2.1 The proof of identities

Now we prove Theorem 1.2 for one case $I=\{1,\cdots ,r\}$ by constructing the following symmetric complex rational function $f(w_{1},\cdots ,w_{d})$ with parameters q and $x_1,\cdots ,x_n$ . We made the following assumptions for parameters:

(2.1) $$ \begin{align} &|q|<1, \nonumber \\ x_i x_j^{-1} \neq q^k, &\qquad \forall i\neq j \in [n],\forall k\in \mathbb{Z}. \end{align} $$

Furthermore, there exists some $\rho>0$ such that

(2.2) $$ \begin{align} \max_{i\in [n]} |x_i|<\rho <\min_{i\in [n]} |q|^{-1}|x_i|, \end{align} $$

where $[n]:=\{1, \cdots , n \}$ and general situations follow from analytic continuation. Let $f(w_{1},\cdots ,w_{d})$ be as follows:

(2.3) $$ \begin{align} f(w_{1},\cdots ,w_{d}) &=\frac{1}{(1-q)^d d!}\prod_{i \neq j}^d \frac{w_{i}-w_{j}}{w_{i}-qw_{j}}\prod_{i=1}^d \frac{w_i^{l-1}}{\prod_{j=1}^{r}(1-x_j/w_{i}) \prod_{j=r+1}^{n}(1-qw_{i}/x_{j})} \end{align} $$

(2.4) $$ \begin{align} &\hspace{12pt}=g(w_{1},\cdots,w_{d})\prod_{i=1}^{d}\left(\prod_{u \in U}\frac{w_i - q^{-1}u}{w_i - u}\prod_{i<j}\frac{(w_i-w_j)^2}{(w_i-qw_j)(qw_i - w_j)}\right), \hspace{-12pt}\end{align} $$

where U is a set of complex numbers all contained in open disk $|w|<\rho $ , at the moment $U = \{x_1,\cdots ,x_r\}$ , and g is a symmetric function of the form

$$ \begin{align*} g(\vec{w})=\frac{1}{(1-q)^d d!}\prod_{i=1}^d \frac{w_i^{l+r-1}}{ \prod_{j=1}^{r}(w_i - q^{-1}x_j)\prod_{j=r+1}^{n}(1-qw_{i}/x_{j})}. \nonumber \end{align*} $$

From the condition in inequality (2.2) and the restriction of l, we know $g(\vec {w})$ is analytical in the polydiscs $\{(w_1,\cdots ,w_n):|w_i| \le \rho , \forall i \in [n]\}$ and g can only have possible zeros for some $w_j = 0$ .

We consider the following integration

(2.5) $$ \begin{align} E_d:=&\int_{C_{\rho}} \frac{dw_{d}}{2\pi\sqrt{-1}} \ldots \int_{C_{\rho}} \frac{dw_{1}}{2\pi\sqrt{-1}} f(\hat{w}_{1},\cdots,\hat{w}_{d}), \end{align} $$

where $(\hat {w}_{1},\cdots ,\hat {w}_{d})$ is any arrangement of $\{w_1,\cdots ,w_d\}$ and the integration contour $C_{\rho }$ for each variable $w_i$ is the circle centred at origin with radius $\rho $ and takes a counterclockwise direction. The condition in inequality (2.2) ensures that there isn’t a pole on the integration contour. By Fubini’s theorem, we could permute the order of integration variables; and since $f(w_{1},\cdots ,w_{d})$ is a symmetric function, we can change $(w_{1},\cdots , w_{d})$ to another order, such as $(\hat {w}_{1},\cdots ,\hat {w}_{d})$ .

Suppose we have the following evaluating sequence for some $S_1 \le d$ by induction,

$$ \begin{align*} \hat{w}_1 = q\hat{w}_2, \hat{w}_2 = q\hat{w}_3, \cdots, \hat{w}_{S_1-1} = q\hat{w}_{S_1}, \nonumber \end{align*} $$

which are all simple poles inside $|w|<\rho $ . Then we have

(2.6) $$ \begin{align} &\underset{\hat{w}_{S_1-1} = q \hat{w}_{S_1}}{\operatorname{Res}} \cdots \underset{\hat{w}_2 = q \hat{w}_3}{\operatorname{Res}}\underset{\hat{w}_1 = q \hat{w}_2}{\operatorname{Res}}f \nonumber \\=& \prod_{i=S_{1}+1}^{d}\left(\prod_{u \in U}\frac{\hat{w}_i - q^{-1}u}{\hat{w}_i - u}\prod_{i<j}\frac{(\hat{w}_i-\hat{w}_j)^2}{(\hat{w}_i-q\hat{w}_j)(q\hat{w}_i - \hat{w}_j)}\right) \nonumber \\ &\times \hat{w}_{S_1}^{S_{1}-1} \prod_{k=0}^{S_{1}-1}\prod_{u \in U}\frac{q^{k}\hat{w}_{S_1} - q^{-1}u}{q^{k}\hat{w}_{S_1} - u} \cdot \prod_{S_1<j}\frac{(\hat{w}_{S_1}-\hat{w}_j)(q^{S_{1}-1}\hat{w}_{S_{1}}-\hat{w}_j)}{(\hat{w}_{S_1}-q\hat{w}_j)(q^{S_1}\hat{w}_{S_1} - \hat{w}_j)} \nonumber \\ &\times \frac{(q-1)^{S_{1}}}{q^{S_{1}}-1} q^{-({S_{1}-1})(d-{S_{1}})}g(q^{{S_{1}-1}}\hat{w}_{S_{1}},q^{{S_{1}}-2}\hat{w}_{S_{1}},\cdots,\hat{w}_{S_1},\hat{w}_{S_1+1},\cdots,\hat{w}_d). \end{align} $$

Now, integrating variable $\hat {w}_{S_1}$ , we pick up the residue as $\hat {w}_{S_1} = q^{-k_1}u_1$ for some $0\le k_1< S_1$ and $u_1 \in U=\{x_1,\cdots ,x_r\}$ because the condition in equation (2.2), $|\hat {w}_{S_1}|< \rho $ , implies that $k_1=0$ . Evaluating $\hat {w}_{S_1} = u_1$ , we obtain

(2.7) $$ \begin{align} & \underset{\hat{w}_{S_1} = u_1}{\operatorname{Res}} \,\underset{\hat{w}_{S_1-1} = q \hat{w}_{S_1}}{\operatorname{Res}} \cdots \underset{\hat{w}_2 = q \hat{w}_3}{\operatorname{Res}} \, \underset{\hat{w}_1 = q \hat{w}_2}{\operatorname{Res}}f \nonumber \\=& \prod_{i=S_{1}+1}^{d}\left( \frac{\hat{w}_i-q^{S_1-1}u_1}{\hat{w}_i-q^{S_1}u_1}\prod_{u \in U\backslash\{u_1\}}\frac{\hat{w}_i - q^{-1}u}{\hat{w}_i - u}\prod_{i<j}\frac{(\hat{w}_i-\hat{w}_j)^2}{(\hat{w}_i-q\hat{w}_j)(q\hat{w}_i - \hat{w}_j)}\right) \nonumber \\ &\times u_{1}^{S_1}\prod_{ k = 0}^{S_1 - 1} \prod_{u \in U \backslash \{u_1\} } \frac{q^k u_1-q^{-1}u}{q^k u_1-u} \cdot (q-1)^{S_1} q^{S_1 (S_1 -1 - d)} \nonumber \\ &\times g(q^{{S_{1}-1}}u_1,q^{{S_{1}}-2}u_1,\cdots,qu_1,u_1,\hat{w}_{S_1+1},\cdots,\hat{w}_d) \end{align} $$

(2.8) $$ \begin{align}=\;& \tilde{g}(\hat{w}_{S_1+1},\cdots,\hat{w}_{d})\prod_{i=S_1+1}^{d}\left(\prod_{u \in \tilde{U}}\frac{\hat{w}_i - q^{-1}u}{\hat{w}_i - u}\prod_{i<j}\frac{(\hat{w}_i-\hat{w}_j)^2}{(\hat{w}_i-q\hat{w}_j)(q\hat{w}_i - \hat{w}_j)}\right), \end{align} $$

where

(2.9) $$ \begin{align} \tilde{U} = U \backslash \{u_1\} \cup \{q^{S_1}u_1\}. \end{align} $$

All elements of $\tilde {U}$ are still in the open disk $|w|<\rho $ , and

(2.10) $$ \begin{align} \tilde{g}(\hat{w}_{S_1+1},\cdots,\hat{w}_{d}) =\;& u_{1}^{S_1}\prod_{ k = 0}^{S_1 - 1} \prod_{u \in U \backslash \{u_1\} } \frac{q^k u_1-q^{-1}u}{q^k u_1-u} \cdot (q-1)^{S_1} q^{S_1 (S_1 -1 - d)} \nonumber \\ & \times g(q^{{S_{1}-1}}u_1,q^{{S_{1}}-2}u_1,\cdots,qu_1,u_1,\hat{w}_{S_1+1},\cdots,\hat{w}_d). \end{align} $$

So we just write $\tilde {f} := \underset {\hat {w}_{S_1} = u}{\operatorname {Res}} \,\underset {\hat {w}_{S_1-1} = q \hat {w}_{S_1}}{\operatorname {Res}} \cdots \underset {\hat {w}_2 = q \hat {w}_3}{\operatorname {Res}} \, \underset {\hat {w}_1 = q \hat {w}_2}{\operatorname {Res}}f$ in the same pattern as in the original form in equation (2.4). One could check that setting $S_1 = 1$ in equation (2.7) is valid.

If one takes the following evaluation sequence of simple poles by induction

(2.11) $$ \begin{align} \hat{w}_1 = u_1, \hat{w}_2 = q u_1, \cdots, \hat{w}_{S_1-1} = q^{S_1-2}u_1,\hat{w}_{S_1} = q^{S_1-1} u_1,\\[-15pt]\nonumber \end{align} $$

we get

(2.12) $$ \begin{align} &\underset{\hat{w}_{S_1} = q^{S_1-1}u_1}{\operatorname{Res}} \cdots \underset{\hat{w}_{2} = qu_1}{\operatorname{Res}}\underset{\hat{w}_{1} = u_1}{\operatorname{Res}}f = \prod_{S_1<i}^{d}\left( \frac{\hat{w}_i - q^{S_1-1} u_1}{\hat{w}_i- q^{S_1} u_1}\prod_{u \in U\backslash\{u_1\}}\frac{\hat{w}_i - q^{-1}u}{\hat{w}_i - u}\prod_{i<j}\frac{(\hat{w}_i-\hat{w}_j)^2}{(\hat{w}_i-q\hat{w}_j)(q\hat{w}_i - \hat{w}_j)}\right) \nonumber \\ & \times u_{1}^{S_1}\prod_{ k = 0}^{S_1 - 1} \prod_{u \in U \backslash \{u_1\} } \frac{q^k u_1-q^{-1}u}{q^k u_1-u} \cdot (q-1)^{S_1} q^{S_1 (S_1 -1 - d)} g(u_1,qu_1,\cdots,q^{S_1 - 1}u_1, \hat{w}_{S_1 + 1},\cdots \hat{w}_d),\\[-15pt]\nonumber \end{align} $$

which agrees with equation (2.7), since $g(\vec {w})$ is a symmetric function. That is to say, we get the same results from two different evaluation sequences

(2.13) $$ \begin{align} \underset{\hat{w}_{S_1} = q^{S_1-1}u_1}{\operatorname{Res}} \cdots \underset{\hat{w}_{2} = qu_1}{\operatorname{Res}}\underset{\hat{w}_{1} = u_1}{\operatorname{Res}}f = \underset{\hat{w}_{S_1} = u_1}{\operatorname{Res}} \,\underset{\hat{w}_{S_1-1} = q \hat{w}_{S_1}}{\operatorname{Res}} \cdots \underset{\hat{w}_2 = q \hat{w}_3}{\operatorname{Res}} \, \underset{\hat{w}_1 = q \hat{w}_2}{\operatorname{Res}}f.\\[-15pt]\nonumber \end{align} $$

As with the evaluation process for the sequence in equation (2.6), we now pick up the residues of $\tilde {f}$ in the following sequence:

(2.14) $$ \begin{align} \hat{w}_{S_{1}+1} = q \hat{w}_{S_1 +2} \quad \hat{w}_{S_{1}+2} = q \hat{w}_{S_1 +3}\quad \cdots \quad \hat{w}_{S_{1}+S_{2}-1} = q \hat{w}_{S_1 +S_2}.\\[-15pt]\nonumber \end{align} $$

Suppose $\hat {w}_{S_1+S_2}=u_2$ . We have two cases here: $u_2 \ne q^{S_1} u_1$ or $u_2 = q^{S_1} u_1$ . With a little computation, we obtain the following.

