When is the Bloch–Okounkov q-bracket modular?

Abstract

We obtain a condition describing when the quasimodular forms given by the Bloch–Okounkov theorem as q-brackets of certain functions on partitions are actually modular. This condition involves the kernel of an operator $\Delta$. We describe an explicit basis for this kernel, which is very similar to the space of classical harmonic polynomials.

1 Introduction

Given a family of quasimodular forms, the question which of its members are modular often has an interesting answer. For example, consider the family of theta series

$$\begin{aligned} \theta _P(\tau ) = \sum _{\underline{x}\in \mathbb {Z}^r} P(\underline{x})q^{x_1^2+\ldots +x_r^2} \quad \quad \quad (q=e^{2\pi i \tau }) \end{aligned}$$

given by all homogeneous polynomials $P\in \mathbb {Z}[x_1,\ldots , x_r]$. The quasimodular form $\theta _P$ is modular if and only if P is harmonic (i.e. $P\in \ker \sum _{i=1}^r \frac{\partial ^2}{\partial x_i^2}$) [10]. (As quasimodular forms were not yet defined, Schoeneberg only showed that $\theta _P$ is modular if P is harmonic. However, for every polynomial P it follows that $\theta _P$ is quasimodular by decomposing P as in Formula (1).) Also, for every two modular forms f, g, one can consider the linear combination of products of derivatives of f and g given by

$$\begin{aligned} \sum _{r=0}^n a_r f^{(r)} g^{(n-r)} \quad \quad (a_r\in \mathbb {C}). \end{aligned}$$

This linear combination is a quasimodular form which is modular precisely if it is a multiple of the Rankin–Cohen bracket $[f,g]_n$ [4, 9]. In this paper, we provide a condition to decide which member of the family of quasimodular forms provided by the Bloch–Okounkov theorem is modular. Let $\mathscr {P}$ denote the set of all partitions of integers and $|\lambda |$ denote the integer that $\lambda$ is a partition of. Given a function $f:\mathscr {P}\rightarrow \mathbb {Q}$, define the q-bracket of f by

$$\begin{aligned} \langle f \rangle _q := \frac{\sum _{\lambda \in \mathscr {P}} f(\lambda ) q^{|\lambda |}}{\sum _{\lambda \in \mathscr {P}} q^{|\lambda |}}. \end{aligned}$$

The celebrated Bloch–Okounkov theorem states that for a certain family of functions $f:\mathscr {P}\rightarrow \mathbb {Q}$ (called shifted symmetric polynomials and defined in Sect. 2) the q-brackets $\langle f \rangle _q$ are the q-expansions of quasimodular forms [2].

Besides being a wonderful result, the Bloch–Okounkov theorem has many applications in enumerative geometry. For example, a special case of the Bloch–Okounkov theorem was discovered by Dijkgraaf and provided with a mathematically rigorous proof by Kaneko and Zagier, implying that the generating series of simple Hurwitz numbers over a torus are quasimodular [5, 7]. Also, in the computation of asymptotics of geometrical invariants, such as volumes of moduli spaces of holomorphic differentials and Siegel–Veech constants, the Bloch–Okounkov theorem is applied [3, 6].

Zagier gave a surprisingly short and elementary proof of the Bloch–Okounkov theorem [13]. A corollary of his work, which we discuss in Sect. 3, is the following proposition:

Proposition 1

There exists actions of the Lie algebra $\mathfrak {sl}_2$ on both the algebra of shifted symmetric polynomials $\varLambda ^*$ and the algebra of quasimodular forms $\widetilde{M}$ such that the q-bracket $\langle \cdot \rangle _q: \varLambda ^*\rightarrow \widetilde{M}$ is $\mathfrak {sl}_2$-equivariant.

The answer to the question in the title is provided by one of the operators $\Delta$ which defines this $\mathfrak {sl}_2$-action on $\varLambda ^*$. Namely letting $\mathcal {H}=\ker \Delta |_{\varLambda ^*}$, we prove the following theorem:

Theorem 1

Let $f\in \varLambda ^*$. Then $\langle f \rangle _q$ is modular if and only if $f=h+k$ with $h\in \mathcal {H}$ and $k\in \ker \langle \cdot \rangle _q$.

The last section of this article is devoted to describing the graded algebra $\mathcal {H}$. We call $\mathcal {H}$ the space of shifted symmetric harmonic polynomials, as the description of this space turns out to be very similar to the space of classical harmonic polynomials. Let $\mathcal {P}_d$ be the space of polynomials of degree d in $m\ge 3$ variables $x_1,\ldots ,x_m$, let $||x||^2=\sum _{i}x_i^2$, and recall that the space $\mathscr {H}_d$ of degree d harmonic polynomials is given by $\ker \sum _{i=1}^r \frac{\partial ^2}{\partial x_i^2}$. The main theorem of harmonic polynomials states that every polynomial $P\in \mathcal {P}_d$ can uniquely be written in the form

$$\begin{aligned} P=h_0+||x||^2h_1+\ldots +||x||^{2d'}h_{d'} \end{aligned}$$

(1)

with $h_i\in \mathscr {H}_{d-2i}$ and $d'=\lfloor d/2\rfloor$. Define $K$, the Kelvin transform, and $D^\alpha$ for $\alpha$ an m-tuple of non-negative integers by

$$\begin{aligned} f(x)\mapsto ||x||^{2-m}f\left( \frac{x}{||x||^2}\right) \quad \text {and} \quad D^\alpha = \prod _i \frac{\partial ^\alpha _i}{\partial x_i^{\alpha _i}}. \end{aligned}$$

