Hilbert Series of ₃-Quasi-Invariant Polynomials in Characteristics 2, 3
We compute the Hilbert series of the space of = 3 variable quasi-invariant polynomials in characteristic 2 and 3, capturing the dimension of the homogeneous components of the space, and explicitly describe the generators in the characteristic 2 case. In doing so, we extend the work of the first aut...
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| author | Wang, Frank Yee, Eric |
| author_facet | Wang, Frank Yee, Eric |
| citation_txt | Hilbert Series of ₃-Quasi-Invariant Polynomials in Characteristics 2, 3. Frank Wang and Eric Yee. SIGMA 21 (2025), 057, 24 pages |
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| description | We compute the Hilbert series of the space of = 3 variable quasi-invariant polynomials in characteristic 2 and 3, capturing the dimension of the homogeneous components of the space, and explicitly describe the generators in the characteristic 2 case. In doing so, we extend the work of the first author in 2023 on quasi-invariant polynomials in characteristic > and prove that a sufficient condition found by Ren-Xu in 2020 on when the Hilbert series differs between characteristic 0 and is also necessary for = 3, = 2,3. This is the first description of quasi-invariant polynomials in the case where the space forms a modular representation over the symmetric group, bringing us closer to describing the quasi-invariant polynomials in all characteristics and numbers of variables.
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| first_indexed | 2026-03-16T09:29:07Z |
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| fulltext |
Symmetry, Integrability and Geometry: Methods and Applications SIGMA 21 (2025), 057, 24 pages
Hilbert Series of S3-Quasi-Invariant Polynomials
in Characteristics 2, 3
Frank WANG and Eric YEE
Department of Mathematics, Massachusetts Institute of Technology,
Cambridge, MA 02139, USA
E-mail: fw0@mit.edu, ericyee@mit.edu
Received January 31, 2025, in final form July 02, 2025; Published online July 13, 2025
https://doi.org/10.3842/SIGMA.2025.057
Abstract. We compute the Hilbert series of the space of n = 3 variable quasi-invariant
polynomials in characteristic 2 and 3, capturing the dimension of the homogeneous com-
ponents of the space, and explicitly describe the generators in the characteristic 2 case.
In doing so we extend the work of the first author in 2023 on quasi-invariant polynomials
in characteristic p > n and prove that a sufficient condition found by Ren–Xu in 2020 on
when the Hilbert series differs between characteristic 0 and p is also necessary for n = 3,
p = 2, 3. This is the first description of quasi-invariant polynomials in the case, where
the space forms a modular representation over the symmetric group, bringing us closer to
describing the quasi-invariant polynomials in all characteristics and numbers of variables.
Key words: quasi-invariant polynomials; modular representation theory
2020 Mathematics Subject Classification: 81R12; 13A50; 20C08
1 Introduction
Let k be a field, and consider the action of the symmetric group Sn on the space k[x1, . . . , xn]
of k-valued polynomials by permuting the variables. A polynomial in k[x1, . . . , xn] is symmet-
ric if it is invariant under this action. Equivalently, since Sn is generated by transpositions,
a polynomial K is symmetric if si1i2K = K or (1 − si1i2)K = 0 for all si1i2 ∈ Sn. One may
consider generalizations of symmetric polynomials in which this condition is relaxed, so that
we only require (1− si1i2)K be divisible by some large polynomial. This leads to the notion of
quasi-invariant polynomials.
Definition 1.1. Let k be a field. For m ∈ Z≥0, n ∈ Z>0, a polynomial K ∈ k[x1, . . . , xn] is
m-quasi-invariant if for all si1i2 ∈ Sn we have that (xi1 − xi2)
2m+1 divides (1 − si1i2)K. We
denote the space of m-quasi-invariants by Qm(n,k).
Note that the symmetric polynomials are exactly the polynomials that are m-quasi-invariant
for all m. For brevity, we also refer to quasi-invariant polynomials as simply quasi-invariants.
Quasi-invariant polynomials were first introduced by Chalykh and Veselov in 1990 [6] to
describe the harmonic, zero eigenvalue eigenfunctions of quantum Calogero–Moser systems.
Calogero–Moser systems are a collection of one-dimensional dynamical particle systems that
were found to be both solvable [4] and integrable [10]. Due to these properties, they have be-
come extensively studied in mathematical physics, with connections to a number of other fields
of mathematics, including representation theory.
Quasi-invariant polynomials were also later found to describe the representation theory of the
spherical subalgebra of the rational Cherednik algebra [3]. This subalgebra is Morita equivalent
to the entire rational Cherednik algebra [7], so quasi-invariants describe representations of ratio-
nal Cherednik algebras as well. Such algebras have connections to combinatorics, mathematical
mailto:fw0@mit.edu
mailto:ericyee@mit.edu
https://doi.org/10.3842/SIGMA.2025.057
2 F. Wang and E. Yee
physics, algebraic geometry, algebraic topology, and more, leading them to become a central
topic in representation theory.
Due to these applications, the quasi-invariant polynomials have been studied extensively in
recent years. Of particular interest are properties such as its freeness as a module over the
symmetric polynomials and the degrees of its generators. To describe these properties, it is
useful to consider the Hilbert series of the quasi-invariants, which encapsulates much of this
information.
Definition 1.2. Let V =
⊕∞
d=0 Vd be a graded vector space. The Hilbert series of V is the
formal power series
H(V ) :=
∞∑
d=0
dim(Vd)t
d.
In 2003, Felder and Veselov found the Hilbert series of the space of quasi-invariants in charac-
teristic zero [9], proving its freeness in the process. Work on quasi-invariants in characteristic p
started in 2020, when Ren and Xu proved a sufficient condition for the Hilbert series ofQm(n,Fp)
to be different from the Hilbert series of Qm(n,Q) [11]. They accomplished this by comput-
ing non-symmetric polynomial “counterexamples” in characteristic p, where the polynomial has
lower degree than any non-symmetric quasi-invariant polynomial in characteristic 0. They also
made several conjectures about quasi-invariants in characteristic p, including that the condition
they found is also sufficient, the quasi-invariants are free, and that the Hilbert polynomial is
palindromic for p > 2. In 2023, the first author proved a general form for the Hilbert series of
the quasi-invariants for n = 3, p > 3, proving freeness and the palindromicity of the Hilbert
polynomial in the process [13].
We expect the development of the theory of quasi-invariants in characteristic p to be useful in
mathematical physics and integrable systems through the theory of q-deformed quasi-invariants.
These are certain deformations of quasi-invariants in characteristic zero introduced by Chalykh
in 2002 [5] used to describe eigenfunctions of Macdonald difference operators, which are a gen-
eralization of elliptic Calogero–Moser systems [12]. We expect the theory of quasi-invariants in
characteristic p to be related to the theory of q-deformed quasi-invariants when q is a root of
unity, in analog to the classical connection between representations of Lie algebras in character-
istic p and quantized enveloping algebras [2]. We note that a few similarities between these two
spaces of quasi-invariants have already been found in [13].
In this paper, we consider the cases n = 3, p = 2, 3. These cases differ from the p > 3 case
studied in [13] since in p = 2, 3 the representations of S3 are modular, i.e., are not completely
reducible. Despite these limitations, we describe the Hilbert series explicitly for all m, proving
the following.
Theorem 1.3. Let k be either F2 or F3. Then the Hilbert series for Qm(3,k) is given by
H(Qm(3,k)) =
1 + 2td + 2t6m+3−d + t6m+3
(1− t)(1− t2)(1− t3)
,
where d = 3m + 1 if there is no Ren–Xu counterexample and d is the degree of the minimal
degree Ren–Xu counterexample otherwise. In particular, the conditions found in [11] for the
Hilbert series of Qm(3,k) to be different from the Hilbert series of Qm(3,Q) are necessary.
Note that this result also implies freeness and the palindromicity of the Hilbert polynomial.
In the case p = 2, we also define m-quasi-invariants in the case where m is a half-integer
and prove an analogous statement to Theorem 1.3 in this case. Using quasi-invariants at half-
integers, we also compute the generators of Qm(3,F2) as an F2[x1, x2, x3]
S3-module explicitly.
Hilbert Series of S3-Quasi-Invariant Polynomials in Characteristics 2, 3 3
In Section 2, we state some of the basic facts about quasi-invariant polynomials and introduce
modular representations of S3. In Section 3, we compute the generators of Qm(3,F2), proving
Theorem 1.3 for p = 2 in the process. In Section 4, we begin discussing p = 3, and show that
some properties of quasi-invariants in 3 variables from [13] carry over to the p = 3 case after
converting from the standard representation to the sign− triv representation. In Section 5,
we show that minimal degree Ren–Xu counterexamples are the lowest degree non-symmetric
generators for Qm(3,F3) and show that there is one other higher degree generator belonging
to the sign− triv representation. Finally, in Section 6, we consider all other indecomposable
representations of S3 in Qm(3,F3), finishing the proof of Theorem 1.3 for p = 3.
2 Preliminaries
We start with some useful properties of the quasi-invariants.
Proposition 2.1 ([8]). Let k be a field.
1. k[x1, x2, x3]
S3 ⊂ Qm(3,k), Q0(3,k) = k[x1, x2, x3], and Qm(3,k) ⊃ Qm′(3,k), where
m′ > m.
2. Qm(3,k) is a ring.
3. Qm(3,k) is a finitely generated k[x1, x2, x3]
S3-module.
Note that [8] proves Proposition 2.1 in the case, where k = C. However, the proofs for the first
two assertions work over any field, and the last assertion follows from the Hilbert basis theorem.
In view of the structure of Qm(3,k) as a module over the symmetric polynomials, given some
K ∈ Qm(3,k), we will frequently refer to quasi-invariant polynomials that can be obtained via
scalar multiplication of Q by a symmetric polynomial. To distinguish these polynomials from
the ordinary k-multiples of Q, we will refer to them as symmetric polynomial multiples of Q.
We consider Qm(3,F2) and Qm(3,F3) as representations of S3, where S3 permutes the vari-
ables x1, x2, x3. Since Qm(3,F2) and Qm(3,F3) are vector spaces over F2 and F3 respectively
and the characteristics 2 and 3 divide |S3|, Qm(3,F2) and Qm(3,F3) are modular representa-
tions of S3.
Proposition 2.2. Qm(3,F2) and Qm(3,F3) are modular representations of S3.
First, we consider characteristic 2.
2.1 Preliminary definitions for p = 2
We describe the indecomposable and irreducible representations of S3 for p = 2.
Proposition 2.3 ([1]). There are 3 irreducible or indecomposable representations of S3 in char-
acteristic 2:
1. triv is the irreducible representation of S3 that is acted on trivially by S3.
2. std is the 2-dimensional irreducible representation of S3 obtained by reducing the standard
representation in characteristic 0 mod 2.
3. triv − triv is the 2-dimensional indecomposable representation that contains a copy of triv
as a subrepresentation such that the quotient of triv − triv by this subrepresentation is triv.
Example 2.4. The polynomial Etriv−triv := x21x2+x22x3+x23x1 ∈ F2[x1, x2, x3] generates a copy
of triv − triv. To see this, note that for any i1, i2, we have
(1− si1i2)Etriv−triv = x21x2 + x1x
2
2 + x21x3 + x1x
2
3 + x22x3 + x2x
2
3 ∈ F2[x1, x2, x3]
S3 .
4 F. Wang and E. Yee
Since the transpositions generate S3, Etriv−triv generates a two-dimensional representation that
contains triv as a subrepresentation. Moreover, since Etriv−triv is not symmetric, this represen-
tation is not triv ⊕ triv, so it must be triv − triv.
We then study the behaviors of each indecomposable representation in the quasi-invariants.
We define Qm(3,F2)triv and Qm(3,F2)std to be the direct sum of all copies of triv and std
respectively in the quasi-invariants. We also define Qm(3,F2)triv−triv to be the direct sum of all
copies of triv and triv − triv.