Case 1: $u_2 \ne q^{S_1} u_1$ ,

(2.15) $$ \begin{align} &\underset{\hat{w}_{S_1+S_2} = u_2}{\operatorname{Res}} \, \underset{\hat{w}_{S_1+S_2-1} = q \hat{w}_{S_1+S_2}}{\operatorname{Res}} \cdots \underset{\hat{w}_{S_{1}+2} = q \hat{w}_{S_{1}+3}}{\operatorname{Res}}\,\underset{\hat{w}_{S_{1}+1} = q \hat{w}_{S_{1}+2}}{\operatorname{Res}} \, \underset{\hat{w}_{S_1} = u_1}{\operatorname{Res}} \,\underset{\hat{w}_{S_1-1} = q \hat{w}_{S_1}}{\operatorname{Res}} \cdots \underset{\hat{w}_2 = q \hat{w}_3}{\operatorname{Res}} \, \underset{\hat{w}_1 = q \hat{w}_2}{\operatorname{Res}}f \nonumber \\ =&\underset{\hat{w}_{S_1+S_2} = u_1}{\operatorname{Res}} \, \underset{\hat{w}_{S_1+S_2-1} = q \hat{w}_{S_1+S_2}}{\operatorname{Res}} \cdots \underset{\hat{w}_{S_{2}+2} = q \hat{w}_{S_{2}+3}}{\operatorname{Res}}\,\underset{\hat{w}_{S_{2}+1} = q \hat{w}_{S_{2}+2}}{\operatorname{Res}} \, \underset{\hat{w}_{S_2} = u_2}{\operatorname{Res}} \,\underset{\hat{w}_{S_2-1} = q \hat{w}_{S_2}}{\operatorname{Res}} \cdots \underset{\hat{w}_2 = q \hat{w}_3}{\operatorname{Res}} \, \underset{\hat{w}_1 = q \hat{w}_2}{\operatorname{Res}}f.\\[-15pt]\nonumber \end{align} $$

Case 2: $u_2 = q^{S_1} u_1$ ,

(2.16) $$ \begin{align} &\underset{\hat{w}_{S_1+S_2} = q^{S_1} u_1}{\operatorname{Res}} \, \underset{\hat{w}_{S_1+S_2-1} = q \hat{w}_{S_1+S_2}}{\operatorname{Res}} \cdots \underset{\hat{w}_{S_{1}+2} = q \hat{w}_{S_{1}+3}}{\operatorname{Res}}\,\underset{\hat{w}_{S_{1}+1} = q \hat{w}_{S_{1}+2}}{\operatorname{Res}} \, \underset{\hat{w}_{S_1} = u_1}{\operatorname{Res}} \,\underset{\hat{w}_{S_1-1} = q \hat{w}_{S_1}}{\operatorname{Res}} \cdots \underset{\hat{w}_2 = q \hat{w}_3}{\operatorname{Res}} \, \underset{\hat{w}_1 = q \hat{w}_2}{\operatorname{Res}}f \nonumber \\ =& \underset{\hat{w}_{S_1+S_2} = q^{S_1 + S_2 -1} u_1}{\operatorname{Res}} \, \underset{\hat{w}_{S_1+S_2-1} = q^{S_1 + S_2 -2} u_1}{\operatorname{Res}} \cdots \underset{\hat{w}_{S_{1}+2} = q^{S_1 + 1} u_1}{\operatorname{Res}}\,\underset{\hat{w}_{S_{1}+1} = q^{S_1} u_1}{\operatorname{Res}} \, \underset{\hat{w}_{S_1} = q^{S_1 -1}u}{\operatorname{Res}} \cdots \underset{\hat{w}_2 = q u_1}{\operatorname{Res}} \, \underset{\hat{w}_1 = u_1}{\operatorname{Res}}f.\\[-15pt]\nonumber \end{align} $$

To summarise all of the above, together with equations (2.13), (2.15) and (2.16), we know that the iterated residue does not depend on the order of the poles we pick but depends on the final set of poles we choose. So we can integrate all variables for the integrand of the form as in equation (2.4) with one less variable each time.

When there is only one variable left

(2.17) $$ \begin{align} f(w) = g(w)\prod_{u \in U} \frac{w-q^{-1}u}{w - u},\\[-15pt]\nonumber \end{align} $$

we still update the set U to $U\backslash \{u\} \cup \{qu\}$ after choosing a pole at $\hat {w} = u \in U$ . Using the same argument to get equations (2.15) and (2.16) after picking up poles for all $w_i, i \in [d]$ , the result only depends on the final set U, which is of the form

(2.18) $$ \begin{align} \{q^{d_1} x_1, \cdots , q^{d_r}x_r\}, \end{align} $$

where $d_1 + \cdots + d_r = d$ , which means for each sequence, the final result can be indexed by a r-tuple partition of d.

Suppose there is a sequence with final set $\{q^{d_1} x_1, \cdots , q^{d_r}x_r\}$ . Then we can compute the result with the following sequence:

(2.19) $$ \begin{align} (\hat{w}_1,\cdots,\hat{w}_d) = (x_1,qx_1\cdots,q^{d_1-1}x_1,x_2,\cdots, q^{d_2-1}x_2,x_r, \cdots,q^{d_r-1}x_r). \end{align} $$

Note that we can do permutations on the left side, so for each partition $|\vec {d}| = d$ , we have $d!$ possible evaluation sequences.

In all, we obtain the following lemma to compute $E_d$ .

Lemma 2.1. We can write E as

(2.20) $$ \begin{align} E_d =\sum_{|\vec{d}|=d}d!E_{\vec{d}} , \end{align} $$

where

(2.21) $$ \begin{align} E_{\vec{d}}= \lim_{w_{d}\rightarrow \hat{w}_{d} }\cdots\lim_{w_{1}\rightarrow \hat{w}_{1} }\left(\prod_{i=1}^{n}(w_i-\hat{w}_{i})f(\vec{w})\right ), \end{align} $$

here

$$ \begin{align*} (\hat{w}_{1},\dots,\hat{w}_{d})=(x_1,qx_1,\dots,q^{d_1-1}x_1,x_2,q x_2,\dots,q^{d_2-1}x_2, \dots,x_r,\dots,q^{d_r -1}x_r), \nonumber \end{align*} $$

and the order in which to take the limit is from $w_1$ to $w_d$ .

We now evaluate one specific configuration of these simple pole residues for given $\vec d$ by changing variables:

$$ \begin{align*} w_{i,n_i}=x_i q^{n_i - 1}z_{i,n_i}, \quad\quad i=1,\dots,r\quad n_i=1,\dots,d_i. \nonumber \end{align*} $$

Notations: From now on, we frequently use the following notations:

(2.22) $$ \begin{align} x_{ij} := x_i / x_j \ \ \ \ \ n_{ij} := n_i-n_j. \end{align} $$

Then

$$ \begin{align*} f(\vec{w})=& \frac{1}{(1-q)^{d} d!\prod_{i,n_i}z_{i,n_i}}\cdot \displaystyle \prod_{i=1}^{r}\prod_{n_i\neq n_j}^{d_i} \frac{1-q^{n_{ij}} z_{i,n_i}/z_{i,n_j}}{1-q^{n_{ij}+1} z_{i,n_i}/z_{i,n_j}} \nonumber \\ & \times \prod_{i,j=1|i\neq j}^{r}\prod_{n_i=1}^{d_i}\prod_{n_j=1}^{d_j}\frac{1-q^{n_{ij}} z_{i,n_i}/z_{j,n_j}x_{ij}}{1-q^{n_{ij}+1} z_{i,n_i}/z_{j,n_j}x_{ij}} \nonumber \\ &\times \frac{\prod_{i}^{r}\prod_{n_i=1}^{d_i}(x_i q^{n_i - 1}z_{i,n_i})^{l}}{ \prod_{i,j=1|i\neq j}^{r}\prod_{n_i=1}^{d_i} (1-x_{ji}q^{1-n_i}/z_{i,n_i})} \nonumber \\ &\times \frac{1}{ \prod_{i=1}^{r}\prod_{n_i=1}^{d_i} (1-q^{1-n_i}/z_{i,n_i})} \cdot \frac{1}{ \prod_{i=1}^{r}\prod_{j=r+1}^{n}\prod_{n_i=1}^{d_i} (1-x_{ij}q^{n_i}z_{i,n_i})}. \nonumber \end{align*} $$

Now we pick up the simple pole terms and evaluate the function with $z_{i,j}=1$ . Note that

$$ \begin{align*} \lim_{z_{i,d_i}\rightarrow 1 }\cdots\lim_{z_{i,1}\rightarrow 1 }\left(\prod_{n_i=1}^{d_i}(z_{i,n_i} - 1 ) \cdot \frac{1}{\left(1-(z_{i,1})^{-1}\right)\left(1-z_{i,1}/z^i_2\right)\ldots \left(1-z_{i,d_i-1}/z_{i,d_i}\right) z_{i,1} \cdots z_{i,d_i}}\right ) =1, \nonumber \end{align*} $$

where the order in which to take the limit is from $z_{i,1}$ to $z_{i,d_i}$ . So this specific configuration of residues is

$$ \begin{align*} &\frac{1}{(1-q)^{d} d!}\cdot \displaystyle \prod_{i=1}^{r}\left(\prod_{n_i\neq n_j|n_{ij}\neq -1}^{d_i} \frac{1-q^{n_{ij}} }{1-q^{n_{ij}+1} } \cdot \prod_{n_i=2}^{d_i}\frac{1-q^{-1}}{1-q^{1-n_i}} \right) \nonumber \\ &\times \prod_{i,j=1|i\neq j}^{r}\prod_{n_i=1}^{d_i}\prod_{n_j=1}^{d_j}\frac{1-q^{n_{ij}} x_{ij}}{1-q^{n_{ij}+1} x_{ij}} \nonumber \\ &\times \frac{\prod_{i}^{r}\prod_{n_i=1}^{d_i}(x_i q^{n_i - 1})^{l}}{ \prod_{i,j=1|i\neq j}^{r}\prod_{n_i=1}^{d_i} (1-x_{ji}q^{1-n_i})} \cdot \frac{1}{ \prod_{i=1}^{r}\prod_{j=r+1}^{n}\prod_{n_i=1}^{d_i} (1-x_{ij}q^{n_i})}, \nonumber \end{align*} $$

and the factor with only $x_{ij}$ for $i=1,\ldots ,r$ and $j=r+1,\ldots ,n$ is

$$ \begin{align*} A:=\frac{1}{ \prod_{i=1}^{r}\prod_{j=r+1}^{n}\prod_{n_i=1}^{d_i} (1-x_{ij}q^{n_i})}=\frac{1}{\prod_{i=1}^{r}\prod_{j=r+1}^{n}(qx_{ij};q)_{d_i}}. \nonumber \end{align*} $$

