# The automorphism group of the quantum grassmannian 111This research was
partly supported by EPSRC grant EP/R009279/1.

S Launois and T H Lenagan

###### Abstract

We calculate the automorphism group of the generic quantum grassmannian.

##  1 Introduction

The quantum grassmannian $\mathcal{O}_{q}(G(k,n))$ is a noncommutative algebra
that is a deformation of the homogeneous coordinate ring of the classical
grassmannian of $k$-planes in $n$-space. In this paper, we calculate the
automorphism group of the quantum grassmannian in the generic case where the
deformation parameter $q$ is not a root of unity.

Typically, quantised coordinate algebras are much more rigid than their
classical counterparts, in the sense that the automorphism group of the
quantum object is much smaller than that of the classical object.
Nevertheless, it has proven difficult to calculate these automorphism groups
and only a few examples are known where the calculation has been completed,
see, for example, [1, 6, 9, 10, 11, 17, 18]. The automorphism group of quantum
matrices [10, 17] will prove crucial in our present work.

The quantum grassmannian $\mathcal{O}_{q}(G(k,n))$ is generated as an algebra
by the $k\times k$ quantum minors of the quantum matrix algebra
$\mathcal{O}_{q}(M(k,n))$. These generators are called quantum Plücker
coordinates and there is a natural partial order on the quantum Plücker
coordinates which is illustrated in the case of $\mathcal{O}_{q}(G(3,6))$ in
Figure 1. There are two obvious sources of automorphisms for
$\mathcal{O}_{q}(G(k,n))$. The first is by restricting column automorphisms of
$\mathcal{O}_{q}(M(k,n))$ to the subalgebra $\mathcal{O}_{q}(G(k,n))$, these
automorphisms are described in Section 4. The second is via studying certain
automorphisms of the (noncommutative) dehomogenisation of
$\mathcal{O}_{q}(G(k,n))$ which is isomorphic to a skew Laurent extension of
$\mathcal{O}_{q}(M(k,p))$, with $p=n-k$, as we shall see in Section 3.

In Section 4 we study these “obvious” automorphisms of
$\mathcal{O}_{q}(G(k,n))$ and consider the relations between them. We then
claim that these provide all of the automorphisms of
$\mathcal{O}_{q}(G(k,n))$, and justify the claim in the following sections.

The quantum grassmannian carries the structure of an ${\mathbb{N}}$-graded
algebra, with each quantum Plucker coordinate having degree one. In Section 5,
we exploit this grading in a series of lemmas to see that we can essentially
fix the minimal and maximal elements in the poset after allowing adjustment by
the automorphisms that we have found in Section 4.

In Section 6 we study these adjusted automorphisms and show that such an
automorphism induces, via the dehomogenisation equality, an automorphism of
$\mathcal{O}_{q}(M(k,n))$. Once this has been done, our main result follows
easily in Section 7 from the known structure of the automorphism group of
quantum matrices.

##  2 Basic definitions

Throughout the paper, we work with a field ${K}$ and a nonzero element
$q\in{K}$ which is not a root of unity.

The algebra of $m\times n$ quantum matrices over ${K}$, denoted by
$\mathcal{O}_{q}(M(m,n))$, is the algebra generated over ${K}$ by $mn$
indeterminates $x_{ij}$, with $1\leq i\leq m$ and $1\leq j\leq n$, which
commute with the elements of ${K}$ and are subject to the relations:

$\begin{array}[]{ll}x_{ij}x_{il}=qx_{il}x_{ij},&\mbox{ for }1\leq i\leq
m,\mbox{ and }1\leq j<l\leq n\>;\\\ x_{ij}x_{kj}=qx_{kj}x_{ij},&\mbox{ for
}1\leq i<k\leq m,\mbox{ and }1\leq j\leq n\>;\\\
x_{ij}x_{kl}=x_{kl}x_{ij},&\mbox{ for }1\leq k<i\leq m,\mbox{ and }1\leq
j<l\leq n\>;\\\ x_{ij}x_{kl}-x_{kl}x_{ij}=(q-q^{-1})x_{il}x_{kj},&\mbox{ for
}1\leq i<k\leq m,\mbox{ and }1\leq j<l\leq n.\end{array}$

It is well known that $\mathcal{O}_{q}(M(m,n))$ is an iterated Ore extension
over ${K}$ with the $x_{ij}$ added in lexicographic order. An immediate
consequence is that $\mathcal{O}_{q}(M(m,n))$ is a noetherian domain.

When $m=n$, the quantum determinant $D_{q}$ is defined by;

$D_{q}:=\sum\,(-q)^{l(\sigma)}x_{1\sigma(1)}\dots x_{n\sigma(n)},$

where the sum is over all permutations $\sigma$ of $\\{1,\dots,n\\}$.

The quantum determinant is a central element in the algebra of quantum
matrices $\mathcal{O}_{q}(M(n,n))$.

If $I$ and $J$ are $t$-element subsets of $\\{1,\dots,m\\}$ and
$\\{1,\dots,n\\}$, respectively, then the quantum minor $[I\mid J]$ is defined
to be the quantum determinant of the $t\times t$ quantum matrix subalgebra
generated by the variables $x_{ij}$ with $i\in I$ and $j\in J$.

The homogeneous coordinate ring of the $k\times n$ quantum grassmannian,
$\mathcal{O}_{q}(G(k,n))$ (informally known as the quantum grassmannian) is
the subalgebra of $\mathcal{O}_{q}(M(k,n))$ generated by the $k\times k$
quantum minors of $\mathcal{O}_{q}(M(k,n))$, see, for example, [7].

The quantum grassmannian $\mathcal{O}_{q}(G(1,n))$ is a quantum affine space,
and, as such, its automorphism group is known, see [1]; so we will assume
throughout this paper that $k>1$. Also, we will see in Proposition 3.4 that
$\mathcal{O}_{q}(G(k,n))\cong\mathcal{O}_{q}(G(n-k,n))$, so in calculating the
automorphism group we will assume that $2k\leq n$ (and so $n\geq 4$, as $k\geq
2$).

A $k\times k$ quantum minor of $\mathcal{O}_{q}(M(k,n))$ must use all of the
$k$ rows, and so we can specify the quantum minor by specifying the columns
that define it. With this in mind, we will write $[J]$ for the quantum minor
$[1,\dots,k\mid J]$, for any $k$-element subset $J$ of $\\{1,\dots,n\\}$.
Quantum minors of this type are called quantum Plücker coordinates. The set of
quantum Plücker coordinates in $\mathcal{O}_{q}(G(k,n))$ is denoted by $\Pi$.
There is a natural partial order on $\Pi$ defined in the following way: if
$I=[i_{1}<\dots<i_{k}]$ and $J=[j_{1}<\dots<j_{k}]$ then $[I]<[J]$ if and only
if $i_{l}\leq j_{l}$ for each $l=1,\dots,k$. This partial order is illustrated
for the case of $\mathcal{O}_{q}(G(3,6))$ in Figure 1. A standard monomial in
the quantum Plücker coordinates is an expression of the form
$[I_{1}][I_{2}]\dots[I_{t}]$ where $I_{1}\leq I_{2}\leq\dots\leq I_{t}$ in
this partial order. The set of all standard monomials forms a vector space
basis of $\mathcal{O}_{q}(G(k,n))$ over ${K}$, see, for example, [7,
Proposition 2.8].

---
$\scriptstyle{[456]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{[356]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{[346]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{[256]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{[345]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{[246]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{[156]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{[245]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{[236]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{[146]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{[235]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{[145]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{[136]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{[234]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{[135]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{[126]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{[134]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{[125]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{[124]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{[123]}$
Figure 1: The partial order on $\Pi$ for $\mathcal{O}_{q}(G(3,6))$.

##  3 Dehomogenisation of $\mathcal{O}_{q}(G(k,n))$

An element $a$ in a ring $R$ is said to be a normal element of $R$ provided
that $aR=Ra$. If $R$ is a domain then a nonzero normal element $a$ may be
inverted, as the Ore condition for the set $S:=\\{a^{n}\\}$ is easily
verified. Standard results for noncommutative noetherian rings can be found in
the books by Goodearl and Warfield [5] and McConnell and Robson [16].

Set $u=\\{1,\dots,k\\}$. Then $[u]$ commutes with all other quantum Plücker
coordinates up to a power of $q$ as the following lemma shows.

######  Lemma 3.1.

Let $[I]$ be a quantum Plücker coordinate in $\mathcal{O}_{q}(G(k,n))$ and set
$d(I):=\\#\left(I\backslash(I\cap u)\right)\geq 0$. Then
$[u][I]=q^{d(I)}[I][u]$.

###### Proof.

This can be obtained from [8] by combining Lemma 3.7 and Theorem 3.4 of that
paper. (Note that [8] uses $q^{-1}$ as the deformation parameter where we use
$q$; so care must be taken in interpreting their results.) It can also be
extracted from [7, Corollary 1.1] by setting $[I]=[u]$ in the statement of the
corollary and noting that the summation on the right hand side of the
displayed equation is then empty. ∎

As $\mathcal{O}_{q}(G(k,n))$ is generated by the quantum Plücker coordinates
it follows from the previous lemma that the element $[u]$ is a normal element
and so we may invert $[u]$ to obtain the overring
$\mathcal{O}_{q}(G(k,n))[[u]^{-1}]$.