Remark 2.5. We cannot define Qm(3,F2)triv−triv to exclude copies of triv since we can add
elements of Qm(3,F2)triv to copies of triv − triv and still obtain a copy of triv − triv. For exam-
ple, F := Etriv−triv + x31 + x32 + x33 still satisfies (1 − si1i2)F = (1 − si1i2)Etriv−triv for all i1, i2,
so it generates a copy of triv − triv by the same argument as Example 2.4.
Proposition 2.6 ([13]). As an F2[x1, x2, x3]
S3-module, Qm(3,F2)triv is freely generated by 1.
Note that by the classification of indecomposables in Proposition 2.3, every extension of
std and every extension of a module by std splits. Thus Qm(3,F2)std is a direct summand
of Qm(3,F2) (whose complement is Qm(3,F2)triv−triv), and we mainly consider Qm(3,F2)std.
Qm(3,F2)std is generated as a F2[x1, x2, x3]
S3-module by homogeneous copies of std, so follow-
ing [13], we consider generating representations of Qm(3,F2)std as homogeneous copies of std in
a generators and relations presentation of Qm(3,F2)std with a minimal generator set.
2.1.1 Quasi-invariants at half-integers
Note that if k is a field with chark ̸= 2 and m ∈ Z≥0, then for any K ∈ k[x1, . . . , xn],
(xi1−xi2)
2m|(1−si1i2)K implies (xi1−xi2)
2m+1|(1−si1i2)K since (1−si1i2)K is si1i2-antiinvariant,
hence the exponent 2m + 1 in the definition of quasi-invariant polynomials. But this does not
hold in characteristic 2, since there is no concept of antiinvariants. Indeed, one can check that
for K = x21 + x22, we have (xi1 − xi2)
2|(1 − si1i2)K for all i1, i2, but (xi1 − xi2)
3 ∤ |(1 − si1i2)K
if i1 = 1, 2, i2 ̸= 1, 2.
We encapsulate this data by extending the definition of quasi-invariants to half-integers when
p = 2. For example, K = x21+x22 is
1
2 -quasi-invariant, and this is in fact the minimal degree non-
symmetric 1
2 -quasi-invariant polynomial. Proposition 2.1 still holds whenm,m′ are half-integers,
and the definitions of Qm(3,F2)triv, Qm(3,F2)std also naturally extend to half-integer m. So
from now on, whenever we refer to quasi-invariants in characteristic 2 we let m be a half-integer.
2.2 Preliminary definitions for p = 3
Next, we define the indecomposable and irreducible representations of S3.
Proposition 2.7 ([1]). There are 6 indecomposable or irreducible representations in S3 in char-
acteristic 3:
1. triv is the irreducible representation of S3 that is acted on trivially by S3.
2. sign is the irreducible representation of S3 that is acted on by negation by the transpositions.
3. sign− triv is the indecomposable representation that contains a copy of triv as a subrep-
resentation, such that the quotient of sign− triv by this subrepresentation is sign.
4. triv − sign is the indecomposable representation that contains a copy of sign as a subrep-
resentation, such that the quotient of triv − sign by this subrepresentation is triv.
5. triv − sign− triv is the indecomposable representation that contains a copy of sign− triv
as a subrepresentation, such that the quotient of triv − sign− triv by this subrepresentation
is triv.
Hilbert Series of S3-Quasi-Invariant Polynomials in Characteristics 2, 3 5
6. sign− triv − sign is the indecomposable representation that contains a copy of triv − sign
as a subrepresentation, such that the quotient of sign− triv − sign by this subrepresentation
is sign.
Provided are some examples of copies of these indecomposable representations:
Example 2.8. The space W ⊂ F3[x1, x2, x3] spanned by x1 + x2 + x3 and x1 − x2 over F3 is
copy of sign− triv. Indeed, the space T ⊂ W spanned by x1+x2+x3 is a copy of triv. One can
check x1−x2 ∈ W/T is acted by negation by all transpositions in S3 and W/T is 1-dimensional
so W/T is a copy of sign. Finally, it is easy to show that there are no copies of triv or sign in
W other than T . Since V has a unique irreducible subrepresentation, it is indecomposable, and
we conclude that it is a copy of sign− triv.
Example 2.9. The space V ⊂ F3[x1, x2, x3] consisting of homogeneous linear polynomials
is a copy of triv − sign− triv. Indeed, W ⊂ V from Example 2.8 is a copy of sign− triv.
Then V/W is one-dimensional, and one can check that it is a copy of triv. Finally, it is easy to
show that there are no copies of triv or sign in V other than T , so V has a unique irreducible
subrepresentation, it is indecomposable, and we conclude that it is a copy of triv − sign− triv.
Example 2.10. Similarly, one may check that the space U spanned by
(x1 − x2)(x1 − x3)(x2 − x3), −x21x2 − x21x3 + x1x
2
2 + x1x
2
3
over F3 is a copy of triv − sign and that the space spanned by
(x1 − x2)(x1 − x3)(x2 − x3), −x21x2 − x21x3 + x1x
2
2 + x1x
2
3, (x1 − x2)x1x2
is a copy of sign− triv − sign.
Similarly to the p = 2 case, we define Qm(3,F3)sign and Qm(3,F3)triv to be the direct sum
of all copies of sign and triv in Qm(3,F3), respectively.
Proposition 2.11 ([13]). As F3[x1, x2, x3]
S3-modules,
1. Qm(3,F3)triv is freely generated by 1.
2. Qm(3,F3)sign is freely generated by
∏
i1<i2
(xi1 − xi2)
2m+1.
Next we define Qm(3,F3)sign−triv as the direct sum of all copies of sign, triv, and sign− triv.
For this paper we consider generators of Qm(3,F3)sign−triv to be homogeneous polynomials other
than 1 and
∏
i1<i2
(xi1 − xi2)
2m+1 such that they are in the (−1)-eigenspace of s12 and are in
a generators and relations presentation of Qm(3,F3)sign−triv as an F3[x1, x2, x3]
S3-module with
the least number of generators. Moreover, if K is a generator of Qm(3,F3)sign−triv then it
necessarily generates a copy of sign− triv since we assumed K neither generates triv nor sign.
Remark 2.12. Similar to in the p = 2 case, we cannot define Qm(3,F3)sign−triv to exclude
copies of sign since we can add elements of Qm(3,F3)sign to copies of sign− triv and still obtain
a copy of sign− triv. For example, the spaces spanned by(
x61 − x62
)
(x1 + x2 + x3)
3,
(
x61 + x62 + x63
)
(x1 + x2 + x3)
3
and ∏
i1<i2
(xi1 − xi2)
3 +
(
x61 − x62
)
(x1 + x2 + x3)
3,
(
x61 + x62 + x63
)
(x1 + x2 + x3)
3
generate two copies of sign− triv in Q1(3,F3), and their sum contains∏
i1<i2
(xi1 − xi2)
3 ∈ Q1(3,F3)sign.
6 F. Wang and E. Yee
Remark 2.13. One could define subspaces ofQm(3,F3) for triv − sign− triv, sign− triv − sign,
triv − sign similar to Qm(3,F3)sign−triv, however this is not particularly helpful, as unlike for
p = 2, we cannot decompose Qm(3,F3) into a direct sum of subspaces of this form. The space
Qm(3,F3)sign−triv is still relevant, as it is the critical piece to understanding quasi-invariants in
characteristic 3, as we see in Sections 4 and 5.
3 Quasi-invariants in characteristic 2
In this section, we write down explicit generators for Qm(3,F2) and prove Theorem 1.3 for p = 2.
Note that we already know the structure of Qm(3,F2)triv from Proposition 2.6. We start by
extending this to Qm(3,F2)triv−triv.
Proposition 3.1. As an F2[x1, x2, x3]
S3-module, Qm(3,F2)triv−triv is freely generated by 1 and
Etriv−triv
∏
(xi1 − xi2)
2m.
Proof. Let K be a nonsymmetric element of Qm(3,F2)triv−triv so that (xi1 − xi2)
2m+1 divides
(1 + si1i2)K. Because
(1 + s12)K = (1 + s13)K = (1 + s23)K,
we have (1 + si1i2)K = P
∏
(xi1 − xi2)
2m+1 for some symmetric polynomial P . Letting G =
Etriv−triv
∏
(xi1−xi2)
2m yields (1+si1i2)G =
∏
(xi1−xi2)
2m+1. Thus (1+si1i2)PG = (1+si1i2)K
and (1+si1i2)(PG−K) = 0, so PG−K is symmetric and K is generated by G and 1. Moreover,
since G is not symmetric, P and G have no relation implying freeness. ■
We have an explicit description of Qm(3,F2)triv−triv, so it remains to compute the generators
and relations of Qm(3,F2)std. A number of the properties of Qm(3,Fp) for p > 3 found in [13]
are true for Qm(3,F2). We prove these first.
If V is a copy of std, then we denote by Vi1i2 the 1-eigenspace of si1i2 in V .
Lemma 3.2. Let V be a copy of std in Qm(3,F2)std, and let K ∈ Vi1i2. Then we have
K + sK + s2K = 0, where s = (1 2 3) ∈ S3 and K = (xi1 − xi2)
2m+1K ′ for some polynomial K ′
that is invariant under the action of si1i2. Conversely, let K ′ be an s12-invariant polynomial
such that
(x1 − x2)
2m+1K ′ + (x2 − x3)
2m+1sK ′ + (x3 − x1)
2m+1s2K ′ = 0.
Then (x1−x2)
2m+1K ′ belongs to the 1-eigenspace of s12 in some copy of std inside Qm(3,F2)std.
Proof. For the first statement, K + sK + s2K = 0 holds for any copy of std. For the next,
suppose {i1, i2, i3} = {1, 2, 3} for some integer i3. Then (1 − si1i3)K = si2i3K, so (xi1 −
xi3)
2m+1|si2i3K, implying (xi1 − xi2)
2m+1|K. The second statement follows from the proof
in [13]. ■
Corollary 3.3. Let V be a generating representation of Qm(3,k)std and let K ∈ Vi1i2. Let us
write K = (xi1−xi2)
2m+1K ′. Then K ′ is not divisible by any nonconstant symmetric polynomial.
The proof of this statement is identical to the one in [13].
Lemma 3.4. Let V , W be distinct generating representations of Qm(3,k)std. Let K ∈ V12,
L ∈ W12. For σKσL := (σK)(σL), we have that KL+ s13Ks23L is a nonsymmetric element of
Qm(3,k)triv−triv and deg V + degW ≥ 6m+ 3.
Hilbert Series of S3-Quasi-Invariant Polynomials in Characteristics 2, 3 7
Proof. KL + s13Ks23L is an element of Qm(3,F2) since the quasi-invariants form a ring by
Proposition 2.1. Using that s12K = K and s12L = L, we have that
(1 + s12)(KL+ s13Ks23L) = s23Ks13L+ s13Ks23L,
(1 + s13)(KL+ s13Ks23L) = KL+ s13Ks23L+ s13Ks13L+Ks23L = Ks13L+ s13KL,
(1 + s23)(KL+ s13Ks23L) = KL+ s13Ks23L+ s23Ks23L+ s13KL = Ks23L+ s23KL.
One can check that each polynomial is a transposition of another and that they are symmet-
ric due to the structure of triv − triv, so they are all the same symmetric polynomial. Thus
KL+ s13Ks23L lies in a quotient of a copy of triv − triv. Note that by the same argument as
in [13], we have Ks23L + s23KL ̸= 0, so KL + s13Ks23L is nonsymmetric and must generate
a copy of triv − triv.
By Proposition 3.1, KL+ s13Ks23L has degree at least 6m+ 3, so deg V + degW ≥ 6m+ 3
as desired. ■
Lemma 3.5. Assume that there exist generating representations V , W of Qm(3,F2)std such
that deg V +degW = 6m+3. Then Qm(3,F2)std is a free module over k[x1, x2, x3]
S3 generated
by V and W .
Proof. Assume for the sake of contradiction there exists another generator U of Qm(3,F2)std.
Supposing degW ≥ deg V , by Lemma 3.4, degU ≥ degW . By Lemma 3.4, if K ∈ V12, L ∈ W12,
and T ∈ U12 then KL + s13Ks23L and KT + s13Ks23T are both nonsymmetric elements of
Qm(3,F2)triv−triv. Moreover, we have
(1 + s12)(KL+ s13Ks23L) = s23Ks13L+ s13Ks23L =
∏
(xi1 − xi2)
2m+1,
and
(1 + s12)(KT + s13Ks23T ) = s23Ks13T + s13Ks23T = Q
∏
(xi1 − xi2)
2m+1
for some symmetric polynomial Q. From there we may proceed identically to [13]. ■
Now, we are ready to prove Theorem 1.3 for p = 2.
Theorem 3.6. Let a be the largest natural number such that 2a < 2m+ 1. Then Qm(3,F2)std
is freely generated by (x1 − x2)
2a+1
and (x1 − x2)
2a
∏
(xi1 − xi2)
2m+1−2a.
Remark 3.7. Note that when m is an integer, the degrees of the generators in this theorem
agree with the degrees conjectured in [11]. In particular, when 2a+1 is one of 3m + 1, 3m+ 2,
we actually have that the Hilbert series of Qm(3,F2) and Qm(3,Q) agree, so (x1 − x2)
2a+1
,
(x1 − x2)
2a
∏
(xi1 − xi2)
2m+1−2a are the reductions modulo 2 of the generators of Qm(3,Q),
when written as integer polynomials with coprime coefficients.
Proof of Theorem 3.6. We prove this by induction on m.
The generators of Q0(3,F2)std are (x1 − x2) and (x1 − x2)
2, completing our base case.
Let j be a half-integer, and suppose thatQj− 1
2
(3,F2)std is freely generated by (x1−x2)
2a+1
and
(x1 − x2)
2a
∏
(xi1 − xi2)
2j−2a , where 2a is the greatest such power of 2 less than 2j. If 2j ̸= 2a+1,
then 2a is the largest power of 2 less than 2j+1, so (x1−x2)
2a+1
and (x1−x2)
2a
∏
(xi1−xi2)
2j+1−2a
are both in Qj(3,F2). Further, (x1 − x2)
2a+1
must be a generator and if (x1 − x2)
2a
∏
(xi1 −
xi2)
2j+1−2a is a not a generator, by Lemma 3.4, (x1 − x2)
2a
∏
(xi1 − xi2)
2j+1−2a is generated by
(x1 − x2)
2a+1
which implies a relation between (x1 − x2)
2a+1
and (x1 − x2)
2a
∏
(xi1 − xi2)
2j−2a .
Because they freely generate Qj− 1
2
(3,F2), this is impossible. Thus (x1 − x2)
2a+1
and (x1 −
x2)
2a
∏
(xi1 − xi2)
2j+1−2a freely generate Qj(3,F2)std by Lemma 3.5.
8 F. Wang and E. Yee
If 2j = 2a+1, then both (x2−x3)
2a+1
and (x2−x3)
2a+2 ∏
(xi1 −xi2)
2j+1−2a lie in Qj(3,F2)std.
The former is a generator by our inductive hypothesis. Since 2a+1 + 2a+2 + 3 = 6j + 3, if the
latter is not a generator, then by Lemma 3.4, (x2 − x3)
2a+2
∏
(xi1
−xi2
)2j+1−2a
is generated by
(x2 − x3)
2a+1
, which is false. Thus (x2 − x3)
2a+1
and (x2 − x3)
2a+2 ∏
(xi1 − xi2)
2j+1−2a freely
generate Qj(3,F2)std by Lemma 3.5 as desired. ■
4 Properties of 3 variable quasi-invariants
Similarly to the p = 2 case, we can adapt many of the properties of Qm(3,Fp) for p > 3 found
in [13] to the p = 3 case. We accomplish this by converting std to sign− triv. For example,
in Q0(3,Fp) for p > 3, the space spanned by x1 − x2, x1 − x3 is a copy of std. However,
in Q0(3,F3), the space spanned by x1 − x2, x1 − x3 becomes a copy of sign− triv. Using this,
we may show that there are equivalents of Lemmas 3.2–3.5 from [13] in characteristic 3.
We define V −
i1i2
to be the (−1)-eigenspace of si1i2 in V , where V is a copy of std or sign− triv.
Note that if v ∈ V −
i1i2
we have v = s23v + s13v. The following lemma and corollary correspond
to Lemma 3.2 and Corollary 3.3 from [13], respectively.
Lemma 4.1. Let V be a copy of sign− triv in Qm(3,F3)sign−triv, and let K ∈ V −
i1i2
. Then
we have K + sK + s2K = 0, where s = (1 2 3) ∈ S3 and K = (xi1 − xi2)
2m+1K ′ for some
polynomial K ′ that is invariant under the action of si1i2. Conversely, let K ′ be an s12-invariant
polynomial such that
(x1 − x2)
2m+1K ′ + (x2 − x3)
2m+1sK ′ + (x3 − x1)
2m+1s2K ′ = 0.
Then (x1 − x2)
2m+1K ′ either belongs to Qm(3,F3)sign or the (−1)-eigenspace of s12 in some
copy of sign− triv inside Qm(3,F3)sign−triv.
Proof. The proof is largely the same as in [13]; the only difference is in the last step. Namely,
now we have 2 2-dimensional indecomposable representations sign− triv and triv − sign, but
an element in the (−1)-eigenspace of s12 in triv − sign must be in a copy of sign. ■
Corollary 4.2. Let K be a generator of Qm(3,F3)sign−triv in V −
i1i2
for some copy V of sign− triv
and write K = (xi1 − xi2)
2m+1K ′. Then K ′ is not divisible by any nonconstant symmetric
polynomial.
The proof of this corollary is identical to the proof of [13, Corollary 3.3].
We define generators of Qm(3,F3) to be “distinct” if they are either in different degrees, or
if no linear combination of them over F3 is generated by lower degree generators.
Lemma 4.3. Let K and L be distinct generators of Qm(3,k)sign−triv, and let V and W be
the copies of sign− triv generated by K and L respectively such that K ∈ V −
i1i2
and L ∈ W−
i1i2
.
Then Ks23L− Ls23K is a nonzero element of Qm(3,F3)sign and deg V + degW ≥ 6m+ 3.
Noting that ∧2(sign− triv) = sign, the proof of this lemma is also identical to the proof
of [13, Lemma 3.4].
Lemma 3.5 from [13] does not completely hold in characteristic 3. A very similar and useful
version does, however, and we have the following.
Lemma 4.4. Assume that there exists generators K and L of Qm(3,F3)sign−triv such that
degK + degL = 6m + 3. Then Qm(3,F3)sign−triv is freely generated by K, L, and 1 over
F3[x1, x2, x3]
S3.
Hilbert Series of S3-Quasi-Invariant Polynomials in Characteristics 2, 3 9
Proof. We note that (L+ s23L)K − (K + s23K)L = c
∏
i1<i2
(xi1 − xi2)
2m+1 for some c ̸= 0 by
Lemma 4.3. Moreover, L+ s23L and K + s23K are symmetric because K and L are both acted
on by negation by s12, so elements in Qm(3,F3)sign are generated by K and L. From there, the
fact that Qm(3,F3)sign−triv is generated by K, L, and 1 over F3[x1, x2, x3]
S3 follows from the
first part of the proof from [13].
To prove freeness, assume for the sake of contradiction that there was a relation PK +QL+
S = 0 for symmetric polynomials P , Q, and S. PK and QL are both in the (−1)-eigenspace
of s12 while S is not, so S = 0. Thus we have PK = −QL and from there we can proceed the
same as [13].
■
5 Ren–Xu counterexamples
We aim to explicitly describe the Hilbert series of Qm(3,F3). To do so we wish to identify the
generators of Qm(3,F3)sign−triv.
In [11], Ren and Xu found polynomials of the form P 3a
k
∏
(xi1 − xi2)
2b in Qm(3,F3) with
degree strictly less than 3m+1, where Pk is the map of the 3k+1 degree generator of Qk(3,Q)
into characteristic 3 and where a, k, and b are natural numbers. We refer to these polynomials
as Ren–Xu counterexamples as they demonstrate the Hilbert series of Qm(3,F3) differs from
that of Qm(3,Q) for certain m.
Definition 5.1. Let Pk be the generator of Qk(3,Q) of degree 3k + 1 in the (−1)-eigenspace
of s12, expressed as an integer polynomial with coprime coefficients. Let Pk be the image of Pk
under the quotient map Z[x1, x2, x3] → F3[x1, x2, x3]. Define the set X as the set of all natural
numbers m such that Qm(3,F3) has a Ren–Xu counterexample. Let Rm be a lowest degree
Ren–Xu counterexample in Qm(3,F3) for all m ∈ X.
A key step in describing the Hilbert series of Qm(3,F3) is proving Ren–Xu’s conjecture [11]
for n = 3 and p = 3.
Conjecture 5.2 ([11]). If the Hilbert series of Qm(n,Fp) differs from that of Qm(n,Q), then
there exists integers a ≥ 0 and k ≥ 0 such that
mn(n− 2) +
(
n
2
)
n(n− 2)k +
(
n
2
)
− 1
≤ pa ≤ mn
nk + 1
.
The main step for proving the conjecture for n = 3, p = 3 is the following theorem.
Theorem 5.3. Qm(3,F3)sign−triv is either freely generated by a generator of degree 3m + 1,
3m + 2, and the polynomial 1, or it is freely generated by Rm, another generator in degree
6m+ 3− degRm, and the polynomial 1.
To prove this theorem, we first describe the Ren–Xu counterexamples.
Lemma 5.4. If m ∈ X, we must have Rm = P 3a
k
∏
(xi1 − xi2)
2b, where a, b, k are natural
numbers and k ̸∈ X.
Proof. Assume for contradiction that there exists a nonnegative integer m ∈ X such that Rm =
P 3a
k
∏
(xi1−xi2)
2b, where a, b, k are natural numbers and k ∈ X. Then ifRk = P 3c
l
∏
(xi1−xi2)
2d,
the polynomial
Rk
3a
∏
(xi1 − xi2)
2b = P 3a+c
l
∏
(xi1 − xi2)
2d·3a+2b
has a strictly smaller degree than Rm since degRk < 3k + 1 = degPk. Moreover, it is at
least m-quasi-invariant, so it is a Ren–Xu counterexample for Qm(3,F3). Yet Rm is a minimal
counterexample, giving a contradiction. ■
10 F. Wang and E. Yee
This lemma allows us to consider only counterexamples P 3a
k
∏
(xi1−xi2)
2b such that Qk(3,F3)
does not contain a Ren–Xu counterexample.
From [11], the Hilbert series for Qm(3,F3) differs from characteristic 0 when there exists
a ∈ N0 such that
1
3
≤
{m
3a
}
≤ 2
3
− 1
3a
.
Notice this is equivalent to m (mod 3a) being in
{
3a−1, 3a−1 + 1, . . . , 2 · 3a−1 − 1
}
.
Lemma 5.5. If m ̸∈ X, then the base 3 representation of m contains no 1’s.
Proof. Suppose m had the digit 1 in the a-th position from the right. Then m (mod 3a) has
a leading digit of 1 if we choose m (mod 3a) to be between 0 and 3a−1 inclusive. However, this
implies that m (mod 3a) is in
{
3a−1, 3a−1 + 1, . . . , 2 · 3a−1 − 1
}
, so m is a counterexample. ■
Corollary 5.6. If m ̸∈ X, then m is even.
Proof. From Lemma 5.5 m has no 1’s in its base 3 representation, so
m =
∑
j=0
cj3
j ,
where cj is 0 or 2. Thus m must be even. ■
Corollary 5.7. For all m ̸∈ X, we have m+ 1 ∈ X.
Proof. By Corollary 5.6, if m ̸∈ X, m is even. Then m+ 1 is odd, so by the contrapositive of
Corollary 5.6, m+ 1 ∈ X. ■
Now we begin describing the degrees of Ren–Xu counterexamples.
Lemma 5.8. If Qm(3,F3)sign−triv has a generator in degree 3m + 1, then m + 1 ∈ X and
degRm+1 = 3m+ 3.
Proof. If m ∈ X, we must have degRm < 3m + 1. This implies a generator in a degree less
than 3m+ 1, violating Lemma 4.3. Thus m ̸∈ X, implying that m+ 1 ∈ X by Corollary 5.7.
Because degRm+1 < 3m+4 and Qm+1(3,F3)sign−triv ⊂ Qm(3,F3)sign−triv, we have 3m+1 ≤
degRm+1 < 3m+ 4. By construction 3|degRm+1, so degRm+1 = 3m+ 3. ■
We now introduce a few useful lemmas.
Lemma 5.9. Suppose Qm(3,F3)sign−triv has a smallest degree generator L in degree 3m + 1.
Assume that for all j < m, if j ̸∈ X, then Qj(3,F3)sign−triv has a degree 3j+1 generator. Then
Qm+1(3,F3)sign−triv has no nonsymmetric degree 3m+ 1 or 3m+ 2 element.
Proof. Any nonsymmetric 3m + 1 degree element in Qm+1(3,F3)sign−triv must be a scalar
multiple of L, so assume for contradiction L is in Qm+1(3,F3). By Lemma 5.8, Rm+1 =
P 3a
k
∏
(xi1 − xi2)
2b is in degree 3m + 3 for natural numbers a, b, k. By Lemma 5.4, k ̸∈ X
implying Pk is a 3k + 1 generator of Qk(3,F3)sign−triv using our assumption. Moreover, with
any other generator in a degree less than 3m + 3 violating Lemma 4.3, Rm+1 must be gener-
ated by L, so P 3a
k
∏
(xi1 − xi2)
2b = SL for some degree 2 symmetric polynomial S. A degree 2
symmetric polynomial divisible by (xi1 − xi2) is impossible, so S|P 3a
k which implies either S|Pk
or (x1 + x2 + x3)|Pk. Since Pk is in the (−1)-eigenspace of s12, Pk is as well and by Lemma 4.1
we have Pk = P ′
k(x1 − x2)
2k+1. In both cases either S|P ′
k or (x1 + x2 + x3)|P ′
k. However, by our
Hilbert Series of S3-Quasi-Invariant Polynomials in Characteristics 2, 3 11
assumption Pk is a generator, so P ′
k is not divisible by any nonconstant symmetric polynomial
by Corollary 4.2.
Similarly, suppose for contradiction that K is a nonsymmetric element of Qm+1(3,F3)sign−triv
of degree 3m + 2. Since Qm+1(3,F3)sign−triv has no nonsymmetric 3m + 1 degree element, K
must be a generator. By Lemma 4.3, K is the only generator in degree less than 3m + 3, so
P 3a
k
∏
(xi1 − xi2)
2b is a symmetric polynomial multiple of K. However, the only symmetric
polynomials of degree 1 are multiples of x1 + x2 + x3, implying (x1 + x2 + x3)|Pk which is
impossible by Corollary 4.2. ■
Note that by [9], Qm(3,Q)std has generators in degree 3m+ 1 and 3m+ 2, and by [13], such
generators with even degree are divisible by x1+x2−2x3. Let π be the canonical mapping from
characteristic 0 to characteristic 3. We then have the following lemma.
Lemma 5.10. Suppose Qm(3,F3)sign−triv has a generator L in degree 3m+ 1. We can choose
the generators of Qm(3,Q)std to be integer polynomials L′ and (x1 + x2 − 2x3)K
′ with π(K ′) =
π(L′) = L. Moreover, if
G = (x1 + x2 + x3)
(
K ′ − L′
3
)
− x3K
′,
then
π(G) = (x1 + x2 + x3)π
(
K ′ − L′
3
)
− x3L
is a degree 3m+ 2 generator for Qm(3,F3)sign−triv.
Proof. Let L′ be an arbitrary 3m + 1 degree generator of Qm(3,Q)std with coprime integer
coefficients in the (−1)-eigenspace of s12. By Lemma 4.1, π(L′) is an element of the (−1)-
eigenspace of s12 in Qm(3,F3)sign−triv and if π(L′) is not a scalar multiple of L then there must
exist some other generator of Qm(3,F3)sign−triv with degree less than or equal to 3m+ 1. That
generator and L would violate Lemma 4.3, so we may set π(L′) = L.
A higher degree generator of Qm(3,Q)std has degree 3m+2. With degL = 3m+1 implying
m ̸∈ X, 3m + 2 is even by Corollary 5.6. Using [13], we let (x1 + x2 − 2x3)K
′ be an arbitrary
degree 3m+2 generator for Qm(3,Q)std with coprime integer coefficients. Similarly, π((x1+x2−
2x3)K
′) = (x1 + x2 + x3)π(K
′) is an element of Qm(3,F3)sign−triv, so π(K ′) is a non-symmetric
polynomial of degree 3m+ 1 in Qm(3,F3)sign−triv. Thus it must be a scalar multiple of L, and
we may set π(K ′) = L.
Let G = (x1 + x2 + x3)
(
K′−L′
3
)
− x3K
′. Since
(x1 + x2 − 2x3)K
′ − (x1 + x2 + x3)L
′ = (x1 + x2 + x3)
(
K ′ − L′)− 3x3K
′
and π(K ′ − L′) = L− L = 0, we have G ∈ Qm(3,Q) ∩ Z[x1, x2, x3]. Then
π(G) = (x1 + x2 + x3)π
(
K ′ − L′
3
)
− x3L.
If π(G) generated by L, we must have π(G) = c(x1+x2+x3)L for some c ∈ F3 since deg (π(G)) =
deg(L)+1. However, x1+x2+x3 does not divide x3L since L is a generator, so x1+x2+x3 ∤ π(G).
Then if π(G) was not a generator, there must be some generator other than L for Qm(3,F3) in
degree less than 3m+ 2 which violates Lemma 4.3. Thus, π(G) is a generator. ■
We aim to prove that minimum Ren–Xu counterexamples are generators and represent the
only cases, where the Hilbert series of the quasi-invariants differs between characteristics 0 and 3.
To this end, we describe the degree of Ren–Xu counterexamples.
12 F. Wang and E. Yee
Example 5.11. We notice a “staircase” pattern for Ren–Xu counterexamples. The following
are counterexamples for m = 3, 4, 5:
(x1 − x2)
9, (x1 − x2)
9, (x1 − x2)
9
∏
(xi1 − xi2)
2.
We note that since (x1 − x2)
9 ∈ Q4(3,F3), (x1 − x2)
9 is the Ren–Xu counterexample for
both m = 3 and m = 4. Moreover, the counterexample in Q5(3,F3) is the previous coun-
terexample (x1 − x2)
9 multiplied by
∏
(xi1 − xi2)
2 to add the extra factor of (x1 − x2)
2. In this
way the degree of counterexample stays constant for the first half of the “staircase” and climbs
by 6 per each increase in m thereafter. Moreover, we note that m = 2, 6 ̸∈ X, so our “staircase”
is surrounded by non-counterexamples. One can also compute another generator for m = 3, 4, 5
in degree 12, 18, and 18 respectively. Since 9+12 = 6·3+3, 9+18 = 6·4+3, and 15+18 = 6·5+3,
Qm(3,F3)sign−triv is freely generated by each of these generators and 1 by Lemma 4.4. This way
we see that the upper degree generators form a complement to the lower degree ones, climbing
by 6 degrees initially and staying constant for the second half of the staircase.
Visually, the following figure shows the degree of the generators for Qm(3,F3) with respect
to m were the staircase pattern and Theorem 5.3 to hold.
0 10 20 30 40 50
0
50
100
150
m
D
e
g
re
e
o
f
g
e
n
e
ra
to
r
Generator 1
Generator 2
Figure 1. Degrees of generators in characteristic 3 with respect to m.
We prove that Ren–Xu counterexamples follow this staircase pattern.
Lemma 5.12. Let m be a natural number not in X and let d be the largest integer such that
Rm+1 lies in Qm+d(3,F3). Suppose that for all k ≤ m, if k ̸∈ X, then Qk(3,F3)sign−triv has
a generator in degree 3k + 1. Then Rm+j = Rm+1 in degree 3m + 3 for 1 ≤ j ≤ d and
Rm+j = Rm+1
∏
(xi1 − xi2)
2(j−d) in degree 3m+ 3 + 6(j − d) for d < j < 2d.
Proof. Let
Rm+1 = P 3a
k
∏
(xi1 − xi2)
2b,
where k is a nonnegative integer, a is a positive integer, and b = max
{
0, 2m+3−3a(2k+1)
2
}
. If b is
positive, the polynomial P 3a
k
∏
(xi1 − xi2)
2(b−1) has degree less than 3m − 2 and is at least m-
quasi-invariant since P 3a
k
∏
(xi1 −xi2)
2b has degree less than 3m+4. Thus P 3a
k
∏
(xi1 −xi2)
2(b−1)
is a Ren–Xu counterexample for Qm(3,F3), a contradiction.
In this way, we have Rm+1 = P 3a
k . Moreover, Qk(3,F3) must be a non-counterexample by
Lemma 5.4, so by our assumption Pk is a generator. By Lemma 5.9, Pk is not in Qk+1(3,F3), so
Hilbert Series of S3-Quasi-Invariant Polynomials in Characteristics 2, 3 13
the largest power of (x1−x2) dividing into Rm+1 must be (x1−x2)
3a(2k+1) andm+d = 3a(2k+1)−1
2
by Lemma 4.1. Then for all 1 ≤ j ≤ d,
2(m+ j) + 1− 3a(2k + 1)
2
≤ 2(m+ d) + 1− 3a(2k + 1)
2
= 0.
Thus Rm+j = P 3a
k = Rm+1 which is indeed in degree 3m+ 3 by Lemma 5.8.
We claim that for d < j < 2d, m+ j ∈ X. Let I be the set of integers h such that a Ren–Xu
counterexample for Qh(3,F3) is P 3a
k
∏
(xi1 − xi2)
2b for some b ∈ Z≥0. By [11], m ∈ I if and
only if
k +
1
3
≤ m
3a
≤ k +
2
3
− 1
3a
,
which implies I is
{
s, s+1, s+2, . . . , s+3a−1−1
}
for some s ≡ 3a−1 (mod 3a). Then note that
m+1 ∈ I, yet m ̸∈ I since m ̸∈ X. Thus m ≡ 3a−1− 1 (mod 3a). Since m+ d = 3a(2k+1)−1
2 ∈ I
as well, we have s = 3ak + 3a−1, m = 3ak + 3a−1 − 1, and d = 3a−1+1
2 . Then
3a(2k + 1)− 1
2
< m+ j <
3a−1 + 1
2
+
3a(2k + 1)− 1
2
= 3ak + 2 · 3a−1,
so m+ j is in I and thus in X.
If Rm+j = P 3a
k
∏
(xi1 − xi2)
2b, where b = 2(m+j)+1−3a(2k+1)
2 for d < j < 2d, then m + d =
3a(2k+1)−1
2 implies b = j − d. Thus Rm+j = P 3a
k
∏
(xi1 − xi2)
2(j−d) has degree 3m+ 3+ 6(j − d)
as desired. ■
In [13], the first author proved that generators of Qm(3,Fp)std for p > 3 lie in Fp[x1 − x3,
x2 − x3] using that Fp[x1 − x3, x2 − x3, x1 + x2 + x3] = Fp[x1, x2, x3]. However, this is not true
for p = 3 since x1 − x3 + x2 − x3 = x1 + x2 + x3 in characteristic 3, so we instead consider the
space F3[x1 − x3, x2 − x3, x3]. From now on, we say a polynomial’s degree in x3 is with respect
to the basis {x1 − x3, x2 − x3, x3}. Moreover, in [13] the first author defined the polynomial
Md = (x1 + x2 − 2x3)
2{ d
2
}(x1 − x3)
⌊ d
2
⌋(x2 − x3)
⌊ d
2
⌋
for natural numbers d and proved that homogeneous s12-invariant elements of Fp[x1 − x3, x2 −
x3]/(x1 − x2)
2 are equal to constant multiples of Md. Extending this gives that elements of
F3[x1 − x3, x2 − x3, x3]/(x1 − x2)
2 are polynomials in x3 with coefficients that are constant
multiples of Md. Some further nice properties of Md are the following.
Lemma 5.13. For any j, j′ ∈ Z≥0,
1. (x1 + x2 + x3)Mj = Mj+1 in F3[x1 − x3, x2 − x3, x3]/(x1 − x2)
2.
2. MjMj′ = Mj+j′ in F3[x1 − x3, x2 − x3, x3]/(x1 − x2)
2.
Proof. 1. In F3[x1 − x3, x2 − x3, x3]/(x1 − x2)
2, for j ∈ Z≥0,
(x1 + x2 + x3)M2j = (x1 + x2 + x3)(x1 − x3)
j(x2 − x3)
j = M2j+1
and
(x1 + x2 + x3)M2j+1 = (x1 + x2 + x3)
2(x1 − x3)
j(x2 − x3)
j
= (x1 − x3)(x2 − x3)M2j = M2j+2.
2. From (1), we have Mj = (x1 + x2 + x3)
j and Mj′ = (x1 + x2 + x3)
j′ in F3[x1 − x3, x2 −
x3, x3]/(x1 − x2)
2, and our equality follows. ■
14 F. Wang and E. Yee
This gives us intuition for the following lemmas.
Lemma 5.14. Let e1, e2, and e3 be the elementary symmetric polynomials for F3[x1, x2, x3]
S3
in degree 1, 2, and 3 respectively. If n is a natural number such that n ̸≡ 0 (mod 3), for all
natural numbers j < n there exists a monomial P in e1, e2, e3 such that P has degree n and
degree j in x3. If n is a natural number such that n ≡ 0 (mod 3), for all natural numbers
j < n− 1 there exists a monomial P in e1, e2, e3 such that P has degree n and degree j in x3.
Proof. We choose e1, e2, and e3 to be
e1 = x1 + x2 + x3 = (x1 − x3) + (x2 − x3),
e2 = x1x2 + x1x3 + x2x3 = (x1 − x3)(x2 − x3) + 2((x1 − x3) + (x2 − x3))x3,
e3 = x1x2x3 = (x1 − x3)(x2 − x3)x3 + ((x1 − x3) + (x2 − x3))x
2
3 + x33.
We prove the lemma by decreasing induction on j.
The base case for n where 3 ∤ n is j = n − 1. If j = n − 1 and n ≡ 1 (mod 3), we can let
P = e
(n−1)/3
3 e1. If n ≡ 2 (mod 3), we let P = e
(n−2)/3
3 e2. The base case when 3|n is j = n−2, so
we can let P = e1e2e
(n/3)−1
3 .
Suppose that, when 3 ∤ n, for all j′ such that n > j′ > j where j ∈ N and 0 ≤ j < n − 1
there exists a monomial in e1, e2, e3 with degree n and degree j′ in x3. Suppose the same for
when 3|n but with n − 1 > j′ > j and j < n − 2. Then there exists a monomial m = ea1e
b
2e
c
3
with degree j+1 in x3 in F3[x1−x3, x2−x3, x3]/(x1−x2)
2. If b ̸= 0 we can take the monomial
ea+2
1 eb−1
2 ec3 to be P since it has degree n and degree j in x3. If b = 0 and a, c > 0, then we take
P = ea−1
1 eb+2
2 ec−1
3 . Finally, we are left with the cases a, b = 0 or b, c = 0. The former would
imply m = e
n
3
3 is our monomial, but 3 ∤ n would imply m is not a polynomial and 3|n implies m
has degree j + 1 = n in x3 and j = n − 1 ̸< n − 2. For the latter case, we have that a = n,
so m = en1 implies that j + 1 = 0 which is below our range for j. ■
Lemma 5.15. For all fj ∈ F3 and n ̸≡ 0 (mod 3), there exists a P ∈ F3[x1, x2, x3]
S3 such that
P = f0Mnx
0
3 + f1Mn−1x
1
3 + · · ·+ fn−2M2x
n−2
3 + fn−1M1x
n−1
3
in F3[x1, x2, x3]
S3/(x1−x2)
2. If n ≡ 0 (mod 3), for all fj ∈ F3 there exists a P ∈ F3[x1, x2, x3]
S3
such that
P = f0Mnx
0
3 + f1Mn−1x
1
3 + · · ·+ fn−2M2x
n−2
3
in F3[x1, x2, x3]
S3/(x1−x2)
2. Moreover, P also satisfies the property that if it has degree k in x3
in F3[x1, x2, x3]
S3/(x1 − x2)
2, then it has degree k in x3 in F3[x1 − x3, x2 − x3, x3].
Proof. A weaker statement is that there exists some fixed c0, c1, . . . , cj ∈ F3 such that for all
fj+1, fj+2, . . . , fn−1 ∈ F3, there exists a symmetric polynomial
P ≡ c0Mnx
0
3 + c1Mn−1x
1
3 + · · ·+ cjMn−jx
j
3
+ fj+1M2x
n−j−1
3 + fj+2M1x
n−j−2
3 + · · ·+ fn−1M1x
n−1
3 (mod (x1 − x2)
2),
when n ̸≡ 3 (mod 3) and j ∈ Z≥0. A similar weaker statement can be made for the n ≡ 0
(mod 3) case. We prove the statement in the lemma by induction on this j.
For the base case when n ̸≡ 0 (mod 3), we claim there exists coefficients cj ∈ F3 such that the
polynomial c0Mnx
0
3 + c1Mn−1x
1
3 + · · ·+ cn−2M2x
n−2
3 + cn−1M1x
n−1
3 is in F3[x1, x2, x3]
S3/(x1 −
x2)
2. The symmetric polynomial 0 satisfies these conditions and has degree 0 in x3. For the base
case when n ≡ 0 (mod 3), we claim there exists coefficients c0, . . . , cn−2 such that the polyno-
mial c0Mnx
0
3 + c1Mn−1x
1
3 + · · ·+ cn−2M2x
n−2
3 is in F3[x1, x2, x3]
S3/(x1 − x2)
2. The symmetric
polynomial 0 satisfies this.
Hilbert Series of S3-Quasi-Invariant Polynomials in Characteristics 2, 3 15
We consider the case where n ̸≡ 0 (mod 3). Suppose that for all n ≥ j′ > j there exists
coefficients c0, . . . , cj′−1 such that for all fj′ , fj′+1, . . . , fn−1 there exists a symmetric polynomial
P such that
P = c0Mnx
0
3 + c1Mn−1x
1
3 + · · ·+ cj′−1Mn−j′+1x
j′−1
3 + fj′Mn−j′x
j′
3 + · · ·+ fn−1M1x
n−1
3
lies in F3[x1, x2, x3]
S3/(x1 − x2)
2, where j ∈ N, 0 ≤ j ≤ n− 1. Moreover, suppose the polyno-
mial P exists such that it has degree in x3 equal to the degree in x3 in F3[x1, x2, x3]
S3/(x1−x2)
2.
Consider arbitrary coefficients fj , fj+1, . . . , fn−1. If they are each 0, then we can take 0 to
be our polynomial just like our base case. Otherwise, let l be the greatest natural number l ≥ j
such that fl ̸= 0. If l = j, by Lemma 5.14 there exists a monomial m in e1, e2, e3 with degree j
in x3 and we may take fjm to be our symmetric polynomial.
If l > j, by assumption there exists coefficients c0, c1, . . . , cj such that
S = c0Mnx
0
3 + c1Mn−1x
1
3 + · · ·+ cjMn−jx
j
3 + fj+1Mn−j−1x
j+1
3 + · · ·+ fn−1M1x
n−1
3
lies in F3[x1, x2, x3]
S3/(x1 − x2)
2. By assumption, S has degree l in x3.
Without loss of generality let the leading coefficient of m be Mn−j , so
S + (fj − cj)m = c′0Mnx
0
3 + c′1Mn−1x
1
3 + · · ·+ c′j−1Mn−j+1x
j−1
3
+ fjMn−jx
j
3 + · · ·+ fn−1M1x
n−1
3
for some coefficients c′0, c
′
1, . . . , c
′
j−1. Moreover, S + (fj − cj)m is still a symmetric polynomial
and m has degree j in x3 while S has degree l, so S + (fj − cj)m has degree l as desired.
An identical argument holds for n ≡ 0 (mod 3). ■
Now we have the tools to prove m ̸∈ X implies m+ 1 begins our staircase.
Lemma 5.16. Suppose that for all k ≤ m, if k ̸∈ X then Qk(3,F3) has a 3k+1 degree generator,
where m is a natural number. Then if Qm(3,F3)sign−triv has a generator in degree 3m+ 1,
Qm+1(3,F3)sign−triv has a generator in degree 3m+ 6.
Proof. By Lemma 5.10, the generators for Qm(3,F3)sign−triv are(
(x1 + x2 + x3)π
(
A′ −B′
3
)
− x3B
)
(x1 − x2)
2m+1
in degree 3m+ 2, and
B(x1 − x2)
2m+1
in degree 3m+1, where (x1−x2)
2m+1(x1+x2− 2x3)A
′ and (x1−x2)
2m+1B′ are the generators
of Qm(3,Q)std, B is an s12-invariant polynomial, and π(A′) = π(B′) = B.
For the greater degree generator, let C =
(
(x1 + x2 + x3)π
(
A′−B′
3
)
− x3B
)
. We would like to
show there exists symmetric polynomials P and Q in degree 4 and 5 respectively such that
PC +QB ≡ 0 (mod (x1 − x2)
2).
Since PC+BQ
(x1−x2)2
is still s12-invariant, this would then imply (PC+QB)(x1−x2)
2m+1 ∈ Qm+1(3,F3)
by Lemma 4.1. Consider writing
P = f0M4x
0
3 + f1M3x
1
3 + f2M2x
2
3 + f3M1x
3
3
and
Q = h0M5x
0
3 + h1M4x
1
3 + h2M3x
2
3 + h3M2x
3
3 + h4M1x
4
3
16 F. Wang and E. Yee
for arbitrary fj and hj . By Lemma 5.15, we know that for any choice of fj and hj , we have
P,Q ∈ F3[x1, x2, x3]
S3/(x1 − x2)
2.
We claim that B|π
(
A′−B′
3
)
in F3[x1, x2, x3]/(x1 − x2)
2. By [13], A′ and B′ are both polyno-
mials in the variables (x1−x2)
2 and (x1−x3)(x2−x3). Moreover, by Lemma 5.9, (x1−x2)
2 ∤ B
so B ≡ cMm (mod (x1−x2)
2) for some c ∈ F3 such that c ̸= 0. Similarly, we know π
(
A′−B′
3
)
≡
c′Mm (mod (x1 − x2)
2) for some c′ ∈ F3. Thus we have π
(
A′−B′
3
)
= dB, where d = c′
c .
We use Lemma 5.13 to expand PC +BQ in F3[x1, x2, x3]/(x1 − x2)
2,
PC +QB =
(
h0M5B + f0(x1 + x2 + x3)M4π
(
A′ −B′
3
)
x03
)
+
3∑
j=1
(
hjM5−jBxj3 + fj(x1 + x2 + x3)M4−jπ
(
A′ −B′
3
)
xj3
− fj−1M5−jBxj3
)
+ h4M1Bx43 − f3M1Bx43
=
(
h0B + f0π
(
A′ −B′
3
))
M5
+
3∑
j=1
((
(hj − fj−1)B + fjπ
(
A′ −B′
3
))
M5−jx
j
3
)
+ (h4 − f3)M1Bx43.
= (h0 + f0d)BM5 +
3∑
j=1
(
(hj − fj−1) + fjd
)
BM5−jx
j
3 + (h4 − f3)M1Bx43.
Letting hj be arbitrary for j > 0, set f3 = h4, fj−1 = hj + fjd for 0 < j < 3 and set h0 = −f0d.
This makes the expression PC +QB = 0.
We claim Qm+1(3,F3)sign−triv has a degree 3m+3 generator, namely Rm+1. From Lemma 5.9,
Qm+1(3,F3)sign−triv has no degree 3m+1 or 3m+2 generator, so it has no generators in degree
less than 3m+3. By Lemma 5.8, Rm+1 is in degree 3m+3 so it must be a generator. Without
loss of generality, we let
Rm+1 = ((x1 + x2 + x3)C + SB) (x1 − x2)
2m+1,
where S is a degree 2 symmetric polynomial.
If (PC + QB)(x1 − x2)
2m+1 were generated by Rm+1, there would exist a symmetric poly-
nomial I such that IRm+1 = (PC +QB)(x1 − x2)
2m+1. This implies (I(x1 + x2 + x3)− P )C +
(IS − Q)B = 0. If I(x1 + x2 + x3) − P ̸= 0 or IS − Q ̸= 0, there is a relation on C and
B over F3[x1, x2, x3]
S3 , but C(x1 − x2)
2m+1 and B(x1 − x2)
2m+1 are generators of Qm(3,F3).
Thus we must have P = I(x1 + x2 + x3), so (x1 + x2 + x3)|P . Now we consider the symmetric
polynomials P ′ = P + e22 + e2e
2
1 + e41 and Q′ = Q + e3e
2
1 + (−d − 1)e22e1 − de31e2 + (−d + 1)e51.
In F3[x1 − x3, x2 − x3, x3]/(x1 − x2)
2, we get that
P ′ = f0M4x
0
3 + f1M3x
1
3 + (f2 + 1)M2x
2
3 + f3M1x
3
3
and
Q′ = h0M5x
0
3 + h1M4x
1
3 + (h2 − d)M3x
2
3 + (h3 + 1)M2x
3
3 + h4M1x
4
3.
Then f2+1 = (h3+ f3d)+ 1 = (h3+1)+ f3d, f1 = h2+ f2d = (h2− d)+ (f2+1)d, and the rest
of the equations necessary for P ′C +Q′B ≡ 0 (mod (x1 − x2)
2) are the same as PC +QB ≡ 0
(mod (x1−x2)
2). Thus P ′C+Q′B ≡ 0 (mod (x1−x2)
2). Moreover, (x1+x2+x3) divides into
P +e2e
2
1+e41 but not e22, so (x1+x2+x3) ∤ P ′. We have shown that if (PC +QB)(x1 − x2)
2m+1
Hilbert Series of S3-Quasi-Invariant Polynomials in Characteristics 2, 3 17
is generated by Rm+1, then (x1 + x2 + x3)|P , implying (P ′C +Q′B)(x1 − x2)
2m+1 is not gen-
erated by Rm+1. If (P ′C + Q′B)(x1 − x2)
2m+1 is not a generator, then whatever generates
it violates Lemma 4.3, so (P ′C + Q′B)(x1 − x2)
2m+1 is indeed a degree 3m + 6 generator
of Qm+1(3,F3). ■
Now we prove that if Rm+1 begins our staircase, then it is the lower degree generator for the
first half of the staircase.
Lemma 5.17. Let m ̸∈ X for some natural number m. Suppose Rm+1 is a degree 3m + 3
generator of Qm+1(3,F3)sign−triv and L is another generator in degree 3m + 6. Further, let
Rm+1 lie in Qm+d(3,F3), where d is maximal. Then L
∏
(xi1 − xi2)
2(j−1), Rm+1, and 1 freely
generate Qm+j(3,F3)sign−triv for 1 ≤ j ≤ d.
Proof. As a generator, L lies in a copy of sign− triv and is divisible by (x1 − x2)
2(m+1)+1
by Lemma 4.1. Since L
∏
(xi1 − xi2)
2(j−1) is divisible by (x1 − x2)
2(m+j)+1, by the second
part of Lemma 4.1, L
∏
(xi1 − xi2)
2(j−1) is in Qm+j(3,F3)sign−triv. If L
∏
(xi1 − xi2)
2(j−1) is not
a generator, Rm+1 must generate L
∏
(xi1 − xi2)
2(j−1), implying a relation between Rm+1 and L.
Thus L
∏
(xi1 − xi2)
2(j−1) is indeed a generator.
Moreover, 3m+3+3m+6j = 6(m+ j)+3 so by Lemma 4.4, L
∏
(xi1 −xi2)
2(j−1) and Rm+1
generate Qm+j(3,F3)sign−triv. ■
Next, we prove that, for all consecutive spaces of quasi-invariants in the second half of the
staircase, the lower degree generator is
∏
(xi1 −xi2)
2 times the previous lower degree generator.
Lemma 5.18. Let m ̸∈ X for some natural number m. Suppose Rm+1 is a degree 3m+ 3 gen-
erator of Qm(3,F3) and L is another generator in degree 3m+6. Let Rm+1 lie in Qm+d(3,F3),
where d is maximal. Further, let L have degree at most 5 in x3. Then for all d ≤ j < 2d,
Qm+j(3,F3)sign−triv is freely generated by a generator in degree 3m+6d, Rm+1
∏
(xi1−xi2)
2(j−d)
in degree 3m+ 6(j − d) + 3, and 1.
Proof. We proceed with induction.
The generator Rm+1 of Qm+d(3,F3) is in degree 3m + 3 = 3m + 6(d − d) + 3, and from
Lemma 5.17 a second generator is L
∏
(xi1 − xi2)
2(d−1) in degree 3m+ 6d. Moreover, these are
the only generators so the claim is true for j = d.
Let k be a natural number with d < k < 2d and suppose Qm+j(3,F3)sign−triv has a generator
in degree 3m + 6d and degree 3m + 6(j − d) + 3 for all d ≤ j < k, where this upper degree
generator is a polynomial of degree at most 5 in x3 and is not generated by Rm+1. Consider
Qm+k(3,F3)sign−triv. We know Rm+1
∏
(xi1−xi2)
2(k−d−1) is an element of Qm+k−1(3,F3)sign−triv
of degree 3m+6(k−d− 1)+3 by Lemma 4.1. Since k− 1 < k, our inductive hypothesis implies
Rm+1
∏
(xi1 − xi2)
2(k−d−1) is a generator for Qm+k−1(3,F3)sign−triv.
Let T be the degree 3m + 6d generator for Qm+k−1(3,F3)sign−triv with degree 5 in x3. We
write Rm+1
∏
(xi1 − xi2)
2(k−d−1) = R′
m+1(x1 − x2)
2(m+k−1)+1 and T = T ′(x1 − x2)
2(m+k−1)+1
for s12 invariant polynomials R′
m+1 and T ′. If o = m + 4k − 6d − 2 and r = m + 6d − 2k + 1,
then degR′
m+1 = o and deg T ′ = r. We want to find a degree r − o symmetric polynomial P
such that
−PR′
m+1 + T ′ ≡ 0 (mod (x1 − x2)
2).
We claim that R′
m+1 has degree 0 in x3. This is because Rm+1
∏
(xi1 − xi2)
2(k−d−1) =
P 3a
l
∏
(xi1 − xi2)
2(k−d−1) as we proved in Lemma 5.12. Since Pl is the map of the gener-
ator of Ql(3,Q) into characteristic 3, Pl must be constant in the variable x3. We can see∏
(xi1 − xi2)
2(k−d−1) is also constant in x3, so Rm+1 and R′
m+1 are constant in x3.
18 F. Wang and E. Yee
Having assumed that T ′ is at most degree 5 in x3,
T ′ = t0Mrx
0
3 + t1Mr−1x
1
3 + t2Mr−2x
2
3 + t3Mr−3x
3
3 + t4Mr−4x
4
3 + t5Mr−5x
5
3
and
R′
m+1 = aMo
for coefficients tj and a in F3. Since Rm+1 is not in Qm+d+1(3,F3)sign−triv, we have a ̸= 0.
We let
P =
t0
a
Mr−ox
0
3 +
t1
a
Mr−o−1x
1
3 +
t2
a
Mr−o−2x
2
3 +
t3
a
Mr−o−3x
3
3
+
t4
a
Mr−o−4x
4
3 +
t5
a
Mr−o−5x
5
3,
so that T ′−PR′
m+1 ≡ 0 (mod (x1−x2)
2) by Lemma 5.13. Since deg(P ) = r−o = 12d−6k+3 ≥
9 > 7, by Lemma 5.15 such a symmetric polynomial P is attainable with P having degree at most
degree 5 in x3. Since T
′ also has at most degree 5 in x3 and R′
m+1 has degree 0, (−PR′
m+1+T ′)
has at most degree 5 in x3. Letting U =
(
−PR′
m+1 + T ′)(x1 − x2)
2(m+k−1)+1, we have U is in
Qm+k(3,F3) with degree 3m+6d and since
(
−PR′
m+1
)
(x1−x2)
2(m+k−1)+1 is generated by Rm+1
and T is not, U is not generated by Rm+1. Finally, we also have Rm+1
∏
(xi1 − xi2)
2(k−d)
is in Qm+k(3,F3)sign−triv with degree 3m + 6(k − d) + 3. Thus what is left is to prove is
Rm+1
∏
(xi1−xi2)
2(k−d) and
(
−PR′
m+1+T ′)(x1−x2)
2(m+k−1)+1 are generators for Qm+k(3,F3).
Assume for sake of contradiction that U and Rm+1
∏
(xi1−xi2)
2(k−d) are not both generators.
If Rm+1
∏
(xi1 − xi2)
2(k−d) is a generator, then any other generator must be of at least degree
3m + 6d by Lemma 4.3. Yet U is not generated by Rm+1
∏
(xi1 − xi2)
2(k−d) since it is not
generated by Rm+1. Thus U must be a generator.
Next, we consider if Rm+1
∏
(xi1 −xi2)
2(k−d) is not a generator. For Rm+1
∏
(xi1 −xi2)
2(k−d)
to not be a generator there must be a generator in a degree less than 3m + 6(k − d) + 3.
Let it be G, and by Lemma 4.3, any other generator must have degree greater than 3m + 6d.
Thus U is not a generator, so U and Rm+1
∏
(xi1 − xi2)
2(k−d) are both generated by G and
specifically U = QG and Rm+1
∏
(xi1 − xi2)
2(k−d) = SG for symmetric polynomials P and Q.
Moreover, Rm+1
∏
(xi1−xi2)
2(k−d−1) is the lowest degree generator for Qm+k−1(3,F3)sign−triv, so
G = CRm+1
∏
(xi1 − xi2)
2(k−d−1) for a symmetric polynomial C. This implies C|
∏
(xi1 − xi2)
2,
and G is not a scalar multiple of Rm+1
∏
(xi1 − xi2)
2(k−d), so C is a constant. We then have U
is a constant multiple of QRm+1
∏
(xi1 − xi2)
2(k−d−1), so U is generated by Rm+1 which is
a contradiction.
Thus U and Rm+1
∏
(xi1 − xi2)
2(k−d) are each generators and together with 1 they freely
generate Qm(3,F3)sign−triv by Lemma 4.4. ■
Finally, we show that after the staircase completes, the next space of quasi-invariants has no
counterexamples.
Lemma 5.19. Let Qm−1(3,F3)sign−triv have generators K in degree 3m−3 and T in degree 3m
such that K is not in Qm(3,F3)sign−triv. If m is even, then Qm(3,F3)sign−triv is freely generated
by a generator in degree 3m+ 1, 3m+ 2, and 1.
Proof. Suppose for the sake of contradiction that Qm(3,F3)sign−triv has a generator U in degree
3m−1 or 3m−2. Then since U is also in the −1 s12 eigenspace of Qm−1(3,F3)sign−triv, U must be
generated by K over F[x1, x2, x3]
S3 . Yet K being divisible by a symmetric polynomial violates
Corollary 4.2.
Suppose for the sake of contradiction that Qm(3,F3)sign−triv has a generator in degree 3m.
Without loss of generality let that generator be T . From [9], we can let L′ be a degree 3m+ 1
Hilbert Series of S3-Quasi-Invariant Polynomials in Characteristics 2, 3 19
generator of Qm(3,Q)std with coprime integer coefficients. Then π(L′) ∈ Qm(3,F3)sign−triv,
so π(L′) must be generated by T since any other generator in degree less than degree 3m + 1
would violate Lemma 4.3. Moreover, the only degree 1 symmetric polynomials are constant
multiples of x1 + x2 + x3, so we can assume without loss of generality that
π(L′) = (x1 + x2 + x3)T.
Note that from [13] all generators of Qm(3,Q)std must lie in Q[x1 − x3, x2 − x3]. Thus (x1 +
x2 + x3)T ∈ F3[x1 − x3, x2 − x3] and so T ∈ F3[x1 − x3, x2 − x3].
We also have T = (x1 − x2)
2m+1T ′ for some s12-invariant polynomial T ′. Thus by the
fundamental theorem of symmetric polynomials T ′ ∈ F3[(x1 − x3)(x2 − x3), x1 + x2 + x3]. Note
that deg T ′ = 3m− 2m− 1 = m− 1 and m is even, so T ′ has an odd degree. However, since it
is generated by (x1 − x3)(x2 − x3) and x1 + x2 + x3, we must have (x1 + x2 + x3)|T ′. This gives
a contradiction because T is a generator. ■
Finally, we have the lemmas to prove Theorem 5.3.
Proof of Theorem 5.3. We prove this using induction on m.
The generators for Q0(3,F3)sign−triv are x1 − x2 and x3(x1 − x2). These generators are in
degree 3 · 0 + 1 and 3 · 0 + 2 so the theorem is true for the base case.
Assume the claim is true when m < j for some j ∈ N. Consider the space Qj(3,F3)sign−triv.
Let t be the largest natural number less than j such that t ̸∈ X. By the inductive hypothesis,
Qt(3,F3) has a generator in degree 3t+1 and 3t+2. By Lemma 5.10, we may let the generators be(
x1 + x2 + x3)π
(
A′ −B′
3
)
− x3B
)
(x1 − x2)
2t+1 and B(x1 − x2)
2t+1,
where (x1 − x2)
2t+1(x1 + x2 − 2x3)A
′ and (x1 − x2)
2t+1B′ are generators for Qt(3,Q)std and
π(A′) = π(B′) = B. From Lemma 5.16, Qt+1(3,F3)sign−triv is generated by a generator in degree
3t+ 6 and 3t + 3. Moreover, Rt+1 is the 3t + 3 degree generator by Lemma 5.8. Let L be the
degree 3t + 6 generator. Suppose Rt+1 lies in Qt+d(3,F3), but not Qt+d+1(3,F3), where d is
a natural number.
First, we consider when t + d ≥ j ≥ t + 1. By Lemma 5.17, Qj(3,F3)sign−triv has gen-
erators Rt+1 and L
∏
(xi1 − xi2)
2(j−t−1). Note that Rt+1 = Rj by Lemma 5.12, and further
deg
(
L
∏
(xi1 − xi2)
2(j−t−1)
)
+ deg(Rt+1) = (6(j − t − 1) + 3t + 6) + 3t + 3 = 6j + 3. By
Lemma 4.4, we then have that Rt+1 and L
∏
(xi1 − xi2)
2(j−t−1) generate Qj(3,F3)sign−triv.
Next, we consider the case where t+2d− 1 ≥ j ≥ t+ d+1. Notice that by our construction
in Lemma 5.16, we can choose L such that it has at most degree 5 in x3. Thus we can apply
Lemma 5.18, which gives us that Qj(3,F3)sign−triv is generated by Rt+1
∏
(xi1 −xi2)
2(j−t−d) and
a generator in degree 3t+ 6d. Note that Rt+1
∏
(xi1 − xi2)
2(j−t−d) is a constant multiple of Rj
by Lemma 5.12. Moreover, the sum of their degrees is 3t+ 6(j − t− d) + 3 + 3t+ 6d = 6j + 3
as desired.
Finally, we consider if j = t+2d. Note that by Lemma 5.18, Qt+2d−1(3,F3) has a generator in
degree 3t+6d and 3t+6(d−1)+3. The degree 3t+6(d−1)+3 generator is Rt+1
∏
(xi1−xi2)
2(d−1),
and Rt+1 is divisible by (x1 − x2)
2(t+d)+1, where d is maximal, so Rt+1
∏
(xi1 − xi2)
2(d−1) does
not lie in Qt+2d(3,F3). Moreover, Qt(3,F3) is a non Ren–Xu counterexample, so t must be even
by Lemma 5.6. Then t+ 2d is even as well, so by Lemma 5.19, Qt+2d(3,F3) has a generator in
degree 3(t+ 2d) + 1 and 3(t+ 2d) + 2.
Now we claim we have exhausted all cases. If we had j > t + 2d, since we just showed
t+ 2d ̸∈ X, we would not have chosen t to be the largest natural number less than j not inX. ■
20 F. Wang and E. Yee
Remark 5.20. We can compute the degrees of generators of Qm(3,F3)sign−triv explicitly. If m
has no digits 1 in its base 3 representation, then the generators have degree 3m+1 and 3m+2.
Otherwise, the lower degree generator is Rm. We can deduce the minimal degree of the Ren–Xu
counterexamples in Qm(3,F3): Let a be the greatest natural number such that the a-th term
from the right in the base 3 representation of m is 1. Then if
⌈ ⌈ m
3a
⌉−1
2
⌉
= k, a minimal deg-
ree Ren–Xu counterexample is P 3a
k
∏
(xi1 − xi2)
2b, where
b = max
{
2m+ 1− 3a(2k + 1)
2
, 0
}
.
The degrees of the generators are then 3a(2k + 1) + 6b and 6m+ 3− 3a(2k + 1)− 6b.
6 Representations of S3 in Qm(3,F3)
Now that we have a complete picture of Qm(3,F3)sign−triv, we consider generators that generate
the other indecomposable modules of S3. We start with triv − sign− triv, which behaves very
similarly to sign− triv.
Proposition 6.1. Suppose that for all j ≤ m, Qj(3,F3)sign−triv has generators in degree d
and 6j + 3 − d respectively for some d. If K, L are distinct generators of Qm(3,F3)sign−triv
then there are two other homogeneous generators K1, L1 of Qm(3,F3) in the same degrees
as K, L, respectively such that as a representation of S3, K1 generates a copy of triv − sign− triv
containing K and L1 generates a copy of triv − sign− triv containing L. Moreover, there are
no relations between K1, L1 over the symmetric polynomials, and there are no other generators
of Qm(3,F3) that generate a copy of triv − sign− triv.
Proof. We prove this by induction on m. For the base case m = 0, note that by Exam-
ple 2.9, for K = x1 − x2 we have that K1 = x1 satisfies the desired conditions. Similarly, for
L = (x1 − x2)x3, we have that L1 = x1(x2 + x3) satisfies the desired conditions. These two are
independent over the symmetric polynomials, as a relation between them would imply a relation
between 1 and x2 + x3.
For the inductive step, let K ′, L′ be the generators of Qm−1(3,F3)sign−triv and let K ′
1, L
′
1 be
the corresponding generators of Qm−1(3,F3). Without loss of generality, we can choose K ′
1, L
′
1
to be s23-invariant with (1 − s12)K
′
1 = K ′, (1 − s12)L
′
1 = L′ (similar to in the base case).
Let K, L be generators of Qm(3,F3)sign−triv. Then since K,L ∈ Qm−1(3,F3)sign−triv, we can
write K = P1K
′ +Q1L
′, L = P2K
′ +Q2L
′ for symmetric polynomials P1, P2, Q1, Q2. Then it
follows that K1 := P1K
′
1 +Q1L
′
1, L1 := P2K
′
1 +Q2L
′
1 each generate a copy of triv − sign− triv
that contains K, L, respectively. Moreover, if there is some relation P3K1 +Q3L1 = 0 for sym-
metric polynomials P3, Q3, then applying 1− s12 to this equation would yield P3K +Q3L = 0,
which violates Lemma 4.4.
Next, we show that K1, L1 are m-quasi-invariants. As the computations are the same for
both polynomials, we give the proof only for K1. First, note that (1 − s23)K1 = 0 since both
K ′
1, L
′
1 are s23-invariant. Next, note that (1 − s12)K1 = K is divisible by (x1 − x2)
2m+1 by
Lemma 4.1. Finally, note that since K1 is s23-invariant, we have
(1− s13)K1 = s23(s23 − s23s13)K1 = s23(1− s23s13s23)K1 = s23(1− s12)K1
is divisible by s23(x1 − x2)
2m+1 = (x1 − x3)
2m+1.
Note that K1, L1 are the minimal degree polynomials such that (1 − s12)K1, (1 − s12)L1
are symmetric polynomial multiples of K, L, respectively, so they cannot be generated by
any other generators and thus must be generators themselves. Then assume for contradiction
that there is some other generator T of Qm(3,F3) that generates a copy of triv − sign− triv.
Hilbert Series of S3-Quasi-Invariant Polynomials in Characteristics 2, 3 21
Then (1 − s12)T is contained in a copy of sign− triv and is s12-antiinvariant, so we can write
(1 − s12)T = S1K + S2L for symmetric polynomials S1, S2. Then T , S1K1 + S2L1 generate
copies of triv − sign− triv with the same sign− triv submodule, so they generate a copy of
(triv − sign− triv ⊕ triv − sign− triv)/sign− triv ∼= triv − sign− triv ⊕ triv.
Thus T is generated by K1, L1, 1, and is not a generator itself. ■
Corollary 6.2. The generators 1, K, K1, L, L1 of Qm(3,F3) defined in Proposition 2.11,
Theorem 5.3 and Proposition 6.1 have no relations between them over the symmetric polynomials.
Proof. Let
P1 + P2K + P3L+ P4K1 + P5L1 = 0
for symmetric polynomials P1, . . . , P5. Then apply 1 + s12 to the equation to yield
2P1 + P4(2K1 −K) + P5(2L1 − L) = 0
since K, L are s12-antiinvariant. Next, apply 1− s23 to this equation to yield
P4(s23 − 1)K + P5(s23 − 1)L = 0.
Note that (s23 − 1)K generates the same copy of sign− triv as K, since s23 − 1 acts bijectively
on sign (and similarly for L). So a relation between (s23 − 1)K, (s23 − 1)L is equivalent to
a relation between K, L, which cannot exist by Lemma 4.4. So we have P4 = P5 = 0.
Now, the result follows from Lemma 4.4. ■
Remark 6.3. In the non-modular case, one has that the polynomial
∏
i1<i2
(xi1 − xi2)
2m+1 is
a generator of Qm(n,k), as it is the lowest degree quasi-invariant in the sign module. However,
from Lemma 4.4 we have that in characteristic 3,
(L+ s23L)K − (K + s23K)L = c
∏
i1<i2
(xi1 − xi2)
2m+1,
so
∏
i1<i2
(xi1 − xi2)
2m+1 is not a generator. We can take this calculation further, and note that
(L+ s23L)K1 − (K + s23K)L1 would then generate a copy of triv − sign, as the quotient of this
module by the space generated by (L+ s23L)K − (K + s23K)L must be a trivial module.
It remains to consider the modules triv − sign, sign− triv − sign. To motivate the results
that follow, we start by considering 0-quasi-invariants.
Example 6.4. Note that from Corollary 6.2 we know that Q0(3,F3) has 5 generators 1, x1−x2,
(x1 − x2)x3, x1, x1(x2 + x3) with no relations between them. By examining the dimension of
the space of all homogeneous degree 3 polynomials, we have that Q0(3,F3)[3] is 10-dimensional.
Since F3[x1, x2, x3]
S3 is 3-dimensional in degree 3, 2-dimensional in degree 2, and 1-dimensional
in degree 1, so far we have accounted for only 3 + 2 + 2 + 1 + 1 = 9 dimensions. Moreover,
every irreducible representation is accounted for, so this extra dimension must be an extension
of an existing indecomposable representation. The only indecomposable representations that
have nontrivial extensions are the triv generated by x1x2x3 and the triv − sign generated by
E := (x1x2 + x1x3 + x2x3)x1 + (x1 + x2 + x3)(x1(x2 + x3))
= −x21x2 − x21x3 + x1x
2
2 + x1x
2
3.
Indeed, the triv − sign generated by E extends to a sign− triv − sign generated by
F := (x1 − x2)x1x2.
22 F. Wang and E. Yee
We will later see that the polynomials E, F defined above are key to understanding triv − sign
and sign− triv − sign in the quasi-invariants.
Proposition 6.5. Q0(3,F3) is freely generated by 1, x1 − x2, (x1 − x2)x3, x1, x1(x2 + x3), F
as a F3[x1, x2, x3]
S3-module.
Proof. We already know that the first 5 polynomials are independent. Now, let
P1 + P2(x1 − x2) + P3(x1 − x2)x3 + P4x1 + P5(x2 + x3)x1 + P6F = 0
for symmetric polynomials Pj . Apply 1− s12 to this equation to get
(P4 − P2)(x1 − x2) + (P5 − P3)(x1 − x2)x3 − P6F = 0.
Next, apply 1 + s23 to get
(P2 − P4)(x1 + x2 + x3) + (P5 − P3)(x1x2 + x1x3 + x2x3) + P6E = 0.
Finally, note that as E can be written in terms of symmetric polynomial multiples of x1,
(x2 + x3)x1, this equation would be a relation between the first 5 generators of Q0(3,F3).
We have seen this is impossible, so we have P6 = 0, and hence all of the Pj must be 0.
Let Q′
0 be the submodule of Q0(3,F3) generated by these 6 polynomials. Then as the poly-
nomials freely generate Q′
0 as a F3[x1, x2, x3]
S3-module, we have that the Hilbert series of Q′
0 is
H(Q′
0) =
(
1 + 2t+ 2t2 + t3
)
H
(
F3[x1, x2, x3]
S3
)
=
1 + 2t+ 2t2 + t3
(1− t)(1− t2)(1− t3)
=
1
(1− t)3
by the fundamental theorem of symmetric polynomials. This is exactly the Hilbert series of
Q0(3,F3), so Q′
0 = Q0(3,F3) and there are no more generators of Q0(3,F3). ■
Similar to how we only considered polynomials in the (−1)-eigenspace of s12 for sign− triv,
we only consider generators in the (−1)-eigenspace of s12 for sign− triv − sign and polynomials
in the 1-eigenspace of s23 for triv − sign. Note that this is sufficient to describe the roles
of sign− triv − sign, triv − sign, as both modules are generated by an element satisfying their
respective constraints.
Lemma 6.6.
1. Let T ∈ Qm(3,F3) generate a copy of triv − sign. Then T is the sum of a symmetric
polynomial multiple of E
∏
i1<i2
(xi1 − xi2)
2m and a symmetric polynomial. Conversely,
any symmetric polynomial multiple of E
∏
i1<i2
(xi1−xi2)
2m generates a copy of triv − sign
in Qm(3,F3).
2. Let T1 ∈ Qm(3,F3) generate a copy of sign− triv − sign. Then T1 is the sum of a symmet-
ric polynomial multiple of F
∏
i1<i2
(xi1 − xi2)
2m and a symmetric polynomial multiple of∏
i1<i2
(xi1 − xi2)
2m+1. Conversely, any symmetric polynomial multiple of F
∏
i1<i2
(xi1 −
xi2)
2m generates a copy of sign− triv − sign in Qm(3,F3).
Proof. 1. We first prove the lemma for m = 0. Consider some T as above, and note that
(1− s12)T is contained in the sign representation, so by Proposition 2.11 we have (1− s12)T =
P (x1−x2)(x1−x3)(x2−x3) for some symmetric polynomial P . Then note that PE, T generate
two copies of triv − sign with the same sign subrepresentation, so they generate a copy of
(triv − sign⊕ triv − sign)/sign ∼= triv − sign⊕ triv.
So T is the sum of PE and a symmetric polynomial, as claimed.
Hilbert Series of S3-Quasi-Invariant Polynomials in Characteristics 2, 3 23
Now, consider general m. By the above we have that any T must be of the form T = PE+Q
for symmetric polynomials P , Q. Then since T is m-quasi-invariant, we have (1 − s12)T =
P (x1−x2)(x1−x3)(x2−x3) is divisible by (x1−x2)
2m+1. So P is divisible by (x1−x2)
2m, and
it must also be divisible by
∏
i1<i2
(xi1 − xi2)
2m since it is symmetric.
The converse is clear.
2. This proof is similar to part (1). For m = 0, any T1 must have that (1 + s23)T1 is in
a triv − sign representation, so (1 + s23)T1 = PE for some P ∈ F3[x1, x2, x3]
S3 . Then T1, PF
generate a copy of
(sign− triv − sign⊕ sign− triv − sign)/triv − sign ∼= sign− triv − sign⊕ sign,
which implies the result for m = 0. Then the extension to general m is the same as in part (1).
The converse is clear, as before. ■
Finally, we can prove Theorem 1.3 for p = 3.
Theorem 6.7. Qm(3,F3) is freely generated by 1, the two generators K, L from Theorem 5.3,
the two generators K1, L1 from Proposition 6.1, and the generator F
∏
i1<i2
(xi1 − xi2)
2m from
Lemma 6.6.
Proof. Let us first show that there are no other generators of Qm(3,F3). Assume for contra-
diction that there is some other generator T of Qm(3,F3). Then T cannot generate a copy of
triv by Proposition 2.11 and it cannot generate a copy of sign− triv or triv − sign− triv by
Theorem 5.3 and Proposition 6.1. If it generates a copy of sign, then by Proposition 2.11 it
must be
∏
i1<i2
(xi1 − xi2)
2m+1, but this polynomial is generated by K, L by Lemma 4.4, so it
cannot be a generator. If it generates a copy of triv − sign, then it is E
∏
i1<i2
(xi1 − xi2)
2m by
Lemma 6.6. But this is generated by K1, L1 by Remark 6.3. Finally, by Lemma 6.6 the only
generator that generates a copy of sign− triv − sign is F
∏
i1<i2
(xi1 − xi2)
2m.
Finally, we show there are no relations between the 6 generators. Note that this also implies
F
∏
i1<i2
(xi1 −xi2)
2m is a generator, since it is not generated by the other 5 generators. But this
is clear: we already know there are no relations between the first 5 generators by Corollary 6.2.
If there was a relation involving F
∏
i1<i2
(xi1 − xi2)
2m, then note that since every generator is
generated by the generators of Q0(3,F3) = F3[x1, x2, x3], this would induce a relation on those
generators. Moreover, the generators other than F
∏
i1<i2
(xi1 − xi2)
2m each generate a copy of
an indecomposable representation that is not sign− triv − sign, so they are each generated by
the first 5 generators of Q0(3,F3). Meanwhile, F
∏
i1<i2
(xi1 − xi2)
2m is the only generator not
generated by the first 5 generators, so the induced relation would be nontrivial. But there is no
such relation by Proposition 6.5. ■
Note that these generators imply a Hilbert series that agrees with Theorem 1.3 since K is
either a minimal degree Ren–Xu counterexample or has degree 3m + 1 if one does not exist.
In this way, the Hilbert series of Qm(3,F3) agrees with that of Qm(3,Q) if and only if there does
not exist a Ren–Xu counterexample. Ren–Xu counterexamples only exist when the conditions
of Conjecture 5.2 are satisfied, so Conjecture 5.2 is also implied.
Acknowledgements
We would like to thank the PRIMES program for making the project possible. We would also
like to thank the referees for their useful comments and suggestions. This material is based
upon work supported under the National Science Foundation Graduate Research Fellowship
under Grant No. 2141064.
24 F. Wang and E. Yee
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https://doi.org/10.1017/CBO9780511623592
https://doi.org/10.1215/S0012-7094-03-11824-4
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http://arxiv.org/abs/math.QA/0212313
https://doi.org/10.1007/BF02125702
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https://doi.org/10.1016/0001-8708(75)90151-6
https://doi.org/10.3842/SIGMA.2020.107
http://arxiv.org/abs/1907.13417
https://doi.org/10.1007/BF01207363
http://arxiv.org/abs/2201.06111
1 Introduction
2 Preliminaries
2.1 Preliminary definitions for p=2
2.1.1 Quasi-invariants at half-integers
2.2 Preliminary definitions for p=3
3 Quasi-invariants in characteristic 2
4 Properties of 3 variable quasi-invariants
5 Ren–Xu counterexamples
6 Representations of S_3 in Q_m(3,F)
References
|
| id | nasplib_isofts_kiev_ua-123456789-213519 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1815-0659 |
| language | English |
| last_indexed | 2026-03-16T09:29:07Z |
| publishDate | 2025 |
| publisher | Інститут математики НАН України |
| record_format | dspace |
| spelling | Wang, Frank Yee, Eric 2026-02-18T11:23:25Z 2025 Hilbert Series of ₃-Quasi-Invariant Polynomials in Characteristics 2, 3. Frank Wang and Eric Yee. SIGMA 21 (2025), 057, 24 pages 1815-0659 2020 Mathematics Subject Classification: 81R12; 13A50; 20C08 arXiv:2412.20673 https://nasplib.isofts.kiev.ua/handle/123456789/213519 https://doi.org/10.3842/SIGMA.2025.057 We compute the Hilbert series of the space of = 3 variable quasi-invariant polynomials in characteristic 2 and 3, capturing the dimension of the homogeneous components of the space, and explicitly describe the generators in the characteristic 2 case. In doing so, we extend the work of the first author in 2023 on quasi-invariant polynomials in characteristic > and prove that a sufficient condition found by Ren-Xu in 2020 on when the Hilbert series differs between characteristic 0 and is also necessary for = 3, = 2,3. This is the first description of quasi-invariant polynomials in the case where the space forms a modular representation over the symmetric group, bringing us closer to describing the quasi-invariant polynomials in all characteristics and numbers of variables. We would like to thank the PRIMES program for making the project possible. We would also like to thank the referees for their useful comments and suggestions. This material is based upon work supported under the National Science Foundation Graduate Research Fellowship under Grant No. 2141064. en Інститут математики НАН України Symmetry, Integrability and Geometry: Methods and Applications Hilbert Series of ₃-Quasi-Invariant Polynomials in Characteristics 2, 3 Article published earlier |
| spellingShingle | Hilbert Series of ₃-Quasi-Invariant Polynomials in Characteristics 2, 3 Wang, Frank Yee, Eric |
| title | Hilbert Series of ₃-Quasi-Invariant Polynomials in Characteristics 2, 3 |
| title_full | Hilbert Series of ₃-Quasi-Invariant Polynomials in Characteristics 2, 3 |
| title_fullStr | Hilbert Series of ₃-Quasi-Invariant Polynomials in Characteristics 2, 3 |
| title_full_unstemmed | Hilbert Series of ₃-Quasi-Invariant Polynomials in Characteristics 2, 3 |
| title_short | Hilbert Series of ₃-Quasi-Invariant Polynomials in Characteristics 2, 3 |
| title_sort | hilbert series of ₃-quasi-invariant polynomials in characteristics 2, 3 |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/213519 |
| work_keys_str_mv | AT wangfrank hilbertseriesof3quasiinvariantpolynomialsincharacteristics23 AT yeeeric hilbertseriesof3quasiinvariantpolynomialsincharacteristics23 |