The factor that does not involve any $x_{ij}$ is

$$ \begin{align*} B:=\frac{1}{(1-q)^{d}}\cdot \displaystyle \prod_{i=1}^{r}\left(\prod_{n_i\neq n_j|n_{ij}\neq -1}^{d_i} \frac{1-q^{n_{ij}} }{1-q^{n_{ij}+1} } \cdot \prod_{n_i=2}^{d_i}\frac{1-q^{-1}}{1-q^{1-n_i}} \right). \nonumber \end{align*} $$

And define $P_d$ as

$$ \begin{align*} P_d: = \left\{ \begin{array}{rcl} & \displaystyle \prod_{i\neq j|i-j\neq -1}^{d} \frac{1-q^{i-j} }{1-q^{i-j+1} } \cdot \prod_{i=2}^{d_i}\frac{1-q^{-1}}{1-q^{1-i}} \quad\quad & d>1 \\ & 1 & d=0,1 \end{array}. \right. \end{align*} $$

By simple induction, it is easy to show that

$$ \begin{align*} \frac{P_d}{(1-q)^d}=\frac{1}{(q;q)_d}, \quad\quad\quad d\geq 0, \nonumber \end{align*} $$

and

$$ \begin{align*} B=\prod_{i=1}^{r}\frac{P_{d_i}}{(1-q)^{d_i}}=\prod_{i=1}^{r}\frac{1}{(q;q)_{d_i} }. \nonumber \end{align*} $$

The factor left is

$$ \begin{align*} C:=& \left( \prod_{i,j=1|i\neq j}^{r}\prod_{n_i=1}^{d_i}\prod_{n_j=1}^{d_j}\frac{1-q^{n_{ij}} x_{ij}}{1-q^{n_{ij}+1} x_{ij}} \right) \frac{\prod_{i=1}^{r}\prod_{n_i=1}^{d_i}(x_i q^{n_i - 1})^l}{ \prod_{i,j=1|i\neq j}^{r}\prod_{n_i=1}^{d_i} (1-x_{ji}q^{1-n_i})} \nonumber \\ =&\prod_{i\neq j}^{r}\prod_{n_j=1}^{d_j}\left( \left(\prod_{n_i=1}^{d_i} \frac{1-q^{n_{ij}} x_{ij}}{1-q^{n_{ij}+1} x_{ij}} \right) \cdot \frac{1}{1-x_{ij}q^{1-n_j}}\right) \cdot \prod_{i=1}^{r}x_{i}^{ld_i}q^{\frac{ld_i(d_i-1)}{2}} \nonumber \\ =&\prod_{i\neq j}^{r}\prod_{n_j=1}^{d_j}\frac{1}{1-q^{d_i-n_j+1}x_{ij}} \cdot \prod_{i=1}^{r}x_{i}^{ld_i}q^{\frac{ld_i(d_i-1)}{2}} \nonumber \\ =&\prod_{i=1}^{r}x_{i}^{ld_i}q^{\frac{ld_i(d_i-1)}{2}}\cdot \prod_{i\neq j}^{r}\frac{1}{(q^{d_{ij}+1}x_{ij};q)_{d_j}}. \nonumber \end{align*} $$

The above equations prove that the summand in equation (2.20) corresponding to a given $\vec d$ equals to one summand in $A_{d}\left (\vec {x},I,l\right )$ . Thus we have

$$ \begin{align*} A_{d}\left(\vec{x},I,l\right) = \sum_{|\vec{d}|=d}d!E_{\vec{d}} =E_d. \nonumber \end{align*} $$

Now calculate the integration in a clockwise direction:

(2.23) $$ \begin{align} E_d^\prime :=\int_{C^{\prime}_{\rho}} \frac{dw_{i_d}}{2\pi\sqrt{-1}} \ldots \int_{C^{\prime}_{\rho}} \frac{dw_{i_1}}{2\pi\sqrt{-1}} f(w_{i_1},\cdots,w_{i_d}). \end{align} $$

The assumption with l ensures that when integrating in any order, for each variable w, the residue at infinity is 0. By definition, we can calculate this integration by taking the sum of residues outside the circle $|w_i|=\rho $ .

The iterated residues, in this case, are similar to the previous counterclockwise direction. Arguments similar to those in equation (2.1) show

$$ \begin{align*} E_d^\prime =\sum_{|\vec{d^\prime}|=d}d!E^{\prime}_{\vec{d^\prime}}, \nonumber \end{align*} $$

where

$$ \begin{align*} E^{\prime}_{\vec{d^\prime}}= \lim_{w_{d}\rightarrow \hat{w}_{d} }\cdots\lim_{w_{1}\rightarrow \hat{w}_{1} }\left(\prod_{i=1}^{n}(w_i-\hat{w}_{i})f(\vec{w})\right ), \nonumber \end{align*} $$

here

$$ \begin{align*} \left\{\hat{w}_{1}, \ldots, \hat{w}_{d}\right\}=\left\{x_{r+1}q^{-1}, x_{r+1}q^{-2}, \ldots, x_{r+1} q^{-d_{r+1}}, \ldots, x_{n}q^{-1}, x_{n} q^{-2}, \ldots, x_{n} q^{-d_{n}} \right\}, \nonumber \end{align*} $$

and the order in which to take the limit is from $w_1$ to $w_d$ .

We now do the following, changing variables and calculating the residues:

$$ \begin{align*} w_{i,n_{i}}=x_{i} q^{-n_{i}} z_{i,n_{i}}, \quad i=r+1, \ldots, n \quad n_{i}=1, \ldots, d_{i}. \nonumber \end{align*} $$

Similarly,

$$ \begin{align*} f(\vec{w})=& \frac{1}{(1-q)^{d} d!\prod_{i,n_i}z_{i,n_i}}\cdot \prod_{i=r+1}^{n}\prod_{n_i\neq n_j}^{d_i} \frac{1-q^{n_{ji}} z_{i,n_i}/z_{i,n_j}}{1-q^{n_{ji}+1} z_{i,n_i}/z_{i,n_j}} \nonumber \\ & \times \prod_{i,j=r+1|i\neq j}^{n}\prod_{n_i=1}^{d_i}\prod_{n_j=1}^{d_j} \frac{1-q^{n_{ji}} z_{i,n_i}/z_{j,n_j}x_{ij}}{1-q^{n_{ji}+1} z_{i,n_i}/z_{j,n_j}x_{ij}} \nonumber \\ &\times \frac{\prod_{i=r+1}^{n}\prod_{n_i=1}^{d_i}(x_i q^{-n_i}z_{i,n_i})^{l-1}}{ \prod_{i,j=r+1|i\neq j}^{n}\prod_{n_i=1}^{d_i} (1-x_{ij}q^{1-n_i}z_{i,n_i})} \nonumber \\ &\times \frac{1}{ \prod_{i=r+1}^{n}\prod_{n_i=1}^{d_i} (1-q^{1-n_i}z_{i,n_i})} \cdot \frac{1}{ \prod_{i=r+1}^{n}\prod_{j=1}^{r}\prod_{n_i=1}^{d_i} (1-x_{ji}q^{n_i}/z_{i,n_i})}. \nonumber \end{align*} $$

Note that

$$ \begin{align*} \lim_{z_{i,d_i}\rightarrow 1 }\cdots\lim_{z_{i,1}\rightarrow 1 }\left(\prod_{n_i=1}^{d_i}(z_{i,n_i} - 1 ) \cdot \frac{1}{\left(1-z_{i,1}\right)\left(1-z_{i,2}/z_{i,1}\right)\ldots \left(1-z_{i,d_i}/z_{i,d_i-1}\right)z_{i,1} \cdots z_{i,d_i} }\right) =(-1)^{d_i}, \nonumber \end{align*} $$

where the order in which to take the limits is from $z_{i,1}$ to $z_{i,d_i}$ . So the residues for one specific configuration of residues of type $\vec {d'}$ are

$$ \begin{align*} &\frac{(-1)^d}{(1-q)^{d} d!}\cdot \displaystyle \prod_{i=r+1}^{n}\left(\prod_{n_i\neq n_j|n_{ji}\neq -1}^{d_i} \frac{1-q^{n_{ji}} }{1-q^{n_{ji}+1} } \cdot \prod_{n_i=2}^{d_i}\frac{1-q^{-1}}{1-q^{1-n_i}} \right) \nonumber \\ &\times \prod_{i,j=r+1|i\neq j}^{n}\prod_{n_i=1}^{d_i}\prod_{n_j=1}^{d_j}\frac{1-q^{n_{ji}} x_{ij}}{1-q^{n_{ji}+1} x_{ij}} \nonumber \\ &\times \frac{\prod_{i=r+1}^{n}\prod_{n_i=}^{d_i}(x_i q^{-n_i})^{l}}{\prod_{i,j=r+1|i\neq j}^{n}\prod_{n_i=1}^{d_i} (1-x_{ij}q^{1-n_i})} \cdot \frac{1}{ \prod_{i=r+1}^{n}\prod_{j=1}^{r}\prod_{n_i=1}^{d_i} (1-x_{ji}q^{n_i})}. \nonumber \end{align*} $$

After almost the same computation as for $E_{\vec {d}}$ , we can simplify the above equation to

$$ \begin{align*} (-1)^d\frac{\prod_{i=r+1}^{n}x_{i}^{ld_i}q^{-\frac{ld_i(d_i+1)}{2}}}{\prod_{i=r+1}^{n}(q; q)_{d_i} \prod_{i \neq j|i,j=r+1 }^{n}\left(q^{d_{ij}+1}x_{ji};q \right)_{d j}\prod_{i=r+1}^{n} \prod_{j=1}^{r} (qx_{ji} ;q)_{d_i}}, \nonumber \end{align*} $$

which proves

$$ \begin{align*} E_d^\prime=(-1)^d B_d\left( \vec{x},I^{\complement } ,-l \right). \nonumber \end{align*} $$

Since the residue at infinity is zero, using the Cauchy Residue Theorem d times,

$$ \begin{align*} \int_{C_{\rho}} \ldots \int_{C_{\rho}} f(\vec w) \frac{dw_1}{2\pi\sqrt{-1}w_1} \ldots \frac{dw_d}{2\pi\sqrt{-1}w_d}\\ =(-1)^d\int_{C^{'}_{\rho}} \ldots \int_{C^{'}_{\rho}} f(\vec w) \frac{dw_1}{2\pi\sqrt{-1}w_1} \ldots \frac{dw_d}{2\pi\sqrt{-1}w_d}, \end{align*} $$

we arrive at equations (2.24), (2.25) and (2.26) of the following theorem stated in the introduction.