An explicit basis for $\mathscr {H}_d$ is given by

$$\begin{aligned} \{KD^\alpha K(1) \mid \alpha \in \mathbb {Z}_{\ge 0}^m, \textstyle \sum _i \alpha _i=d, \alpha _1\le 1\}, \end{aligned}$$

see for example [1]. We prove the following analogous results for the space of shifted symmetric polynomials:

Theorem 2

For every $f\in \varLambda ^*_n$ there exists unique $h_{i}\in \mathcal {H}_{n-2i}$ ($i=0,1,\ldots , n'$ and $n'={\lfloor \tfrac{n}{2}\rfloor }$) such that

$$\begin{aligned} f= h_0+Q_2h_1+\ldots +Q_2^{n'}h_{n'}, \end{aligned}$$

where $Q_2$ is an element of $\varLambda ^*_2$ given by $Q_2(\lambda )=|\lambda |-\frac{1}{24}$. $\square$

Theorem 3

The set

$$\begin{aligned} \{\mathrm {pr}\, K\, \Delta _\lambda \, K(1) \mid \lambda \in \mathscr {P}(n), \text {all parts are } \ge 3\} \end{aligned}$$

is a vector space basis of $\mathcal {H}_n$, where $\mathrm {pr}, K$, and $\Delta _\lambda$ are defined by (4), Definition 4, respectively, Definition 6.

The action of $\mathfrak {sl}_2$ given by Proposition 1 makes $\varLambda ^*$ into an infinite-dimensional $\mathfrak {sl}_2$-representation for which the elements of $\mathcal {H}$ are the lowest weight vectors. Theorem 2 is equivalent to the statement that $\varLambda ^*$ is a direct sum of the (not necessarily irreducible) lowest weight modules

$$\begin{aligned} \displaystyle V_n=\bigoplus _{m=0}^\infty Q_2^m \mathcal {H}_n \quad \quad (n\in \mathbb {Z}). \end{aligned}$$

2 Shifted symmetric polynomials

Shifted symmetric polynomials were introduced by Okounkov and Olshanski as the following analogue of symmetric polynomials [8]. Let $\varLambda ^*(m)$ be the space of rational polynomials in m variables $x_1,\ldots , x_m$ which are shifted symmetric, i.e. invariant under the action of all $\sigma \in \mathfrak {S}_m$ given by $x_i\mapsto x_{\sigma (i)}+i-\sigma (i)$ (or more symmetrically $x_i-i\mapsto x_{\sigma (i)}-\sigma (i)$). Note that $\varLambda ^*(m)$ is filtered by the degree of the polynomials. We have forgetful maps $\varLambda ^*(m)\rightarrow \varLambda ^*({m-1})$ given by $x_m\mapsto 0$, so that we can define the space of shifted symmetric polynomials $\varLambda ^*$ as $\displaystyle \varprojlim _m \varLambda ^*(m)$ in the category of filtered algebras. Considering a partition $\lambda$ as a non-increasing sequence $(\lambda _1,\lambda _2,\ldots )$ of non-negative integers $\lambda _i$, we can interpret $\varLambda ^*$ as being a subspace of all functions $\mathscr {P}\rightarrow \mathbb {Q}$.

One can find a concrete basis for this abstractly defined space by considering the generating series

$$\begin{aligned} w_\lambda (T):=\sum _{i=1}^\infty T^{\lambda _i-i+\tfrac{1}{2}} \in T^{1/2}\mathbb {Z}[T][[T^{-1}]] \end{aligned}$$

(2)

for every $\lambda \in \mathscr {P}$ (the constant $\tfrac{1}{2}$ turns out to be convenient for defining a grading on $\varLambda ^*$). As $w_\lambda (T)$ converges for $T>1$ and equals

$$\begin{aligned} \frac{1}{T^{1/2}-T^{-1/2}} + \sum _{i=1}^{\ell (\lambda )}\left( T^{\lambda _i-i+\tfrac{1}{2}}-T^{-i+\tfrac{1}{2}}\right) \end{aligned}$$

one can define shifted symmetric polynomials $Q_i(\lambda )$ for $i\ge 0$ by

$$\begin{aligned} \sum _{i=0}^\infty Q_i(\lambda ) z^{i-1} := w_\lambda (e^z) \quad \quad (0<|z|<2\pi ). \end{aligned}$$

(3)

The first few shifted symmetric polynomials $Q_i$ are given by

$$\begin{aligned} Q_0(\lambda )=1,\quad Q_1(\lambda )=0,\quad Q_2(\lambda )=|\lambda |-\tfrac{1}{24}. \end{aligned}$$

The $Q_i$ freely generate the algebra of shifted symmetric polynomials, i.e. $\varLambda ^*=\mathbb {Q}[Q_2,Q_3,\ldots ]$. It is believed that $\varLambda ^*$ is maximal in the sense that for all $Q:\mathscr {P}\rightarrow \mathbb {Q}$ with $Q\not \in \varLambda ^*$ it holds that $\langle \varLambda ^*[Q]\rangle _q \not \subseteq \widetilde{M}$.

