear
in the study of Chevalley involutions and Chevalley bases for extended affine
Lie algebras. We show that for almost all extended affine Lie algebras, any
finite order Cartan automorphism is diagonal, and its corresponding
combinatorial map is a character.

###### Key words and phrases:

Extended affine Lie algebra, Character, Cartan automorphism, Chavalley
involution, involutive Lie algebra

###### 2020 Mathematics Subject Classification:

17B67, 17B22, 16W10, 17B65

This work is based upon research funded by Iran National Science Foundation
(INSF) under project No. 4001480.

This research was in part carried out in IPM-Isfahan Branch.

##  1\. Introduction

Extended affine Lie algebras are a class of Lie algebras that are defined by a
set of axioms. They share many common features with finite and affine Lie
algebras, but also possess unique characteristics that make them fascinating
objects of study, see [H-KT90], [AABGP97] and [Neh11]. For instance, like
finite and affine Lie algebras, an extended affine Lie algebra is equipped
with a symmetric invariant non-degenerate bilinear form, and has a finite-
dimensional Cartan subalgebra that splits the Lie algebra into a direct sum of
root spaces.

The set of roots of an extended affine Lie algebra is called an extended
affine root system. Similar to affine Lie algebras, a root is called isotropic
if any integral multiple of it is also a root. The dimension of the vector
space spanned by isotropic roots is known as the nullity of an extended affine
Lie algebra or its root system. The finite and affine Lie algebras correspond
to nullities 0 and 1, respectively.

Concerning the structure theory, two of the most important objects that are
associated with an extended affine Lie algebra are the core which is the
subalgebra generated by non-isotropic root spaces, and the centerless core
which is the quotient algebra of the core over its center. Similarly,
associated with an extended affine root system $R$ there are some objects
which are essential in its study, we mention some here which are more related
to our work. The image $\bar{R}$ of $R$ in the quotient space of the real span
of roots over the span of isotropic roots is an irreducible finite root system
whose rank and type are turned to be isomorphism invariants of $R$. These are
called the rank and the nullity of $R$, respectively. In this work, we assume
that $R$ is reduced. The root system $R$ can be then described with three
ingredients; an irreducible finite root system $\dot{R}$ isomorphic to
$\bar{R}$, and two discrete subsets $S$ and $L$ of isotropic roots. The
interrelations between $S$ and $L$, called semilattices, reflect the
interaction of reflections based on short and long roots of $R$.

In this study, we investigate the concepts of ”characters” and “diagonal
automorphisms” for extended affine Lie algebras. These concepts exhibit
characteristics in higher nullities which are absent in finite or affine
cases. Our study of characters stems from our interest in Chevalley
involutions. These are finite order automorphisms that act as minus identity
on the Cartan subalgebra. Chevalley involutions are essential in the study of
Chevalley bases, and more generally in the modular theory of extended affine
Lie algebras, see [AFI22] and [AI23]. The Composition of two Chevalley
involutions results in an automorphism that stabilizes pointwise the Cartan
subalgebra, which we refer to as a Cartan automorphism. Involutions, in a more
broad context have been studied in the context of involutive Lie algebras, see
for example [Gao96] and [Cal04].

Let $E$ be an extended affine Lie algebra with Cartan subalgebra
${\mathcal{H}}$ and root system $R$. Since a Cartan automorphism $\Psi$
stabilizes each root space, it acts on the root space $E_{\alpha}$, $\alpha$
non-isotropic, as $\psi(\alpha)\hbox{id}$ where $\psi(\alpha)$ is a non-zero
complex number. If $\Psi$ is diagonal, i.e., if it also acts as a scalar
multiple of identity on each isotropic root space, then the map
$\psi:R\rightarrow{\mathbb{C}}^{\times}$ is called a core-character associated
with $\Psi$ for $R$. If $\psi(\alpha+\beta)=\psi(\alpha)\psi(\beta)$ for
$\alpha,\beta,\alpha+\beta\in R$, then $\psi$ is called a character, and if
$\psi$ is extendable to the root lattice $\langle R\rangle$, we say that
$\psi$ is a root-lattice character. While any character for $R$ leads to a
diagonal automorphism, the converse is not necessarily true. In this work, we
investigate the interrelations between Chevalley involutions, Cartan
automorphisms, and characters. The paper is organized as follows.

In Section 2, we provide the definition of an extended affine Lie algebra
denoted by $E$ and report some fundamental facts about $E$ and its root
system. In particular, we explain the concepts of “index” and “reflectable
base” which are essential to extended affine root systems.

In Section 3, we establish certain basic facts about Chevalley involutions.
Specially, we prove that any finite order Chevalley involution has order $2$,
as stated in Lemma 3.1.2, and that a finite-order automorphism whose
restriction to the core is a Chevalley involution for the core is also a
Chevalley involution for $E$ only if it stabilizes the Cartan subalgebra, see
Corollary 3.1.5. Furthermore, we show that the existence of Chevalley
involutions does not depend on the choice of the Cartan subalgebra, see Lemma
3.1.6.

In Section 4, the core part of the paper, we consider an extended affine Lie
algebra $E$ of type $X$ and $\hbox{rank}>1$, with root system $R$; also
consider a finite order Cartan automorphism $\Psi$ of $E$ and let
$\psi:R\rightarrow{\mathbb{K}}^{\times}$ be the core-character associated with
$R$, see Definition 4.1.7. We proceed with a type-dependent combinatorial
approach to show that if $X\not=B_{\ell}$, or if $X=B_{\ell}$ of index $0$,
then $\Psi$ is diagonal and $\psi$ is a character, see Propositions 4.2.1,
4.3.4, 4.4.5, and 4.4.10. The connections among Chevalley involutions, Cartan
automorphisms, and characters are summarized in Theorem 4.4.11.

In Section 5, the final section, we show that in contrast to the affine case,
not every character for an extended affine root system is expendable to a
root-lattice character, see Definition 4.1.2 and Corollary 5.2.2.

##  2\. Preliminaries

Throughout this work ${\mathbb{K}}$ is the field of complex numbers, and all
vector spaces are over ${\mathbb{K}}$ unless otherwise is specified. We set
${\mathbb{K}}^{\times}={\mathbb{K}}\setminus\\{0\\}$. We denote the dual space
of a vector space ${\mathcal{U}}$ with ${\mathcal{U}}^{\star}$. The notation
$R_{1}\uplus R_{2}$ means the union of two disjoint sets $R_{1}$ and $R_{2}$.
We denote by $\langle S\rangle$, the ${\mathbb{Z}}$-span of a subset $S$ of a
vector space.

###  2.1. Extended affine Lie algebras and root systems

We begin by recalling some terminologies and basic facts about extended affine
Lie algebras and extended affine root systems. For more details, we refer the
reader to [AABGP97] and [Neh11].

######  2.1.1.

An extended affine Lie algebra is a triple $(E,(\cdot,\cdot),{\mathcal{H}})$
where $E$ is a Lie algebra over ${\mathbb{K}}$, ${\mathcal{H}}$ is a non-
trivial finite dimensional subalgebra of $E$, and $(\cdot,\cdot)$ is a
${\mathbb{K}}$-valued symmetric bilinear form on $E$ satisfying axioms
(A1)-(A6) below:

(A1) The bilinear form $(\cdot,\cdot)$ is invariant and non-degenerate.

(A2) ${\mathcal{H}}$ is a splitting Cartan subalgebra of $E$ meaning that
$E=\sum_{\alpha\in R}E_{\alpha}$ with $E_{\alpha}=\\{x\in
E\mid[h,x]=\alpha(h)x\hbox{ for all }h\in{\mathcal{H}}\\},$
$R=\\{\alpha\in{\mathcal{H}}^{\star}\mid E_{\alpha}\not=\\{0\\}\\}$, and
$E_{0}={\mathcal{H}}.$ The set $R$ is called the root system of $E$.

From axioms (A1)-(A2), we see that the form $(\cdot,\cdot)$ on $E$ restricted
to ${\mathcal{H}}$ is non-degenerate. For $\alpha\in{\mathcal{H}}^{\star}$ let
$t_{\alpha}\in{\mathcal{H}}$ be the unique element satisfying
$\alpha(h)=(h,t_{\alpha})$, $h\in{\mathcal{H}}$, and consider the form on
${\mathcal{H}}^{\star}$ defined by $(\alpha,\beta):=(t_{\alpha},t_{\beta})$.
The root $\alpha$ is called isotropic if $(\alpha,\alpha)=0$ and non-isotropic
otherwise. We denote by $R^{0}$ the set of isotropic roots and by $R^{\times}$
the set of non-isotropic roots of $R$. The subalgebra $E_{c}$ of $E$ generated
by non-isotropic root spaces is called the core of $E$.

(A3) For $\alpha\in R^{\times}$ and $x\in E_{\alpha}$, $\hbox{ad}(x)$ is
locally nilpotent on $E$.

(A4) $E$ is tame, that is, the centralizer of $E_{c}$ in $E$ is contained in
$E_{c}$.

(A5) The ${\mathbb{Z}}$-span of $R$ in ${\mathcal{H}}^{\star}$ is a free
abelian group of finite rank.

(A6) $R^{\times}$ is indecomposable meaning that if $R^{\times}=R_{1}\uplus
R_{2}$ with $R_{1}\not=\emptyset$ and $R_{2}\not=\emptyset,$ then
$R_{1}\not\perp R_{2}$.

######  Remark 2.1.2.

In order to better achieve our goals, we have chosen to include the tameness
condition (axiom (A4)) in the definition of an extended affine Lie algebra,
despite it is typically not included in the definition. It follows from the
tameness condition that isotropic roots are non-isolated, meaning that for
$\sigma\in R^{0}$, there exists $\alpha\in R^{\times}$ with $\alpha+\sigma\in
R$.

Assume that $(E,(\cdot,\cdot),{\mathcal{H}})$ is an extended affine Lie
algebra. The term

$h_{\alpha}:=\frac{2t_{\alpha}}{(\alpha,\alpha)}\qquad(\alpha\in R^{\times}),$

will be used frequently in the sequel. It follows that for $\alpha\in
R^{\times}$, there exists $x_{\pm\alpha}\in E_{\pm\alpha}$ such that
$(x_{\alpha},h_{\alpha},x_{-\alpha})$ forms an $\mathfrak{sl}_{2}$-triple.
Also we recall from [Az06, Remark 1.5(ii)] and [AABGP97, Chapter I] that

(2.1) $[E_{\alpha},E_{\beta}]\not=\\{0\\}\qquad(\alpha\in
R^{\times},\;\beta,\alpha+\beta\in R),$ (2.2)
$[x_{\alpha},x_{-\alpha}]=(x_{\alpha},x_{-\alpha})t_{\alpha}\qquad(\alpha\in
R,{x_{\pm\alpha}\in E_{\pm\alpha}}).$

######  2.1.3.

Next, we recall some facts about the root system $R$ of $E$ which is called an
extended affine root system. Set ${\mathcal{V}}:=\hbox{span}_{{\mathbb{R}}}R$
and ${\mathcal{V}}^{0}:=\hbox{span}_{{\mathbb{R}}}R^{0}$. It turns out that
the form on ${\mathcal{H}}^{\star}$ restricted to ${\mathcal{V}}$ can be
assumed to be positive semidefinite, $2\alpha\not\in R^{\times}$ for
$\alpha\in R^{\times}$, and that the root-string property holds, namely for
$\alpha\in R^{\times}$ and $\beta\in R$, there exist non-negative integers
$d,u$ such that

(2.3) $\\{n\alpha+\beta\in R\mid
n\in{\mathbb{Z}}\\}=\\{\beta-d\alpha,\ldots,\beta,\ldots\beta+u\alpha\\}$

with $2(\beta,\alpha^{\vee})/(\alpha,\alpha)=d-u.$ Moreover, the image
$\bar{R}$ of $R$ in $\bar{\mathcal{V}}:={\mathcal{V}}/{\mathcal{V}}^{0}$ is an
irreducible finite root system in $\bar{\mathcal{V}}$. One can find a preimage
$\dot{R}$ of $\bar{R}$ in ${\mathcal{V}}$, under $\bar{\;}$, such that
$\dot{R}$ is an irreducible finite root system in its real span, isomorphic to
$\bar{R}$. Then $R\subseteq\dot{R}+\Lambda$, where $\Lambda=\langle
R^{0}\rangle$ is a free abelian group of rank equal to
$\dim{\mathcal{V}}^{0}$.

The rank and the type of $E$ or $R$ is by definition the rank and the type of
$\dot{R}$. The dimension of ${\mathcal{V}}^{0}$ is called the nullity of $R$.
We denote the rank and the nullity of $R$ by $\hbox{rank}(R)$ and
$\hbox{null}(R)$, respectively. We note that $\hbox{rank}\langle
R\rangle=\hbox{rank}(R)+\hbox{null}(R)$. In this work, we assume that $E$ is
of reduced type.

###  2.2. Semilattices

The internal structure of an extended affine root system $R$ can be described
in terms the finite root system $\dot{R}$ and certain subsets of $R^{0}$
called ”semilattices”.

######  2.2.1.

A semilattice in a $\nu$-dimensional real vector space ${\mathcal{U}}$ is a
discrete spanning subset $S$ containing $0$ with $S\pm 2S\subseteq S$. It
turns out that $\Lambda:=\langle S\rangle$ is a lattice in ${\mathcal{U}}$,
and $S$ is the union of a set of cosets of $2\Lambda$ in $\Lambda$ including
the trivial coset $2\Lambda$. Moreover,
$S=\uplus_{i=0}^{m}(\tau_{i}+2\Lambda)$, where $\tau_{i}$’s can be chosen such
that, $\tau_{i}$’s represent distinct cosets of $2\Lambda$, $\tau_{0}=0$, and
$\Lambda=\sum_{i=1}^{\nu}{\mathbb{Z}}\tau_{i}$. The integer $m$, the number of
non-trivial cosets of $2\Lambda$ in $S$, is called the index of $S$, denoted
by $\hbox{ind}(S)$. We call $\nu$ the rank of $S$. Note that
$\nu\leq\hbox{ind}(S)\leq 2^{\nu}$.

######  2.2.2.

Now, we can give the following description of $R$, in terms of the length of
roots and two involved semilattices,

(2.4) $\begin{array}[]{c}R^{0}=S+S\quad\hbox{and}\quad
R=R^{0}\cup(\dot{R}_{sh}+S)\cup(\dot{R}_{lg}+L),\\\ S+L=S,\quad
kS+L=L,\end{array}$

where $\dot{R}_{sh}$ and $\dot{R}_{lg}$ are the sets of short and long roots
of $\dot{R}$, respectively, and $k$ is the maximum number of edges appearing
between two consecutive roots in the Dynkin diagram of $\dot{R}$. Here if
$\dot{R}_{lg}=\emptyset$, we ignore in (2.4) all terms in which $\dot{R}_{lg}$
or $L$ appear. It turns out that

\- If $R$ is simply laced of rank $>1$, or is of type $F_{4}$ or $G_{2}$, then
$S$ and $L$ are lattices.

\- If $R$ is of type $B_{\ell}$ (respectively $C_{\ell}$), $\ell\geq 3$, the
$L$ (respectively $S$) is a lattice.

To have a more precise description of $R$ when it is of one of non-simply
laced types, we need to introduce another isomorphism invariant of $R$ called
the twist number, given by $k^{t}=|\langle L\rangle/\langle S\rangle|$. If $R$
has twist number $t$, then there exists semilattices $S_{1}$ of rank $t$ and
$S_{2}$ of rank $\nu-t$, such that

(2.5) $S=S_{1}\oplus\langle S_{2}\rangle\quad\hbox{and}\quad L=k\langle
S_{1}\rangle\oplus S_{2}.$

Next, we explain briefly about the Weyl group of $R$. Set
$\tilde{\mathcal{V}}:={\mathcal{V}}\oplus({\mathcal{V}}^{0})^{\star}$, where
$({\mathcal{V}}^{0})^{\star}$ is the dual space of ${\mathcal{V}}^{0}$. The
form $(\cdot,\cdot)$ on ${\mathcal{V}}$ extends to a non-degenerate form on
$\tilde{\mathcal{V}}$ by

$(\dot{\mathcal{V}},({\mathcal{V}}^{0})^{\star})=(({\mathcal{V}}^{0})^{\star},({\mathcal{V}}^{0})^{\star})=\\{0\\},\;(\gamma,\sigma):=\gamma(\sigma),\quad(\gamma\in({\mathcal{V}}^{0})^{\star},\sigma\in{\mathcal{V}}^{0}).$

For $\alpha\in R^{\times}$ the reflection $w_{\alpha}\in
GL(\tilde{\mathcal{V}})$ is defined by
$w_{\alpha}(\beta)=\beta-(2(\alpha,\beta)/(\alpha,\alpha))\alpha.$ The Weyl
group ${\mathcal{W}}$ of $R$ is the subgroup of $GL(\tilde{\mathcal{V}})$
generated by reflections $w_{\alpha}$, $\alpha\in R^{\times}.$

######  2.2.3.

For $\mathcal{P}\subseteq R^{\times}$, denote by ${\mathcal{W}}_{\mathcal{P}}$
the subgroup of ${\mathcal{W}}$ generated by reflections $w_{\alpha}$,
$\alpha\in\mathcal{P}$. A reflectable base for $R$ is a subset $\mathcal{P}$
of $R^{\times}$ such that ${\mathcal{W}}_{\mathcal{P}}\mathcal{P}=R^{\times}$
and no proper subset of $\mathcal{P}$ has this property. According to [APT23,
Corollary 4.2.2], any reflectable base is finite.