Theorem 2.2. Denoted by $[n]$ , the set of elements $\{ 1, \ldots ,n \}$ , let $ \emptyset \neq I \subsetneq [n] $ be a subset of $[n]$ , $|I|$ be its cardinality, and denoted by $I^\complement $ , the complementary set of I in $[n]$ . Then for constant positive integers d, n and integer l with restriction $1-|I|\leq l\leq n-|I|-1$ , let $A_d\left ( \vec {x} , I ,l\right )$ and $B_d\left ( \vec {x}, I,l\right )$ be two rational functions in $\vec {x}$ and q with an extra data l:

(2.24) $$ \begin{align} A_d\left( \vec{x}, I, l\right)=&\sum_{| {d}_I|=d}\frac{\left( \prod_{i \in I}x_{i}^{d_i}q^{\frac{d_i(d_i-1)}{2}}\right)^l}{\prod_{i ,j \in I}\left( q^{d_{ij}+1}x_{ij};q \right)_{d j}\prod_{i \in I } \prod_{j \in I^\complement } (qx_{ij};q )_{d_i}}, \end{align} $$

(2.25) $$ \begin{align} B_d\left( \vec{x}, I, l\right)=&\sum_{|\vec{d}_{I}|=d}\frac{\left(\prod_{i \in I }x_{i}^{-d_i}q^{\frac{d_i(d_i+1)}{2}}\right)^l}{ \prod_{i , j \in I }\left( q^{d_{ij}+1}x_{ji};q \right)_{d j}\prod_{i \in I } \prod_{j \in I^\complement } (qx_{ji} ;q)_{d_i}},\\[-15pt]\nonumber \end{align} $$

where $\vec {d}_I$ is an $|I|$ -tuple of nonnegative integers and $|\vec {d}_I|:=\sum _{i \in I} d_i$ . $x_i, i=1,$ … $,n$ are parameters. For convenience, we use the notation $x_{ij}:=x_{i}/x_{j}$ and $d_{ij}:=d_i-d_j$ . Then we have

(2.26) $$ \begin{align} A_d\left( \vec{x},I,l\right)= B_d\left( \vec{x},I^\complement,-l\right). {}\\[-15pt]\nonumber \end{align} $$

2.2 Examples

In the following two examples, we show how the proof of Theorem 2.2 works.

Example 2.3 d=1

For the case l=0, d=1, r=2, n=3, equation (2.3) becomes the following simple form:

$$ \begin{align*} f(w)=&\frac{1}{(1-q) } \frac{w^{-1}}{(1-x_1/w)(1-x_2/w) (1-qw/x_{3})}.\\[-15pt] \end{align*} $$

Consider the integration in equation (2.5). Then there are simple poles of type $(1,0)$ and $(0,1)$ in the counter $C_{\rho }$ :

• ○ type (1,0): $ w = x_1 $ ,

• ○ type (0,1): $ w = x_2 $ .

Then the residue for each type is as follows:

• ○ type ${(1,0)}$ :

  $$ \begin{align*} E_{(1,0)}=\underset{\hat{w} = x_1}{\operatorname{Res}}f = \frac{1}{(1-q)(1-x_{21})(1-qx_{13})},\\[-15pt] \end{align*} $$

• ○ type ${(0,1)}$ :

  $$ \begin{align*} E_{(0,1)}=\underset{\hat{w} = x_2}{\operatorname{Res}}f= \frac{1}{(1-q)(1-x_{12})(1-qx_{23})}.\\[-15pt] \end{align*} $$

And there is only one simple pole $w=q^{-1}x_3 $ in the counter $C^{\prime }_{\rho }$ , so

• ○ type $1$ :

  $$ \begin{align*} E^{\prime}_{1}= \underset{\hat{w} = q^{-1}x_3}{\operatorname{Res}}f = \frac{-1}{(1-q)(1-qx_{13})(1-qx_{23})}.\\[-15pt] \end{align*} $$

It is easy to obtain

$$ \begin{align*} \frac{1}{(1-q)(1-x_{21})(1-qx_{13})} +\frac{1}{(1-q)(1-x_{12})(1-qx_{23})} = \frac{1}{(1-q)(1-qx_{13})(1-qx_{23})},\\[-15pt] \end{align*} $$

which agrees with equation (2.26).

Example 2.4 d=2

For the case l=0, d=2, r=2, n=3, equation (2.3) becomes the following simple form:

$$ \begin{align*} f(\vec w)=&\frac{1}{2(1-q)^2 }\prod_{i \neq j}^2 \frac{1-w_{i}/w_{j}}{1-qw_{i}/w_{j}}\prod_{i=1}^2 \frac{w_i^{-1}}{\prod_{j=1}^{2}(1-x_j/w_{i}) \cdot (1-qw_{i}/x_{3})}. \end{align*} $$

Consider the integration in equation (2.5). Then there are simple poles of type $(2,0)$ , $(1,1)$ and $(0,2)$ in the counter $C_{\rho _i}$ :

• ○ type (2,0): $\{ w_1, w_2 \} = \{ x_1, x_1 q \} $ ,

• ○ type (1,1): $\{ w_1, w_2 \} = \{ x_1, x_2 \} $ ,

• ○ type (0,2): $\{ w_1, w_2 \} = \{ x_2, x_2 q \} $ .

Then the residue for each type is as follows:

• ○ type ${(2,0)}$ :

  $$ \begin{align*} 2!E_{(2,0)} &= \underset{\hat{w}_{2}= qx_1}{\operatorname{Res}}\underset{\hat{w}_{1} = x_1}{\operatorname{Res}}f+\underset{\hat{w}_{2}= x_1}{\operatorname{Res}}\underset{\hat{w}_{1} = qw_2}{\operatorname{Res}}f \\ &=\frac{1}{2(1-q)^2} \frac{1}{(1+q)(1-q x_{13})(1-q^2x_{13})(1-x_{21})(1-q^{-1}x_{21})} \\ &\quad + \frac{1}{2(1-q)^2} \frac{1}{(1+q)(1-q^2x_{13})(1-qx_{13})(1-q^{-1}x_{21})(1-x_{21})} \\ &= \frac{1}{(1-q)^2} \frac{1}{(1+q)(1-q^2 x_{13})(1-q x_{13})(1-q^{-1} x_{21})(1- x_{21})}, \end{align*} $$

• ○ type ${(1,1)}$ :

  $$ \begin{align*} 2!E_{(1,1)} &= \underset{\hat{w}_{2} = x_2}{\operatorname{Res}}\underset{\hat{w}_{1} = x_1}{\operatorname{Res}}f +\underset{\hat{w}_{2} = x_1}{\operatorname{Res}}\underset{\hat{w}_{1} = x_2}{\operatorname{Res}}f \\ &= \frac{1}{2(1-q)^2} \frac{1}{(1-qx_{12})(1-qx_{21})(1-qx_{13})(1-qx_{23})} \\ &\quad + \frac{1}{2(1-q)^2} \frac{1}{(1-qx_{21})(1-qx_{12})(1-qx_{23})(1-qx_{13})} \\ &= \frac{1}{(1-q)^2} \frac{1}{(1-q x_{21})(1-q x_{12})(1-q x_{23})(1-q x_{13})}, \end{align*} $$

• ○ type ${(0,2)}$ :

  $$ \begin{align*} 2!E_{(0,2)} &= \underset{\hat{w}_{2} = qx_2}{\operatorname{Res}}\underset{\hat{w}_{1} = x_2}{\operatorname{Res}}f+\underset{\hat{w}_{2} = x_2}{\operatorname{Res}}\underset{\hat{w}_{1} = qw_2}{\operatorname{Res}}f \\ &= \frac{1}{2(1-q)^2} \frac{1}{(1+q)(1-x_{12})(1-qx_{23})(1-q^{-1}x_{12})(1-q^2x_{23})} \\ &\quad + \frac{1}{2(1-q)^2}\frac{1}{(1+q)(1-q^{-1}x_{12})(1-q^2x_{23})(1-x_{12})(1-qx_{23})} \\ &= \frac{1}{(1-q)^2}\frac{1}{(1+q)(1-q^{-1} x_{12})(1-q^2 x_{23})(1- x_{12})(1-q x_{23})}. \end{align*} $$

Consider the integration in equation (2.23). Then there are simple poles of type 2 in the counter $C^{\prime }_{\rho _i}$ :

• ○ type 2: $\{ w_1, w_2 \} = \{ q^{-1}x_3, q^{-2}x_3 \} $ .

Then the residue for each type 2 is as follows:

• ○ type $ {2}$ :

  $$ \begin{align*} (-1)^22!E^{\prime}_{2} &= \underset{\hat{w}_{2} = q^{-2}x_3}{\operatorname{Res}}\underset{\hat{w}_{1} = q^{-1}x_3}{\operatorname{Res}}f+\underset{\hat{w}_{2} = q^{-1}x_3}{\operatorname{Res}}\underset{\hat{w}_{1} = q^{-1}w_2}{\operatorname{Res}}f \\ &=\frac{1}{(1+q)(1-q)^2(1-q^2 x_{13})(1-q x_{13})(1-q^2 x_{23})(1-q x_{23})}. \end{align*} $$

By a little computation, we have

$$ \begin{align*} E_2&=2!E_{(2,0)}+2!E_{(1,1)}+2!E_{(0,2)} = E^{\prime}_2. \end{align*} $$

Example 2.5. From Proposition 1.2, if we take $n=3$ , $l=0$ and $I=[2]$ , we know that $A_d(\vec {x},[2],0)=B_d(\vec {x},[3]\backslash [2],0 )$ . By the following computation, there is a phenomenon that we can extract from $A_d(\vec {x},[2],0)$ to get $B_d(\vec {x},[3]\backslash [2],0 )$ times another factor when $d=1,2$ : that is, $A_d(\vec {x},[2],0)=B_d(\vec {x},[3]\backslash [2],0 ) \times G(\vec {x}), \ d=1,2 $ . Thus we can conclude that $G(\vec {x})=1$ . Furthermore, this is a general phenomenon for all d; see the following Corollary 2.1.

By definition, $\vec {x}=\{x_1,x_2,x_3 \}$ , so

(2.27) $$ \begin{align} &A_d(\vec{x},[2],0)= \sum_{d_1+d_2=d}\frac{1}{(q;q)_{d_1}(q;q)_{d_2}(q^{d_{12}+1}x_{12};q)_{d_2}(q^{d_{21}+1}x_{12};q)_{d_1} (qx_{13};q)_{d_1}(qx_{23};q)_{d_2} }, \end{align} $$

(2.28) $$ \begin{align} &B_d(\vec{x},[3]\backslash [2],0) = \frac{1}{(q;q)_d(qx_{13};q)_d(qx_{23};q)_d}. \end{align} $$

For $d=1$ – that is, $(d_1,d_2)=(1,0)$ or $(0,1)$ – we have

$$ \begin{align*} & A_1(\vec{x},[2],0) =\sum_{d_1+d_2=1}\frac{1}{(q;q)_{d_1}(q;q)_{d_2}(q^{d_{12}+1}x_{12};q)_{d_2}(q^{d_{21}+1}x_{12};q)_{d_1} (qx_{13};q)_{d_1}(qx_{23};q)_{d_2} } \nonumber \\ =&\sum_{d_1+d_2=1}\frac{1}{(q;q)_1(qx_{13};q)_1(qx_{23};q)_1} \\&\quad \cdot\frac{(q;q)_1(qx_{13};q)_1(qx_{23};q)_1}{(q;q)_{d_1}(q;q)_{d_2}(q^{d_{12}+1}x_{12};q)_{d_2}(q^{d_{21}+1}x_{12};q)_{d_1} (qx_{13};q)_{d_1}(qx_{23};q)_{d_2} } \nonumber \\ =&B_1(\vec{x},[3] \backslash [2] ,0) \times\sum_{(d_1,d_2)=(1,0),(0,1)}\frac{(q;q)_1}{(q;q)_{d_1}(q;q)_{d_2}}\prod_{i=1}^2 \left( \prod_{j \neq i}^{2} \frac{(q^{d_i+1}x_{i 3};q)_{1-d_i}}{(q^{d_{ij}+1}x_{i j};q)_{d_j} }\right) \nonumber \\ =& B_1(\vec{x},[3] \backslash [2] ,0) \times \left( \frac{1-qx_{23}}{1-x_{21}}+ \frac{1-qx_{13}}{1-x_{12}} \right) \nonumber\\ =& B_1(\vec{x},[3] \backslash [2] ,0). \nonumber \end{align*} $$

For $d=2$ – that is, $(d_1,d_2)=(2,0),(1,1)$ or $(0,2)$ – similarly we have

$$ \begin{align*} &A_2(\vec{x},[2],0) = B_2(\vec{x},[3] \backslash [2] ,0) \times \sum_{(d_1,d_2)=(2,0),(1,1),(0,2)}\frac{(q;q)_2}{(q;q)_{d_1}(q;q)_{d_2}}\prod_{i=1}^2 \left( \prod_{j \neq i}^{2} \frac{(q^{d_i+1}x_{i 3};q)_{2-d_i}}{(q^{d_{ij}+1}x_{i j};q)_{d_j} }\right) \\ &= B_2(\vec{x},[3] \backslash [2] ,0)\\ &\quad \times \left( \frac{(1-qx_{13})(1-q^2x_{13})}{(1-q^{-1}x_{21})(1-x_{21})}+\frac{(1+q)(1-q^2x_{13})(1-q^2x_{23})}{(1-qx_{12})(1-qx_{21})} + \frac{(1-qx_{13})(1-q^2x_{13})}{(1-q^{-1}x_{12})(1-x_{12})} \right) \\ &= B_2(\vec{x},[3] \backslash [2] ,0). \end{align*} $$

More generally, we have the following corollary.