Remark 1

The space $\varLambda ^*$ can equally well be defined in terms of the Frobenius coordinates. Given a partition with Frobenius coordinates $(a_1,\ldots , a_r,b_1,\ldots , b_r)$, where $a_i$ and $b_i$ are the arm and leg lengths of the cells on the main diagonal, let

$$\begin{aligned} C_\lambda =\left\{ -b_1-\tfrac{1}{2},\ldots , -b_r-\tfrac{1}{2},a_r+\tfrac{1}{2},\ldots ,a_1+\tfrac{1}{2} \right\} . \end{aligned}$$

Then

$$\begin{aligned} Q_k(\lambda ) =\beta _k + \frac{1}{(k-1)!}\sum _{c\in C_{\lambda }} {{\,\mathrm{sgn}\,}}(c)c^{k-1}, \end{aligned}$$

where $\beta _k$ is the constant given by

$$\begin{aligned} \sum _{k\ge 0}\beta _k z^{k-1} = \frac{1}{2\sinh (z/2)} = w_\emptyset (e^z). \end{aligned}$$

We extend $\varLambda ^*$ to an algebra where $Q_1\not \equiv 0$. Observe that a non-increasing sequence $(\lambda _1,\lambda _2,\ldots )$ of integers corresponds to a partition precisely if it converges to 0. If, however, it converges to an integer n, Eqs. (2) and (3) still define $Q_k(\lambda )$. In fact, in this case

$$\begin{aligned} Q_k(\lambda ) = (e^{n\varvec{\partial }})Q_k(\lambda -n) \end{aligned}$$

by [13, Proposition 1] where $\varvec{\partial } Q_0=0$, $\varvec{\partial } Q_k = Q_{k-1}$ for $k\ge 1$, and $\lambda -n=(\lambda _1-n,\lambda _2-n,\ldots )$ corresponds to a partition (i.e. converges to 0). In particular, $Q_1(\lambda )=n$ equals the number the sequence $\lambda$ converges to. We now define the Bloch–Okounkov ring $\mathcal {R}$ to be $\varLambda ^*[Q_1]$, considered as a subspace of all functions from non-increasing eventually constant sequences of integers to $\mathbb {Q}$. It is convenient to work with $\mathcal {R}$ instead of $\varLambda ^*$ to define the differential operators $\Delta$ and more generally $\Delta _\lambda$ later. Both on $\varLambda ^*$ and $\mathcal {R}$, we define a weight grading by assigning to $Q_i$ weight i. Denote the projection map by

$$\begin{aligned} \mathrm {pr}:\mathcal {R}\rightarrow \varLambda ^*. \end{aligned}$$

(4)

We extend $\langle \cdot \rangle _q$ to $\mathcal {R}$.

The operator $E=\sum _{m=0}^\infty Q_m \frac{\partial }{\partial Q_{m}}$ on $\mathcal {R}$ multiplies an element of $\mathcal {R}$ by its weight. Moreover, we consider the differential operators

$$\begin{aligned} \varvec{\partial } = \sum _{m=0}^\infty Q_m \frac{\partial }{\partial Q_{m+1}} \quad \text {and} \quad \mathscr {D} = \sum _{k,\ell \ge 0} \left( {\begin{array}{c}k+\ell \\ k\end{array}}\right) Q_{k+\ell } \frac{\partial ^2}{\partial Q_{k+1} \partial Q_{\ell +1}}. \end{aligned}$$

Let $\Delta =\tfrac{1}{2}(\mathscr {D}-\varvec{\partial }^2)$, i.e.

$$\begin{aligned} 2\Delta = \sum _{k,\ell \ge 0}\left( \left( {\begin{array}{c}k+\ell \\ k\end{array}}\right) Q_{k+\ell }-Q_kQ_\ell \right) \frac{\partial ^2}{\partial Q_{k+1} \partial Q_{\ell +1}}-\sum _{k\ge 0} Q_k \frac{\partial }{\partial Q_{k+2}}. \end{aligned}$$

In the following (antisymmetric) table, the entry in the row of operator A and column of operator B denotes the commutator [A, B], for proofs see [13, Lemma 3].

$$\begin{aligned} \begin{array}{l|lllll} &{}\quad \Delta &{}\quad \varvec{\partial } &{}\quad E&{}\quad Q_1 &{}\quad Q_2 \\ \hline \Delta &{}\quad 0 &{}\quad 0 &{}\quad 2\Delta &{}\quad 0 &{}\quad E-Q_1\varvec{\partial }-\tfrac{1}{2} \\ \varvec{\partial } &{}\quad 0 &{}\quad 0 &{}\quad \varvec{\partial }&{}\quad 1 &{}\quad Q_1 \\ E &{}\quad -2\Delta &{}\quad -\varvec{\partial } &{}\quad 0 &{}\quad Q_1 &{}\quad 2Q_2 \\ Q_1 &{}\quad 0 &{}\quad -1 &{}\quad -Q_1 &{}\quad 0 &{}\quad 0 \\ Q_2 &{}\quad -E+Q_1\varvec{\partial }+\tfrac{1}{2} &{}\quad -Q_1 &{}\quad -2 Q_2 &{}\quad 0 &{}\quad 0 \end{array} \end{aligned}$$

Definition 1

A triple (X, Y, H) of operators is called an $\mathfrak {sl}_2$-triple if

$$\begin{aligned}{}[H,X]=2X, \quad [H,Y]=-2Y, \quad [Y,X]=H. \end{aligned}$$

Let $\hat{Q}_2:=Q_2-\tfrac{1}{2}Q_1^2$ and $\hat{E}:=E-Q_1\varvec{\partial }-\frac{1}{2}.$ The following result follows by a direct computation using the above table:

Proposition 2

The operators $(\hat{Q}_2,\Delta ,\hat{E})$ form an $\mathfrak {sl}_2$-triple. $\square$

For later reference, we compute $[\Delta ,Q_2^n]$. This could be done inductively by noting that $[\Delta ,Q_2^n] = Q_2^{n-1}[\Delta ,Q_2]+[\Delta ,Q_2^{n-1}]Q_2$ and using the commutation relations in the above table. The proof below is a direct computation from the definition of $\Delta$.

