Quasi-invariants in characteristic p and twisted quasi-invariants

The spaces of quasi-invariant polynomials were introduced by Feigin and Veselov, where their Hilbert series over fields of characteristic 0 were computed. In this paper, we show some partial results and make two conjectures on the Hilbert series of these spaces over fields of positive characteristic. On the other hand, Braverman, Etingof, and Finkelberg introduced the spaces of quasi-invariant polynomials twisted by a monomial. We extend some of their results to the spaces twisted by a smooth function.


Introduction
A polynomial in variables x 1 , . . . , x n is symmetric if permuting the variables does not change it. Another way to view symmetric polynomials is as invariant polynomials under the action of the symmetric group. A natural generalization of symmetric polynomials then arises: if s i,j is the operator on polynomials that swaps the variables x i and x j , then we may consider polynomials F such that F − s i,j (F ) vanishes to some order at x i = x j . Notably, if F is symmetric in x i and x j , then F −s i,j (F ) vanishes to infinite order. These polynomials may be viewed as quasi-invariant polynomials of the symmetric group.
Definition. Let k be a field, n be a positive integer, and m be a nonnegative integer. We say that a polynomial F ∈ k[x 1 , . . . , x n ] is m-quasi-invariant if (x i − x j ) 2m+1 | F (x 1 , . . . , x i , . . . , x j , . . . , x n ) − F (x 1 , . . . , x j , . . . , x i , . . . , x n ) for all 1 ≤ i, j ≤ n. Denote by Q m (n) the set of all m-quasi-invariant polynomials over k in n variables.
Note that Q m (n) is a module over the ring of symmetric polynomials over k in n variables. Also, as Q m (n) is a space of polynomials, it has an grading by degree. Thus, we may define a Hilbert series and a Hilbert polynomial to encapsulate the structure of Q m (n).
The motivation for studying quasiivariant polynomials arises from their relation with integrable systems. In 1971, Calogero first solved the problem in mathematical physics of determining the energy spectrum of a one-dimensional system of quantum-mechanical particles with inversely quadratic potentials [2]. Moser later on connected the classical variant of his problem with integrable Hamiltonian systems and showed that the classical analogue is indeed integrable [6]. These so-called Calogero-Moser systems have been of great interest to mathematicians as they connect many different fields including algebraic geometry, representation theory, deformation theory, homological algebra, and Poisson geometry. See e.g., [8] and the references therein.
Quasi-invariant polynomials are deeply related with solutions of quantum Calogero-Moser systems as well as representations of Cherednik algebras [4]. As such, the structure of Q m (n), in particular freeness as a module, and its corresponding Hilbert series and polynomials have been extensively investigated by mathematicians. Introduced by Feigin and Veselov in 2001, their Hilbert series and lowest degree non-symmetric elements have subsequently been computed by Felder and Veselov [5]. In 2010, Berest and Chalykh generalized the idea to quasi-invariant polynomials over an arbitrary complex reflection group [9]. Recently in 2016, Braverman, Etingof, and Finkelberg proved freeness results and computed the Hilbert series of a generalization of Q m (n) twisted by monomial factors [1]. Our goal is to extend the investigation of Q m (n) and its various generalizations.
In Section 2, we investigate quasi-invariant polynomials over finite fields. In particular, we provide sufficient conditions for which the Hilbert series in characteristic p is greater than in characteristic 0. We conjecture that our sufficient conditions are also necessary. We also make conjectures about the properties of the Hilbert series over finite fields.
In Section 3, we investigate a generalization of the twisted quasi-invariants. In [1], Braverman, Etingof and Finkelberg introduced the space of quasi-invariants twisted by monomial factors, again a module over the ring of symmetric polynomials. They proved freeness results and computed the corresponding Hilbert series. We generalize their work to the space of quasi-invariants twisted by arbitrary smooth functions and determine the Hilbert series in certain cases when there are two variables.
In Section 4, we discuss future directions for our research, in particular considering spaces of polynomial differential operators and q-deformations.
