Dominance Regions for Rank Two Cluster Algebras

We study the polygons defining the dominance order on $\mathbf{g}$-vectors in cluster algebras of rank 2.


Introduction
Cluster algebras were introduced by Fomin and Zelevinsky as a tool in the study of Lusztig's dual canonical bases.Since their inception they have found application in a variety of different areas in mathematics, nevertheless a fundamental problem in the theory remains constructing bases with "good" properties.
Over time several bases for cluster algebras have been described in different generalities [BZ14, Cer12, DT13, Dup11, Dup12, GHKK18, LLZ14, MSW13, Pla13, SZ04, Thu14].All of their elements share the property of being "pointed"; this turned out to be a desirable feature for a basis to have and a natural question is to find all bases enjoying this property.Recently Qin studied the deformability of pointed bases whenever the cluster algebra has full rank [Qin21].As no explicit calculation is carried out in his work, we aim here at describing explicitly the space of deformability for pointed bases in rank two, i.e. when clusters contain a pair of mutable cluster variables.In this setting, frozen variables do not carry any additional information so we will work in the coefficient-free case.
Fix integers b, c > 0. The cluster algebra A(b, c) is the Z-subalgebra of Q(x 0 , x 1 ) generated by the cluster variables x m , m ∈ Z, defined recursively by x b m + 1 if m is even; x c m + 1 if m is odd.By the Laurent Phenomenon [FZ02], each x m is actually an element of Z[x ±1 k , x ±1 k+1 ] for any k ∈ Z.Moreover, these Laurent polynomials are known to have positive coefficients [GHKK18,LLZ14,LS15].
An element of A(b, c) has g-vector λ = (λ 0 , λ 1 ) ∈ Z 2 , with respect to the embedding A(b, c) ֒→ Z[x ±1 0 , x ±1 1 ], if it can be written in the form (1) x λ0 0 x λ1 1 α0,α1≥0 ρ α0,α1 x −bα0 0 x cα1 1 with ρ 0,0 = 1.An element of A(b, c) is pointed if an analogous structure reproduces after expanding in terms of any pair {x k , x k+1 }; a basis is pointed if it consists entirely of pointed elements.Important examples of pointed elements are the cluster monomials of A(b, c), i.e. the elements of the form x α k k x α k+1 k+1 for some k ∈ Z and α k , α k+1 non-negative integers.Indeed, by [Qin21, Lemma 3.4.12]cluster monomials are part of any pointed basis in any (upper) cluster algebra of full rank.We achieve the same conclusion in our setting by elementary calculations (cf.Lemma 4.1).
Salvatore Stella was partially supported by the University of Leicester and the University of Rome "La Sapienza".
1 Elements of any pointed basis are parametrized by Z 2 thought of as the collection of possible g-vectors.Qin introduced a partial order on g-vectors called the dominance order, refining the order used in [CILFS15, Proposition 4.3], and showed that it provides a characterization of pointed bases.We restate his results in the generality needed for this paper and using our notation.
Theorem 1.1.[Qin21, Theorem 1.2.1]Let {x λ } and {y λ } be pointed bases of A(b, c).Then for each λ ∈ Z 2 , there exist scalars q λ,µ for µ ≺ λ such that Moreover, having fixed a reference pointed basis {x λ }, any choice of scalars q λ,µ as above provides a pointed basis of A(b, c).
When bc ≤ 3, the cluster algebra A(b, c) will be of finite-type and cluster monomials form its only pointed basis.We therefore assume that bc ≥ 4. Write I ⊂ R 2 for the imaginary cone (positively) spanned by the vectors 2b, −bc ± bc(bc − 4) .Lattice points outside of I are precisely the g-vectors of cluster monomials in A(b, c).
We give an explicit description of the dominance relation among g-vectors.Specifically, we show that the g-vector λ dominates the collection of g-vectors of the form λ + (bα 0 , cα 1 ), α 0 , α 1 ∈ Z, inside its dominance region P λ (cf.Definition 3.1).
Theorem 1.2.If λ lies outside of I, then the dominance region P λ is the point λ.Otherwise the dominance region P λ of the g-vector λ = (λ 0 , λ 1 ) is the polygon consisting of those µ = (µ 0 , µ 1 ) ∈ R 2 satisfying the following inequalities: Remark 1.3.Explicit calculations reveal that for certain pairs of notable bases (e.g.greedy and triangular) most of the coefficients q λ,µ are zero.Understanding this phenomenon might be worth further study.
Corollary 1.4.There are six classes of dominance polygons.
(1) If λ lies outside of I, then P λ is the point λ.
(2) If λ lies in the cone spanned by the vectors (2b, −bc − bc(bc − 4)) and (2, −c), then P λ is the trapezoid with vertices λ, 0, bc+ √ bc(bc−4) 2b (3) If λ lies on the ray spanned by (2, −c), then P λ is the triangle with vertices λ, 0, Remark 1.5.Note that the rays which separate the regions inside I correspond exactly to the columns of the associated Cartan matrix.Moreover the expression for the dominance regions along those rays coincide.These unexpected coincidences are one of our reasons for deciding to write down these results.