Lemma 1

For all $n\in \mathbb {N}$, the following relation holds

$$\begin{aligned}{}[\Delta ,Q_2^n]=-\frac{n(n-1)}{2}Q_1^2Q_2^{n-2}-nQ_1Q_2^{n-1} \varvec{\partial }+nQ_2^{n-1}(E+n-\tfrac{3}{2}). \end{aligned}$$

Proof

Let $f\in \mathbb {Q}[Q_1,Q_2]$, $g\in \mathscr {R}$, and $n\in \mathbb {N}$. Then

$$\begin{aligned} \Delta (fg)&= \Delta (f) g + \frac{\partial f}{\partial Q_2}(Eg-Q_1 \varvec{\partial } g)+f\Delta (g), \end{aligned}$$

(5)

$$\begin{aligned} \Delta (Q_2^n)&=n(n-\tfrac{3}{2})Q_2^{n-1}-\frac{n(n-1)}{2}Q_2^{n-2}Q_1^2. \end{aligned}$$

(6)

By (5) and (6), we find

$$\begin{aligned} \Delta (Q_2^ng)&= \big (n(n-\tfrac{3}{2})Q_2^{n-1}-\frac{n(n-1)}{2}Q_1^2Q_2^{n-2}\big )g\\&+\,nQ_2^{n-1}(Eg-Q_1 \varvec{\partial } g)+Q_2^{n}\Delta (g). \end{aligned}$$

$\square$

3 An $\mathfrak {sl}_2$-equivariant mapping

The space of quasimodular forms for $\mathrm {SL}_2(\mathbb {Z})$ is given by $\widetilde{M} =\mathbb {Q}[P,Q,R]$, where P, Q, and R are the Eisenstein series of weight 2, 4, and 6, respectively (in Ramanujan’s notation). We let $\widetilde{M}^{(\le p)}_k$ be the space of quasimodular forms of weight k and depth $\le p$ (the depth of a quasimodular form written as a polynomial in P, Q, and R is the degree of this polynomial in P). See [12, Section 5.3] or [13, Section 2] for an introduction into quasimodular forms.

The space of quasimodular forms is closed under differentiation, more precisely the operators $D=q\frac{d}{dq}$, $\mathfrak {d}=12\frac{\partial }{\partial P}$, and the weight operator W given by $Wf=kf$ for $f\in \widetilde{M}_k$ preserve $\widetilde{M}$ and form an $\mathfrak {sl}_2$-triple. In order to compute the action of D in terms of the generators P, Q, and R, one uses the Ramanujan identities

$$\begin{aligned} D(P)=\frac{P^2-Q}{12}, \quad D(Q)=\frac{PQ-R}{3}, \quad D(R)= \frac{PR-Q^2}{2}. \end{aligned}$$

In the context of the Bloch–Okounkov theorem, it is more natural to work with $\hat{D} := D - \frac{P}{24}$, as for all $f\in \varLambda ^*$ one has $\langle Q_2 f\rangle _q = \hat{D} \langle f \rangle _q$. Moreover, $\hat{D}$ has the property that it increases the depth of a quasimodular form by 1, in contrast to D for which $D(1)=0$ does not have depth 1:

Lemma 2

Let $f\in \widetilde{M}$ be of depth r. Then $\hat{D}f$ is of depth $r+1$.

Proof

Consider a monomial $P^aQ^bR^c$ with $a,b,c\in \mathbb {Z}_{\ge 0}$. By the Ramanujan identities, we find

$$\begin{aligned} D(P^aQ^bR^c) = \left( \frac{a}{12}+\frac{b}{3}+\frac{c}{2}\right) P^{a+1}Q^bR^c + O(P^{a}), \end{aligned}$$

where $O(P^{a})$ denotes a quasimodular form of depth at most a. The lemma follows by noting that $\frac{a}{12}+\frac{b}{3}+\frac{c}{2}-\frac{1}{24}$ is non-zero for $a,b,c\in \mathbb {Z}$. $\square$

Moreover, letting $\hat{W}=W-\frac{1}{2}$, the triple ($\hat{D}, \mathfrak {d}, \hat{W})$ forms an $\mathfrak {sl}_2$-triple as well. With respect to these operators, the q-bracket becomes $\mathfrak {sl}_2$-equivariant. The following proposition is a detailed version of Proposition 1:

Proposition 3

(The $\mathfrak {sl}_2$-equivariant Bloch–Okounkov theorem) The mapping $\langle \cdot \rangle _q: \mathcal {R} \rightarrow \widetilde{M}$ is $\mathfrak {sl}_2$-equivariant with respect to the $\mathfrak {sl}_2$-triple $(\hat{Q}_2, \Delta ,\hat{E})$ on $\mathcal {R}$ and the $\mathfrak {sl}_2$-triple $(\hat{D}, \mathfrak {d}, \hat{W})$ on $\widetilde{M}$, i.e. for all $f\in \mathcal {R}$, one has

$$\begin{aligned} \hat{D}\langle f\rangle _q = \langle \hat{Q}_2f\rangle _q,\quad \mathfrak {d}\langle f\rangle _q=\langle \Delta f\rangle _q, \quad \hat{W}\langle f\rangle _q= \langle \hat{E}f\rangle _q. \end{aligned}$$

Proof

This follows directly from [13, Equation (37)] and the fact that for all $f\in \mathcal {R}$ one has $\langle Q_1 f \rangle _q=0$. $\square$

4 Describing the space of shifted symmetric harmonic polynomials

In this section, we study the kernel of $\Delta$. As $[\Delta ,Q_1]=0$, we restrict ourselves without loss of generality to $\varLambda ^*$. Note, however, that $\Delta$ does not act on $\varLambda ^*$ as, for example, $\Delta (Q_3)=-\tfrac{1}{2}Q_1$. However, $\mathrm {pr}\Delta$ does act on $\varLambda ^*$.

Definition 2

Let

$$\begin{aligned} \mathcal {H} = \{f \in \varLambda ^* \mid \Delta f\in Q_1\mathcal {R}\}=\ker \mathrm {pr}\Delta , \end{aligned}$$

be the space of shifted symmetric harmonic polynomials.