Corollary 2.1.

$$ \begin{align*} \sum_{d_1+d_2=d} \frac{(q;q)_d}{(q;q)_{d_1}(q;q)_{d_2}} \prod_{j \neq i}^{2} \frac{(q^{d_i+1}x_{i 3};q)_{d-d_i}}{(q^{d_i-d_j+1}x_{i j};q)_{d_j}} =1. \end{align*} $$

Proof. Set $l=0$ , $r=2,n=3$ in equation (2.26). We have

$$ \begin{align*} A_d(\vec{x},[2],0) =& \sum_{d_1+d_2= d} \prod^2_{i,j=1}\frac{1}{(q^{d_{ij}+1}x_{ij};q)_{d_j}} \prod^2_{i=1}\frac{1}{(qx_{i3};q)_{d_i}} \nonumber \\ =& \sum_{d_1+d_2= d} \prod^2_{i=1}\left( \frac{1}{(q;q)_{d_i}}\prod_{j \neq i}^2 \frac{1}{(q^{d_{ij}+1}x_{ij};q)_{d_j}} \cdot \frac{1}{(qx_{i3};q)_{d_i}} \right) \nonumber \\ =& \sum_{d_1+d_2 = d} \frac{(q^{d_1+1}x_{1 3};q)_{d-d_1}(q^{d_2+1}x_{2 3};q)_{d-d_2}}{(qx_{1 3};q)_{d}(qx_{2 3};q)_{d}} \prod_{j \neq i}^2 \left( \frac{1}{(q;q)_{d_i}} \frac{1}{(q^{d_i-d_j+1}x_{i j};q)_{d_{j}}}\right) \nonumber \\ =& \sum_{d_1+d_2=d} \frac{(q;q)_d}{(q;q)_d(qx_{1 3};q)_{d}(qx_{2 3};q)_{d}} \prod_{j \neq i}^2 \left( \frac{(q^{d_i+1}x_{i 3};q)_{d-d_i}}{(q;q)_{d_i}(q^{d_{ij}+1}x_{i j};q)_{d_j} }\right) \nonumber \\ =& \sum_{d_1+d_2=d}B_d(\vec{x},[n] \backslash [2] ,0) \cdot \frac{(q;q)_d}{(q;q)_{d_1}(q;q)_{d_2}} \prod_{j \neq i}^2 \left( \frac{(q^{d_i+1}x_{i 3};q)_{d-d_i}}{(q^{d_{ij}+1}x_{i j};q)_{d_j} }\right). \nonumber \end{align*} $$

Since we know $A_d(\vec {x},I ,0)$ equals $B_d(\vec {x},I^{\complement } ,0)$ , we get the conclusion.

2.3 Boundary cases

For $l=-|I|$ , $l=n-|I|$ , equation (2.26) no longer holds, since the residue at infinity is nonzero, but we can compare the behaviour of some special limit in equation (2.26) to obtain following results.

Proof. Consider $[n+1]$ , $I \subsetneq [n+1] $ , $\{n+1\} \notin I $ , $l=n-|I|$ in equation (2.26). Then we have

(2.31) $$ \begin{align} & \sum_{|\vec{d}_I|=d}\frac{\left( \prod_{i \in I}x_{i}^{d_i}q^{\frac{d_i(d_i-1)}{2}}\right)^l}{\prod_{i ,j \in I}\left( q^{d_{ij}+1}x_{ij};q \right)_{d j}\prod_{i \in I } \prod_{j \in I^\complement } (qx_{ij};q )_{d_i}} \end{align} $$

(2.32) $$ \begin{align} =& \sum_{|\vec{d}_{I^\complement}|=d}\frac{\left(\prod_{i \in I^\complement }x_{i}^{-d_i}q^{\frac{d_i(d_i+1)}{2}}\right)^{-l}}{ \prod_{i , j \in I^\complement }\left(q^{d_{ij}+1}x_{ji};q \right)_{d j}\prod_{i \in I^\complement } \prod_{j \in I } (qx_{ji} ;q)_{d_i}}. \end{align} $$

It is easy to see that taking $\lim _{ x_{n+1} \rightarrow \infty }$ in equation (2.31), we obtain

$$ \begin{align*} \lim_{ x_{n+1} \rightarrow \infty }(2.31)=A_d\left( \vec{x},I,l \right), \ \ \text{for} \ \ l=n-|I|. \end{align*} $$

Now let’s take limit $ \lim _{x_{n+1} \rightarrow \infty } $ in equation (2.32):

(2.33) $$ \begin{align} &\sum_{|\vec{d}_{I^\complement}|=d}\frac{\left(\prod_{i \in I^\complement }x_{i}^{-d_i}q^{\frac{d_i(d_i+1)}{2}}\right)^{-l}}{ \prod_{i , j \in I^\complement }\left(q^{d_{ij}+1}x_{ji};q \right)_{d j}\prod_{i \in I^\complement } \prod_{j \in I } (qx_{ji};q )_{d_i}} \nonumber \\ =&\sum_{|\vec{d}_{I^\complement}|=d} \frac{1}{(q;q)_{d_{n+1}}} \cdot \frac{1}{\prod_{j\in I}(qx_{j,n+1};q)_{d_{n+1}}} \cdot \frac{1}{\prod_{j \in \{ [n] \backslash I \} }(q^{d_{n+1}-d_j+1}x_{j,n+1};q)_{d_j}} \end{align} $$

(2.34) $$ \begin{align} \times &\; \frac{(x^{d_{n+1}}_{n+1}q^{-\frac{d_{n+1}(d_{n+1}+1)}{2}})^l}{\prod_{i \in \{[n] \backslash I \} }(q^{d_i-d_{n+1}+1}x_{n+1,i};q)_{d_{n+1}}} \\ \times & \prod_{i \in \{[n] \backslash I \} } \left( \frac{1}{\prod_{j \in \{[n] \backslash I \} } (q^{d_i-d_j+1}x_{j i};q)_{d_{j}}} \cdot \frac{(x^{d_{i}}_{i}q^{\frac{-d_{i}(d_{i}+1)}{2}})^{l}}{ \prod_{j \in I }(qx_{j i};q)_{d_i}} \right). \nonumber \end{align} $$

The limits of the last two terms in equation (2.33) equal 1, and by a little computation, we obtain that equation (2.34) equals

$$ \begin{align*} (-1)^{d_{n+1}(n-r)} \cdot q^{-(\sum_{i \in \{[n] \backslash I \} } d_i)d_{n+1}-(n-r)d_{n+1}} \prod_{i \in \{[n] \backslash I \}}x^{d_{n+1}}_i. \end{align*} $$

Then we obtain

$$ \begin{align*} \lim_{x_{n+1} \rightarrow \infty} & \sum_{|\vec{d}_{I^\complement}|=d} \prod_{i \in I^\complement } \left(\frac{1}{ \prod_{j \in I^\complement}(q^{d_i-d_j+1}x_{j i};q)_{d_{j}}} \cdot \frac{(x^{d_{i}}_{i}q^{\frac{-d_{i}(d_{i}+1)}{2}})^{l}}{ \prod_{j \in I}(qx_{j i};q)_{d_i}} \right) \nonumber \\ =& \sum_{|\vec{d}_{I^\complement}|=d} \frac{1}{(q;q)_{d_{n+1}}} \cdot \frac{(-1)^{d_{n+1}(n-|I|)}}{q^{(\sum_{i \in \{[n] \backslash I \}} d_i)d_{n+1}+(n-|I|)d_{n+1}}} \cdot \frac{1}{\prod_{i \in \{[n] \backslash I \}}x^{-d_{n+1}}_i} \nonumber \\ \times & \prod_{i \in \{[n] \backslash I \}} \left( \frac{1}{\prod_{j \in \{[n] \backslash I \}}(q^{d_i-d_j+1}x_{j i};q)_{d_{j}}} \cdot \frac{(x^{d_{i}}_{i}q^{\frac{-d_{i}(d_{i}+1)}{2}})^{l}}{ \prod_{j \in I}(qx_{j i};q)_{d_i}} \right) \nonumber \\ =& \sum_{\alpha=0}^d \frac{1}{(q;q)_{d-\alpha}} \cdot \frac{(-1)^{(d-\alpha)(n-|I|)}}{q^{(n-|I|+\alpha)(d-\alpha)}} \cdot \frac{1}{\prod_{i=r+1}^nx^{-(d-\alpha)}_i} \cdot B_\alpha \left( \vec{x},I^\complement,-l\right). \nonumber \end{align*} $$

We obtain the conclusion.

Similarly, consider $A_d\left ( \vec {x}\cup x_{n+1} ,\tilde {I},l\right )$ and $B_d\left ( \vec {x} \cup x_{n+1},\tilde {I} ^\complement ,-l\right )$ . For $\tilde {I}= I \cup \{n+1\}$ and $l=-|I|$ , from equation (2.26), we have

(2.35) $$ \begin{align} & \sum_{|\vec{d}_{\tilde{I}}| = d} \prod_{i \in \tilde{I} } \left( \frac{1}{\prod_{j \in \tilde{I} }(q^{d_i-d_j+1}x_{i j};q)_{d_{j}}} \cdot \frac{(x^{d_{i}}_{i}q^{\frac{d_{i}(d_{i}-1)}{2}})^{l}}{ \prod_{j \in \tilde{I}^\complement }(qx_{i j};q)_{d_i}} \right) \end{align} $$

(2.36) $$ \begin{align} =& \sum_{|\vec{d}_{\tilde{I}^\complement}| = d} \prod_{i \in \tilde{I}^\complement }\left( \frac{1}{\prod_{j \in \tilde{I}^\complement }(q^{d_i-d_j+1}x_{j i};q)_{d_{j}}} \cdot \frac{(x^{d_{i}}_{i}q^{\frac{-d_{i}(d_{i}+1)}{2}})^{-l}}{ \prod_{j\in \tilde{I}}(qx_{j i};q)_{d_i}} \right). \end{align} $$

It is easy to see that after taking $\operatorname{lim}_{x_{n+1} \rightarrow 0} $ in equation (2.36), we obtain

$$ \begin{align*} B_d\left( \vec{x},I^\complement,l\right) , \ \ \text{for} \ \ l=-|I|. \end{align*} $$

