The automorphism group of the quantum grassmannian

The automorphism group of a quantised coordinate algebra is usually much smaller than that of its classical counterpart. Nevertheless, these automorphism groups are often very difficult to calculate. In this paper, we calculate the automorphism group of the quantum grassmannian in the case that the deformation parameter is not a root of unity. The main tool employed is the dehomogenisation equality which shows that a localisation of the quantum grassmannian is equal to a skew Laurent extension of quantum matrices. This equality is used to connect the automorphism group of the quantum grassmannian with that of quantum matrices, where the automorphism group is known.


Introduction
The quantum grassmannian 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 O q (G(k, n)) is generated as an algebra by the k×k quantum minors of the quantum matrix algebra 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 O q (G(3, 6)) in Figure 1.There are two obvious sources of automorphisms for O q (G(k, n)).The first is by restricting column automorphisms of O q (M(k, n)) to the subalgebra O q (G(k, n)), these automorphisms are described in Section 4. The second is via studying certain automorphisms of the (noncommutative) dehomogenisation of O q (G(k, n)) which is isomorphic to a skew Laurent extension of 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 O q (G(k, n)) and consider the relations between them.We then claim that these provide all of the automorphisms of O q (G(k, n)), and justify the claim in the following sections.
The quantum grassmannian carries the structure of an 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 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.

Basic definitions
Throughout the paper, we work with a field K and a nonzero element q ∈ K which is not a root of unity.
The algebra of m × n quantum matrices over K, denoted by O q (M(m, n)), is the algebra generated over K by mn indeterminates x ij , with 1 ≤ i ≤ m and 1 ≤ j ≤ n, which commute with the elements of K and are subject to the relations: for 1 ≤ i ≤ m, and 1 ≤ j < l ≤ n ; x ij x kj = qx kj x ij , for 1 ≤ i < k ≤ m, and 1 ≤ j ≤ n ; It is well known that 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 O q (M(m, n)) is a noetherian domain.
When m = n, the quantum determinant D q is defined by; where the sum is over all permutations σ of {1, . . ., n}.
The quantum determinant is a central element in the algebra of quantum matrices O q (M(n, n)).
If I and J are t-element subsets of {1, . . ., m} and {1, . . ., n}, respectively, then the quantum minor [I | J] is defined to be the quantum determinant of the t × t quantum matrix subalgebra generated by the variables x ij with i ∈ I and j ∈ J.
The homogeneous coordinate ring of the k × n quantum grassmannian, O q (G(k, n)) (informally known as the quantum grassmannian) is the subalgebra of O q (M(k, n)) generated by the k × k quantum minors of O q (M(k, n)), see, for example, [7].
The quantum grassmannian 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 , so in calculating the automorphism group we will assume that 2k ≤ n (and so n ≥ 4, as k ≥ 2).
A k ×k quantum minor of 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, . . ., k | J], for any k-element subset J of {1, . . ., n}.Quantum minors of this type are called quantum Plücker coordinates.The set of quantum Plücker coordinates in O q (G(k, n)) is denoted by Π.There is a natural partial order on Π defined in the following way: if if and only if i l ≤ j l for each l = 1, . . ., k.This partial order is illustrated for the case of O q (G(3, 6)) in Figure 1.A standard monomial in the quantum Plücker coordinates is an expression of the form this partial order.The set of all standard monomials forms a vector space basis of O q (G(k, n)) over K, see, for example, [7,Proposition 2.8].
3 Dehomogenisation of 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, . . ., k}.Then [u] commutes with all other quantum Plücker coordinates up to a power of q as the following lemma shows.
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. [456] [356] As 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 The case a = 1 of [15, Theorem 2.2] shows that the elements x ij generate an algebra R, say, that sits inside (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 = 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 The equality above says that where σ is the automorphism of O q (M(k, p)) such that σ(x ij ) = qx ij for each i = 1, . . ., k and j = 1, . . ., 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.
(ii) Let [L] be a quantum Plücker coordinate in O q (G(k, n)) and write L = L ≤k ∩L >k where L ≤k = L ∩ {1, . . ., k} and L >k = L ∩ {k + 1, . . ., n}.Then The following corollary to the above lemma will be useful in later calculations.
We will use dehomogenisation, see Equation ( 1), in the next three sections to transfer the problem of finding automorphisms of O q (G(k, n)) to that of finding automorphisms of 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 O q (G(k, n)) in the case that n = 2k.
First, note that there is an isomorphism τ between the quantum matrix algebras ij this is automorphism of quantum matrices given by transposition and so we will refer to τ as the transpose isomorphism.)This automorphism extends to an isomorphism from We also denote this extension by τ .
, where L is the complement of L in {1, . . ., n} and w 0 is the longest element of the symmetric group on {1, . . ., n}; that is, w 0 reverses the order of {1, . . ., n}.
Proof.Recall that we are assuming that 2k ≤ n.There is an isomorphism where We need to track the destination of an arbitrary quantum Plücker coordinate of [L] of 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.
As the quantum Plücker coordinates of O q (G(k, n)) generate O q (G(k, n)) as an algebra, and their images generate 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 An immediate corollary of this result is the following.Corollary 3.5.When 2k = n, there is an automorphism of O q (G(k, n)) which sends the quantum Plücker coordinate [I] to [w 0 ( I )], where w 0 is the longest element of the symmetric group on {1, . . ., 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 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 ( 126 )] = [w 0 (345)] = [234].There is a diagram automorphism for O q (G(k, n)) only in the case that n = 2k.Note that both the diagram automorphism and the transpose automorphism τ extend to O q (G(k, 2k)([u] −1 ) = O q (M(k, k))[y, y −1 ; σ] and they agree on this common overring, so we denote the diagram automorphism by τ also.