Definition. Let the Hilbert series of Q m (n) be where Q m,d (n) is the k vector subspace of Q m (n) consisting of polynomials with degree d.
By the Hilbert basis theorem, Q m (n) is a finitely generated module over the ring of symmetric polynomials. Thus, we may write is the Hilbert polynomial associated with H m (t) and the terms in the denominator correspond to the elementary symmetric polynomials that generate the ring of symmetric polynomials in n variables.
We are mainly concerned with the difference between the Hilbert series of Q m (n) in characteristic p and characteristic 0. The following proposition states that the Hilbert series of Q m (n) is at least as large in the former case as in the latter case. Proposition 1. dim Q m,d (n) in F p is at least as large as in C for each choice of m, n, and d.
Then, either d < 2m + 1, in which case we must have F symmetric or (x i − x j ) 2m+1 would divide a nonzero polynomial with degree d for some choice of i and j, a contradiction. This means that the dimensions are equal in either characteristic. Otherwise, we have that for each pair i, j. These yield a system of linear equations in the undetermined coefficients of F and , which is with integral coefficients we are considering. It then follows from considering the null-space that the dimension of the solution space over a field of characteristic p is at least the dimension over a field of characteristic 0.
However, there are only finitely many primes for which it is strictly greater for each m.
Proposition 2. For any fixed m and n, there are only finitely many primes p for which the Hilbert series of Q m (n) is greater in F p than in C.
Proof. Let P = Z[x 1 , . . . , x n ], Q = 1≤i<j≤n P/(x i − x j ) 2m+1 P , and h be the linear map from P to Q defined as Note that Ker(h) coincides with Q m (n) by definition. Set M = Coker(h) as the cokernal of h in Q and note that if Q m (n) over F p has a higher dimension than Q m (n) over C for some degree of the polynomials, then M must have p-torsion. To prove that there are only finitely many such primes p, we use the following generic freeness lemma, see e.g., [3], Theorem 14.4.
Lemma 3. For a Noetherian integral domain A, a finitely generated A-algebra B, and a finitely generated B-module M , there exists a nonzero element r of A such that the localization M r is a free A r module.
We apply this in the case where A = Z, B = Z[x 1 , . . . , x n ] Sn and M = Coker(h). It is easy to see that these satisfy the conditions for A, B, and M in the lemma. Thus there exists an integer r ∈ Z \ {0} such that M r is free over Z[1/r]. As M has no p-torsion for any p ∤ r, M has no p-torsion for all but finitely many primes p so the Hilbert series in F p is the same as in C.
We now determine the primes for which Q m (n) is greater. First, we examine the case when n = 2.
Proposition 4. When n = 2, the Hilbert series for Q m (2) over characteristic p coincides with that of characteristic 0. It is 1+t 2m+1 (1−t)(1−t 2 ) over all fields. Proof. We claim that the dimension of Q m,d (2) over C is equal to the dimension of Q m,d (2) over F p . By Proposition 1, it suffices to show that for each m and d, the for all i. This means that F 1 , . . . , F k are linearly independent, as otherwise there exist relatively prime integers n 1 , . . . , n k with n 1 F 1 +· · ·+n k F k = 0. Taking the equation modulo p yields n 1 f 1 +· · ·+n k f k ≡ 0 (mod p), a contradiction with f 1 , . . . , f k forming a basis of Q m,d (2) as not all of n 1 , . . . , n k are divisible by p.
To do so, let f = f i for a fixed i and suppose that f ( by definition, so we are done. Hence, the dimension, and thus the series, is invariant. It is known from [ , as desired. When n > 2, the series differs greatly for many primes. In this case, we have found a sufficient condition for when the Hilbert series in characteristic p is greater.
Theorem 5. Let m ≥ 0 and n ≥ 3 be integers. Let p be a prime such that there exist integers a ≥ 0 and k ≥ 0 with Then the Hilbert series of Q m (n) with n variables in F p is different from the Hilbert series in C.