For $1\leq i\leq k$ and $1\leq j\leq n-k$, set

$x_{ij}:=[1\dots,\widehat{k+1-i},\dots
k,j+k][u]^{-1}\in\mathcal{O}_{q}(G(k,n))[[u]^{-1}].$

The case $a=1$ of [15, Theorem 2.2] shows that the elements $x_{ij}$ generate
an algebra $R$, say, that sits inside $\mathcal{O}_{q}(G(k,n))[[u]^{-1}]$ and
is isomorphic to ${\mathcal{O}}_{q}(M(k,n-k))$. Also,

$\mathcal{O}_{q}(G(k,n))[[u]^{-1}]=R[[u],[u]^{-1}]$

(NB. The way we’ve fixed things, that really is an equality in the above
display, rather than just an isomorphism.) In the rest of this note, we will
write $R={\mathcal{O}}_{q}(M(k,p))$ where $p:=n-k$ and when we are operating
on the right hand side of this equality, we will write $y$ for $[u]$.

As $d([1\dots,\widehat{k+1-i},\dots k,j+k])=1$, it follows from Lemma 3.1 that
$[u]x_{ij}=qx_{ij}[u]$ and that $yx_{ij}=qx_{ij}y$.

The equality above says that

$\displaystyle\mathcal{O}_{q}(G(k,n))[[u]^{-1}]=\mathcal{O}_{q}(M(k,p))[y,y^{-1};\sigma]$
(1)

where $\sigma$ is the automorphism of $\mathcal{O}_{q}(M(k,p))$ such that
$\sigma(x_{ij})=qx_{ij}$ for each $i=1,\dots,k$ and $j=1,\dots,p$. We will
refer to Equation (1) as the dehmogenisation equality.

The next lemma gives the formulae for passing between quantum minors and
Plücker coordinates in the above equality.

######  Lemma 3.2.

(i) Let $[I\mid J]$ be a quantum minor in $R=\mathcal{O}_{q}(M(k,p))$. Then

$[I\mid J]=[\\{1\dots
k\\}\backslash(k+1-I)\sqcup(k+J)][u]^{-1}\in\mathcal{O}_{q}(G(k,n))[[u]^{-1}]\,.$

(ii) Let $[L]$ be a quantum Plücker coordinate in $\mathcal{O}_{q}(G(k,n))$
and write $L=L_{\leq k}\cap L_{>k}$ where $L_{\leq k}=L\cap\\{1,\dots,k\\}$
and $L_{>k}=L\cap\\{k+1,\dots,n\\}$. Then

$[L]=[(k+1)-\left(\\{1,\dots,k\\}\backslash L_{\leq k}\right)\mid
L_{>k}-k]y\in\mathcal{O}_{q}(M(k,p))[y,y^{-1};\sigma]$

###### Proof.

(i) This formula occurs as the case $a=1$ of [15, Proposition 3.3] which gives
the formula for general quantum minors of $R=\mathcal{O}_{q}(M(k,p))$ in terms
of quantum Plücker coordinates of $\mathcal{O}_{q}(G(k,n))$.
(ii) Let $[L]$ be a quantum Plücker coordinate in $\mathcal{O}_{q}(G(k,n))$.
Set $I=(k+1)-(\\{1,\dots,k\\}\backslash L_{\leq k})$ and $J=L_{>k}-k$. Note
that $|I|=|J|=|L_{>k}|$ and so we can form the quantum minor $[I\mid J]$.
Apply (i) to $[I\mid J]$ to see that

$\displaystyle[I\mid J][u]$ $\displaystyle=$
$\displaystyle[\\{1,\dots,k\\}\backslash\\{(k+1)-\\{(k+1)-(\\{1,\dots,k\\}\backslash
L_{\leq k})\\}\sqcup k+(L_{>k}-k)]$ $\displaystyle=$
$\displaystyle[\\{1,\dots,k\\}\backslash(\\{1,\dots,k\\}\backslash L_{\leq
k})\sqcup L_{>k}]=[L_{\leq k}\sqcup L_{>k}]=[L],$

so that (ii) is established. ∎

The following corollary to the above lemma will be useful in later
calculations.

######  Corollary 3.3.

Suppose that $[L]$ is a quantum Plücker coordinate in
$\mathcal{O}_{q}(G(k,n))$ and that $[L]=[I\mid J][u]$ for some $[I\mid
J]\in\mathcal{O}_{q}(M(k,p))$.
(i) Let $i\in\\{1,\dots,k\\}$. Then $i\in I$ if and only if $(k+1)-i\not\in
L$.
(ii) Let $j\in\\{1\dots,p\\}$. Then $j\in J$ if and only if $j+k\in L$.

We will use dehomogenisation, see Equation (1), in the next three sections to
transfer the problem of finding automorphisms of $\mathcal{O}_{q}(G(k,n))$ to
that of finding automorphisms of $\mathcal{O}_{q}(M(k,p))$ where the problem
has been solved in [10] and [17]. Before doing that, we illustrate the
usefulness of dehomogenisation by the following two results, the second of
which identifies an extra automorphism of $\mathcal{O}_{q}(G(k,n))$ in the
case that $n=2k$.

First, note that there is an isomorphism $\tau$ between the quantum matrix
algebras $\mathcal{O}_{q}(M(k,n-k))=K(x_{ij})$ and
$\mathcal{O}_{q}(M(n-k,k))=K(x^{\prime}_{ij})$ that sends $x_{ij}$ to
$x^{\prime}_{ji}$ and more generally sends a quantum minor $[I\mid J]$ of
$\mathcal{O}_{q}(M(k,n-k))$ to the quantum minor $[J\mid I]$ of
$\mathcal{O}_{q}(M(n-k,k))$. Note that, with a slight abuse of notation, we
denote quantum minors of $\mathcal{O}_{q}(M(n-k,k))$ without dashes to
differentiate them from quantum minors from $\mathcal{O}_{q}(M(k,n-k))$. (When
$k=n-k$ and $x_{ij}=x^{\prime}_{ij}$ this is automorphism of quantum matrices
given by transposition and so we will refer to $\tau$ as the transpose
isomorphism.) This automorphism extends to an isomorphism from
$\mathcal{O}_{q}(M(k,n-k))[y,y^{-1};\sigma]$ to
$\mathcal{O}_{q}(M(n-k,k))[y^{\prime},y^{\prime-1};\sigma^{\prime}]$, where
$\sigma(x_{ij})=qx_{ij}$ and
$\sigma^{\prime}(x^{\prime}_{ij})=qx^{\prime}_{ij}$. We also denote this
extension by $\tau$.

######  Proposition 3.4.

$\mathcal{O}_{q}(G(k,n))\cong\mathcal{O}_{q}(G(n-k,n))$ via an automorphism
that sends the quantum Plücker coordinate $[L]$ of $\mathcal{O}_{q}(G(k,n))$
to the quantum Plücker coordinate $[w_{0}(\,\widehat{L}\,)]$ of
$\mathcal{O}_{q}(G(n-k,n))$, where $\widehat{L}$ is the complement of $L$ in
$\\{1,\dots,n\\}$ and $w_{0}$ is the longest element of the symmetric group on
$\\{1,\dots,n\\}$; that is, $w_{0}$ reverses the order of $\\{1,\dots,n\\}$.

###### Proof.

Recall that we are assuming that $2k\leq n$. There is an isomorphism

$\displaystyle\mathcal{O}_{q}(G(k,n))[[u]^{-1}]=\mathcal{O}_{q}(M(k,n-k))[y,y^{-1};\sigma]\cong^{1}$
$\displaystyle\mathcal{O}_{q}(M(n-k,k)[y^{\prime},y^{\prime-1};\sigma^{\prime}]\cong\mathcal{O}_{q}(G(n-k,n))[[u^{\prime}]^{-1}],$

where $[u^{\prime}]=[1,\dots,n-k]\in\mathcal{O}_{q}(G(n-k,n))$ and
$\sigma^{\prime}(x_{ij}^{\prime})=qx_{ij}^{\prime}$. (Note that $\cong^{1}$ is
given by applying the transpose isomorphism $\tau$ that sends $[I\mid J]$ to
$[J\mid I]$ and sends $y=[u]=[1,\dots,k]$ to
$y^{\prime}=[u^{\prime}]=[1,\dots,n-k]$.)

We need to track the destination of an arbitrary quantum Plücker coordinate of
$[L]$ of $\mathcal{O}_{q}(G(k,n))$ under this isomorphism, using the formulae
that we have developed above for translating between quantum Plücker
coordinates and quantum minors.