Obvious automorphisms of O q (G(k, n))
There are two obvious sources of automorphisms of The second is by using the the dehomogenisation equality introduced in Section 3: where p = n − k and u = {1, . . ., k} while σ is the automorphism of O q (M(k, p)) such that σ(x ij ) = qx ij for each i = 1, . . ., k and j = 1, . . ., p.
In this section, we introduce these automorphisms and consider the connections between them.
First, O q (G(k, n)) is a subalgebra of O q (M(k, n)) by definition.The torus H 0 := (K * ) n acts by column multiplication on O q (M(k, n)) and this induces an action on O q (G(k, n)) defined on quantum Plücker coordinates by This is the torus action on 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 O q (M(k, p)) which operates by row and column scaling, so that (α 1 , . . ., α k ; β 1 , . . ., β p ) • x ij = α i β j x ij .As n = k + p, we can extend this to an action of the torus where α I := α i 1 . . .α i k when I = [α i 1 , . . ., α i k ], and The dehomogenisation equality induces an action of As the quantum Plücker coordinates generate O q (G(k, n)) and are sent to scalar multiples of themselves by each h ∈ H 1 , such h act as automorphisms of O q (G(k, n)).
We now consider connections between the actions of H 0 and H 1 on O q (G(k, n)).
Lemma 4.1.For every automorphism g ∈ H 0 acting on O q (G(k, n)) there is an automorphism f ∈ H 1 which has the same action on O q (G(k, n)).
Proof.Let g = (a 1 , . . ., a n ) ∈ H 0 .We seek f = (α 0 ; α 1 , . . ., α k ; β 1 , . . ., β p ) ∈ H 1 such that the actions of g and f on O q (G(k, n)) are the same.As g We seek an f ∈ H which has the same effect on the generators x ij and y.Set f = (a 1 . . .a k ; a −1 k , . . ., a −1 1 ; a k+1 , . . ., a n ) ∈ H. Then f • y = a 1 . . .a k y = g • y.Also, the entry in f multiplying the ith row is a −1 (k+1)−i and the entry multiplying the jth column is a k+j so f Hence, f and g agree on the generators x ij and y ±1 of ; so the actions of f and g on 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 Q) and consider O q (G(2, 4)) over this field.Let f = (1; 2, 1; 1, 1) ∈ H 1 .Then there is no element g = (a 1 , a 2 , a 3 , a 4 ) ∈ H 0 whose action on O q (G(2, 4)) coincides with the action of f on O q (G(2, 4)).To verify this claim, we use the formulae in Lemma 3.2 to see that These equations lead to the following actions of f on the quantum Plücker coordinates: Next, for g = (a 1 , a 2 , a 3 , a 4 ) ∈ H 0 we see that so for this g to act in the same way as f we require that Equations 4 and 5 immediately above show that we need a 3 = a 4 (= b say) and then equation 6 gives b 2 = 2, which is not possible for any b ∈ 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 O q (G(k, n)) and O q (M(k, p)) over K.For every automorphism f ∈ H 1 acting on O q (G(k, n)) there is an automorphism g ∈ H 0 which has the same action on O q (G(k, n)).
Proof.Consider the set S of elements f = (α 0 ; α 1 , . . ., α k ; β 1 , . . ., β p ) ∈ H 1 that are equal to 1 in all positions except one.The set S generates 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 H 0 .We look at three cases separately.
The first case we consider is when one of the β i terms is not equal to 1. Suppose that the element β 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 By using Corollary 3.3(ii) we see that f An element of g ∈ H 0 that has the same effect on quantum Plücker coordinates is the g = (g 1 , . . ., g n ) where g j+k = β j while all other g i = 1.
Then each of the g l j is equal to b and so g This shows that the action of g coincides with the action of f , as required.
The final case to consider is f = (α 0 ; α 1 , . . ., α k ; β 1 , . . ., β p ) ∈ H 1 where α 0 is not equal to 1 but all other entries are equal to 1. Thus, f and the actions of f and g coincide.
The action of H 1 on O q (G(k, n)) is not faithful, as we will see in the next proposition.Let H denote H 1 factored by this kernel of this action.Then H acts faithfully on O q (G(k, n)).The next result shows that H is isomorphic to a torus (K * ) n .Proposition 4.4.The group H is isomorphic to a torus (K * ) n .
Proof.The kernel of the action of H 1 on O q (G(k, n)) is the same as the kernel of the action of Using the right hand side, it is easy to check that this kernel is {(1; λ, . . ., λ; λ −1 , . . ., λ −1 ) | λ ∈ K * }.Hence, choosing λ = β p , we see that h = (α 0 ; α 1 , . . ., α k ; β 1 , . . ., β p ) ∈ H 1 has the same action as It is also easy to check that two distinct elements in H 1 that each have 1 in the final place act differently on O q (M(k, p))[y ±1 ], and so the claim is established.
In view of this result, we refer to the actions on O q (G(k, n)) provided by H as the torus automorphisms of O q (G(k, n)).
In the case that n = 2k, so that k = p, the dehomogenisation equality states that In this case, a simple analysis using the formula shows that the extra automorphism of O q (M(k, k)) given by transposition of the x ij variables extends to an automorphism of ] which, when restricted to O q (G(k, n)), gives rise to the diagram automorphism τ of Corollary 3.5.We will denote this automorphism by τ for each of the three algebras In the case where 2k = n, let (α 0 ; α 1 , . . ., α k ; so that τ acts on H 1 .Also, if (1; λ, . . ., λ; λ −1 , . . ., λ −1 ) is in the kernel of the action of is also in the kernel of this action and so τ acts on H.The analysis above shows that the elements of A act naturally as automorphisms of O q (G(k, n)) via the dehomogenisation equality.
Claim 4.6.The automorphism group of O q (G(k, n)) is A.
We will prove this claim in the following sections.

