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

Abstract

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.

1 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 $\mathbb {F}_q$. Suppose the eigenvalues of the Frobenius acting on the first cohomology of C are $\alpha _1,\ldots ,\alpha _{2g}$ with $\alpha _{i+g}=q\alpha _i^{-1}$ for $i=1,\ldots ,g$. This means that for all $k\ge 1$ we have

$$\begin{aligned} \# C(\mathbb {F}_{q^k}) = 1+q^k-\sum _{i=1}^{2g} \alpha _i^k. \end{aligned}$$

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

$$\begin{aligned} A_{g,r,d}(q,\alpha _1,\ldots ,\alpha _g) \in \mathbb {Z}[q,\alpha _1^{\pm 1}, \ldots ,\alpha _g^{\pm 1}], \end{aligned}$$

symmetric in $\alpha _i$ and invariant under $\alpha _i\rightarrow q\alpha _i^{-1}$.

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 $\mathbb {C}$. The moduli space of stable Higgs bundles $\mathcal {M}_{g,r,d}(C)$ is a quasi-projective variety, and by a theorem of Katz [9] its E-polynomial defined as

$$\begin{aligned} E_{g,r,d}(x,y) = \sum _{i,j,k} (-1)^k x^i y^j \dim \mathrm {Gr}_F^i \mathrm {Gr}^W_{i+j} H^k_c(\mathcal {M}_{g,r,d}(C), \mathbb {C}) \end{aligned}$$

is given by $(xy)^{1+(g-1)r^2} A_{g,r,d}(xy,x,\ldots ,x)$. It is known [7] that this moduli space has pure cohomology. In particular, the Poincaré polynomial

$$\begin{aligned} P_{q,r,d}(q) = \sum _i (-1)^i q^{\frac{i}{2}} \dim H^i_c(\mathcal {M}_{g,r,d}(C)) \end{aligned}$$

is the following specialization:

$$\begin{aligned} P_{q,r,d}(q) = E_{g,r,d}(q^{\frac{1}{2}},q^{\frac{1}{2}}) = q^{1+(g-1)r^2} A_{g,r,d}(q,q^{\frac{1}{2}},\ldots ,q^{\frac{1}{2}}). \end{aligned}$$

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:

Theorem 1.1

Let $g\ge 1$. Let $\Omega _g$ denote the series

$$\begin{aligned} \Omega _g = \sum _{\mu \in \mathcal {P}} T^{|\mu |} \prod _{\square \in \mu } \frac{\prod _{i=1}^g (z^{a(\square )+1} - \alpha _i q^{l(\square )}) (z^{a(\square )} - \alpha _i^{-1} q^{l(\square )+1})}{(z^{a(\square )+1} - q^{l(\square )}) (z^{a(\square )} - q^{l(\square )+1})}, \end{aligned}$$

and let

$$\begin{aligned} H_g = -(1-q)(1-z) {{\,\mathrm{Log}\,}}\Omega _g,\qquad H_g = \sum _{r=1}^\infty {H_{g,r} T^r}. \end{aligned}$$

Then for all $r\ge 1$, $H_{g,r}$ is a Laurent polynomial in q, z and $\alpha _1,\ldots ,\alpha _g$, and for all d, $A_{g,r,d}$ is obtained by setting $z=1$ in $H_{g,r}$:

$$\begin{aligned} A_{g,r,d}(q,\alpha _1,\ldots ,\alpha _g) = H_{g,r}(q,1,\alpha _1,\ldots ,\alpha _g). \end{aligned}$$

As a corollary, we obtain the GL-version of the conjecture of Hausel and Thaddeus (see Conjecture 3.2 in [7]):

Corollary 1.2

For $r,d,d'$ satisfying $(r,d)=(r,d')=1$, the E-polynomials of $\mathcal {M}(g,r,d)$ and $\mathcal {M}(g,r,d')$ coincide.

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].

2 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 $\lambda$ and any cell $\square$ we denote by $a_\lambda (\square )$ and $l_\lambda (\square )$ the arm and leg lengths of $\square$ with respect to $\lambda$. These numbers are non-negative when $\square \in \lambda$ and negative otherwise. For partitions $\mu , \nu$ define

$$\begin{aligned} E_{\mu ,\nu } = \sum _{\square \in \mu } q^{-a_\nu (\square )} t^{l_\mu (\square )+1} + \sum _{\square \in \nu } q^{a_\mu (\square )+1} t^{-l_\nu (\square )}. \end{aligned}$$

For any partition $\mu$, let

$$\begin{aligned} B_\mu = \sum _{\square \in \mu } q^{c(\square )} t^{r(\square )}, \end{aligned}$$

where $c(\square ), r(\square )$ denote the column and row indices. For any f let $f^*$ be obtained from f by the substitution $q\rightarrow q^{-1}$, $t\rightarrow t^{-1}$.

Lemma 2.1

For any partitions $\mu , \nu$, we have

$$\begin{aligned} E_{\mu , \nu } = qt B_\mu + B_{\nu }^* - (q-1)(t-1) B_\mu B_\nu ^*. \end{aligned}$$

(1)

Proof

We prove by induction on the largest part $\mu _1$ of $\mu$ (defined to be 0 if $\mu =\varnothing$). If $\mu =\varnothing$, we have $a_\mu (\square )=-1-c(\square )$. Therefore

$$\begin{aligned} E_{\varnothing ,\nu } = \sum _{\square \in \mu } q^{-c(\square )}t^{-l_\nu (\square )}. \end{aligned}$$

For each fixed value of $c(\square )$, the numbers $l_\nu (\square )$ go over the same range as the numbers $r(\square )$. Thus we obtain

$$\begin{aligned} E_{\mu ,\varnothing } = B_\nu ^*. \end{aligned}$$

This establishes the case $\mu _1=0$.

For the induction step, let $\mu '$ be obtained from $\mu$ by removing the first column, i.e. $\mu '=(\mu _1-1,\mu _2-1,\ldots )$. Splitting the sum according to whether $\square$ is in the first column, we obtain

$$\begin{aligned} \sum _{\square \in \mu } q^{-a_\nu (\square )} t^{l_\mu (\square )+1} = q \sum _{\square \in \mu '} q^{-a_\nu (\square )} t^{l_\mu (\square )+1} + \sum _{i=1}^{l(\mu )} q^{1-\nu _i} t^{l(\mu )-i+1}. \end{aligned}$$

For any cell $\square$, we have

$$\begin{aligned} a_\mu (\square ) = {\left\{ \begin{array}{ll} a_{\mu '}(\square )+1 &{} \text {if } r(\square )<l(\mu ),\\ -1-c(\square ) &{} \text {otherwise.}\end{array}\right. } \end{aligned}$$

This implies

$$\begin{aligned} \sum _{\square \in \nu } q^{a_\mu (\square )+1} t^{-l_\nu (\square )}= & {} q \sum _{\square \in \nu } q^{a_{\mu }'(\square )+1} t^{-l_\nu (\square )} \\&+ (1-q)\sum _{\square \in \nu \,:\, r(\square )\ge l(\mu )} q^{-c(\square )} t^{-l_\nu (\square )}. \end{aligned}$$

In the last sum for each fixed value of $c(\square )$ the numbers $l_\nu (\square )$ go over the same range as the numbers $r(\square )-l(\mu )$, so we have

$$\begin{aligned} \sum _{\square \in \nu \,:\, r(\square )\ge l(\mu )} q^{-c(\square )} t^{-l_\nu (\square )}= & {} \sum _{\square \in \nu \,:\, r(\square )\ge l(\mu )} q^{-c(\square )} t^{l(\mu )-r(\square )} \\= & {} \sum _{i=l(\mu )+1}^\infty t^{l(\mu )-i+1} \frac{1-q^{-\nu _i}}{1-q^{-1}}. \end{aligned}$$

Putting things together, we have

$$\begin{aligned} E_{\mu ,\nu } - q E_{\mu ',\nu } = \sum _{i=1}^\infty t^{l(\mu )-i+1} (q^{1-\nu _i}-q) + q \sum _{i=1}^{l(\mu )} t^{l(\mu )-i+1}. \end{aligned}$$

The first sum reduces to

$$\begin{aligned} \sum _{i=1}^\infty t^{l(\mu )-i+1} (q^{1-\nu _i}-q) = q t^{l(\mu )} (q^{-1}-1) B_\nu ^* = (1-q) t^{l(\mu )} B_\nu ^*. \end{aligned}$$

The second sum becomes

$$\begin{aligned} q \sum _{i=1}^{l(\mu )} t^{l(\mu )-i+1} = qt \frac{t^{l(\mu )}-1}{t-1}. \end{aligned}$$

This implies

$$\begin{aligned} E_{\mu ,\nu } - q E_{\mu ',\nu } = (1-q) t^{l(\mu )} B_\nu ^* + qt \frac{t^{l(\mu )}-1}{t-1} \end{aligned}$$