$\displaystyle[L]$ $\displaystyle=$ $\displaystyle[L_{\leq k}\sqcup L_{>k}]$
$\displaystyle=$ $\displaystyle[(k+1)-(\\{1,\dots,k\\}\backslash L_{\leq
k})\mid L_{>k}-k]y\qquad(\in\mathcal{O}_{q}(M(k,n-k))y)$
$\displaystyle\overset{1}{\mapsto}$
$\displaystyle[L_{>k}-k\mid(k+1)-(\\{1,\dots,k\\}\backslash L_{\leq
k})]y^{\prime}\qquad(\in\mathcal{O}_{q}(M(n-k,k))y^{\prime})$ $\displaystyle=$
$\displaystyle[\\{1,\dots,n-k\\}\backslash(n+1-L_{>k})\sqcup(n-k)+((k+1)-(\\{1,\dots,k\\}\backslash
L_{\leq k}))]$ $\displaystyle=$
$\displaystyle[\\{1,\dots,n-k\\}\backslash(n+1-L_{>k})\sqcup((n+1)-(\\{1,\dots,k\\}\backslash
L_{\leq k}))]$ $\displaystyle=$ $\displaystyle[\\{1,\dots,n-k\\}\backslash
w_{0}(L_{>k})\sqcup w_{0}(\\{1,\dots,k\\}\backslash L_{\leq k})]$
$\displaystyle=$ $\displaystyle[\\{1,\dots,n-k\\}\backslash
w_{0}(L_{>k})\sqcup\\{n-k+1,\dots n\\}\backslash w_{0}(L_{\leq k})]$
$\displaystyle=$
$\displaystyle[\\{1,\dots,n\\}\backslash\\{w_{0}(L_{>k})\sqcup w_{0}(L_{\leq
k})\\}]$ $\displaystyle=$ $\displaystyle[\\{1,\dots,n\\}\backslash w_{0}(L)]$
$\displaystyle=$
$\displaystyle[\,\widehat{w_{0}(L)}\,]=[w_{0}(\,\widehat{L}\,)].$

As the quantum Plücker coordinates of $\mathcal{O}_{q}(G(k,n))$ generate
$\mathcal{O}_{q}(G(k,n))$ as an algebra, and their images generate
$\mathcal{O}_{q}(G(n-k,n))$ as an algebra, this calculation shows that the
isomorphism displayed at the beginning of the proof restricts to an
isomorphism between $\mathcal{O}_{q}(G(k,n))$ and $\mathcal{O}_{q}(G(n-k,n))$.
∎

An immediate corollary of this result is the following.

######  Corollary 3.5.

When $2k=n$, there is an automorphism of $\mathcal{O}_{q}(G(k,n))$ which sends
the quantum Plücker coordinate $[I]$ to $[w_{0}(\,\widehat{I}\,\,)]$, where
$w_{0}$ is the longest element of the symmetric group on $\\{1,\dots,n\\}$.

######  Remark 3.6.

The automorphism in the previous corollary will be called the diagram
automorphism. In Figure 1, which shows the standard poset for
$\mathcal{O}_{q}(G(3,6))$, the effect of this automorphism on the quantum
Plücker coordinates is seen by reflection of the poset in the vertical. For
example, $[126]$ is sent to $[w_{0}(\,\widehat{126}\,)]=[w_{0}(345)]=[234]$.
There is a diagram automorphism for $\mathcal{O}_{q}(G(k,n))$ only in the case
that $n=2k$. Note that both the diagram automorphism and the transpose
automorphism $\tau$ extend to
$\mathcal{O}_{q}(G(k,2k)([u]^{-1})=\mathcal{O}_{q}(M(k,k))[y,y^{-1};\sigma]$
and they agree on this common overring, so we denote the diagram automorphism
by $\tau$ also.

##  4 Obvious automorphisms of $\mathcal{O}_{q}(G(k,n))$

There are two obvious sources of automorphisms of $\mathcal{O}_{q}(G(k,n))$.
The first is via the inclusion
$\mathcal{O}_{q}(G(k,n))\subseteq\mathcal{O}_{q}(M(k,n))$. The second is by
using the the dehomogenisation equality introduced in Section 3:

${\mathcal{O}}_{q}(M(k,p))[y^{\pm
1};\sigma]=\mathcal{O}_{q}(G(k,n))[[u]^{-1}],$

where $p=n-k$ and $u=\\{1,\dots,k\\}$ while $\sigma$ is the automorphism of
$\mathcal{O}_{q}(M(k,p))$ such that $\sigma(x_{ij})=qx_{ij}$ for each
$i=1,\dots,k$ and $j=1,\dots,p$.

In this section, we introduce these automorphisms and consider the connections
between them.

First, $\mathcal{O}_{q}(G(k,n))$ is a subalgebra of $\mathcal{O}_{q}(M(k,n))$
by definition. The torus ${\mathcal{H}}_{0}:=({K}^{*})^{n}$ acts by column
multiplication on $\mathcal{O}_{q}(M(k,n))$ and this induces an action on
$\mathcal{O}_{q}(G(k,n))$ defined on quantum Plücker coordinates by

$(\beta_{1},\dots,\beta_{n})\cdot[i_{1},\dots,i_{k}]=\beta_{i_{1}}\dots\beta_{i_{k}}[i_{1},\dots,i_{k}].$

This is the torus action on $\mathcal{O}_{q}(G(k,n))$ that is considered in
papers such as [12, 14].

Secondly, there is an action of the torus $({K}^{*})^{k+p}$ on
$\mathcal{O}_{q}(M(k,p))$ which operates by row and column scaling, so that
$(\alpha_{1},\dots,\alpha_{k};\beta_{1},\dots,\beta_{p})\cdot
x_{ij}=\alpha_{i}\beta_{j}x_{ij}$. As $n=k+p$, we can extend this to an action
of the torus ${\mathcal{H}}_{1}:=({K}^{*})^{n+1}$ on
$\mathcal{O}_{q}(M(k,p))[y^{\pm 1};\sigma]$ by setting

$(\alpha_{0};\alpha_{1},\dots,\alpha_{k};\beta_{1},\dots,\beta_{p})\cdot
x_{ij}=\alpha_{i}\beta_{j}x_{ij},\quad(\alpha_{0};\alpha_{1},\dots,\alpha_{k};\beta_{1},\dots,\beta_{p})\cdot
y=\alpha_{0}y.$

Set
$h=(\alpha_{0};\alpha_{1},\dots,\alpha_{k};\beta_{1},\dots,\beta_{p})\in{\mathcal{H}}_{1}$.
It is easy to check that $h\cdot[I\mid J]=\alpha_{I}\beta_{J}[I\mid J]$, where
$\alpha_{I}:=\alpha_{i_{1}}\dots\alpha_{i_{k}}$ when
$I=[\alpha_{i_{1}},\dots,\alpha_{i_{k}}]$, and
$\beta_{J}=\beta_{j_{1}}\dots\beta_{j_{p}}$ when
$J=[\beta_{j_{1}},\dots,\beta_{j_{k}}]$.

The dehomogenisation equality induces an action of ${\mathcal{H}}_{1}$ on
$\mathcal{O}_{q}(G(k,n))[[u]^{-1}]$. For
$h=(\alpha_{0};\alpha_{1},\dots,\alpha_{k};\beta_{1},\dots,\beta_{p})\in{\mathcal{H}}_{1}$
we have $h\cdot[u]=\alpha_{0}[u]$. If $[L]$ is any other quantum Plücker
coordinate then we may write $[L]=[I\mid J][u]$, where
$I=(k+1)-\left(\\{1,\dots,k\\}\backslash L_{\leq k}\right)$ and $J=L_{>k}-k$,
by Lemma 3.2. Then

$h\cdot[L]=h\cdot[I\mid J]\times h\cdot[u]=\alpha_{I}\beta_{J}[I\mid
J]\times\alpha_{0}[u]=\alpha_{0}\alpha_{I}\beta_{J}[I\mid
J][u]=\alpha_{0}\alpha_{I}\beta_{J}[L].$

As the quantum Plücker coordinates generate $\mathcal{O}_{q}(G(k,n))$ and are
sent to scalar multiples of themselves by each $h\in{\mathcal{H}}_{1}$, such
$h$ act as automorphisms of $\mathcal{O}_{q}(G(k,n))$.

We now consider connections between the actions of ${\mathcal{H}}_{0}$ and
${\mathcal{H}}_{1}$ on $\mathcal{O}_{q}(G(k,n))$.

######  Lemma 4.1.

For every automorphism $g\in{\mathcal{H}}_{0}$ acting on
$\mathcal{O}_{q}(G(k,n))$ there is an automorphism $f\in{\mathcal{H}}_{1}$
which has the same action on $\mathcal{O}_{q}(G(k,n))$.