First, rewrite equation (2.35) as follows:

$$ \begin{align*} & \sum_{|\vec{d}_{\tilde{I}}| = d} \prod_{i \in \tilde{I} } \left( \frac{1}{ \prod_{j \in \tilde{I} }(q^{d_i-d_j+1}x_{i j};q)_{d_{j}}} \cdot \frac{(x^{d_{i}}_{i}q^{\frac{d_{i}(d_{i}-1)}{2}})^{-|I|}}{ \prod_{j\in \tilde{I}^\complement }(qx_{i j};q)_{d_i}} \right) \nonumber \\ =&\sum_{|\vec{d}_{\tilde{I}}| = d} \frac{(\prod_{i \in {I} }x_i^{d_i}q^{\frac{d_i(d_i-1)}{2}})^{-|I|}}{\prod_{i , j \in {I} }(q^{d_{ij}+1} x_{ij};q)_{d_j} \prod_{i \in {I} }\prod_{j\in \{ [n] \backslash I\} }(qx_{ij};q)_{d_i}} \nonumber \\ \times & \frac{(x^{d_{n+1}}_{n+1}q^{\frac{d_{n+1}(d_{n+1}-1)}{2}})^{-|I|}}{(q;q)_{d_{n+1}} \prod_{i \in I }(q^{d_i-d_{n+1}+1}x_{i,n+1};q)_{d_{n+1}} \prod_{j \in I}(q^{d_{n+1}-d_{j}+1}x_{n+1,j};q)_{d_{j}} \prod_{j \in \{ [n] \backslash I \} }(q x_{n+1,j};q)_{d_{n+1}}}. \nonumber \end{align*} $$

Now let’s take limit $ \lim _{x_{n+1} \rightarrow 0 } $ in the above formula. We obtain

$$ \begin{align*} & \lim_{x_{n+1} \rightarrow 0 } \sum_{|\vec{d}_{\tilde{I}}| = d} \prod_{i \in \tilde{I}} \left( \frac{1}{ \prod_{j \in \tilde{I}}(q^{d_i-d_j+1}x_{i j};q)_{d_{j}}} \cdot \frac{(x^{d_{i}}_{i}q^{\frac{d_{i}(d_{i}-1)}{2}})^{-|I|}}{ \prod_{j\in \tilde{I}^\complement}(qx_{i j};q)_{d_i}} \right) \nonumber \\ =& \sum_{|\vec{d}_{\tilde{I}}| = d} \frac{(\prod_{i \in I }x_i^{d_i}q^{\frac{d_i(d_i-1)}{2}})^{-|I|}}{\prod_{i , j \in I }(q^{d_{ij}+1}x_{ij};q)_{d_j}\prod_{i\in I }\prod_{j \in \{[n] \backslash I \} }(qx_{ij};q)_{d_i}} \nonumber \\ & \times \frac{(-1)^{|I| \cdot d_{n+1}}}{(q;q)_{d_{n+1}} q^{d_{n+1}(d-d_{n+1})} \prod_{i \in I } x^{d_{n+1}}_{i}} \nonumber \\ =& \sum_{s=0}^d \frac{(-1)^{|I| \cdot s}}{(q;q)_{s} q^{s(d-s)} \prod_{i \in I } x^{s}_{i}} \times A_{d-s}\left( \vec{x} ,I, -|I| \right). \nonumber\\[-36pt] \end{align*} $$

3.1 Definitions

Let X be a GIT quotient $V/\!/_\theta G$ , where V is a vector space and ${G}$ is a connected reductive complex Lie group. Let ${\mathcal {Q}}^{\epsilon }_{g,n}(X,\beta )$ be the moduli stack of $\epsilon $ -stable quasimaps [Reference Ciocan-Fontanine and Kim3] parametrising data $(C,p_1,$ … $,p_n,\mathcal {P},s)$ , where C is an n-pointed genus g Riemann surface, $\mathcal {P}$ is a principal G-bundle over C, s is a section and $\beta \in \mathrm {Hom}(\mathrm {Pic}^G(V))$ . There are natural maps

$$ \begin{align*} e v_{i}: {\mathcal{Q}}^{\epsilon}_{g,n}(X,d) \rightarrow X, \quad i=1, \ldots, n \end{align*} $$

given by evaluation at the ith marked point: that is,

$$ \begin{align*} ev_i (C,p_1,\ldots,p_n,\mathcal{P},s) = s(p_i) \in X. \end{align*} $$

There are line bundles

$$ \begin{align*} L_{i} \rightarrow {\mathcal{Q}}^{\epsilon}_{g,n}(X,d) , \quad i=1, \ldots, n, \end{align*} $$

which are called universal cotangent line bundles. The fibre of $L_i$ over the point $(C^{\epsilon },p_1,$ … $,p_n,\mathcal {P},s)$ is the cotangent line to C at the point $p_i$ .

The permutation-equivariant K-theoretic quasimap invariants with level structures [Reference Ruan and Zhang14] are holomorphic Euler characteristics over ${\mathcal {Q}}^{\epsilon }_{g,n}(X,d)$ of the sheaves

(3.1) $$ \begin{align} \left\langle \mathbf{t}(L), \ldots, \mathbf{t}(L) \right\rangle_{g, n, d}^{R,l,S_n,\epsilon}:=\chi \left({\mathcal{Q}}^{\epsilon}_{g,n}(X,d); \mathcal{O}_{g, n, d}^{v i r t} \otimes \prod_{m,i} L_{i}^{k} t_{k,i} \mathrm{ev}_{i}^{*}\left(\phi_{i}\right) \otimes \mathcal{D}^{R,l} \right), \end{align} $$

where $\mathcal {O}^{v i r}_{g,n,d}$ is called the virtual structure sheaf [Reference Lee9]. $\mathbf {t}(q)$ is defined as follows:

$$ \begin{align*} \mathbf{t}(q)=\sum_{m \in \mathbb{Z}} t_{m} q^{m}, \quad t_{m}=\sum_{\alpha} t_{m, \alpha} \phi_{\alpha}, \end{align*} $$

$\{\phi _\alpha \}$ is a basis in $K^0(X)\otimes Q$ and $t_{k,\alpha }$ are formal variables. The last term in equation (3.1) is the level l determinant line bundle over $\mathcal {Q}^{\epsilon }_{g,n}(X,d)$ defined as

$$ \begin{align*} \mathcal{D}^{R,l} :=(\mathrm{det}R^{\bullet}\pi_*(\mathcal{P} \times _G R ))^{-l}, \end{align*} $$

where $\pi $ is the forgetful map from the universal curve: that is,

$$ \begin{align*} \pi: \mathcal{C} \rightarrow {\mathcal{Q}}^{\epsilon}_{g,n}(X,d). \end{align*} $$

The bundle $\mathcal {P}$ is the universal principal bundle over the universal curve, and R is a G-representation.

Similarly, we can define a quasimap graph space $ \mathcal { QG }^{\epsilon } _ { 0 , n } ( X , \beta ) $ , which parametrises quasimaps with parametrised component $\mathbb {P}^1$ , so there is a natural $\mathbb {C}^*$ -action on the quasimap graph space. It is denoted by $\mathrm {F}_{0,\beta }$ , the special fixed loci in $ ( \mathcal { QG }^{\epsilon } _ { 0 , n } ( X , \beta ) )^{\mathbb {C}^*}$ , and denoted by q, the weight of the cotangent bundle at $0 :=[1,0] $ of $\mathbb {P}^1$ ; for details, see [Reference Ciocan-Fontanine and Kim3].

Definition 3.1 [Reference Ruan and Zhang14]

The permutation-equivariant K-theoretic $\mathcal {J}^{\hspace{1.5pt}R,l,\epsilon }$ -function of $V // {G}$ of level l is defined as

$$ \begin{align*} \mathcal { J } _ { S _ { \infty } } ^ { R , l , \epsilon } ( \mathbf { t } ( q ) , Q ) &:= \sum_{k \geq 0, \beta \in {\operatorname { Eff } ( V , \mathbf { G } , \theta )} } Q^{\beta} (ev_{\bullet})_{*} [\operatorname{Res}_{\operatorname{F}_{0,\beta}}( \mathcal{QG} _ { 0 , n } ^ { \epsilon } ( V / / \mathbf { G } , \beta )_{0})^{\mathrm{vir}} \otimes \mathcal { D } ^ { R , l } \otimes _ { i = 1 } ^ { n } \mathbf { t } ( L _ { i } ) ]^{S_n} \\ &:= 1 + \frac { \mathbf { t } ( q ) } { 1 - q } +\sum _ { a } \sum _ { \beta \neq 0 } Q ^ { \beta } \chi \left( \operatorname{F} _ { 0 , \beta } , \mathcal { O } _ { \operatorname{F} _ { 0 , \beta } } ^ { \mathrm { vir } } \otimes { ev }_{\bullet} ^ { * } ( \phi _ { a } ) \otimes \left( \frac { \operatorname { t r } _ { \mathbb { C } ^ { * } } \mathcal { D } ^ { R , l } } { \lambda_{-1}^{\mathbb{C}^*} N _ { \operatorname{F} _ { 0 , \beta } } ^ { \vee } } \right) \right) \phi ^ { a } \\ &+ \sum_a \sum _ { n \geq 1 or \beta(L_\theta) \geq \frac{1}{\epsilon} \atop ( n , \beta ) \neq ( 1,0 ) } Q ^ { \beta } \left\langle \frac { \phi _ { a } } { ( 1 - q ) ( 1 - q L ) } , \mathbf { t } ( L ) , \ldots , \mathbf { t } ( L ) \right\rangle _ { 0 , n + 1 , \beta } ^ { R , l , \epsilon , S _ { n } } \phi ^ { a }, \end{align*} $$

where $\{ \phi _\alpha \}$ is a basis of $K^0(V/\!/{G})$ and $\{ \phi ^\alpha \}$ is the dual basis with respect to twisted pairing $( \ \ , \ \ )^{R,l}$ : that is,

$$ \begin{align*} (u,v)^{R,l}:=\chi\left( X,u \otimes v \otimes \mathrm{det}^{-l}(V^{ss} \times_{G} R ) \right). \end{align*} $$

Definition 3.2 [Reference Ruan and Zhang14]

When taking $\epsilon $ small enough, denoted by $\epsilon =0^{+}$ , we call $\mathcal {J}^{R,l,0^{+}}(0)$ the small I-function of level l: that is,

$$ \begin{align*} {I}^{R,l}(q;Q):= \mathcal { J } _ { S _ { \infty } } ^ { R , l , 0^{+} } ( 0 , Q ) = 1 + \sum _ { \beta \geq 0 } Q ^ { \beta } (ev_{\bullet})_{*} \left( \mathcal { O } _ { \operatorname{F} _ { 0 , \beta } } ^ { \mathrm { vir } } \otimes \left( \frac { \operatorname { t r } _ { \mathbb { C } ^ { * } } \mathcal { D } ^ { R , l } } { \lambda_{-1}^{\mathbb{C}^*} N _ { \operatorname{F} _ { 0 , \beta } } ^ { \vee } } \right) \right) \cdot \mathrm{det}^l(V^{ss} \times_{G} R ). \end{align*} $$

3.2 Level correspondence in Grassmann duality

Let V be $r \times n$ matrixes $M_{r \times n}$ , G be the general linear group $GL_r$ and $ \theta $ be the $\mathbf {det}: GL_r \rightarrow \mathbb {C}^*$ . Then we have

$$ \begin{align*} V/\!/_{\mathbf{det}}G = M_{r \times n}/\!/_{\mathbf{det}} G = Gr(r,n). \end{align*} $$

There is a natural $T=(\mathbb {C}^*)^n$ -action on $\mathbb {C}^n$ with weights $\mathbb {C}^n=\Lambda _1 + \cdots + \Lambda _n $ . Then deducing an action on $Gr(r,n)$ by $T \cdot A = AT$ , $A \in M_{r\times n}$ . Using general abelian/nonabelian correspondence in [Reference Wen20] for $Gr(r,n)$ , we have

$$ \begin{align*} I^{Gr(r,n)}_{T} =& 1+ \sum _{d} \sum_{|\vec{d}|= d} \sum_{ \omega \in S_{r} / S_{r_{1}} \times \cdots \times S_{r_{h+1}}} \\ &\omega \left[\frac{\prod_{1 \leqslant j<i \leqslant r} \prod_{ 1 \leqslant m \leqslant d_{i}-d_{j}}\left(1-L_{i} L_{j}^{-1} q^{m}\right)}{ \prod_{1 \leqslant i<j \leqslant r_{j} \atop 1 \leqslant m \leqslant d_{j}-d_{i}-1}\left(1-L_{i} L_{j}^{-1} q^{-m}\right) \prod_{1 \leqslant i<j \leqslant r}\left(1-L_{i}^{-1} L_{j}\right)}\prod_{i=1}^{r}\prod_{k=1}^{d_i}\prod_{m=1}^{n}\frac{1}{(1-q^kL_i\Lambda^{-1}_m)} \right] Q^d, \end{align*} $$

where $\vec {d}=\{d_1 \leq d_2 \leq \cdots \leq d_r \}$ such that $d_1 = d_2 = \cdots = d_{r_1} < d_{r_1+1}= \cdots = d_{r_1+r_2} < d_{r_1+\cdots +r_h}\cdots = d_{r_1+\cdots +r_h+r_{h+1}}$ : that is, $r_1+\cdots +r_{h+1}=r$ . $\omega $ is the Weyl group acting on $L_i$ to change the index, $\{ L_i \}^r_{i=1}$ come from the filtration of tautological bundle $\mathcal {S}_r$ of $Gr(r,n)$ . We could rewrite the equivariant I-function in the following way

(3.2) $$ \begin{align} I^{Gr(r,n)}_{T} =& 1+ \sum _{d} \sum_{|\vec{d}|= d} \sum_{ \omega \in S_{r} / S_{r_{1}} \times \cdots \times S_{r_{h+1}}}\omega\left[\prod_{i,j=1}^r \frac{\prod^{d_i-d_j}_{k=-\infty}(1-q^kL_iL^{-1}_j)}{\prod^{0}_{k=-\infty}(1-q^kL_iL^{-1}_j)} \prod_{i=1}^{r}\prod_{k=1}^{d_i}\prod_{m=1}^{n}\frac{1}{(1-q^kL_i\Lambda^{-1}_m)} \right] Q^d. \end{align} $$

Suppose $\omega $ changes $i_1$ to $i_2$ and $j_1$ to $j_2$ . Then one of the factors changes from

(3.3) $$ \begin{align} \frac{\prod^{d_{i_1}-d_{j_1}}_{k=-\infty}(1-q^kL_{i_1}L^{-1}_{j_1})}{\prod^{0}_{k=-\infty}(1-q^kL_{i_1}L^{-1}_{j_1})}\cdot \frac{\prod^{d_{i_2}-d_{j_2}}_{k=-\infty}(1-q^kL_{i_2}L^{-1}_{j_2})}{\prod^{0}_{k=-\infty}(1-q^kL_{i_2}L^{-1}_{j_2})}, \end{align} $$

to

(3.4) $$ \begin{align} \frac{\prod^{d_{i_1}-d_{j_1}}_{k=-\infty}(1-q^kL_{i_2}L^{-1}_{j_2})}{\prod^{0}_{k=-\infty}(1-q^kL_{i_2}L^{-1}_{j_2})} \cdot \frac{\prod^{d_{i_2}-d_{j_2}}_{k=-\infty}(1-q^kL_{i_1}L^{-1}_{j_1})}{\prod^{0}_{k=-\infty}(1-q^kL_{i_1}L^{-1}_{j_1})}. \end{align} $$

Since $\omega \in S_{r} / S_{r_{1}} \times \cdots \times S_{r_{h+1}}$ , we have $d_{i_1} \neq d_{i_2}, d_{j_1} \neq d_{j_2}$ . In equation (3.2), we have an order of partition $\vec {d}$ ; one can see from equation (3.3) to equation (3.4) that $\omega $ -action is just $\{ d_i \}$ rearranged without changing the form. There is a unique $\omega \in S_{r} /\left (S_{r_{1}} \times \ldots \times S_{r_{h+1}}\right )$ whose inverse $\omega ^{-1}$ arranges $\left (d_{1}, \ldots , d_{r}\right )$ in nondecreasing order $d_{1} \leq d_{2} \leq \ldots \leq d_{r}$ . Then we have

$$ \begin{align*} I^{Gr(r,n)}_{T}=\sum_d \sum_{d_1+d_2+ \cdots + d_r = d} Q^d \prod_{i,j=1}^r \frac{\prod^{d_i-d_j}_{k=-\infty}(1-q^kL_iL^{-1}_j)}{\prod^{0}_{k=-\infty}(1-q^kL_iL^{-1}_j)}\prod_{i=1}^{r}\prod_{k=1}^{d_i}\prod_{m=1}^{n}\frac{1}{(1-q^kL_i\Lambda^{-1}_m)}. \nonumber \end{align*} $$

Note that in [Reference Taipale15], the author claimed a version of the mirror theorem with a different I-function.

If we consider the standard representation of $GL_r$ , denoted by $E_r$ , then the associated bundle $\mathcal {P} \times _G R|_{F_{0,\beta }}$ can be identified with $\oplus _{i=1}^{r}L_i \otimes \mathcal {O}_{\mathbb {P}^1}(-d_i) $

$$ \begin{align*} \mathcal{D}^{E_r,l}|_{F_{0,\beta}} &= \mathrm{det}^{-l}R^{\bullet}\pi_*( \oplus_{i=1}^{r} L_i \otimes \mathcal{O}_{\mathbb{P}^1}(-d_i) ) \\ &= \mathrm{det}^{-l}( \oplus_{i=1}^{r} [L_i \otimes R^1\pi_*(\mathcal{O}_{\mathbb{P}^1}(-d_i))]^{-1} ) \\ &= \otimes_{i=1}^r \left( L_i^{d_i-1} \cdot q^{\frac{d_i(d_i-1)}{2}} \right)^l. \end{align*} $$

Similarly, if we take a dual standard representation, denoted by $E_r^{\vee }$ , then

$$ \begin{align*} \mathcal{D}^{E_r^\vee,l}|_{F_{0,\beta}} &= \mathrm{det}^{-l}( \oplus_{i=1}^{r} L^{-1}_i \otimes R^{0}\pi_*(\mathcal{O}_{\mathbb{P}^1}(d_i)) ) \\ &= \otimes_{i=1}^r \left( L_i^{d_i+1} \cdot q^{\frac{d_i(d_i+1)}{2}} \right)^l. \end{align*} $$

So the equivariant I-function of $Gr(r,n)$ with a level structure is as follows:

(3.5) $$ \begin{align} I^{Gr(r,n),E_r,l}_{T,d}=\sum_{d_1+d_2+ \cdots + d_r = d} Q^d \prod_{i,j=1}^r \frac{\prod^{d_i-d_j}_{k=-\infty}(1-q^kL_iL^{-1}_j)}{\prod^{0}_{k=-\infty}(1-q^kL_iL^{-1}_j)}\prod_{i=1}^{r}\frac{(L^{d_{i}}_{i}q^{\frac{d_{i}(d_{i}-1)}{2}})^l}{\prod_{k=1}^{d_i}\prod_{m=1}^{n}(1-q^kL_i\Lambda^{-1}_m)}, \end{align} $$

and

(3.6) $$ \begin{align} I^{Gr(r,n),E_r^\vee, l}_{T,d}=\sum_{d_1+d_2+ \cdots + d_r = d} Q^d \prod_{i,j=1}^r \frac{\prod^{d_i-d_j}_{k=-\infty}(1-q^kL_iL^{-1}_j)}{\prod^{0}_{k=-\infty}(1-q^kL_iL^{-1}_j)}\prod_{i=1}^{r}\frac{(L^{d_{i}}_{i}q^{\frac{d_{i}(d_{i}+1)}{2}})^l}{\prod_{k=1}^{d_i}\prod_{m=1}^{n}(1-q^kL_i \Lambda^{-1}_m)}. \end{align} $$

Remark. For the dual Grassmannian $Gr(n-r,n)$ , the $(\mathbb {C}^*)^n$ -action on $\mathbb {C}^n$ is the dual action, so the weights are $\mathbb {C}^n=\Lambda _1^{-1}+ \cdots + \Lambda _n^{-1}$ . The deduced action on $Gr(n-r,n)$ is as follows: $T \cdot B = BT$ , $B \in M_{n-r \times n}$ . So the corresponding equivariant I-function is as follows

$$ \begin{align*} I^{Gr(n-r,n),E_{n-r},l}_{T,d}=\sum_{d_1+d_2+ \cdots + d_{n-r} = d} Q^d \prod_{i,j=1}^{n-r} \frac{\prod^{d_i-d_j}_{k=-\infty}(1-q^k\tilde{L}_i\tilde{L}^{-1}_j)}{\prod^{0}_{k=-\infty}(1-q^k\tilde{L}_i\tilde{L}^{-1}_j)}\prod_{i=1}^{n-r}\frac{(\tilde{L}^{d_{i}}_{i}q^{\frac{d_{i}(d_{i}-1)}{2}})^l}{\prod_{k=1}^{d_i}\prod_{m=1}^{n}(1-q^k\tilde{L}_i\Lambda _m)} \nonumber \end{align*} $$

and

$$ \begin{align*} I^{Gr(n-r,n),E_{n-r}^\vee, l}_{T,d}=\sum_{d_1+d_2+ \cdots + d_{n-r} = d} Q^d \prod_{i,j=1}^{n-r} \frac{\prod^{d_i-d_j}_{k=-\infty}(1-q^k\tilde{L}_i\tilde{L}^{-1}_j)}{\prod^{0}_{k=-\infty}(1-q^k\tilde{L}_i\tilde{L}^{-1}_j)}\prod_{i=1}^{n-r}\frac{(\tilde{L}^{d_{i}}_{i}q^{\frac{d_{i}(d_{i}+1)}{2}})^l}{\prod_{k=1}^{d_i}\prod_{m=1}^{n}(1-q^k\tilde{L}_i \Lambda _m)}, \nonumber \end{align*} $$

where $\tilde {L}_i$ for $i=1,\ldots ,n-r$ come from the filtration of tautological bundle $\mathcal {S}_{n-r}$ over $Gr(n-r,n)$ .

Let T act on the Grassmannian $Gr(r,n)$ as before. Then there are ${n \choose r} $ fixed points: that is, denoted by $\{e_1, \ldots , e_n \}$ , the basis of $\mathbb {C}^n$ . Then the subspace V spanned by $\{e_{i_1}, \ldots , e_{i_r} \}$ is a T-fixed point for $\{i_1, \cdots , i_r \} \subset [n] $ . Let

$$ \begin{align*} \mathfrak{l}_{*}: {K}_{{T}}\left({Gr(r,n)}^{{T}}\right) \rightarrow {K}_{{T}}({Gr(r,n)}) \end{align*} $$

be the map induced from the close embedding $\mathfrak {l}: Gr(r,n)^T \hookrightarrow Gr(r,n)$ . The kernel and cokernel are $K_T(pt)$ -modules and have some support in the torus T. From a very general localisation theorem of Thomason [Reference Thomason16], we know

$$ \begin{align*} \operatorname{supp} \text { Coker } \mathfrak{l}_{*} \subset \bigcup_{\mu}\left\{\mathfrak{t}^{\mu}=1\right\}, \end{align*} $$

where the union is over finitely many nontrivial characters $\mu $ . The same is true of $\text { ker } \mathfrak {l}_{*}$ , but since

$$ \begin{align*} {K}_{{T}}\left({Gr(r,n)}^{{T}}\right)={K}({Gr(r,n)}) \otimes_{\mathbb{Z}} {K}_{{T}}({pt}) \end{align*} $$

has no such torsion, this forces $\text { ker } \mathfrak {l}_{*}=0$ , so after inverting finitely many coefficients of the form $t^{\mu }-1$ , we obtain an isomorphism: that is,

$$ \begin{align*} K_T^{loc}(Gr(r,n)^T) \cong K^{loc}_T(Gr(r,n)). \end{align*} $$

We denote $K^{loc}_{T}(-)$ by

$$ \begin{align*} K^{loc}_{T}(-) = K_T(-) \otimes_{R(T)} \mathcal{R}, \end{align*} $$

where $\mathcal {R} \cong \mathbb {Q}(t_1,\ldots ,t_n)$ and $\{t_i\}$ are the charaters of torus T.