On the other hand we have

$$\begin{aligned} B_\mu - q B_{\mu '} = \sum _{i=1}^{l(\mu )} t^{i-1} = \frac{t^{l(\mu )}-1}{t-1}. \end{aligned}$$

Therefore if we denote the right hand side of (1) by $E'_{\mu ,\nu }$, we obtain

$$\begin{aligned} \begin{aligned} E'_{\mu ,\nu } - q E'_{\mu ',\nu }=&{} qt \frac{t^{l(\mu )}-1}{t-1} + (1-q) B_\nu ^* - (q-1)(t-1)B_{\nu }^* \frac{t^{l(\mu )}-1}{t-1}\\ =&{} qt \frac{t^{l(\mu )}-1}{t-1} + (1-q)t^{l(\mu )} B_{\nu }^*. \end{aligned} \end{aligned}$$

So $E_{\mu ',\nu }=E'_{\mu ',\nu }$ implies $E_{\mu ,\nu }=E'_{\mu ,\nu }$ and the induction step is established. $\square$

For a partition $\mu$, we define $z_i(\mu )$ to match $z_i$ in [17]:

$$\begin{aligned} z_i(\mu ) = t^{-l(\mu )+i} q^{\mu _i}\qquad (i=1,2,\ldots ,l(\mu )). \end{aligned}$$

Our notations match after the substitution $(q,z)\rightarrow (t,q)$. Note the following generating series identity:

$$\begin{aligned} \sum _{i=1}^{l(\mu )} z_i(\mu ) = t^{-l(\mu )+1} \left( (q-1)B_\mu + \frac{t^{l(\mu )}-1}{t-1}\right) . \end{aligned}$$

(2)

What we will actually need is the following generating series:

$$\begin{aligned} K_\mu := (1-t) \sum _{i<j} \frac{z_i(\mu )}{z_j(\mu )}. \end{aligned}$$

It can be obtained as follows. Note that the sum $K_\mu$ contains only terms with non-positive powers of t. So we can start with

$$\begin{aligned} \tilde{K}_{\mu }:=(1-t) \sum _{i=1}^{l(\mu )} z_i(\mu ) \sum _{i=1}^{l(\mu )} z_i(\mu )^{-1}=(1-t) \sum _{i,j=1}^{l(\mu )} \frac{z_i(\mu )}{z_j(\mu )}, \end{aligned}$$

and take only non-positive powers of t. Let L be the operator

$$\begin{aligned} L(t^i q^j) = {\left\{ \begin{array}{ll}t^i q^j &{} (i\le 0)\\ 0 &{} (i>0)\end{array}\right. }. \end{aligned}$$

Then

$$\begin{aligned} K_\mu = L(\tilde{K}_{\mu }) - l(\mu ). \end{aligned}$$

Note that we had to subtract $l(\mu )$ to cancel the contribution from the terms $i=j$ appearing in $\tilde{K}_\mu$. We can calculate $\tilde{K}_{\mu }$ using Lemma 2.1 and (2):

$$\begin{aligned} \tilde{K}_{\mu }= & {} (1-t) \left( (q-1)B_\mu + \frac{t^{l(\mu )}-1}{t-1}\right) \left( (q^{-1}-1)B_\mu ^* + \frac{t^{-l(\mu )}-1}{t^{-1}-1}\right) \\= & {} (q^{-1}-1)E_{\mu ,\mu } - t^{l(\mu )} (q^{-1}-1) B_\mu ^* + t^{1-l(\mu )}(q-1) B_\mu \\&- (t^{l(\mu )}-1)\sum _{i=0}^{l(\mu )-1} t^{-i}, \end{aligned}$$

from which it is clear that

$$\begin{aligned} L(\tilde{K}_{\mu })= & {} (q^{-1}-1) L(E_{\mu ,\mu }) + t^{1-l(\mu )} (q-1) B_\mu + \sum _{i=0}^{l(\mu )-1} t^{-i}.\\= & {} (q^{-1}-1) \sum _{\square \in \mu } q^{a_\mu (\square )+1} t^{-l_\mu (\square )} + \sum _{i=1}^{l(\mu )} z_i(\mu ). \end{aligned}$$

The conclusion is the following

Proposition 2.2

For any partition $\mu$ we have

$$\begin{aligned} (1-t) \sum _{i<j} \frac{z_i(\mu )}{z_j(\mu )} = (q^{-1}-1) \sum _{\square \in \mu } q^{a_\mu (\square )+1} t^{-l_\mu (\square )} + \sum _{i=1}^{l(\mu )} (z_i(\mu )-1). \end{aligned}$$

Converting additive generating functions to multiplicative with an extra variable u we obtain

Corollary 2.3

For any partition $\mu$ we have

$$\begin{aligned} \prod _{i<j} \frac{1-t u \frac{z_i(\mu )}{z_j(\mu )}}{1-u \frac{z_i(\mu )}{z_j(\mu )}} = \prod _{\square \in \mu } \frac{1-u q^{a_\mu (\square )+1} t^{-l_\mu (\square )}}{1-u q^{a_\mu (\square )} t^{-l_\mu (\square )}} \prod _{i=1}^{l(\mu )} \frac{1-u}{1-u z_i(\mu )}. \end{aligned}$$

Note that the left hand side contains “non-symmetric” ratios $\frac{z_i(\mu )}{z_j(\mu )}$ for $i<j$, while the right hand side contains “simple terms” $z_i(\mu )$ and 1, “correct arm-leg terms” $q^{a_\mu (\square )+1} t^{-l_\mu (\square )}$ and “incorrect arm-leg terms” $q^{a_\mu (\square )} t^{-l_\mu (\square )}$. Our strategy is to trade incorrect arm-leg terms in Schiffmann’s formula for non-symmetric ratios, which will complement or cancel other non-symmetric ratios so that the result contains only correct arm-leg terms and something symmetric.

3 Schiffmann’s terms

Let X be a smooth projective curve over $\mathbb {F}_q$ of genus g with zeta function

$$\begin{aligned} \zeta _X(x) = \frac{\prod _{i=1}^{2g} (1-\alpha _i x)}{(1-x)(1-qx)}. \end{aligned}$$

Let us order $\alpha _i$ in such a way that $\alpha _{i+g}=\frac{q}{\alpha _i}$ holds. We will treat $\alpha _1, \alpha _2, \ldots , \alpha _g$ as formal variables and set $\alpha _{i+g}=\frac{q}{\alpha _i}$. An alternative way to think of the parameters $\alpha _i$ is to view them as the exponentials of the chern roots of the Hodge bundle on the moduli space of curves times $q^{\frac{1}{2}}$. The expressions we will be writing will depend on $q,z,\alpha _1,\ldots ,\alpha _g$. There is a correspondence between these variables and the variables from [14] given as follows:

$$\begin{aligned} q,z,\alpha _1,\ldots ,\alpha _g \rightarrow t,q,u_1^{-1},\ldots ,u_g^{-1}. \end{aligned}$$

(3)

The formula of Schiffmann (see [16, 17]) involves a sum over partitions

$$\begin{aligned} \Omega ^\mathrm {Sch}:= \sum _{\mu } \Omega _\mu T^{|\mu |}. \end{aligned}$$

For each partition $\mu$ the corresponding coefficient is

$$\begin{aligned} \Omega _\mu := q^{(g-1)\langle \mu ,\mu \rangle } J_\mu H_\mu . \end{aligned}$$

Here $\langle \mu ,\mu \rangle =\sum _i \mu _i'^2$ where $\mu '$ is the conjugate partition of $\mu$. We will proceed defining $J_\mu$ and $H_\mu$ and taking them apart in the process. We have

$$\begin{aligned} J_\mu = \prod _{\square \in \mu } \frac{\prod _{i=1}^{2g} (1-\alpha _i q^{-1-l(\square )} z^{a(\square )})}{(1- q^{-1-l(\square )} z^{a(\square )})(1-q^{-l(\square )} z^{a(\square )})_{\ne 0}}. \end{aligned}$$

The notation $(-)_{\ne 0}$ means we omit the corresponding factor if it happens to be zero. This naturally splits as follows:

$$\begin{aligned} J_\mu = \prod _{\square \in \mu } \frac{\prod _{i=1}^{g} (1-\alpha _i q^{-1-l(\square )} z^{a(\square )})}{1- q^{-1-l(\square )} z^{a(\square )}} \prod _{\square \in \mu } \frac{\prod _{i=1}^{g} (1-\alpha _i^{-1} q^{-l(\square )} z^{a(\square )})}{(1- q^{-l(\square )} z^{a(\square )})_{\ne 0}}. \end{aligned}$$

Applying Corollary 2.3 we obtain

$$\begin{aligned} J_\mu= & {} \prod _{\square \in \mu } \frac{\prod _{i=1}^{g} (1-\alpha _i q^{-1-l(\square )} z^{a(\square )})}{1- q^{-1-l(\square )} z^{a(\square )}} \times \prod _{\square \in \mu } \frac{\prod _{i=1}^{g} (1-\alpha _i^{-1} q^{-l(\square )} z^{a(\square )+1})}{1- q^{-l(\square )} z^{a(\square )+1}}\\&\times \prod _{i<j} \prod _{k=1}^g \frac{1-\alpha _k^{-1} \frac{z_i(\mu )}{z_j(\mu )}}{1-q \alpha _k^{-1} \frac{z_i(\mu )}{z_j(\mu )}} \times \prod _{i=1}^{l(\mu )} \prod _{k=1}^g \frac{1-\alpha _k^{-1}}{1-\alpha _k^{-1}z_i(\mu )}\\&\times \prod _{i<j} \frac{\left( 1-q\frac{z_i(\mu )}{z_j(\mu )}\right) _{\ne 0}}{1-\frac{z_i(\mu )}{z_j(\mu )}} \times \prod _{i=1}^{l(\mu )} (1-z_i(\mu )), \end{aligned}$$

where $z_i(\mu )=q^{-l(\mu )+i} z^{\mu _i}$ 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 $\sum l(\square ) + \sum (l(\square )+1) = \langle \mu ,\mu \rangle$, so $q^{\langle \mu ,\mu \rangle }$ together with the first two products produce

$$\begin{aligned} \frac{\prod _{i=1}^g N_\mu (\alpha _i^{-1})}{N_\mu (1)}, \end{aligned}$$

where $N_\mu$ is the arm-leg product as in [14]:

$$\begin{aligned} N_\mu (u)=\prod _{\square \in \mu }(z^{a(\square )}-u q^{1+l(\square )})(z^{a(\square )+1}-u^{-1} q^{l(\square )}). \end{aligned}$$

(4)

So we have

$$\begin{aligned} q^{\langle \mu ,\mu \rangle } J_\mu = \frac{\prod _{i=1}^g N_\mu (\alpha _i^{-1})}{N_\mu (1)} ABCD, \end{aligned}$$

where

$$\begin{aligned} A= & {} \prod _{i<j} \prod _{k=1}^g \frac{1-\alpha _k^{-1} \frac{z_i(\mu )}{z_j(\mu )}}{1-q \alpha _k^{-1} \frac{z_i(\mu )}{z_j(\mu )}},\qquad B=\prod _{i=1}^{l(\mu )} \prod _{k=1}^g \frac{1-\alpha _k^{-1}}{1-\alpha _k^{-1}z_i(\mu )}\\ C= & {} \prod _{i<j} \frac{\left( 1-q\frac{z_i(\mu )}{z_j(\mu )}\right) _{\ne 0}}{1-\frac{z_i(\mu )}{z_j(\mu )}} \qquad D = \prod _{i=1}^{l(\mu )} (1-z_i(\mu )). \end{aligned}$$

We proceed by defining $H_\mu$. Let¹

$$\begin{aligned} \tilde{\zeta }'(x) = x^{1-g} \zeta (x) = \frac{\prod _{k=1}^{g} x^{-1} (1-\alpha _k x)(1-q \alpha _k^{-1} x)}{x^{-1}(1-x)(1-qx)}. \end{aligned}$$

Let $L(z_1,\ldots ,z_{l(\mu )})$ be the rational function (note that we reversed the order of $z_i$)

$$\begin{aligned}&L(z_1,\ldots ,z_{l(\mu )}) \\&\quad = \frac{1}{\prod _{i>j} \tilde{\zeta }'\left( \frac{z_i}{z_j}\right) } \sum _{\sigma \in S_{l(\mu )}} \sigma \left\{ \prod _{i<j} \tilde{\zeta }'\left( \frac{z_i}{z_j}\right) \frac{1}{\prod _{i<l(\mu )} \left( 1-q \frac{z_{i+1}}{z_{i}}\right) } \frac{1}{1-z_1}\right\} . \end{aligned}$$

Note that $\tilde{\zeta }'$ appears in the numerator as many times as in the denominator, so it can be multiplied by a constant without changing L. So we replace $\tilde{\zeta }'$ with something more resembling the other products we have seen:

$$\begin{aligned} \tilde{\zeta }(x) = \frac{\prod _{k=1}^{g} (1-\alpha _k^{-1} x^{-1})(1-q \alpha _k^{-1} x)}{(1-x^{-1})(1-qx)}. \end{aligned}$$

$H_\mu$ is defined as the iterated residue (remember that our ordering of $z_i$ is the opposite of Schiffman’s)

$$\begin{aligned} H_\mu = {{\,\mathrm{res}\,}}_{z_i=z_i(\mu )} L(z_1,\ldots ,z_{l(\mu )}) \prod _{i\,:\, \mu _i=\mu _{i+1}} \frac{d z_{i+1}}{z_{i+1}}. \end{aligned}$$

Note that the only poles L can have at $z_i=z_i(\mu )$ are coming from factors of the form $1-q \frac{z_i}{z_{i+1}}$ for i such that $\mu _i=\mu _{i+1}$. Each such factor can appear at most once in the denominator of L. We have

$$\begin{aligned} {{\,\mathrm{res}\,}}_{z_{i+1}=q z_{i}} \frac{1}{1-q\frac{z_i}{z_{i+1}}} \frac{d z_{i+1}}{z_{i+1}} = 1. \end{aligned}$$

Thus we will obtain the same result if we multiply L by the product of these factors and then evaluate at $z_i=z_i(\mu )$. Note that C has precisely the same factors removed. Therefore we have

$$\begin{aligned} C H_\mu = \left( \prod _{i<j} \frac{1-q\frac{z_i}{z_j}}{1-\frac{z_i}{z_j}} L\right) (z_1(\mu ),\ldots ,z_{l(\mu )}(\mu )). \end{aligned}$$

Putting in A as well we obtain a nice expression:

$$\begin{aligned} AC H_\mu = \left( \prod _{i\ne j} \frac{ 1-q\frac{z_i}{z_j}}{\prod _{k=1}^g 1-q \alpha _k^{-1} \frac{z_i}{z_j}} \sum _{\sigma \in S_{l(\mu )}} \sigma \left\{ \cdots \right\} \right) (z_1(\mu ),\ldots ,z_{l(\mu )}(\mu )). \end{aligned}$$

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

$$\begin{aligned}&f(z_1,\ldots ,z_n) = \prod _i \prod _{k=1}^g \frac{1-\alpha _k^{-1}}{1- \alpha _k^{-1} z_i}\nonumber \\&\qquad \times \sum _{\sigma \in S_n}\sigma \left\{ \prod _{i>j}\left( \frac{1}{1-\frac{z_i}{z_j}} \prod _{k=1}^g \frac{1-\alpha _k^{-1} \frac{z_i}{z_j}}{1{-}q \alpha _k^{-1} \frac{z_i}{z_j}}\right) \prod _{i> j+1} \left( 1{-}q \tfrac{z_i}{z_j}\right) \prod _{i\ge 2} (1-z_i) \right\} .\nonumber \\ \end{aligned}$$

(5)

Then

$$\begin{aligned} ABCD H_\mu = f(z_1(\mu ),\ldots ,z_{l(\mu )}(\mu )). \end{aligned}$$

Summarizing we obtain

Proposition 3.1

For any partition $\mu$ the term $\Omega _\mu$ is given by

$$\begin{aligned} \Omega _\mu = \frac{f_\mu \prod _{i=1}^g N_\mu (\alpha _i^{-1})}{N_\mu (1)},\qquad f_\mu =f(z_1(\mu ),\ldots ,z_{l(\mu )}(\mu )), \end{aligned}$$

where $z_i(\mu )=q^{-l(\mu )+i} z^{\mu _i}$, and N, f are defined in (4), (5).

Example 3.1

Let us calculate f in a few cases. It is convenient to set

$$\begin{aligned} P(x) = \prod _{i=1}^g (1-\alpha _i^{-1} x). \end{aligned}$$

We have

$$\begin{aligned}&f(z_1) = \frac{P(1)}{P(z_1)}\\&f(z_1,z_2) {=} \frac{P(1)^2}{P(z_1)P(z_2)(z_1-z_2)}\left( z_1(1-z_2) \frac{P(\tfrac{z_2}{z_1})}{P(q\tfrac{z_2}{z_1})} {-} z_2(1-z_1) \frac{P(\tfrac{z_1}{z_2})}{P(q\tfrac{z_1}{z_2})}\right) \end{aligned}$$

Note that the denominator of this expression is $P(z_1) P(z_2) P(q\tfrac{z_1}{z_2})P(q\tfrac{z_2}{z_1})$ if no cancellations happen. If $z_2=q z_1$, the denominator reduces to $P(z_1) P(z_2) P(q^2)$, so it has only 3 P-factors instead of 4.

4 Combinatorics of the function f

4.1 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

$$\begin{aligned} \prod _{i} P(z_i) \prod _{i\ne j} P(q \tfrac{z_i}{z_j}), \end{aligned}$$

where $P(x)=\prod _{k=1}^g (1-\alpha _k^{-1} x)$. Pick numbers $r_1, r_2,\ldots$ such that $\sum _m r_m=n$. Split $z_1, z_2, \ldots , z_n$ into a union of subsequences of sizes $r_1$, $r_2$, .... Let $j_m=1+\sum _{i<m} r_i$. For each m the m-th subsequence looks like $z_{j_m}, z_{j_m+1}, \ldots , z_{j_m+r_m-1}$. Suppose each subsequence forms a geometric progression with quotient q:

$$\begin{aligned} \begin{aligned} z_{j_m+i} = q^i z_{j_m}\qquad (i< r_m). \end{aligned} \end{aligned}$$

Then f can be viewed as a function of variables $z_{j_m}$. The denominator can be bounded as follows

Proposition 4.1

The following expression is a Laurent polynomial:

$$\begin{aligned} f \prod _{i=1}^n \left( P(z_i)\prod _{m:\, j_m+r_m> i} P\left( q^{r_m} \tfrac{z_{j_m}}{z_i}\right) \prod _{m:\, j_m> i} P\left( q \tfrac{z_i}{z_{j_m}}\right) \right) \end{aligned}$$

Proof

First write the definition of f as follows:

$$\begin{aligned} f = \prod _i \frac{P(1)}{P(z_i)} \sum _{\sigma \in S_n} \prod _{\sigma (i)>\sigma (j)}\frac{P\left( \frac{z_i}{z_j}\right) }{\left( 1-\frac{z_i}{z_j}\right) P\left( q\frac{z_i}{z_j}\right) } \prod _{\sigma (i)> \sigma (j)+1} \left( 1-q \tfrac{z_i}{z_j}\right) \prod _{\sigma (i)\ge 2} (1-z_i). \end{aligned}$$

Note that $1-\frac{z_i}{z_j}$ 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 $1-q\frac{z_i}{z_j}=0$. So all summands with $\sigma (i)>\sigma (j)+1$ vanish. So it is enough to sum only over those $\sigma$ which satisfy the condition

$$\begin{aligned} \sigma (i+1)\ge \sigma (i)-1 \quad \text {whenever } i, i+1 \text { are in the same subsequence.} \end{aligned}$$

(6)

So in each sequence $\sigma (j_m),\ldots ,\sigma (j_m+r_m-1)$ if there is a drop, the size of the drop is 1. Now for each such $\sigma$ we look at the product

$$\begin{aligned} \prod _{\sigma (i)>\sigma (j)}\frac{P\left( \frac{z_i}{z_j}\right) }{P\left( q\frac{z_i}{z_j}\right) } = \prod _{i<j,\;\sigma (i)>\sigma (j)}\frac{P\left( \frac{z_i}{z_j}\right) }{P\left( q\frac{z_i}{z_j}\right) } \prod _{i<j,\;\sigma (i)<\sigma (j)}\frac{P\left( \frac{z_j}{z_i}\right) }{P\left( q\frac{z_j}{z_i}\right) }. \end{aligned}$$

It is enough to show that for each value of i and each $\sigma$ the following expressions are Laurent polynomials:

$$\begin{aligned}&P(1) \prod _{m:\, j_m> i} P\left( q \tfrac{z_i}{z_{j_m}}\right) \times \prod _{i<j,\;\sigma (i)>\sigma (j)}\frac{P\left( \frac{z_i}{z_j}\right) }{P\left( q\frac{z_i}{z_j}\right) },\\&\prod _{m:\, j_m+r_m> i} P\left( q^{r_m} \tfrac{z_{j_m}}{z_i}\right) \times \prod _{i<j,\;\sigma (i)<\sigma (j)}\frac{P\left( \frac{z_j}{z_i}\right) }{P\left( q\frac{z_j}{z_i}\right) }. \end{aligned}$$

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 $j_m\le i<j_m+r_m$ (when they are in the same subsequence). So it is enough to show that the following products are Laurent polynomials:

$$\begin{aligned}&P\left( q \tfrac{z_i}{z_{j_m}}\right) \prod _{k<r_m,\;\sigma (i)>\sigma (j_m+k)} \frac{P\left( \frac{z_i}{z_{j_m+k}}\right) }{P\left( q\frac{z_i}{z_{j_m+k}}\right) } \qquad (j_m>i), \end{aligned}$$

(7)

$$\begin{aligned}&P\left( q^{r_m} \tfrac{z_{j_m}}{z_i}\right) \prod _{k<r_m,\;\sigma (i)<\sigma (j_m+k)} \frac{P\left( \frac{z_{j_m+k}}{z_i}\right) }{P(q\frac{z_{j_m+k}}{z_i})} \qquad (j_m>i),\end{aligned}$$

(8)

$$\begin{aligned}&P(1) \prod _{i-j_m<k<r_m,\;\sigma (i)>\sigma (j_m+k)} \frac{P\left( \frac{z_i}{z_{j_m+k}}\right) }{P\left( q\frac{z_i}{z_{j_m+k}}\right) } \qquad (j_m>i),\end{aligned}$$

(9)

$$\begin{aligned}&P\left( q^{r_m} \tfrac{z_{j_m}}{z_i}\right) \prod _{i-j_m<k<r_m,\;\sigma (i)<\sigma (j_m+k)} \frac{P\left( \frac{z_{j_m+k}}{z_i}\right) }{P\left( q\frac{z_{j_m+k}}{z_i}\right) } \qquad (j_m>i). \end{aligned}$$

(10)

Observe that because of the condition (6) in each of the cases (7)–(10) the values of k from a contiguous set $k_{min},\ldots ,k_{max}$ (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 $P(q\frac{z_i}{z_{j_m+k_{min}}})$ in cases (7) and (9), and $P(q\frac{z_i}{z_{j_m+k_{max}}})$ 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). $\square$

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 $z_2=q z_1$ 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 $z_i=z_i(\mu )=z^{\mu _i} q^{i-l(\mu )}$ for a partition $\mu$ we obtain

Proposition 4.2

The following product is a Laurent polynomial for any partition $\mu$:

$$\begin{aligned} f_\mu \prod _{\square \in \mu } P(z^{a(\square )+1} q^{-l(\square )}) P(z^{-a(\square )} q^{l(\square )+1}). \end{aligned}$$

(11)

Proof

Recall that $f_\mu$ is a shorthand for $f(z_1(\mu ),\ldots ,z_{l(\mu )}(\mu ))$ where $z_i(\mu ) = z^{\mu _i} q^{i-l(\mu )}$. In view of Proposition 4.1 it is enough to show that for each i the product

$$\begin{aligned} P(z_i)\prod _{m:\, j_m+r_m> i} P\left( q^{r_m} \tfrac{z_{j_m}}{z_i}\right) \prod _{m:\, j_m> i} P\left( q \tfrac{z_i}{z_{j_m}}\right) \end{aligned}$$

(12)

divides the arm-leg product in (11) for cells $\square \in \mu$ occurring in the row i. Note that our subsequences of geometric progressions in $z_i$ simply correspond to repeated parts of $\mu$. Let $\square$ be the cell in row i and column $\mu _{j_m}$ ($j_m+r_m>i$). Then we have $a(\square )=\mu _i-\mu _{j_m}$, $l(\square )=j_m+r_m-1-i$. Therefore

$$\begin{aligned} z^{-a(\square )} q^{l(\square )+1} = q^{r_m} \tfrac{z_{j_m}}{z_i}. \end{aligned}$$

Let $\square$ be the cell in row i and column $\mu _{j_m}+1$ ($j_m>i$). Then $a(\square )=\mu _i-\mu _{j_m}-1$, $l(\square )=j_m-1-i$. Therefore

$$\begin{aligned} z^{a(\square )+1} q^{-l(\square )} = q \tfrac{z_i}{z_{j_m}}. \end{aligned}$$

For the cell in column 1 we have $a(\square )=\mu _i-1$, $l(\square )=l(\mu )-i$, so

$$\begin{aligned} z^{a(\square )+1} q^{-l(\square )} = z_i. \end{aligned}$$

Thus the factors of (12) form a sub-multiset of the factors of the arm-leg product (11), and the claim follows. $\square$

Corollary 4.3

For any partition $\mu$, the product $N_\mu (1)\Omega _\mu$ is a Laurent polynomial.

Proof

We have

$$\begin{aligned} N_\mu (1)\Omega _\mu = \prod _{i=1}^g N_\mu (\alpha _i^{-1}) f_\mu \end{aligned}$$

and

$$\begin{aligned} \prod _{i=1}^g N_\mu (\alpha _i^{-1}) = \prod _{\square \in \mu } P(z^{a(\square )+1} q^{-l(\square )}) P(z^{-a(\square )} q^{l(\square )+1}) \times \pm \;\text {a monomial}. \end{aligned}$$

$\square$

4.2 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:

Proposition 4.4

For any n we have

$$\begin{aligned} f(1,z_1,\ldots ,z_n) = f(q z_1, \ldots , q z_n). \end{aligned}$$

Proof

Note that because of the product $\prod _{i=2}^n (1-z_i)$ in the definition of $f(1, z_1, \ldots , z_n)$, only the terms with $\sigma (1)=1$ survive. So we can reduce the summation over $S_{n+1}$ to a summation over $S_n$. After cancellation of $\prod _i (1-z_i)$ we obtain

$$\begin{aligned}&f(1, z_1,\ldots ,z_n) = \prod _i \frac{P(1)}{P(z_i)}\\&\quad \times \sum _{\sigma \in S_n}\sigma \left\{ \prod _{i>j}\frac{P\left( \frac{z_i}{z_j}\right) }{\left( 1-\frac{z_i}{z_j}\right) P\left( q \frac{z_i}{z_j}\right) } \prod _{i> j+1} \left( 1-q \tfrac{z_i}{z_j}\right) \prod _{i>1} (1-q z_i) \prod _{i}\frac{P(z_i)}{P(q z_i)} \right\} , \end{aligned}$$

which coincides with $f(q z_1, \ldots , q z_n)$. $\square$

Corollary 4.5

Let $\mu$ be a partition and let $n\ge l(\mu )$. Define $z_{n,i}(\mu )=z^{\mu _i} q^{i-n}$ for $i=1,\ldots ,n$. Then

$$\begin{aligned} f(z_{n,1}(\mu ), \ldots , z_{n,n}(\mu )) = f(z_1(\mu ), \ldots , z_{l(\mu )}(\mu )). \end{aligned}$$

Thus, instead of having a separate function for each value of $l(\mu )$ we can use the same function of n arguments if n is big enough.

5 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 $\mathbb {Q}(t)[q^{\pm 1}]$. We denote by $R^*$ the tensor product $R\otimes _{\mathbb {Q}(t)[q^{\pm 1}]} \mathbb {Q}(q,t)$ and assume $R\subset R^*$.

Definition 5.1

A regular function of $z_i$ is a sequence of Laurent polynomials

$$\begin{aligned} f_n\in R[z_1^{\pm 1},\ldots ,z_n^{\pm 1}] \qquad (n\ge 0) \end{aligned}$$

such that

• (i) $f_n$ is symmetric in $z_1, \ldots , z_n$,

• (ii) $f_{n+1}(1,z_1,\ldots ,z_n)=f_n(t z_1, \ldots , t z_n)$.

For a regular function f and a partition $\mu$, we set

$$\begin{aligned} f_\mu = f_{l(\mu )}(z_1(\mu ), \ldots , z_{l(\mu )}(\mu )),\qquad z_i(\mu ) = q^{\mu _i} t^{i-l(\mu )}. \end{aligned}$$

We will use the modified Macdonald polynomials $\tilde{H}_\mu [X;q,t]$, see [6, 14].

Lemma 5.1

Let $f(u)=1+f^{(1)} u + f^{(2)} u^2+\cdots$ be a power series whose coefficients $f^{(i)}$ are regular functions in the above sense. Let

$$\begin{aligned} \Omega [X] = \sum _{\mu \in \mathcal {P}} c_\mu \tilde{H}_\mu [X;q,t] \end{aligned}$$

be a series with $c_\mu \in R^*$, $c_{\varnothing }=1$ such that all coefficients of

$$\begin{aligned} \mathbb {H}[X] = (q-1) {{\,\mathrm{Log}\,}}\Omega [X] \end{aligned}$$

are in R. Let

$$\begin{aligned} \Omega _f[X,u] = \sum _{\mu } c_\mu \tilde{H}_\mu [X;q,t] f_\mu (u),\qquad \mathbb {H}_f[X,u] = (q-1) {{\,\mathrm{Log}\,}}\Omega _f[X,u]. \end{aligned}$$

Consider the expansion

$$\begin{aligned} \mathbb {H}_f[X,u]= \mathbb {H}[X] + u \mathbb {H}_{f,1}[X] + u^2 \mathbb {H}_{f,2}[X] + \cdots . \end{aligned}$$

Then all coefficients of $\mathbb {H}_{f,i}[X]$ for $i\ge 1$ are in $(q-1) R$. In other words, the specialization $q=1$ of $\mathbb {H}_f[X,u]$ is independent of u.

Proof

Let $S=-(q-1)(t-1)$. Recall the notation $\int ^S_X F[X,X^*]$ (see [14]). This is a linear operation such that

$$\begin{aligned} \int ^S_X G[X] F[X^*] = (G[X], F[X])_X^S =(G[X], F[SX])_X, \end{aligned}$$

and $(-,-)_X$ is the standard Hall scalar product,

$$\begin{aligned} (s_\mu [X], s_{\lambda }[X])_X = \delta _{\mu ,\lambda }. \end{aligned}$$

Recall that modified Macdonald polynomials are orthogonal with respect to $(-,-)_X^S$. In this proof we call an expression F admissible if $(q-1){{\,\mathrm{Log}\,}}F$ has all of its coefficients in R. It was proved in [14] that $\int ^S_X$ preserves admissibility. By the assumption $\Omega [X]$ is admissible. We will “construct” $\mathbb {H}_f[X,u]$ 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 $i\ge 1$ let $\tilde{f}^{(i)}\in R[Z,Z^*]$ be any element such that

$$\begin{aligned} \tilde{f}^{(i)}\left[ \sum _{i=1}^N z_i, \sum _{i=1}^N z_i^{-1}\right] = f^{(i)}_N(z_1,\ldots ,z_N). \end{aligned}$$

One way to construct such an element is to find $m\ge 0$ such that $(z_1\cdots z_N)^m f^{(i)}_N(z_1,\ldots ,z_N)=p(z_1,\ldots ,z_n)$ does not contain negative powers of $z_i$, then lift p to a symmetric function $\tilde{p}\in R[Z]$ and set

$$\begin{aligned} \tilde{f}^{(i)}[Z,Z^*] = \tilde{p}[Z] e_N[Z^*]^m. \end{aligned}$$

Then set

$$\begin{aligned} \tilde{f}(u) = 1 + \tilde{f}^{(1)}u + \tilde{f}^{(2)} u^2 +\cdots \in R[Z,Z^*][[u]]. \end{aligned}$$

We can take plethystic logarithm:

$$\begin{aligned} {{\,\mathrm{Log}\,}}\tilde{f}(u) = g(u)= g^{(1)} u + g^{(2)} u^2 + \cdots \in u R[Z,Z^*][[u]]. \end{aligned}$$

For any partition $\mu$ satisfying $l(\mu )\le N$ by regularity of f we have

$$\begin{aligned} f_\mu = f_{l(\mu )}(q^{\mu _1} t^{1-l(\mu )}, q^{\mu _2} t^{2-l(\mu )}, \ldots , q^{\mu _{l(\mu )}}) = f_N(q^{\mu _1} t^{1-N}, q^{\mu _2} t^{2-N}, \ldots , q^{\mu _N}). \end{aligned}$$

Thus we can obtain $f_\mu$ from $\tilde{f}$ by specializing at

$$\begin{aligned} Z= & {} Z_\mu =\sum _{i=1}^N q^{\mu _i} t^{i-N} = t^{1-N} (q-1) B_\mu + \sum _{i=1}^N t^{i-N} \\= & {} \frac{t^{1-N}}{1-t} S B_\mu + \frac{t^{-N}-1}{t^{-1}-1}, \end{aligned}$$

and similarly for $Z^*$. Hence there exists a series

$$\begin{aligned} g'(u)\in u R[Z,Z^*][[u]] \end{aligned}$$

such that for any partition $\mu$ with $l(\mu )\le N$ we have

$$\begin{aligned} f_\mu = {{\,\mathrm{Exp}\,}}[g'(u)[S B_\mu , S B_\mu ^*]]. \end{aligned}$$

This $g'$ is obtained from g by the lambda ring homomorphism which sends Z to $\frac{t^{1-N}}{1-t} Z + \frac{t^{-N}-1}{t^{-1}-1}$ and similarly for $Z^*$. Specialization can be replaced by scalar product using the identity

$$\begin{aligned} F[S Y] = (F[X], {{\,\mathrm{Exp}\,}}[SXY])_X = (F[X], {{\,\mathrm{Exp}\,}}[XY])_X^*, \end{aligned}$$

and we obtain

$$\begin{aligned} f_\mu = \int ^S_{Z,V} {{\,\mathrm{Exp}\,}}[g'(u)[Z, V]] {{\,\mathrm{Exp}\,}}[Z^* B_\mu + V^* B_\mu ^*]\qquad (l(\mu )\le N). \end{aligned}$$

(13)

Let us show that the sum

$$\begin{aligned} \tilde{\Omega }[X,Z,V]=\sum _{\mu \in \mathcal {P}} c_\mu \tilde{H}_\mu [X;q,t] {{\,\mathrm{Exp}\,}}[Z B_\mu + V B_\mu ^*] \end{aligned}$$

(14)

is admissible. Begin with the series

$$\begin{aligned} \sum _{\mu \in \mathcal {P}} \frac{\tilde{H}_\mu [X] \tilde{H}_\mu [Y] \tilde{H}_\mu [Z] \tilde{H}_\mu [V]}{(\tilde{H}_\mu , \tilde{H}_\mu )^S}, \end{aligned}$$

(15)

which is admissible by the main theorem of [14]. Recall the nabla operator $\nabla$, the shift operator $\tau$ and the multiplication by ${{\,\mathrm{Exp}\,}}\left[ \frac{X}{S}\right]$ operator $\tau ^*$, and Tesler’s identity

$$\begin{aligned} \nabla \tau \tau ^* \tilde{H}_\mu [X] = {{\,\mathrm{Exp}\,}}\left[ \frac{D_\mu X}{S}\right] , \end{aligned}$$

where $D_\mu = -1-S B_\mu$. This implies

$$\begin{aligned} \tau ^* \nabla \tau \tau ^* \tilde{H}_\mu [X] = {{\,\mathrm{Exp}\,}}[-X B_\mu ]. \end{aligned}$$

All of the operators involved preserve admissibility (Corollary 6.3 from [14]). In particular, we see that the operator that sends $\tilde{H}_\mu [X]$ to ${{\,\mathrm{Exp}\,}}[X B_\mu ]$ preserves admissibility. Let $\omega$ be the operator that sends q, t, X to $q^{-1}, t^{-1}, -X$. Then using $\omega \nabla =\nabla ^{-1}\omega$, $\omega \tilde{H}_\mu [X] = \frac{\tilde{H}_\mu [X]}{\tilde{H}_\mu [-1]}$ and the fact that $\nabla ^{-1}$ preserves admissibility (Corollary 6.4 from [14]) we see that the operator that sends $\tilde{H}_\mu [X]$ to ${{\,\mathrm{Exp}\,}}[X B_\mu ^*]$ preserves admissibility too. Applying these operators to (15) in the variables Z, V we obtain that the following series is admissible:

$$\begin{aligned} \sum _{\mu \in \mathcal {P}} \frac{\tilde{H}_\mu [X] \tilde{H}_\mu [Y] {{\,\mathrm{Exp}\,}}[Z B_\mu + V B_\mu ^*]}{(\tilde{H}_\mu , \tilde{H}_\mu )^S}. \end{aligned}$$

Finally, pairing this series with $\Omega [X]$ we obtain admissibility of (14).

Because of (13) we have

$$\begin{aligned} \Omega _f(u) = \int ^S_{Z,V} {{\,\mathrm{Exp}\,}}[g'(u)[Z, V]] \tilde{\Omega }[X, Z^*, V^*]\qquad \text {up to terms of degree } >N \text { in } X. \end{aligned}$$

In what follows we ignore the terms of degree $>N$ in X. Since N can be chosen as large as possible, this is enough. Notice that ${{\,\mathrm{Exp}\,}}[g'(u)[Z, V]]$ is “more” than admissible in the following sense. Introduce a new free (in the lambda ring sense) variable W. Then ${{\,\mathrm{Exp}\,}}[\tfrac{W}{S} g'(u)[Z, V]]$ is admissible. Therefore the following is admissible:

$$\begin{aligned} \Omega _f[X,W,u] := \int ^S_{Z,V} {{\,\mathrm{Exp}\,}}[\tfrac{W}{S} g'(u)[Z, V]] \tilde{\Omega }[X, Z^*, V^*]. \end{aligned}$$

So we have

$$\begin{aligned} \mathbb {H}_f[X,W,u] = (q-1){{\,\mathrm{Log}\,}}\Omega _f[X,W,u] = \sum _{i\ge 0} \mathbb {H}_{f,i}[X,W] u^i \end{aligned}$$

with $\mathbb {H}_{f,i}[X,W]\in R[X,W]$. Finally notice that

$$\begin{aligned} \mathbb {H}_{f,i}[X] = \mathbb {H}_{f,i}[X,S]\equiv \mathbb {H}_{f,i}[X,0] \pmod {(q-1)R[X]}, \end{aligned}$$

and

$$\begin{aligned} \mathbb {H}_{f}[X,0,u]= & {} (q-1){{\,\mathrm{Log}\,}}\Omega _f[X,0,u],\\ \Omega _f[X,0,u]= & {} \int ^S_{Z,V} \tilde{\Omega }[X, Z^*, V^*] = \tilde{\Omega }[X, 0, 0] = \Omega [X]. \end{aligned}$$

This implies

$$\begin{aligned} \mathbb {H}_{f,i}[X] \equiv 0 \pmod {(q-1)R[X]} \qquad (i\ge 1). \end{aligned}$$

$\square$

Then our main result is

Theorem 5.2

For any $g\ge 0$ let

$$\begin{aligned} \Omega (T, q, t,\alpha _1,\ldots ,\alpha _g) = \sum _{\mu \in \mathcal {P}} \frac{\prod _{i=1}^g N_\mu (\alpha _i^{-1})}{N_\mu (1)} T^{|\mu |}, \end{aligned}$$

where

$$\begin{aligned} N_\mu (u)=\prod _{\square \in \mu }(q^{a(\square )}-u t^{1+l(\square )})(q^{a(\square )+1}-u^{-1} t^{l(\square )}). \end{aligned}$$

Let

$$\begin{aligned} \Omega ^\mathrm {Sch}(T, q, t,\alpha _1,\ldots ,\alpha _g) = \sum _{\mu \in \mathcal {P}} \Omega _\mu (q,t,\alpha _1,\ldots ,\alpha _g) T^{|\mu |}, \end{aligned}$$

where $\Omega _\mu$ are the Schiffmann’s terms defined in Sect. 3. Let

$$\begin{aligned} \mathbb {H}(T, q, t,\alpha _1,\ldots ,\alpha _g)= & {} -(q-1)(t-1){{\,\mathrm{Log}\,}}\Omega (T, q, t,\alpha _1,\ldots ,\alpha _g),\\ \mathbb {H}^\mathrm {Sch}(T, q, t,\alpha _1,\ldots ,\alpha _g)= & {} -(q-1)(t-1){{\,\mathrm{Log}\,}}\Omega ^\mathrm {Sch}(T, q, t,\alpha _1,\ldots ,\alpha _g), \end{aligned}$$

and let $\mathbb {H}(q, t,\alpha _1,\ldots ,\alpha _g)_k$ denote the k-th coefficient of $\mathbb {H}(T, q, t,\alpha _1,\ldots ,\alpha _g)$ viewed as a power series in T, and similarly for $\mathbb {H}^\mathrm {Sch}$. Then we have

• (i) $\mathbb {H}^\mathrm {Sch}(q, t,\alpha _1,\ldots ,\alpha _g)_k \in \mathbb {Q}(t)[q^{\pm 1}, \alpha _1^{\pm 1},\ldots ,\alpha _g^{\pm 1}]$,

• (ii) $\mathbb {H}^\mathrm {Sch}(1, t,\alpha _1,\ldots ,\alpha _g)_k=\mathbb {H}(1, t,\alpha _1,\ldots ,\alpha _g)_k$.

Proof

By the main result of [14] we have

$$\begin{aligned} \mathbb {H}(q, t,\alpha _1,\ldots ,\alpha _g)_k \in \mathbb {Q}[t,q, \alpha _1^{\pm 1},\ldots ,\alpha _g^{\pm 1}]. \end{aligned}$$

By Corollary 4.3 we have

$$\begin{aligned} \mathbb {H}^\mathrm {Sch}(q, t,\alpha _1,\ldots ,\alpha _g)_k \in \mathbb {Q}(t,q)[\alpha _1^{\pm 1},\ldots ,\alpha _g^{\pm 1}]. \end{aligned}$$

So we can pass to the ring of Laurent series in $\alpha _1^{-1},\ldots ,\alpha _g^{-1}$ and it is enough to prove the corresponding statements (i) and (ii) for the coefficients in front of monomials of the form $\prod _{i=1}^g \alpha _i^{m_i}$. Let us apply Lemma 5.1 for the ring $R=\mathbb {Q}(t)[q^{\pm 1}, \alpha _1^{\pm 1},\ldots ,\alpha _g^{\pm 1}]$, series

$$\begin{aligned} \Omega [X, q, t,\alpha _1,\ldots ,\alpha _g] = \sum _{\mu \in \mathcal {P}} \frac{\prod _{i=1}^g N_\mu (\alpha _i^{-1}) \tilde{H}_\mu [X,q,t]}{N_\mu (1)}, \end{aligned}$$

and the regular function f(u) obtained from f (see (5) and Proposition 4.4) by setting $u \alpha _i^{-1}$ in place of $\alpha _i^{-1}$, 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

$$\begin{aligned} \sum _{\sigma \in S_n}\sigma \left\{ \prod _{i>j}\left( \frac{1}{1-\frac{z_i}{z_j}}\right) \prod _{i> j+1} \left( 1-q \tfrac{z_i}{z_j}\right) \prod _{i\ge 2} (1-z_i) \right\} =1. \end{aligned}$$

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 $\prod _{i=1}^n (z_i-1)$. 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 identity $\tilde{H}_\mu [T;q,t]=T^{|\mu |}$. Let

$$\begin{aligned} \mathbb {H}^\mathrm {Sch}(T,q,t,u) = -(q-1)(t-1){{\,\mathrm{Log}\,}}\left[ \sum _{\mu \in \mathcal {P}} \frac{\prod _{i=1}^g N_\mu (\alpha _i^{-1})}{N_\mu (1)} T^{|\mu |} f_\mu (u) \right] . \end{aligned}$$

Lemma 5.1 says that

$$\begin{aligned} \mathbb {H}^\mathrm {Sch}(T,q,t,u)-\mathbb {H}(T,q,t) \in (q-1) \mathbb {Q}(t)[q^{\pm 1}, \alpha _1^{\pm 1},\ldots ,\alpha _g^{\pm 1}][[T,u]]. \end{aligned}$$

On the other hand, the coefficient in front of any monomial in $\alpha _1, \ldots ,\alpha _g, T$ has bounded degree in u, wo we can set $u=1$ and obtain a statement about Laurent series in $\alpha _i^{-1}$:

$$\begin{aligned} \mathbb {H}^\mathrm {Sch}(T,q,t,1)-\mathbb {H}(T,q,t) \in (q-1) \mathbb {Q}(t)[q^{\pm 1}]((\alpha _1^{-1},\ldots ,\alpha _g^{-1}))[[T]]. \end{aligned}$$

Finally we remember that $\mathbb {H}^\mathrm {Sch}(T,q,t)=\mathbb {H}^\mathrm {Sch}(T,q,t,1)$ and remember that the coefficients of $\mathbb {H}^\mathrm {Sch}(T,q,t)$ are Laurent polynomials in $\alpha _i$ to obtain

$$\begin{aligned} \mathbb {H}^\mathrm {Sch}(T,q,t)-\mathbb {H}(T,q,t) \in (q-1) \mathbb {Q}(t)[q^{\pm 1}, \alpha _1^{\pm 1},\ldots ,\alpha _g^{\pm 1}][[T]]. \end{aligned}$$

$\square$

Remark 5.1

We have been using the substitution $(q,z)\rightarrow (t,q)$ to relate Schiffmann’s variables to the HLV variables. Note that the Hausel–Villegas functions $\Omega$ and $\mathbb {H}$ are symmetric in q, t because replacing $\mu$ by the conjugate partition interchanges arms and legs. So we could have worked with the substitution $(q,z)\rightarrow (q,t)$, 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].

6 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 $\mathrm {Mot}$ 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 $\mathrm {Mot}$ is denoted by [X] and is called the motivic class of X. The dimensional completion of $\mathrm {Mot}$ is denoted by $\overline{\mathrm {Mot}}$. Fix a curve C and assume it has a divisor of degree 1. The Grothendieck ring of stacks $\mathrm {Mot}$ contains natural elements: the class of $\mathbb {A}^1$ denoted by $\mathbb {L}$, and the class of the curve [C]. There is a natural operation of symmetric power which for a stack X is defined by

$$\begin{aligned} S^n[X] = [X^n/S_n]. \end{aligned}$$

This operation satisfies the condition

$$\begin{aligned} S^n[X+Y]=\sum _{i=0}^n S^i[X] S^{n-i}[Y], \end{aligned}$$

which means that $\mathrm {Mot}$ is a pre-$\lambda$-ring. $S^n$ extends to $\overline{\mathrm {Mot}}$ and makes it into a pre-$\lambda$-ring too. It is not clear² whether $\mathrm {Mot}$ or $\overline{\mathrm {Mot}}$ are $\lambda$-rings, which would mean that $S^n(xy)$ and $S^n(S^m(x)))$ can be expressed as certain prescribed polynomials in $x, S^2(x),S^3(x),\ldots$ and $y, S^2(y), S^3(y),\ldots$. Nevertheless, Totaro’s lemma [5] tells us that $S^n(\mathbb {L}X) = \mathbb {L}^n S^n(X)$ holds for any X and n. The formal sum of the form

$$\begin{aligned} 1+zS^1[X]+z^2S^2[X]+\cdots = {{\,\mathrm{Exp}\,}}[zX] \end{aligned}$$

is called plethystic exponential (a.k.a. motivic zeta function). The inverse operation is called the plethystic logarithm${{\,\mathrm{Log}\,}}$.

Denote by $R_g$ the ring of polynomials in $\alpha _1^{\pm 1}$, ..., $\alpha _{g}^{\pm 1}$ and $q^{\pm 1}$ invariant under permutations of $\alpha _i$ and substitutions of the form $\alpha _i \rightarrow q \alpha _i^{-1}$. As a ring $R_g$ is the polynomial ring in the first g elementary symmetric functions in $\alpha _1,\ldots ,\alpha _g,q \alpha _1^{-1}, \ldots ,q\alpha _g^{-1}$ over $\mathbb {Z}[q,q^{-1}]$. Using Kapranov’s results [10] on the motivic zeta function of a curve one can show that there is a ring homomorphism

$$\begin{aligned} \mathrm {ev}_C:R_g \rightarrow \mathrm {Mot} \end{aligned}$$

which sends q to $\mathbb {L}$ and when extended to $R_g[[z]]\rightarrow \mathrm {Mot}[[z]]$ sends the formal zeta function to the motivic zeta function:

$$\begin{aligned} \mathrm {ev}_C \left( \frac{\prod _{i=1}^g (1-z \alpha _i)(1-z q \alpha _i^{-1})}{(1-z)(1-qz)}\right) = \sum _{n=0}^\infty z^n S^n[C]=\zeta _C(z) \in \mathrm {Mot}[[z]]. \end{aligned}$$

The ring $R_g$ is a $\lambda$-ring. $R_g$ has a filtration by the total degree in $q,\alpha _1,\ldots ,\alpha _g$. Denote the corresponding completion by $\overline{R}_g$. An infinite sum converges in the completion if degrees of the summands tend to $-\infty$. The homomorphism $\mathrm {ev}_C$ extends to the completions.

Consider Schiffmann’s generating function (see Sect. 3 for details)

$$\begin{aligned} \Omega ^{\mathrm {Sch}}(T,z,q,\alpha _1,\ldots ,\alpha _g) = \sum _\mu \Omega _\mu (z,q,\alpha _1,\ldots ,\alpha _g) T^{|\mu |}. \end{aligned}$$

In [4] it is explained how the definition of $\Omega ^\mathrm {Sch}$ can be recast to produce an element $\Omega ^{\mathrm {Mot}}\in \overline{\mathrm {Mot}}[[T,z]]$. Analysing their construction it is easy to see that in fact we have

$$\begin{aligned} \Omega ^{\mathrm {Mot}} = \mathrm {ev}_C(\Omega ^\mathrm {Sch}), \end{aligned}$$

where we extend $\mathrm {ev}_C$ 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:

$$\begin{aligned} \sum _{r,d} B_{r,d} T^r z^d = \mathbb {L}{{\,\mathrm{Log}\,}}\Omega ^{\mathrm {Mot}}(T,z), \end{aligned}$$

then for any rational slope $\tau$ compute plethystic exponential

$$\begin{aligned} \sum _{d/r=\tau } \mathbb {L}^{(1-g)r^2} H_{r,d}^\mathrm {Mot}T^r z^d = {{\,\mathrm{Exp}\,}}\sum _{d/r=\tau } B_{r,d} T^r z^d, \end{aligned}$$

and finally recover the motivic class of the moduli stack as

$$\begin{aligned}{}[\mathcal {M}_{r,d}^{\mathrm {ss}}] = H_{r,d+er}^\mathrm {Mot}\qquad (e\gg 0) \end{aligned}$$

for sufficiently large e.

This recipe can be reformulated as follows. Let us decompose $\Omega ^\mathrm {Mot}$ into an infinite product according to slopes first:

$$\begin{aligned} \Omega ^\mathrm {Mot}= \prod _{\tau \in \mathbb {Q}_{\ge 0}} \sum _{d/r=\tau } \widetilde{H}_{r,d}^\mathrm {Mot}T^r z^d. \end{aligned}$$

Then we have

$$\begin{aligned} \sum _{d/r=\tau } \mathbb {L}^{(1-g)r^2} H_{r,d}^\mathrm {Mot}T^r z^d = {{\,\mathrm{Exp}\,}}\mathbb {L}{{\,\mathrm{Log}\,}}\sum _{d/r=\tau } \widetilde{H}_{r,d}^\mathrm {Mot}T^r z^d. \end{aligned}$$

Analogously, we can decompose

$$\begin{aligned} \Omega ^\mathrm {Sch}(T,z,q,\alpha _1,\ldots ,\alpha _g) = \prod _{\tau \in \mathbb {Q}_{\ge 0}} \sum _{d/r=\tau } \widetilde{H}_{r,d}^\mathrm {Sch}(q,\alpha _1,\ldots ,\alpha _g) T^r z^d, \end{aligned}$$

and define $H_{r,d}^\mathrm {Sch}$ by

$$\begin{aligned} \sum _{d/r=\tau } q^{(1-g)r^2} H_{r,d}^\mathrm {Sch}T^r z^d = {{\,\mathrm{Exp}\,}}q {{\,\mathrm{Log}\,}}\sum _{d/r=\tau } \widetilde{H}_{r,d}^\mathrm {Sch}T^r z^d. \end{aligned}$$

We make two observations. If $\overline{\mathrm {Mot}}$ was a $\lambda$-ring, it would follow that $\mathrm {ev}_C$ commutes with ${{\,\mathrm{Exp}\,}}$ and ${{\,\mathrm{Log}\,}}$, and we would have $H_{r,d}^\mathrm {Mot}=\mathrm {ev}_C(H_{r,d}^\mathrm {Sch})$. So any result about $\Omega ^\mathrm {Sch}$ could have been directly translated to $H_{r,d}^\mathrm {Mot}$. For instance, we would have

$$\begin{aligned} \sum _{d/r=\tau } \mathbb {L}^{(1-g)r^2} [\mathcal {M}_{r,d}^\mathrm {ss}] T^r z^d = \left( {{\,\mathrm{Exp}\,}}\frac{\mathbb {L}\sum _{d/r=\tau } \mathrm {ev}_C(A_{g,r}) T^r z^d}{\mathbb {L}-1}\right) \quad \text {(in a } \lambda \text {-ring)}. \end{aligned}$$

(16)

In particular, in the universal $\lambda$-ring quotient of $\overline{\mathrm {Mot}}$ 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 $\lambda$-ring property because in all the expansions of ${{\,\mathrm{Exp}\,}}$ and ${{\,\mathrm{Log}\,}}$ above we use only the first term. So we directly obtain

$$\begin{aligned}&H_{r,d}^\mathrm {Mot}= \mathbb {L}^{(g-1)r^2+1} \widetilde{H}_{r,d}^\mathrm {Mot}= \mathbb {L}^{(g-1)r^2} B_{r,d} = \mathrm {ev}_C(H_{r,d}^\mathrm {Sch}),\nonumber \\&{[}\mathcal {M}_{r,d}^\mathrm {s}] = \frac{\mathbb {L}^{(g-1)r^2+1} \mathrm {ev}_C(A_{g,r})}{\mathbb {L}-1} \qquad ((r,d)=1). \end{aligned}$$

(17)

So in the above two situations no information is lost when passing from motivic invariants to functions in $q,\alpha _1,\ldots ,\alpha _g$.

Next we would like to connect the motivic formula to the Hausel–Rodriguez-Villegas generating function $\Omega _g$. Unfortunately, the function is not invariant under the map $\alpha _i\rightarrow q\alpha _i^{-1}$, so the function is not in $R_g$. Following Mozgovoy’s approach [15], we apply change of variables $q\rightarrow q z^{-1}$:

$$\begin{aligned} \Omega _g^\mathrm {Moz}= & {} \sum _{\mu \in \mathcal {P}} T^{|\mu |} z^{(g-1)\langle \mu ',\mu '\rangle } \prod _{\square \in \mu } \frac{\prod _{i=1}^g (1 - \alpha _i q^{l(\square )} z^{-h(\square )}) (1 - \alpha _i^{-1} q^{l(\square )+1}z^{-h(\square )})}{(1 - q^{l(\square )}z^{-h(\square )}) (1 - q^{l(\square )+1}z^{-h(\square )})}\\= & {} \sum _{\mu \in \mathcal {P}} T^{|\mu |} (q/z)^{(g-1)\langle \mu ,\mu \rangle } \prod _{\square \in \mu } \frac{\prod _{i=1}^g (1 - \alpha _i q^{-l(\square )-1} z^{h(\square )}) (1 - \alpha _i^{-1} q^{-l(\square )}z^{h(\square )})}{(1 - q^{-l(\square )-1}z^{h(\square )}) (1 - q^{-l(\square )}z^{h(\square )})}, \end{aligned}$$

where $h(\square )=a(\square )+l(\square )+1$ and $\langle \mu ',\mu '\rangle =\sum _i \mu _i^2$. In this way we obtain $\Omega _g^\mathrm {Moz}\in R_g[[T,z]]$. We apply $\mathrm {ev}_C$ to define the motivic Mozgovoy function:

$$\begin{aligned} \Omega _g^{\mathrm {Mot}\mathrm {Moz}} = \mathrm {ev}_C(\Omega _g^\mathrm {Moz}) = \sum _{\mu \in \mathcal {P}} T^{|\mu |} (q/z)^{(g-1)\langle \mu ,\mu \rangle } \prod _{\square \in \mu } \zeta _C(q^{-l(\square )-1} z^{h(\square )}), \end{aligned}$$

where $\zeta _C(z)={{\,\mathrm{Exp}\,}}[C z]$ is the motivic zeta function of C. Theorem 1.1 implies

$$\begin{aligned} \Omega _g^\mathrm {Moz}= {{\,\mathrm{Exp}\,}}\left[ \frac{z/q \sum _{r=1}^\infty H_{g,r}(q/z,z,\alpha _1,\ldots ,\alpha _g) T^r}{(1-z/q)(1-z)}\right] . \end{aligned}$$

Using the evaluations $H_{g,r}(q,1,\ldots )=H_{g,r}(1,q,\ldots )=A_{g,r}(q,\ldots )$, we obtain

$$\begin{aligned} \frac{z/q H_{g,r}(q/z,z,\alpha _1,\ldots ,\alpha _g)}{(1-z/q)(1-z)}= & {} \frac{1}{q-1}\left( \frac{1}{1-z}-\frac{1}{1-z/q}\right) H_{g,r}(q/z,z,\alpha _1,\ldots ,\alpha _g)\\= & {} \sum _{d=1}^\infty \frac{A_{g,r}(q,\alpha _1,\ldots ,\alpha _g)}{q-1}(1-q^{-d})z^d\\&+\text {Laurent polynomial in } z. \end{aligned}$$

So the coefficients do not stabilize like they do for ${{\,\mathrm{Log}\,}}\Omega ^\mathrm {Sch}$, but nevertheless tend to $\frac{A_{g,r}}{q-1}$ in $\overline{R_g}$ as d goes to $\infty$. Therefore in the product expansion of the ratio

$$\begin{aligned} \frac{\Omega ^\mathrm {Sch}(T,z,q,\alpha _1,\ldots ,\alpha _g)}{\Omega ^\mathrm {Moz}(T,z,q,\alpha _1,\ldots ,\alpha _g)} = \prod _{\tau \in \mathbb {Q}} \sum _{d/r=\tau } D_{r,d} T^r z^d \end{aligned}$$

the coefficients $D_{r,d}\in \overline{R_g}$ tend to 0 when r is fixed and d goes to $\infty$. Applying $\mathrm {ev}_C$ to both sides we conclude that

$$\begin{aligned} \frac{\Omega ^\mathrm {Mot}(T,z)}{\Omega ^{\mathrm {Mot}\mathrm {Moz}}(T,z)} = \prod _{\tau \in \mathbb {Q}} \sum _{d/r=\tau } D_{r,d}^\mathrm {Mot}T^r z^d, \end{aligned}$$

where $D_{r,d}^\mathrm {Mot}$ tend to 0 in $\overline{\mathrm {Mot}}$ when r is fixed and d goes to $\infty$. So constructing $H_{r,d}^{\mathrm {Mot}\mathrm {Moz}}$ from $\Omega ^{\mathrm {Mot}\mathrm {Moz}}$ in the same way as $H_{r,d}^\mathrm {Mot}$ was constructed from $\Omega ^\mathrm {Mot}$ in [4], we obtain for $\tau \in \mathbb {Q}$

$$\begin{aligned} \frac{\sum _{d/r=\tau } \mathbb {L}^{(1-g)r^2} H_{r,d+er}^\mathrm {Mot}T^r}{\sum _{d/r=\tau } \mathbb {L}^{(1-g)r^2} H_{r,d+er}^{\mathrm {Mot}\mathrm {Moz}} T^r} \rightarrow 1\qquad (e\rightarrow \infty ) \end{aligned}$$

coefficientwise. This immediately implies a version of [4] for Mozgovoy’s function:

Corollary 6.1

For any r, d we have

$$\begin{aligned}{}[\mathcal {M}_{r,d}^\mathrm {ss}] = \lim _{e\rightarrow \infty } H_{r,d+er}^{\mathrm {Mot}\mathrm {Moz}} \in \overline{\mathrm {Mot}}. \end{aligned}$$