###### Proof.

Let $g=(a_{1},\dots,a_{n})\in{\mathcal{H}}_{0}$. We seek
$f=(\alpha_{0};\alpha_{1},\dots,\alpha_{k};\beta_{1},\dots,\beta_{p})\in{\mathcal{H}}_{1}$
such that the actions of $g$ and $f$ on $\mathcal{O}_{q}(G(k,n))$ are the
same. As $g\cdot[u]=a_{1}\dots a_{k}[u]$, we may extend $g$ to act on
$\mathcal{O}_{q}(G(k,n))[[u]^{-1}]$. The dehomogenisation equality then
transfers this action to ${\mathcal{O}}_{q}(M(k,p))[y^{\pm 1};\sigma]$. We
calculate the action of $g$ on the generators $x_{ij}$ and $y^{\pm 1}$ of
${\mathcal{O}}_{q}(M(k,p))[y^{\pm 1};\sigma]$. As $y$ corresponds to $[u]$ in
the dehomogenisation equality, $g\cdot y=a_{1}\dots a_{k}y$ and $g\cdot
y^{-1}=(a_{1}\dots a_{k})^{-1}y^{-1}$. Now,
$x_{ij}:=[1\dots,\widehat{k+1-i},\dots k,j+k][u]^{-1}$; so

$\displaystyle g\cdot x_{ij}$ $\displaystyle=$ $\displaystyle
g\cdot[1\dots,\widehat{k+1-i},\dots k,j+k]\times g\cdot[u]^{-1}$
$\displaystyle=$ $\displaystyle a_{1}\dots
a_{k}a_{k+1-i}^{-1}a_{j+k}[1\dots,\widehat{k+1-i},\dots
k,j+k]\times(a_{1}\dots a_{k})^{-1}[u]^{-1}$ $\displaystyle=$ $\displaystyle
a_{j+k}a_{(k+1)-i}^{-1}[1\dots,\widehat{k+1-i},\dots k,j+k][u]^{-1}$
$\displaystyle=$ $\displaystyle a_{j+k}a_{(k+1)-i}^{-1}x_{ij}$

We seek an $f\in{\mathcal{H}}$ which has the same effect on the generators
$x_{ij}$ and $y$. Set $f=(a_{1}\dots
a_{k};a_{k}^{-1},\dots,a_{1}^{-1};a_{k+1},\dots,a_{n})\in{\mathcal{H}}$. Then
$f\cdot y=a_{1}\dots a_{k}y=g\cdot y$. Also, the entry in $f$ multiplying the
$i$th row is $a_{(k+1)-i}^{-1}$ and the entry multiplying the $j$th column is
$a_{k+j}$ so $f\cdot x_{ij}=a_{(k+1)-i}^{-1}a_{k+j}x_{ij}=g\cdot x_{ij}$.
Hence, $f$ and $g$ agree on the generators $x_{ij}$ and $y^{\pm 1}$ of
${\mathcal{O}}_{q}(M(k,p))[y^{\pm
1};\sigma]=\mathcal{O}_{q}(G(k,n))[[u]^{-1}]$; so the actions of $f$ and $g$
on $\mathcal{O}_{q}(G(k,n))$ are the same.∎

The converse question is more delicate, as the following example shows.

######  Example 4.2.

Let $K$ be a field in which there is no element $b$ such that $b^{2}=2$ (eg
${\mathbb{Q}}$) and consider $\mathcal{O}_{q}(G(2,4))$ over this field. Let
$f=(1;2,1;1,1)\in{\mathcal{H}}_{1}$. Then there is no element
$g=(a_{1},a_{2},a_{3},a_{4})\in{\mathcal{H}}_{0}$ whose action on
$\mathcal{O}_{q}(G(2,4))$ coincides with the action of $f$ on
$\mathcal{O}_{q}(G(2,4))$. To verify this claim, we use the formulae in Lemma
3.2 to see that

$[12]=[u],\;[13]=x_{11}[u],\;[14]=x_{12}[u],\;[23]=x_{21}[u],\;[24]=x_{22}[u],\;[34]=[12\mid
12][u].$

These equations lead to the following actions of $f$ on the quantum Plücker
coordinates:

$f\cdot[12]=[12],\;f\cdot[13]=2[13],\;f\cdot[14]=2[14],\;f\cdot[23]=[23],\;f\cdot[24]=[24],\;f\cdot[34]=2[34].$

Next, for $g=(a_{1},a_{2},a_{3},a_{4})\in{\mathcal{H}}_{0}$ we see that

$g\cdot[12]=a_{1}a_{2}[12],\;g\cdot[13]=a_{1}a_{3}[13],\;g\cdot[14]=a_{1}a_{4}[14],\;$
$g\cdot[23]=a_{2}a_{3}[23],\;g\cdot[24]=a_{2}a_{4}[24],\;g\cdot[34]=a_{3}a_{4}[34];$

so for this $g$ to act in the same way as $f$ we require that

$a_{1}a_{2}=1,\;a_{1}a_{3}=2,\;a_{1}a_{4}=2,\;a_{2}a_{3}=1,\;a_{2}a_{4}=1,\;a_{3}a_{4}=2.$

Equations 4 and 5 immediately above show that we need
$a_{3}=a_{4}\;(=b{\rm~{}say})$ and then equation 6 gives $b^{2}=2$, which is
not possible for any $b\in K$.

In view of this example, it is appropriate to assume that $K$ is algebraically
closed in the following result.

######  Lemma 4.3.

Suppose that $K$ is algebraically closed. Consider $\mathcal{O}_{q}(G(k,n))$
and $\mathcal{O}_{q}(M(k,p))$ over $K$. For every automorphism
$f\in{\mathcal{H}}_{1}$ acting on $\mathcal{O}_{q}(G(k,n))$ there is an
automorphism $g\in{\mathcal{H}}_{0}$ which has the same action on
$\mathcal{O}_{q}(G(k,n))$.

###### Proof.

Consider the set $S$ of elements
$f=(\alpha_{0};\alpha_{1},\dots,\alpha_{k};\beta_{1},\dots,\beta_{p})\in{\mathcal{H}}_{1}$
that are equal to $1$ in all positions except one. The set $S$ generates
${\mathcal{H}}_{1}$ so it is enough to show that the action of each member of
$S$ can be realised via the action of an element of ${\mathcal{H}}_{0}$. We
look at three cases separately.

The first case we consider is when one of the $\beta_{i}$ terms is not equal
to $1$. Suppose that the element $\beta_{j}$ in position $k+1+j$ of $f$ is not
equal to $1$ but that all other positions of $f$ contain the element $1$. Note
that $f\cdot[u]=[u]$. Let $[L]$ be any other quantum Plücker coordinate in
$\mathcal{O}_{q}(G(k,n))$ with $[L]=[I\mid J][u]$ for some quantum minor
$[I\mid J]\in\mathcal{O}_{q}(M(k,p))$. Then $f\cdot[L]=(f\cdot[I\mid
J])(f\cdot[u])=(f\cdot[I\mid J])[u]$ so that $f\cdot[L]=[L]$ when $j\not\in J$
and $f\cdot[L]=\beta_{j}[L]$ when $j\in J$.

By using Corollary 3.3(ii) we see that $f\cdot[L]=[L]$ if $j+k\not\in L$ and
that $f\cdot[L]=\beta_{j}[L]$ when $j+k\in L$. An element of
$g\in{\mathcal{H}}_{0}$ that has the same effect on quantum Plücker
coordinates is the $g=(g_{1},\dots,g_{n})$ where $g_{j+k}=\beta_{j}$ while all
other $g_{i}=1$.

The next case that we consider is
$f=(\alpha_{0};\alpha_{1},\dots,\alpha_{k};\beta_{1},\dots,\beta_{p})\in{\mathcal{H}}_{1}$
where $\alpha_{i}$ is not equal to $1$ for some $i\in\\{1,\dots,k\\}$ but all
other entries of $f$ are equal to $1$. Again, $f\cdot[u]=[u]$. Let $[L]$ be
any other quantum Plücker coordinate in $\mathcal{O}_{q}(G(k,n))$ with
$[L]=[I\mid J][u]$ for some quantum minor $[I\mid
J]\in\mathcal{O}_{q}(M(k,p))$. Then $f\cdot[L]=(f\cdot[I\mid
J])(f\cdot[u])=(f\cdot[I\mid J])[u]$ so that $f\cdot[L]=[L]$ when $i\not\in I$
and $f\cdot[L]=\alpha_{i}[L]$ when $i\in I$. By using Corollary 3.3(i) we see
that $f\cdot[L]=[L]$ if $(k+1)-i\in L$ and $f\cdot[L]=\alpha_{i}[L]$ when
$(k+1)-i\not\in L$. Let $b$ be an element in $K$ such that $b^{k}=\alpha_{i}$.
Let $g=(g_{1},\dots,g_{n})\in{\mathcal{H}}_{0}$ be such that
$g_{k+1-i}=b^{-(k-1)}$ while every other entry $g_{j}=b$. Let
$[L]=[l_{1},\dots,l_{k}]$. Then $g\cdot[L]=g_{l_{1}}\dots g_{l_{k}}[L]$.

Suppose that $k+1-i\in L$. Then one of $g_{l_{1}},\dots,g_{l_{k}}$ is equal to
$b^{-(k-1)}$ while the other $k-1$ are equal to $b$. Thus, $g_{l_{1}}\dots
g_{l_{k}}=b^{k-1}b^{-(k-1)}=1$ and so $g\cdot[L]=[L]$.

Now assume that $k+1-i\not\in L$. Then each of the $g_{l_{j}}$ is equal to $b$
and so $g\cdot[L]=b^{k}[L]=\alpha_{i}[L]$.

This shows that the action of $g$ coincides with the action of $f$, as
required.

