Poincaré polynomials of moduli spaces of Higgs bundles and character varieties (no punctures)

Using our earlier results on polynomiality properties of plethystic logarithms of generating series of certain type, we show that Schiffmann’s formulas for various counts of Higgs bundles over finite fields can be reduced to much simpler formulas conjectured by Mozgovoy. In particular, our result implies the conjecture of Hausel and Rodriguez-Villegas on the Poincaré polynomials of twisted character varieties and the conjecture of Hausel and Thaddeus on independence of E-polynomials on the degree.


Introduction
Schiffmann [17] computed the number of absolutely indecomposable vector bundles of rank r and degree d over a compete curve C of genus g over F q . Suppose the eigenvalues of the Frobenius acting on the first cohomology of C are α 1 , . . . , α 2g with α i+g = qα −1 i for i = 1, . . . , g. This means that for all k ≥ 1 we have Schiffmann's result says that the number of absolutely indecomposable vector bundles of rank r and degree d on C is given by a Laurent polynomial independent of C A g,r,d (q, α 1 , . . . , α g ) ∈ Z[q, α ±1 1 , . . . , α ±1 g ], symmetric in α i and invariant under α i → qα −1 i . Suppose (r, d) = 1. Schiffmann showed that the number of stable Higgs bundles of rank r and degree d is given by q 1+(g−1)r 2 A g,r,d . Let C be a curve over C. The moduli space of stable Higgs bundles M g,r,d (C) is a quasiprojective variety, and by a theorem of Katz [9] its E-polynomial defined as is given by (x y) 1+(g−1)r 2 A g,r,d (x y, x, . . . , x). It is known [7] that this moduli space has pure cohomology. In particular, the Poincaré polynomial is the following specialization: P q,r,d (q) = E g,r,d (q 1 2 , q 1 2 ) = q 1+(g−1)r 2 A g,r,d (q, q 1 2 , . . . , q 1 2 ).
Since twisted character varieties are diffeomorphic to the moduli spaces of stable Higgs bundles (see [9]), their Poincaré polynomials coincide. The formula of Schiffmann was difficult to work with. In particular, it was not clear that his formula is equivalent to a much simpler formula conjectured earlier by Hausel and Rodriguez-Villegas for Poincaré polynomials [9], and then extended by Mozgovoy for the polynomials A g,r,d [15].
Here we study Schiffmann's formula from the combinatorial point of view and establish these conjectures. Our main result is: and let H g,r T r .
As a corollary, we obtain the G L-version of the conjecture of Hausel and Thaddeus (see Conjecture 3.2 in [7]): Davesh Maulik and Aaron Pixton announced an independent proof of Theorem 1.1. Their approach is to make rigorous the physical considerations of [1]. They claim that their work will settle the more general conjectures about Higgs bundles with parabolic structures.
In the next paper [13], we extend Schiffmann's [17] and Schiffmann-Mozgovoy's [16] methods to the parabolic case. Combined with the results of the present work, we obtain a proof of the conjecture of Hausel et al. [8] on the Poincaré polynomials of character varieties with punctures.
For a more precise technical version of the main result the reader is referred to Theorem 5.2. We warn the reader that variables q, t in Sect. 5 correspond to z, q in the rest of the paper and apologize for the inconvenience.
In Sect. 6 we discuss motivic classes of moduli stacks and connect our results to the work of Fedorov, Soibelman and Soibelman [4].

Arms and legs
We begin by stating an elementary formula which relates the generating series of arms and legs and the generating series of weights of partitions, proved in [2] (we follow notations from [3]). For a partition λ and any cell we denote by a λ ( ) and l λ ( ) the arm and leg lengths of with respect to λ. These numbers are non-negative when ∈ λ and negative otherwise. For partitions μ, ν define For any partition μ, let where c( ), r ( ) denote the column and row indices. For any f let f * be obtained from f by the substitution q → q −1 , t → t −1 .