Given a Laurent polynomial in Z[x ±1
0 , x ±1 1 ], its support is the set of its exponent vectors inside Z 2 .Given a g-vector λ, in the next result we identify a polygon S λ whose lattice points of the form λ + (bα 0 , cα 1 ) for α 0 , α 1 ∈ Z give the maximum possible support of a pointed basis element x λ .
The maximal supports S λ as described in Theorem 1.6.
The paper is organized as follows.In Section 2, we collect useful results related to two-parameter Chebyshev polynomials which support our main calculations.Section 3 contains calculations related to the transformation of g-vectors under mutations.Section 4 proves Theorem 1.2.Section 5 proves Corollary 1.4.Section 6 proves Theorem 1.6.The paper ends with Section 7 interpreting the dominance polygons in terms of generalized minors in the cases where b = c = 2.

Chebyshev Polynomials
Define two-parameter Chebyshev polynomials u ε i for i ∈ Z and ε ∈ {±} recursively by u ε 0 = 0, u ε 1 = 1, and Remark 2.1.Observe that, by easy inductions, we have u ε −i = −u ε i for i ∈ Z and u + 2j+1 = u − 2j+1 for j ∈ Z. Lemma 2.2.For i, ℓ ∈ Z and ε ∈ {±}, we have Proof.We work by induction on ℓ for all i simultaneously, the cases ℓ = 0, 1 being tautological and reproducing the defining recursions, respectively.Using the claim for ℓ > 0 and then the defining recursion twice, u ε i+ℓ+1 can be rewritten for ℓ odd as and for ℓ even as This gives the claimed recursion for ℓ + 1.These calculations can be reversed to show the result for ℓ < 0.
Lemma 2.3.We have Moreover, the limits in the first line converge monotonically from above and the limits in the second line converge monotonically from below.
Remark 2.4.The analogous limits with ε = + and ε = − reversed are obtained from these by interchanging the roles of b and c.
Proof.When bc = 4 it is easy to compute closed formulas for u ε i and the claimed limits follow; we thus concentrate on the case bc > 4.
The standard Chebyshev polynomials (normalized, of the second kind) are defined by the recursion u 0 = 0, u 1 = 1, u i+1 = ru i − u i−1 , which can be computed explicitly as An induction on i shows that It follows that u ε i can be computed explicitly as which is equivalent to the desired expression.Similarly, for i = 0 we have which is again equivalent to the desired expression.This proves the claim for i → ∞, the cases i → −∞ follow from these using u ε −i = −u ε i .For the final claim, observe that This inequality is then immediate, for i > 0, from the following inductions:

g-vector Mutations
We begin by studying transformations of R 2 which determine the change of g-vectors when expanding an expression of the form (1) in terms of a cluster {x k , x k+1 }.This adapts the notation from [Qin21, Definition 2.1.4]to our setting, see also [Rea14, Definition 4.1].
Write φ 0 : R 2 → R 2 for the identity map and define piecewise-linear maps φ ±1 : R 2 → R 2 as follows: (3) For k ∈ Z with |k| > 1, define piecewise-linear maps The dominance region P λ is the intersection k∈Z φ −1 k C k (φ k λ).When µ ∈ P λ , we say λ dominates µ.We record here a few useful calculations relating to the tropical transformations φ k .First observe the following explicit expression for φ 2 : By an eigenvector of a piecewise-linear map φ, we will mean a vector λ such that there exists a positive scalar ν so that φ(λ) = νλ.Lemma 3.2.Any nonzero eigenvector of φ 2 is a positive multiple of one of the vectors 2b, −bc± bc(bc − 4) .
Observe that the imaginary cone I is spanned by the eigenvectors of φ 2 and that φ 2 is linear in I.
Proof.Since φ 2j = φ j 2 , the result follows from the case j = 1 which is immediate.It will be useful to have explicit expressions for φ k (λ) for λ ∈ I.
Lemma 3.4.For j ∈ Z and λ ∈ I, we have We work by induction on j, the case j = 0 being clear from the definitions.For λ ∈ I, the action of φ 2 from (4) can be rewritten as . Therefore, after applying the equivalences for odd Chebyshev polynomials from Remark 2.1, we have , where the last equality uses (2) with i = 2j + 1 and ℓ = 2. Using Remark 2.1 again, this is equivalent to the desired expression.

Proof of Theorem 1.2
Here we explicitly compute the dominance regions P λ .The following Lemma proves Theorem 1.2 for λ ∈ I.
Proof.Any such λ is the g-vector of a cluster monomial, say k+1 .In this case, the intersection ) is a cone containing λ directed away from the origin (with walls parallel to the boundary of the domain of linearity containing λ).Next, again for k odd, the cone C k+1 (φ k+1 λ) intersects the domain of linearity for φ −1 k+1 containing φ k+1 λ in a (possibly degenerate) convex quadrilateral with corners at the origin and φ k+1 λ.In particular, the intersection of φ −1 k+1 C k+1 (φ k+1 λ) with the cone containing λ is a (possibly degenerate) convex quadrilateral with corners at the origin and λ.Combining these observations proves the result for k odd, the case of even k is similar.
Next we aim to understand how inequalities transform under the action of a piecewise-linear map.The following well known fact about linear maps will suffice.
Lemma 4.2.Let M be an invertible 2 × 2 matrix.Under the left action of M , say M µ = µ ′ for µ, µ ′ ∈ R 2 , the region inside R 2 defined by the inequality α, µ ≤ t with α ∈ R 2 and t ∈ R is transformed into the region defined by the inequality Proof.This is immediate from the equalities The following calculation is key to our main result.
Lemma 4.3.For k ∈ Z, k = 0, and λ ∈ I, the region φ −1 k C k (φ k λ) ⊂ R 2 is determined by the following inequalities when k > 0: and by the following inequalities when k < 0: Proof.We prove the claim for k > 0, the proof for k < 0 is similar or can be deduced from the other case by a symmetry argument.Following Lemma 3.4, we consider even and odd sequences of mutations separately.For k = 2j, j > 0, and λ ∈ I, we observe that C k (φ k λ) ⊂ R 2 is given by the inequalities We compute the region φ −1 2j C 2j (φ 2j λ) using Lemma 4.2 and the equality φ −1 2j = φ −1 2 j .First observe that φ −1 2 = φ −2 is given as follows: (9) if λ 1 > 0 and λ 0 > 0.
Claim: For i > 0, φ −1 2i C 2j (φ 2j λ) is the region determined by the inequalities We prove the claim by induction on i.We see from (9), that λ ∈ I implies the boundary ray for C 2j (φ 2j λ) corresponding to ( ‡) lies entirely in the region in which φ , which is the inequality (a) for i + 1 by Lemma 2; while applying this to (d) gives the inequality λ 1 , which is the inequality (d) for i + 1 again by Lemma 2; finally applying this to (b) gives the inequality λ 1 , which is the inequality (b) for i + 1.Similarly, the boundary segment corresponding to (c) lies entirely in the region where φ −1 2 acts according to the matrix −1 0 c −1 .Thus applying Lemma 4.2 to (c) gives the inequality Applying Lemma 4.2 to (b) with the first matrix gives the inequality , which is the inequality (c) for i + 1 by Lemma 2, while applying Lemma 4.2 to (b) with the second matrix gives the inequality λ 1 , which again reproduces the inequality (d) and aligns with the previous segment and ray in the image.This completes the induction on i, proving the Claim and the result for k even.
We rewrite the inequalities from Lemma 4.3 using the Chebyshev recursion.For k > 0, we get The first inequality in each list is redundant so we drop them.Moreover, for k = 1 the third and fourth inequalities in the first list are the same so we can increment k in the last equality without losing any information and similarly for k = −1 in the second list.This gives Then, using that −u ε k < 0 for k > 0 and u ε k < 0 for k < 0, we rewrite the inequalities again as for k < 0. We now study each sequence of inequalities in turn.
For µ 1 − λ 1 ≥ 0, the inequalities (10) become more restrictive as k gets larger since the sequence of negative slopes is monotonically increasing (cf.Lemma 2.3).Thus, following Lemma 2.3, in the limit as k → ∞, we obtain the inequality below determining a boundary of P λ : Similarly, the inequalities (11) become more restrictive for µ 0 ≤ 0 and µ 1 ≥ 0 as k gets larger since the sequence of positive slopes is monotonically increasing and the intersection with (10) moves lower on the µ 1 -axis as k increases.Thus, following Lemma 2.3, in the limit as k → ∞, we obtain the inequality determining a boundary of P λ .Observe further that the inequalities µ 0 − λ 0 ≤ 0 and 0 ≤ µ 1 − λ 1 allow to strengthen this as Finally, the inequalities (12) become more restrictive for µ 1 ≤ 0 as k gets larger since the sequence of negative slopes is monotonically increasing (cf.Lemma 2.3) and the intersection with (11) (for k + 1) moves to the right on the µ 0 -axis as k increases.Thus, following Lemma 2.3, in the limit as k → ∞, we obtain the inequality below determining a boundary of P λ : Similar arguments using (13)-(15) lead to the remaining inequalities determining the boundary of These can easily be seen to be equivalent to the remaining inequalities from Theorem 1.2 and this complete the proof.