Adjusting automorphisms
The quantum grassmannian O q (G(k, n)) carries the structure of an 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 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 O q (G(k, n)) in a series of results to see that given an arbitrary automorphism of 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 H. Lemma 5.1.Let A = ⊕ ∞ i=0 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 + • • • + a m is a normal element with a i ∈ 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 = 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 ∈ A 1 .Then ba 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 ∈ a 1 A, and this demonstrates that a 1 is normal.
Lemma 5.2.Let A = ⊕ ∞ i=0 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 O q (G(k, n)).As in earlier sections, let [u] = [1, . . ., k].Proof.Since a = 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 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 = [u]K, and so [u] ∈ aO 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 = 0. We will show that d(I) = d(J) for all I such that a I = 0. Now by using Lemma 3.1, and so Thus, there is a scalar λ ∈ K with b = λa.If λ = 0 then this is a contradiction, as [J] occurs nontrivially in λa 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.
Proof.The element ρ([u]) is a normal element, and the degree zero term of ρ([u]) is equal to 0, by [10,Proposition 4.2].Suppose that the degree of ρ([u]) is t and that ρ( The degree of r must be one.Assume r = r 0 + r 1 with r i having degree i.Thus, 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 λ Write ρ([u]) in terms of the standard basis for O q (G(k, n)), as in [7], say ρ([u]) = α i S i , where α i is in the field and each S i is a standard monomial.In any case, note that [u]S is a standard monomial, as [u] is the unique minimal quantum Plücker coordinate.Hence, As the extreme left and right terms in the above display are in the standard basis, this forces d(S i ) = 0 whenever α i = 0. Hence, ρ([u]) must be a polynomial in [u].
The same argument applies to the automorphism ρ −1 ; so ρ −1 ([u]) must also be a polynomial in [u].
The above result refers to [u] = [1, . . ., k], the extreme leftmost quantum Plücker coordinate.We want to establish a similar result for [w] := [n − k + 1, . . ., n], the extreme rightmost quantum Plücker coordinate.In order to do this we employ an antiautomorphism of 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 θ : O q (G(k, n)) −→ O q (G(k, n)) given by θ( In what follows, we will often replace the original automorphism ρ by h • ρ so that we may assume that ρ([u]) = [u] and ρ([w]) = [w] in calculations.

Transfer to quantum matrices
Recall from the discussion in Section 2 that when discussing O q (G(k, n)) we are assuming that 1 < k and that 2k ≤ n.
Let ρ be an automorphism of O q (G(k, n)).Set We will show that such a ρ sends O q (M(k, p)) to itself.Once we have done this, we will know how ρ acts on each quantum minor in O q (M(k, p)) as we know the automorphism group of O q (M(k, p)).We can then calculate how ρ acts on arbitrary quantum Plücker coordinates of 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

Lemma 5 . 3 .
Suppose that a = a I [I] = 0, with a I ∈ 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 = 0.
If S = [I 1 ] . . .[I m ] is such a standard monomial, then set d(S) := d(I i ) and note that each d(I i ) ≥ 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).