Lemma 2.1 For any partitions μ, ν, we have
Proof We prove by induction on the largest part μ 1 of μ (defined to be 0 if For each fixed value of c( ), the numbers l ν ( ) go over the same range as the numbers r ( ). Thus we obtain This establishes the case μ 1 = 0. For the induction step, let μ be obtained from μ by removing the first column, i.e. μ = (μ 1 −1, μ 2 −1, . . .). Splitting the sum according to whether is in the first column, we obtain For any cell , we have In the last sum for each fixed value of c( ) the numbers l ν ( ) go over the same range as the numbers r ( ) − l(μ), so we have Putting things together, we have The first sum reduces to The second sum becomes This implies On the other hand we have Therefore if we denote the right hand side of (1) by E μ,ν , we obtain So E μ ,ν = E μ ,ν implies E μ,ν = E μ,ν and the induction step is established.
For a partition μ, we define z i (μ) to match z i in [17]: Our notations match after the substitution (q, z) → (t, q). Note the following generating series identity: What we will actually need is the following generating series: It can be obtained as follows. Note that the sum K μ contains only terms with non-positive powers of t. So we can start with and take only non-positive powers of t. Let L be the operator Note that we had to subtract l(μ) to cancel the contribution from the terms i = j appearing inK μ . We can calculateK μ using Lemma 2.1 and (2): from which it is clear that The conclusion is the following Proposition 2.2 For any partition μ we have Converting additive generating functions to multiplicative with an extra variable u we obtain

Corollary 2.3 For any partition
Note that the left hand side contains "non-symmetric" ratios z i (μ) z j (μ) for i < j, while the right hand side contains "simple terms" z i (μ) and 1, "correct armleg terms" q a μ ( )+1 t −l μ ( ) and "incorrect arm-leg terms" q a μ ( ) t −l μ ( ) . Our strategy is to trade incorrect arm-leg terms in Schiffmann's formula for nonsymmetric ratios, which will complement or cancel other non-symmetric ratios so that the result contains only correct arm-leg terms and something symmetric.

Schiffmann's terms
Let X be a smooth projective curve over F q of genus g with zeta function Let us order α i in such a way that α i+g = q α i holds. We will treat α 1 , α 2 , . . . , α g as formal variables and set α i+g = q α i . An alternative way to think of the parameters α i is to view them as the exponentials of the chern roots of the Hodge bundle on the moduli space of curves times q 1 2 . The expressions we will be writing will depend on q, z, α 1 , . . . , α g . There is a correspondence between these variables and the variables from [14] given as follows: The formula of Schiffmann (see [16,17]) involves a sum over partitions For each partition μ the corresponding coefficient is Here μ, μ = i μ 2 i where μ is the conjugate partition of μ. We will proceed defining J μ and H μ and taking them apart in the process. We have =0 .
The notation (−) =0 means we omit the corresponding factor if it happens to be zero. This naturally splits as follows: Applying Corollary 2.3 we obtain where z i (μ) = q −l(μ)+i z μ i coincides with Schiffmann's z n−i+1 . Denote the last four products in the above right hand side by A, B, C, D. Note that l( )+ (l( )+1) = μ, μ , so q μ,μ together with the first two products produce where N μ is the arm-leg product as in [14]: So we have We proceed by defining H μ . Let 1 .
Let L(z 1 , . . . , z l(μ) ) be the rational function (note that we reversed the order of z i ) Note thatζ appears in the numerator as many times as in the denominator, so it can be multiplied by a constant without changing L. So we replaceζ with something more resembling the other products we have seen: .
H μ is defined as the iterated residue (remember that our ordering of z i is the opposite of Schiffman's) Note that the only poles L can have at z i = z i (μ) are coming from factors of the form 1 − q z i z i+1 for i such that μ i = μ i+1 . Each such factor can appear at most once in the denominator of L. We have Thus we will obtain the same result if we multiply L by the product of these factors and then evaluate at z i = z i (μ). Note that C has precisely the same factors removed. Therefore we have Putting in A as well we obtain a nice expression: We see that the product is symmetric in z i , so it can be moved inside the summation. Since B and D are symmetric, they can also be moved inside the summation. After some cancellations we arrive at the following. Define for any n Then Summarizing we obtain Proposition 3.1 For any partition μ the term μ is given by (1) , (4), (5).
Let us calculate f in a few cases. It is convenient to set We have Note that the denominator of this expression is P(z 1 )P(z 2 )P(q z 1 z 2 )P(q z 2 z 1 ) if no cancellations happen. If z 2 = qz 1 , the denominator reduces to P(z 1 )P(z 2 )P(q 2 ), so it has only 3 P-factors instead of 4.