Proof. The following formula, due to [4], gives the Hilbert polynomial for Q m (n) in C: Here, the sum is over Young diagrams with n boxes, a i denotes the number of boxes to the right of the ith box, ℓ i denotes the number of boxes below the ith box, and It is not hard to see that the formula gives that the Hilbert polynomial is of the form 1 + (n − 1)t mn+1 + . . ., where the exponents are sorted in ascending order. This implies that all polynomials in Q m (n) with degree at most mn are symmetric, and Q m (n) as a module over symmetric polynomials has a generator of degree mn + 1. For any m, denote this generator in Q m (n) by P m . In the following construction, we will use the generator P k of Q k (n) for certain k < m.
To show that the Hilbert series is different in F p , we consider the following nonsymmetric polynomial: , and a, k are integers such that the above inequalities are satisfied. So deg F = p a (nk+1)+2b n 2 = p a (nk+1)+ n 2 (2m+1−p a (2k+1)) = n 2 (2m+ 1) + p a (1 − n 2 − n(n − 2)k) ≤ n 2 (2m + 1) − (mn(n − 2) + n 2 ) = mn < deg P m . Hence, if we show that F ∈ Q m (n), then as deg F < deg P m we obtain a different Hilbert series in F p , in particular in the coefficient of t deg F . To do that, note that b is an integer when p is odd and a half-integer when p = 2. Either way, ( , so this produces a generator of Q m (n) of lower degree in F p and thus a different Hilbert series in F p , as desired.
Remark. Let us write the inequalities in Theorem 5 in the form In this way, we can rewrite it as where {−} denotes fractional part, which eliminates k. Also from this form it is clear that a cannot be zero, i.e., a ≥ 1.
Remark. Let k = 0 in the inequality in Theorem 5. Then, we see that all primes p with a power between roughly 2m and mn satisfy the inequality. These primes p satisfy the property that Q m (n) in F p has a different Hilbert series than in C. Conjecture 1. The sufficient condition we have given in Theorem 5 is also necessary. That is if the Hilbert series of Q m (n) in F p is different from the Hilbert series in C, then there exist integers a ≥ 0 and k ≥ 0 such that In particular, if p > mn, then the Hilbert series in F p is the same as in C.
This is supported by computer calculations, especially in the case of n = 3, 4. They suggest that the Hilbert series takes a form depending on the smallest non-symmetric element of Q m (n) which is described by the proof of Theorem 5 and hence follows the inequality. The following table summarizes the results of our computer program verification for n = 3, m ≤ 15 and p ≤ 50. Each box in which the series is greater in F p than in C is labeled with its integers a, k that make the inequality hold. Through our programs, we have found that when n = 3, the Hilbert series takes the form for small p, where d is the degree of the smallest non-symmetric generator of Q m in F p . In particular, this smallest non-symmetric polynomial in Q m is of the form P p a k (x i −x j ) 2b where the P k are as described in Theorem 5. Furthermore, we conjecture that Q m is a free module over the ring of symmetric polynomials for n = 3 in any field.
In [4], the authors prove some properties of Hilbert series and polynomials in C, specifically their maximal term and symmetry. We believe that similar results still hold in F p , and this is supported by our computer calculations for n = 3, 4.
Conjecture 2. The largest degree term in the Hilbert polynomial is always t ( n 2 )(2m+1) . Furthermore, when p is an odd prime Q m is a free module over the ring of the symmetric polynomials of rank n!, and the the Hilbert polynomial is palindromic.

A generalization of quasi-invariants
In [1], Braverman, Etingof and Finkelberg introduced quasi-invariants twisted by a monomial x a 1 1 . . . x an n where a 1 . . . , a n ∈ C. We further generalize this by allowing the twist to be a product of general functions. To be more precise, let m be a nonnegative integer. Fix one-variable meromorphic functions f 1 , f 2 , . . . , f n , and denote by D ⊂ C n the domain where the product f 1 (x 1 )f 2 (x 2 ) . . . f n (x n ) and its inverse are smooth.
Definition. We define Q m (f 1 , . . . , f n ) to be the space of polynomials F ∈ C[x 1 , . . . , x n ] for which ( is smooth on D for all 1 ≤ i < j ≤ n.
In the following, for simplicity when we say smooth functions we always mean smooth on D.
Remark. In [1], the authors studied the case f i (x) = x a i for a i ∈ C, and denoted Q m (f 1 , . . . , f n ) by Q m (a 1 , . . . , a n ). In cases of unambiguous use, we will shorten this to Q m (n).
Similar to [1], we believe that Q m (f 1 , . . . , f n ) is in general a free module. For generic f 1 , . . . , f n , in particular when f i f j is not a monomial in x 1 , . . . , x n , Q m (f 1 , . . . , f n ) is a free module over the ring of symmetric polynomials in x 1 , . . . , x n .

Rationality of the logarithmic derivative
Note that for any f 1 , . . . , f n and i, j, any F ∈ (x i − x j ) 2m C[x 1 , ..., x n ] has the property that is smooth. This is a trivial case, thus we want to find out which choices of the f i , f j yield polynomials in Q m (n) that are not in ( is smooth then (here and for the rest of this section, g (with a possible subscript) denotes a function smooth on D) and setting x i = x j = x gives The second part of the proposition follows by definition as Proof. The proposition is trivial for m = 0. For m > 0, note that F ∈ Q m (n) if and only if Here, we treat the rest of the functions and variables as constants. Differentiating with respect to x i , we have that where for a function F (x, y) we define F 1 = ∂F ∂x and F 2 = ∂F ∂y . Setting x i = x j = x, we have that which is a contradiction as the right hand side is a rational function. Hence, ( , which implies the desired result by a straightforward induction.

Hilbert series in two variables
Let n = 2, x = x 1 , and y = x 2 . Note that scaling f 1 and f 2 by some smooth function does not affect Q m (2). Hence, we may multiply them both by 1 f 2 and let f = f 1 f 2 . For convenience, we use Q m (f ) to denote the space of quasi-invariants. Throughout this section, we will let dlog(f (x)) = p(x) q(x) for relatively prime p, q ∈ C[x], as we have from Section 3.2 that either Q m = (x − y) 2m C[x, y] or dlog(f (x)) is a rational function. For convenience, we will also set F x = ∂F ∂x , F y = ∂F ∂y , and F xy = ∂ 2 F ∂x∂y .
Proof. We begin with our quasi-invariant condition, which in our case of n = 2 is Differentiating by x and then y, we obtain By the definition of p and q, this is equivalent to which is exactly the quasi-invariant condition that is desired. Dividing our quasi-invariant condition by f (x)f (y) gives (Note that g 3 is a function smooth on D, because 1 f (x)f (y) is smooth on D by assumption.) Thus, −p(x)F x (y, x) + q(x)F xy (y, x) ∈ Q m−1 1 f q by the above. Expanding the quasi-invariant condition and multiplying by f (x)f (y)q(x)q(y), we obtain the equivalent statement −p(y)F x (x, y) + q(y)F xy (x, y) ∈ Q m−1 (f q), as desired. Now, we specialize to the case in which x−a i . Definition. For a nonnegative integer m and a complex number z, denote for a 1 , . . . , a k , b 1 , . . . , b k ∈ C with a 1 , . . . , a k pairwise distinct.
Proof. We proceed using induction, with the base case of m = 0 clearly true. It suffices to prove this divisibility for each (x − a i ) dm(b i ) . By the inductive hypothesis and Lemma 8, dx . As p and q are relatively prime, we must have ( In fact, this lemma is sharp in the sense that there exists F such that the divisibility becomes equality. To prove that, we utilize the following lemma: Lemma 10. If F ∈ Q m (f ) and G ∈ Q m (g), then F G ∈ Q m (f g).

Proof.
We have that is smooth.
Lemma 11. There exists P m ∈ Q m (f ) with Proof. Note that by Lemma 10 it suffices to show this when k = 1 as for k > 1 we can take the product of all such Shifting, we may also assume that a 1 = 0. Now, note that if z = b 1 is an integer less than m, we can simply take P m = y z . Otherwise, we claim that we can take Indeed, note that we have that P m (x, x) = x m by Vandermonde's identity, so it suffices to show that P m , or equivalently the numerator of P m , is in Q m . We proceed using induction, with the base case of m = 0 obvious. For the inductive step, note that we wish to show that vanishes at x = y to order 2m + 1. Differentiating with respect to x and setting x = y, we would like to show that by Vandermonde's identity, the expression reduces to Differentiating by both x and y, it suffices to show that vanishes at x = y to order 2m − 1. But note that this expression is which vanishes at x = y to order 2m − 1 by the inductive hypothesis on Q m−1 (x z−1 ), as desired.
Lemma 12. Let R denote the ring of symmetric polynomials in x and y. Then, for all m > 0 we have that Proof. Let F (x, y) be an element of Q m . By Lemma 9, which is in Q m as F, P m ∈ Q m and g x+y 2 ∈ R. But now note that F ′ (x, x) = F (x, x) − P m (x, x)g(x) = 0, so by Proposition 6, F ′ ∈ (x − y) 2 Q m−1 , which immediately implies the desired.
Corollary. We have that Now, we are finally ready to prove our main result of this section.
Theorem 13. The Hilbert series for Q m (f ) is .
Proof. Note that Q 0 = C[x, y], which is generated by P 0 = 1 and x − y. By the corollary of Lemma 12, Q m is generated by Let g m,i = (x − y) 2(m−i) P i and g m = (x − y) 2m+1 . We claim that Q m is generated by g m , g m,0 , . . . , g m,m and m independent relations of the form (x − y) 2 g m,m = r m,m−1 g m,m−1 + · · · + r m,0 g m,0 + r m g m (x − y) 2 g m,m−1 = r m−1,m−2 g m,m−2 + · · · + r m−1,0 g m,0 + r m−1 g m · · · (x − y) 2 g m,1 = r 1,0 g m,0 + r 1 g m for some r i , r i,j ∈ R. We proceed using induction, noting that Q 0 is generated by 1 and Let q m = (x − y) 2 q ′ m . Then, subtracting q ′ m times the first relation from qg m + q 0 g m,0 + . . . + q m g m,m = 0, we obtain a relation of the form q ′ g m + q ′ 0 g m,0 + . . . + q ′ m−1 g m,m−1 = 0 with q ′ , q ′ 0 , . . . , q ′ m−1 ∈ R. Note that this relation is uniquely determined by the first generating relation we have so the first generating relation is independent of the rest of the relations. Furthermore, this relation is (x−y) 2 times a relation among the generators of Q m−1 . By the inductive hypothesis, such a relation is generated by (x − y) 2 times the m − 1 independent generating relations of Q 1 , which are by definition the last m − 1 generating relations on our list. Hence, Q m is generated by those m + 2 elements and m independent relations among those elements, as desired.
For the Hilbert polynomial, note that the generators have degrees

Future prospects
It would be interesting to study Conjectures 1, 2, and 3. As with our current results, we expect to make extensive use of computer programs to discover key properties of quasiinvariant polynomials and their Hilbert series. We expect that resolving Conjecture 1 will require studying the modular representation theory of S n . A possible approach to Conjecture 3 is to adapt the approach of the authors of [1], namely to construct a Cherednik-like algebra related to f 1 , . . . , f n and the quasi-invariant polynomials. Along the way, one may also find the formula of the Hilbert series.
Finally, it would be interesting to study q-deformations of the spaces of twisted quasiinvariant polynomials. In [1], Braverman, Etingof, and Finkelberg study q-deformations of their special case and show that when Q m is free, its q-deformation is a flat deformation. They conjecture that it is a flat deformation in general even when Q m is not a free module. Here, Q m,q (f 1 , . . . , f n ) is defined as the set of polynomials F for which is a smooth function for all 1 ≤ i < j ≤ n. It would be interesting to resolve this in the case that Braverman, Etingof, and Finkelberg consider, as well as the general case we have presented. We believe that q-analogues of some of our results hold. For example, the q-analogue of Theorem 5 would be that Q m,q ⊂ m k=−m ( is not rational.