Proposition 4

If $f\in Q_2\varLambda ^*$ is non-zero, then $f \not \in \mathcal {H}$.

Proof

Write $f=Q_2^n f'$ with $f'\in \varLambda ^*$ and $f'\not \in Q_2\varLambda ^*$. Then

$$\begin{aligned} \mathrm {pr}\Delta (f)=Q_2^{n-1}(n(n+k-\tfrac{3}{2})f'+Q_2\mathrm {pr}\Delta f') \end{aligned}$$

by Lemma 1. As $f'$ is not divisible by $Q_2$, it follows that $\mathrm {pr}\Delta (f) =0$ precisely if $f'=0$. $\square$

Proposition 5

For all $n\in \mathbb {Z}$, one has

$$\begin{aligned} \varLambda _n^* = \mathcal {H}_n \oplus Q_2 \varLambda _{n-2}^*. \end{aligned}$$

Proof

For uniqueness, suppose $f=Q_2g+h$ and $f=Q_2g'+h'$ with $g,g'\in \varLambda _{n-2}^*$ and $h,h' \in \mathcal {H}_n$. Then, $Q_2(g-g')=h'-h\in \mathcal {H}$. By Proposition 4 we find $g=g'$ and hence $h=h'$.

Now, define the linear map $T:\varLambda _n^*\rightarrow \varLambda _n^*$ by $f\mapsto \mathrm {pr}\Delta (Q_2f).$ By Proposition 4 we find that T is injective, which by finite dimensionality of $\varLambda _n^*$ implies that T is surjective. Hence, given $f\in \varLambda _{n}^*$ let $g\in \varLambda _{n-2}^*$ be such that $T(g)=\mathrm {pr}\Delta (f) \in \varLambda _{n-2}^*$. Let $h=f-Q_2g$. As $f=Q_2g+h$, it suffices to show that $h\in \mathcal {H}$. That holds true because $\mathrm {pr}\Delta (h)=\mathrm {pr}\Delta (f)-\mathrm {pr}\Delta (Q_2g)=0$. $\square$

Proposition 5 implies Theorem 2 and the following corollary. Denote by p(n) the number of partitions of n.

Corollary 1

The dimension of $\mathcal {H}_n$ equals the number of partitions of n in parts of size at least 3, i.e.

$$\begin{aligned} \dim \mathcal {H}_n = p(n)-p(n-1)-p(n-2)+p(n-3). \end{aligned}$$

Proof

Observe that $\dim \varLambda _n^*$ equals the number of partitions of n in parts of size at least 2. Hence, $\dim \varLambda _n^*=p(n)-p(n-1)$ and the Corollary follows from Proposition 5. $\square$

Proof of Theorem 1

If $\langle f \rangle _q$ is modular, then $\langle \Delta f \rangle _q=\mathfrak {d}\langle f \rangle _q =0$. Write $f=\sum _{r=0}^{n'} Q_2^rh_{r}$ as in Theorem 2 with $n'=\lfloor \tfrac{n}{2}\rfloor$. Then by Lemma 1 it follows that $\mathrm {pr}\Delta f = \sum _{r=0}^{n'} r(n-r-\frac{3}{2})Q_2^{r-1}h_r.$ Hence,

$$\begin{aligned} \sum _{r=1}^{n'} r(n-r\tfrac{3}{2})\hat{D}^{r-1}\langle h_r\rangle _q=0. \end{aligned}$$

(7)

As $\langle h_r\rangle _q$ is modular, either it is equal to 0 or it has depth 0. Suppose the maximum m of all $r\ge 1$ such that $\langle h_r\rangle _q$ is non-zero exists. Then, by Lemma 2 it follows that the left-hand side of (7) has depth $m-1$, in particular is not equal to 0. So, $h_1,\ldots , h_{n'}\in \ker \langle \cdot \rangle _q$. Note that $f\in \ker \langle \cdot \rangle _q$ implies that $Q_2f\in \ker \langle \cdot \rangle _q$. Therefore, $k:=\sum _{r=1}^{n'} Q_2^rh_{r} \in \ker \langle \cdot \rangle _q$ and $f=h+k$ with $h=h_0$ harmonic.

The converse follows directly as $\mathfrak {d}\langle h+k\rangle _q = \mathfrak {d}\langle h \rangle _q=\langle \Delta h \rangle _q = 0.$$\square$

Remark 2

A description of the kernel of $\langle \cdot \rangle _q$ is not known.

Another corollary of Proposition 5 is the notion of depth of shifted symmetric polynomials which corresponds to the depth of quasimodular forms:

Definition 3

The space $\varLambda _k^{*(\le p)}$ of shifted symmetric polynomials of depth $\le p$ is the space of $f\in \varLambda _k^*$ such that one can write

$$\begin{aligned} f = \sum _{r=0}^{p} Q_2^rh_{r}, \end{aligned}$$

with $h_r\in \mathcal {H}_{k-2r}$.

Theorem 4

If $f\in \varLambda _k^{*(\le p)}$, then $\langle f \rangle _q \in \widetilde{M}_k^{(\le p)}$.