Bounding denominators
First we analyse denominators of f defined in (5). For generic values of z i , the denominator of f can be as bad as the full product . Pick numbers r 1 , r 2 , . . . such that m r m = n. Split z 1 , z 2 , . . . , z n into a union of subsequences of sizes r 1 , r 2 , …. Let j m = 1 + i<m r i . For each m the m-th subsequence looks like z j m , z j m +1 , . . . , z j m +r m −1 . Suppose each subsequence forms a geometric progression with quotient q: Then f can be viewed as a function of variables z j m . The denominator can be bounded as follows The following expression is a Laurent polynomial: First write the definition of f as follows: Note that 1− z i z j does not contribute to the denominator because of symmetrization. Next note that if j = i + 1 and j, i belong to the same subsequence, then 1 − q z i z j = 0. So all summands with σ (i) > σ ( j) + 1 vanish. So it is enough to sum only over those σ which satisfy the condition σ (i + 1) ≥ σ (i) − 1 whenever i, i + 1 are in the same subsequence. (6) So in each sequence σ ( j m ), . . . , σ ( j m + r m − 1) if there is a drop, the size of the drop is 1. Now for each such σ we look at the product It is enough to show that for each value of i and each σ the following expressions are Laurent polynomials: Further, let us split the product over all j > i into products over our subsequences. We only need to consider values of m such that j m > i (when j and i are in different subsequences) or j m ≤ i < j m + r m (when they are in the same subsequence). So it is enough to show that the following products are Laurent polynomials: P q r m z jm P q r m z jm Observe that because of the condition (6) in each of the cases (7)-(10) the values of k from a contiguous set k min , . . . , k max (if non-empty). So the arguments to P from a geometric progression with ratio q or q −1 . Hence the product collapses and the only remaining denominator is P(q z i z jm +k min ) in cases (7) and (9), and P(q z i z jm +kmax ) in cases (8) and (10). Further analysis leads to k min = 0 in (7), k max = r m − 1 in (8), k min = i − j m + 1 in (9) and k max = r m − 1 in (10).

Example 4.1
In the situation of n = 1 we obtain that f P(z 1 )P(q) is a Laurent polynomial. For n = 2 and z 2 = qz 1 we obtain f P(z 1 )P(z 2 )P(q 2 )P(q) is a Laurent polynomial. Comparing with Example 3.1 one can notice that our denominator bound is not optimal.
For the case when z i = z i (μ) = z μ i q i−l(μ) for a partition μ we obtain Proposition 4.2 The following product is a Laurent polynomial for any partition μ: Proof Recall that f μ is a shorthand for f (z 1 (μ), . . . , z l(μ) (μ)) where z i (μ) = z μ i q i−l(μ) . In view of Proposition 4.1 it is enough to show that for each i the product  (12) divides the arm-leg product in (11) for cells ∈ μ occurring in the row i. Note that our subsequences of geometric progressions in z i simply correspond to repeated parts of μ. Let be the cell in row i and column μ j m ( j m + r m > i).
Let be the cell in row i and column μ j m + 1 ( j m > i).
For the cell in column 1 we have a( ) Thus the factors of (12) form a sub-multiset of the factors of the arm-leg product (11), and the claim follows.

Corollary 4.3
For any partition μ, the product N μ (1) μ is a Laurent polynomial.
Proof We have

Interpolation
We remind the reader that f is not a function in fixed number of variables, but a sequence of functions: a function in n variables for each n. A nice property of f is that the substitution z 1 = 1 into the function in n + 1 variables leads to essentially the same function in n variables: Proof Note that because of the product n i=2 (1 − z i ) in the definition of f (1, z 1 , . . . , z n ), only the terms with σ (1) = 1 survive. So we can reduce the summation over S n+1 to a summation over S n . After cancellation of i (1−z i ) we obtain which coincides with f (qz 1 , . . . , qz n ).

Corollary 4.5 Let μ be a partition and let n ≥ l(μ). Define z n,i
Thus, instead of having a separate function for each value of l(μ) we can use the same function of n arguments if n is big enough.