The final case to consider is
$f=(\alpha_{0};\alpha_{1},\dots,\alpha_{k};\beta_{1},\dots,\beta_{p})\in{\mathcal{H}}_{1}$
where $\alpha_{0}$ is not equal to $1$ but all other entries are equal to $1$.
Thus, $f\cdot[u]=\alpha_{0}[u]$. Let $[L]$ be any other quantum Plücker
coordinate in $\mathcal{O}_{q}(G(k,n))$ with $[L]=[I\mid J][u]$ for some
quantum minor $[I\mid J]\in\mathcal{O}_{q}(M(k,p))$. Note that $f\cdot[I\mid
J]=[I\mid J]$ for all quantum minors $[I\mid J]$. Hence,
$f\cdot[L]=(f\cdot[I\mid J])(f\cdot[u])=[I\mid
J].\alpha_{0}[u]=\alpha_{0}[L]$. Thus, $f\cdot[L]=\alpha_{0}[L]$ for all
quantum minors $[L]$ of $\mathcal{O}_{q}(G(k,n))$. Let $b\in K$ be such that
$b^{k}=\alpha_{0}$ and set $g=(b,\dots,b)$. Then
$g\cdot[L]=b^{k}[L]=\alpha_{0}[L]$ for all $[L]$ and the actions of $f$ and
$g$ coincide. ∎

The action of ${\mathcal{H}}_{1}$ on $\mathcal{O}_{q}(G(k,n))$ is not
faithful, as we will see in the next proposition. Let ${\mathcal{H}}$ denote
${\mathcal{H}}_{1}$ factored by this kernel of this action. Then
${\mathcal{H}}$ acts faithfully on $\mathcal{O}_{q}(G(k,n))$. The next result
shows that ${\mathcal{H}}$ is isomorphic to a torus $(K^{*})^{n}$.

######  Proposition 4.4.

The group ${\mathcal{H}}$ is isomorphic to a torus $(K^{*})^{n}$.

###### Proof.

The kernel of the action of ${\mathcal{H}}_{1}$ on $\mathcal{O}_{q}(G(k,n))$
is the same as the kernel of the action of ${\mathcal{H}}_{1}$ on
$\mathcal{O}_{q}(G(k,n))[[u]^{-1}]=\mathcal{O}_{q}(M(k,p))[y^{\pm 1}]$. Using
the right hand side, it is easy to check that this kernel is
$\\{(1;\lambda,\dots,\lambda;\lambda^{-1},\dots,\lambda^{-1})\mid\lambda\in{K}^{*}\\}$.
Hence, choosing $\lambda=\beta_{p}$, we see that
$h=(\alpha_{0};\alpha_{1},\dots,\alpha_{k};\beta_{1},\dots,\beta_{p})\in{\mathcal{H}}_{1}$
has the same action as
$h^{\prime}:=(\alpha_{0};\alpha_{1}\beta_{p},\dots,\alpha_{k}\beta_{p};\beta_{1}\beta_{p}^{-1},\dots,\beta_{p-1}\beta_{p}^{-1},1)\in{\mathcal{H}}_{1}$.
It is also easy to check that two distinct elements in ${\mathcal{H}}_{1}$
that each have $1$ in the final place act differently on
$\mathcal{O}_{q}(M(k,p))[y^{\pm 1}]$, and so the claim is established. ∎

In view of this result, we refer to the actions on $\mathcal{O}_{q}(G(k,n))$
provided by ${\mathcal{H}}$ as the torus automorphisms of
$\mathcal{O}_{q}(G(k,n))$.

In the case that $n=2k$, so that $k=p$, the dehomogenisation equality states
that

${\mathcal{O}}_{q}(M(k,k))[y^{\pm
1};\sigma]=\mathcal{O}_{q}(G(k,n))[[u]^{-1}].$

In this case, a simple analysis using the formula $[L]=[I\mid J][u]$ shows
that the extra automorphism of ${\mathcal{O}}_{q}(M(k,k))$ given by
transposition of the $x_{ij}$ variables extends to an automorphism of
${\mathcal{O}}_{q}(M(k,k))[y^{\pm
1};\sigma]=\mathcal{O}_{q}(G(k,n))[[u]^{-1}]$ which, when restricted to
$\mathcal{O}_{q}(G(k,n))$, gives rise to the diagram automorphism $\tau$ of
Corollary 3.5. We will denote this automorphism by $\tau$ for each of the
three algebras $\mathcal{O}_{q}(G(k,n))$, ${\mathcal{O}}_{q}(M(k,k))$ and
${\mathcal{O}}_{q}(M(k,k))[y^{\pm
1};\sigma]=\mathcal{O}_{q}(G(k,n))[[u]^{-1}]$.

In the case where $2k=n$, let
$(\alpha_{0};\alpha_{1},\dots,\alpha_{k};\beta_{1},\dots,\beta_{k})\in{\mathcal{H}}_{1}$.
It is easy to check that

$\tau\circ(\alpha_{0};\alpha_{1},\dots,\alpha_{k};\beta_{1},\dots,\beta_{k})\circ\tau=(\alpha_{0};\beta_{1},\dots,\beta_{k};\alpha_{1},\dots,\alpha_{k})\in{\mathcal{H}}_{1},$

so that $\left<\tau\right>$ acts on ${\mathcal{H}}_{1}$. Also, if
$(1;\lambda,\dots,\lambda;\lambda^{-1},\dots,\lambda^{-1})$ is in the kernel
of the action of ${\mathcal{H}}_{1}$ on $\mathcal{O}_{q}(G(k,n))[[u]^{-1}]$
then

$\displaystyle\tau\circ(1;\lambda,\dots,\lambda;\lambda^{-1},\dots,\lambda^{-1})\circ\tau$
$\displaystyle=(1;\lambda^{-1},\dots,\lambda^{-1};\lambda,\dots,\lambda)$
$\displaystyle=(1;\lambda^{-1},\dots,\lambda^{-1};(\lambda^{-1})^{-1},\dots,(\lambda^{-1})^{-1})$

is also in the kernel of this action and so $\left<\tau\right>$ acts on
${\mathcal{H}}$.

######  Definition 4.5.

Set ${\mathcal{A}}:={\mathcal{H}}$ when $2k\neq n$ and
${\mathcal{A}}:={\mathcal{H}}\rtimes\langle\tau\rangle$ when $2k=n$.

The analysis above shows that the elements of ${\mathcal{A}}$ act naturally as
automorphisms of $\mathcal{O}_{q}(G(k,n))$ via the dehomogenisation equality.

######  Claim 4.6.

The automorphism group of $\mathcal{O}_{q}(G(k,n))$ is ${\mathcal{A}}$.

We will prove this claim in the following sections.

##  5 Adjusting automorphisms

The quantum grassmannian $\mathcal{O}_{q}(G(k,n))$ carries the structure of an
${\mathbb{N}}$-graded algebra generated in degree one when we give degree one
to each of the quantum Plücker coordinates. In addition, [13, Theorem 5.3]
shows that $\mathcal{O}_{q}(G(k,n))$ is a unique factorisation domain in the
sense of Chatters [3]. According to Chatters, an element $p$ of a noetherian
domain $R$ is said to be prime if (i) $pR=Rp$, (ii) $pR$ is a height one prime
ideal of $R$, and (iii) $R/pR$ is an integral domain. A noetherian domain R is
then said to be a unique factorisation domain if $R$ has at least one height
one prime ideal, and every height one prime ideal is generated by a prime
element.

In this section, we exploit these properties of $\mathcal{O}_{q}(G(k,n))$ in a
series of results to see that given an arbitrary automorphism of
$\mathcal{O}_{q}(G(k,n))$ we can essentially fix the minimal and maximal
elements in the poset after allowing adjustment of the automorphism by
elements of ${\mathcal{H}}$.

######  Lemma 5.1.

Let $A=\oplus_{i=0}^{\infty}\,A_{i}$ be a graded algebra that is a domain with
$A_{0}$ equal to the base field and $A$ generated in degree one. Suppose that
$a=a_{1}+\dots+a_{m}$ is a normal element with $a_{i}\in A_{i}$ for each $i$.
Then $a_{1}$ is a normal element.

###### Proof.

If $a_{1}=0$ then there is nothing to prove; so assume that $a_{1}\neq 0$. As
$A$ is generated in degree one, it is enough to check normality with respect
to homogeneous elements of degree one; so suppose that $b\in A_{1}$. Then
$ba=ba_{1}+\dots+ba_{m}=ac=(a_{1}+\dots+a_{m})(c_{0}+c_{1}+\dots+c_{t})$ for
some $c=c_{0}+c_{1}+\dots+c_{t}\in A$ with $c_{i}\in A_{i}$. Comparing degree
one terms gives $0=a_{1}c_{0}$; so $c_{0}=0$. The degree two terms then show
that $ba_{1}=a_{1}c_{1}\in a_{1}A$, and this demonstrates that $a_{1}$ is
normal. ∎

######  Lemma 5.2.

Let $A=\oplus_{i=0}^{\infty}\,A_{i}$ be a graded algebra that is a domain with
$A_{0}$ equal to the base field. Suppose also that $A$ is a unique
factorisation domain.

Let $a$ be a homogeneous element of degree one that is normal. Then $a$
generates a prime ideal of height one.

###### Proof.

Let $P$ be a prime that is minimal over the ideal $aR$. By the noncommutative
principal ideal theorem [16, Theorem 4.1.11], the height of $P$ is one. Hence,
$P=pA$ for some normal element $p$, as $A$ is a UFD. Thus, $a$ is a (right)
multiple of $p$. By degree considerations, $p$ must have degree one and $a$
must be a scalar multiple of $p$. Thus, $a$ and $p$ generate the same ideal,
which is the prime ideal $P$. This establishes the claim. ∎

The remaining results in this section all deal with $\mathcal{O}_{q}(G(k,n))$.
As in earlier sections, let $[u]=[1,\dots,k]$.

######  Lemma 5.3.

Suppose that $a=\sum a_{I}[I]\neq 0$, with $a_{I}\in{K}$, is a linear
combination of quantum Plücker coordinates that is a normal element. Then
$d(I)$ is the same for each $I$ that has $a_{I}\neq 0$.