Proof

Expanding f as in Definition 3 we find

$$\begin{aligned} \langle f \rangle _q = \sum _{k=0}^{p} \langle Q_2^kh_{k} \rangle _q = \sum _{k=0}^{p} \hat{D}^k\langle h_{k}\rangle _q. \end{aligned}$$

By Lemma 2, we find that the depth of $\langle f\rangle _q$ is at most p. $\square$

Next, we set up notation to determine the basis of $\mathcal {H}$ given by Theorem 3. Let $\tilde{\mathcal {R}}=\mathcal {R}[Q_2^{-1/2}]$ and $\tilde{\varLambda }=\varLambda ^*[Q_2^{-1/2}]$ be the formal polynomial algebras graded by assigning to $Q_k$ weight k (note that the weights are—possibly negative—integers). Extend $\Delta$ to $\tilde{\varLambda }$ and observe that $\Delta (\tilde{\varLambda })\subset \tilde{\varLambda }$. Also extend $\mathcal {H}$ by setting

$$\begin{aligned} \tilde{\mathcal {H}}=\{f \in \tilde{\varLambda }\mid \Delta f\in Q_1\tilde{\mathcal {R}}\}=\ker \mathrm {pr}\Delta |_{\tilde{\varLambda }}. \end{aligned}$$

Definition 4

Define the partition-Kelvin transform$K:\tilde{\varLambda }_n\rightarrow \tilde{\varLambda }_{3-n}$ by

$$\begin{aligned} K(f)=Q_2^{3/2-n}f. \end{aligned}$$

Note that $K$ is an involution. Moreover, f is harmonic if and only if K(f) is harmonic, which follows directly from the computation

$$\begin{aligned} \Delta K(f)=Q_2^{3/2-n}\Delta f-(\tfrac{3}{2}-n)Q_1Q_2^{\tfrac{1}{2}-n}\varvec{\partial }{f}-\tfrac{1}{2}(\tfrac{3}{2}-n)(\tfrac{1}{2}-n)Q_1^2Q_2^{-\tfrac{1}{2}-n}f. \end{aligned}$$

Example 1

As $K(1)=Q_2^{3/2}$, it follows that $Q_2^{3/2}\in \tilde{\mathcal {H}}$.

Definition 5

Given $\underline{i}\in \mathbb {Z}_{\ge 0}^n$, let

$$\begin{aligned} |\underline{i}|=i_1+i_2+\ldots + i_n, \quad \quad \partial _{\underline{i}} = \frac{\partial ^n}{\partial Q_{i_1+1} \partial Q_{i_2+1}\cdots \partial Q_{i_n+1}}. \end{aligned}$$

Define the nth order differential operators $\mathscr {D}_n$ on $\tilde{\mathcal {R}}$ by

$$\begin{aligned} \mathscr {D}_n = \sum _{\underline{i}\in \mathbb {Z}_{\ge 0}^n} \left( {\begin{array}{c}|\underline{i}|\\ i_1,i_2,\ldots ,i_n\end{array}}\right) Q_{|\underline{i}|}\partial _{\underline{i}}, \end{aligned}$$

where the coefficient is a multinomial coefficient.

This definition generalises the operators $\varvec{\partial }$ and $\mathscr {D}$ to higher weights: $\mathscr {D}_1=\varvec{\partial }$, $\mathscr {D}_2=\mathscr {D}$, and $\mathscr {D}_n$ reduces the weight by n.

Lemma 3

The operators $\{\mathscr {D}_n\}_{n\in \mathbb {N}}$ commute pairwise.

Proof

Set $I=|\underline{i}|$ and $J=|\underline{j}|$. Let $\underline{a}^{\hat{k}}=(a_1,\ldots ,a_{k-1},a_{k+1},\ldots , a_n)$. Then

$$\begin{aligned} \begin{aligned}&\left[ \left( {\begin{array}{c}I\\ i_1,i_2,\ldots ,i_n\end{array}}\right) Q_{I}\partial _{\underline{i}}, \left( {\begin{array}{c}J\\ j_1,j_2,\ldots ,j_m\end{array}}\right) Q_{J}\partial _{\underline{j}} \right] \\&\quad = \sum _{k=1}^n \delta _{i_k,J-1}J \left( {\begin{array}{c}I\\ i_1,i_2,\ldots ,\hat{i_k},\ldots ,i_n,j_1,j_2,\ldots ,j_m\end{array}}\right) Q_I \partial _{\underline{i}^{\hat{k}}}\partial _{\underline{j}} +\\ {}&\qquad -\sum _{l=1}^m \delta _{j_l,I-1} I \left( {\begin{array}{c}J\\ i_1,i_2,\ldots ,i_n,j_1,j_2,\ldots ,\hat{j_l},\ldots ,j_m\end{array}}\right) Q_J \partial _{\underline{i}}\partial _{\underline{j}^{\hat{l}}}. \end{aligned} \end{aligned}$$

(8)

Hence, $[\mathscr {D}_n,\mathscr {D}_m]$ is a linear combination of terms of the form $Q_{|\underline{a}|+1} \partial _{\underline{a}},$ where $\underline{a}\in \mathbb {Z}_{\ge 0}^{n+m-1}$. We collect all terms for different vectors $\underline{a}$ which consists of the same parts (i.e. we group all vectors $\underline{a}$ which correspond to the same partition). Then, the coefficient of such a term equals

$$\begin{aligned}&\sum _{k=1}^n \sum _{\sigma \in S_{m+n-1}} (a_{\sigma (1)}+\ldots + a_{\sigma (m)})\left( {\begin{array}{c}|\underline{a}|+1\\ a_1,a_2,\ldots , a_{n+m-1}\end{array}}\right) \\&\quad - \sum _{l=1}^m \sum _{\sigma \in S_{m+n-1}} (a_{\sigma (1)}+\ldots + a_{\sigma (n)})\left( {\begin{array}{c}|\underline{a}|+1\\ a_1,a_2,\ldots , a_{n+m-1}\end{array}}\right) \\&\quad = (mn-mn) \sum _{\sigma \in S_{m+n-1}} a_{\sigma (1)}\left( {\begin{array}{c}|\underline{a}|+1\\ a_1,a_2,\ldots , a_{n+m-1}\end{array}}\right) =0. \end{aligned}$$