Polynomiality and the main result
In this section we return to variables q, t which correspond to Schiffmann's variables z, q respectively. First we prove the following statement. The proof is straightforward using methods of [14], but tedious. Let R be a lambda ring containing Q(t)[q ±1 ]. We denote by R * the tensor product R⊗ Q(t)[q ±1 ] Q(q, t) and assume R ⊂ R * .
We will use the modified Macdonald polynomialsH μ [X ; q, t], see [6,14]. (2) u 2 + · · · be a power series whose coefficients f (i) are regular functions in the above sense. Let be a series with c μ ∈ R * , c ∅ = 1 such that all coefficients of Consider the expansion Then all coefficients of H f,i [X ] for i ≥ 1 are in (q − 1)R. In other words, the specialization q = 1 of H f [X, u] is independent of u.
Proof Let S = −(q − 1)(t − 1). Recall the notation S X F[X, X * ] (see [14]). This is a linear operation such that Recall that modified Macdonald polynomials are orthogonal with respect to (−, −) S X . In this proof we call an expression F admissible if (q − 1) Log F has all of its coefficients in R. It was proved in [14] that S X preserves admissibility. By the assumption [X ] is admissible. We will "construct" H f [X, u] from admissible building parts.
Let R[Z , Z * ] be the free lambda ring over R with two generators Z and Z * . Fix a large integer N . For each i ≥ 1 letf (i) ∈ R[Z , Z * ] be any element such thatf . . . , z N ).
One way to construct such an element is to find m ≥ 0 such that (z 1 · · · z N ) m f (i) N (z 1 , . . . , z N ) = p(z 1 , . . . , z n ) does not contain negative powers of z i , then lift p to a symmetric functionp ∈ R[Z ] and set Then setf We can take plethystic logarithm: For any partition μ satisfying l(μ) ≤ N by regularity of f we have Thus we can obtain f μ fromf by specializing at and similarly for Z * . Hence there exists a series such that for any partition μ with l(μ) ≤ N we have This g is obtained from g by the lambda ring homomorphism which sends Z to t 1−N 1−t Z + t −N −1 t −1 −1 and similarly for Z * . Specialization can be replaced by scalar product using the identity and we obtain N ). (13) Let us show that the sum is admissible. Begin with the series which is admissible by the main theorem of [14]. Recall the nabla operator ∇, the shift operator τ and the multiplication by Exp X S operator τ * , and Tesler's identity All of the operators involved preserve admissibility (Corollary 6.3 from [14]).

By Corollary 4.3 we have
So we can pass to the ring of Laurent series in α −1 1 , . . . , α −1 g and it is enough to prove the corresponding statements (i) and (ii) for the coefficients in front of monomials of the form g i=1 α m i i . Let us apply Lemma 5.1 for the ring and the regular function f (u) obtained from f (see (5) and Proposition 4.4) by setting uα −1 i in place of α −1 i , so that f (u) becomes a power series in u with coefficients in R.
To be able to apply Lemma 5.1, we need to show that the constant coefficient of f (u) is 1, in other words we need to check that We do this by induction. Denote the left hand side by L n . Notice that L n is a polynomial. Suppose we know that L n−1 = 1. Then by Proposition 4.4 we know that L n − 1 is divisible by z 1 − 1. Since it is a symmetric polynomial, it must be divisible by n i=1 (z i − 1). On the other hand, the degree of L n is at most n − 1, so necessarily L n − 1 = 0.
After applying Lemma 5.1 we can set X = T , where T is the variable from the statement of the Theorem. In particular, T is assumed to satisfy p k [T ] = T k and we can use the identityH μ [T ; q, t] = T |μ| . Let Lemma 5.1 says that On the other hand, the coefficient in front of any monomial in α 1 , . . . , α g , T has bounded degree in u, wo we can set u = 1 and obtain a statement about Laurent series in α −1 i : Finally we remember that H Sch (T, q, t) = H Sch (T, q, t, 1) and remember that the coefficients of H Sch (T, q, t) are Laurent polynomials in α i to obtain Remark 5. 1 We have been using the substitution (q, z) → (t, q) to relate Schiffmann's variables to the HLV variables. Note that the Hausel-Villegas functions and H are symmetric in q, t because replacing μ by the conjugate partition interchanges arms and legs. So we could have worked with the substitution (q, z) → (q, t), but then we would need to replace a partition by the conjugate partition somewhere in the argument. Theorem 1.1 is a direct corollary of Theorem 5.2 and [17].