###### Proof.

Since $a=\sum a_{I}[I]$ has degree one, $a$ is irreducible, as well as being
normal. Hence, the ideal $P$ generated by $a$ is a height one prime ideal of
$\mathcal{O}_{q}(G(k,n))$, by Lemma 5.2.

If $a=a_{I}[I]$ for some $[I]$ then the result holds. Thus we may assume that
at least two scalars $a_{I}$ are nonzero. In particular, $aK\neq[u]K$, and so
$[u]\not\in a\mathcal{O}_{q}(G(k,n))=P$.

Let $J$ be such that $d(J)$ is as small as possible among those $d(I)$ for
which $a_{I}\neq 0$. We will show that $d(I)=d(J)$ for all $I$ such that
$a_{I}\neq 0$.

Now

$[u]a=[u]\left(\sum_{I\neq J}\,a_{I}[I]+a_{J}[J]\right)=\left(\sum_{I\neq
J}\,a_{I}q^{d(I)}[I]+a_{J}q^{d(J)}[J]\right)[u],$

by using Lemma 3.1, and so

$\left(\sum_{I\neq
J}\,a_{I}q^{d(I)}[I]+a_{J}q^{d(J)}[J]\right)[u]-q^{d(J)}\left(\sum_{I\neq
J}\,a_{I}[I]+a_{J}[J]\right)[u]=[u]a-q^{d(J)}a[u]\in P;$

that is,

$b:=\left(\sum_{I\neq J}\,(q^{d(I)}-q^{d(J)})a_{I}[I]\right)[u]\in P$

As $[u]\not\in P$ this gives

$b=\left(\sum_{I\neq J}\,(q^{d(I)}-q^{d(J)})a_{I}[I]\right)\in
P=a\mathcal{O}_{q}(G(k,n)).$

Thus, there is a scalar $\lambda\in K$ with $b=\lambda a$. If $\lambda\neq 0$
then this is a contradiction, as $[J]$ occurs nontrivially in $\lambda a$ and
not in $b$. Therefore, $b=0$ and so each $q^{d(I)}-q^{d(J)}=0$. This forces
$d(I)=d(J)$, since $q$ is not a root of unity. ∎

In order to prove the next lemma, we need to check that we can apply [10,
Proposition 4.2] to the quantum grassmannian. In order to do this, we need to
observe that for each quantum Plücker coordinate $[L]$ there exists a quantum
Plücker coordinate $[L^{\prime}]$ such that
$[L][L^{\prime}]=\alpha[L^{\prime}][L]$ for some $1\neq\alpha\in K$. We use
Lemma 3.1 to do this. If $[L]=[u]=[1,\dots,k]$ then set
$[L^{\prime}]=[1,\dots,k-1,k+1]$ so that $[L][L^{\prime}]=q[L^{\prime}][L]$
while if $[L]\neq[u]$ set $[L^{\prime}]=[u]$ so that
$[L][L^{\prime}]=q^{-d(L^{\prime})}[L^{\prime}][L]$.

######  Lemma 5.4.

Let $\rho$ be an automorphism of $\mathcal{O}_{q}(G(k,n))$. Then
$\rho([u])_{1}=\lambda[u]$, for some $\lambda\in K^{*}$.

###### Proof.

Note that $\rho([u])$ is a normal element of $\mathcal{O}_{q}(G(k,n))$; so
Lemma 5.1 applied to $\rho([u])$ shows that $\rho([u])_{1}$ is a normal
element. If $\rho([u])_{1}=\lambda[u]$, for some $\lambda\in K^{*}$, then we
are finished; so assume that $\rho([u])_{1}\neq\lambda[u]$. Then
$\rho([u])_{1}=\sum a_{I}[I]$ for some $[I]$ with each $d(I)=d>0$, by Lemma
5.3.

Let $w\in\mathcal{O}_{q}(G(k,n))$ be such that $\rho(w)=[u]$, or,
equivalently, $\rho^{-1}([u])=w$. Note that the degree zero term of $w$ must
be zero. Write $w=w_{1}+w_{>1}$. As $\rho(w_{>1})_{1}=0$, by [10, Proposition
4.2], we see that $[u]=\rho(w)=\rho(w_{1})+\rho(w_{>1})$; so that
$[u]=\rho(w)_{1}=\rho(w_{1})_{1}$. If $w_{1}=\lambda[u]$ then
$[u]=\rho(w_{1})_{1}=\rho(\lambda[u])_{1}=\lambda\rho([u])_{1}$ and so
$\rho([u])_{1}=\lambda^{-1}[u]$, a contradiction; so we have $w_{1}\neq\lambda
u$, for any $\lambda\in K^{*}$. By applying Lemma 5.3 to $w_{1}$, we may write
$w_{1}=\sum b_{J}[J]$ for some $[J]$ with each $d(J)=e>0$, say. Hence,
$[u]w_{1}=q^{e}w_{1}[u]$.

Consider the degree two term in $\rho([u]w)$. We know that
$\rho([u]w)=\rho([u])\rho(w)$ and that the degree zero terms of $\rho([u])$
and $\rho(w)$ are both zero, by [10, Proposition 4.2]. Hence,

$\rho([u]w)_{2}=\rho([u])_{1}\rho(w)_{1}=\left(\sum
a_{I}[I]\right)[u]=q^{-d}[u]\left(\sum a_{I}[I]\right).$

On the other hand,
$\rho([u]w)=\rho([u]w_{1}+[u]w_{>1})=\rho([u]w_{1})+\rho([u]w_{>1})$. Now,
$[u]w_{>1}$ has no term in degree less than three. Hence, the same is true for
$\rho([u]w_{>1})$, by [10, Proposition 4.2], and, in particular,
$\rho([u]w_{>1})_{2}=0$. Thus, $\rho([u]w)_{2}=\rho([u]w_{1})_{2}$. However,
$\rho([u]w_{1})=\rho(q^{e}w_{1}[u])=q^{e}\rho(w_{1})\rho([u])$. Therefore,

$\rho([u]w)_{2}=\rho([u]w_{1})_{2}=\rho(q^{e}w_{1}[u])_{2}=q^{e}\rho(w_{1})_{1}\rho([u])_{1}=q^{e}[u]\left(\sum
a_{I}[I]\right)$

The two expressions we have obtained for $\rho([u]w)_{2}$ must be equal; so
$q^{-d}[u](\sum a_{I}[I])=q^{e}[u](\sum a_{I}[I])$. Hence, $q^{-d}=q^{e}$; so
that $q^{e+d}=1$. As $q$ is not a root of unity and $e+d>0$, this is a
contradiction, and our lemma is proved. ∎

######  Lemma 5.5.

Let $\rho$ be an automorphism of $\mathcal{O}_{q}(G(k,n))$. Then
$\rho([u])=\lambda[u]$, for some $\lambda\in K^{*}$.

###### Proof.

The element $\rho([u])$ is a normal element, and the degree zero term of
$\rho([u])$ is equal to $0$, by [10, Proposition 4.2]. Suppose that the degree
of $\rho([u])$ is $t$ and that $\rho([u])=a_{1}+\dots+a_{t}$ with
$\deg(a_{i})=i$. Recall that $a_{1}=\lambda[u]$, for some $\lambda\in K^{*}$,
by Lemma 5.4. There is an element $r\in\mathcal{O}_{q}(G(k,n))$ with
$[u]\rho([u])=\rho([u])r$. The degree of $r$ must be one. Assume
$r=r_{0}+r_{1}$ with $r_{i}$ having degree $i$. Thus,

$[u](\lambda[u]+a_{2}+\dots+a_{t})=(\lambda[u]+a_{2}+\dots+a_{t})(r_{0}+r_{1}).$

As there is no term in degree one on the left hand side of the above equation,
we must have $r_{0}=0$. Looking at terms in degree two, we then see that
$\lambda[u]^{2}=\lambda[u]r_{1}$; so that $r_{1}=[u]$ and
$[u]\rho([u])=\rho([u])[u]$.

Write $\rho([u])$ in terms of the standard basis for
$\mathcal{O}_{q}(G(k,n))$, as in [7], say $\rho([u])=\sum\alpha_{i}S_{i}$,
where $\alpha_{i}$ is in the field and each $S_{i}$ is a standard monomial. If
$S=[I_{1}]\dots[I_{m}]$ is such a standard monomial, then set $d(S):=\sum
d(I_{i})$ and note that each $d(I_{i})\geq 0$ with $d(I_{i})=0$ if and only if
$[I_{i}]=[u]$. Then, $S[u]=q^{d(S)}[u]S$, and so $S[u]=[u]S$ if and only if
$d(S)=0$ (in which case $S=[u]^{m}$ for some $m$).

In any case, note that $[u]S$ is a standard monomial, as $[u]$ is the unique
minimal quantum Plücker coordinate. Hence,

$\sum\alpha_{i}[u]S_{i}=[u]\rho([u])=\rho([u])[u]=\left(\sum\alpha_{i}S_{i}\right)[u]=[u]\left(\sum\alpha_{i}q^{d(S_{i})}S_{i}\right)=\sum\alpha_{i}q^{d(S_{i})}[u]S_{i}$

As the extreme left and right terms in the above display are in the standard
basis, this forces $d(S_{i})=0$ whenever $\alpha_{i}\neq 0$. Hence,
$\rho([u])$ must be a polynomial in $[u]$.