We define $\hbox{refl}(R)$ to be the smallest cardinality of a reflectable
base for $R$, and $\hbox{ind}(R)$ as
$\hbox{refl}(R)-\hbox{rank}(R)-\hbox{null}(R)=\hbox{refl}(R)-\hbox{rank}\langle
R\rangle$. These terms, along with $\hbox{rank}(R)$ and $\hbox{null}(R)$, are
all isomorphism invariants for $R$. We may also describe the $\hbox{ind}(R)$
in terms of rank $\ell$, nullity $\nu$, and semilattices $S,$ $S_{1}$ and
$S_{2}$ given in (2.4) and (2.5). In fact, depending on type $X$ of $R$, we
have

(2.6)
$\hbox{ind}(R)=\left\\{\begin{array}[]{ll}\hbox{ind}(S)-\ell-\nu&X=A_{1},\\\
0&X=A_{\ell}\;(\ell\geq 2),D_{\ell},E_{6,7,8},F_{4},G_{2},\\\
\hbox{ind}(S_{1})+\hbox{ind}(S_{2})-\nu&X=B_{2},\\\
\hbox{ind}(S_{1})-t&X=B_{\ell},\;\ell\geq 3,\\\
\hbox{ind}(S_{2})-\nu-t&X=C_{\ell},\;\ell\geq 3.\end{array}\right.$

Note that in type $B_{\ell}$, $\ell\geq 3$, if $\hbox{ind}(R)=0$, then
$\hbox{ind}(S_{1})=t=\hbox{rank}(S_{1})$, so as we saw in 2.2.1, in this case
we have $S_{1}=\cup_{i=0}^{t}(\tau_{i}+2\langle S_{1}\rangle)$ with $\langle
S_{1}\rangle=\sum_{i=1}^{t}{\mathbb{Z}}\tau_{i}$.

Extended affine root systems of index $0$ are more similar to finite and
affine root systems than those with a non-zero index, as per [APT23]. We note
from [Az99, §4] that $\hbox{ind}(R)=0$ if $R$ is finite or affine (i.e., if
$\hbox{null}(R)=0$ or $1$) or if $R$ has one of the types $A_{\ell}$
($\ell\geq 2$), $D_{\ell}$ ($\ell\geq 4$), $F_{4}$, $G_{2}$, $E_{6,7,8}$.
Therefore, the extended affine root systems with a non-zero index occur only
in types $A_{1}$, $B_{\ell}$, and $C_{\ell}$.

##  3\. Chevalley involutions

We begin the section by establishing some fundamental facts regarding
Chevalley involutions for an extended affine Lie algebra
$E=(E,(\cdot,\cdot),{\mathcal{H}})$. We set $E_{c,\alpha}:=E_{c}\cap
E_{\alpha}$ for $\alpha\in R$. When we refer to an involution on a vector
space, we are referring to an automorphism of order $2$.

###  3.1. Chevalley involutions

Although we have assumed that all extended affine Lie algebras in this work
are reduced, the results in this subsection are also valid for non-reduced
ones.

######  Lemma 3.1.1.

Suppose $\tau$ is a finite order automorphism of an extended affine Lie
algebra. Then $\tau=\hbox{id}$ if and only the restriction of $\tau$ to the
core is identity.

Proof. Suppose $\tau_{|_{E_{c}}}=\hbox{id}_{E_{c}}$. Let $x\in E$ and $y\in
E_{c}$. Since $E_{c}$ is an ideal of $E$, we get
$[x,y]=\tau[x,y]=[\tau(x),y]$. Therefore, $\tau(x)-x$ is in centralizer of the
core. Since $E$ is tame, $\tau(x)-x\in E_{c}$ and so
$\tau(\tau(x)-x)=\tau(x)-x,$ for each $x\in E$. Thus $\tau$ satisfies the
polynomial $(t-1)^{2}=0$. Since $\tau$ is of finite order, we get
$\tau=\hbox{id}$. $\Box$

######  Lemma 3.1.2.

Let $\tau$ be a finite order automorphism of $E$. Fix $\epsilon\in\\{1,-1\\}$.

(i) If $\tau(E_{\alpha})=E_{\epsilon\alpha}$ for all $\alpha\in R^{\times}$,
then

\- $\tau_{|_{{\mathcal{H}}\cap E_{c}}}=\epsilon\hbox{id},$

\- if $\epsilon=-1$ then $\tau^{2}=\hbox{id}$,

\- $\tau_{|_{\mathcal{H}}}=\epsilon\hbox{id}_{{\mathcal{H}}}$ mod $Z(E_{c})$.

(ii) $\tau(E_{\alpha})=E_{\epsilon\alpha}$ for all $\alpha\in R$ if and only
if $\tau(h)=\epsilon h$ for all $h\in{\mathcal{H}}$.

(iii) If $\tau(E_{\alpha})=E_{\epsilon\alpha}$ for all $\alpha\in R$, then
$\tau$ preserves the form.

Proof. (i) Assume that $\tau(E_{\alpha})=E_{\epsilon\alpha}$ for each
$\alpha\in R^{\times}$. For $\alpha\in R^{\times}$, fix $x_{\pm\alpha}\in
E_{\pm\alpha}$ such that $[x_{\alpha},x_{-\alpha}]=h_{\alpha}$. Then
$\tau(x_{\alpha})=\mu_{\alpha}x_{\epsilon\alpha}$ for some
$\mu_{\alpha}\in{\mathbb{K}}^{\times}$. So

$\tau(h_{\alpha})=\tau[x_{\alpha},x_{-\alpha}]=\mu_{\alpha}\mu_{-\alpha}[x_{\epsilon\alpha},x_{-\epsilon\alpha}]=\mu_{\alpha}\mu_{-\alpha}h_{\epsilon\alpha}=\epsilon\mu_{\alpha}\mu_{-\alpha}h_{\alpha}.$

Now by applying $\tau$ to the both sides of
$2x_{\alpha}=[h_{\alpha},x_{\alpha}]$, we get

$2\mu_{\alpha}x_{\epsilon\alpha}=\epsilon\mu_{\alpha}\mu_{-\alpha}[h_{\alpha},\mu_{\alpha}x_{\epsilon\alpha}]=\epsilon^{2}2\mu_{\alpha}^{2}\mu_{-\alpha}x_{\epsilon\alpha}=2\mu_{\alpha}^{2}\mu_{-\alpha}x_{\epsilon\alpha}.$

Thus

(3.1)
$\mu_{\alpha}\mu_{-\alpha}=1\quad\hbox{and}\quad\tau(h_{\alpha})=\epsilon
h_{\alpha},\hbox{ for each }\alpha\in R^{\times}.$

Since ${\mathcal{H}}\cap E_{c}=\sum_{\alpha\in R^{\times}}h_{\alpha}$, we get

(3.2) $\tau_{|_{{\mathcal{H}}\cap E_{c}}}=\epsilon\hbox{id}.$

Note that if $\epsilon=-1$ then form (3.1) for $\alpha\in R^{\times}$ we have
$\tau^{2}(x_{\alpha})=\mu_{\alpha}\mu_{-\alpha}x_{\alpha}=x_{\alpha}$. Since
$E_{c}$ is generated by $x_{\alpha}$, $\alpha\in R^{\times}$, we get
$\tau^{2}=\hbox{id}_{|_{E_{c}}}$, and so by Lemma 3.1.1, $\tau^{2}=\hbox{id}$.

Next for $h\in{\mathcal{H}}$ and $\alpha\in R^{\times}$, we have

$\displaystyle[h-\epsilon\tau(h),x_{\alpha}]$ $\displaystyle=$
$\displaystyle[h,x_{\alpha}]-\epsilon\tau[h,\mu_{\epsilon\alpha}^{-1}x_{\epsilon\alpha}]$
$\displaystyle=$
$\displaystyle\alpha(h)x_{\alpha}-\epsilon\epsilon\mu_{\epsilon\alpha}^{-1}\alpha(h)\tau(x_{\epsilon\alpha})$
$\displaystyle=$
$\displaystyle\alpha(h)x_{\alpha}-\alpha(h)\mu_{\epsilon\alpha}^{-1}\mu_{\epsilon\alpha}x_{\alpha}=0.$

Therefore, $h-\epsilon\tau(h)$ is in the centralizer of $E_{c}$ in $E$ for
each $h\in{\mathcal{H}}$. Since $E$ is tame, $h-\epsilon\tau(h)\in Z(E_{c})$.

(ii) Assume first that $\tau(E_{\alpha})=E_{\epsilon\alpha}$ for all $\alpha$;
in particular, $\tau({\mathcal{H}})=\tau(E_{0})=E_{0}={\mathcal{H}}$. So from
part (i), we have $h-\epsilon\tau(h)\in{\mathcal{H}}\cap Z(E_{c})$. Then from
(3.2) we have $\tau(h-\epsilon\tau(h))=\epsilon(h-\epsilon\tau(h))$ for each
$h\in{\mathcal{H}}$. Thus $\epsilon\tau$ restricted to ${\mathcal{H}}$
satisfies the equation $(t-1)^{2}=0$. Since $\tau$ is of finite order, it
follows that $\tau_{|_{\mathcal{H}}}=\epsilon\hbox{id}$.

Conversely, assume that $\tau_{|_{\mathcal{H}}}=\epsilon\hbox{id}.$ Then for
$\alpha\in R$, $h\in{\mathcal{H}}$ and $x\in E_{\alpha}$, we have

$[h,\tau(x)]=\tau[\epsilon h,x]=\epsilon\alpha(h)\tau(x),$

and so $\tau(x)\in E_{\epsilon\alpha}.$

(iii) By (ii), for $h,h^{\prime}\in{\mathcal{H}}=E_{0}$, we have
$(\tau(h),\tau(h^{\prime}))=(\epsilon h,\epsilon h^{\prime})=(h,h^{\prime})$.
Assume now that $0\not=\alpha\in R$, $\beta\in R$. Let $x_{\alpha}\in
E_{\alpha},x_{\beta}\in E_{\beta}$. If $\alpha+\beta\not=0$, then
$(\tau(x_{\alpha}),\tau(x_{\beta}))=0=(x_{\alpha},x_{\beta})$. If
$\beta=-\alpha$, then by (ii) and (2.2),

$t_{\epsilon\alpha}(\tau(x_{\alpha}),\tau(x_{-\alpha}))=\tau[x_{\alpha},x_{-\alpha}]=t_{\epsilon\alpha}(x_{\alpha},x_{-\alpha}).$

Therefore, $(x_{\alpha},x_{\beta})=(\tau(x_{\alpha}),\tau(x_{\beta}))$ for
$\alpha,\beta\in R$ and $x_{\alpha}\in E_{\alpha},x_{\beta}\in E_{\beta}$.
$\Box$

######  Definition 3.1.3.

(i) We call a finite order automorphism $\tau$ of $E$ a Chevalley involution
if $\tau(E_{\alpha})=E_{-\alpha}$ for all $\alpha\in R$, or equivalently (see
3.1.2(ii)) if $\tau(h)=-h$ for all $h\in{\mathcal{H}}$.

(ii) We call a finite order automorphism $\tau$ of $E_{c}$ a Chevalley
involution for $E_{c}$ if $\tau(E_{\alpha})=E_{-\alpha}$ for all $\alpha\in
R^{\times}$.

######  Remark 3.1.4.

Since by [CNPY16, Proposition 3.4] any automorphism $\tau$ for $E$ stabilizes
the core, it is immediate from Definition 3.1.3 that any Chevalley involution
for $E$ restricts to a Chevalley involution for $E_{c}$. Also, we have from
Lemma 3.1.2(iii) that any finite order Chevalley involution preserves the
form.

######  Corollary 3.1.5.

Let $\tau$ be a finite order automorphism of $E$ such that the restriction of
$\tau$ to $E_{c}$ is a Chevalley involution for $E_{c}$. Then $\tau$ is a
Chevalley involution for $E$ if and only if $\tau$ stabilizes ${\mathcal{H}}$.
In particular, if $E$ is an affine Kac-Moody Lie algebra, then $\tau$ is a
Chevalley involution for $E$ if and only if its restriction to $E_{c}$ is a
Chevalley involution.

Proof. Since $\tau$ is a Chevalley involution for $E_{c}$, we have
$\tau(E_{\alpha})=E_{-\alpha}$ for $\alpha\in R^{\times}$. By assumption
$\tau$ stabilizes ${\mathcal{H}}=E_{0}$, so we have
$\tau(E_{\alpha})=E_{-\alpha}$ for all $\alpha\in R$. Thus the first part of
the statement is an immediate consequence of Lemma 3.1.2(ii). For the second
part, we note that if $E$ is an affine Kac–Moody Lie algebra, one knows that
$Z(E_{c})\subseteq{\mathcal{H}}$, see [Kac90], thus $\tau$ stabilizes
${\mathcal{H}}$ and we are done. $\Box$

######  Lemma 3.1.6.

The existence of Chevalley involutions for an extended affine Lie algebra does
not depend on the choice of the Cartan subalgebra.

Proof. Suppose ${\mathcal{H}}$ and ${\mathcal{H}}^{\prime}$ are two Cartan
subalgebras for $E$, and $\tau$ is a Chevalley involution for
$(E,(\cdot,\cdot),{\mathcal{H}})$. By conjugacy theorem for extended affine
Lie algebras (see [CNPY16, Theorem 7.6] and [CNP17]), there exists an
automorphism $\sigma$ of $E$ which maps ${\mathcal{H}}$ onto
${\mathcal{H}}^{\prime}$. Then $\bar{\tau}:=\sigma\tau\sigma^{-1}$ is a
Chevalley involution for $(E,(\cdot,\cdot),{\mathcal{H}}^{\prime})$. $\Box$

##  4\. Cartan automorphisms, and characters

Assume that $E=(E,(\cdot,\cdot),{\mathcal{H}})$ is an extended affine Lie
algebra with root system $R$. The composition of two Chevalley involutions for
$E$ is referred to as a ”Cartan automorphism” since it stabilizes pointwise
the elements of the Cartan subalgebra ${\mathcal{H}}$ and thus, stabilizes
each root space. Moreover, the composition of a Cartan automorphism and a
Chevalley involution results in another Chevalley involution. In turn, Cartan
automorphisms can be constructed using the ”characters” of the root system,
and vice versa, to each Cartan automorphism one associates a map that is
perceivable to be generally a character. This section studies the
relationships between Chevalley involutions, Cartan automorphisms, and
characters.

###  4.1. Characters

We begin by giving definitions of Cartan automorphisms, characters, and some
related terms.

######  Definition 4.1.1.

Let $(E,(\cdot,\cdot),{\mathcal{H}})$ be an extended affine Lie algebra and
$R$ be an extended affine root system.

(i) A Cartan automorphism $\psi$ for $E$ is the one that satisfies $\psi(h)=h$
for all $h\in{\mathcal{H}}$. It is evident that Cartan automorphisms stabilize
each root space.

(ii) We call a Cartan automorphism diagonal if it acts as a scalar multiple of
identity when restricted to each root space.

######  Definition 4.1.2.

(i) We call a map $\psi:R\rightarrow{\mathbb{K}}^{\times}$, a core-character
for $R$ if $\psi(\alpha+\beta)=\psi(\alpha)\psi(\beta)$, for $\alpha\in
R^{\times},$ $\beta,\alpha+\beta\in R$. A core-character $\psi$ is called of
finite-order if there exists $m\in{\mathbb{Z}}_{>0}$ with $\psi(\alpha)^{m}=1$
for all $\alpha\in R$. If $R$ has type $X$, we call a core-character for $R$ a
core-character of type $X$.

(ii) A core-character $\psi$ is called a character if
$\psi(\alpha+\beta)=\psi(\alpha)\psi(\beta)$ for $\alpha,\beta,\alpha+\beta\in
R$. A root-lattice character for $R$ is a group homomorphism $\psi:\langle
R\rangle\rightarrow{\mathbb{K}}^{\times}$.

