# Euler characteristics of tautological bundles over Quot schemes of curves

Dragos Oprea  Department of Mathematics, University of California, San Diego
<EMAIL_ADDRESS>and  Shubham Sinha  Department of Mathematics, University
of California, San Diego<EMAIL_ADDRESS>
###### Abstract.

We compute the Euler characteristics of tautological vector bundles and their
exterior powers over the Quot schemes of curves. We give closed-form
expressions over punctual Quot schemes in all genera. For higher rank
quotients of a trivial vector bundle, we obtain answers in genus $0$. We also
study the Euler characteristics of the symmetric powers of the tautological
bundles, for rank $0$ quotients.

##  1\. Introduction

In this paper, we prove several closed-form expressions for the holomorphic
Euler characteristics of tautological vector bundles and their exterior and
symmetric powers over the Quot schemes of curves.

###  1.1. Punctual Quot schemes

To set the stage, let $E\to C$ be a locally free sheaf over a smooth
projective curve $C$. Let $\operatorname{\mathsf{Quot}}_{d}(E)$ denote the
Quot scheme parameterizing rank $0$ degree $d$ quotients of $E$:

$0\to S\to E\to Q\to 0,\quad\text{rank }Q=0,\quad\text{deg }Q=d.$

It is easy to see that $\operatorname{\mathsf{Quot}}_{d}(E)$ is smooth of
dimension $Nd$ where $N=\text{rank }E.$

We write

$0\to\mathcal{S}\to p^{*}E\to\mathcal{Q}\to 0$

for the universal exact sequence over
$C\times\operatorname{\mathsf{Quot}}_{d}(E),$ and we let $p$ and $\pi$ denote
the two projections over the factors of
$C\times\operatorname{\mathsf{Quot}}_{d}(E)$. For any line bundle $L\to C$,
there is an induced tautological rank $d$ vector bundle over
$\operatorname{\mathsf{Quot}}_{d}(E)$ given by

$L^{[d]}=\pi_{*}(p^{*}L\otimes\mathcal{Q}).$

We first study the holomorphic Euler characteristics of all exterior powers
$\wedge^{k}L^{[d]}$. To this end, for any vector bundle $V$ over a scheme $Y$,
we set

$\wedge_{y}V:=\sum_{k}y^{k}\wedge^{k}V.$

We show

######  Theorem 1.

Let $E\to C$ be a vector bundle over a smooth projective curve, and let $L\to
C$ be a line bundle. Then

$\sum_{d=0}^{\infty}q^{d}\chi(\operatorname{\mathsf{Quot}}_{d}(E),\wedge_{y}L^{[d]})={(1-q)^{-\chi(\operatorname{\mathcal{O}}_{C})}}{(1+qy)^{\chi(E\otimes
L)}}.$

The same methods will establish a slightly stronger result:

######  Theorem 2.

For any line bundles $M_{1},M_{2},\dots,M_{r}$ and $L$ over $C$, where $0\leq
r\leq\text{rk }E-1$, we have

$\sum_{d=0}^{\infty}q^{d}\chi\left(\operatorname{\mathsf{Quot}}_{d}(E),\wedge_{y}L^{[d]}\otimes_{i=1}^{r}\big{(}\wedge_{x_{i}}M_{i}^{[d]}\big{)}^{\vee}\right)=(1-q)^{-\chi(\operatorname{\mathcal{O}}_{C})}(1+qy)^{\chi(E\otimes
L)}\prod_{i=1}^{r}(1-qx_{i}y)^{-\chi(M_{i}^{\vee}\otimes L)}.$

Our proofs rely on universality arguments in the spirit of [EGL] to reduce to
the case of genus $0$, and equivariant localization in genus $0$. These are
well-established techniques. In general however, the ensuing localization sums
are not immediately expressed in closed form. The novelty here is that we show
how to overcome the combinatorial difficulties by using Lagrange-Bürmann
inversion and considerations involving Schur polynomials and Jacobi-Trudi
determinants. These lead to drastic simplifications of the answers.

In a similar context, the prior work [MO1] also makes use of localization
techniques to calculate a large part of the intersection theory of Quot
schemes of curves (for quotients of arbitrary ranks of a trivial vector
bundle), in particular to derive and extend the Vafa-Intriligator formula. The
current $K$-theoretic setup requires that we handle the localization sums
differently. We believe that the simplicity of the final formulas makes the
calculations worthwhile to be recorded.

###  1.2. Analogies with surfaces

Theorems 1 and 2 suggest an unexpected analogy between $\mathsf{Quot}_{d}(E)$
and the Hilbert scheme of points $X^{[d]}$ over a smooth projective surface
$X$. Specifically, Theorem 1 can be compared with the calculations of [A1, Da,
Sc1]:

$\sum_{d=0}^{\infty}q^{d}\chi(X^{[d]},\wedge_{y}L^{[d]})=(1-q)^{-\chi(\mathcal{O}_{X})}(1+qy)^{\chi(L)}.$

This is proved in [Sc1] by passing to the derived category and computing the
image of $\wedge^{\bullet}L^{[d]}$ under the Bridgeland-King-Reid equivalence

${\mathbf{D}}^{\textrm{b}}(X^{[d]})\simeq\mathbf{D}^{\textrm{b}}_{S_{d}}(X^{d})$

established in [BKR, H]. The same result is obtained by studying different
equivariant limits in the Donaldson-Thomas theory of toric Calabi-Yau
$3$-folds in [A1].

In the same vein, Theorem 2 when $r=1$ mirrors the following result of [WZ]
and its strengthening by [K1]:

$\displaystyle\sum_{d=0}^{\infty}q^{d}\chi\left(X^{[d]},\wedge_{y}L^{[d]}\otimes(\wedge_{x}M^{[d]})^{\vee}\right)$
$\displaystyle=(1-q)^{-\chi(\mathcal{O}_{X})}(1+qy)^{\chi(L)}(1+qx)^{\chi(M^{\vee})}(1-qxy)^{-\chi(L\otimes
M^{\vee})}.$

While the latter equality is conjectured to hold in all dimensions $\geq 2$,
it was noted in [K1] that the naive analogue for symmetric products of curves
fails. Theorem 2 can be viewed as a remedy: a similar but different formula
holds for curves, and only when $N\geq 2$. Theorem 2 furthermore allows for
multiple dualized factors.

###  1.3. Higher rank

Turning to higher rank quotients, while we do not obtain a closed-form
expression for the corresponding generating series, our techniques yield the
result below. We restrict to the case $C=\mathbb{P}^{1}$, and $E$ a trivial
rank $N$ vector bundle. The Quot scheme
$\operatorname{\mathsf{Quot}}_{d}(E,r)$ parametrizes short exact sequences

$0\to S\to E\to Q\to 0,\quad\text{rank }Q=r,\quad\deg Q=d.$

######  Theorem 3.

Let $\deg L=\ell$ and $0<r<N$. We have

$\displaystyle\chi(\operatorname{\mathsf{Quot}}_{d}(E,r),\wedge_{y}L^{[d]})=(-1)^{(N-r-1)d}\left[q^{d}\right]\frac{\det(f_{i}(z_{j}))}{\det(z_{i}^{N-j})}.$

In the numerator $(f_{i}(z_{j}))$ is the $N\times N$ matrix with

$\displaystyle f_{i}(z)=\begin{cases}z^{\ell+d+N-i+1}&\text{ if }1\leq i\leq
N-r\\\ z^{N-i}(z+y)^{\ell+1}&\text{ if }N-r+1\leq i\leq N\end{cases}$

and $z_{1},\dots,z_{N}$ are the distinct roots of the equation
$(z-1)^{N}-q(z+y)z^{r-1}=0$. The denominator is the Vandermonde determinant.

In the statement above, the brackets indicate taking the coefficient of the
corresponding power of $q$.

In Corollary 15 of Section 4, we note a connection between
$\chi\left(\operatorname{\mathsf{Quot}}_{d}(E,r),\det L^{[d]}\right)$ and
Schur polynomials. We also work out several specializations of the
corresponding formula.