Proof of Corollary 1.4
This follows from basic manipulations finding the intersection points of the boundary segments determined by the inequalities from Theorem 1.2.Note that in each of the cases (2), (4), and (6) there are only four inequalities to consider while in cases (3) and (5) there are only three inequalities to consider.We leave the details as an exercise for the reader.

Proof of Theorem 1.6
We begin observing that, by Lemma 4.1, the dominance region P λ of any g-vector λ ∈ I is just the point λ.Therefore, by Theorem 1.1 the corresponding pointed element is the cluster monomial whose g-vector is λ.The support of this element is exactly S λ by [LLZ14, Proposition 4.1].
Every pointed basis element for A(b, c) admits an opposite g-vector arising by interchanging the roles of b, c and x 0 , x 1 in (1).One can easily compute the following correspondence.Lemma 6.1.Given a g-vector λ, the opposite g-vector λ ′ is obtained as follows: In particular, we obtain an opposite dominance region P ′ λ for each g-vector λ.
For λ ∈ I, the region S λ from Theorem 1.6 is determined by the following inequalities: To begin proving that this is the maximal support we claim that P λ , P ′ λ ⊂ S λ .Observe that the last inequality can be rewritten as bc which together with the second to last inequality already gives two of the boundaries for the dominance region P λ .Note then that, under the assumption µ 0 ≤ λ 0 , the inequality defining another boundary of P λ is more restrictive than the inequality λ 1 ≤ µ 1 bounding S λ .It follows from Corollary 1.4 that the minimum value for µ 0 inside P λ occurs when µ 1 = 0, i. < b and λ 1 < 0, we see that the minimum value of µ 0 inside P λ satisfies the inequality λ ′ 0 = λ 0 + bλ 1 ≤ µ 0 .In particular, combining with the observation above, we see that P λ ⊂ S λ .
Similarly, the second to last inequality can be rewritten as bc cλ ′ 0 , which together with the last inequality already gives two of the boundaries for the opposite dominance region P ′ λ .Note then that, under the assumption defining another boundary of P λ is more restrictive than the inequality λ ′ 0 ≤ µ 0 bounding S λ .As above, the minimum value of µ 1 inside P ′ λ occurs when µ 0 = 0, i.e. either at the point 0, The second point can be rewritten as < c and λ ′ 0 < 0, we see that the minimum value of µ 1 inside P ′ λ satisfies the inequality λ 1 = cλ ′ 0 + λ ′ 1 ≤ µ 1 and so P ′ λ ⊂ S λ .To continue, we compare the support for an arbitrary basis element pointed at λ ∈ I with the greedy basis elements having g-vector inside the dominance polygon P λ .
The support region G λ of the greedy basis element with g-vector λ is well-known [CGM + 17, LLZ14].This is indicated by the solid and dotted lines in our figures where dotted lines indicate points that are excluded from the support.(The actual support of the greedy basis element consists of the points of the form λ + (bα 0 , cα 1 ), α 0 , α 1 ∈ Z, inside G λ .)For our purposes, it is enough to observe the following closure property for the greedy support region which is evident from the figure in Theorem 1.6.Definition 6.3.A subset S ⊂ R 2 is called -closed if λ, µ ∈ S implies the segments joining λ and µ to min(λ 0 , µ 0 ), min(λ 1 , µ 1 ) are contained in S.
As we saw above, for any g-vector µ ∈ P λ its opposite g-vector µ ′ is also contained in S λ .But S λ is -closed and closed under downward scaling.It follows that G µ ⊂ S λ for any µ ∈ P λ and hence, following Theorem 1.1, the support of any basis element pointed at λ is contained in S λ .

The untwisted affine case
In this section we compare our main result with the construction of [RSW19] in the case b = c = 2.To match conventions, in this section, we work over an algebraically closed field k of characteristic 0. Because the exchange matrix is full rank there is no loss of generality in continuing to work in the coefficient-free case.We will identify the family of bases in Theorem 1.1 with the continuous family of bases of A(2, 2) constructed in [RSW19] from generalized minors, which we recast here with the current notation.
Theorem 7.1 ([RSW19, Theorem 4.6]).Choose a point a (n) = (a 1 , . . ., a n ) ∈ (k × ) n for each n ≥ 1.Then, together with all cluster monomials, the elements We begin by noting that when b = c = 2 the imaginary cone I degenerates to the ray spanned by (1, −1).For any λ ∈ I, the dominance region P λ is the segment connecting λ to the origin.It follows that g-vectors dominated by λ = (n, −n) are of the form (n − 2r, −n + 2r) for 0 ≤ r ≤ n/2.
In order to use Theorem 1.1 we need to fix a reference pointed basis of A(2, 2); to simplify our computations, we choose to work with the generic basis.This basis consists of the cluster monomials of A(2, 2) together with elements Proof.To begin, we observe that (16) may be rewritten as .
The first binomial coefficient above is zero if 0 ≤ ℓ − r < k while the second is zero for ℓ < r ≤ ⌊n/2⌋ or if n − r < ℓ, therefore x a (n) (n,−n) can be expressed as evaluated at a (n) .The analogous expansion coefficients when x a (n) (n,−n) is expressed in terms of the triangular basis (resp.greedy basis) are the ratios of Schur functions s 2 (r) ,1 (n−2r) s 1 (n) (resp.ratios of elementary symmetric functions en−r,r en ) evaluated at a (n) .We leave the details to the reader.

( 5 )
If λ lies on the ray spanned by (b, −2), then P λ is the triangle with vertices λ, 0, bc+ √ bc(bc−4) 2b λ 0 +λ 1 , and λ 0 + bc+ √ bc(bc−4) 2c λ 1 , 0 .(6) If λ lies in the cone spanned by the vectors (b, −2) and (2b, −bc+ bc(bc − 4)), then P λ is the trapezoid with vertices λ, bc+ √ bc(bc−4) 2 √ bc(bc−4) 2, the inequality ( ‡) transforms into the inequality cµ 0 + µ 1 ≤ u − 2j λ 0 + u + 2j−1 λ 1 which corresponds to (a) with i = 1.Similarly, the boundary ray for C 2j (φ 2j λ) corresponding to ( †) intersects the three domains of linearity in which φ −1 2 acts according to the matrices −the inequality ( †) can be seen to transform by these into each of inequalities (b), (c), (d) with i = 1.This establishes the base of our induction.Assuming the inequalities (a)-(d) hold for i, we apply Lemma 4.2 for φ −1 2 .Both of the boundary rays corresponding to the inequalities (a) and (d) lie entirely in the region where φ −boundary segment corresponding to (b) intersects this region.Thus applying Lemma 4.2 to (a) gives the inequality which is the inequality (d) for i + 1 by Lemma 2, in particular we see that the segment determined by (c) and the ray determined by (d) align in the image.Lastly, the boundary segment corresponding to (b) also intersects the regions where φ −1 2 acts according to the matrices −

.
But then, rearranging terms and replacing ℓ by ℓ + r, this becomesx a (n)Remark 7.3.Observe that the expansion coefficients S a (n) ,r can be expressed as the ratio of monomial symmetric functions m 2 (r) ,1 (n−2r) m 1 (n) the first coordinate is greater than