######  Example 4.1.3.

(i) Given an extended affine Lie algebra $E$ with root system $R$ and a
(finite-)order character $\Phi:R\rightarrow{\mathbb{K}}^{\times}$, the linear
map defined by $\bar{\Phi}(x)=\Phi(\alpha)x$, for $x\in E_{\alpha}$,
$\alpha\in R$, defines a (finite order) diagonal automorphism for $E$.

(ii) Let $m$ be a positive integer. Given an extended affine root system $R$,
fix a basis $\mathcal{P}$ for the free abelian group $\langle R\rangle$. For
$\alpha\in\mathcal{P}$ fix an $m$-th root of unity $\eta_{\alpha}$. Then the
group homomorphism $\Phi:\langle R\rangle\longrightarrow{\mathbb{K}}^{\times}$
induced by $\Phi(\alpha)=\eta_{\alpha}$, $\alpha\in\mathcal{P}$, defines a
finite-order root-lattice character for $R$.

######  Remark 4.1.4.

(i) It is immediate that a Cartan automorphism stabilizes each root space.
Conversely, suppose $\psi$ is a finite order automorphism of $E$ that
stabilizes each root space. Then by Lemma 3.1.2(ii), $\psi$ is a Cartan
automorphism.

(ii) In [L88], the author classifies involutive automorphisms (automorphisms
of order 2) for affine Kac-Moody Lie algebras by dividing them into two
categories: first and second kind automorphisms. Chevalley involutions belong
to the second kind, while Cartan involutions belong to the first kind.

(iii) By Lemma 3.1.2(iii), any finite order Cartan automorphism preserves the
form.

Previously, in Example 4.1.3, we learned about producing non-trivial finite
order diagonal automorphisms for the extended affine Lie algebra $E$ using a
finite-order character for the root system $R$. But does every finite order
diagonal automorphism arise this way? We will find out in the following
sections that the answer is ”almost” affirmative, with a few exceptions in
types $A_{1}$ and $B_{\ell}$.

From now on we assume that $E$ is an extended affine Lie algebra with root
system $R$, and we fix a Cartan automorphism $\Psi$. Hereafter, whenever we
write $x_{\alpha}$, $\alpha\in R$, we mean $x_{\alpha}$ is a non-zero element
of $E_{\alpha}$.

For $\alpha\in R^{\times}$, we define $\eta_{\alpha}$ by

(4.1) $\Psi(x_{\alpha})=\eta_{\alpha}x_{\alpha},\quad(0\not=x\in
E_{\alpha},\alpha\in R^{\times}).$

Now let $\sigma\in R^{0}$. Since isotropic roots are non-isolated, there
exists $\alpha\in R^{\times}$ with $\alpha+\sigma\in R^{\times}$. We define

$\eta_{\sigma}^{\alpha}:=\eta_{\alpha+\sigma}\eta_{-\alpha},\quad(\sigma\in
R^{0},\alpha,\alpha+\sigma\in R^{\times}).$

######  Convention 4.1.5.

Whenever we use the notation $\eta^{\alpha}_{\sigma}$, we understand that
$\alpha\in R^{\times}$, $\sigma\in R^{0}$, $\alpha+\sigma\in R^{\times}$.

######  Lemma 4.1.6.

Let $\Psi$ be a finite order Cartan automorphism. Then $\Psi$ is diagonal if
and only if $\eta^{\alpha}_{\sigma}=\eta^{\beta}_{\sigma}$, $\alpha,\beta\in
R^{\times}$, $\sigma\in R^{0}$.

Proof. Assume first that $\Psi$ is diagonal. Then
$\Psi_{|_{E_{\alpha}}}=\eta_{\alpha}\hbox{id}_{|_{E_{\alpha}}}$, $\alpha\in
R$, for some $\eta_{\alpha}\in{\mathbb{K}}^{\times}$. Let $\sigma\in R^{0}$
and $\alpha,\alpha+\sigma\in R^{\times}$. By (2.1), there exist $x_{\alpha}\in
E_{\alpha}$ and $x_{\sigma}\in E_{\sigma}$ such that
$[x_{\alpha},x_{\sigma}]\not=0$. Again by (2.1),
$X_{\sigma}:=[x_{-\alpha},[x_{\alpha},x_{\sigma}]]\not=0$. Then

$\displaystyle\eta_{\sigma}[x_{-\alpha},[x_{\alpha},x_{\sigma}]]$
$\displaystyle=$ $\displaystyle\Psi([x_{-\alpha},[x_{\alpha},x_{\sigma}]])$
$\displaystyle=$
$\displaystyle[\Psi(x_{-\alpha}),\Psi[x_{\alpha},x_{\sigma}]]$
$\displaystyle=$
$\displaystyle\eta_{-\alpha}\eta_{\alpha+\sigma}[x_{-\alpha},[x_{\alpha},x_{\sigma}]]$
$\displaystyle=$
$\displaystyle{\eta_{\sigma}^{\alpha}[x_{-\alpha},[x_{\alpha},x_{\sigma}]].}$

This shows that the definition of $\eta^{\alpha}_{\sigma}$ is independent of
the choice of $\alpha\in R^{\times}.$

Conversely, assume that $\eta_{\sigma}^{\alpha}=\eta_{\sigma}^{\beta}$ for
$\sigma\in R^{0}$, $\alpha,\beta\in R^{\times}.$ So for $\sigma$ in $R^{0}$,
the term $\eta_{\sigma}:=\eta_{\sigma}^{\alpha}$, for some $\alpha\in
R^{\times}$, is well-defined. Since for $\alpha\in R^{\times}$, $\dim
E_{\alpha}=1$, we have
$\Psi_{|_{E_{\alpha}}}=\eta_{\alpha}\hbox{id}_{|_{E_{\alpha}}}$. Now let
$\sigma\in R^{0}$. Then if $\alpha,\alpha+\sigma\in R^{\times}$, we have

$\Psi[x_{\alpha+\sigma},x_{-\alpha}]=\eta_{\alpha+\sigma}\eta_{-\alpha}[x_{\alpha+\sigma},x_{-\alpha}]=\eta_{\sigma}[x_{\alpha+\sigma},x_{-\alpha}].$

Since $E_{c}\cap E_{\sigma}$ is spanned by $1$-dimensional spaces
$[E_{\alpha+\sigma},E_{\alpha}]$, $\alpha\in R^{\times}$, we get

(4.2) $\Psi_{|_{E_{c}\cap E_{\alpha}}}=\eta_{\sigma}\hbox{id}_{|_{E_{c}\cap
E_{\alpha}}},\qquad(\alpha\in R).$

Now, consider an arbitrary element $x_{\sigma}\in E_{\sigma}$, $\sigma\in
R^{0}$. Then for each $\alpha\in R^{\times}$ such that $\alpha+\sigma\in R$ we
have

$\eta_{\alpha+\sigma}[x_{\alpha},x_{\sigma}]=\Psi[x_{\alpha},x_{\sigma}]=\eta_{\alpha}[x_{\alpha},\Psi(x_{\sigma})].$

So $[x_{\alpha},\eta_{\sigma}x_{\sigma}-\Psi(x_{\sigma})]=0$. We note that
this equality holds even if $\alpha+\sigma\not\in R$, since in this case
$[x_{\alpha},x_{\sigma}]=0=[x_{\alpha},\Psi(x_{\sigma})].$ Thus
$\eta_{\sigma}x_{\sigma}-\Psi(x_{\sigma})$ is in the centralizer of $E_{c}$.
Since $E$ is tame, $\eta_{\sigma}x_{\sigma}-\Psi(x_{\sigma})\in E_{c}\cap
E_{\sigma}$. Then by (4.2),
$\Psi(\eta_{\sigma}x_{\sigma}-\Psi(x_{\sigma}))=\eta_{\sigma}(\eta_{\sigma}x_{\sigma}-\Psi(x_{\sigma})).$
Therefore, $\eta_{\sigma}^{-1}\Psi_{|_{E_{\sigma}}}$ satisfies the equation
$(t-1)^{2}=0$. Since $\Psi$ is of finite order, we get
$\Psi(x_{\sigma})=\eta_{\sigma}x_{\sigma}$ for each $x_{\sigma}\in
E_{\sigma}$, and so
$\Psi_{|_{E_{\alpha}}}=\eta_{\alpha}\hbox{id}_{|_{E_{\alpha}}},$ $\alpha\in
R$, as required. $\Box$

######  Definition 4.1.7.

Let $\Psi$ be a finite order diagonal automorphism for $E$. For $\sigma\in
R^{0}$, we set $\eta_{\sigma}:=\eta_{\alpha+\sigma}\eta_{-\alpha}$, for some
$\alpha\in R^{\times}.$ By Lemma 4.1.6, $\eta_{\sigma}$ is well-defined.
Proposition 4.1.8 below shows that the map
$\psi:R\rightarrow{\mathbb{K}}^{\times}$ defined by
$\psi(\alpha)=\eta_{\alpha}$, $\alpha\in R$, is a core-character, hereafter,
we refer to it as the core-character associated to $\Psi$.

######  Proposition 4.1.8.

Let $\Psi$ be a finite-order diagonal automorphism. The map
$\psi:R\rightarrow{\mathbb{K}}^{\times}$ defined by
$\psi(\alpha)=\eta_{\alpha}$, $\alpha\in R$ is a finite-order core-character
for $R$, satisfying $\psi(-\alpha)=\psi(\alpha)^{-1}$, $\alpha\in R$.

Proof. Let $\psi$ be the map associated to $\Psi$, given in Definition 4.1.7.
Since $\Psi$ stabilizes pointwise ${\mathcal{H}}$, we note that if
$0\not=x_{\pm\alpha}\in E_{\pm\alpha}$, $\alpha\in R^{\times}$, we have

$[x_{\alpha},x_{-\alpha}]=\Psi[x_{\alpha},x_{-\alpha}]=\eta_{\alpha}\eta_{-\alpha}[x_{\alpha},x_{-\alpha}],$

implying that $\psi(-\alpha)=\psi(\alpha)^{-1}$. Also if $\sigma\in R^{0}$ and
$\alpha,\alpha+\sigma\in R^{\times}$, then

$\psi(\sigma)\psi(-\sigma)=\eta_{\sigma}\eta_{-\sigma}=\eta_{\alpha+\sigma}\eta_{-\alpha}\eta_{-\alpha-\sigma}\eta_{\alpha}=1.$

Next, we show that $\psi$ is a core-character. If
$\alpha,\beta,\alpha+\beta\in R^{\times}$ and $x_{\alpha},x_{\beta}$ are non-
zero elements of $E_{\alpha}$ and $E_{\beta}$, then
$[x_{\alpha},x_{\beta}]\not=0$, and

$\eta_{\alpha+\beta}[x_{\alpha},x_{\beta}]=\Psi[x_{\alpha},x_{\beta}]=\eta_{\alpha}\eta_{\beta}[x_{\alpha},x_{\beta}].$

Thus
$\psi(\alpha+\beta)=\eta_{\alpha+\beta}=\eta_{\alpha}\eta_{\beta}=\psi(\alpha)\psi(\beta).$
Also if $\beta\in R^{0},\alpha,\alpha+\beta\in R^{\times}$, then by the way
$\eta_{\beta}$ is defined, we have

$\psi(\alpha+\beta)=\eta_{\alpha+\beta}=\eta_{\alpha+\beta}\eta_{-\alpha}\eta_{\alpha}=\eta_{\beta}\eta_{\alpha}=\psi(\alpha)\psi(\beta).$

Thus $\psi$ is a core-character. Since $\Psi$ is of finite-order, so is
$\psi$. $\Box$

We proceed with stating a few lemmas which will be used frequently in the
sequel.

######  Lemma 4.1.9.

Let $\sigma\in R^{0}$.

(i) If $\alpha,\beta,\alpha+\sigma,\beta+\sigma,\alpha-\beta\in R^{\times}$,
then $\eta^{\alpha}_{\sigma}=\eta^{\beta}_{\sigma}$.

(ii) If $\alpha,\beta,\alpha+\beta,\alpha+\beta+\sigma,\beta+\sigma\in R$,
then $\eta^{\alpha+\beta}_{\sigma}=\eta^{\beta}_{\sigma}$.

Proof. (i) Applying $\Psi$ to the non-zero element
$[x_{\alpha+\sigma},x_{-\beta-\sigma}]$, gives
$\eta_{\alpha+\sigma}\eta_{-\beta-\sigma}=\eta_{\alpha-\beta}=\eta_{\alpha}\eta_{-\beta}$,
as required.

(ii) Set $\gamma=\alpha+\beta$. Then the elements $\gamma$, $\beta$ and
$\sigma$ satisfy conditions of part (i), with $\gamma$ in place of $\alpha$.
$\Box$

######  Lemma 4.1.10.

Let $\sigma\in R^{0}$ and $\alpha,\beta,\alpha+\sigma,\beta+\sigma\in
R^{\times}$. For $\gamma\in\\{\pm\alpha,\pm\beta,\pm(\alpha+\sigma)\\}$, fix
$0\not=x_{\gamma}\in E_{\gamma}$. If $\beta+\alpha+\sigma\not\in R$ but
$\beta-\alpha\in R^{\times}$, then
$[x_{\beta},[x_{\alpha+\sigma},x_{-\alpha}]]\not=0$.

Proof. By assumptions, $[x_{\alpha+\sigma},[x_{\beta},x_{-\alpha}]]\not=0$ and
$[x_{-\alpha},[x_{\alpha+\sigma},x_{\beta}]]=0$. Therefore, we are done by the
Jacobi identity. $\Box$

######  Lemma 4.1.11.

Let $\sigma\in R^{0}$ and $\alpha,\alpha+\sigma\in R^{\times}$. Then

(i) $[E_{\alpha},[E_{\alpha},E_{\sigma-\alpha}]]\not=\\{0\\}$.

(ii) $\eta_{\alpha}^{2}=\eta_{\alpha+\sigma}\eta_{\alpha-\sigma}$.

Proof. (i) Consider the $\alpha$-string through $\sigma$ (see (2.3)), namely

$\\{\sigma-d\alpha,\ldots,\sigma,\ldots\sigma+u\alpha\\}=R\cap\\{\sigma+n\alpha\mid
n\in{\mathbb{Z}}\\},\quad(u,d\geq 0).$

Since $(\alpha,\sigma)=0$, $1\leq u=d\leq 2$. Fix $e_{\pm\alpha}\in
E_{\pm\alpha}$ with $[e_{\alpha},e_{-\alpha}]=h_{\alpha}$, and let
$0\not=x_{\sigma-d\alpha}\in E_{\sigma-d\alpha}.$ Then
$[e_{-\alpha},x_{\sigma-d\alpha}]=0$. Therefore by [AABGP97, Lemma I.1.21],

$\hbox{ad}(e_{\alpha})^{N}(x_{\sigma-d\alpha})\not=0\quad\hbox{and}\quad\hbox{ad}(e_{\alpha})^{N+1}(x_{\sigma-d\alpha})=0,$

where $N:=(\sigma-d\alpha,-\alpha^{\vee})=2d\geq 2$. Then,
$z:=[e_{\alpha},[e_{\alpha},(\hbox{ad}e_{\alpha})^{d-1}(x_{\sigma-d\alpha})]]\not=0$,
as $d+1\leq 2d=N$. Now $z\in[E_{\alpha},[E_{\alpha},E_{\sigma-\alpha}]]$ and
we are done.

(ii) Applying $\Psi$ to the element
$[x_{\alpha},[x_{\alpha},x_{-\alpha+\sigma}]$ which is non-zero by part (i),
we get $\eta_{\alpha}\eta_{\alpha}\eta_{-\alpha+\sigma}=\eta_{\alpha+\sigma}$,
as required. $\Box$

###  4.2. Simply laced Cartan automorphisms

We study the class of finite-order Cartan automorphisms, depending on type $X$
of $E$. We begin with simply laced types.

######  Proposition 4.2.1.

Let $X$ be simply laced of rank $>1$, and $\Psi$ a finite order Cartan
automorphism for $E$. Then

(i) $\Psi$ is diagonal,

(ii) the core-character $\psi$ associated to $\Psi$ is a character.

Proof. Under the assumptions, we know from 2.5 that $S=\Lambda=R^{0}$ is a
lattice, and $R^{\times}=\dot{R}^{\times}+\Lambda$. By Lemma 4.1.6, it is
enough to show that for $\lambda\in R^{0}$ and $\alpha,\beta\in R^{\times}$,
$\eta^{\alpha}_{\lambda}=\eta^{\beta}_{\lambda}$. We first claim that,

(4.3)
$\eta^{\dot{\alpha}}_{\lambda}=\eta^{\dot{\beta}}_{\lambda},\quad\dot{\alpha},\dot{\beta}\in\dot{R}^{\times},\lambda\in
R^{0}.$

To prove this, we primarily note that if
$\dot{\alpha},\dot{\beta},\dot{\alpha}+\dot{\beta}\in\dot{R}^{\times}$, then
by applying $\Psi$ on the non-zero element
$[x_{-\dot{\alpha}},[x_{\dot{\alpha}+\lambda},x_{\dot{\beta}}]]$, we obtain
$\eta_{\dot{\beta}+\lambda}=\eta_{-\dot{\alpha}}\eta_{\dot{\alpha}+\lambda}\eta_{\dot{\beta}}$,
or equivalently

(4.4)
$\eta^{\dot{\alpha}}_{\lambda}=\eta^{\dot{\beta}}_{\lambda},\quad(\dot{\alpha},\dot{\beta},\dot{\alpha}+\dot{\beta}\in\dot{R}^{\times}).$

Also, we see from Lemma 4.1.11(ii) that

(4.5)
$\eta_{\lambda}^{\dot{\alpha}}=\eta_{\lambda}^{-\dot{\alpha}},\quad(\dot{\alpha}\in\dot{R}^{\times},\lambda\in
R^{0}).$

Now let $\dot{\alpha},\dot{\beta}\in\dot{R}^{\times}$. By (4.4), we may assume
that $\dot{\alpha}\not=\pm\dot{\beta}$. Since $R$ is irreducible, we can find
a sequence
$\dot{\alpha}_{1}=\dot{\alpha},\dot{\alpha}_{2},\ldots\,\dot{\alpha}_{n-1},\dot{\alpha}_{n}=\dot{\beta}$
of roots in $\dot{R}^{\times}$ with
$(\dot{\alpha}_{i},\dot{\alpha}_{i+1})\not=0$, for all $i$. We may assume also
that $\dot{\alpha}_{i}\not=\pm\dot{\alpha}_{i+1}.$ By the root string
property, either $\dot{\alpha}_{i}+\dot{\alpha}_{i+1}\in\dot{R}^{\times}$ or
${\dot{\alpha}}_{i}-{\dot{\alpha}}_{i+1}\in\dot{R}^{\times}$. Therefore, by
changing signs if necessary, we may assume that
${\dot{\alpha}}_{i}+{\dot{\alpha}}_{i+1}\in\dot{R}^{\times}$ for all $i$. Then
$\eta_{\lambda}^{\dot{\alpha}}{=\eta_{\lambda}^{\dot{\alpha}_{1}}=}\eta_{\lambda}^{\dot{\alpha}_{2}}=\cdots=\eta_{\lambda}^{\dot{\alpha}_{n}}=\eta_{\lambda}^{\dot{\beta}}.$
This proves (4.3).

Next let $\alpha\in R^{\times}$, $\lambda\in R^{0}$. Then
$\alpha=\dot{\alpha}+\sigma$ for some $\dot{\alpha}\in\dot{R}$, $\sigma\in
R^{0}$. Since $\hbox{rank}R>1$, we can choose $\dot{\beta}\in\dot{R}^{\times}$
such that $\dot{\alpha}+\dot{\beta}\in R^{\times}$. Then

$\displaystyle\eta^{\alpha}_{\lambda}$ $\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}+\sigma+\lambda}\eta_{-\dot{\alpha}-\sigma}$
$\displaystyle=$
$\displaystyle\eta_{(\dot{\alpha}+\dot{\beta}+\sigma)+(-\dot{\beta}+\lambda)}\eta_{-\dot{\alpha}-\sigma}$
$\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}+\dot{\beta}+\sigma}\eta_{-\dot{\beta}+\lambda}\eta_{-\dot{\alpha}-\sigma}$
$\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}+\sigma}\eta_{\dot{\beta}}\eta_{-\dot{\beta}+\lambda}\eta_{-\dot{\alpha}-\sigma}$
$\displaystyle=$ $\displaystyle\eta^{-\dot{\beta}}_{\lambda}$
$\displaystyle(\hbox{by (\ref{tempaug31-1}})$ $\displaystyle=$
$\displaystyle\eta^{\dot{\alpha}}_{\lambda}.$

Therefore, we see from this by again applying (4.3) that if
$\beta=\dot{\beta}+\sigma^{\prime}\in R^{\times}$,
$\dot{\beta}\in\dot{R}^{\times}$, $\sigma^{\prime}\in R^{0}$, we have
$\eta^{\alpha}_{\lambda}=\eta^{\dot{\alpha}}_{\lambda}=\eta^{\dot{\beta}}_{\lambda}=\eta^{\beta}_{\lambda}.$
This completes the proof of (i).

(ii) By part (i) and Proposition 4.1.7, $\psi$ is a core-character. To see
that $\psi$ is a character, let $\sigma,\sigma^{\prime}\in S=R^{0}$. Then for
$\dot{\alpha}\in\dot{R}^{\times}$ we have

$\displaystyle\psi(\sigma+\sigma^{\prime})$ $\displaystyle=$
$\displaystyle\psi(\dot{\alpha}+\sigma-\dot{\alpha}+\sigma^{\prime})$
$\displaystyle=$
$\displaystyle\psi(\dot{\alpha}+\sigma)\psi(-\dot{\alpha}+\sigma^{\prime})$
$\displaystyle=$
$\displaystyle\psi(\dot{\alpha})\psi(\sigma)\psi(-\dot{\alpha})\psi(\sigma^{\prime})$
$\displaystyle=$ $\displaystyle\psi(\sigma)\psi(\sigma^{\prime}).$

$\Box$

###  4.3. Non-simply laced Cartan automorphisms

We continue with the study of Cartan automorphisms, for non-simply laced
extended affine Lie algebras. We assume that $E$ is a non-simply laced reduced
extended affine Lie algebra of type $X$, with root system $R$. We recall from
2.5 that $R=R^{0}\cup(\dot{R}_{sh}+S)\cup(\dot{R}_{lg}+L)$, with
$R^{0}=(S+S)$. Also, we recall that
$\dot{R}=\dot{R}_{sh}\cup\dot{R}_{lg}\cup\\{0\\}$ is an irreducible finite
root system which has one of the types $X=B_{\ell},$ $C_{\ell}$, $F_{4}$ or
$G_{2}$. We denote the highest short and long roots of $\dot{R}$ with
$\dot{\theta}_{s}$ and $\dot{\theta}_{\ell}$, respectively. As before $\Psi$
is a finite order Cartan automorphism.

######  Lemma 4.3.1.

We have

(4.6)
$\eta^{\dot{\theta}_{\ell}}_{\lambda}=\eta^{\dot{\theta}_{s}}_{\lambda},\quad(\lambda\in
L),$

and

(4.7)
$\eta^{\dot{\alpha}}_{\sigma}=\eta^{\dot{\alpha}^{\prime}}_{\sigma},\quad(\dot{\alpha},\dot{\alpha}^{\prime}\in\dot{R}_{sh},\sigma\in
S).$

Proof. Let $\lambda\in L$. Take $\alpha=\dot{\theta}_{\ell}+\lambda$ and
$\beta=\dot{\theta}_{s}+\lambda$. Then
$\alpha-\beta=\dot{\theta}_{\ell}-\dot{\theta}_{s}$, which is a root.Thus
applying Lemma 4.1.9 we get (4.6).

Next, let $\dot{\alpha},\dot{\alpha}^{\prime}\in\dot{R}_{sh}$ and $\sigma\in
S$. If $X$ is of type $B_{\ell}$, then
$\dot{\alpha}\pm\dot{\alpha}^{\prime}\in\dot{R}_{lg}$ and so we are done by
Lemma 4.1.9. If $X=C_{\ell}$ $(\ell\geq 3)$, $F_{4}$ or $G_{2}$, then by
[AABGP97, Theorem II.2.27], $R^{\prime}:=S\cup(\dot{R}_{sh}+S)$ is an extended
affine root system of simply laced type, so the Proposition 4.2.1 applies.
$\Box$

######  Lemma 4.3.2.

(i) If $X=B_{\ell}$, then $\eta^{\alpha}_{\lambda}=\eta^{\beta}_{\lambda}$,
$\alpha,\beta\in R^{\times}$, $\lambda\in L$.

(ii) If $X=C_{\ell},\ell\geq 3$, $F_{4}$ or $G_{2}$, then
$\eta^{\alpha}_{\sigma}=\eta^{\beta}_{\sigma}$, $\alpha,\beta\in R^{\times}$,
$\sigma\in R^{0}$.

Proof. (i) First, we claim that

(4.8) $\eta^{\alpha}_{\lambda}=\eta^{\beta}_{\lambda},\quad(\lambda\in
L,\alpha,\beta,\alpha+\lambda,\beta+\lambda\in R_{\lg}).$

To see this, we consider two cases $\ell\geq 3$ and $\ell=2$, separately.
Assume first that $\ell\geq 3$. Then $L$ is a lattice and so
$\alpha+\lambda\in R^{\times}$ for any $\alpha\in R^{\times}$ and $\lambda\in
L$. We also note that by [AABGP97, Theorem II.2.37],
$R^{\prime}:=L\cup(\dot{R}_{lg}+L)$ is a simply laced extended affine root
system of $\hbox{rank}>1$. Therefore, by Proposition 4.2.1(i), the equality
(4.8) holds.

Next, assume that $\ell=2$. Let $\lambda\in L$, $\beta\in R_{lg}$ and
$\beta+\lambda\in R$. Then $\beta=\dot{\beta}+\lambda_{\ell}$, for some
$\dot{\beta}\in\dot{R}_{lg}$ and some $\lambda_{\ell}\in L$. From the
realization of $\dot{R}$, we know that
$\dot{\beta}=\dot{\alpha}_{1}+\dot{\alpha}_{2}$ for some short roots
$\dot{\alpha}_{1},\dot{\alpha}_{2}\in\dot{R}_{sh}$. Since $\lambda\in
L\subseteq S$, we have
$[x_{\dot{\alpha_{1}}+\dot{\alpha_{2}}+\lambda_{\ell}+\lambda},[x_{-\dot{\alpha_{1}}-\dot{\alpha_{2}}},x_{\dot{\alpha}_{1}-\lambda_{\ell}}]]\not=0$.
Applying $\Psi$ to this, we get

$\eta_{\dot{\alpha}_{1}+\dot{\alpha}_{2}+\lambda_{\ell}+\lambda}\eta_{-\dot{\alpha}_{1}-\dot{\alpha}_{2}}\eta_{\dot{\alpha}_{1}-\lambda_{\ell}}=\eta_{\dot{\alpha}_{1}+\lambda},$

or equivalently,

$\eta_{\dot{\alpha}_{1}+\dot{\alpha}_{2}+\lambda_{\ell}+\lambda}\eta_{-\dot{\alpha}_{2}-\lambda_{\ell}}=\eta_{\dot{\alpha}_{1}+\lambda}.$

Now inserting $\eta_{-\dot{\alpha}_{1}}$ to the both sides of this last
equality we get
$\eta^{\beta}_{\lambda}=\eta^{\dot{\alpha}_{1}+\dot{\alpha}_{2}+\lambda_{\ell}}_{\lambda}=\eta^{\dot{\alpha}_{1}}_{\lambda}$,
and so the claim (4.8) holds by (4.7).

Now combining (4.8) with Lemma 4.3.1 gives

(4.9)
$\eta^{\gamma}_{\lambda}=\eta_{\lambda}^{\dot{\alpha}},\quad(\dot{\alpha}\in\dot{R}^{\times},\gamma\in
R_{lg},\lambda\in L).$

Next, let $\alpha\in R_{sh}$ and $\lambda\in L$. Then
$\alpha=\dot{\alpha}_{s}+\lambda_{s}$ for some
$\dot{\alpha}_{s}\in\dot{R}_{sh}$ and $\lambda_{s}\in S$. Choose
$\dot{\beta}_{s}\in\dot{R}_{sh}$ such that
$\dot{\alpha}_{s}-\dot{\beta}_{s}\in\dot{R}_{lg}$. Then

$\displaystyle\eta_{\lambda}^{\alpha}$ $\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}_{s}+\lambda_{s}+\lambda}\eta_{-\dot{\alpha}_{s}-\lambda_{s}}$
$\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}_{s}+\dot{\beta}_{s}-\dot{\beta}_{s}+\lambda_{s}+\lambda}\eta_{-\dot{\alpha}_{s}-\lambda_{s}}$
$\displaystyle=$
$\displaystyle\eta_{(\dot{\alpha}_{s}-\dot{\beta}_{s})+\lambda}\eta_{\dot{\beta}_{s}+\lambda_{s}}\eta_{-\dot{\alpha}_{s}-\lambda_{s}}$
$\displaystyle=$
$\displaystyle\eta_{(\dot{\alpha}_{s}-\dot{\beta}_{s})+\lambda}\eta_{-\dot{\alpha}_{s}+\dot{\beta}_{s}}\eta_{\dot{\alpha}_{s}-\dot{\beta}_{s}}\eta_{\dot{\beta}_{s}+\lambda_{s}}\eta_{-\dot{\alpha}_{s}-\lambda_{s}}$
$\displaystyle=$
$\displaystyle\eta^{\dot{\alpha}_{s}-\dot{\beta}_{s}}_{\lambda}\eta_{-\dot{\beta}_{s}}\eta_{\dot{\beta}_{s}+\lambda_{s}}\eta_{-\dot{\alpha}_{s}-\lambda_{s}}\eta_{\dot{\alpha}_{s}}$
$\displaystyle=$
$\displaystyle\eta^{\dot{\alpha}_{s}-\dot{\beta}_{s}}_{\lambda}\eta^{-\dot{\alpha}_{s}}_{-\lambda_{s}}\eta^{\dot{\beta}_{s}}_{\lambda_{s}}.$

Now from (4.7), we have
$\eta_{-\lambda_{s}}^{-\dot{\alpha}_{s}}=\eta^{\dot{\beta}_{s}}_{-\lambda_{s}}$.
Therefore,
$\eta_{\lambda}^{\alpha}=\eta^{\dot{\alpha}_{s}-\dot{\beta}_{s}}_{\lambda}\eta^{\dot{\beta}_{s}}_{-\lambda_{s}}\eta^{\dot{\beta}_{s}}_{\lambda_{s}}=\eta^{\dot{\alpha}_{s}-\dot{\beta}_{s}}_{\lambda}.$
This together with (4.9) completes the proof of (i).

(ii) First we consider type $X=C_{\ell}$, $\ell\geq 3$. In this case $R^{0}=S$
is a lattice. Now by [AABGP97, Theorem II.2.37],
$R^{\prime}=R^{0}\cup(\dot{R}_{sh}+S)$ is a simply laced extended affine root
system. In fact if $\ell=3$, then $R^{\prime}$ is of type $A_{3}$ and
otherwise is of type $D_{\ell}$. Then by Proposition 4.2.1(i),

(4.10) $\eta_{\sigma}^{\alpha}=\eta_{\sigma}^{\beta},\qquad(\sigma\in
R^{0},\alpha,\beta\in R_{sh}=\dot{R}_{sh}+S).$

Next, let $\sigma\in R^{0}=S$, $\gamma,\gamma+\sigma\in
R_{\lg}=\dot{R}_{lg}+L$. Then $\gamma=\dot{\gamma}+\lambda$ for some
$\dot{\gamma}\in\dot{R}_{lg}$, $\lambda\in L$. Since $\gamma+\sigma\in R$, we
have $\lambda+\sigma\in L$. From realizations of finite root systems of type
$C_{\ell}$, we know that there exist roots,
$\dot{\alpha}_{1},\dot{\alpha}_{2}\in\dot{R}_{sh}$ such that
$\dot{\alpha}_{1}+\dot{\alpha}_{2}=\dot{\gamma}$. Therefore,

$\displaystyle\eta_{\sigma}^{\gamma}$ $\displaystyle=$
$\displaystyle\eta_{\dot{\gamma}+\lambda+\sigma}\eta_{-\dot{\gamma}-\lambda}$
$\displaystyle=$
$\displaystyle\eta_{(\dot{\alpha}_{1}+\sigma)+(\dot{\alpha}_{2}+\lambda)}\eta_{-\dot{\alpha}_{1}+(-\dot{\alpha}_{2}-\lambda)}$
$\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}_{1}+\sigma}\eta_{\dot{\alpha}_{2}+\lambda}\eta_{-\dot{\alpha}_{1}}\eta_{-\dot{\alpha}_{2}-\lambda}$
$\displaystyle=$ $\displaystyle\eta^{\dot{\alpha}_{1}}_{\sigma}.$

This together with (4.10) completes the proof for type $C_{\ell}$.

Finally, let $X=F_{4}$ or $G_{2}$, and $\sigma\in R^{0}$. In this case both
$S$ and $L$ are lattices. Moreover, both $R^{\prime}=S\cup(\dot{R}_{sh}+S)$
and $R^{\prime\prime}=L\cup(\dot{R}_{lg}+L)$ are simply laced extended affine
root systems of rank $>1$. Therefore, by Proposition 4.2.1,

(4.11) $\eta_{\sigma}^{\alpha}=\eta_{\sigma}^{\beta},\quad(\alpha,\beta\in
R_{sh},\sigma\in S),\hbox{ or }(\alpha,\beta\in R_{\lg},\sigma\in L).$

Now if $\sigma\in S=R^{0}$, $\alpha\in R_{sh},\beta\in R_{lg}$ and
$\alpha+\sigma,\beta+\sigma\in R$, then since $L$ is a lattice, we get
$\sigma\in L$, so by (4.11) and (4.6),

$\eta_{\sigma}^{\beta}=\eta_{\sigma}^{\dot{\theta}_{l}}=\eta_{\sigma}^{\dot{\theta}_{s}}=\eta_{\sigma}^{\alpha}.$

This completes the proof. $\Box$

######  Corollary 4.3.3.

For $\sigma\in S$ and $\dot{\alpha},\dot{\alpha}^{\prime}\in\dot{R}^{\times}$,
$\eta^{\dot{\alpha}}_{\sigma}=\eta^{\dot{\alpha}^{\prime}}_{\sigma}$.

Proof. Combining (4.7) with Lemma 4.3.2, we get the result. $\Box$

######  Proposition 4.3.4.

Any finite-order Cartan automorphism $\Psi$ of type $C_{\ell}$, $\ell\geq 3$,
$F_{4}$ or $G_{2}$, is diagonal, and the corresponding map
$\psi:R\rightarrow{\mathbb{K}}^{\times}$ is a character.

Proof. By Lemma 4.3.2, $\Psi$ is diagonal. From Proposition 4.1.8, we know
that $\psi$ is a core-character. Now any two isotropic roots
$\sigma_{1},\sigma_{2}\in R^{0}=S$ can be considered as roots of the simply
laced extended affine root system $R^{\prime}:=S\cup(\dot{R}_{sh}+S)$. Since
$\hbox{rank}R^{\prime}>1$, we have from Proposition 4.2.1 that
$\psi(\sigma_{1}+\sigma_{2})=\psi(\sigma_{1})\psi(\sigma_{2})$. Thus $\psi$ is
a character. $\Box$

###  4.4. Cartan automorphisms of Type $B_{\ell}$, $\hbox{ind}(R)=0$

Throughout this section $R$ is an extended affine root system of type
$B_{\ell}$, with root system $R$, such that $\hbox{ind}(R)=0$. Let $t$ be the
twist number of $R$. From 2.2.3, we see that $S=S_{1}\oplus\Lambda_{2}$,
$L=2\Lambda_{1}\oplus S_{2}$, where $\Lambda_{1}=\langle S_{1}\rangle$ and
$\Lambda_{2}=\langle S_{2}\rangle.$ Moreover, we have $\hbox{ind}(S_{1})=t$,
and $S_{1}$ can be written in the form
$S_{1}=\uplus_{i=1}^{t}(\tau_{i}+2\Lambda_{1})$, where $\Lambda_{1}=\langle
S_{1}\rangle=\sum_{i=1}^{t}{\mathbb{Z}}\tau_{i}$, and $\tau_{i}$’s are
distinct modulo $2\Lambda_{1}$. For the rest of this section, we fix
$\tau_{i}$’s as above. Recall that $\Psi$ is a finite order Cartan
automorphism.

######  Lemma 4.4.1.

Let $\alpha,\in R_{sh}$ and $\lambda_{1},\ldots,\lambda_{k}\in L$, then

$\eta^{\alpha}_{\lambda_{1}+\cdots+\lambda_{k}}=\eta^{\alpha}_{\lambda_{1}}\cdots\eta^{\alpha}_{\lambda_{k}}=\eta^{\dot{\theta}_{s}}_{\lambda_{1}}\cdots\eta^{\dot{\theta}_{s}}_{\lambda_{k}}=\eta^{{\dot{\theta}}_{s}}_{\lambda_{1}+\cdots+\lambda_{k}}.$

In particular,
$\eta^{\alpha}_{\lambda+\lambda^{\prime}}=\eta^{\alpha}_{\lambda}\eta^{\alpha}_{\lambda^{\prime}}=\eta^{\alpha^{\prime}}_{\lambda}\eta^{\alpha^{\prime}}_{\lambda^{\prime}}=\eta^{\alpha^{\prime}}_{\lambda+\lambda^{\prime}}$,
$\alpha,\alpha^{\prime}\in R_{sh}$, $\lambda,\lambda^{\prime}\in\langle
L\rangle$.

Proof. The proof will make frequent use of the fact that $R_{sh}+\langle
L\rangle\subseteq R_{sh}$, without explicitly stating it every time. We have
$\alpha=\dot{\alpha}+\sigma$, where $\dot{\alpha}\in\dot{R}_{sh}$ and
$\sigma\in S$. Take $\dot{\beta}\in\dot{R}_{sh}$ with
$\dot{\beta}\not=\pm\dot{\alpha}$. Then
$\dot{\alpha}\pm\dot{\beta}\in\dot{R}_{lg}$. Now

$\displaystyle\eta^{\alpha}_{\lambda_{1}+\lambda_{2}}$ $\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}+\sigma+\lambda_{1}+\lambda_{2}}\eta_{-\dot{\alpha}-\sigma}$
$\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}+\dot{\beta}-\dot{\beta}+\sigma+\lambda_{1}+\lambda_{2}}\eta_{-\dot{\alpha}-\sigma}$
$\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}-\dot{\beta}+\lambda_{1}}\eta_{\dot{\beta}+\sigma+\lambda_{2}}\eta_{-\dot{\alpha}-\sigma}$
$\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}-\dot{\beta}+\lambda_{1}}\eta_{\dot{\beta}-\dot{\alpha}+\lambda_{2}}$
$\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}+\lambda_{1}}\eta_{-\dot{\beta}}\eta_{\dot{\beta}}\eta_{-\dot{\alpha}+\lambda_{2}}$
$\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}+\lambda_{1}}\eta_{-\dot{\alpha}}\eta_{\dot{\alpha}}\eta_{-\dot{\alpha}+\lambda_{2}}$
$\displaystyle=$
$\displaystyle\eta^{\dot{\alpha}}_{\lambda_{1}}\eta^{-\dot{\alpha}}_{\lambda_{2}}$
$\displaystyle(\hbox{by Lemma \ref{lemsept1a}(i)})$ $\displaystyle=$
$\displaystyle\eta^{{\dot{\theta}}_{s}}_{\lambda_{1}}\eta^{{\dot{\theta}}_{s}}_{\lambda_{2}}.$

Now using induction on $k$, we have

$\displaystyle\eta^{\alpha}_{\lambda_{1}+\cdots+\lambda_{k}}$ $\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}+\sigma+\lambda_{1}+\cdots+\lambda_{k}}\eta_{-\dot{\alpha}-\sigma}$
$\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}-\dot{\beta}+\lambda_{1}}\eta_{\dot{\beta}+\sigma+\lambda_{2}+\cdots+\lambda_{k}}\eta_{-\dot{\alpha}-\sigma}$
$\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}-\dot{\beta}+\lambda_{1}}\eta_{\dot{\beta}+\sigma+\lambda_{2}+\cdots+\lambda_{k}}\eta_{-\dot{\beta}-\sigma}\eta_{\dot{\beta}+\sigma}\eta_{-\dot{\alpha}-\sigma}$
$\displaystyle=$
$\displaystyle\eta^{\dot{\alpha}-\dot{\beta}}_{\lambda_{1}}\eta_{\dot{\alpha}-\dot{\beta}}\eta^{\dot{\beta}+\sigma}_{\lambda_{2}+\cdots+\lambda_{k}}\eta_{\dot{\beta}+\sigma}\eta_{-\dot{\alpha}-\sigma}$
$\displaystyle(\hbox{induction steps})$ $\displaystyle=$
$\displaystyle\eta^{\dot{\theta}_{s}}_{\lambda_{1}}\eta^{\dot{\theta}_{s}}_{\lambda_{2}}\cdots\eta^{\dot{\theta}_{s}}_{\lambda_{k}}\eta^{\dot{\beta}}_{\sigma}\eta^{-\dot{\alpha}}_{-\sigma}$
$\displaystyle(\hbox{by Corollary \ref{bim5}})$ $\displaystyle=$
$\displaystyle\eta^{\dot{\theta}_{s}}_{\lambda_{1}}\eta^{\dot{\theta}_{s}}_{\lambda_{2}}\cdots\eta^{\dot{\theta}_{s}}_{\lambda_{k}}\eta^{\dot{\alpha}}_{\sigma}\eta^{\dot{\alpha}}_{-\sigma}$
$\displaystyle=$
$\displaystyle\eta^{\dot{\theta}_{s}}_{\lambda_{1}}\eta^{\dot{\theta}_{s}}_{\lambda_{2}}\cdots\eta^{\dot{\theta}_{s}}_{\lambda_{k}}.$

This completes the proof. $\Box$

######  Corollary 4.4.2.

Let $\alpha,\beta\in R_{sh}$, $\lambda\in L$ and $\sigma\in R^{0}$. Then

$\eta^{\alpha}_{\lambda+2\sigma}=\eta^{\beta}_{\lambda+2\sigma}\Longleftrightarrow\eta^{\alpha}_{\lambda}=\eta^{\beta}_{\lambda}.$

Proof. First, we note that $\alpha+\lambda+2\sigma\subseteq
R_{sh}+L+2\Lambda\subseteq R_{sh}$. By Lemma 4.4.1, we have
$\eta^{\alpha}_{\lambda+2\sigma}=\eta^{\alpha}_{\lambda}\eta^{\alpha}_{2\sigma}$,
and
$\eta^{\beta}_{\lambda+2\sigma}=\eta^{\beta}_{\lambda}\eta^{\beta}_{2\sigma}$.
So, we are done by Lemma 4.3.2(i),
$\eta^{\alpha}_{2\sigma}=\eta^{\beta}_{2\sigma}$. $\Box$

######  Lemma 4.4.3.

For $\sigma\in S$ and $\alpha,\beta\in R^{\times}$,
$\eta^{\alpha}_{\sigma}=\eta^{\beta}_{\sigma}$.

Proof. We begin by proving that

(4.12) $\eta^{\alpha}_{\sigma}=\eta^{\beta}_{\sigma},\quad(\sigma\in
S,\alpha,\beta\in R_{sh}).$

First, assume that $\sigma\in S_{1}$. Let $\alpha\in R_{sh}$. Using Corollary
4.4.2, we may assume that
$\alpha=\dot{\alpha}+\epsilon_{i}\tau_{i}+\lambda_{2}$, and $\sigma=\tau_{j}$,
for some $\dot{\alpha}\in\dot{R}_{sh}$, $\lambda_{2}\in\Lambda_{2}$, $1\leq
i,j\leq t$, $\epsilon_{i}\in\\{0,1\\}$. If $\alpha+\sigma\in R$, then
$\epsilon_{i}\tau_{i}+\tau_{j}+\lambda_{2}\in S$. Since $\hbox{ind}(R)=0$,
either $i=0$, or $\epsilon_{i}=1$ and $i=j$. In the latter case, we have

$\displaystyle\eta^{\alpha}_{\sigma}$ $\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}+2\tau_{i}+\lambda_{2}}\eta_{-\dot{\alpha}-\tau_{i}-\lambda_{2}}$
$\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}+2\tau_{j}+\lambda_{2}}\eta_{-\dot{\alpha}-2\tau_{j}}\eta_{\dot{\alpha}+2\tau_{j}}\eta_{-\dot{\alpha}-\tau_{j}}\eta_{\dot{\alpha}+\tau_{j}}\eta_{-\dot{\alpha}-\tau_{j}-\lambda_{2}}$
$\displaystyle=$
$\displaystyle\eta^{\dot{\alpha}+2\tau_{j}}_{\lambda_{2}}\eta_{\dot{\alpha}+2\tau_{j}}\eta^{-\dot{\alpha}-\tau_{j}}_{\lambda_{2}}\eta_{-\dot{\alpha}-\tau_{j}}$
$\displaystyle(\hbox{by Lemma \ref{lemsept1a}(i)})$ $\displaystyle=$
$\displaystyle\eta^{\dot{\alpha}}_{\lambda_{2}}\eta_{\dot{\alpha}+2\tau_{j}}\eta_{-\dot{\alpha}-\tau_{j}}\eta^{-\dot{\alpha}}_{-\lambda_{2}}$
$\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}+2\tau_{j}}\eta_{-\dot{\alpha}-\tau_{j}}$
$\displaystyle(\hbox{by Lemma \ref{Aug23a}})$ $\displaystyle=$
$\displaystyle(\eta_{\dot{\alpha}+\tau_{j}})^{2}\eta_{-\dot{\alpha}}\eta_{-\dot{\alpha}-\tau_{j}}$
$\displaystyle=$ $\displaystyle\eta^{\dot{\alpha}}_{\tau_{j}}.$

If $\epsilon_{i}=0$, then

$\displaystyle\eta^{\alpha}_{\sigma}$ $\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}+\tau_{j}+\lambda_{2}}\eta_{-\dot{\alpha}-\lambda_{2}}$
$\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}+\tau_{j}+\lambda_{2}}\eta_{-\dot{\alpha}-\tau_{j}}\eta_{\dot{\alpha}+\tau_{j}}\eta_{-\dot{\alpha}-\lambda_{2}}$
$\displaystyle=$
$\displaystyle\eta^{\dot{\alpha}+\tau_{j}}_{\lambda_{2}}\eta_{\dot{\alpha}+\tau_{j}}\eta_{-\dot{\alpha}}\eta_{\dot{\alpha}}\eta_{-\dot{\alpha}-\lambda_{2}}$
$\displaystyle(\hbox{by Lemma \ref{lemsept1a}(i)})$ $\displaystyle=$
$\displaystyle\eta^{\dot{\alpha}}_{\lambda_{2}}\eta^{\dot{\alpha}}_{\tau_{j}}\eta^{-\dot{\alpha}}_{-\lambda_{2}}$
$\displaystyle=$ $\displaystyle\eta^{\dot{\alpha}}_{\tau_{j}}.$

Thus in either cases $\eta^{\alpha}_{\sigma}=\eta^{\dot{\alpha}}_{\sigma}$,
and so by Corollary 4.3.3, $\eta^{\alpha}_{\sigma}=\eta^{\beta}_{\sigma}$, for
any $\alpha,\beta\in R_{sh}$, $\sigma\in S_{1}$.

Generally, assume that $\sigma$ is an arbitrary element of $S$. Then
$\sigma=\sigma_{1}+\lambda_{2}$, where $\sigma_{1}\in S_{1}$ and
$\lambda_{2}\in\Lambda_{2}$. Then

$\displaystyle\eta^{\alpha}_{\sigma}$ $\displaystyle=$
$\displaystyle\eta_{\alpha+\sigma_{1}+\lambda_{2}}\eta_{-\alpha}$
$\displaystyle=$
$\displaystyle\eta_{\alpha+\sigma_{1}+\lambda_{2}}\eta_{-\alpha-\sigma_{1}}\eta_{\alpha+\sigma_{1}}\eta_{-\alpha}$
$\displaystyle=$
$\displaystyle\eta^{\alpha+\sigma_{1}}_{\lambda_{2}}\eta^{\alpha}_{\sigma_{1}}$
$\displaystyle=$
$\displaystyle\eta^{\alpha}_{\lambda_{2}}\eta^{\alpha}_{\sigma_{1}}.$

Thus by Lemma 4.3.2(i),

(4.13)
$\eta^{\alpha}_{\sigma_{1}+\lambda_{2}}=\eta^{\alpha}_{\sigma_{1}}\eta^{\alpha}_{\lambda_{2}}=\eta^{\beta}_{\sigma_{1}}\eta^{\beta}_{\lambda_{2}}=\eta^{\beta}_{\sigma_{1}+\lambda_{2}},\quad(\alpha,\beta\in
R_{sh},\sigma_{1}\in S_{1},\lambda_{2}\in\Lambda_{2})$

This completes the proof of (4.12).

Next, let $\beta\in R_{lg}$ and $\sigma\in S$ with $\beta+\sigma\in R$. By
Corollary 4.4.2, we may assume that $\beta=\dot{\beta}+\sigma_{2}$, and
$\sigma=\sigma_{1}+\lambda_{2}$, where $\dot{\beta}\in\dot{R}_{lg}$,
$\sigma_{1}\in S_{1}$, $\sigma_{2}\in S_{2}$ and $\lambda_{2}\in\Lambda_{2}$.
Since $\alpha+\sigma\in R$, we get $\sigma_{1}+\sigma_{2}+\lambda_{2}\in
L\subseteq 2\Lambda_{1}\oplus\Lambda_{2}$, and so $\sigma_{1}\in
2\Lambda_{1}$. Also $\sigma_{2}\in L+L+2\Lambda_{2}\subseteq\langle L\rangle$.
Take $\dot{\alpha},\dot{\gamma}\in\dot{R}_{sh}$ such that
$\dot{\beta}=\dot{\alpha}+\dot{\gamma}$. Then

$\displaystyle\eta^{\beta}_{\sigma}$ $\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}+\dot{\gamma}+\sigma_{1}+\sigma_{2}+\lambda_{2}}\eta_{-\dot{\alpha}-\dot{\gamma}-\sigma_{2}}$
$\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}+\sigma_{1}+\lambda_{2}}\eta_{\dot{\gamma}+\sigma_{2}}\eta_{-\dot{\alpha}}\eta_{-\dot{\gamma}-\sigma_{2}}$
$\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}+\sigma_{1}+\lambda_{2}}\eta_{-\dot{\alpha}-\sigma_{1}}\eta_{\dot{\alpha}+\sigma_{1}}\eta_{-\dot{\alpha}}$
$\displaystyle=$
$\displaystyle\eta^{\dot{\alpha}+\sigma_{1}}_{\lambda_{2}}\eta^{\dot{\alpha}}_{\sigma_{1}}$
$\displaystyle(\hbox{by (\ref{red4})})$ $\displaystyle=$
$\displaystyle\eta^{\dot{\alpha}}_{\lambda_{2}}\eta^{\dot{\alpha}}_{\sigma_{1}}=\eta^{\dot{\alpha}}_{\sigma}.$

Since $\dot{\alpha}$ is a short root, the proof is completed by (4.12). $\Box$

######  Definition 4.4.4.

For $\sigma\in S$, we define $\eta_{\sigma}=\eta^{\alpha}_{\sigma}$, for some
$\alpha\in R^{\times}.$ By Lemma 4.4.3, this is well-defined.

######  Lemma 4.4.5.

(i) $\eta_{\lambda+2\sigma}=\eta_{\lambda}\eta_{2\sigma}$, $\lambda\in S$,
$\sigma\in R^{0}$.

(ii) $\eta_{\lambda_{1}+\lambda_{2}}=\eta_{\lambda_{1}}\eta_{\lambda_{2}}$,
$\lambda_{1}\in S_{1}$, $\lambda_{2}\in\langle S_{2}\rangle$.

Proof. Take $\dot{\alpha}_{1},\dot{\alpha}_{2}\in\dot{R}_{sh}$ with
$\dot{\alpha}_{1}\not=\pm\dot{\alpha}_{2}$. Then
$\dot{\alpha}_{1}\pm\dot{\alpha}_{2}\in\dot{R}_{lg}$.

(i) For $\sigma\in R^{0}$, we have $2\sigma\in 2R^{0}\subseteq 2\langle
S\rangle\subseteq L\subseteq S$. Now

$\displaystyle\eta_{\lambda+2\sigma}$ $\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}_{1}+\lambda+2\sigma}\eta_{-\dot{\alpha}_{1}}$
$\displaystyle=$
$\displaystyle\eta_{(\dot{\alpha}_{2}+\lambda)+(\dot{\alpha}_{1}-\dot{\alpha}_{2}+2\sigma)}\eta_{-\dot{\alpha}_{1}}$
$\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}_{2}+\lambda}\eta_{\dot{\alpha}_{1}-\dot{\alpha}_{2}+2\sigma}\eta_{-\dot{\alpha}_{1}}$
$\displaystyle=$
$\displaystyle(\eta_{\dot{\alpha}_{2}+\lambda}\eta_{-\dot{\alpha}_{2}})\eta_{\dot{\alpha}_{2}}\eta_{\dot{\alpha}_{1}-\dot{\alpha}_{2}+2\sigma}\eta_{-\dot{\alpha}_{1}}$
$\displaystyle=$
$\displaystyle\eta_{\lambda}\eta_{\dot{\alpha}_{1}-\dot{\alpha}_{2}+2\sigma}\eta_{\dot{\alpha}_{2}-\dot{\alpha}_{1}}$
$\displaystyle=$ $\displaystyle\eta_{\lambda}\eta_{2\sigma}.$

(ii) We have

$\displaystyle\eta_{\lambda_{1}+\lambda_{2}}$ $\displaystyle=$
$\displaystyle\eta^{\dot{\alpha}_{1}}_{\lambda_{1}+\lambda_{2}}$
$\displaystyle(\hbox{by Lemma \ref{b100}})$ $\displaystyle=$
$\displaystyle\eta^{\dot{\alpha}_{1}}_{\lambda_{1}}\eta^{\dot{\alpha}_{1}}_{\lambda_{2}}$
$\displaystyle=$ $\displaystyle\eta_{\lambda_{1}}\eta_{\lambda_{2}}.$

$\Box$

######  Corollary 4.4.6.

We have

(i) $\eta_{\lambda}^{2}=\eta_{\lambda+\sigma}\eta_{\lambda-\sigma}$,
$\sigma\in R^{0}$,$\lambda,\sigma+\lambda\in S$,

(ii) $\eta_{n\lambda}=\eta^{n}_{\lambda}$, $\lambda\in S$, $n\in{\mathbb{Z}}$,

(iii) $\eta_{\lambda+\sigma}=\pm\eta_{\lambda}\eta_{\sigma}$,
$\lambda,\sigma,\lambda+\sigma\in S.$

Proof. (i) Let $\sigma\in R^{0}$, $\lambda,\lambda+\sigma\in S$. Let
$\dot{\alpha}\in\dot{R}^{\times}$ and set $\alpha=\dot{\alpha}+\lambda$. Then
$\alpha,\alpha+\sigma\in R^{\times}$, and so by Lemma 4.1.11(ii),
$\eta_{\dot{\alpha}+\lambda}^{2}=\eta_{\dot{\alpha}+\lambda+\sigma}\eta_{\dot{\alpha}+\lambda-\sigma}.$
By inserting $\eta_{\dot{\alpha}}\eta_{-\dot{\alpha}}$ to both sides, we
obtain
$\eta_{\lambda}\eta_{\lambda}=\eta_{\lambda+\sigma}\eta_{\lambda-\sigma}$, as
required.

(ii) Apply part (i) with $\lambda=\sigma\in S$, to get
$\eta_{2\lambda}=\eta^{2\lambda}$. Then by Lemma 4.4.5(i),
$\eta_{3\lambda}=\eta_{\lambda}\eta_{2\lambda}=\eta^{3}_{\lambda}$. Since
$\eta^{-1}_{\lambda}=\eta_{-\lambda},$ the result follows inductively.

(iii) Let $\lambda,\sigma,\lambda+\sigma\in S$. Replacing $\lambda$ with
$\lambda+\sigma$ in (i), and applying (ii), we get
$\eta^{2}_{\lambda+\sigma}=\eta_{\lambda+2\sigma}\eta_{\lambda}=(\eta_{\lambda}\eta_{\sigma})^{2}.$
$\Box$

######  Proposition 4.4.7.

Any finite order Cartan automorphism $\Psi$ for $E$ is diagonal.

Proof. To prove that $\Psi$ is diagonal, we fix
$\sigma=\sigma^{\prime}+\sigma^{\prime\prime}\in R^{0}$,
$\sigma^{\prime},\sigma^{\prime\prime}\in S$. By Lemma 4.1.6, we must show
that $\eta_{\sigma}^{\alpha}=\eta_{\sigma}^{\beta}$ for $\alpha,\beta\in
R^{\times}$. Now $\sigma^{\prime}=\tau^{\prime}+\lambda^{\prime}$ and
$\sigma^{\prime\prime}=\tau^{\prime\prime}+\lambda^{\prime\prime}$ for some
$\tau^{\prime},\tau^{\prime\prime}\in S_{1}$,
$\lambda^{\prime},\lambda^{\prime\prime}\in\langle S_{2}\rangle$. Let
$\alpha\in R^{\times}$ with $\alpha+\sigma\in R$. Then
$\alpha=\dot{\alpha}+\sigma_{s}+\lambda$, for some
$\dot{\alpha}\in\dot{R}^{\times}$, $\sigma_{s}\in S_{1}$, $\lambda\in\langle
S_{2}\rangle$. Therefore,
$\alpha+\sigma=\dot{\alpha}+\sigma_{s}+\tau^{\prime}+\tau^{\prime\prime}+\lambda+\lambda^{\prime}+\lambda^{\prime\prime}\in
R$. This implies that $\sigma_{s}+\tau^{\prime}+\tau^{\prime\prime}\in S_{1}$
and $\lambda+\lambda^{\prime}+\lambda^{\prime\prime}\in\langle S_{2}\rangle.$
Thus

$\displaystyle\eta^{\alpha}_{\sigma}$ $\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}+\sigma_{s}+\tau^{\prime}+\tau^{\prime\prime}+\lambda+\lambda^{\prime}+\lambda^{\prime\prime}}\eta_{-\dot{\alpha}-\sigma_{s}-\lambda}$
$\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}+\sigma_{s}+\tau^{\prime}+\tau^{\prime\prime}+\lambda+\lambda^{\prime}+\lambda^{\prime\prime}}\eta_{-\dot{\alpha}-\sigma_{s}-\tau^{\prime}-\tau^{\prime\prime}}\eta_{\dot{\alpha}+\sigma_{s}+\tau^{\prime}+\tau^{\prime\prime}}\eta_{-\dot{\alpha}-\sigma_{s}-\lambda}$
$\displaystyle=$
$\displaystyle\eta^{\dot{\alpha}+\sigma_{s}+\tau^{\prime}+\tau^{\prime\prime}}_{\lambda+\lambda^{\prime}+\lambda^{\prime\prime}}\eta_{\dot{\alpha}+\sigma_{s}+\tau^{\prime}+\tau^{\prime\prime}}\eta_{-\dot{\alpha}-\sigma_{s}-\lambda}\eta_{\dot{\alpha}+\sigma_{s}}\eta_{-\dot{\alpha}-\sigma_{s}}$
$\displaystyle=$
$\displaystyle\eta^{\dot{\alpha}+\sigma_{s}+\tau^{\prime}+\tau^{\prime\prime}}_{\lambda+\lambda^{\prime}+\lambda^{\prime\prime}}\eta_{\dot{\alpha}+\sigma_{s}+\tau^{\prime}+\tau^{\prime\prime}}\eta^{-\dot{\alpha}-\sigma_{s}}_{-\lambda}\eta^{\dot{\alpha}}_{\sigma_{s}}\eta_{-\dot{\alpha}}$
$\displaystyle(\hbox{by Lemma \ref{red2}})$ $\displaystyle=$
$\displaystyle\eta^{\dot{\theta}_{s}}_{\lambda+\lambda^{\prime}+\lambda^{\prime\prime}}\eta^{\dot{\theta}_{s}}_{\sigma_{s}+\tau^{\prime}+\tau^{\prime\prime}}\eta^{-\dot{\theta}_{s}}_{-\lambda}\eta^{\dot{\theta}_{s}}_{\sigma_{s}}$
$\displaystyle(\hbox{by Lemma \ref{b100}})$ $\displaystyle=$
$\displaystyle\eta_{\lambda^{\prime}}\eta_{\lambda^{\prime\prime}}\eta_{\sigma_{s}+\tau^{\prime}+\tau^{\prime\prime}}\eta_{\sigma_{s}}.$

Thus, we are done if we show that
$\eta_{\sigma_{s}+\tau^{\prime}+\tau^{\prime\prime}}=\eta_{\sigma_{s}}\eta_{\tau^{\prime}}\eta_{\tau^{\prime\prime}}$.

Since $\hbox{ind}(R)=0$, we have $\hbox{ind}(S_{1})=t$. Then as we have
already mentioned, $S_{1}$ can be written as
$S_{1}=\uplus_{i=1}^{t}(\tau_{i}+2\Lambda_{1})$, where
$\Lambda_{1}=\sum_{i=1}^{t}{\mathbb{Z}}\tau_{i}$, and $\tau_{i}$’s are
distinct modulo $2\Lambda_{1}$. Now, there exist $1\leq i,j,k\leq t$ such that
$\sigma_{s}=\epsilon_{i}\tau_{i}+2\lambda,$
$\tau^{\prime}=\epsilon_{j}\tau_{j}+2\lambda^{\prime}$,
$\tau^{\prime\prime}=\epsilon_{k}\tau_{k}+2\lambda^{\prime\prime}$, for some
$\epsilon_{i},\epsilon_{j},\epsilon_{k}\in\\{0,1\\}$,
$\lambda,\lambda^{\prime},\lambda^{\prime\prime}\in\Lambda_{1}$. Depending on
whether $\epsilon_{t}$’s are $0$ or $1$, we must consider different
possibilities. We examine one of these possibilities, as the other cases can
be treated similarly. Let for example
$\epsilon_{i}=\epsilon_{j}=\epsilon_{k}=1$. Then since
$\sigma_{s}+\tau^{\prime}+\tau^{\prime\prime}\equiv\tau_{i}+\tau_{j}+\tau_{k}\in
S_{1}$, modulo $2\Lambda_{1}$, and $\hbox{ind}(S_{1})=t$, we conclude that
$|\\{i,j,k\\}|\leq 2$. Assume for example that
$\tau_{i}=\tau_{j}\not=\tau_{k}$. Then, using Lemma 4.4.5(i) and Corollary
4.4.6(iii), we have

$\displaystyle\eta_{\sigma_{s}+\tau^{\prime}+\tau^{\prime\prime}}$
$\displaystyle=$
$\displaystyle\eta_{2\tau_{i}}\eta_{\tau_{k}}\eta_{2\lambda}\eta_{2\lambda^{\prime}}\eta_{2\lambda^{\prime\prime}}$
$\displaystyle=$
$\displaystyle\eta_{\tau_{i}}^{2}\eta_{\tau_{k}}\eta_{2\lambda}\eta_{2\lambda^{\prime}}\eta_{2\lambda^{\prime\prime}}$
$\displaystyle(\hbox{since $\tau_{i}=\tau_{j}$})$ $\displaystyle=$
$\displaystyle\eta_{\tau_{i}+2\lambda}\eta_{\tau_{j}+2\lambda^{\prime}}\eta_{\tau_{k}+2\lambda^{\prime\prime}}$
$\displaystyle=$
$\displaystyle\eta_{\sigma_{s}}\eta_{\tau^{\prime}}\eta_{\tau^{\prime\prime}},$

as required. This completes the proof. $\Box$

######  Definition 4.4.8.

Let $R$ be of type $B_{\ell}$ with $\hbox{ind}(R)=0$. For $\sigma\in R^{0}$,
we define $\eta_{\sigma}:=\eta^{\alpha}_{\sigma}$, for some $\alpha\in
R^{\times}$ with $\alpha+\sigma\in R^{\times}$. By Proposition 4.4.7,
definition of $\eta_{\sigma}$ is independent of the choice of $\alpha$.

######  Lemma 4.4.9.

The following holds:

(i) $\eta_{\sigma+\sigma^{\prime}}=\eta_{\sigma}\eta_{\sigma^{\prime}}$,
$\sigma,\sigma^{\prime}\in S$.

(ii) For $\sigma,\sigma^{\prime}\in S$ and $\sigma^{\prime\prime}\in R^{0}$,
$\eta_{\sigma+\sigma^{\prime}+2\sigma^{\prime\prime}}=\eta_{\sigma}\eta_{\sigma^{\prime}}\eta_{2\sigma^{\prime\prime}}.$

(iii) For $\sigma_{1},\sigma^{\prime}_{1}\in S_{1}$ and $\lambda_{2}\in\langle
S_{2}\rangle,$ we have
$\eta_{\sigma_{1}+\sigma^{\prime}_{1}+\lambda_{2}}=\eta_{\sigma_{1}}\eta_{\sigma^{\prime}_{1}}\eta_{\lambda_{2}}$.

Proof. (i) We have $S=S_{1}\oplus\langle S_{2}\rangle$ with
$\hbox{ind}(S_{1})=t$. Let
$\sigma=\epsilon_{i}\tau_{i}+2\sigma_{i}+\lambda,\sigma^{\prime}=\epsilon_{j}\tau_{j}+2\sigma_{j}+\lambda^{\prime}\in
S$, where $1\leq i,j\leq t$, $\sigma_{i},\sigma_{j}\in\langle S_{1}\rangle$,
$\lambda,\lambda^{\prime}\in\langle S_{2}\rangle$ and
$\epsilon_{i},\epsilon_{j}\in\\{0,1\\}$. By definition,
$\eta_{\sigma+\sigma^{\prime}}=\eta_{\sigma+\sigma^{\prime}}^{\alpha}$, for
some $\alpha\in R^{\times}$. Let $\dot{\alpha}\in\dot{R}_{sh}$ and set
$\alpha:=\dot{\alpha}-\epsilon_{i}\tau_{i}$. Then
$\alpha+\sigma+\sigma^{\prime}=\dot{\alpha}+2\sigma_{i}+\lambda+\epsilon_{j}\tau_{j}+2\sigma_{j}+\lambda^{\prime}\in
R$. Thus

$\displaystyle\eta_{\sigma+\sigma^{\prime}}$ $\displaystyle=$
$\displaystyle\eta^{\dot{\alpha}-\epsilon_{i}\tau_{i}}_{\epsilon_{i}\tau_{i}+\epsilon_{j}\tau_{j}+2\sigma_{i}+2\sigma_{j}+\lambda+\lambda^{\prime}}$
$\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}+\epsilon_{j}\tau_{j}+2\sigma_{i}+2\sigma_{j}+\lambda+\lambda^{\prime}}\eta_{-\dot{\alpha}+\epsilon_{i}\tau_{i}}$
$\displaystyle=$
$\displaystyle\eta_{\dot{\alpha}+\epsilon_{j}\tau_{j}+2\sigma_{i}+2\sigma_{j}+\lambda+\lambda^{\prime}}\eta_{\dot{\alpha}}\eta_{-\dot{\alpha}}\eta_{-\dot{\alpha}+\epsilon_{i}\tau_{i}}$
$\displaystyle=$
$\displaystyle\eta^{\dot{\alpha}}_{\epsilon_{j}\tau_{j}+2\sigma_{i}+2\sigma_{j}+\lambda+\lambda^{\prime}}\eta^{-\dot{\alpha}}_{\epsilon_{i}\tau_{i}}$
$\displaystyle=$
$\displaystyle\eta_{\epsilon_{j}\tau_{j}+2\sigma_{i}+2\sigma_{j}+\lambda+\lambda^{\prime}}\eta_{\epsilon_{i}\tau_{i}}$
$\displaystyle(\hbox{by Lemma \ref{bim15}})$ $\displaystyle=$
$\displaystyle\eta_{\epsilon_{j}\tau_{j}}\eta_{2\sigma_{i}}\eta_{2\sigma_{j}}\eta_{\lambda}\eta_{\lambda^{\prime}}\eta_{\epsilon_{i}\tau_{i}}.$

On the other hand

$\eta_{\sigma}\eta_{\sigma^{\prime}}=\eta_{\epsilon_{i}\tau_{i}+2\tau_{i}+\lambda}\eta_{\epsilon_{j}\tau_{j}+2\sigma_{j}+\lambda^{\prime}}=\eta_{\epsilon_{i}\tau_{i}}\eta_{2\sigma_{i}}\eta_{\lambda}\eta_{\epsilon_{j}\tau_{j}}\eta_{2\sigma_{j}}\eta_{\lambda^{\prime}},$

so $\eta_{\sigma+\sigma^{\prime}}=\eta_{\sigma}\eta_{\sigma^{\prime}}$.

(ii) Let $\sigma,\sigma^{\prime},\sigma^{\prime\prime}$ be as in the
statement. Then, we have
$2\sigma^{\prime\prime},\sigma^{\prime}+2\sigma^{\prime\prime}\in S$, and the
results easily follows from part (i).

(iii) Let $\sigma_{1},\sigma^{\prime}_{1},\lambda_{2}$ be as in the statement.
Then as $\sigma^{\prime}_{1}+\lambda_{2}\in S$, we have from part (i) and
Lemma 4.4.5(ii) that

$\eta_{\sigma_{1}+\sigma^{\prime}_{1}+\lambda_{2}}=\eta_{\sigma_{1}}\eta_{\sigma^{\prime}_{1}+\lambda_{2}}=\eta_{\sigma_{1}}\eta_{\sigma^{\prime}_{2}}\eta_{\lambda_{2}}.$

$\Box$

######  Proposition 4.4.10.

Let $R$ be of type $B_{\ell}$ with $\hbox{ind}(R)=0$. Assume that $\Psi$ is a
finite order Cartan automorphism for $E$. Then the corresponding map
$\psi:R\rightarrow{\mathbb{K}}^{\times}$ is a character.

Proof. By Proposition 4.4.7, $\Psi$ is diagonal. So by Lemma 4.1.8, $\psi$ is
a core-character. Therefore, we only need to show that
$\eta_{\sigma+\sigma^{\prime}}=\eta_{\sigma}\eta_{\sigma^{\prime}}$ for
$\sigma,\sigma^{\prime},\sigma+\sigma^{\prime}\in R^{0}$. If
$\sigma,\sigma^{\prime}\in S$, we are done by Lemma 4.4.9(i). Otherwise, using
Lemma 4.4.9(ii)-(iii), we may assume that $\sigma=\tau_{i}+\tau_{j}$ with
$1\leq i\not=j\leq t$, and
$\sigma^{\prime}=\epsilon_{k}\tau_{k}+\epsilon_{r}\tau_{r}$, with $1\leq
k\not=r\leq t$, $\epsilon_{k},\epsilon_{r}\in\\{0,1\\}$. If
$\epsilon_{k}=\epsilon_{r}=0$, then $\sigma^{\prime}=0$ and we are done. If
$\epsilon_{k}=1$ and $\epsilon_{r}=0$, we have
$\sigma+\sigma^{\prime}=\tau_{i}+\tau_{j}+\tau_{k}\in R^{0}$. Since
$\hbox{ind}(S_{1})=t$, we get $k=i$ or $k=j$, say $k=j$. Then

$\displaystyle\eta_{\sigma+\sigma^{\prime}}$ $\displaystyle=$
$\displaystyle\eta_{\tau_{i}+\tau_{j}+\tau_{k}}=\eta_{\tau_{i}+2\tau_{j}}$
$\displaystyle(\hbox{Lemma \ref{bim19}})$ $\displaystyle=$
$\displaystyle\eta_{\tau_{i}}\eta_{2\tau_{j}}=\eta_{\tau_{i}}\eta_{\tau_{j}}^{2}=\eta_{\tau_{i}+\tau_{j}}\eta_{\tau_{j}}=\eta_{\sigma}\eta_{\sigma^{\prime}}.$

Finally, if $\epsilon_{k}=\epsilon_{r}=1$, then as $i\not=j$, $k\not=r$ and
$\sigma+\sigma^{\prime}\in S$, we conclude that $\\{i,j\\}=\\{k,r\\}$, say
$i=k,j=r$. Then

$\eta_{\sigma+\sigma^{\prime}}=\eta_{2\tau_{i}+2\tau_{j}}=\eta_{\tau_{i}}^{2}\eta_{\tau_{j}}^{2}=\eta^{2}_{\tau_{i}+\tau_{j}}=\eta_{\sigma}\eta_{\sigma^{\prime}}.$

$\Box$

To conclude the section, we introduce a few more notations here. Let
${\mathcal{C}}$ be the group of Cartan automorphisms for $E$, $\chi$ be the
group of characters for $R$, and set
$\mathcal{I}:=\\{\psi\in\hbox{Aut}(E)\mid\psi_{|_{\mathcal{H}}}=-\hbox{id}_{\mathcal{H}}\\}$.
Also, we let ${\mathcal{C}}^{\hbox{fin}}$, $\chi^{\hbox{fin}}$, and
$\mathcal{I}^{\hbox{fin}}$ be the finite order elements of ${\mathcal{C}}$,
$\chi$, and $\mathcal{I}$, respectively. It is worth noting that
$\mathcal{I}^{\hbox{fin}}$ corresponds to the set of Chevalley involutions for
$E$. The following theorem summarizes the connections among Chevalley
involutions, Cartan automorphisms, and characters.

######  Theorem 4.4.11.

Let $E$ be an extended affine Lie algebra with root system $R$ such that
$\hbox{rank}\;R>1$ and $\hbox{ind}(R)=0$. Then

(i) ${\mathcal{C}}^{\hbox{fin}}\cong\chi^{\hbox{fin}}$.

(ii) For any $\tau\in{\mathcal{I}}^{\hbox{fin}}$,
$\tau{\mathcal{C}}^{\hbox{fin}}\subseteq{\mathcal{I}}^{\hbox{fin}}$.

Proof. Part (i) follows from Propositions 4.2.1, 4.3.4 and 4.4.10. Next let
$\tau\in{\mathcal{I}}^{\hbox{fin}}$ and let
$\Psi\in{\mathcal{C}}^{\hbox{fin}}$. Let $\alpha\in R$ and $x_{\alpha}\in
E_{\alpha}$. By part (i), $\Psi(x_{\alpha})=\eta_{\alpha}x_{\alpha}$ for some
$\eta_{\alpha}\in{\mathbb{K}}^{\times}$. Then by Lemma 3.1.2(i),
$(\tau\Psi)^{2}(x_{\alpha})=\tau\Psi(\eta_{\alpha}\tau(x_{\alpha}))=\eta_{\alpha}\eta_{-\alpha}\tau^{2}(x_{\alpha})=x_{\alpha}$.
Therefore, $\tau\Psi$ has order $2$ and acts as $-\hbox{id}$ on
${\mathcal{H}}$, so $\tau\Psi\in{\mathcal{I}}^{\hbox{fin}}$. $\Box$

######  Remark 4.4.12.

(i) Let $E$ be an extended affine root system of rank $>1$. Assume that $E$ is
equipped with a Chevalley involution $\tau$. In [AFI22], the authors show that
the core $E_{c}$ of $E$ admits a Chevalley system with respect to $\tau$,
namely a set $\\{x_{\alpha}\in E_{\alpha}\mid\alpha\in R^{\times}\\}$ such
that for $\alpha\in R^{\times}$, $[x_{\alpha},x_{-\alpha}]=h_{\alpha}$ and
$\tau(x_{\alpha})=x_{-\alpha}$. Let $\\{\bar{x}_{\alpha}\in
E_{\alpha}\mid\alpha\in R^{\times}\\}$ be another Chevalley system with
respect to a Chevalley involution, say $\bar{\tau}$. Define
$\lambda_{\alpha}\in{\mathbb{K}}^{\times}$ by
$\bar{x}_{\alpha}=\lambda_{\alpha}x_{\alpha}$. Suppose $\hbox{ind}(R)=0$ and
the Cartan automorphism $\Psi:=\bar{\tau}\tau$ is of finite order. By Theorem
4.4.11, $\Psi$ is diagonal and the associated map $\psi$ is a character. Now
let $\delta\in R^{0}$ and $\alpha,\alpha+\delta\in R^{\times}$. Then

$\displaystyle[x_{-\alpha},x_{-\delta+\alpha}]$ $\displaystyle=$
$\displaystyle\tau[x_{\alpha},x_{\delta-\alpha}]$ $\displaystyle=$
$\displaystyle\bar{\tau}\Psi[x_{\alpha},x_{\delta-\alpha}]$ $\displaystyle=$
$\displaystyle\psi(\delta)\lambda_{-\alpha}\lambda_{-\delta+\alpha}[\bar{x}_{-\alpha},\bar{x}_{-\delta+\alpha}]$
$\displaystyle=$
$\displaystyle\psi(\delta)(\lambda_{-\alpha}\lambda_{-\delta+\alpha})^{2}[x_{-\alpha},x_{-\delta+\alpha}].$

Therefore, $(\lambda_{\alpha}\lambda_{\delta-\alpha})^{2}=\psi(\delta)$. This
shows that for $\delta\in R^{0}$, $\alpha,\beta,\alpha+\delta,\beta+\delta\in
R^{\times}$, we have
$\lambda_{\alpha}\lambda_{\delta-\alpha}=\pm\lambda_{\beta}\lambda_{\delta-\beta},$
see [AFI22, Lemma 6.1(ii)].

(ii) The content of Section 4 shows that the techniques used in the proofs are
heavily reliant on the assumption that the rank of $E$ is greater than $1$. As
seen in [AFI22], the same situation happens when one deals with constructing
Chevalley bases for extended affine Lie algebras. This suggests that a
different approach should be considered to investigate the rank $1$ situation,
namely the type $A_{1}$. The same holds for type $B_{\ell}$, as our proof
relies heavily on the index of the involved semilattices in the structure of
the root system. We note that in this case the short roots can be described as
(non-isotropic roots of) a disjoint union of $A_{1}$-type extended affine root
systems.

##  5\. Characters vs. root-lattice characters

In this section, we discuss the possibility of extending a character for an
extended affine root system to a root-lattice character.

###  5.1. Extendability in index zero

We begin with the following lemma for extended affine roots systems which
resembles and extends a familiar result for finite root systems, see [Hum72,
Corollary 10.2].

######  Lemma 5.1.1.

Let $R$ be an extended affine root system, and
$\alpha_{1},\ldots,\alpha_{n+1}\in R^{\times}$. Then $w_{\alpha_{1}}\cdots
w_{\alpha_{n}}(\alpha_{n+1})=\epsilon_{1}\alpha_{i_{1}}+\cdots+\epsilon_{m}\alpha_{i_{m}}$
for some $m\in{\mathbb{Z}}_{>0}$,
$\alpha_{i_{j}}\in\\{\alpha_{1},\ldots,\alpha_{n+1}\\}$ and
$\epsilon_{i}\in\\{\pm 1\\}$, such that
$\epsilon_{1}\alpha_{i_{1}}+\cdots\epsilon_{k}\alpha_{i_{k}}\in R$ for $1\leq
k\leq m$.

Proof. Assume $\beta=w_{\alpha_{1}}\cdots w_{\alpha_{n}}(\alpha_{n+1})$. We
use induction on $n$. Assume first that $n=1$, namely
$\beta=w_{\alpha_{1}}(\alpha_{2})$. If $(\alpha_{1},\alpha_{2})=0$, or
$\alpha_{1}$ and $\alpha_{2}$ are linearly dependent, we are clearly done.
Otherwise, there exists $\epsilon_{1},\epsilon_{2}\in\\{\pm 1\\}$ such that
$\\{\epsilon_{1}\alpha_{1},\epsilon_{2}\alpha_{2}\\}$ is a base for the rank
$2$ finite root system
$R\cap({\mathbb{Z}}\alpha_{1}\oplus{\mathbb{Z}}\alpha_{2})$. Then the claim is
known from finite dimensional theory, see [Hum72]. Assume next that $n>1$. Set
$\gamma:=w_{\alpha_{2}}\cdots w_{\alpha_{n}}(\alpha_{n+1})$. By induction
steps, $\gamma=\epsilon_{2}\alpha_{i_{2}}+\cdots+\epsilon_{p}\alpha_{i_{p}}$,
where $p\in{\mathbb{Z}}_{>0}$,
$\alpha_{i_{j}}\in\\{\alpha_{2},\ldots,\alpha_{n+1}\\}$,
$\epsilon_{i}\in\\{\pm 1\\}$ and
$\epsilon_{2}\alpha_{i_{2}}+\cdots+\epsilon_{k}\alpha_{i_{k}}\in R$ for $2\leq
k\leq p.$

Now, $\beta=w_{\alpha_{1}}(\gamma)$. If $(\gamma,\alpha_{1})=0$, or $\gamma$
and $\alpha_{1}$ are linearly dependent we are done. Otherwise, we consider
the rank $2$ finite root system
$R\cap({\mathbb{Z}}\alpha_{1}\oplus{\mathbb{Z}}\gamma)$ and we may assume
that, up to a plus-minus sign of $\gamma$, $\\{\alpha_{1},\gamma\\}$ is a base
for this root system. Then using the realization of rank $2$ finite root
systems, we have

$\beta=w_{\alpha_{1}}(\gamma)\in\\{\gamma\pm\alpha_{1},\gamma\pm
2\alpha_{1},\gamma\pm 3\alpha_{1}\\}.$

Correspondingly, $\beta=\gamma\pm\alpha_{1}$,
$\beta=\gamma\pm\alpha_{1}\pm\alpha_{1}$ or
$\beta=\gamma\pm\alpha_{1}\pm\alpha_{1}\pm\alpha_{1}$, where $\gamma\pm
j\alpha_{1}\in R^{\times}$ if $\gamma\pm(j+1)\alpha_{1}\in R^{\times}$. Thus
we are done by induction steps. $\Box$

We recall from 2.2.3 that if $R$ is simply laced of rank $>1$ or if $R$ is of
type $F_{4}$ or $G_{2}$, then $\hbox{ind}(R)=0$.

######  Lemma 5.1.2.

Let $R$ be an extended affine root system and
$\Phi:R\rightarrow{\mathbb{K}}^{\times}$ be a finite-order character for $R$.
If $ind(R)=0$, then $\Phi$ can be extended to a finite-order root-lattice
character.

Proof. By [Az99, Proposition 4.41], $R^{\times}$ contains a basis
$\mathcal{P}$ for the root lattice $\langle R\rangle$ such that
${\mathcal{W}}_{\mathcal{P}}\mathcal{P}=R^{\times}$. Consider the group
homomorphism $\widehat{\Phi}:\langle R\rangle\rightarrow{\mathbb{K}}^{\times}$
for $\langle R\rangle$ given by $\widehat{\Phi}(\alpha)=\Phi(\alpha)$ for
$\alpha\in\mathcal{P}$, see Example 4.1.3. Using Lemma 5.1.1, for $\alpha\in
R^{\times}$ there exists $\alpha_{1},\ldots,\alpha_{n}\in\mathcal{P}$ and
$\epsilon_{1},\ldots,\epsilon_{n}\in\\{\pm 1\\}$ such that
$\alpha=\epsilon_{1}\alpha_{1}+\cdots+\epsilon_{n}\alpha_{n}$, and
$\epsilon_{1}\alpha_{1}+\cdots+\epsilon_{k}{\alpha_{k}}\in R^{\times}$ for
each $1\leq k\leq n$. Then

$\displaystyle\widehat{\Phi}(\alpha)=\widehat{\Phi}(\epsilon_{1}\alpha_{1}+\cdots+\epsilon_{n}\alpha_{n})$
$\displaystyle=$
$\displaystyle\prod_{i=1}^{n}\widehat{\Phi}(\epsilon_{i}\alpha_{i})$
$\displaystyle=$ $\displaystyle\prod_{i=1}^{n}\Phi(\epsilon_{i}\alpha_{i})$
$\displaystyle(\hbox{since $\epsilon_{1}\alpha_{1}+\epsilon_{2}\alpha_{2}\in
R^{\times}$})$ $\displaystyle=$
$\displaystyle\Phi(\epsilon_{1}\alpha_{1}+\epsilon_{2}\alpha_{2})\prod_{i=3}^{n}\Phi(\epsilon_{i}\alpha_{i})$
$\displaystyle(\hbox{since
$\epsilon_{1}\alpha_{1}+\epsilon_{2}\alpha_{2}+\epsilon_{3}\alpha_{3}\in
R^{\times}$})$ $\displaystyle=$
$\displaystyle\Phi(\epsilon_{1}\alpha_{1}+\epsilon_{2}\alpha_{2}+\epsilon_{3}\alpha_{3})\prod_{i=4}^{n}\Phi(\epsilon_{i}\alpha_{i})$
$\displaystyle\vdots$ $\displaystyle=$
$\displaystyle\Phi(\epsilon_{1}\alpha_{1}+\cdots+\epsilon_{n}\alpha_{n})=\Phi(\alpha).\qquad\hbox{\hfill$\Box$}$

######  Proposition 5.1.3.

Let $E$ be an extended affine Lie algebra with root system $R$. If
$\hbox{ind}(R)=0$ then any Cartan automorphism $\Psi$ for $E$ corresponds to a
finite-order root-lattice character. In particular, if $R$ is a finite or
affine root system, then any character for $R$ extends to a root-lattice
character.

Proof. By Propositions 4.4.10, there exists a finite-order character
$\psi:R\rightarrow{\mathbb{K}}^{\times}$ for $R$ such that
$\Psi_{|_{E_{\alpha}}}=\psi(\alpha)\hbox{id}_{|_{E_{\alpha}}}$, $\alpha\in R$.
By Lemma 5.1.2, $\psi$ can be extended to a root-lattice character and so we
are done. $\Box$

######  Remark 5.1.4.

We may give an alternative proof for Proposition 5.1.3 as follows. By
Proposition 4.4.10, there exists a finite-order character
$\psi:R\rightarrow{\mathbb{K}}^{\times}$ such that
$\Psi_{|_{E_{\alpha}}}=\psi(\alpha)\hbox{id}_{|_{E_{\alpha}}}$, $\alpha\in R$.
By [APT23, Lemma 6.2.4], $R^{\times}$ contains a basis $\mathcal{P}$ of the
root lattice $\langle R\rangle$ such that the root spaces $E_{\pm\alpha}$,
$\alpha\in\mathcal{P}$, generate $E_{c}$ as a Lie algebra. Consider the group
homomorphism $\widehat{\psi}:\langle R\rangle\rightarrow{\mathbb{K}}^{\times}$
given by $\widehat{\psi}(\alpha)=\psi(\alpha)$ for $\alpha\in\mathcal{P}$. Let
$\widehat{\Psi}$ be the corresponding Cartan automorphism given in Example
4.1.3. Since $E_{\pm\alpha}$, $\alpha\in\mathcal{P}$ generate $E_{c}$, $\Psi$
and $\widehat{\Psi}$ coincide on the core. Moreover,
$\widehat{\Psi}\Psi=\Psi\widehat{\Psi}$. Thus by Lemma 3.1.1,
$\Psi=\widehat{\Psi}$, and so ${\widehat{\psi}}_{|_{R}}=\psi.$

###  5.2. Characters of type $\mathbf{A_{1}}$

Lemma 5.1.2 demonstrates that in order to examine the existence of characters
lacking extensions to the root lattice, the type $A_{1}$ is a suitable
candidate. To begin with, we present an extended affine root system of type
$A_{1}$. Consider a one-dimensional real vector space
$\dot{\mathcal{V}}:={\mathbb{R}}\dot{\alpha}$, and a $\nu$-dimensional real
vector space ${\mathcal{V}}^{0}:=\sum_{i=1}^{\nu}{\mathbb{R}}\sigma_{i}$. Set
${\mathcal{V}}:=\dot{\mathcal{V}}\oplus{\mathcal{V}}^{0}$ and
$\Lambda=\sum_{i=1}^{\nu}{\mathbb{Z}}\sigma_{i}$. In what follows, for
$\sigma,\sigma^{\prime}\in\Lambda$, we employ the notation
$\sigma\equiv\sigma^{\prime}$ to signify that $\sigma-\sigma^{\prime}\in
2\Lambda$. We define a symmetric bilinear map $(\cdot,\cdot)$ on
${\mathcal{V}}$ by

$(\dot{\alpha},\dot{\alpha})=2,\quad\hbox{and}\quad({\mathcal{V}}^{0},{\mathcal{V}})=\\{0\\}.$

Let $m$ be a non-negative integer and fix

(5.1) $0=\tau_{0},\tau_{1},\ldots,\tau_{m}\in\Lambda\hbox{ such that for
}i\not=j,\;\tau_{i}\not\equiv\tau_{j}.$

Set

(5.2) $S=\cup_{i=0}^{m}(\tau_{i}+2\Lambda)\quad\hbox{and}\quad
R=(S+S)\cup(\pm\dot{\alpha}+S).$

Then $S$ is a semilattice in ${\mathcal{V}}^{0}$ and $R$ is an extended affine
root system of type $A_{1}$ in ${\mathcal{V}}$ with $R^{0}=S+S$, moreover, any
extended affine root system of type $A_{1}$ can be described this way, see 2.5
and [AABGP97, Theorem II.2.37].

######  Lemma 5.2.1.

Consider the extended affine root system $R=R^{0}\cup(\pm\dot{\alpha}+S)$ of
type $A_{1}$, where $S=\cup_{i=0}^{m}(\tau_{i}+2\Lambda)$ is given by (5.1)
and (5.2). Assume further that

($\star$) $\tau_{i_{1}}+\cdots+\tau_{i_{k}}\not\equiv 0\hbox{ for }3\leq k\leq
6\hbox{ and distinct }1\leq i_{1},\ldots,i_{k}\leq m.$

Define $\Phi:R^{\times}\rightarrow\\{\pm 1\\}$ by

$\Phi(\pm\dot{\alpha}+2\Lambda)=1,\;\Phi(\pm\dot{\alpha}+\tau_{i}+2\Lambda)=-1\hbox{
for }1\leq i\leq m,$

and $\Phi:R^{0}\rightarrow\\{\pm 1\\}$ by

$\Phi(\sigma)=\left\\{\begin{array}[]{ll}1&\hbox{if }\sigma\in
2\Lambda\cup((S+S)\setminus S),\\\ -1&S\setminus 2\Lambda.\end{array}\right.$

Then $\Phi:R\rightarrow\\{\pm 1\\}$ is a character for $R$.

Proof. First we note that the assumption ($\star$) implies that

(5.3) $\begin{array}[]{c}\tau_{i_{1}}+\tau_{i_{2}}\not\in
S,\;\tau_{i_{1}}+\tau_{i_{2}}+\tau_{i_{3}}\not\in
R^{0},\;\tau_{i_{1}}+\tau_{i_{2}}+\tau_{i_{3}}+\tau_{i_{4}}\not\in
R^{0},\vspace{2mm}\\\ \hbox{ if $\tau_{i_{j}}$'s are distinct}.\end{array}$

Now since $\tau_{i_{1}}+\tau_{i_{2}}\not\in S$ for $1\leq i_{1}\not=i_{2}\leq
m$, we get $R^{0}=(S+S)=2\Lambda\uplus(S\setminus
2\Lambda)\uplus((S+S)\setminus S).$ In fact if $\sigma\equiv\tau_{i}$ and
$\sigma^{\prime}\equiv\tau_{j}$ are elements of $S$, then

$\sigma+\sigma^{\prime}\equiv\left\\{\begin{array}[]{ll}\tau_{i}+\tau_{j}\in(S+S)\setminus
S&\hbox{if }1\leq i\not=j\leq m,\\\ 2\tau_{i}\in 2\Lambda&\hbox{if }i=j,\\\
\tau_{j}\in S\setminus 2\Lambda&\hbox{if }i=0,\;j\not=0.\end{array}\right.$

Therefore $\Phi:R\rightarrow\\{\pm 1\\}$ is a well defined map. From
definition of $\Phi$ it follows that $\Phi(\alpha)=\Phi(\beta)$ if
$\alpha\equiv\beta$. We proceed to show that $\Phi$ is a character. In the
following, when there is symmetry in the indices, we present the argument for
one and leave the others.

Let $\alpha,\beta\in R$ with $\alpha+\beta\in R$. If $\alpha,\beta\in
R^{\times}$, then there exist $i,j$ and $\epsilon\in\\{\pm 1\\}$ such that
$\alpha\equiv\epsilon\dot{\alpha}+\tau_{i}$,
$\beta\equiv-\epsilon\dot{\alpha}+\tau_{j}$. Then

$\displaystyle\Phi(\alpha+\beta)=\Phi(\tau_{i}+\tau_{j})$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ll}1&\hbox{if }i=j\hbox{ or }1\leq
i\not=j\leq m,\\\ -1&i=0,\;j>0.\\\ \end{array}\right.$ $\displaystyle=$
$\displaystyle\Phi(\alpha)\Phi(\beta).$

If $\alpha\in R^{\times}$ and $\beta\in R^{0}$, then
$\alpha\equiv\epsilon\dot{\alpha}+\tau_{i}$ and $\beta\equiv\tau_{j}+\tau_{k}$
for some $\epsilon\in\\{\pm 1\\}$ and some $0\leq i,j,k\leq m$. If
$|\\{i,j,k\\}|=3$, then as $\alpha+\beta\in R$ and $\tau_{s}+\tau_{t}\not\in
S$ if $s\not=t$, we get $1\leq i,j,k\leq m$ and $\tau_{i}+\tau_{j}+\tau_{k}\in
S\setminus 2\Lambda$. Thus $\Phi(\alpha)=-1$, $\Phi(\beta)=1$ and
$\Phi(\alpha+\beta)=-1$.

Next, assume that $|\\{i,j,k\\}|=2$. We have the following possibilities, up
to symmetry between $j,k$:

$j=k,\;i=0\Rightarrow\Phi(\alpha)=1,\;\Phi(\beta)=1,\;\Phi(\alpha+\beta)=1,$
$j=k,\;i\not=0\Rightarrow\Phi(\alpha)=-1,\;\Phi(\beta)=1,\;\Phi(\alpha+\beta)=-1,$
$j\not=k,\;i=j=0\Rightarrow\Phi(\alpha)=1,\;\Phi(\beta)=-1,\;\Phi(\alpha+\beta)=-1,$
$j\not=k,\;i=j\not=0\Rightarrow\Phi(\alpha)=-1,\;\Phi(\beta)=1,\;\Phi(\alpha+\beta)=-1.$

If $|\\{i,j,k\\}|=1$, we have the following possibilities:

$i=j=k=0\Rightarrow\Phi(\alpha)=1,\;\Phi(\beta)=1,\;\Phi(\alpha+\beta)=1,$
$i=j=k\not=0\Rightarrow\Phi(\alpha)=-1,\;\Phi(\beta)=1,\;\Phi(\alpha+\beta)=-1.$

The above computations show that $\Phi(\alpha+\beta)=\Phi(\alpha)\Phi(\beta)$
for $\alpha\in R^{\times}$ $\beta\in R^{0}$, $\alpha+\beta\in R$.

Finally, we consider the case $\alpha,\beta,\alpha+\beta\in R^{0}=S+S$. We
have $\alpha\equiv\tau_{i}+\tau_{j}$, $\beta\equiv\tau_{k}+\tau_{t}$ for some
$0\leq i,j,k,t\leq m$. We have the following possibilities, up to symmetry
between $i,j$ or $k,t$:

$i=j,\;k=t\Rightarrow\Phi(\alpha)=1,\;\Phi(\beta)=1,\;\Phi(\alpha+\beta)=1,$
$i=j=k\not=t=0\Rightarrow\Phi(\alpha)=1,\Phi(\beta)=-1,\;\Phi(\alpha+\beta)=-1,$
$i=j=k=0,\;t\not=0\Rightarrow\Phi(\alpha)=1,\;\Phi(\beta)=-1,\;\Phi(\alpha+\beta)=-1,$
$i=j=0,\;k=t\not=0\Rightarrow\Phi(\alpha)=1,\;\Phi(\beta)=1,\;\Phi(\alpha+\beta)=1,$
$i=j=0,\;k,t\not=0,\;k\not=t\Rightarrow\Phi(\alpha)=1,\;\Phi(\beta)=1,\;\Phi(\alpha+\beta)=1,$
$i=0,\;j=k=t\not=0\Rightarrow\Phi(\alpha)=-1,\;\Phi(\beta)=1,\;\Phi(\alpha+\beta)=-1,$
$i=0,\;j,k,t\not=0,\;j=k\not=t\Rightarrow\Phi(\alpha)=-1,\;\Phi(\beta)=1,\;\Phi(\alpha+\beta)=-1,$
$i=0,\;j,k,t\not=0,\;|\\{j,k,t\\}|=3\Rightarrow\Phi(\alpha)=-1,\;\Phi(\beta)=1,\;\alpha+\beta\equiv\tau_{j}+\tau_{k}+\tau_{t}\not\in
R^{0},$

see (5.3) for the conclusion that $\alpha+\beta\not\in R^{0}$ in the last
implication. The cases we discussed exhaust situations in which $i=j,\;k=t$ or
at least one of $i,j,k,t$ is zero.

Next we consider the remaining cases, namely $1\leq i,j,k,t\leq m$, and
$|\\{i,j,k,t\\}|\geq 2$:

$i=j=k\not=t\Rightarrow\Phi(\alpha)=1,\;\Phi(\beta)=1,\;\Phi(\alpha+\beta)=1,$
$i=j,k\not=t,\;\\{i,j,k\\}|=3\Rightarrow\Phi(\alpha)=1,\;\Phi(\beta)=1,\;\Phi(\alpha+\beta)=1,$
$i\not=j=k\not=t,\;\\{i,j,t\\}|=3\Rightarrow\Phi(\alpha)=1,\;\Phi(\beta)=1,\;\Phi(\alpha+\beta)=1,$
$|\\{i,j,k,t\\}|=4\Rightarrow\Phi(\alpha)=1,\;\Phi(\beta)=1,\;\alpha+\beta\equiv\tau_{i}+\tau_{j}+\tau_{k}+\tau_{t}\not\in
R^{0},$

see (5.3) for the conclusion that $\alpha+\beta\not\in R^{0}$ in the last
implication. $\Box$

We continue the section by showing that a character for $R$ may fail to extend
to a character for $\langle R\rangle$.

######  Corollary 5.2.2.

There exist certain characters for extended affine root systems that are not
extendable to root-lattice character.

Proof. Let $R$ be an extended affine root system of type $A_{1}$ as in (5.2)
with $\nu\geq 6$, $\tau_{i}=\sigma_{i}$ for $1\leq i\leq 6$ and
$\tau_{7}=\sigma_{1}+\cdots+\sigma_{6}$. Note that with our choices of
$\tau_{i}$’s, the conditions ($\star$) in the statement of Lemma 5.2.1 is
fulfilled. Now if the character $\Phi$ is extendable to a character
$\bar{\Phi}$ for $\langle R\rangle$, then we would have
$\bar{\Phi}(\tau_{7})=\prod_{i=1}^{6}\bar{\Phi}(\sigma_{i})=\prod_{i=1}^{6}\Phi(\sigma_{i})=1$.
On the other hand by definition of $\Phi$, we have
$\bar{\Phi}(\tau_{7})=\Phi(\tau_{7})=-1$, a contradiction. $\Box$

## References