Hence, $[\mathscr {D}_n,\mathscr {D}_m]=0$. $\square$

It does not hold true that $[\mathscr {D}_n,Q_1]=0$ for all $n\in \mathbb {N}$. Therefore, we introduce the following operators:

Definition 6

Let

$$\begin{aligned} \Delta _n=\sum _{i=0}^n (-1)^i \left( {\begin{array}{c}n\\ i\end{array}}\right) \mathscr {D}_{n-i} \varvec{\partial }^i . \end{aligned}$$

For $\lambda \in \mathscr {P}$ let

$$\begin{aligned} \Delta _\lambda = \left( {\begin{array}{c}|\lambda |\\ \lambda _1,\ldots ,\lambda _{\ell (\lambda )}\end{array}}\right) \prod _{i=1}^\infty \Delta _{\lambda _i}. \end{aligned}$$

(Note that $\Delta _0=\mathscr {D}_0=1$, so this is in fact a finite product.)

Remark 3

By Möbius inversion

$$\begin{aligned} \mathscr {D}_n=\sum _{i=0}^n \left( {\begin{array}{c}n\\ i\end{array}}\right) \Delta _{n-i}\varvec{\partial }^i. \end{aligned}$$

The first three operators are given by

$$\begin{aligned} \Delta _0=1,\quad \Delta _1=0,\quad \Delta _2=\mathscr {D}-\varvec{\partial }^2=2\Delta . \end{aligned}$$

Proposition 6

The operators $\Delta _\lambda$ satisfy the following properties: for all partitions $\lambda ,\lambda '$

• (a) the order of $\Delta _{|\lambda |}$ is $|\lambda |;$

• (b) $[\Delta _\lambda ,\Delta _{\lambda '}]=0;$

• (c) $[\Delta _\lambda ,Q_1]=0$.

Proof

Property (a) follows by construction and (b) is a direct consequence of Lemma 3. For property (c), let $f\in \tilde{\varLambda }$ be given. Then

$$\begin{aligned} \Delta _n(Q_1 f)&=\sum _{i=0}^n (-1)^i \left( {\begin{array}{c}n\\ i\end{array}}\right) \mathscr {D}_{n-i} \varvec{\partial }^i(Q_1 f) \\&=\sum _{i=0}^n (-1)^i \left( {\begin{array}{c}n\\ i\end{array}}\right) \left( (n-i)\mathscr {D}_{n-i-1}\varvec{\partial }^{i} f + Q_1\mathscr {D}_{n-i}\varvec{\partial }^{i} f + i \mathscr {D}_{n-i}\varvec{\partial }^{i-1} f\right) \\&= Q_1 \Delta _n(f)+\sum _{i=0}^n (-1)^i \left( {\begin{array}{c}n\\ i\end{array}}\right) \left( (n-i)\mathscr {D}_{n-i-1}\varvec{\partial }^{i} f + i\mathscr {D}_{n-i}\varvec{\partial }^{i-1} f\right) . \end{aligned}$$

Observe that by the identity

$$\begin{aligned} (n-i)\left( {\begin{array}{c}n\\ i\end{array}}\right) =(i+1)\left( {\begin{array}{c}n\i +1\end{array}}\right) , \end{aligned}$$

the sum in the last line is a telescoping sum, equal to zero. Hence $\Delta _n(Q_1f)=Q_1\Delta _n(f)$ as desired. $\square$

In particular, the above proposition yields $[\Delta _\lambda , \Delta ]=0$ and $[\Delta _\lambda ,\mathrm {pr}]=0$.

Denote by $(x)_{n}$ the falling factorial power$(x)_n = \prod _{i=0}^{n-1} (x-i)$ and for $\lambda \in \mathscr {P}_n$ define $Q_\lambda =\prod _{i=1}^\infty Q_{\lambda _i}.$ Let

$$\begin{aligned} h_\lambda =\mathrm {pr}K\Delta _\lambda K(1). \end{aligned}$$

Observe that $h_\lambda$ is harmonic, as $\mathrm {pr}\Delta$ commutes with $\mathrm {pr}$ and $\Delta _\lambda$.

Proposition 7

For all $\lambda \in \mathscr {P}_n$ there exists an $f\in \varLambda ^*_{n-2}$ such that

$$\begin{aligned} h_\lambda = (\tfrac{3}{2})_n n! Q_\lambda + Q_2 f. \end{aligned}$$

Proof

Note that the left-hand side is an element of $\varLambda ^*$ of which the monomials divisible by $Q_2^i$ correspond precisely to terms in $\Delta _\lambda$ involving precisely $n-i$ derivatives of $K(1)$ to $Q_2$. Hence, as $\Delta _\lambda$ has order n all terms not divisible by $Q_2$ correspond to terms in $\Delta _\lambda$ which equal $\frac{\partial ^{n} }{\partial Q_2^{n}}$ up to a coefficient. There is only one such term in $\Delta _{\lambda }$ with coefficient $\left( {\begin{array}{c}|\lambda |\\ \lambda _1,\ldots ,\lambda _r\end{array}}\right) \lambda _1!\ldots \lambda _r!Q_{\lambda }$. $\square$

For $f\in \mathcal {R}$, we let $f^\vee$ be the operator where every occurrence of $Q_i$ in f is replaced by $\Delta _i$. We get the following unusual identity:

Corollary 2

If $h\in \mathcal {H}_n$, then

$$\begin{aligned} h=\frac{\mathrm {pr}Kh^\vee K(1)}{n!(\tfrac{3}{2})_n}. \end{aligned}$$

(9)

Proof

By Proposition 7, we know that the statement holds true up to adding $Q_2 f$ on the right-hand side for some $f\in \varLambda ^*_{n-2}$. However, as both sides of (9) are harmonic and the shifted symmetric polynomial $Q_2f$ is harmonic precisely if $f=0$ by Proposition 4, it follows that $f=0$ and (9) holds true. $\square$