Similarly, $T=\left (\mathbb {C}^{*}\right )^{n}$ -action on $Gr(n-r,n)$ also has $\binom {n}{n-r} = \binom {n}{r}$ isolated fixed points, which are indexed by $(n-r)$ -element subsets of $[n]$ , so identification of $Gr(r,n)^{T}$ with $Gr(n-r,n)^{T}$ gives an $\mathcal { R }$ -module isomorphism of $K^{loc}_{T} ( Gr(r,n) )$ with $K^{loc}_{T} ( Gr(n-r,n) )$ . Indeed, suppose W is a subspace of dimension r in a vector space V of dimension n. Then we have a natural short exact sequence

$$ \begin{align*} 0 \rightarrow W \rightarrow V \rightarrow V / W \rightarrow 0. \end{align*} $$

Taking the dual of this short exact sequence yields an inclusion of $(V / W)^{*}$ in $V^{*}$ with quotient $W^{*}$

$$ \begin{align*} 0 \rightarrow(V / W)^{*} \rightarrow V^{*} \rightarrow W^{*} \rightarrow 0, \end{align*} $$

so $ \psi : W \mapsto (V/W)^{*}$ gives a cannocial equivariant isomorphism $Gr(r, V) \cong Gr(n-r, V^{*})$ , where the action of $T=\left (\mathbb {C}^{*}\right )^{n}$ on $V^{*}$ is induced from the action of T on V. Thus, $\psi $ gives the canonical identification of fixed points

(3.7) $$ \begin{align} \psi : Gr(r,n)^{T} \longrightarrow Gr(n-r,n)^{T}, \qquad <e_j>_{j \in I} \longmapsto <e^{j}>_{j \in I^\complement}, \end{align} $$

where I is a set of $[n]$ with $|I|=r$ and $\{e^i\}_{i = 1} ^{n}$ is the dual basis of $\{e_i\}_{i = 1} ^{n}$ . Now we can state the following Level correspondence in Grassmann duality.

Theorem 3.1 Level correspondence

For the Grassmannian $Gr(r,n)$ and its dual Grassmannian $Gr(n-r,n)$ with standard $T=(\mathbb {C}^*)^n$ torus action, let $E_r$ , $E_{n-r}$ be the standard representation of $\operatorname {GL}(r,\mathbb {C})$ and $\operatorname {GL}(n-r,\mathbb {C})$ , respectively. Consider the following equivariant I-function:

$$ \begin{align*} I^{Gr(r,n),E_r,l}_T =& 1 + \sum_{d=1}^\infty I^{Gr(r,n),E_r,l}_{T,d} Q^d ,\\ I^{Gr(n-r,n),E_{n-r}^{\vee},-l}_T =& 1 + \sum_{d=1}^\infty I^{Gr(n-r,n),E^{\vee}_{n-r},-l}_{T,d} Q^d. \end{align*} $$

Then we have the following relations between $ I^{Gr(r,n),E_r,l}_{T,d} $ and $ I^{Gr(n-r,n),E^\vee _{n-r},-l}_{T,d} $ in $ K^{loc}_T(Gr(r,n)) \otimes \mathbb {C}(q) $ (which equals $ K^{loc}_T(Gr(n-r,n)) \otimes \mathbb {C}(q) $ ):

• ○ For $1-r\leq l\leq n-r-1$ , we have

  $$ \begin{align*} I_{T,d}^{Gr(r,n),E_{r},l} = I_{T,d}^{Gr(n-r,n),E_{n-r}^\vee,-l}. \nonumber \end{align*} $$

• ○ For $l=n-r$ , we have

  $$ \begin{align*} I_{T,d}^{Gr(r,n),E_{r},l}= \sum_{s=0}^d C_{s}( n-r,d) I_{T,d-s}^{Gr(n-r,n),E_{n-r}^\vee,-l}, \nonumber \end{align*} $$
  where $C_{s}(k,d) $ is defined as
  $$ \begin{align*} C_{s}(k,d)=\frac{(-1)^{k s}}{(q; q)_{s} q^{s(d-s+k)} \left(\bigwedge^{top}\mathcal{S}_{n-r} \right)^s}, \end{align*} $$
  and $\mathcal {S}_{n-r}$ is the tautological bundle of $Gr(n-r,n)$ .

• ○ For $l=-r$ , we have

  $$ \begin{align*} I_{T,d}^{Gr(n-r,n),E_{n-r}^\vee,-l}= \sum_{s=0}^d D_{s}( r, d) I_{T,d-s}^{Gr(r,n),E_{r},l}, \nonumber \end{align*} $$
  where
  $$ \begin{align*} D_{s}(r,d)=\frac{(-1)^{rs}}{ (q;q)_sq^{s(d-s)}\left( \bigwedge^{top}\mathcal{S}_{r} \right)^s}, \end{align*} $$
  and $\mathcal {S}_r$ is the tautological bundle of $Gr(r,n)$ .

Proof. From the discussion above, we prove the above identity by comparing $i^*_I I^{E_r, l}_{T}$ and $i^*_{I^\complement } I^{E_{n-r}^\vee , -l}_{T}$ ; here $i_{I}$ and $i_{I^\complement }$ are inclusion maps from the corresponding fixed points: that is, we compare two I-functions by restricting them to corresponding fixed points. Let $I= (j_1, \cdots , j _r)$ be the subset of $[n]=\{1, \ldots , n \}$ , with $|I|=r$ . Denote $v_1, v_2,\cdots , v_r$ , the fibre coordinates in the fibre of $\mathcal {S}$ at fixed point $<e_j>_{j \in I}$ , $\forall ( t_1, \cdots , t_n ) \in \left (\mathbb {C}^{*}\right )^{n}$ , with weights $\mathbb {C}^n= \Lambda _1 + \cdots + \Lambda _n $ and

$$ \begin{align*} & ( t_1, \cdots , t_n ) \cdot (e_{j_1}, \cdots, e_{j_r} ;v_1, v_2,\cdots, v_r ) = (t_{j_1}e_{j_1}, \cdots, t_{j_r} e_{j_r} ;v_1, v_2,\cdots, v_r ) \\ & \sim \operatorname{diag}(t_{j_1},\cdots, t_{j_r}) \cdot (t_{j_1} e_{j_1}, \cdots, t_{j_r} e_{j_r} ;v_1, v_2,\cdots, v_r ) = (e_{j_1}, \cdots, e_{j_r}; t_{j_1} v_1, t_{j_2} v_2,\cdots, t_{j_r} v_r ). \end{align*} $$

So the weights of $i^*_{I}\mathcal {S}_r$ are $\{\Lambda _i \}_{i \in I}$ and the weights of $i^*_{I^\complement }\mathcal {S}_{n-r}$ are $\{\Lambda ^{-1}_i \}_{i \in I^\complement }$ . Since the I-function is symmetric with respect to $\{L_i\}$ , we can take any choice of weights

$$ \begin{align*} i^*_I I^{Gr(r,n),E_r, l}_{T,d} = \sum_{|\vec{d}_I| = d} \prod_{i,j \in I } \frac{\prod^{d_i-d_j}_{k=-\infty}(1-q^k \Lambda_i \Lambda^{-1}_j)}{\prod^{0}_{k=-\infty}(1-q^k \Lambda_i \Lambda^{-1}_j)}\prod_{i \in I}\frac{(\Lambda^{d_{i}}_{i}q^{\frac{d_{i}(d_{i}-1)}{2}})^l}{\prod_{k=1}^{d_i}\prod_{m \in [n]}(1-q^k \Lambda_i\Lambda^{-1}_m)}, \nonumber \end{align*} $$

and

$$ \begin{align*} i^*_{I^{\complement}} I^{Gr(n-r,n),E_{n-r}^\vee, -l}_{T,d}=\sum_{|\vec{d}_{I^{\complement}}| = d} \prod_{i,j \in I^{\complement}} \frac{\prod^{d_i-d_j}_{k=-\infty}(1-q^k\Lambda^{-1}_i \Lambda_j)}{\prod^{0}_{k=-\infty}(1-q^k\Lambda^{-1}_i \Lambda_j)} \prod_{i \in I^{\complement}} \frac{(\Lambda^{-d_{i}}_{i}q^{\frac{d_{i}(d_{i}+1)}{2}})^{-l}}{\prod_{k=1}^{d_i}\prod_{m \in [n]}(1-q^k\Lambda^{-1}_i \Lambda_m)}. \nonumber \end{align*} $$

Using notation $\Lambda _{ij}=\Lambda _i \Lambda _j^{-1}$ and the following Lemma 3.2, we obtain

(3.8) $$ \begin{align} i^*_I I^{Gr(r,n),E_r, l}_{T,d} = \sum_{|\vec{d}_I| = d} \prod_{i \in I} \left(\frac{1}{ \prod_{j \in I}(q^{d_i-d_j+1}\Lambda_{i j};q)_{d_{j}}} \cdot \frac{(\Lambda^{d_{i}}_{i}q^{\frac{d_{i}(d_{i}-1)}{2}})^l}{ \prod_{j \in I^{\complement} }(q\Lambda_{i j};q)_{d_i}} \right), \end{align} $$

and

(3.9) $$ \begin{align} i^*_{I^{\complement}} I^{Gr(n-r,n),E_{n-r}^\vee,-l}_{T,d} = \sum_{|\vec{d}_{I^{\complement}}| = d} \prod_{i \in I^{\complement} } \left( \frac{1}{\prod_{j \in I^{\complement} }(q^{d_i-d_j+1}\Lambda_{j i};q)_{d_{j}}} \cdot \frac{(\Lambda^{d_{i}}_{i}q^{\frac{-d_{i}(d_{i}+1)}{2}})^{l}}{ \prod_{j \in I }(q\Lambda_{j i};q)_{d_i}} \right). \end{align} $$

Comparing equations (3.8) and (3.9) with equations (2.24) and (2.25), we obtain the conclusion.

Lemma 3.2. Let I be the subset of $[n]=\{1, \ldots , n \}$ . We have

$$ \begin{align*} \prod_{i,j \in I} \left( \frac{\prod^{d_{ij}}_{k=-\infty}(1-q^k x_{ij} )}{\prod^{0}_{k=-\infty}(1-q^k x_{ij} )} \frac{1}{\prod_{k=1}^{d_i}(1-q^k x_{i j})} \right) = \prod_{i,j \in I}\frac{1}{(q^{d_{ij}+1} x_{i j};q)_{d_{j}}}. \end{align*} $$

Proof. It is sufficient to consider one term. If $d_i \geq d_j $ , then

$$ \begin{align*} LHS = \frac{\prod_{k=1}^{d_{ij}}(1-q^k x_{ij})}{\prod_{k=1}^{d_i}(1-q^k x_{ij})} = \frac{1}{\prod_{k=d_{ij}+1}^{d_i}(1-q^k x_{ij})} =RHS. \end{align*} $$

If $d_i \leq d_j$ , then

$$ \begin{align*} LHS = \frac{1}{\prod_{k=d_{ij}+1}^{0}(1-q^k x_{ij}) \prod_{k=1}^{d_i}(1-q^k x_{ij})} =\frac{1}{\prod_{k=d_{ij}+1}^{d_i}(1-q^k x_{ij})} =RHS. \end{align*} $$