The same argument applies to the automorphism $\rho^{-1}$; so $\rho^{-1}([u])$
must also be a polynomial in $[u]$.

Suppose that $\rho([u])=\sum_{i=1}^{t}\alpha_{i}[u]^{i}$ with $\alpha_{t}\neq
0$, and, similarly, suppose that
$\rho^{-1}([u])=\sum_{i=1}^{s}\beta_{i}[u]^{i}$ with $\beta_{s}\neq 0$

Then,

$[u]=\rho^{-1}\rho([u])=\rho^{-1}\left(\sum_{i=1}^{t}\alpha_{i}[u]^{i}\right)=\sum_{i=1}^{t}\alpha_{i}\rho^{-1}([u])^{i}=\alpha_{1}\beta_{1}[u]+\dots+\alpha_{t}\beta_{s}[u]^{st}$

Therefore, $s=t=1$, and $\rho([u])=\lambda[u]$, for some $\lambda\in K^{*}$,
as required. ∎

The above result refers to $[u]=[1,\dots,k]$, the extreme leftmost quantum
Plücker coordinate. We want to establish a similar result for
$[w]:=[n-k+1,\dots,n]$, the extreme rightmost quantum Plücker coordinate. In
order to do this we employ an antiautomorphism of $\mathcal{O}_{q}(G(k,n))$
which we now describe.

Let $w_{0}$ denote the longest element on the symmetric group on $n$ elements;
that is, $w_{0}(i)=n+1-i$. The discussion immediately before Proposition 2.12
of [2] shows that the map
$\theta:\mathcal{O}_{q}(G(k,n))\longrightarrow\mathcal{O}_{q}(G(k,n))$ given
by $\theta([I])=[w_{0}(I)]$ for each quantum Plücker coordinate $[I]$ is an
antiautomorphism. Note that
$\theta([u])=\theta([1,\dots,k])=[n+1-k,\dots,n]=[w]$.

######  Corollary 5.6.

Let $\rho$ be an automorphism of $\mathcal{O}_{q}(G(k,n))$. Then there exists
$h\in{\mathcal{H}}$ such that $(h\circ\rho)([u])=u$ and
$(h\circ\rho)([w])=[w]$.

###### Proof.

The map $\theta\rho\theta$ is an automorphism of $\mathcal{O}_{q}(G(k,n))$. By
Lemma 5.5, there is an element $\mu\in K^{*}$ such that
$\theta\rho\theta([u])=\mu[u]$. Apply $\theta$ to both sides of this equality
to obtain $\rho\theta([u])=\mu\theta([u])$; that is, $\rho([w])=\mu[w]$. We
also know that $\rho([u])=\lambda[u]$ for some $\lambda\in K^{*}$. Set
$h:=(\lambda^{-1},1,\dots,1,\mu^{-1})\in{\mathcal{H}}_{0}$. Then
$(h\circ\rho)([u])=[u]$ and $(h\circ\rho)([w])=[w]$. Lemma 4.1 shows that the
action of an element of ${\mathcal{H}}_{0}$ is realised by the action of an
element of ${\mathcal{H}}_{1}$ and hence of ${\mathcal{H}}$, so the result
follows. ∎

In what follows, we will often replace the original automorphism $\rho$ by
$h\circ\rho$ so that we may assume that $\rho([u])=[u]$ and $\rho([w])=[w]$ in
calculations.

##  6 Transfer to quantum matrices

Recall from the discussion in Section 2 that when discussing
$\mathcal{O}_{q}(G(k,n))$ we are assuming that $1<k$ and that $2k\leq n$.

Let $\rho$ be an automorphism of $\mathcal{O}_{q}(G(k,n))$. Set $[u]=[1\dots
k]$ and $[w]=[n-k+1,\dots,n]$. By using Corollary 5.6, at the expense of
adjusting $\rho$ by an element of ${\mathcal{H}}$, we can, and will, assume
that $\rho([u])=[u]$ and $\rho([w])=[w]$. The automorphism $\rho$ now extends
to $\mathcal{O}_{q}(G(k,n))[[u]^{-1}]$, and so to
${\mathcal{O}}_{q}(M(k,p))[y^{\pm 1};\sigma]$, by the dehomogenisation
equality of Section 3, and we know that $\rho(y)=y$.

We will show that such a $\rho$ sends ${\mathcal{O}}_{q}(M(k,p))$ to itself.
Once we have done this, we will know how $\rho$ acts on each quantum minor in
${\mathcal{O}}_{q}(M(k,p))$ as we know the automorphism group of
${\mathcal{O}}_{q}(M(k,p))$. We can then calculate how $\rho$ acts on
arbitrary quantum Plücker coordinates of $\mathcal{O}_{q}(G(k,n))$, by using
the formulae of Lemma 3.2.

From the discussion in Section 3, we know that the quantum matrix generators
$x_{ij}$ are defined by

$x_{ij}:=[1,\dots,\widehat{k+1-i},\dots k,j+k][u]^{-1},$

for $1\leq i\leq k$ and $1\leq j\leq p$. In the following calculations, all
quantum minors $[-\mid-]$ are formed from the generators $x_{ij}$ of
${\mathcal{O}}_{q}(M(k,p))$.

As $2k\leq n$, we know that $k\leq n-k=p$. In this case, the quantum minor
$[I\mid J]:=[1,\dots,k\mid p+1-k,\dots,p]$ is defined (we are using all the
rows of ${\mathcal{O}}_{q}(M(k,p))$ and the last $k$ columns (and there are at
least $k$ columns, by the assumption)).

Now, Lemma 3.2 shows that $[I\mid J]=[p+1\dots n][1\dots k]^{-1}=[w][u]^{-1}$,
and it follows this that $\rho([I\mid J])=[I\mid J]$ for any element
$\rho\in{\mathcal{H}}$ such that $\rho([u])=[u]$ and $\rho([w])=[w]$.

We can calculate how $[I\mid J]$ commutes with $[u]=[1\dots k]$. Note that
$k<n-k+1$, as $2k\leq n$. Thus the index sets $\\{1,\dots,k\\}$ and
$\\{p+1,\dots,n\\}$ do not overlap, and

$[u][I\mid
J]=[u][w][u]^{-1}=q^{k}[w][u][u]^{-1}=q^{k}[w][u]^{-1}[u]=q^{k}[I\mid
J]\,[u],$

where the second equality comes from Lemma 3.1.

Also, we know how $[I\mid J]$ commutes with the $x_{ij}$ by the following
lemma.

######  Lemma 6.1.

(i) If $j\geq n+1-2k=p+1-k$ then $x_{ij}[I\mid J]=[I\mid J]\,x_{ij}$.
(ii) If $j<n+1-2k=p+1-k$ then $x_{ij}[I\mid J]=q[I\mid J]\,x_{ij}$.

###### Proof.

(i) In this case, $x_{ij}$ is in the quantum matrix algebra determined by the
rows from $I$ and the columns from $J$ and $[I\mid J]$ is the quantum
determinant of this algebra, so the claim follows as $[I\mid J]$ is central in
this algebra.
(ii) This result is obtained from the first equation in E(1.3c) in [4, Section
1.3]. ∎

We define two gradings on $T:={\mathcal{O}}_{q}(M(k,p))[y^{\pm
1};\sigma]=\mathcal{O}_{q}(G(k,n))[[u]^{-1}]$ which grade $T$ according to how
elements commute with $y=u$ and $[I\mid J]=[1,\dots,k\mid p+1-k,\dots,p]$.

First, set $T_{i}:=\\{a\in T\mid yay^{-1}=q^{i}a\\}$.

######  Lemma 6.2.

(i) $T=\bigoplus_{i=1}^{\infty}\,T_{i}$,
(ii) $\rho(T_{i})=T_{i}$.

###### Proof.

Note that $T$ is generated by $y^{\pm 1}$ and the $x_{ij}$, and that $y^{\pm
1}\in T_{0}$, while $x_{ij}\in T_{1}$, as $yx_{ij}=qx_{ij}y$. As
$\mathcal{O}_{q}(M(k,p))$ is an iterated Ore extension with the elements
$x_{ij}$ added lexicographically, the elements of the form
$x_{11}^{a_{11}}\dots x_{kp}^{a_{kp}}y^{s}$ with $a_{ij}\geq 0$ and
$s\in{\mathbb{Z}}$ form a basis for $T$. Part (i) now follows as
$yx_{ij}y^{-1}=qx_{ij}$. Part (ii) follows from the fact that $\rho(y)=y$. ∎

Next, set $T^{(i)}:=\\{a\in T\mid[I\mid J]a[I\mid J]^{-1}=q^{-i}a\\}$. The
commutation rules given in Lemma 6.1 show that $x_{ij}\in T^{(0)}\cup
T^{(1)}$. Also, note that $y[I\mid J]=[u][I\mid J]=q^{k}[I\mid
J][u]=q^{k}[I\mid J]y$, so $y=u\in T^{(k)}$ and $y^{-1}\in T^{(-k)}$.

######  Lemma 6.3.

(i) $T=\bigoplus_{i\in{\mathbb{Z}}}\,T^{(i)}$,
(ii) $\rho(T^{(i)})=T^{(i)}$.

###### Proof.

Part (i) is proved as in the previous lemma, and Part (ii) follows from the
fact that $\rho([I\mid J])=[I\mid J]$.
∎

######  Lemma 6.4.

$\left(T^{(0)}\cup T^{(1)}\right)\cap T_{1}\subseteq\mathcal{O}_{q}(M(k,p))$.

###### Proof.

Suppose that $a\in\left(T^{(0)}\cup T^{(1)}\right)\cap T_{1}$. Then $a$ is a
sum of scalar multiples of monomials of the form $m:=x_{11}^{a_{11}}\dots
x_{kp}^{a_{kp}}y^{s}$ with each $a_{ij}\geq 0$ and $s\in{\mathbb{Z}}$. Such a
monomial is in $T_{f}$, where $f=\sum a_{ij}$ and so we must have $\sum
a_{ij}=1$, as $a\in T_{1}$. Thus, only one $x_{kl}$ can occur in each monomial
and $a$ is a sum of scalar multiples of monomials of the form
$m:=x_{kl}y^{b}$. Such an $m$ is in $T^{(e+bk)}$, where $e=0$ or $e=1$. As
each $m$ must be in $T^{(0)}\cup T^{(1)}$ we must have $e+bk=0$ or $e+bk=1$.
The only possible solutions to these restrictions is that $b=0$, as $k\geq 2$.
Hence, $a$ is a sum of scalar multiples of monomials of the form $x_{kl}$
which means that $a\in\mathcal{O}_{q}(M(k,p))$. ∎

######  Theorem 6.5.

Let $\rho$ be an automorphism of $\mathcal{O}_{q}(G(k,n))$ with $1<k$ and
$2k\leq n$ that satisfies $\rho([u])=[u]$ and $\rho([w])=[w]$. Then $\rho$
extends naturally to an automorphism of
$T=\mathcal{O}_{q}(G(k,n))[[u]^{-1}]=\mathcal{O}_{q}(M(k,p))[y^{\pm
1};\sigma]$ such that $\rho(y)=y$ and
$\rho(\mathcal{O}_{q}(M(k,p)))=\mathcal{O}_{q}(M(k,p))$.

###### Proof.

We know that $\rho(y)=y$ and so only need to show that
$\rho(\mathcal{O}_{q}(M(k,p)))=\mathcal{O}_{q}(M(k,p))$. To do this, it is
sufficient to prove that $\rho(x_{ij})\in\mathcal{O}_{q}(M(k,p))$ for each
generator $x_{ij}$.

Note that $x_{ij}\in\left(T^{(0)}\cup T^{(1)}\right)\cap T_{1}$ for each
$i,j$, and so $\rho(x_{ij})\in\left(T^{(0)}\cup T^{(1)}\right)\cap T_{1}$ for
each $i,j$.

Hence, $\rho(x_{ij})\in\mathcal{O}_{q}(M(k,p))$ by the previous lemma. ∎

##  7 The automorphism group of $\mathcal{O}_{q}(G(k,n))$

In Definition 4.5 and Claim 4.6 we identified a group of automorphisms of
$\mathcal{O}_{q}(G(k,n))$ that we claimed would give us all the automorphisms.
We are now in a position to justify this claim.

Recall from Definition 4.5 that ${\mathcal{A}}={\mathcal{H}}$ when $2k\neq n$
and ${\mathcal{A}}:={\mathcal{H}}\rtimes\langle\tau\rangle$ when $2k=n$.

######  Theorem 7.1.

The automorphism group ${\rm Aut}(\mathcal{O}_{q}(G(k,n)))$ of
$\mathcal{O}_{q}(G(k,n))$ is isomorphic to ${\mathcal{A}}$.

###### Proof.

Let $\rho$ be an arbitrary automorphism of $\mathcal{O}_{q}(G(k,n))$. By
Corollary 5.6, there is an automorphism $h\in{\mathcal{H}}$ such that
$h\cdot\rho([u])=[u]$ and $h\cdot\rho([w])=[w]$. It is enough to prove that
this adjusted automorphism is in ${\mathcal{H}}$ or in
${\mathcal{H}}\rtimes\langle\tau\rangle$; so we may assume that
$\rho([u])=[u]$ and $\rho([w])=[w]$. This adjusted automorphism satisfies the
hypothesis for Theorem 6.5 and so $\rho$ extends naturally to an automorphism
of $T=\mathcal{O}_{q}(G(k,n))[[u]^{-1}]=\mathcal{O}_{q}(M(k,p))[y^{\pm
1};\sigma]$ such that $\rho(y)=y$ and
$\rho(\mathcal{O}_{q}(M(k,p)))=\mathcal{O}_{q}(M(k,p))$. Given this, the
action of $\rho$ is completely determined by its restriction to
$\mathcal{O}_{q}(M(k,p))$; so it is enough to show that $\rho$ restricted to
$\mathcal{O}_{q}(M(k,p))$ is realised by an element of ${\mathcal{H}}$ or, if
$2k=n$, that either $\rho$ or $\rho\circ\tau$ is realised by an element of
${\mathcal{H}}$.

If $2k\neq n$, then by [10, Corollary 4.11 and its proof ] $\rho$ is
determined on $\mathcal{O}_{q}(M(k,p))$by row and column operations and so is
in ${\mathcal{H}}$, as required.

If $2k=n$, then by [17, Theorem 3.2] either $\rho$ or $\rho\circ\tau$ is
determined by row and column operations and so is in ${\mathcal{H}}$. In
either case, $\rho\in{\mathcal{H}}\rtimes\langle\tau\rangle$, as required. ∎

## References

  * [1] J Alev and M Chamarie, Derivations et automorphismes de quelques algebras quantiques, Comm Alg 20 (1992), 1787 – 1802.
  * [2] J M Allman and J E Grabowski, A quantum analogue of the dihedral action on Grassmannians, Journal of Algebra 359 (2012), 49 – 68.
  * [3] A W Chatters, Non-commutative unique factorisation domains, Math. Proc. Cambridge Philos. Soc. 95 (1984), 4 – 54.
  * [4] K R Goodearl and T H Lenagan, Primitive ideals in quantum $SL_{3}$ and $GL_{3}$, New Trends in Noncommutative Algebra, Contemporary Mathematics 562 (2012), 115 – 140.
  * [5] K R Goodearl and R B Warfield Jr., An Introduction to Noncommutative Noetherian Rings. Second edition. London Mathematical Society Student Texts, 61. Cambridge University Press, Cambridge, 2004. xxiv+344 pp.
  * [6] K R Goodearl and M T Yakimov, Unipotent and Nakayama automorphisms of quantum nilpotent algebras, Commutative algebra and noncommutative algebraic geometry. Vol. II, 181 – 212, Math. Sci. Res. Inst. Publ., 68, Cambridge Univ. Press, New York, 2015.
  * [7] A C Kelly, T H Lenagan and L Rigal, Ring theoretic properties of quantum grassmannians, J. Algebra Appl. 3 (2004), 9 – 30.
  * [8] D Krob and B Leclerc, Minor identities for Quasi-Determinants and Quantum Determinants, Commun. Math. Phys. 169 (1995), 1 – 23.
  * [9] S Launois, Primitive ideals and automomorphism group of $U_{q}^{+}(B_{2})$, J. Algebra Appl. 6 (2007), no. 1, 21 – 47.
  * [10] S Launois and T H Lenagan, Primitive ideals and automorphisms of quantum matrices, Algebr. Represent. Theor. 10 (4), (2007), 339 – 365.
  * [11] S Launois and T H Lenagan, Automorphisms of quantum matrices, Glasgow Mathematical Journal 55A (2013), 89 – 100. Special issue in honour of 60th birthdays of Kenny Brown and Toby Stafford.
  * [12] S Launois, T H Lenagan and B M Nolan, Total positivity is a quantum phenomenon: the grassmannian case, arXiv:1906.06199, to appear in Memoirs of the American Mathematical Society.
  * [13] S Launois, T H Lenagan and L Rigal, Quantum unique factorisation domains, J. London Math. Soc. (2) 74 (2006), no. 2, 321 – 340.
  * [14] S Launois, T H Lenagan and L Rigal, Prime ideals in the quantum Grassmannian, Selecta Math. (N.S.) 13 (4) (2008), 697 – 725.
  * [15] T H Lenagan and E J Russell, Cyclic orders on the quantum Grassmannian, Arab. J. Sci. Eng. Sect. C Theme Issues 33 (2) (2008), 337 – 350.
  * [16] J C McConnell and J C Robson, Noncommutative noetherian rings. With the cooperation of L. W. Small. Revised edition. Graduate Studies in Mathematics, 30. American Mathematical Society, Providence, RI, 2001. xx+636 pp.
  * [17] M Yakimov, The Launois–Lenagan conjecture, J Algebra 392 (2013), 1 – 9.
  * [18] M Yakimov, Rigidity of quantum tori and the Andruskiewitsch–Dumas conjecture, Selecta Math. 20 (2014), 421 – 464.

S Launois
School of Mathematics, Statistics and Actuarial Science,
University of Kent
Canterbury, Kent, CT2 7FS,
UK
email<EMAIL_ADDRESS>

T H Lenagan
Maxwell Institute for Mathematical Sciences,
School of Mathematics, University of Edinburgh,
James Clerk Maxwell Building
The King’s Buildings
Peter Guthrie Tait Road
Edinburgh EH9 3FD
UK
email<EMAIL_ADDRESS>