Proof of Theorem 3

Let $\mathcal {B}_n=\{h_\lambda \mid \lambda \in \mathscr {P}_n \text { all parts are } \ge 3\}$. First of all, observe that by Corollary 1 the number of elements in $\mathcal {B}_n$ is precisely the dimension of $\mathcal {H}_n$. Moreover, the weight of an element in $\mathcal {B}_n$ equals $|\lambda |=n$. By Proposition 7 it follows that the elements of $\mathcal {B}_n$ are linearly independent harmonic shifted symmetric polynomials. $\square$

Appendix: Tables of shifted symmetric harmonic polynomials up to weight 10

We list all harmonic polynomials $h_{\lambda }$ of even weight at most 10. The corresponding q-brackets $\langle h_\lambda \rangle _q$ are computed by the algorithm prescribed by Zagier [13] using SageMath [11].

$\lambda$	$h_{\lambda }$	$\langle h_{\lambda }\rangle _q$
()	1	1
(4)	$\frac{27}{4} \left( Q_2^2+2 Q_4\right)$	$\frac{9}{320}Q$
(6)	$\frac{225}{4} \left( 63 Q_6+9 Q_2 Q_4+Q_2^3\right)$	$-\frac{55}{384}R$
(3,3)	$\frac{225}{4} \left( 63 Q_3^2-108 Q_2 Q_4+2 Q_2^3\right)$	$\frac{115}{384}R$
(8)	$\frac{19845}{16} \left( 3960 Q_8+360 Q_2 Q_6+20 Q_2^2 Q_4+Q_2^4\right)$	$\frac{19173}{4096} Q^2$
(5,3)	$\frac{19845}{2} \left( 495 Q_3Q_5 + 45 Q_2Q_3^2-1350 Q_2 Q_6 -50 Q_2^2Q_4 +2 Q_2^4\right)$	$-\frac{2415}{128} Q^2$
(4,4)	$\frac{297675}{8} \left( 132 Q_4^2 + 24 Q_2 Q_3^2-440Q_2 Q_6-28 Q_2^2Q_4 +Q_2^4\right)$	$-\frac{38241}{2048} Q^2$
(10)	$\frac{382725}{8} \left( 450450 Q_{10} + 30030 Q_2 Q_8 + 1155 Q_2^2 Q_6 + 35 Q_2^3 Q_4 + Q_2^5\right)$	$-\frac{2053485}{4096} QR$
(7,3)	$\frac{1913625}{8} \Big (90090 Q_3 Q_7 + 6006 Q_2 Q_3 Q_5 - 336336 Q_2 Q_8 + 231 Q_2 Q_3^2 +$
	$\quad \quad \quad \quad - 12936 Q_2^2Q_6 -112 Q_2^3Q_4 + 10 Q_2^5\big )$	$\frac{11975985}{4096} QR$
(6,4)	$\frac{13395375}{8} \Big ( 12870 Q_4 Q_6 + 1716 Q_2 Q_3 Q_5 + 858 Q_2 Q_4^2 -96096 Q_2 Q_8 +$
	$\quad \quad \quad \quad + 132 Q_2^2Q_3^2 - 6501 Q_2^2 Q_6 -89 Q_2^3 Q_4+5 Q_2^4\Big )$	$\frac{21255885}{4096} QR$
(5,5)	$\frac{8037225}{4} \Big (10725 Q_5^2+1430 Q_2Q_3Q_5 + 1430 Q_2 Q_4^2-10010 Q_2 Q_8 +$
	$\quad \quad \quad \quad + 165 Q_2^2Q_3^2 - 7700 Q_2^2 Q_6 -120 Q_2^3Q_4+6 Q_2^5\Big )$	$\frac{7759395}{1024} QR$
(4,3,3)	$\frac{13395375}{8} \Big (12870 Q_3^2Q_4-34320 Q_2Q_3Q_5+10296Q_2 Q_4^2+363Q_2^2Q_3^2+$
	$\quad \quad \quad \quad +55440Q_2^2 Q_6-376Q_2^3Q_4+10Q_2^5\Big )$	$-\frac{16583805}{4096} QR$

In case $|\lambda |$ is odd, the harmonic polynomials $h_{\lambda }$ up to weight 9 are given in the following table. The q-bracket of odd degree (harmonic) polynomials is zero, hence trivially modular.

$\lambda$	$h_{\lambda }$
(3)	$-\frac{9}{4}Q_3$
(5)	$-\frac{135}{4}\left( 5Q_5+Q_2Q_3\right)$
(7)	$-\frac{14175}{16}\left( 126Q_7+14Q_2Q_5+Q_2^2Q_3\right)$
(4, 3)	$-\frac{99225}{16}\left( 18Q_3Q_4 - 40 Q_2 Q_5+Q_2^2Q_3\right)$
(9)	$-\frac{297675}{8}\left( 7722 Q_9 + 594Q_2 Q_7 +27 Q_2^2Q_5+Q_2^3Q_3\right)$
(6, 3)	$-\frac{893025}{4}\left( 1287 Q_3 Q_6 +99 Q_2Q_3Q_4-4158 Q_2Q_7-162 Q_2^2Q_5+5Q_2^3Q_3\right)$
(5, 4)	$- \frac{8037225}{8}\left( 286 Q_4Q_5+66 Q_2Q_3Q_4-1540 Q_2Q_7 -117 Q_2^2Q_5+3Q_2^3Q_3\right)$
(3, 3, 3)	$-\frac{893025}{4}\left( 1287 Q_3^3-3564Q_2Q_3Q_4+3240Q_2^2Q_5+10Q_2^3Q_3\right)$