Motivic interpretation
To answer some of the questions asked by Yan Soibelman, Davesh Maulik and an anonymous referee, we sketch an approach connecting the present work to the work of Fedorov et al. [4]. There they explain how to recast Schiffmann's formula to compute motivic classes of moduli spaces in the Grothendieck group of stacks over a field of characteristic zero. It is natural to ask if results of the present paper can be used to improve our understanding of motivic classes, and if Hausel-Rodriguez-Villegas formula can be recast in a similar way. The Grothendieck group of stacks is denoted by Mot and consists of formal linear combinations of Artin stacks of finite type modulo cut-and-paste relations. For a stack X the corresponding element of Mot is denoted by [X ] and is called the motivic class of X . The dimensional completion of Mot is denoted by Mot. Fix a curve C and assume it has a divisor of degree 1. The Grothendieck ring of stacks Mot contains natural elements: the class of A 1 denoted by L, and the class of the curve [C]. There is a natural operation of symmetric power which for a stack X is defined by This operation satisfies the condition which means that Mot is a pre-λ-ring. S n extends to Mot and makes it into a preλ-ring too. It is not clear 2 whether Mot or Mot are λ-rings, which would mean that S n (x y) and S n (S m (x))) can be expressed as certain prescribed polynomials in x, S 2 (x), S 3 (x), . . . and y, S 2 (y), S 3 (y), . . .. Nevertheless, Totaro's lemma [5] tells us that S n (LX ) = L n S n (X ) holds for any X and n. The formal sum of the form is called plethystic exponential (a.k.a. motivic zeta function). The inverse operation is called the plethystic logarithm Log.
Denote by R g the ring of polynomials in α ±1 1 , …, α ±1 g and q ±1 invariant under permutations of α i and substitutions of the form α i → qα −1 i . As a ring R g is the polynomial ring in the first g elementary symmetric functions in The ring R g is a λ-ring. R g has a filtration by the total degree in q, α 1 , . . . , α g . Denote the corresponding completion by R g . An infinite sum converges in the completion if degrees of the summands tend to −∞. The homomorphism ev C extends to the completions. Consider Schiffmann's generating function (see Sect. 3 for details) In [4] it is explained how the definition of Sch can be recast to produce an element Mot ∈ Mot[[T, z]]. Analysing their construction it is easy to see that in fact we have where we extend ev C to formal power series in z and T . To obtain the motivic class of the moduli stack of semistable Higgs bundles of rank r and degree d one needs to first compute plethystic logarithm: (16) In particular, in the universal λ-ring quotient of Mot we conclude that the above formula holds.
On the other hand, if we are only interested in the case of coprime r, d, then we do not need the λ-ring property because in all the expansions of Exp and Log above we use only the first term. So we directly obtain So in the above two situations no information is lost when passing from motivic invariants to functions in q, α 1 , . . . , α g . Next we would like to connect the motivic formula to the Hausel-Rodriguez-Villegas generating function g . Unfortunately, the function is not invariant under the map α i → qα −1 i , so the function is not in R g . Following Mozgovoy's approach [15], we apply change of variables q → qz −1 : where h( ) = a( ) + l( ) + 1 and μ , μ = i μ 2 i . In this way we obtain We apply ev C to define the motivic Mozgovoy function: where ζ C (z) = Exp[Cz] is the motivic zeta function of C. Theorem 1.1 implies Using the evaluations H g,r (q, 1, . . .) = H g,r (1, q, . . .) = A g,r (q, . . .), we obtain z/q H g,r (q/z, z, α 1 , . . . , α g ) A g,r (q, α 1 , . . . , α g ) So the coefficients do not stabilize like they do for Log Sch , but nevertheless tend to A g,r q−1 in R g as d goes to ∞. Therefore in the product expansion of the ratio Sch (T, z, q, α 1 , . . . , α g ) Moz (T, z, q, α 1  Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder.