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# Automorphisms of quantum polynomials

Ashish Gupta  Department of Mathematics
Ramakrishna Mission Vivekananda Educational and Research Institute
Belur,
Howrah, WB 711202
India<EMAIL_ADDRESS>
###### Abstract.

An important step in the determination of the automorphism group of the
quantum torus of rank $n$ (or twisted group algebra of $\mathbb{Z}^{n}$) is
the determination of its so-called non-scalar automorphisms. We present a new
algorithimic approach towards this problem based on the bivector
representation
$\mathop{\bigwedge\nolimits^{\\!2}}:\operatorname{GL}(n,Z)\rightarrow\operatorname{GL}(\binom{n}{2},\mathbb{Z})$
of $\operatorname{GL}(n,\mathbb{Z})$ and thus compute the non-scalar
automorphism group $\operatorname{\mathrm{Aut}}(\mathbb{Z}^{n},\lambda)$ in
several new cases. As an application of our ideas we show that the quantum
polynomial algebra (multiparameter quantum affine space of rank $n$) has only
scalar (or toric) automorphisms provided that the torsion-free rank of the
subgroup generated by the defining multiparameters is no less than
$\binom{n-1}{2}+1$ thus improving an earlier result. We also investigate the
question: when is a multiparameter quantum affine space free of so-called
linear automorphisms other than those arising from the action of the $n$-torus
${(\mathbb{F}^{\ast})}^{n}$.

###### Contents

  1. 1 Introduction
  2. 2 Automorphisms of quantum tori
  3. 3 Automorphisms of quantum affine spaces - preliminary facts
  4. 4 Automorphisms of quantum affine spaces–Main results
    1. 4.1 Examples

##  1\. Introduction

Quantum polynomials are non-commutative versions of polynomial and Laurent
polynomial algebras. The difference with the ordinary polynomials lies in the
fact that the commutativity of the variables is replaced by quasi-
commutativity, that is, $X_{i}X_{j}=q_{ij}X_{j}X_{i}$ for non-zero scalars
$q_{ij}$. This relation is the Weyl form of the cannonical commutation
relation of quantum mechanics. For the Laurent case (where each variable
$X_{j}$ has an inverse) there other terms in use, for example, quantum torus,
twisted group algebra, McConnell–Pettit algebra etc.

Quantum polynomials play a key role in the theory of quantum groups [Ar] and
also in non-commutative geometry [M]. Their Laurent versions arise in Lie
theory as coordinate structures of extended affine Lie algebras [Ne] and also
in the representation theory of nilpotent groups [Br]. The question of
automorphisms of polynomial algebras is a still open one (the Jacobian
conjecture) and it is therefore of interest to investigate the automorphisms
of quantum polynomial algebras. It is generally believed that these algebras
have relatively fewer automorphisms, that is, they are rigid. However, the
complete understanding of their automorphisms groups is yet to be had and the
results in this paper are expected to be a step in this direction.

We briefly recall the definitions. A quantum affine space
$\mathcal{O}_{\mathfrak{q}}(\mathbb{F}^{n})$ over a field $\mathbb{F}$ is
defined as

(1) $\displaystyle\mathcal{O}_{\mathfrak{q}}(\mathbb{F}^{n})$
$\displaystyle:=\mathbb{F}\langle
X_{1},X_{2},\cdots,X_{n}\rangle/(X_{i}X_{j}-q_{ij}X_{j}X_{i})$

where $\mathfrak{q}$ stands for the matrix of _multiparameters_ $q_{ij}$, that
is, $\mathfrak{q}=(q_{ij})$ and

$q_{ij}\in\mathbb{F}^{\ast}:=\mathbb{F}\smallsetminus\\{0\\}.$

The matrix $\mathfrak{q}$ is assumed to be multiplicatively anti-symmetric,
that is, $q_{ii}=1$ and $q_{ji}=q_{ij}^{-1}.$ A given quantum affine space
$\mathcal{O}_{\mathfrak{q}}(\mathbb{F}^{n})$ can be embedded in a _quantum
torus_ $\mathcal{O}_{\mathfrak{q}}({(\mathbb{F}^{\ast})}^{n})$ by means of
localization. The latter type of algebra is generated by the indeterminates
$X_{1},\cdots,X_{n}$ together with their inverses subject to the quantum
commutation relations as in (1). The following notion plays an important role
in the theory of quantum polynomials.

######  Definition 1.1.

For the quantum polynomial algebras
$\mathcal{O}_{\mathfrak{q}}({\mathbb{F}}^{n})$ and
$\mathcal{O}_{\mathfrak{q}}({(\mathbb{F}^{\ast})}^{n})$ the $\lambda$-group
(denoted $\Lambda$) is the subgroup of $\mathbb{F}^{\ast}$ generated by the
multiparameters $q_{ij}$.

For the quantum tori the monomials
$\mathbf{X}^{\mathbf{m}}:=X_{1}^{m_{1}}\cdots X_{n}^{m_{n}}$ are units and the
group-theoretic commutator
$[\mathbf{X}^{\mathbf{m}},\mathbf{X}^{\mathbf{m}^{\prime}}]$ defined as

$[\mathbf{X}^{\mathbf{m}},\mathbf{X}^{\mathbf{m}^{\prime}}]:=\mathbf{X}^{\mathbf{m}}\mathbf{X}^{\mathbf{m}^{\prime}}{(\mathbf{X}^{\mathbf{m}})}^{-1}{(\mathbf{X}^{\mathbf{m}^{\prime}})}^{-1}$

yields an alternating $\mathbb{Z}$-bilinear function

(2) $\lambda:\Gamma\times\Gamma\rightarrow\mathbb{F}^{\ast},\ \ \ \ \ \ \ \ \
\
\lambda(\mathbf{m},\mathbf{m}^{\prime})=[\mathbf{X}^{\mathbf{m}},\mathbf{X}^{\mathbf{m}^{\prime}}],\
\ \ \ \ \ \ \ \ \
\forall\mathbf{m},\mathbf{m}^{\prime}\in\Gamma:=\mathbb{Z}^{n}.$

whose image is contained in the group $\Lambda$ (e.g., [OP, Section 1]). Let

(3) $\Lambda:=\langle p_{1}\rangle\times\langle
p_{2}\rangle\times\cdots\times\langle p_{l}\rangle,\ \ \ \ \ \ \ \ \
l\in\mathbb{N}.$

It was observed in [OP, Lemma 3.3(ii)] (see Section 2) that in the study of
the autmorphisms of a quantum torus the crucial case is where the group
$\Lambda$ is torsion-free. We may thus assume that each direct summand
$\langle p_{i}\rangle\ (i=1,\cdots,l)$ in (3) is an infinite cycle. For

$\mathbf{m},\mathbf{m}^{\prime}\in\Gamma=\mathbb{Z}^{n}$

we thus have

(4)
$\lambda(\mathbf{m},\mathbf{m}^{\prime})=p_{1}^{e_{1}(\mathbf{m},\mathbf{m}^{\prime})}p_{2}^{e_{2}(\mathbf{m},\mathbf{m}^{\prime})}\cdots
p_{l}^{e_{l}(\mathbf{m},\mathbf{m}^{\prime})}$

where each exponent map $e_{i}:\Gamma\times\Gamma\rightarrow\mathbb{Z}$ yields
an alternating bilinear form on $\Gamma$.

It is known (e.g., [MP]) that the units of a quantum torus algebra are
trivial, that is, are of the form $\alpha\mathbf{X}^{\mathbf{m}}$, where
$\alpha\in\mathbb{F}^{\ast}$. Let us denote by $\mathscr{A}$ the group
$\operatorname{\mathrm{Aut}}_{\mathbb{F}}(\mathcal{O}_{\mathfrak{q}}({(F^{\ast})}^{n}))$
of all $\mathbb{F}$-automorphisms and by $\mathscr{U}$ the group of trivial
units of the algebra $\mathcal{O}_{\mathfrak{q}}({(F^{\ast})}^{n})$. It is
easily seen (e.g., [OP]) that the action of the group $\mathscr{A}$ on the
quantum torus $\mathcal{O}_{\mathfrak{q}}({(F^{\ast})}^{n})$ induces an action
of this same group on the group $\mathscr{U}$ of trivial units fixing
$\mathbb{F}^{\ast}$ elementwise. There is thus an action of the group
$\mathscr{A}$ on the quotient group $\mathscr{U}/\mathbb{F}^{\ast}\cong\Gamma$
yielding a homomorphism

(5)
$\mathscr{A}\longrightarrow\operatorname{\mathrm{Aut}}\Gamma=\operatorname{GL}(n,\mathbb{Z})$

whose kernel is the group $\mathscr{S}$ of all _scalar automorphisms_ defined
by $\psi(\mathbf{X}^{\mathbf{m}})=\phi(\mathbf{m})(\mathbf{X}^{\mathbf{m}})$
for $\phi\in\operatorname{Hom}(\Gamma,\mathbb{F}^{\ast})$ [OP]. Thus
$\mathscr{S}\cong(\mathbb{F}^{\ast})^{n}$. Furthermore by [OP, Lemma 3.3(iii)]
the image of the map in (5) coincides with the group
$\operatorname{\mathrm{Aut}}(\mathbb{Z}^{n},\lambda)\leq\operatorname{GL}(n,\mathbb{Z})$
of all _non-scalar automorphisms_ $\sigma$ of $\Gamma$ satisfying

(6)
$\lambda(\sigma\mathbf{m},\sigma\mathbf{m^{\prime}})=\lambda(\mathbf{m},\mathbf{m^{\prime}})\
\ \ \ \ \ \ \ \forall\bf{m},\bf{m^{\prime}}\in\mathbb{Z}^{n}.$

We thus have the following exact sequence for the group $\mathscr{A}$ as noted
in [Ne]

(7)
$1\rightarrow\mathscr{S}\rightarrow\mathscr{A}\rightarrow\operatorname{\mathrm{Aut}}(\mathbb{Z}^{n},\lambda)\rightarrow
1.$

In terms of alternating bilinear forms $e_{i}$ defined above the following
characterization of the non-scalar automorphism group
$\operatorname{\mathrm{Aut}}(\mathbb{Z}^{n},\lambda)$ (as noted in [OP]) is
immediate.

######  Theorem 1.1 (Theorem 3.4 of [OP]).

Let $\mathrm{Sp}(\mathbb{Z},e_{i})$ be the symplectic group associated with
the form $e_{i}$. Then

(8)
$\operatorname{\mathrm{Aut}}(\mathbb{Z}^{n},\lambda)=\bigcap_{i=1}^{l}\mathrm{Sp}(\mathbb{Z},e_{i}).$

Thus a non-scalar automorphism must preserve each of the forms $e_{i}$ and so
must stabilize the radicals of each of these forms. This fact has been
fruitfully used in [OP] towards determining the non-scalar automorphism group
$\operatorname{\mathrm{Aut}}(\mathbb{Z}^{n},\lambda)$ in certain cases.
However, notwithstanding the nicety of the preceding theorem from a
computational viewpoint (8) seems to entail some difficulties limiting the
cases where it can be directly applied.

In the following we present a new approach towards the computation of this
group. Before stating it we fix some notation.

######  Notation 1.

In view of $\eqref{dcpms-Lmbda}$ let

(9) $\lambda(\mathbf{e}_{i},\mathbf{e}_{j})=p_{1}^{m_{1}^{(ij)}}\cdots
p_{l}^{m_{l}^{(ij)}},\ \ \ \ \ \ \ \ \ \ \ \ \forall 1\leq i<j\leq n,$

where $\mathbf{e}_{i},\mathbf{e}_{j}$ are standard basis vectors of the free
$\mathbb{Z}$-module $\Gamma$. On the $\binom{n}{2}$ pairs $(ij),\ (i<j)$ we
assume the lexicographic order. Let
$\mathsf{M}\in\mathrm{Mat}_{\binom{n}{2}\times l}(\mathbb{Z})$ be the matrix
whose $((ij),s)$ entry is the exponent $m_{s}^{(ij)}$ of $p_{s}$ in (9)
($\forall s\in\\{1.\cdots,l\\}$).

With this notation we have the following.

######  Theorem A.

For a quantum torus $\mathcal{O}_{\mathfrak{q}}(\mathbb{(}F^{\ast})^{n})$
suppose that the group $\Lambda$ generated by the multiparameters $q_{ij}$ is
torsion-free. Set $N=\binom{n}{2}$ and let $\mathsf{M}$ be as defined in
Notation 1 above. Then

$\operatorname{\mathrm{Aut}}(\mathbb{Z}^{n},\lambda)=\bigl{(}\operatorname{Stab}_{\operatorname{GL}(n,\mathbb{Z})}(\mathsf{M})\bigr{)}^{t}$

where $t$ denotes transposition and
$\operatorname{Stab}_{\operatorname{GL}(n,\mathbb{Z})}(\mathsf{M})$ the
stabilizer of $\mathsf{M}$ in $\operatorname{GL}(n,\mathbb{Z})$ with respect
to the bivector representation

$\displaystyle\mathop{\bigwedge\nolimits^{\\!2}}:\operatorname{GL}(n,\mathbb{Z})\rightarrow\operatorname{GL}(N,\mathbb{Z}),\
\ \ \ \ \ A\rightarrow\wedge^{2}(A)$

of $\operatorname{GL}(n,\mathbb{Z})$ and the action of
$\operatorname{GL}(N,\mathbb{Z})$ on the space
$\mathrm{Mat}_{N,l}(\mathbb{Z})$ by left multiplication. In other words

$\operatorname{\mathrm{Aut}}(\mathbb{Z}^{n},\lambda)=\\{A^{t}\mid
A\in\operatorname{GL}(n,\mathbb{Z})\ \mathrm{and}\
(\wedge^{2}A)\mathsf{M}=\mathsf{M}\\}.$

######  Remark 1.1.

Although the matrix $\mathsf{M}$ depends on the choice of a $\mathbb{Z}$-basis
in the group $\Lambda$ the group
$\operatorname{Stab}_{\operatorname{GL}(N,\mathbb{Z})}(\mathsf{M})$ (and
consequently
$\operatorname{Stab}_{\operatorname{GL}(n,\mathbb{Z})}(\mathsf{M})$) is
independent of such a choice. Indeed for a change of basis matrix
$P\in\operatorname{GL}(l,\mathbb{Z})$ we have
$\mathsf{M}^{\prime}=\mathsf{M}P$ and then
$X\mathsf{M}^{\prime}=\mathsf{M}^{\prime}$ if and only if
$X\mathsf{M}=\mathsf{M}$ for $X\in\operatorname{GL}(N,\mathbb{Z})$.

Theorem A allows us to determine the non-scalar automorphism group
$\operatorname{\mathrm{Aut}}(\mathbb{Z}^{n},\lambda)$ thus:

_Step 1_ : We first determine the group
$\mathscr{N}:=\operatorname{Stab}_{\operatorname{GL}(N,\mathbb{Z})}(\mathsf{M})$
(where $N=\binom{n}{2}$). This is easily done if $\mathsf{M}$ is brought to
the Smith normal form.

_Step 2_ : Next we find the subgroup
$\mathscr{N}_{1}:=\mathscr{N}\cap\mathop{\bigwedge\nolimits^{\\!2}}(\operatorname{GL}(n,\mathbb{Z}))$.
This can be done by noting as in [Nem] that the projective transformations of
the projective space $\mathrm{P}(\wedge^{2}{\bar{\mathbb{Q}}}^{n})$ (here
$\bar{\mathbb{Q}}$ stands for the algebraic closure of $\mathbb{Q}$) arising
from the matrices in
$\mathop{\bigwedge\nolimits^{\\!2}}(\operatorname{GL}(n,\mathbb{Z}))$ preserve
the projective grassmannian variety $\mathrm{Gr}(2,n)$. More algebraically,
the group
$\mathop{\bigwedge\nolimits^{\\!2}}(\operatorname{GL}(n,\mathbb{Z}))$ is
identified in [VP] with the connected component of the subgroup of
$\operatorname{GL}(N,\mathbb{Z})$ preserving the ideal
$\operatorname{Pl\ddot{u}}$ generated by all the Plücker polynomials defining
the variety $\mathrm{Gr}(2,n)$.

_Step 3_ : Once $\mathscr{N}_{1}$ has thus been determined we must calculate
$N=\bigl{(}\mathop{\bigwedge\nolimits^{\\!2}}\bigr{)}^{-1}(\mathscr{N}_{1})$.
This amounts to calculating the exterior root of a matrix in $\mathscr{N}_{1}$
which is determined uniquely up to minus sign [LoLu] and for which a
polynomial-time algorithm (implemented in GAP) exists [Grn99].

######  Remark 1.2.

The algorithm mentioned in Step 3 can also determine (in polynomial time) if a
matrix $A\in\operatorname{GL}(N,\mathbb{Z})$ has an exterior root and
consequently Step 2 may not be necessary. However it may still be helpful in
the determination of the non-scalar automorphism group.

Until now the nonscalar automorphism group has been calculated ([OP, Ne]) in
the case $n\geq 3$ or in the case when the multiparameters are
(multiplicatively) independent. Examples of the above approach applied to the
calculation of the non-scalar automorphism group of quantum tori for $n=4$ are
given in Section 2. One such example is as follows.

######  Example.

For $n=4$ suppose that the group $\Lambda$ is freely generated by the
commutators $q_{14}$, $q_{23}$, $q_{24}$ and $q_{34}$, while

$q_{12}=q_{13}=1.$

Then

$\operatorname{\mathrm{Aut}}(\mathbb{Z}^{4},\lambda)=\Biggl{\\{}f_{b,\epsilon}:=\begin{pmatrix}\epsilon&0&0&-b\\\
0&\epsilon&0&0\\\ 0&0&\epsilon&0\\\ 0&0&0&\epsilon\end{pmatrix},\ \
b\in\mathbb{Z},\epsilon\in\\{\pm 1\\}\Biggr{\\}}.$

The following proposition is also established in Section 2.

######  Proposition 3.3.

Let $n\geq 5$. For a quantum torus
$\mathcal{O}_{\mathfrak{q}}(\mathbb{(}F^{\ast})^{n})$ such that

$q_{12}=q_{13}=\cdots=q_{1(n-2)}=1$

and the remaining multiparameters independent in $\mathbb{F}^{\ast}$ we have

$\operatorname{\mathrm{Aut}}(\mathbb{Z}^{n},\lambda)\cong\mathbb{Z}_{2}.$

We hope that the non-scalar automorphism group can similarly be determined in
many new and interesting situations using the approach described above. Using
Theorem A we can also easily recover the group
$\operatorname{\mathrm{Aut}}(\mathbb{Z}^{n},\lambda)$ in the limited number of
known cases [KPS94, OP, Ne] dealing with small values of $n$ and the case
where the group $\Lambda$ has the maximal possible rank.

We next consider the automorphism groups of quantum affine spaces. Until now
in the study of automorphisms of quantum affine spaces the following important
cases have been dealt with: (i) the uniparameter case, that is, $q_{ij}=q$ and
the group $\langle q\rangle$ is infinite cyclic [AC] and (ii) the case in
which the multiparameters are in general position [Ar], that is, the group
$\Lambda$ has maximum possible (torsion-free) rank. The main conclusions here
are that under these assumptions the automorphism group shrinks to the image
of the action of the torus ${(\mathbb{F}^{\ast})}^{n}$. We show the same
conclusion remains true in somewhat more general situations:

######  Theorem B.

A quantum affine space $\mathcal{O}_{\mathfrak{q}}(\mathbb{F}^{n})$ such that
the subgroup $\Lambda$ has rank no smaller than $\binom{n-1}{2}+1$ satisfies

$\operatorname{\mathrm{Aut}}(\mathcal{O}_{\mathfrak{q}}(\mathbb{F}^{n}))=(\mathbb{F}^{\ast})^{n}.$

An automorphism of the algebra $\mathcal{O}_{\mathfrak{q}}$ is called linear
if it stabilizes the subspace spanned by $X_{1},\cdots,X_{n}$. The following
proposition gives a criterion for a multiparameter quantum affine space to be
free of linear automorphisms other than those arising from the action of the
torus ${(\mathbb{F}^{\ast})}^{n}$.

######  Proposition 4.1.

Suppose that $\operatorname{\mathrm{char}}(\mathbb{F})\neq 2$ and $n\geq 3$.
Let $\mathfrak{q}=(q_{ij})$ be a multiplicatively antisymmetric matrix whose
entries satisfy

$\displaystyle\mbox{(i)}\ q_{ij}\neq q_{kl}\ \ \ \ \ \ \ \ \forall i<j,\ \
\forall k<l,$ $\displaystyle\mbox{(ii)}\ q_{ij}q_{kl}\neq 1\ \ \ \ \ \ \ \ \
\forall i<j,\ \ \forall k<l,\ (i,j)\neq(k,l).$

The group of linear automorphisms of the quantum affine space
$\mathcal{O}_{\mathfrak{q}}(\mathbb{F}^{n})$ coincides with the torus
${(\mathbb{F}^{\ast})}^{n}$.

######  Notation 2.

We will use the short forms $\mathcal{O}_{\mathfrak{q}}$ for the quantum
affine space and $\widehat{\mathcal{O}}_{\mathfrak{q}}$ for quantum torus
obtained from it by localizing at the multiplicative subset generated by the
indeterminates $X_{i}$ ($1\leq i\leq n$) and refer to it as the “corresponding
quantum torus”. Automorphisms of the algebras we consider will always be
$\mathbb{F}$-automorphisms.

##  2\. Automorphisms of quantum tori

As noted above in (7) the essential question here is the determination of the
group
$\operatorname{\mathrm{Aut}}(\mathbb{Z}^{n},\lambda)\leq\operatorname{GL}(n,\mathbb{Z})$
of non-scalar automorphisms. The following fact shown in [OP] is the basis for
our assumption that the group $\Lambda$ is torsion-free.

######  Lemma 2.1 (Lemma 3.3(ii)).

Let $p$ denote the size of the torsion subgroup of $\Lambda$. The subalgebra
$\widehat{\mathcal{O}^{\prime}}$ of $\widehat{\mathcal{O}}_{\mathfrak{q}}$
generated by the powers $X_{i}^{\pm p}$ of the indeterminates $X_{i}$ is a
characteristic sub-algebra of the same rank. Moreover
$\widehat{\mathcal{O}}_{\mathfrak{q}}$ is free left
$\widehat{\mathcal{O}^{\prime}}$-module of finite rank and the corresponding
$\lambda$-group $\Lambda^{\prime}$ associated with
$\widehat{\mathcal{O}^{\prime}}$ is torsion free.

In view of Lemma 2.1 we will assume throughout this section that the group
$\Lambda$ is torsion-free. We recall that for a given matrix
$A\in\operatorname{GL}(n,\mathbb{Z})$ the exterior square $\wedge^{2}A$ of $A$
is the $\binom{n}{2}\times\binom{n}{2}$-matrix whose rows and columns are
indexed by the pairs $(ij)\ (1\leq i<j\leq n)$ ordered lexicographically and
whose $((ij),(kl))$ entry is the $2\times 2$-minor corresponding to rows $i,j$
and columns $k,l$.

By the well-known Cauchy-Binet formula the map $A\mapsto\wedge^{2}A$ is
multiplicative and satisfies $\det(\wedge^{2}A)=(\det A)^{n-1}$ [HJ]. We also
have $\wedge^{2}A^{t}=(\wedge^{2}A)^{t}$ whete $t$ denotes transposition. With
$\mathsf{M}$ as defined in Notation 1 we have the following.

######  Theorem A.

For a quantum torus $\mathcal{O}_{\mathfrak{q}}(\mathbb{(}F^{\ast})^{n})$
suppose that the group $\Lambda$ generated by the multiparameters $q_{ij}$ is
torsion-free. Set $N=\binom{n}{2}$ and let $\mathsf{M}$ be as defined in
Notation 1 above. Then

$\operatorname{\mathrm{Aut}}(\mathbb{Z}^{n},\lambda)=\bigl{(}\operatorname{Stab}_{\operatorname{GL}(n,\mathbb{Z})}(\mathsf{M})\bigr{)}^{t}$

where $t$ denotes transposition and
$\operatorname{Stab}_{\operatorname{GL}(n,\mathbb{Z})}(\mathsf{M})$ the
stabilizer of $\mathsf{M}$ in $\operatorname{GL}(n,\mathbb{Z})$ with respect
to the bivector representation

$\displaystyle\mathop{\bigwedge\nolimits^{\\!2}}:\operatorname{GL}(n,\mathbb{Z})\rightarrow\operatorname{GL}(N,\mathbb{Z}),\
\ \ \ \ \ A\rightarrow\wedge^{2}(A)$

of $\operatorname{GL}(n,\mathbb{Z})$, that is,

$\operatorname{Stab}_{\operatorname{GL}(n,\mathbb{Z})}(\mathsf{M})=\\{A\in\operatorname{GL}(n,Z)\mid(\wedge^{2}A)\mathsf{M}=\mathsf{M}\\}.$

###### Proof.

Writing the group $\Lambda\leq\mathbb{F}^{\ast}$ additively, in view of
Notation 1 we have

(10)
$\lambda(\mathbf{e}_{i},\mathbf{e}_{j})={m_{1}}^{(ij)}p_{1}+\cdots+{m_{l}}^{(ij)}p_{l},\
\ \ \ \ \ \ \ \forall 1\leq i<j\leq n,$

where $m_{k}^{(ij)}\in\mathbb{Z}$. Thus $\mathsf{M}=(a_{(ij)s})$ where
$a_{(ij)s}=m_{s}^{(ij)}$. Now let

$\mathscr{M}=(m_{ij})\in\operatorname{GL}(n,\mathbb{Z})$

be such that
$\mathscr{M}^{t}\in\operatorname{\mathrm{Aut}}(\mathbb{Z}^{n},\lambda)$.
Setting

$\mathbf{e}_{j}^{\prime}=\mathscr{M}^{t}\mathbf{e}_{j}=m_{j1}\mathbf{e}_{1}+\cdots+m_{jn}\mathbf{e}_{n}$

we note that since $\lambda$ is an alternating function therefore
$\lambda(\mathbf{e}_{i}^{\prime},\mathbf{e}_{j}^{\prime})$ may be expressed
(e.g., [MP, Section 1.2]) in terms of $\lambda(\mathbf{e}_{i},\mathbf{e}_{j})$
where $i<j$ as follows:

(11)
$\lambda(\mathbf{e}_{i}^{\prime},\mathbf{e}_{j}^{\prime})=m_{ij,12}\lambda(\mathbf{e}_{1},\mathbf{e}_{2})+{m_{ij,13}}\lambda(\mathbf{e}_{1},\mathbf{e}_{3})+\cdots+m_{ij,(n-1)n}\lambda(\mathbf{e}_{n-1},\mathbf{e}_{n}).$

We also note that the coefficients appearing in the RHS of the above
expression constitute row (ij) of the matrix $\wedge^{2}(\mathscr{M})$. Since
$\mathscr{M}^{t}$ is $\lambda$-preserving by (6) we have

$\lambda(\mathbf{e}_{i}^{\prime},\mathbf{e}_{j}^{\prime})=\lambda(\mathbf{e}_{i},\mathbf{e}_{j})\
\ \ \ \ \ \ \forall 1\leq i<j\leq n.$

Expanding and comparing the coefficients of $p_{s}(s=1,\cdots,l)$ in both
sides of the last equation we get noting

(12)
$m_{ij,12}m_{s}^{(12)}+m_{ij,13}m_{s}^{(13)}+\cdots+m_{ij,(n-1)n}m_{s}^{((n-1)n)}=m_{s}^{(ij)}\
\ \ \ \ \forall s=1,\cdots,l.$

Letting $(ij)$ take values in the set

(13) $\\{(12),(13),\cdots,(n-1)n\\}$

in equation (12) and setting

$\mathsf{M}^{(s)}=(m_{s}^{(12)},m_{s}^{(13)},\cdots,m_{s}^{((n-1)n)})^{t}$

we thus get

$\wedge^{2}(\mathscr{M})\mathsf{M}^{(s)}=\mathsf{M}^{(s)}\ \ \ \ \ \ \forall
s\in 1,\cdots,l.$

As $\mathsf{M}^{(s)}$ is column $s$ of $\mathsf{M}$ (as defined in Notation 1)
it follows that:

$\wedge^{2}(\mathscr{M})\mathsf{M}=\mathsf{M}.$

Clearly the above reasoning is reversible. This establishes the assertion of
the theorem. ∎

######  Remark 2.1.

In the situation of Theorem A since the $N$ elements
$\lambda(\mathbf{e}_{i},\mathbf{e}_{j})$ ($1\leq i<j\leq n$) generate the
group $\Lambda$ therefore the columns of $\mathsf{M}$ must span a free
$\mathbb{Z}$-submodule of $\mathbb{Z}^{N}$ of rank $l$. Thus $\mathsf{M}$ must
fix a free submodule of $\mathbb{Z}^{N}$ of rank $l$.

The remainder of this section is devoted to the determination of the
automorphism group of the quantum torus defined by multiparameters satisfying
a given set of conditions. Our line of approach as suggested by Theorem A for
determining the matrices $A\in\operatorname{GL}(n,\mathbb{Z})$ whose transpose
lies in the group $\operatorname{\mathrm{Aut}}(\mathbb{Z}^{n},\lambda)$ can be
summarized as follows.

(14) $\boxed{\mbox{Relations matrix
$\mathsf{M}$}}\longrightarrow\boxed{\mbox{Submodule Fixed by
$\wedge^{2}A$}}\longrightarrow\boxed{\mbox{$\wedge^{2}A$}}\longrightarrow\boxed{A}.$

######  Example 1.

For $n=2$, the group $\Lambda$ is infinite cyclic with the generator
$q=q_{i2}$ and $\mathsf{M}$ is the $1\times 1$ matrix $1$. Clearly, the
stabilizer of $\mathsf{M}$ is $\\{1\\}$ and thus
$\operatorname{\mathrm{Aut}}(\mathbb{Z}^{2},\lambda)=\mathrm{SL}(2,\mathbb{Z})$.

######  Example 2 ([OP]).

For $n=3$ the group $\operatorname{\mathrm{Aut}}(\mathbb{Z}^{3},\lambda)$ was
discussed in [OP]. The most non-trivial situation is obtained when
$\Lambda\cong\mathbb{Z}^{2}$. As shown in the proof of [OP, Proposition 3.7]
using a change of variables we may suppose in this case that $q_{12}=1$ and
the set $\\{q_{13},q_{23}\\}$ is independent where $q_{ij}=[X_{i},X_{j}]$ and
moreover

$\operatorname{\mathrm{Aut}}(\mathbb{Z}^{3},\lambda)=\
\Biggl{\\{}\begin{pmatrix}\epsilon&0&a\\\ 0&\epsilon&b\\\
0&0&\epsilon\end{pmatrix},\ a,b\in\mathbb{Z},\ \epsilon\in\\{\pm
1\\}\Biggr{\\}}.$

Note that in this case

$\mathsf{M}:=\begin{pmatrix}0&0\\\ 1&0\\\ 0&1\end{pmatrix}\ \ \mathrm{and\
therefore\ for}\ A\in\operatorname{\mathrm{Aut}}(\mathbb{Z}^{3},\lambda),\ \
\wedge^{2}A^{t}=\begin{pmatrix}1&0&0\\\ \ast&1&0\\\ \ast&0&1\end{pmatrix}.$

We also note that

$\wedge^{2}\begin{pmatrix}\epsilon&0&a\\\ 0&\epsilon&b\\\
0&0&\epsilon\end{pmatrix}^{t}=\begin{pmatrix}1&0&0\\\ b\epsilon&1&0\\\
-a\epsilon&0&1\end{pmatrix}$

as would be expected from Theorem A.

As just seen in Example 2 for $n=3$ and $\Lambda\cong\mathbb{Z}^{2}$ a change
of variables leads to a simple set of relations for the commutators
$[X_{i},X_{j}]$. In turn this means a simple form for the relations matrix
$\mathsf{M}$ in Theorem A thus facilitating the computation of the non-scalar
automorphism group in this case.

The situation is more complex for $n=4$. For example, if $n=4$ and
$\Lambda\cong\mathbb{Z}^{5}$ one msy expect that a suitable change of
variables will allow us to assume, for example, that $q_{12}=1$. However this
is not true by the example of [MP, Section 3.11] and is still not true when
$\Lambda\cong\mathbb{Z}^{4}$ [GQ].

######  Proposition 2.2.

For $n=4$ suppose that the group $\Lambda$ is freely generated by the
commutators $[X_{1},X_{4}]$, $[X_{2},X_{3}]$,$[X_{2},X_{4}]$ and
$[X_{3},X_{4}]$, while

$[X_{1},X_{2}]=[X_{1},X_{3}]=1.$

Then

$\operatorname{\mathrm{Aut}}(\mathbb{Z}^{4},\lambda)=\Biggl{\\{}f_{b,\epsilon}:=\begin{pmatrix}\epsilon&0&0&-b\\\
0&\epsilon&0&0\\\ 0&0&\epsilon&0\\\ 0&0&0&\epsilon\end{pmatrix},\ \
b\in\mathbb{Z},\epsilon\in\\{\pm 1\\}\Biggr{\\}}.$

###### Proof.

In this case for $A^{t}\in\operatorname{\mathrm{Aut}}(\mathbb{Z}^{4},\lambda)$
using Theorem A we see that the columns $C^{(i)},\ i=1,\cdots,6$ of
$\wedge^{2}A$ satisfy $C^{(i)}=I^{(i)},\ \forall i=3,\cdots,6$ where $I^{(i)}$
stands for the $i$-th column of the identity matrix $I_{6}$. Thus only the
first two columns need to be determined. To this end we note that by [Nem] the
image of $\wedge^{2}A$ under the projection
$\rho:\operatorname{GL}(6,\bar{\mathbb{Q}})\rightarrow\operatorname{PGL}(6,\bar{\mathbb{Q}})$
preserves the projective grassmannian variety $\mathrm{Gr}(2,4)$ embedded in
the projective space $\mathrm{P}(\wedge^{2}\bar{\mathbb{Q}}^{4})$ where
$\bar{\mathbb{Q}}$ stands for the algebraic closure of $\mathbb{Q}$.

By a well-known fact [Be, Section 14.7] we know that the group of the
projective quadric $\mathrm{Gr}(2,4)$ is the image of the isometry group
$\mathrm{O}(\beta)$ in the projective group
$\operatorname{PGL}(6,\bar{\mathbb{Q}})$ where $\beta$ is the polarization of
the Plücker quadratic form

$q(\xi_{12},\xi_{13},\xi_{14},\xi_{23},\xi_{24},\xi_{34})=\xi_{12}\xi_{34}-\xi_{13}\xi_{24}+\xi_{14}\xi_{23}.$

It is easily checked that the matrix of the function $\beta$ with respect to
the basis $\mathbf{e}_{i}\wedge\mathbf{e}_{j}\ (i<j)$ is the matrix

$P:=\begin{pmatrix}0&0&0&0&0&1\\\ 0&0&0&0&-1&0\\\ 0&0&0&1&0&0\\\
0&0&1&0&0&0\\\ 0&-1&0&0&0&0\\\ 1&0&0&0&0&0\end{pmatrix}.$

Thus $(\wedge^{2}A)C\in\mathrm{O}(\beta)$ for some scalar matrix
$C\in\operatorname{GL}(6,\bar{\mathbb{Q}})$. Therefore $\wedge^{2}A$ must
satisfy

(15) $C^{2}(\wedge^{2}A)^{t}P\wedge^{2}A=P.$

As

$\det\wedge^{2}A=(\det A)^{3}=\pm 1$

it is clear from the preceding equation that $\det C=\pm 1$. Writing
$C=\operatorname{diag}(\lambda,\cdots,\lambda)$ this means that either
$\lambda$ is root of the polynomial $\Phi_{-}:=Y^{6}-1$ or
$\Phi_{+}:=Y^{6}+1=0$. Since the matrices $\wedge^{2}A$ and $P$ have integer
entries it is clear from (15) that $\lambda=\pm 1$ and thus $C=\pm I$. This
means that $\wedge^{2}A\in\mathrm{O}(\beta)$. Direct calculation using (15)
reveals that $\wedge^{2}A$ has the form

$\wedge^{2}A=\begin{pmatrix}1&0&0&0&0&0\\\ 0&1&0&0&0&0\\\ 0&0&1&0&0&0\\\
0&0&0&1&0&0\\\ b&0&0&0&1&0\\\ 0&b&0&0&0&1\end{pmatrix},\ \ \ \ \ \ \ \ \
b\in\mathbb{Z}.$

As is readily checked the last equation means noting [LoLu, Corollary 2] that

$A=\epsilon\begin{pmatrix}1&0&0&0\\\ 0&1&0&0\\\ 0&0&1&0\\\
-b&0&0&1\end{pmatrix},\ \ \ \ \ \epsilon=\pm 1.$

and thus

$\operatorname{\mathrm{Aut}}(\mathbb{Z}^{4},\lambda)=\Biggl{\\{}f_{b,\epsilon}:=\begin{pmatrix}\epsilon&0&0&b\\\
0&\epsilon&0&0\\\ 0&0&\epsilon&0\\\ 0&0&0&\epsilon\end{pmatrix},\ \ \ \
b\in\mathbb{Z},\epsilon\in\\{\pm 1\\}\Biggr{\\}}.$

∎

######  Example 3.

Following a similar approach as in Proposition 2.2 we can show that for $n=4$
assuming that the group $\Lambda$ is freely generated by the commutators
$[X_{2},X_{3}]$,$[X_{2},X_{4}]$ and $[X_{3},X_{4}]$, while

$[X_{1},X_{2}]=[X_{1},X_{3}]=[X_{1},X_{4}]=1$

the non-scalar automorphism group is given in this case by

$\operatorname{\mathrm{Aut}}(\mathbb{Z}^{n},\lambda)=\Biggl{\\{}\phi_{a,b,\epsilon}:=\begin{pmatrix}\epsilon&a&b&0\\\
0&\epsilon&0&0\\\ 0&0&\epsilon&0\\\ 0&0&0&\epsilon\end{pmatrix},\ \
a,b\in\mathbb{Z},\epsilon\in\\{\pm 1\\}\Biggr{\\}}.$

######  Remark 2.2.

In a more general situation where $\mathsf{M}$ does not have a simple form as
seen in the above examples it is easily checked that

$\operatorname{Stab}_{\operatorname{GL}(N,\mathbb{Z})}(\mathsf{M})=U^{-1}(\operatorname{Stab}_{\operatorname{GL}(N,\mathbb{Z})}(U\mathsf{M}V)U$

where $U\mathsf{M}V$ is the Smith normal form
$\operatorname{\mathcal{SNF}}(\mathsf{M})$ of $\mathsf{M}$.

######  Proposition 2.3.

For $n=4$ suppose that the multiparamters $q_{ij}\ (i,j)\neq(1,2)$ are
independent for $i<j$ and

$q_{12}=\prod_{i<j,\ (i,j)\neq(1,2)}q_{ij}.$

Then $\operatorname{\mathrm{Aut}}(\mathbb{Z}^{4},\lambda)=\\{\pm I_{4}\\}$.

###### Proof.

Clearly, in this case the relations matrix $\mathsf{M}$ has the form

$\mathsf{M}=\begin{pmatrix}1&1&1&1&1\\\ 1&0&0&0&0\\\ 0&1&0&0&0\\\ 0&0&1&0&0\\\
0&0&0&1&0\\\ 0&0&0&0&1\end{pmatrix}$

Using the smith form calculator [Ma] we find

$\operatorname{\mathcal{SNF}}(\mathsf{M})=U\mathsf{M}V=\begin{pmatrix}1&0&0&0&0\\\
0&1&0&0&0\\\ 0&0&1&0&0\\\ 0&0&0&1&0\\\ 0&0&0&0&1\\\ 0&0&0&0&0\end{pmatrix},\ \
\ \ \ \ U=\begin{pmatrix}0&1&0&0&0&0&\\\ 0&0&1&0&0&0\\\ 0&0&0&1&0&0\\\
0&0&0&0&1&0\\\ 0&0&0&0&0&1\\\ -1&1&1&1&1&1\end{pmatrix},\ \ \ V=I_{5}.$

Clearly,

$\operatorname{Stab}_{\operatorname{GL}(N,\mathbb{Z})}(U\mathsf{M}V)=\begin{pmatrix}1&0&0&0&0&x\\\
0&1&0&0&0&y\\\ 0&0&1&0&0&z\\\ 0&0&0&1&0&u\\\ 0&0&0&0&1&v\\\
0&0&0&0&0&1\end{pmatrix},\ \ \ \ \ \ x,y,z,u,v\in\mathbb{Z}.$

Using Remark 2.2 and calculating with the help of SageMath [SM] we find that

$\operatorname{Stab}_{\operatorname{GL}(N,\mathbb{Z})}(\mathsf{M})=\begin{pmatrix}1-S&S&S&S&S&S\\\
-x&x+1&x&x&x&x\\\ -y&y&y+1&y&y&y\\\ -z&z&z&z+1&z&z\\\ -u&u&u&u&u+1&u\\\
-v&v&v&v&v&v+1\end{pmatrix},\ \ \ S=u+v+x+y+z.$

For a matrix
$B\in\mathop{\bigwedge\nolimits^{\\!2}}(\operatorname{GL}(n,\mathbb{Z})\cap\operatorname{Stab}_{\operatorname{GL}(N,\mathbb{Z})}(\mathsf{M})$
by the same reasoning as in Proposition 2.2 we obtain $B^{t}PB=P$. Comparing
the first rows in both sides we obtain

$\displaystyle T-2v$ $\displaystyle=0$ $\displaystyle u+v-T$ $\displaystyle=0$
$\displaystyle v-z-T$ $\displaystyle=0$ $\displaystyle v-y-T$
$\displaystyle=0$ $\displaystyle v+x-T$ $\displaystyle=0$ $\displaystyle
u+x+y+z+T$ $\displaystyle=0$

where $T=2uv+2v^{2}-2ux+2vx+2vy+2vz+2yz$. It is easily seen that this system
has a unique solution $x=y=z=u=v=0$. This completes our proof.

∎

It was shown in [OP] that if the $\binom{n}{2}$ multiparameters $q_{ij}(1\leq
i<j\leq n)$ are independent in $\mathbb{F}^{\ast}$ then
$\operatorname{\mathrm{Aut}}(\mathbb{Z}^{n},\lambda)\cong\mathbb{Z}_{2}$. With
the help of Theorem A we show in the next proposition that the same conclusion
remains valid under a somewhat weaker hypothesis.

######  Proposition 2.4.

For a quantum torus $\mathcal{O}_{\mathfrak{q}}(\mathbb{(}F^{\ast})^{n})$ with
the $n-3$ multiparameters

$q_{12},q_{13},\cdots,q_{1(n-2)}$

set to one and the remaining multiparameters

$q_{1(n-1)},q_{1n},q_{23},q_{24},\cdots,q_{(n-1)n}$

independent in $\mathbb{F}^{\ast}$

$\operatorname{\mathrm{Aut}}(\mathbb{Z}^{n},\lambda)\cong\mathbb{Z}_{2}$

and consequently by (7) we have

$1\rightarrow\operatorname{Hom}(\mathbb{Z}^{n},{\mathbb{F}}^{\ast})\rightarrow\operatorname{\mathrm{Aut}}(\mathcal{O}_{\mathfrak{q}}({F^{\ast}}^{n}))\rightarrow\mathbb{Z}_{2}\rightarrow
1.$

###### Proof.

From the theorem hypothesis it is evident that in this case the relations
matrix $\mathsf{M}$ of Theorem A is obtained from the identity matrix
$I_{\binom{n}{2}}$ by deleting the first $n-3$ columns. Consequently, any
matrix $B\in\operatorname{Stab}(\mathsf{M})$ must coincide with
$I_{\binom{n}{2}}$ in all but the first $n-3$ columns, that is, $B$ must have
the form

(16) $B=\left(\BMAT(c){ccc}{ccc}\ast\cdots\ast&\\\
\ast\cdots\ast\BMAT(c){ccc}{ccc}\\\ 1\\\ 1\\\ \ast\cdots\ast
I_{\binom{n-1}{2}}\right).$

Suppose that $B=(b_{ij,kl})$ is induced from a matrix
$A:=(a_{uv})\in\operatorname{GL}(n,\mathbb{Z})$, that is, $B=\wedge^{2}(A)$.
The $(n-1)\times(n-1)$-submatrix $A^{+}$ of $A$ formed by the rows
$2,\cdots,n$ and columns $2,\cdots,n$ satisfies
$\wedge^{2}(A^{+})=I_{\binom{n-1}{2}}$. As is well known

$\det(\wedge^{2}(A^{+}))=\det(A^{+})^{n-2}$

and hence $A^{+}$ is also nonsingular. Applying [LoLu, Corollary 2] we obtain

(17) $A^{+}=\epsilon I_{n-1},\ \ \ \ \ \ \ \ \epsilon\in\\{-1,+1\\}.$

In view of (16) we clearly have

$b_{12,2j}=0\ \ \ \ \ \ \ \ \ \ \forall j=3,\cdots,n.$

Computing the $2\times 2$ minors of $A$ corresponding to the $(12,2j)$ entries
of $B$ with the help of (17) it is immediately seen that the $1j$ entry
$a_{1j}$ of $A$ equals to zero. Similarly, $b_{13,23}=0$ implies $a_{12}=0$.
By the same token using the evident fact that

$b_{jn,1n}=0\ \ \ \ \ \ \ \ \ \ \forall j=2,\cdots,n-1$

we obtain $a_{j1}=0$. Again $b_{(n-1)n,1(n-1)}=0$ implies $a_{n1}=0$.

It now follows that $A$ is diagonal and moreover
$A\in\operatorname{GL}(n,\mathbb{Z})$ implies

(18) $a_{11}=\epsilon^{\prime},\ \ \ \ \ \ \ \ \ \
\epsilon^{\prime}\in\\{-1,+1\\}$

But $\epsilon=\epsilon^{\prime}$ as otherwise

$b_{(1n),(1n)}=\epsilon^{\prime}\epsilon=-1$

which contradicts the form of $B$ as noted in (16). Thus $A=\epsilon I_{n}$
and the proof is complete. ∎

##  3\. Automorphisms of quantum affine spaces - preliminary facts

Automorphisms of the quantum affine spaces $\mathcal{O}_{\mathfrak{q}}$ were
considered in [AC, Ar, OP]. In our theorems to follow we shall be utilizing
the definitions, facts and results in these articles which we briefly recall
here and also add a few easy propositions.

######  Definition 3.1 (Section 1.4 of [AC]).

An automorphism $\sigma$ of $\mathcal{O}_{\mathfrak{q}}$ is called linear if
it has the form

(19) $\sigma(X_{i})=\sum_{j=1}^{n}\alpha_{ij}X_{j}\ \ \ \ \ \ \ \forall
i\in\\{1,\cdots,n\\},\ \ \ \ (\alpha_{ij})\in\operatorname{GL}(n,\mathbb{F}).$

In other words, an automorphism is linear if and only if it preserves the
degree one component of the algebra $\mathcal{O}_{\mathfrak{q}}$ with respect
to the $\mathbb{N}_{0}$-grading by total degree. The subgroup of linear
automorphisms is denoted as
$\operatorname{\mathrm{Aut}}_{\mathrm{L}}(\mathcal{O}_{\mathfrak{q}})$. For a
matrix $(\alpha_{ij})\in\operatorname{GL}(n,\mathbb{F})$ to define an
automorphism as in (19) the following necessary and sufficient conditions must
hold ([AC]):

(20)
$\alpha_{ik}\alpha_{jl}(1-q_{ij}q_{lk})=\alpha_{il}\alpha_{jk}(q_{ij}-q_{lk})\
\ \ \ \ \ \ \forall i<j,\ \ \forall k\leq l.$

The last equation may be re-written as

(21)
$\alpha_{ik}\alpha_{jl}(q_{kl}-q_{ij})=\alpha_{il}\alpha_{jk}(q_{kl}q_{ij}-1)\
\ \ \ \ \ \forall i<j,\ \ \forall k\leq l.$

Setting $k=l$ in the last equation we obtain

(22) $\alpha_{ik}\alpha_{jk}(1-q_{ij})=\alpha_{ik}\alpha_{jk}(q_{ij}-1)\ \ \ \
\ \ \forall i<j,\ \ \forall k\in\\{1,\cdots,n\\}.$

######  Observation.

Clearly, the last equation means that if
$\operatorname{\mathrm{char}}(\mathbb{F})\neq 2$ and none of the
multiparameters $q_{ij}(i<j)$ equals to unity then at least one of the
coefficients $\alpha_{ik}$ and $\alpha_{jk}$ vanishes. It is immediate that in
this case the nonsingular matrix $(\alpha_{ij})$ has exactly one nonzero entry
in each row and each column.

The next proposition is an easy consequence of the preceding observation.

######  Proposition 3.1.

Suppose that $\operatorname{\mathrm{char}}(\mathbb{F})\neq 2$ and the entries
of $\mathfrak{q}$ satisfy

($\blacklozenge$) $q_{ij}\neq 1\ \ \ \ \ \ \ \ \ \ \ \ \forall 1\leq i<j\leq
n.$

Then

(23)
$\operatorname{\mathrm{Aut}}_{\mathrm{L}}(\mathcal{O}_{\mathfrak{q}})\cong{\mathbb{F}^{\ast}}^{n}\ltimes\mathcal{P}$

for a subgroup $\mathcal{P}$ of $S_{n}$.

Besides the rather simple condition in ($\blacklozenge$ ‣ 3.1) there are other
situations where (23) holds. Indeed as shown in [OP] this is the case if the
following condition is satisfied:

($\lozenge$) $\mathrm{The\ \ localization}\ \
\widehat{\mathcal{O}}_{\mathfrak{q}}\ \ \mathrm{of}\ \
\mathcal{O}_{\mathfrak{q}}\ \ \mathrm{has\ \ center}\ \ \mathbb{F}.$

In view of the foregoing we make the following definition.

######  Definition 3.2.

Given a quantum affine space $\mathcal{O}_{\mathfrak{q}}$ a permutation
$\pi\in S_{n}$ is said to be _admissible_ if
$\pi\in\operatorname{\mathrm{Aut}}(\mathcal{O}_{\mathfrak{q}})$.

By the remark following [OP, Proposition 3.2] a permutation $\pi$ is
admissible if and only if the any of the following equivalent pair of
conditions holds for the corresponding permutation matrix $\mathfrak{p}$.

(24) $\mathfrak{p}\mathfrak{q}\mathfrak{p}^{t}=\mathfrak{q}\ \ \ \ \ \
\Leftrightarrow\ \ \ \ \ \ \mathfrak{p}\mathfrak{q}=\mathfrak{q}\mathfrak{p}.$

The next lemma and proposition record some consequences of a permutation $\pi$
being admissible.

######  Lemma 3.2.

For a given quantum affine space $\mathcal{O}_{\mathfrak{q}}$ if a permutation
$\pi$ is admissible then

  * (i)

for each $r$-cycle $(j_{1}j_{2}\cdots j_{r})$, where $r\geq 2$ in the
decomposition of $\pi$

$q_{j_{r}j_{1}}=q_{j_{1}j_{2}}=q_{j_{2}j_{3}}=\cdots=q_{j_{r-2}j_{r-1}}=q_{j_{r-1}j_{r}},\
\ \ \ \mbox{and}$

  * (ii)

for each fixed point $k$ of $\pi$ and each $r$-cycle $(j_{1}j_{2}\cdots
j_{r})$ in the the decomposition of $\pi$

$q_{j_{1}k}=q_{j_{2}k}=\cdots=q_{j_{r}k}.$

###### Proof.

(i) This follows easily by applying $\pi$ to

$X_{j_{t}}X_{j_{t\dot{+}1}}=q_{j_{t}j_{t\dot{+}1}}X_{j_{t\dot{+}1}}X_{j_{t}}$

where $\dot{+}$ denotes addition modulo $r$. As a result we obtain

$X_{j_{t\dot{+}1}}X_{j_{t\dot{+}2}}=q_{j_{t}j_{t\dot{+}1}}X_{j_{t\dot{+}2}}X_{j_{t\dot{+}1}}$

for all $t\geq 1$ running modulo $r$. Part (ii) is similar to (i). ∎

######  Proposition 3.3.

Suppose that the group
$\operatorname{\mathrm{Aut}}(\mathcal{O}_{\mathfrak{q}})$ contains a non-
trivial permutation $\pi$. Then there exists a relation

(25) $q_{ij}q_{kl}=1,\ \ \ \ \ (i,j),(k,l)\in\\{1,\cdots,n\\}^{2}$

such that $i<j$ and $k<l$.

###### Proof.

If $(j_{1}j_{2}\cdots j_{r})$ is a cycle in the decomposition of $\pi$ then by
Lemma 3.2 we have

(26)
$q_{j_{r}j_{1}}=q_{j_{1}j_{2}}=q_{j_{2}j_{3}}=\cdots=q_{j_{r-2}j_{r-1}}=q_{j_{r-1}j_{r}},$

Evidently the sequence of differences $j_{t\dot{+}1}-j_{t}$ in (26) where $t$
varies modulo $r$ in the set $\\{1,\cdots,r\\}$ has both positive as well as
negative terms. We can thus find $u,v$ ($u\neq v$) and such that
$j_{u\dot{+}1}-j_{u}$ and $j_{v\dot{+}1}-j_{v}$ have opposite signs. Without
loss of generality we may assume that $j_{u+1}-j_{u}<0$. In view of (26) we
now obtain:

$q_{j_{v}j_{v+1}}q_{j_{u+1}j_{u}}=1.$

∎

Combining the above facts immediately yields a criterion for ensuring that all
linear automorphisms of a quantum affine space result from the action of the
torus.

######  Corollary 3.4.

Suppose that $\operatorname{\mathrm{char}}(\mathbb{F})\neq 2$ and the
multiparameters satisfy:

($\clubsuit$) $q_{ij}q_{kl}\neq 1\ \mbox{for all}\ i<j\ \mbox{and}\ k<l.$

Then

$\operatorname{\mathrm{Aut}}_{\mathrm{L}}(\mathcal{O}_{\mathfrak{q}})=(\mathbb{F}^{\ast})^{n}.$

###### Proof.

The condition ($\clubsuit$ ‣ 3.4) means that $q_{ij}\neq\pm 1$ whenever $i<j$.
Thus Proposition 3.1 applies and we conclude by Proposition 3.3. ∎

In the next section we will show that the same conclusion remains valid under
a different hypothesis which allows $1$ to be included as an (off-diagonal)
entry of $\mathfrak{q}$.

##  4\. Automorphisms of quantum affine spaces–Main results

######  Proposition 4.1.

Suppose that $\operatorname{\mathrm{char}}(\mathbb{F})\neq 2$ and $n\geq 3$.
Let $\mathfrak{q}=(q_{ij})$ be a multiplicatively antisymmetric matrix whose
entries satisfy

$\displaystyle\mbox{(i)}\ q_{ij}\neq q_{kl}\ \ \ \ \ \ \ \ \forall i<j,\ \
\forall k<l,$ $\displaystyle\mbox{(ii)}\ q_{ij}q_{kl}\neq 1\ \ \ \ \ \ \ \ \
\forall i<j,\ \ \forall k<l,\ (i,j)\neq(k,l).$

The group of linear automorphisms of the quantum affine space
$\mathcal{O}_{\mathfrak{q}}(\mathbb{F}^{n})$ coincides with the torus
${(\mathbb{F}^{\ast})}^{n}$.

###### Proof.

Note that the hypothesis means that there is at most one entry
$q_{i^{\prime}j^{\prime}}$ with $i^{\prime}<j^{\prime}$ that is equal to $1$.
Similarly, there is at most one entry $q_{i^{\prime\prime}j^{\prime\prime}}$
with $i^{\prime\prime}<j^{\prime\prime}$ that is equal to $-1$. Suppose that

$A=(\alpha_{ij})\in\operatorname{GL}(n,\mathbb{F})$

induces a linear automorphism of the given quantum affine space. It suffices
to show that $A$ is a diagonal matrix. Using (22) it is easily seen that in
any column of $A$ at most two entries can be nonzero and in the case there are
two nonzero entries, these must be in the $i^{\prime}$-th and $j^{\prime}$-th
rows.

We claim that there can be at most two columns in $A$ that have two nonzero
entries. Indeed suppose that $2+s$ columns have two nonzero entries
necessarily in the $i^{\prime}$-th and $j^{\prime}$-th rows. As just noted the
remaining $n-2-s$ columns each has exactly one non-zero entry. Clearly, if
$s>0$ these $n-2-s$ nonzero entries in $A$ cannot fulfill the requirement of a
non-zero entry in each of the $n-2$ rows other than the $i^{\prime}$-th and
$j^{\prime}$-th rows. This shows that $s=0$.

Next we note that at least one of the entries $\alpha_{i^{\prime}i^{\prime}}$
and $\alpha_{j^{\prime}i^{\prime}}$ is non-zero. To see this we pick $m<p$ in
the range $1,\cdots,n$. By (21) we have noting that
$q_{i^{\prime}j^{\prime}}=1$

(27) $\displaystyle\alpha_{i^{\prime}m}\alpha_{j^{\prime}p}(q_{mp}-1)$
$\displaystyle=\alpha_{i^{\prime}p}\alpha_{j^{\prime}m}(q_{mp}-1).$

If $(m,p)\neq(i^{\prime},j^{\prime})$ then as noted above $q_{mp}\neq 1$ and
(27) reveals that the minor of $A$ corresponding to the $2\times 2$ submatrix
formed by the $i^{\prime}$-th and $j^{\prime}$-th rows and the $m$-th and
$p$-th columns is equal to zero. If
$\alpha_{i^{\prime}i^{\prime}}=\alpha_{j^{\prime}i^{\prime}}=0$ then the minor
corresponding to the $2\times 2$ submatrix $K$ defined by the $i^{\prime}$-th
and $j^{\prime}$-th rows and the $i^{\prime}$-th and $j^{\prime}$-th columns
is also equal to zero. This would mean that a row of the exterior square
$\wedge^{2}A$ of $A$ is the zero row contradicting the assumption $A$ is non-
singular. By the same token at least one of the entries
$\alpha_{i^{\prime}j^{\prime}}$ and $\alpha_{j^{\prime}j^{\prime}}$ in column
$j$ is nonzero. Moreover, the non-zero entries in the two columns, namely,
$i^{\prime}$ and $j^{\prime}$ cannot be in only one of the rows $i^{\prime}$
or $j^{\prime}$ as in this case the determinant of $K$ will be zero.

We now claim that if a column of $A$ has two non-zero entries then it must be
the $i^{\prime}$-th or the $j^{\prime}$-th column. Indeed let $h$ be a column
of $A$ having two non-zero entries. As noted above these non-zero entries of
$h$ must be rows $i^{\prime}$ and $j^{\prime}$. Thus the three columns
$i^{\prime}$, $j^{\prime}$ and $h$ have non-zero entries only in rows
$i^{\prime}$ and $j^{\prime}$. Consequently there can be at most $n-3$ nonzero
entries in columns other than $i^{\prime},j^{\prime}$ and $h$ that are
contained in the $n-2$ rows other than $i^{\prime}$ and $j^{\prime}$. But this
means there is a row with no non-zero entry contradicting the assumption that
$A$ is non-singular.

We now consider a column of $A$, say the $k$-th, that has only one non-zero
entry, say in the $u$-th row. This is possible in view of the foregoing noting
that $n\geq 3$ buy the theorem hypothesis.

Let $\alpha_{uk}$ be the unique nonzero entry in the $k$-th column. In view of
the foregoing observations it is easily seen that $\alpha_{uk}$ is also the
unique nonzero entry in the $u$-th row. We note as $n\geq 3$ (by the
hypothesis) it is always possible to choose an $l$ ($1\leq l\neq k\leq n$)
such that either

$\displaystyle(i)\ k<l\ \mathrm{and}\
(k,l)\neq(i^{\prime\prime},j^{\prime\prime});\ \mathrm{or},$ (28)
$\displaystyle(ii)\ l<k\ \mathrm{and}\
(l,k)\neq(i^{\prime\prime},j^{\prime\prime}).$

note that we do not assume that the column $l$ has only one non-zero entry.
Step 1: In case (i) we suppose that the $l$-th column has a nonzero entry,
namely, $\alpha_{vl}$ in the $v$-th row. We claim that $v>u$. Indeed if $v<u$
then (21) yields

(29)
$\alpha_{vk}\alpha_{ul}(q_{kl}-q_{vu})=\alpha_{uk}\alpha_{vl}(q_{kl}q_{vu}-1).$

By the preceding paragraph the LHS of (29) vanishes (as $\alpha_{uk}$ is the
only non-zero entry in column $k$) and (noting theorem hypothesis) the RHS is
nonzero, unless,

(30) $(k,l)=(v,u)=(i^{\prime\prime},j^{\prime\prime}).$

But by the choice of the pair $(k,l)$ (30) cannot hold. Our assertion, namely,
$u<v$ now follows. We now apply (21) to $u<v$ and $k<l$ and thus obtain

$\alpha_{uk}\alpha_{vl}(q_{kl}-q_{uv})=\alpha_{vk}\alpha_{ul}(q_{kl}q_{uv}-1).$

Clearly in the last equation the RHS is equal to zero, but the LHS can vanish
only if $q_{kl}=q_{uv}$ and by the theorem hypothesis this is possible only if
$k=u$ and $l=v$. The $k$-th and $l$-th columns of $A$ must therefore coincide
with the corresponding columns of a suitable diagonal matrix $D$.

Step 2: We now suppose that case (ii) holds in (28). As before let
$\alpha_{vl}\neq 0$ for some $v$. We claim that in this case $v<u$. For
assuming the contrary and applying (21) leads to a contradiction quite
similarly to that resulting from (29) above. Arguing as in case (i) and
applying (21) to $v<u$ and $l<k$ we obtain

$\alpha_{vl}\alpha_{uk}(q_{lk}-q_{vu})=\alpha_{ul}\alpha_{vk}(q_{lk}q_{vu}-1)$

and the last equation means that $l=v$ and $k=u$.

Conclusion of proof: Suppose that $k\neq i^{\prime\prime},j^{\prime\prime}$.
Then Steps 1 and 2 above immediately yield the fact that the matrix $A$ is a
diagonal. Otherwise if $k=i^{\prime\prime}$ (resp. $k=j^{\prime\prime}$) by
the above reasoning we obtain that the $s$-th column of $A$ has the form
$\alpha_{s}\mathbf{e}_{s}$ (where $\mathbf{e}_{s}$ stands for the $s$-th
column of the identity matrix and $\alpha_{s}\in\mathbb{F}^{\ast}$) except
possibly when $s=j^{\prime\prime}$ (resp. $s=i^{\prime\prime}$). But we may
now redefine $k$ to be any natural number in
$\\{1,\cdots,n\\}\setminus\\{i^{\prime\prime},j^{\prime\prime}\\}$ and apply
one of the Steps 1 and 2 depending on whether $k<j^{\prime\prime}$ or
$k>j^{\prime\prime}$ (resp. $k<i^{\prime\prime}$ or $k>i^{\prime\prime}$).

∎

As before, let $\\{\mathbf{e}_{1},\cdots,\mathbf{e}_{n}\\}$ denote the
standard basis in $\Gamma:=\mathbb{Z}^{n}$. Then
$\\{\mathbf{e}_{i}\wedge\mathbf{e}_{j}\mid 1\leq i<j\leq n\\}$ is a basis in
$\wedge^{2}\Gamma$. As usual, for a permutation $\pi\in S_{n}$ let
$P\in\operatorname{GL}(n,\mathbb{Z})$ denote the corresponding permutation
matrix. Clearly $S_{n}$ acts on the $\mathbb{Z}$-module $\wedge^{2}\Gamma$ via

(31) $\pi(e_{i}\wedge e_{j})=\wedge^{2}P(e_{i}\wedge e_{j})=e_{\pi(i)}\wedge
e_{\pi(j)}\ \ \ \forall\pi\in S_{n}.$

_In this action the image of $\pi$ in
$\operatorname{End}_{\mathbb{Z}}(\wedge^{2}\Gamma)$ will be denoted as
$\wedge^{2}\pi$._ By restriction we get an action of $S_{n}$ on the subset

$\bar{B}=\\{\epsilon e_{i}\wedge e_{j}\mid i<j,\ \mathrm{and}\
\epsilon\in\\{-1,1\\}\\}.$

We note the following lemma that will be used in the proof of Theorem B.

######  Lemma 4.2.

Let $\pi\in S_{n}\ (n\geq 3)$ and $\mathrm{Fix}(\wedge^{2}\pi)$ stand for the
sub-module of $\wedge^{2}\Gamma$ left fixed by the permutation $\pi$ under the
action (31). Then

$\operatorname{\mathrm{rk}}(\mathrm{Fix}(\wedge^{2}\pi))\leq\binom{n-1}{2}.$

with equality holding whenever $\pi$ a transposition.

###### Proof.

It is seen without difficulty that the submodule $\mathrm{Fix}(\wedge^{2}\pi)$
is generated by the sums of elements in the orbits of the cyclic subgroup
$C_{\pi}$ generated by $\pi$ acting on the set $\bar{B}$ as defined in the
discussion preceding the theorem. Needless to say such “orbit sums” need not
be linearly independent (over $\mathbb{Z}$) or even be non-zero. For example
the $C_{(ij)}$-orbit
$\\{\mathbf{e}_{i}\wedge\mathbf{e}_{j},-\mathbf{e}_{i}\wedge\mathbf{e}_{j}\\}$
has zero sum. Let $\mathcal{N}_{\pi}$ the number of $C_{(ij)}$-orbits in this
action on the set $\bar{B}$. By an application of the well-known formula of
Burnside it can be shown ([Si]) that the number $N_{\pi}$ is maximal when
$\pi$ is a transposition $(ij)$ and in this case

$\mathcal{N}_{(ij)}=(n-2)(n-3)+(2n-3).$

It is also clear that the orbit
$C_{(ij)}(\mathbf{e}_{i^{\prime}}\wedge\mathbf{e}_{j^{\prime}})$ has a nonzero
sum whenever $(i,j)\neq(i^{\prime},j^{\prime})$ and that this sum is the
negative of the sum of the orbit
$C_{(ij)}(-\mathbf{e}_{i^{\prime}}\wedge\mathbf{e}_{j^{\prime}})$. We thus
obtain

$\operatorname{\mathrm{rk}}\bigl{(}\mathrm{Fix}(\wedge^{2}(ij))\bigr{)}=\frac{\mathcal{N}_{(ij)}-1}{2}=\binom{n-1}{2}$

$\mathbb{Z}$-independent orbit-sums that constitute a $\mathbb{Z}$-basis for
$\mathrm{Fix}(\wedge^{2}\pi)$. In general the group $C_{\pi}$ will have some
orbits whose sum is zero. Clearly these orbits are precisely the orbits
$C_{\pi}(\mathbf{e}_{i}\wedge\mathbf{e}_{j})$ such that
$C_{\pi}(\pm\mathbf{e}_{i}\wedge\mathbf{e}_{j})\ni\mp\mathbf{e}_{i}\wedge\mathbf{e}_{j}$
and the orbits $C_{\pi}$ with non-zero sum are precisely those for which

$C_{\pi}(\mathbf{e}_{i}\wedge\mathbf{e}_{j})\cap
C_{\pi}(-\mathbf{e}_{i}\wedge\mathbf{e}_{j})=\emptyset.$

It is immediate from this that the number of independent orbit-sums for the
permutation $\pi$ are bounded above by
$\left\lfloor\frac{\mathcal{N}_{\pi}}{2}\right\rfloor$.

Evidently for any non-identity permutation $\pi\in S_{n}$ the number of fixed
points $\mathrm{Fix}(\pi)$ is bounded above by $n-2$ and this bound in
attained only by a transposition $(ij)$. Using this fact and the Burnside
formula an upper bound for $\mathcal{N}_{\pi}$ was obtained in [Si] as follows

$\displaystyle\mathcal{N}_{\pi}$ $\displaystyle=\frac{1}{\left\lvert
C_{\pi}\right\rvert}\Biggl{(}n(n-1)+\sum_{\phi\in C_{\pi},\phi\neq
1}2\binom{\mathrm{Fix}(\phi)}{2}\Biggr{)}\leq\frac{n(n-1)}{\left\lvert
C_{\pi}\right\rvert}+\frac{(\left\lvert C_{\pi}\right\rvert-1)}{\left\lvert
C_{\pi}\right\rvert}2\binom{n-2}{2}$
$\displaystyle=(n-2)(n-3)+\frac{4n-6}{\left\lvert C_{\pi}\right\rvert}.$

This yields

$\mathcal{N}_{(ij)}-\mathcal{N}(\pi)\geq(4n-6)\Biggl{(}\frac{1}{2}-\frac{1}{\left\lvert
C_{\pi}\right\rvert}\Biggr{)}$

and thus for $n\geq 2$

$\displaystyle\operatorname{\mathrm{rk}}(\mathrm{Fix}\bigl{(}\wedge^{2}(ij))\bigr{)}-\operatorname{\mathrm{rk}}(\mathrm{Fix}\bigl{(}\wedge^{2}\pi)\bigr{)}$
$\displaystyle=\frac{\mathcal{N}_{(ij)}-1}{2}-\left\lfloor\frac{\mathcal{N}_{\pi}}{2}\right\rfloor$
$\displaystyle\geq-\frac{1}{2}+\frac{\mathcal{N}_{(ij)}-\mathcal{N}_{\pi}}{2}\geq-\frac{1}{2}+(2n-3)\Biggl{(}\frac{1}{2}-\frac{1}{\left\lvert
C_{\pi}\right\rvert}\Biggr{)}\geq-\frac{1}{2}.$

As
$\frac{\mathcal{N}_{(ij)}-1}{2},\left\lfloor\frac{\mathcal{N}_{\pi}}{2}\right\rfloor\in\mathbb{Z}$
it clearly follows that

$\operatorname{\mathrm{rk}}(\mathrm{Fix}\bigl{(}\wedge^{2}(ij))\bigr{)}\geq\operatorname{\mathrm{rk}}(\mathrm{Fix}\bigl{(}\wedge^{2}\pi)\bigr{)}.$

∎

######  Theorem B.

A quantum affine space $\mathcal{O}_{\mathfrak{q}}(\mathbb{F}^{n})$ such that
the subgroup $\Lambda$ has rank no smaller than $\binom{n-1}{2}+1$ satisfies

$\operatorname{\mathrm{Aut}}(\mathcal{O}_{\mathfrak{q}}(\mathbb{F}^{n}))=(\mathbb{F}^{\ast})^{n}.$

###### Proof.

As usual we write
$\mathcal{O}_{\mathfrak{q}}=\mathcal{O}_{\mathfrak{q}}(\mathbb{F}^{n})$. The
assumption concerning the rank of the group $\Lambda$ means that the
corresponding quantum torus $\widehat{\mathcal{O}}_{\mathfrak{q}}$ (arising
from localization) of the given quantum affine space
$\mathcal{O}_{\mathfrak{q}}(\mathbb{F}^{n})$ has center $\mathcal{Z}$ no
bigger than the ground field $\mathbb{F}$. Indeed viewing the latter algebra
as a crossed product $\mathbb{F}\ast\Gamma$ ($\Gamma=\mathbb{Z}^{n}$) as in
[OP] it is easily seen that for any subgroup $\Gamma^{\prime}$ of $\Gamma$ of
finite index the corresponding $\lambda$-group $\Lambda^{\prime}$ of the sub-
quantum torus $\mathbb{F}\ast\Gamma^{\prime}$ has the same rank as that of the
group $\Lambda$.

It is well known (e.g., [OP]) that the center of a quantum torus is a
generated by monomials. Given a central monomial
$\mathbf{z}:=\mathbf{X}^{\mathbf{m}}$ in the algebra
$\widehat{\mathcal{O}}_{\mathfrak{q}}$ we can clearly extend the set
$\\{\mathbf{z}\\}$ to a set of $n$ monomials which together with their
inverses generate a subalgebra of the form $\mathbb{F}\ast\Gamma^{\prime}$ as
described in the previous paragraph. But the central monomial $\mathbf{z}$
evidently reduces $\operatorname{\mathrm{rk}}(\Lambda^{\prime})$ by $n-1$
implying $\operatorname{\mathrm{rk}}(\Lambda)\leq\binom{n-1}{2}$ and thus
contradicting the theorem hypothesis. We also note that by the theorem of
Section 1.3 of [MP] the algebra $\mathcal{O}_{\mathfrak{q}}$ is simple.

Let $\pi$ be a permutation of the generators $X_{1},\cdots X_{n}$ of the
generators of the algebra $\widehat{\mathcal{O}}_{\mathfrak{q}}$. To show the
assertion in the theorem it suffices in view of [OP, Propositon 3.2] to show
that if $\pi\in\operatorname{\mathrm{Aut}}(\mathcal{O}_{\mathfrak{q}})$ then
$\pi=\mathrm{id}$. To this end we note that by [OP, Proposition 1.5] the
permutation $\pi$ extends to an automorphism of the quantum torus
$\widehat{\mathcal{O}}_{\mathfrak{q}}$. Moreover this extension induces on
$\Gamma$ the automorphism which is given by the permutation matrix $P$
corresponding to $\pi$. By Remark 2.1

$\mathrm{Fix}\wedge^{2}\pi\geq\binom{n-1}{2}+1$

contradicting Lemma 4.2 ∎

###  4.1. Examples

######  Example 4.

In the situation of Example 2 of Section 2 we consider the quantum affine
space defined by the same matrix $\mathfrak{q}$ of multiparameters. In this
case the $\lambda$-group has rank $2$ and so Theorem B applies. In particular
this quantum space is a simple ring and whose automorphisms (permutations of
generators) lift to automorphisms of the corresponding quantum torus
$\widehat{\mathcal{O}}_{\mathfrak{q}}$ . As seen in the same example referred
to above these automorphisms induce on $\Gamma$ automorphisms having the form
$\begin{pmatrix}\epsilon&0&a\\\ 0&\epsilon&b\\\ 0&0&\epsilon\end{pmatrix}$ But
the only permutation matrix which has this form is the identity. In other
words the assertion of Theorem B holds true in this case.

######  Example 5.

Consider the multiplicatively antisymmetric matrices $\mathfrak{q}\in
M_{5}(\mathbb{F})$ defined by the $5\times 5$ grid of Figure 1

Figure 1. Multiplicatively antisymmetric matrix $\mathfrak{q}$ with circulant
symmetry

where the boxes with the same pattern indicate that the corresponding entries
of $\mathfrak{q}$ are equal. Furthermore, the (equal) entries in the boxes
with northeast hatching are inverses of the (equal) entries in the boxes with
northwest hatching. Similarly for the boxes with vertical and horizontal
hatching. The unhatched boxes correspond to $1\in\mathbb{F}$.

Thus the matrices arising in this way are multiplicatively antisymmetric as
well as circulant. Evidently, the rank of the $\lambda$-group in this case is
at most $2$.

Using (24) it is easily seen that the the 5-cycle $(12345)$ is an automorphism
of $\mathcal{O}_{\mathfrak{q}}$ which clearly lifts to an automorphism of the
corresponding quantum torus $\widehat{\mathcal{O}}_{\mathfrak{q}}$ (and thus
induces an automorphism of $\mathbb{Z}^{5}$). As
$\operatorname{\mathrm{rk}}(\mathrm{Fix}(\wedge^{2}(12345))=2$ this is
agreement with Remark 2.1.

## References

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Note that $|\theta_{1}|$ in Eq. (VI.5) is not bounded for large $\zeta=t+x$.
This indicates that the perturbation with respect to $\alpha$ becomes invalid
at $k_{0}=1$, leading to the discrepancy. The divergences at $k_{0}=1$ can be
regarded as the resonance of the inhomogeneous linearized differential
equation [53], and one can expect that the SOC has non-perturbative effects
around $k_{0}=1$.

To understand the origin of the non-perturbative effect of $\alpha$, let us
consider a set of solutions of Eq. (III.1) that depends on $x$ and $t$ only
through $x-vt$. For the later comparison to a special solution developed in
Sec. III, $\theta_{0}+\theta_{1}+{\cal O}(\alpha^{2})$, let $v$ to be $k_{0}$,

$\displaystyle\theta(x,t)=\theta(x-k_{0}t),\quad\theta^{\prime}\equiv\partial_{x}\theta=-\frac{1}{k_{0}}\partial_{t}\theta.$
(VI.6)

Here, the prime denotes an $x$-derivative. Eq. (III.1) effectively becomes an
ordinary differential equation,

$\displaystyle(k_{0}^{2}-1)\theta^{\prime\prime}=-\alpha\theta^{\prime\prime}\mathrm{cos}(2\theta)+\alpha\theta^{\prime
2}\mathrm{sin}(2\theta).$ (VI.7)

To solve this equation, use its analogy to one-dimensional classical
mechanics, where the phase $\theta(x)$ as a function of $x$ corresponds to a
one-dimensional coordinate as a function of time. The classical mechanics for
the one-dimensional coordinate has a Lagrangian whose variation gives Eq.
(VI.7) as a classical EOM,

$\displaystyle L_{\rm 1D}=\frac{1}{2}(k_{0}^{2}-1)\theta^{\prime
2}+\frac{\alpha}{2}\theta^{\prime 2}\mathrm{cos}(2\theta).$ (VI.8)

The classical mechanics has a canonical momentum conjugate to the coordinate,

$\displaystyle\pi=\frac{\partial L_{\rm
1D}}{\partial\theta^{\prime}}=(k_{0}^{2}-1)\theta^{\prime}+\alpha\theta^{\prime}\mathrm{cos}(2\theta),$
(VI.9)

as well as a conserved Hamiltonian,

$\displaystyle H_{\rm 1D}=\pi\theta^{\prime}-L_{\rm
1D}=\frac{1}{2}(k_{0}^{2}-1)\theta^{\prime 2}+\frac{\alpha}{2}\theta^{\prime
2}\mathrm{cos}(2\theta).$ (VI.10)

Utilizing the $x$-independence (“time”-independence) of $H_{\rm 1D}$, we can
solve the EOM from Eq. (VI.10),

$\displaystyle\frac{d\theta}{dx}=\pm\sqrt{\frac{2H_{\rm
1D}}{k_{0}^{2}-1+\alpha\mathrm{cos}(2\theta)}}.$ (VI.11)

Its formal solution is given by

$\displaystyle\pm(x-x_{0})=\int_{\theta(x_{0})}^{\theta(x)}\sqrt{\frac{k_{0}^{2}-1+\alpha\mathrm{cos}(2\theta)}{2H_{\rm
1D}}}d\theta.$ (VI.12)

With Eq. (VI.6), we get a set of (1+1)-dimensional solutions,

$\displaystyle\pm(x-k_{0}t-C_{0})=\int_{\theta(x_{0},t_{0})}^{\theta(x,t)}\sqrt{\frac{k_{0}^{2}-1+\alpha\mathrm{cos}(2\theta)}{2H_{\rm
1D}}}d\theta,$ (VI.13)

where $C_{0}=x_{0}-k_{0}t_{0}$.

Eq. (VI.13) is inclusive of those perturbative solutions in Sec. III that
depend on $x$ and $t$ only through $x-k_{0}t$, i.e. $\theta_{0}+\theta_{1}$
given by Eqs. (III.8,III.13). Namely, when $\alpha\ll|k_{0}^{2}-1|$, we can
apply an expansion in $\alpha$,

$\displaystyle\pm(x-k_{0}t-C_{0})=\int_{\theta(x_{0},t_{0})}^{\theta(x,t)}\sqrt{\frac{k_{0}^{2}-1}{2H_{\rm
1D}}}[1+\frac{\alpha}{2(k_{0}^{2}-1)}\mathrm{cos}(2\theta)]d\theta+\mathcal{O}(\alpha^{2}).$
(VI.14)

From this, we obtain

$\displaystyle\theta(x,t)=$ $\displaystyle\pm\sqrt{\frac{2H_{\rm
1D}}{k_{0}^{2}-1}}(x-k_{0}t-C_{0}^{\prime})-\frac{\alpha}{4(k_{0}^{2}-1)}\mathrm{sin}(2\theta)+\mathcal{O}(\alpha^{2})$
$\displaystyle=$
$\displaystyle\theta_{0}-\frac{\alpha}{4(k_{0}^{2}-1)}\mathrm{sin}(2\theta_{0})+\mathcal{O}(\alpha^{2}),$
(VI.15)

where

$\displaystyle\theta_{0}=\pm\sqrt{\frac{2H_{\rm
1D}}{k_{0}^{2}-1}}(x-k_{0}t-C_{0}^{\prime}),$ (VI.16) $\displaystyle
C_{0}^{\prime}=C_{0}\mp\sqrt{\frac{k_{0}^{2}-1}{2H_{\rm
1D}}}\\{\theta(x_{0},t_{0})+\frac{\alpha}{4(k_{0}^{2}-1)}\mathrm{sin}[2\theta(x_{0},t_{0})]\\}.$
(VI.17)

Note that due to the absence of $\theta_{2}(x,t)$ in its $\alpha$-expansion,
Eq. (VI.1) does not satisfy the boundary conditions in the spin-injection
model in general.

Nonetheless, Eq. (VI.13) is still useful to see that the expansion in $\alpha$
is invalid when $|k_{0}^{2}-1|<\mathcal{O}(\alpha)$. Take $k_{0}=1$ in Eq.
(VI.13) as an example. Thereby, the sign of $\mathrm{cos}(2\theta)$ is
conserved from Eq. (VI.10). Then, the integral of Eq. (VI.13) gives

$\displaystyle\pm\sqrt{\frac{2H_{\rm
1D}}{\alpha}}(x-k_{0}t-C_{0})=E(\theta(x,t),2)-E(\theta(x_{0},t_{0}),2),$
(VI.18)

where $E(\theta,m)$ is the second-kind elliptic integral. This special
solution can be made independent from $\alpha$ because $\alpha$ can be
absorbed into another parameter $H_{\rm 1D}$. The $\alpha$-independence as
well as the conserved sign of $\cos(2\theta)$ are not consistent with
$\theta_{0}(x,t)+\theta_{1}(x,t)$ in Eqs. (III.8,III.13). This suggests that
the expansion in $\alpha$ is invalid, particularly at $k_{0}=1$.

###  VI.2 $k_{0}=0$ (straight geometry)

Another special parameter point is $k_{0}=0$. $k_{0}=0$ is a limit where
$\chi$ in Eq. (III.21) is much smaller than
$\chi^{\prime}\sqrt{\frac{D_{s}}{T_{1}^{\prime}}}$ and $\beta_{t}$, and
$k_{c}$ goes to zero for any $c$;

$\displaystyle
k_{c}\equiv\frac{\chi}{\chi^{\prime}}\sqrt{\frac{1}{D_{s}(\frac{1}{T^{\prime}_{1}}+ic)}}+\frac{\chi}{\beta}\rightarrow
0.$ (VI.19)

We first consider $k_{0}=0$ in the straight geometry ($r^{-1}=0$).
$\theta_{0}$ at $k_{0}=0$ has no time dependence in Eq. (III.8), so that the
phase $F_{0}$ in Eq. (III.8) cannot be absorbed into the time. Meanwhile,
$s(x=L-)=0$ because $k_{c}=0$ for any $c$, and
$s(x)\equiv\partial_{t}\theta=0$ can be always satisfied by a time-independent
$\theta$. Thus, we have only to make Eqs. (III.8,III.10) with an additional
$F_{0}$ to satisfy the other boundary condition, $j^{s}_{x}(x=0+)=j_{0}$.
Firstly, let us choose $\theta_{0}=-j_{0}x+F_{0}$ that satisfies the boundary
condition. The substitution into Eq. (III.11) gives
$\theta_{1}=\frac{\alpha}{4}\sin(-2j_{0}x+F_{0})$. To satisfy the boundary
condition up to the 1st order in $\alpha$ (see also Eq. (III.4)),

$j^{s}_{x}(x=0+)=j_{0}-\frac{j_{0}\alpha}{2}\mathrm{cos}(2F_{0})-\partial_{x}\theta_{2}+\mathcal{O}(\alpha^{2})=j_{0},$
(VI.20)

we have

$\theta_{2}(x)=-\frac{j_{0}\alpha}{2}\mathrm{cos}(2F_{0})x.$ (VI.21)

Here, $|\theta_{2}(x)|$ for large $x$ is not bounded, where the perturbation
in $\alpha$ breaks down. An alternative way to get a consistent perturbative
solution is to require $\theta_{2}(x)=0$ and take
$\theta_{0}(x)=-\tilde{j}_{0}x+F_{0}$ where $\tilde{j}_{0}\neq j_{0}$. Then we
have

$\displaystyle
j^{s}_{x}(x=0+)=\tilde{j}_{0}-\frac{\tilde{j}_{0}\alpha}{2}\mathrm{cos}(2F_{0})-\partial_{x}\theta_{2}+\mathcal{O}(\alpha^{2})=j_{0}.$
(VI.22)

From this, we get

$\displaystyle\tilde{j}_{0}=j_{0}[1-\frac{\alpha}{2}\mathrm{cos}(2F_{0})]^{-1}=j_{0}[1+\frac{\alpha}{2}\mathrm{cos}(2F_{0})]+\mathcal{O}(\alpha^{2}).$
(VI.23)

At ${\cal O}(\alpha)$, this solution is equivalent to absorbing
$\theta_{2}(x)$ in Eq. (VI.21) into $\theta_{0}(x)$. Note also that from Eq.
(IV), $\theta_{2}(x)=0$ implies that the steady (but not uniform) current
without spin accumulation has no energy dissipation. However, a higher-order
calculation in $\alpha$ suggests that $\delta J$ can be non-zero and negative.
This is because $\delta\theta_{1}$ and $\theta_{1}$ can share identical
Fourier components (see Eq. (B) in Appendix B).

###  VI.3 $k_{0}=0$ (circular geometry)

Let us next consider the circular geometry with $k_{0}=0$. In a similar way as
in the previous section, take $\theta_{2}(\ell)=0$ and
$\theta_{0}(\ell)=-\tilde{j}_{0}\ell+F_{0}$ with $\tilde{j}_{0}\neq j_{0}$.
Then, similar to Eq. (20) in the main text, we obtain

$\displaystyle-\partial_{\ell}^{2}\theta_{1}=-\alpha\tilde{j}_{0}(\tilde{j}_{0}+\frac{2}{r})\mathrm{sin}(-2\tilde{j}_{0}\ell-\frac{2}{r}\ell+2F_{0}).$
(VI.24)

When $\tilde{j}_{0}\neq-\frac{1}{r}$, Eq. (VI.24) leads to

$\displaystyle\theta_{1}(\ell)=\frac{\alpha\tilde{j}_{0}(\tilde{j}_{0}+\frac{2}{r})}{4(\tilde{j}_{0}+\frac{1}{r})^{2}}\mathrm{sin}(2\tilde{j}_{0}\ell+\frac{2}{r}\ell-2F_{0}).$
(VI.25)

Note that when $\tilde{j}_{0}=-\frac{2}{r}$, $\theta_{1}(\ell)=0$, and
$\theta(\ell)=\frac{2\ell}{r}+F_{0}$ ($F_{0}\in\mathbb{R}$) becomes an “exact”
solution of Eq. (III.2). The solution is not exact in the presence of higher-
order derivatives in the model Eq. (5) in the main text.

From Eq. (VI.25) and $\theta_{2}=0$, the boundary condition at $\ell=0+$ reads

$\displaystyle j^{s}_{\ell}(\ell=0+)=j_{0}=$
$\displaystyle\tilde{j}_{0}+\alpha\tilde{j}_{0}\mathrm{cos}(2F_{0})-\frac{\alpha\tilde{j}_{0}(\tilde{j}_{0}+\frac{2}{r})}{2(\tilde{j}_{0}+\frac{1}{r})}\mathrm{cos}(2F_{0})+\mathcal{O}(\alpha^{2})$
$\displaystyle=$
$\displaystyle\tilde{j}_{0}+\frac{\alpha\tilde{j}_{0}^{2}}{2(\tilde{j}_{0}+\frac{1}{r})}\mathrm{cos}(2F_{0})+\mathcal{O}(\alpha^{2}).$
(VI.26)

So we have

$\displaystyle\tilde{j}_{0}=j_{0}-\frac{\alpha\tilde{j}_{0}^{2}}{2(\tilde{j}_{0}+\frac{1}{r})}\mathrm{cos}(2F_{0})+\mathcal{O}(\alpha^{2}).$
(VI.27)

As in the previous case, the steady current without spin accumulation has an
energy dissipation.

####  VI.3.1 $k_{0}=0$, $\tilde{j}_{0}=-\frac{1}{r}$ (circular geometry)

When $\tilde{j}_{0}=-\frac{1}{r}$, Eq. (VI.24) leads to

$\displaystyle\theta_{1}=-\frac{\ell^{2}}{2}\alpha\tilde{j}_{0}(\tilde{j}_{0}+\frac{2}{r})\mathrm{sin}(2F_{0})=\frac{\ell^{2}\alpha}{2r^{2}}\mathrm{sin}(2F_{0}).$
(VI.28)

When $F_{0}\neq\frac{N}{2}\pi$, the perturbation of $\alpha$ becomes invalid
for larger $\ell$, where $|\theta_{1}(\ell)|$ is not bounded. When
$F_{0}=\frac{N}{2}\pi$, $\theta_{1}(\ell)=0$, and the boundary condition
requires

$\displaystyle
j_{0}=\tilde{j}_{0}[1+\alpha(-1)^{N}]=-\frac{1}{r}[1+\alpha(-1)^{N}].$ (VI.29)

This solution ($\theta(\ell,t)=\frac{\ell}{r}+\frac{N\pi}{2}$) can exist only
at $k_{0}=0$, while it is not continuously connected to a solution at finite
small $k_{0}$ (see below).

####  VI.3.2 $k_{0}\rightarrow 0$, $j_{0}=-\frac{1}{r}+\mathcal{O}(\alpha)$
(circular geometry)

To see that the solution Eq. (VI.29) is not continuously connected to a
perturbative solution at finite small $k_{0}$, let us keep a finite $k_{0}$
and choose $j_{0}=-\frac{1}{r}+\mathcal{O}(\alpha)$ in Sec. III;
$\theta_{0}(\ell)=\frac{1}{r}(k_{0}t-\ell)+\mathcal{O}(\alpha)$. In the
perturbation theory, we should neglect $\mathcal{O}(\alpha^{2})$ contributions
to $\theta_{1}$, so $\theta_{1}$ is not affected by the $\mathcal{O}(\alpha)$
component in $j_{0}$. Then Eq. (20) in the main text becomes

$\displaystyle\partial_{t}^{2}\theta_{1}-\partial_{\ell}^{2}\theta_{1}=-\frac{\alpha}{r^{2}}\mathrm{sin}(-\frac{2k_{0}}{r}t),$
(VI.30)

which leads to a solution,

$\displaystyle\theta_{1}=-\frac{\alpha}{4k_{0}^{2}}\mathrm{sin}(\frac{2k_{0}}{r}t).$
(VI.31)

When $k_{0}\rightarrow 0$, the solution has the divergence. Physically
speaking, when $k_{0}\rightarrow 0$, $F_{0}$ changes slowly with respect to
time, so we cannot fix the phase $F_{0}$ in Eq. (VI.28).

##  Appendix A Derivation of the Landau criterion

In this section, we use the same framework as in Sec. IV and derive the Landau
criterion. A similar argument can be found in Ref. [13], while the derivation
here is more formal than Ref. [13]. We begin with a one-dimensional superfluid
model,

$\displaystyle\tilde{\mathcal{L}}_{\phi}=i\hbar\phi^{\dagger}\partial_{t}\phi-\frac{\hbar^{2}}{2m}(\partial_{x}\phi^{\dagger})(\partial_{x}\phi)-\frac{U}{2}(\phi^{\dagger}\phi)^{2}+\mu\phi^{\dagger}\phi,$
(A.1)

with its classical equation of motion,

$\displaystyle
i\hbar\partial_{t}\phi=(-\frac{\hbar^{2}}{2m}\partial_{x}^{2}+U\phi^{\dagger}\phi-\mu)\phi.$
(A.2)

Thanks to the Galilean covariance of the Lagrangian Eq. (A.1), our discussion
will be easier than Sec. IV, and we do not need to expand the variation of the
motion as in Sec. IV. Namely, using the Galilean covariance, we can directly
obtain two motions that are close to each other, and compare their energies
from the corresponding Hamiltonian,

$\displaystyle\tilde{H}_{\phi}[\phi]=\int
dx[\frac{\hbar^{2}}{2m}(\partial_{x}\phi^{\dagger})(\partial_{x}\phi)+\frac{U}{2}(\phi^{\dagger}\phi)^{2}-\mu\phi^{\dagger}\psi].$
(A.3)

Consider a steady flow $\phi_{0}(x,t)$,

$\displaystyle\phi_{0}(x,t)=\sqrt{\rho_{0}}\exp[\frac{i}{\hbar}(mvx-\frac{mv^{2}}{2}t)].$
(A.4)

Let us assume that the following motion $\phi(x,t)$, as well as
$\phi_{0}(x,t)$, satisfies the EOM, Eq. (A.2),

$\displaystyle\phi(x,t)=\phi^{\prime}(x^{\prime},t^{\prime})\exp[\frac{i}{\hbar}(mvx-\frac{mv^{2}}{2}t)]=\phi^{\prime}(x-vt,t)\exp[\frac{i}{\hbar}(mvx-\frac{mv^{2}}{2}t)],$
(A.5)

with $x^{\prime}=x-vt$ and $t^{\prime}=t$. Here
$|\phi^{\prime}(x^{\prime},t^{\prime})-\sqrt{\rho_{0}}|\ll\sqrt{\rho_{0}}$ and
$\phi(x,t)$ is close to $\phi_{0}(x,t)$. Using a Galilean transformation,

$\displaystyle x^{\prime}=x-vt,\quad t^{\prime}=t,$ (A.6)
$\displaystyle\partial_{x}=\partial_{x^{\prime}},\quad\partial_{t}=\partial_{t^{\prime}}-v\partial_{x},$
(A.7)

we can see that $\phi^{\prime}(x^{\prime},t^{\prime})$ must satisfy a similar
equation as Eq. (A.2),

$\displaystyle(i\hbar\partial_{t}+\frac{1}{2}mv^{2})\phi^{\prime}(x^{\prime},t^{\prime})=[\frac{1}{2m}(-i\hbar\partial_{x}+mv)^{2}+U\phi^{\prime\dagger}\phi^{\prime}-\mu]\phi^{\prime}(x^{\prime},t^{\prime}),$
(A.8) $\displaystyle
i\hbar\partial_{t^{\prime}}\phi^{\prime}(x^{\prime},t^{\prime})=(-\frac{\hbar^{2}}{2m}\partial_{x^{\prime}}^{2}+U\phi^{\prime\dagger}\phi^{\prime}-\mu)\phi^{\prime}(x^{\prime},t^{\prime}).$
(A.9)

Now we compare the energies of $\phi(x,t)$ and $\phi_{0}(x,t)$. The energy of
$\phi_{0}(x,t)$ is,

$\displaystyle\tilde{H}_{\phi}[\phi_{0}]=\int
dx[\frac{\hbar^{2}}{2m}(\partial_{x}\phi_{0}^{\dagger})(\partial_{x}\phi_{0})+\frac{U}{2}(\phi_{0}^{\dagger}\phi_{0})^{2}-\mu\phi_{0}^{\dagger}\phi_{0}]=E_{0}+\frac{1}{2}mv^{2}Q_{0},$
(A.10)

where

$\displaystyle E_{0}=\int
dx(\frac{U}{2}\rho_{0}^{2}-\mu\rho_{0})=-\frac{\mu}{2}\int dx\rho_{0},\quad
Q_{0}=\int dx\rho_{0}.$ (A.11)

The energy of $\phi(x,t)$ is,

$\displaystyle\tilde{H}_{\phi}[\phi]=$ $\displaystyle\int
dx[\frac{\hbar^{2}}{2m}(\partial_{x}\phi^{\dagger})(\partial_{x}\phi)+\frac{U}{2}(\phi^{\dagger}\phi)^{2}-\mu\phi^{\dagger}\phi]$
$\displaystyle=$ $\displaystyle\int
dx\phi^{\prime\dagger}(x-vt,t)[\frac{1}{2m}(-i\hbar\partial_{x}+mv)^{2}+\frac{U}{2}\phi^{\prime\dagger}\phi^{\prime}-\mu]\phi^{\prime}(x-vt,t)$
$\displaystyle=$ $\displaystyle\int
dx\phi^{\prime\dagger}(x,t)[\frac{1}{2m}(-i\hbar\partial_{x}+mv)^{2}+\frac{U}{2}\phi^{\prime\dagger}\phi^{\prime}-\mu]\phi^{\prime}(x,t)$
$\displaystyle=$ $\displaystyle\int
dx\phi^{\prime\dagger}(x,t)[-\frac{\hbar^{2}}{2m}\partial_{x}^{2}-iv\hbar\partial_{x}+\frac{1}{2}mv^{2}+\frac{U}{2}\phi^{\prime\dagger}\phi^{\prime}-\mu]\phi^{\prime}(x,t)$
$\displaystyle=$
$\displaystyle\tilde{H}_{\phi}[\phi^{\prime}]+vP_{\phi}[\phi^{\prime}]+\frac{1}{2}mv^{2}Q_{\phi}[\phi^{\prime}],$
(A.12)

where

$\displaystyle P_{\phi}[\phi^{\prime}]=\int
dx\phi^{\prime\dagger}(x,t)(-i\hbar\partial_{x})\phi^{\prime}(x,t),\quad
Q_{\phi}[\phi^{\prime}]=\int dx\phi^{\prime\dagger}(x,t)\phi^{\prime}(x,t).$
(A.13)

Thus, the energy difference between $\phi(x,t)$ and $\phi_{0}(x,t)$ is

$\displaystyle\Delta
E_{\phi}[\phi,\phi_{0}]=\tilde{H}_{\phi}[\phi]-\tilde{H}_{\phi}[\phi_{0}]=(\tilde{H}_{\phi}[\phi^{\prime}]-E_{0})+vP_{\phi}[\phi^{\prime}]+\frac{1}{2}mv^{2}(Q_{\phi}[\phi^{\prime}]-Q_{0}).$
(A.14)

For $\phi^{\prime}$ that satisfies the equation of motion Eq. (A.9), energy
$H_{\phi}[\phi^{\prime}]$, momentum $P_{\phi}[\phi^{\prime}]$, and U(1) charge
$Q_{\phi}[\phi^{\prime}]$ must be all conserved. This is because from Eq.
(A.9), $\phi^{\prime}(x,t)$ is a solution of Eq. (A.2).

The average velocity of $\phi(x,t)$ is not small, while $\phi^{\prime}(x,t)$
can be assumed at the near-equilibrium limit [44]. In this limit, the EOM of
$\phi^{\prime}=\sqrt{\rho_{0}+\delta\rho^{\prime}}e^{i\theta^{\prime}}$ can be
described by a wave equation of $\delta\rho^{\prime}$ and $\theta^{\prime}$,

$\displaystyle\partial_{t}^{2}\theta^{\prime}(x,t)-\frac{\rho_{0}U}{m}\partial_{x}^{2}\theta^{\prime}(x,t)=0,$
(A.15)
$\displaystyle\delta\rho^{\prime}(x,t)=-\frac{\hbar}{U}\partial_{t}\theta^{\prime}(x,t).$
(A.16)

Thereby, $\theta^{\prime}(x,t)$ is given by a superposition of oscillations,

$\displaystyle\theta^{\prime}(x,t)=\frac{1}{\sqrt{L}}\sum_{q}[f_{q}e^{iq(x-v_{c}t)}+f_{q}^{\prime}e^{iq(x+v_{c}t)}],\quad
v_{c}=\sqrt{\frac{\rho_{0}U}{m}}.$ (A.17)

In the near-equilibirum limit of $\phi^{\prime}$, we evaluate the energy
difference in the leading order in small $f_{q}$ and $f^{\prime}_{q}$,

$\displaystyle\tilde{H}_{\phi}[\phi^{\prime}]-E_{0}=$ $\displaystyle\int
dx[\frac{\hbar^{2}\rho_{0}}{2m}(\partial_{x}\theta^{\prime})^{2}+\frac{U}{2}(\delta\rho^{\prime})^{2}]=\int
dx[\frac{\hbar^{2}\rho_{0}}{2m}(\partial_{x}\theta^{\prime})^{2}+\frac{\hbar^{2}}{2U}(\partial_{t}\theta^{\prime})^{2}]$
$\displaystyle=$
$\displaystyle\frac{\hbar^{2}\rho_{0}}{m}\sum_{q}q^{2}(|f_{q}|^{2}+|f^{\prime}_{q}|^{2}),$
(A.18) $\displaystyle P_{\phi}[\phi^{\prime}]=\int
dx\hbar\delta\rho^{\prime}(\partial_{x}\theta^{\prime})=-\frac{\hbar^{2}}{U}\int
dx(\partial_{t}\theta^{\prime})(\partial_{x}\theta^{\prime})=\frac{\hbar^{2}\rho_{0}}{mv_{c}}\sum_{q}q^{2}(|f_{q}|^{2}-|f^{\prime}_{q}|^{2}),$
(A.19) $\displaystyle Q_{\phi}[\phi^{\prime}]-Q_{0}=\int
dx\delta\rho^{\prime}=0.$ (A.20)

Taking Eqs. (A-A.20) into Eq. (A.14), we have

$\displaystyle\Delta
E_{\phi}=\frac{\hbar^{2}\rho_{0}}{m}\sum_{q}q^{2}[(1+\frac{v}{v_{c}})|f_{q}|^{2}+(1-\frac{v}{v_{c}})|f_{q}^{\prime}|^{2}].$
(A.21)

To make $\Delta E_{\phi}\geq 0$ for any (small) $f_{q}$ and $f_{q}^{\prime}$,
we obtain the Landau criterion,

$\displaystyle|v|\leq v_{c}=\sqrt{\frac{\rho_{0}U}{m}}.$ (A.22)

##  Appendix B  Local deformations of a classical solution of the EOM

The superfluid state with a finite supercurrent is characterized by the
solution $\theta(x,t)$ of the classical EOM in the one-dimensional spin-
injection model (e.g. with the straight geometry). In Sec. IV, we introduced
its local deformation $\theta(x,t)+\delta\theta(x,t)$ as another solution of
the EOM with different boundary conditions. We regarded that
$\delta\theta(x,t)$, as well as $\theta(x,t)$, can be determined
perturbatively in the SOC ($\alpha$). At the zeroth order in SOC,
$\theta+\delta\theta$, as well as $\theta$, is a solution of
$\partial^{2}_{t}\theta-\partial^{2}_{x}\theta=0$, and so is $\delta\theta$.
Since $\delta\theta(x,t)$ is a local deformation and its spacetime derivatives
should not contain any uniform components in space, the zeroth order of
$\delta\theta(x,t)$ must be given by Eq. (IV.6);

$\displaystyle\delta\theta_{0}(x,t)=\frac{1}{\sqrt{L}}\sum_{q}\big{[}\delta
d_{q}e^{iq(x-t)}+\delta d^{\prime}_{q}e^{iq(x+t)}\big{]}.$ (B.1)

In this appendix, for a given form of Eq. (IV.6) as the zeroth order, we will
show how to determine the first order of $\delta\theta(x,t)$;

$\displaystyle\delta\theta(x,t)=\delta\theta_{0}(x,t)+\delta\theta_{1}(x,t)+{\cal
O}(\alpha^{2}).$ (B.2)

We first give a general framework to determine $\delta\theta$.
$\theta+\delta\theta$, as well as $\theta$, is a local minimum of the action,
$S=\int d^{3}r\mathcal{L}\equiv\int dtd^{2}r\mathcal{L}$, and $\delta\theta$
is infinitesimally small. Thus, we take a $\delta\theta$-variation of $S$;

$\displaystyle S[\theta]\equiv
S_{xx}[\theta]+S_{yy}[\theta]+S_{xy}[\theta]+\frac{1}{2}\int
d^{3}r(\partial_{t}\theta)^{2},$ (B.3) $\displaystyle
S_{xx}\equiv-\frac{1}{2}\int
d^{3}r(\partial_{x}\theta)^{2}[1-\alpha\cos(2\theta)],\\!\ \\!\
S_{yy}\equiv\frac{1}{2}\int
d^{3}r(\partial_{y}\theta)^{2}[1+\alpha\cos(2\theta)],\\!\ \\!\ S_{xy}$
$\displaystyle\equiv\alpha(\partial_{x}\theta)(\partial_{y}\theta)\sin(2\theta).$
(B.4)

The first-order variation just gives the equation of motion Eq. (II),

$\displaystyle\delta S=\delta S_{xx}+\delta S_{yy}+\delta S_{xy}-\int
d^{3}r(\delta\theta)(\partial_{t}^{2}\theta),$ (B.5) $\displaystyle\delta
S_{xx}\equiv\int
d^{3}r(\delta\theta)\\{(\partial_{x}^{2}\theta)[1-\alpha\mathrm{cos}(2\theta)]+\alpha(\partial_{x}\theta)^{2}\mathrm{sin}(2\theta)\\},$
(B.6) $\displaystyle\delta S_{yy}\equiv\int
d^{3}r(\delta\theta)\\{(\partial_{y}^{2}\theta)[1+\alpha\mathrm{cos}(2\theta)]-\alpha(\partial_{y}\theta)^{2}\mathrm{sin}(2\theta)\\},$
(B.7) $\displaystyle\delta S_{xy}\equiv-2\alpha\int
d^{3}r(\delta\theta)[(\partial_{x}\partial_{y}\theta)\mathrm{sin}(2\theta)+(\partial_{x}\theta)(\partial_{y}\theta)\mathrm{cos}(2\theta)].$
(B.8)

$\delta S$ vanishes since $\theta$ is an extremum or a saddle point of $S$.
The second-order variation $\delta^{2}S$ determines small deformation
$\delta\theta$ in such a way that $\theta+\delta\theta$ is an extremum or a
saddle point of $S$. $S_{xx}$ gives

$\displaystyle\delta^{2}S_{xx}=$ $\displaystyle\int
d^{3}r(\delta\theta)\\{(\partial_{x}^{2}\delta\theta)[1-\alpha\mathrm{cos}(2\theta)]+2\alpha(\partial_{x}^{2}\theta)\mathrm{sin}(2\theta)(\delta\theta)\\}$
$\displaystyle+2\alpha(\partial_{x}\theta)(\partial_{x}\delta\theta)\mathrm{sin}(2\theta)+2\alpha(\partial_{x}\theta)^{2}\mathrm{cos}(2\theta)(\delta\theta)\\}$
$\displaystyle=$ $\displaystyle-\int
d^{3}r[(\partial_{x}\delta\theta)^{2}+\alpha\mathrm{cos}(2\theta)(\delta\theta)\partial_{x}^{2}(\delta\theta)-2\alpha\mathrm{sin}(2\theta)(\partial_{x}^{2}\theta)(\delta\theta)^{2}$
$\displaystyle-2\alpha(\partial_{x}\theta)(\delta\theta)(\partial_{x}\delta\theta)\mathrm{sin}(2\theta)-2\alpha(\partial_{x}\theta)^{2}\mathrm{cos}(2\theta)(\delta\theta)^{2}],$
(B.9)

where

$\displaystyle\int
d^{3}r\alpha\mathrm{cos}(2\theta)(\delta\theta)\partial_{x}^{2}(\delta\theta)=-\int
d^{3}r\alpha\partial_{x}[\mathrm{cos}(2\theta)(\delta\theta)](\partial_{x}\delta\theta)$
$\displaystyle=$ $\displaystyle\int
d^{3}r[2\alpha\mathrm{sin}(2\theta)(\partial_{x}\theta)(\delta\theta)(\partial_{x}\delta\theta)-\alpha\mathrm{cos}(2\theta)(\partial_{x}\delta\theta)^{2}].$
(B.10)

Taking Eq. (B) into Eq. (B), we get

$\displaystyle\delta^{2}S_{xx}=-\int
d^{3}r\\{(\partial_{x}\delta\theta)^{2}[1-\alpha\mathrm{cos}(2\theta)]-2\alpha(\delta\theta)^{2}[\mathrm{cos}(2\theta)(\partial_{x}\theta)^{2}+\mathrm{sin}(2\theta)(\partial_{x}^{2}\theta)]\\}.$
(B.11)

Similarly, we get from $S_{yy}$

$\displaystyle\delta^{2}S_{yy}=-\int
d^{3}r\\{(\partial_{y}\delta\theta)^{2}[1+\alpha\mathrm{cos}(2\theta)]+2\alpha(\delta\theta)^{2}[\mathrm{cos}(2\theta)(\partial_{y}\theta)^{2}+\mathrm{sin}(2\theta)(\partial_{y}^{2}\theta)]\\}.$
(B.12)

$S_{xy}$ gives

$\displaystyle\delta^{2}S_{xy}=$ $\displaystyle-2\alpha\int
d^{3}r(\delta\theta)[(\partial_{x}\partial_{y}\delta\theta)\mathrm{sin}(2\theta)+2(\delta\theta)\mathrm{cos}(2\theta)(\partial_{x}\partial_{y}\theta)$
$\displaystyle+(\partial_{x}\delta\theta)(\partial_{y}\theta)\mathrm{cos}(2\theta)+(\partial_{y}\delta\theta)(\partial_{x}\theta)\mathrm{cos}(2\theta)-2(\delta\theta)\mathrm{sin}(2\theta)(\partial_{x}\theta)(\partial_{y}\theta)],$
(B.13)

where

$\displaystyle-\alpha\int
d^{3}r(\delta\theta)(\partial_{x}\partial_{y}\delta\theta)\mathrm{sin}(2\theta)=$
$\displaystyle\alpha\int
d^{3}r(\partial_{y}\delta\theta)[(\partial_{x}\delta\theta)\mathrm{sin}(2\theta)+2(\delta\theta)\mathrm{cos}(2\theta)(\partial_{x}\theta)]$
$\displaystyle=$ $\displaystyle\alpha\int
d^{3}r(\partial_{x}\delta\theta)[(\partial_{y}\delta\theta)\mathrm{sin}(2\theta)+2(\delta\theta)\mathrm{cos}(2\theta)(\partial_{y}\theta)].$
(B.14)

Taking Eq. (B) into Eq. (B), we get from $S_{xy}$

$\displaystyle\delta^{2}S_{xy}=2\alpha\int
d^{3}r\\{(\partial_{x}\delta\theta)(\partial_{y}\delta\theta)\mathrm{sin}(2\theta)+2(\delta\theta)^{2}[\mathrm{sin}(2\theta)(\partial_{x}\theta)(\partial_{y}\theta)-\mathrm{cos}(2\theta)(\partial_{x}\partial_{y}\theta)]\\}.$
(B.15)

Besides, we have

$\displaystyle-\delta\int d^{3}r(\delta\theta)(\partial_{t}^{2}\theta)=\int
d^{3}r(\delta\partial_{t}\theta)^{2}.$ (B.16)

Combining Eqs. (B.11,B.12,B.15,B.16) together, we obtain the second-order
variation of $S$ with respect to small $\delta\theta$,

$\displaystyle S_{\delta\theta}=\frac{1}{2}\delta^{2}S=$ $\displaystyle\int
d^{3}r\big{\\{}\frac{1}{2}(\partial_{t}\delta\theta)^{2}-\frac{1}{2}(\partial_{x}\delta\theta)^{2}[1-\alpha\mathrm{cos(2\theta)}]$
$\displaystyle-\frac{1}{2}(\partial_{y}\delta\theta)^{2}[1+\alpha\mathrm{cos}(2\theta)]+\alpha(\partial_{x}\delta\theta)(\partial_{y}\delta\theta)\mathrm{sin}(2\theta)$
$\displaystyle+\alpha(\delta\theta)^{2}\\{2\mathrm{sin}(2\theta)(\partial_{x}\theta)(\partial_{y}\theta)+\mathrm{cos}(2\theta)[(\partial_{x}\theta)^{2}-(\partial_{y}\theta)^{2}]$
$\displaystyle+\mathrm{sin}(2\theta)(\partial_{x}^{2}\theta-\partial_{y}^{2}\theta)-2\mathrm{cos}(2\theta)(\partial_{x}\partial_{y}\theta)\\}\big{\\}}.$
(B.17)

Here $\theta$ is a solution of the classical EOM. Given such $\theta$, we have
only to find those $\delta\theta$ that makes $\delta
S_{\delta\theta}[\delta\theta]=0$.

In the following, we focus on the one-dimensional solution
($\partial_{y}\theta=\partial_{y}\delta\theta=0$) for simplicity, and neglect
the integral over $y$, while the following derivation can be generalized to
$\partial_{y}\delta\theta\neq 0$. The action becomes in the one-dimensional
model

$\displaystyle K\equiv$ $\displaystyle
S_{\delta\theta}|_{\partial_{y}\theta=\partial_{y}\delta\theta=0}$
$\displaystyle=$ $\displaystyle\int^{\infty}_{-\infty}dt\int_{L}dx\\!\
\\{\frac{1}{2}(\partial_{t}\delta\theta)^{2}-\frac{1}{2}(\partial_{x}\delta\theta)^{2}[1-\alpha\mathrm{cos}(2\theta)]+\alpha(\delta\theta)^{2}[(\partial_{x}\theta)^{2}\mathrm{cos}(2\theta)+(\partial^{2}_{x}\theta)\mathrm{sin}(2\theta)]\\}.$
(B.18)

We substitute into $K$ a perturbative solution of $\theta$ in $\alpha$, e.g.
Eqs. (III.1), and expand $K$ in powers of $\alpha$. This gives

$\displaystyle K=K_{0}+K_{1}+K_{2}+\mathcal{O}(\alpha^{3}),$ (B.19)

where $K_{n}=\mathcal{O}(\alpha^{n})$. $K$ is a quadratic function of
$\delta\theta$. In terms of a Fourier transform of $\delta\theta_{q,\omega}$,

$\displaystyle\delta\theta_{q,\omega}=\frac{1}{\sqrt{L}}\int_{L}dx\int^{\infty}_{-\infty}dt\\!\
\\!\ \delta\theta(x,t)\\!\ e^{-iqx+i\omega t},$ (B.20)

the quadratic function can be characterized by matrix elements among
wavenumber $q$ and frequency $\omega$;

$\displaystyle
K=\frac{1}{L}\sum_{q,q^{\prime}}\int\frac{d\omega}{2\pi}\int\frac{d\omega^{\prime}}{2\pi}\\!\
\delta\theta_{q,\omega}^{\dagger}K_{q,\omega;q^{\prime},\omega^{\prime}}[\theta]\\!\
\delta\theta_{q^{\prime},\omega^{\prime}}.$ (B.21)

Let us take a solution $\theta(x,t)$ for the spin-injection model with the
straight geometry as an example. We first take a part of $\theta(x,t)$ with
only one spatial wavelength from Eqs. (III.8,III.13),

$\displaystyle\theta(x,t)=\theta_{0}(x,t)+\theta_{1}(x,t)+\mathcal{O}(\alpha^{2}),$
(B.22)
$\displaystyle\theta_{0}(x,t)=-j_{0}x+j_{0}k_{0}t,\quad\theta_{1}(x,t)=\frac{\alpha}{4(k_{0}^{2}-1)}\mathrm{sin}(2j_{0}x-2j_{0}k_{0}t).$
(B.23)

An inclusion of $\theta_{2}(x,t)$ shall be given later. The zeroth order of
$K$ is given by

$\displaystyle K_{0}=\frac{1}{2}\int dt\int_{L}dx\\!\
[(\partial_{t}\delta\theta)^{2}-(\partial_{x}\delta\theta)^{2}]=\frac{1}{2}\sum_{q}\int\frac{d\omega}{2\pi}\\!\
\delta\theta_{q,\omega}^{\dagger}(\omega^{2}-q^{2})\delta\theta_{q,\omega},$
(B.24)

A substitution of Eq. (B.23) into ${\cal O}(\alpha)$-terms in Eq. (B) gives
$K_{1}$ and $K_{2}$,

$\displaystyle\frac{\alpha}{2}(\partial_{x}\delta\theta)^{2}$
$\displaystyle\mathrm{cos}(2\theta_{0}+2\theta_{1})+\alpha(\delta\theta)^{2}(\partial_{x}\theta_{0}+\partial_{x}\theta_{1})^{2}\mathrm{cos}(2\theta_{0}+2\theta_{1})$
$\displaystyle+\alpha(\delta\theta)^{2}$
$\displaystyle(\partial_{x}^{2}\theta_{0}+\partial_{x}^{2}\theta_{1})\mathrm{sin}(2\theta_{0}+2\theta_{1})$
$\displaystyle=$
$\displaystyle\frac{\alpha}{2}(\partial_{x}\delta\theta)^{2}\mathrm{cos}(2\theta_{0})+\alpha(\delta\theta)^{2}(\partial_{x}\theta_{0})^{2}\mathrm{cos}(2\theta_{0})$
$\displaystyle-\alpha(\partial_{x}\delta\theta)^{2}\mathrm{sin}(2\theta_{0})\theta_{1}-2\alpha(\delta\theta)^{2}(\partial_{x}\theta_{0})(\partial_{x}\theta_{1})\mathrm{cos}(2\theta_{0})$
$\displaystyle-2\alpha(\delta\theta)^{2}(\partial_{x}\theta_{0})^{2}\mathrm{sin}(2\theta_{0})\theta_{1}-\alpha(\delta\theta)^{2}(\partial_{x}^{2}\theta_{1})\mathrm{sin}(2\theta_{0})+\mathcal{O}(\alpha^{3})$
$\displaystyle=$
$\displaystyle\frac{\alpha}{2}(\partial_{x}\delta\theta)^{2}\mathrm{cos}(2j_{0}x-2k_{0}j_{0}t)+\alpha
j_{0}^{2}(\delta\theta)^{2}\mathrm{cos}(2j_{0}x-2j_{0}k_{0}t)$
$\displaystyle+\frac{\alpha^{2}(\partial_{x}\delta\theta)^{2}}{4(k_{0}^{2}-1)}\mathrm{sin}^{2}(2j_{0}x-2j_{0}k_{0}t)-\frac{\alpha^{2}j_{0}^{2}(\delta\theta)^{2}}{k_{0}^{2}-1}\mathrm{cos}^{2}(2j_{0}x-2j_{0}k_{0}t)$
$\displaystyle+\frac{\alpha^{2}j_{0}^{2}(\delta\theta)^{2}}{2(k_{0}^{2}-1)}\mathrm{sin^{2}}(2j_{0}x-2j_{0}k_{0}t)+\frac{\alpha^{2}j_{0}^{2}(\delta\theta)^{2}}{k_{0}^{2}-1}\mathrm{sin}^{2}(2j_{0}x-2j_{0}k_{0}t)+\mathrm{O}(\alpha^{3}).$
(B.25)

Equivalently, we have

$\displaystyle K_{1}=$ $\displaystyle\alpha\int dt\int_{L}dx\\!\
[\frac{1}{2}(\partial_{x}\delta\theta)^{2}\mathrm{cos}(2j_{0}x-2j_{0}k_{0}t)+j_{0}^{2}(\delta\theta)^{2}\mathrm{cos}(2j_{0}x-2j_{0}k_{0}t)]$
$\displaystyle=$ $\displaystyle\alpha\int dt\int_{L}dx\\!\
[\frac{1}{4L}\sum_{q,q^{\prime}}\int_{\omega,\omega^{\prime}}qq^{\prime}\delta\theta_{q,\omega}^{\dagger}\delta\theta_{q^{\prime},\omega^{\prime}}+\frac{j_{0}^{2}}{2L}\sum_{q,q^{\prime}}\int_{\omega,\omega^{\prime}}\\!\
\delta\theta_{q,\omega}^{\dagger}\delta\theta_{q^{\prime},\omega^{\prime}}]e^{-iqx+i\omega
t}e^{iq^{\prime}x-i\omega^{\prime}t}e^{2ij_{0}x-2ij_{0}k_{0}t}+\mathrm{H.c.}$
$\displaystyle=$ $\displaystyle\frac{\alpha}{4}\sum_{q}\int_{\omega}\\!\
\delta\theta^{\dagger}_{q+2j_{0},\omega+2j_{0}k_{0}}\delta\theta_{q,\omega}[q(q+2j_{0})+2j_{0}^{2}]+\mathrm{H.c.}$
$\displaystyle=$ $\displaystyle\frac{\alpha}{4}\sum_{q}\int_{\omega}\\!\
\delta\theta^{\dagger}_{q+j_{0},\omega+j_{0}k_{0}}\delta\theta_{q-j_{0},\omega-
j_{0}k_{0}}(q^{2}+j_{0}^{2})e^{-2ij_{0}k_{0}t}+\mathrm{H.c.},$ (B.26)
$\displaystyle K_{2}=$ $\displaystyle\alpha^{2}\int dt\int_{L}dx\\!\
\Big{\\{}\frac{(\partial_{x}\delta\theta)^{2}}{8(k_{0}^{2}-1)}[1-\mathrm{cos}(4j_{0}x-4j_{0}k_{0}t)]-\frac{j_{0}^{2}(\delta\theta)^{2}}{k_{0}^{2}-1}\mathrm{cos}(4j_{0}x-4j_{0}k_{0}t)$
$\displaystyle+\frac{j_{0}^{2}(\delta\theta)^{2}}{4(k_{0}^{2}-1)}[1-\mathrm{cos}(4j_{0}x-4j_{0}k_{0}t)]\Big{\\}}$
$\displaystyle=$
$\displaystyle\frac{\alpha^{2}}{8(k_{0}^{2}-1)}\sum_{q}\int\frac{d\omega}{2\pi}\\!\
\delta\theta^{\dagger}_{q,\omega}\delta\theta_{q,\omega}q^{2}+\frac{\alpha^{2}j_{0}^{2}}{4(k_{0}^{2}-1)}\sum_{q}\int\frac{d\omega}{2\pi}\\!\
\delta\theta^{\dagger}_{q,\omega}\delta\theta_{q,\omega}$
$\displaystyle-\frac{\alpha^{2}}{16(k_{0}^{2}-1)}\sum_{q}\int\frac{d\omega}{2\pi}\\!\
\Big{\\{}\delta\theta^{\dagger}_{q+2j_{0},\omega+2j_{0}k_{0}}\delta\theta_{q-2j_{0},\omega-2j_{0}k_{0}}\big{[}(q^{2}-4j_{0}^{2})+8j_{0}^{2}+2j_{0}^{2}\big{]}e^{-4ij_{0}k_{0}t}+\mathrm{H.c.}\Big{\\}}$
$\displaystyle=$
$\displaystyle\frac{\alpha^{2}}{8(k_{0}^{2}-1)}\sum_{q}\int\frac{d\omega}{2\pi}\\!\
\delta\theta^{\dagger}_{q,\omega}\delta\theta_{q,\omega}(2j_{0}^{2}+q^{2})$
$\displaystyle-\frac{\alpha^{2}}{16(k_{0}^{2}-1)}\sum_{q}\int\frac{d\omega}{2\pi}\\!\
\big{[}\delta\theta^{\dagger}_{q+2j_{0},\omega+2j_{0}k_{0}}\delta\theta_{q-2j_{0},\omega-2j_{0}k_{0}}(q^{2}+6j_{0}^{2})+\mathrm{H.c.}\big{]}.$
(B.27)

Taking Eqs. (B.24,B,B) into Eq. (B.20), we obtain the matrix elements,

$\displaystyle\frac{1}{2\pi}K_{q+p,\omega+\nu;q-p,\omega-\nu}=(\omega^{2}-q^{2})\delta_{p,0}\delta(\nu)+\frac{\alpha(q^{2}+j_{0}^{2})}{2}(\delta_{p,j_{0}}\delta(\nu-
j_{0}k_{0})+\delta_{p,-j_{0}}\delta(\nu+j_{0}k_{0}))$
$\displaystyle-\frac{\alpha^{2}}{8(k_{0}^{2}-1)}[-2(2j_{0}^{2}+q^{2})\delta_{p,0}\delta(\nu)+(q^{2}+6j_{0}^{2})(\delta_{p,2j_{0}}\delta(\nu-2j_{0}k_{0})+\delta_{p,-2j_{0}}\delta(\nu+2j_{0}k_{0}))]+\mathcal{O}(\alpha^{3}).$
(B.28)

To find $\delta\theta$ that satisfies $\delta
S_{\delta\theta}[\delta\theta]=0$, we only have to find an eigenmode of $K$ in
Eq. (B.21) that belongs to zero eigenvalue (“eigenenergy”). At the zeroth
order in $\alpha$, eigenmodes of $K$ are characterized by $q$ and $\omega$,
and the “zero-energy” eigenmodes are obtained by setting $\omega$ to be $q$
(on-shell condition). When $\alpha$ is included perturbatively, eigenmodes at
$q$ and $\omega$ hybridize with eigenmodes at $q\pm 2j_{0}$ and $\omega\pm
2j_{0}k_{0}$ as well as eigenmodes at $q\pm 4j_{0}$ and $\omega\pm
4j_{0}k_{0}$ in terms of off-diagonal mixing terms. Due to the off-diagonal
mixing terms, eigenmodes of $K$ are characterized by $q$ and $\omega$ modulo
$2j_{0}$ and $2j_{0}k_{0}$ respectively, and
$(q,\omega)\in[-j_{0},j_{0}]\times[-k_{0}j_{0},k_{0}j_{0}]$ plays a role of a
first Brillouin zone. In the Brillouin zone, eigenmodes at the same
$(q,\omega)$ are distinguished by a band index $n$,

$\displaystyle K=\frac{1}{2}\sum_{n\in\mathbb{N}}\sum_{-j_{0}\leq
q<j_{0}}\int^{k_{0}j_{0}}_{-k_{0}j_{0}}\frac{d\omega}{2\pi}\\!\
\delta\varphi^{\dagger}_{q,\omega,n}\Lambda_{q,\omega,n}\delta\varphi_{q,\omega,n}+\mathcal{O}(\alpha^{3}),$
(B.29)

with

$\displaystyle\delta\theta_{q+2j_{0}m_{1},\omega+2k_{0}j_{0}m_{2}}=\sum_{n\in\mathbb{N}}c_{q,\omega,n;m_{1},m_{2}}\delta\varphi_{q,\omega,n},$
(B.30)

$m_{1}\in\mathbb{Z}$, $m_{2}\in\mathbb{Z}$. Here, $c_{q,\omega,n;m_{1},m_{2}}$
is analogous to the periodic part of a Bloch wavefunction in the band theory.
From Eq. (B) $K_{q,\omega;q,\omega^{\prime}}$ is real symmetric, so that
$c_{q,\omega,n;m_{1},m_{2}}$ are real. An eigenstate of the lowest energy, say
$\delta\varphi_{q,\omega,n=0}$, must approach $\delta\theta_{q,\omega}$ in the
limit of $\alpha\rightarrow 0$. Such lowest eigenmode ($n=0$) is calculated up
to the 2nd order in $\alpha$ as follows,

$\displaystyle c_{q,\omega,0;m_{1},m_{2}}=$
$\displaystyle\delta_{m_{1},0}\delta_{m_{2},0}-\frac{\alpha[\frac{(q+j_{0})^{2}+j_{0}^{2}}{2}]}{(\omega+2j_{0}k_{0})^{2}-(q+2j_{0})^{2}-\omega^{2}+q^{2}}\delta_{m_{1},1}\delta_{m_{2},1}$
$\displaystyle-\frac{\alpha[\frac{(q-j_{0})^{2}+j_{0}^{2}}{2}]}{(\omega-2j_{0}k_{0})^{2}-(q-2j_{0})^{2}-\omega^{2}+q^{2}}\delta_{m_{1},-1}\delta_{m_{2},-1}+\mathcal{O}(\alpha^{2})$
$\displaystyle=$
$\displaystyle\delta_{m_{1},0}\delta_{m_{2},0}-\alpha\frac{1+\mathcal{O}(q,\omega)}{4(k_{0}^{2}-1)}\delta_{m_{1},1}\delta_{m_{2},1}-\alpha\frac{1+\mathcal{O}(q,\omega)}{4(k_{0}^{2}-1)}\delta_{m_{1},-1}\delta_{m_{2},-1}+\mathcal{O}(\alpha^{2}).$
(B.31)

A corresponding “eigenenergy” is calculated up to the 2nd order,

$\displaystyle\Lambda_{q,\omega,0}=$
$\displaystyle\omega^{2}-q^{2}+\frac{\alpha^{2}(2j_{0}^{2}+q^{2})}{4(k_{0}^{2}-1)}$
$\displaystyle-\frac{\alpha^{2}[\frac{(q+j_{0})^{2}+j_{0}^{2}}{2}]^{2}}{(\omega+2j_{0}k_{0})^{2}-(q+2j_{0})^{2}-\omega^{2}+q^{2}}-\frac{\alpha^{2}[\frac{(q-j_{0})^{2}+j_{0}^{2}}{2}]^{2}}{(\omega-2j_{0}k_{0})^{2}-(q-2j_{0})^{2}-\omega^{2}+q^{2}}+\mathcal{O}(\alpha^{3})$
$\displaystyle=$
$\displaystyle\omega^{2}-q^{2}+\frac{\alpha^{2}(2j_{0}^{2}+q^{2})}{4(k_{0}^{2}-1)}-\frac{\alpha^{2}(2j_{0}^{2}+2qj_{0}+q^{2})^{2}}{4[4j_{0}^{2}(k_{0}^{2}-1)+4qj_{0}-4\omega
j_{0}k_{0}]}$
$\displaystyle-\frac{\alpha^{2}(2j_{0}^{2}-2qj_{0}+q^{2})^{2}}{4[4j_{0}^{2}(k_{0}^{2}-1)-4qj_{0}+4\omega
j_{0}k_{0}]}+\mathcal{O}(\alpha^{3}),$ (B.32)

namely

$\displaystyle\Lambda_{q,\omega,0}=$
$\displaystyle\omega^{2}+\frac{\alpha^{2}j_{0}^{2}}{2}\frac{1}{k_{0}^{2}-1}-q^{2}[1-\frac{\alpha^{2}}{4(k_{0}^{2}-1)}]$
$\displaystyle-\frac{\alpha^{2}}{16j_{0}^{2}(k_{0}^{2}-1)}(4j_{0}^{4}+8qj_{0}^{3}+8q^{2}j_{0}^{2})[1-\frac{q-\omega
k_{0}}{j_{0}(k_{0}^{2}-1)}+\frac{(q-\omega
k_{0})^{2}}{j_{0}^{2}(k_{0}^{2}-1)^{2}}+\mathcal{O}((q+\omega k_{0})^{3})]$
$\displaystyle-\frac{\alpha^{2}}{16j_{0}^{2}(k_{0}^{2}-1)}(4j_{0}^{4}-8qj_{0}^{3}+8q^{2}j_{0}^{2})[1+\frac{q-\omega
k_{0}}{j_{0}(k_{0}^{2}-1)}+\frac{(q-\omega
k_{0})^{2}}{j_{0}^{2}(k_{0}^{2}-1)^{2}}+\mathcal{O}((q+\omega
k_{0})^{3})]+\mathcal{O}(\alpha^{3})$ $\displaystyle=$
$\displaystyle\omega^{2}+\frac{\alpha^{2}j_{0}^{2}}{2}\frac{1}{k_{0}^{2}-1}-q^{2}[1-\frac{\alpha^{2}}{4(k_{0}^{2}-1)}]-\frac{\alpha^{2}j_{0}^{2}}{2(k_{0}^{2}-1)}+\mathcal{O}(\alpha^{2}\omega^{2},\alpha^{2}q^{2},\alpha^{2}\omega
q,\alpha^{3})$ $\displaystyle=$
$\displaystyle\omega^{2}-q^{2}[1-\frac{\alpha^{2}}{4(k_{0}^{2}-1)}]+\mathcal{O}(\alpha^{2}\omega^{2},\alpha^{2}q^{2},\alpha^{2}\omega
q,\alpha^{3}).$ (B.33)

The lowest energy band indicates that $\delta\theta$ evaluated on shell,
$\Lambda_{q,\omega,0}=0$, behaves like a gapless classical wave. This is
because the original theory, Eq. (5) in the main text, has a spacetime
translational symmetry. Thus, for any $\theta$, one can choose $\delta\theta$
as a translation of $\theta$, and such $\delta\theta$ does not change the
Lagrangian. For a general $k_{0}$ (off the resonance point; $k_{0}\neq 1$),
$\alpha$ can be treated perturbatively, and the classical wave up to the 2nd
order in $\alpha$ has a well-defined (i.e. real-valued) velocity $v$,

$\displaystyle
1-\frac{\alpha^{2}}{4(k_{0}^{2}-1)}=1+\mathcal{O}(\alpha^{2})\equiv v^{2}>0.$
(B.34)

By evaluating the eigenmode on shell ($|\omega|=|q|+{\cal O}(\alpha^{2})|q|$),
we finally determine the first-order $\delta\theta_{1}$ for an arbitrary form
of $\delta\theta_{0}$ given by Eq. (B.1),

$\displaystyle\delta\theta_{1}(x,t)$
$\displaystyle=\frac{1}{\sqrt{L}}\sum^{|q|<j_{0}}_{q}\int^{k_{0}j_{0}}_{-k_{0}j_{0}}d\omega\Big{[}\delta(q-\omega)d_{q}+\delta(q+\omega)d^{\prime}_{q}\Big{]}$
$\displaystyle\hskip
28.45274pt\times\Big{[}-\alpha\frac{1+\mathcal{O}(q)}{4(k_{0}^{2}-1)}e^{i(q+2j_{0})x-i(\omega+2j_{0}k_{0})t}-\alpha\frac{1+\mathcal{O}(q)}{4(k_{0}^{2}-1)}e^{i(q-2j_{0})x-i(\omega-2j_{0}k_{0})t}\Big{]}.$
(B.35)

Finally, let us include $\theta_{2}(x,t)$ into Eq. (B.22),

$\displaystyle\theta(x,t)=\theta_{0}(x,t)+\theta_{1}(x,t)+\theta_{2}(x,t)+\mathcal{O}(\alpha^{2}),$
(B.36) $\displaystyle\theta_{2}(x,t)=$ $\displaystyle
2\alpha\mathrm{Re}(g)\mathrm{cos}[2k_{0}j_{0}(t-x)]+2\alpha\mathrm{Re}(g^{\prime})\mathrm{cos}[2k_{0}j_{0}(t+x)]$
$\displaystyle-2\alpha\mathrm{Im}(g)\mathrm{sin}[2k_{0}j_{0}(t-x)]-2\alpha\mathrm{Im}(g^{\prime})\mathrm{sin}[2k_{0}j_{0}(t+x)].$
(B.37)

Eq. (B) has an additional $\mathcal{O}(\alpha^{2})$ contribition,

$\displaystyle\Delta K_{2}=$ $\displaystyle\int
dtdx[-\alpha(\partial_{x}\delta\theta)^{2}\mathrm{sin}(2\theta_{0})\theta_{2}+2\alpha(\delta\theta)^{2}(\partial_{x}\theta_{0})(\partial_{x}\theta_{2})\mathrm{cos}(2\theta_{0})$
$\displaystyle-2\alpha(\delta\theta)^{2}(\partial_{x}\theta_{0})^{2}\mathrm{sin}(2\theta_{0})\theta_{2}+\alpha(\delta\theta)^{2}(\partial_{x}^{2}\theta_{2})\mathrm{sin}(2\theta_{0})].$
(B.38)

For $k_{0}\neq 1$, $\Delta K_{2}$ contributes only to off-diagonal matrix
elements of $K_{q,\omega;q,\omega}$, so that it changes neither Eq. (B) nor
Eq. (B) at their respective sub-leading order.