###  1.4. Symmetric products

For rank zero quotients, we similarly study the series of symmetric powers of
the tautological bundles

$\sum_{d=0}^{\infty}q^{d}\chi\left(\operatorname{\mathsf{Quot}}_{d}(E),\mathsf{Sym}_{y}L^{[d]}\right).$

Here, for any vector bundle $V$, we write

$\quad\mathsf{Sym}_{y}V=\sum_{k=0}^{\infty}y^{k}\,\mathsf{Sym}^{k}V.$

######  Theorem 4.

For $C=\mathbb{P}^{1}$ and $d\geq k$, we have

$\chi\left(\operatorname{\mathsf{Quot}}_{d}(E),\mathsf{Sym}^{k}L^{[d]}\right)=\binom{\chi(E\otimes
L)+k-1}{k}.$

The answer is independent of $d\geq k$. The question was also studied over
Hilbert schemes of points on surfaces [A1, Sc2]. Just as for exterior powers,
the analogy with the Hilbert scheme of surfaces persists here as well. Indeed,
the following result was established in [A1] via the Donaldson-Thomas theory
of toric Calabi-Yau $3$-folds

$\chi\left(X^{[d]},\mathsf{Sym}^{k}L^{[d]}\right)=\binom{\chi(L)+k-1}{k},$

whenever $d\geq k$ and $\chi(\mathcal{O}_{X})=1$.

Theorem 12 in Section 3 gives a more general result than Theorem 4 above,
covering the case $d<k$ as well.

In arbitrary genus, universality arguments as in [EGL] show that the
factorization

$\mathsf{W}=\sum_{d=0}^{\infty}q^{d}\chi\left(\operatorname{\mathsf{Quot}}_{d}(E),\mathsf{Sym}_{y}L^{[d]}\right)=\mathsf{A}^{\chi(\mathcal{O}_{C})}\cdot\mathsf{B}^{\chi(E\otimes
L)}$

holds true, for two universal power series
$\mathsf{A},\mathsf{B}\in\mathbb{Q}(y)[[q]]$ that depend on $N$, but not on
the triple $(C,E,L)$. Our results give precise information about the series
$\mathsf{B}$. While we can determine $\mathsf{A}$ in principle, we do not have
a closed-form expression.

######  Theorem 5.

We have

$\mathsf{B}=f\left(\frac{qy}{(1-y)^{N+1}}\right)$

where $f(z)$ is the solution to the equation

$f(z)^{N}-f(z)^{N+1}+z=0,\,\quad f(0)=1.$

For instance, in the special case $N=2$, we obtain

$f(z)=1+\frac{4}{3}\sinh^{2}\left(\frac{1}{3}\operatorname{arcsinh}\left(\frac{3\sqrt{3z}}{2}\right)\right).$

In the last section, we raise a few questions related to the results in this
paper.

###  1.5. Previous work.

Several aspects of the geometry of the punctual Quot schemes
$\mathsf{Quot}_{d}(E)$ are already understood. Since it is difficult to be
exhaustive here, we mention only a few selected results. The Poincaré
polynomials and the motives of $\mathsf{Quot}_{d}(E)$ are computed in [Bi,
BFP, C, R, Str], the cohomology and stabilization phenomena are considered in
[Mo], automorphisms and Torelli type results are studied in [BDH, Ga], various
cones of divisors are investigated in [GS, Str], positivity results for the
tautological vector bundles are obtained in [O], the cohomology of the tangent
bundle is studied in [BGS]. The punctual Quot schemes bear close connections
to the moduli space of bundles over curves, and in fact, the study of the
Poincaré polynomials and motives of the latter can be undertaken in this
context [BD, BGL, HPL]. As alluded to above, the (virtual) intersection theory
of Quot schemes (of quotients of arbitrary rank) over curves was studied in a
wider context, see [Ber1, Ber2, MO1]. Furthermore, in this wider context,
connections with the intersection theory of the moduli of bundles were proven
in [BDW, M, MO2].

Regarding the $K$-theory of Quot schemes, fewer direct calculations were
available. 111Over surfaces, the virtual $K$-theory of the punctual Quot
schemes $\mathsf{Quot}_{d}(E)$ was thoroughly studied in [AJLOP, Bo]. Via
cosection localization to a canonical curve, these results can be used to
compute certain twisted holomorphic Euler characteristics of tautological
bundles over the punctual Quot schemes. These calculations are indirect, and
more importantly, due to the aforementioned twists, the results of our paper
are not accessible via these methods.  We note however the work of [RZ] which
connects certain holomorphic Euler characteristics of line bundles over
certain Quot schemes (parametrizing rank 2 subbundles of a trivial high rank
bundle) with the Verlinde numbers over the moduli of rank $2$ bundles, via a
wallcrossing analysis as in [BDW, T]. 222It is natural to ask whether the
Euler characteristic $\chi(\wedge^{d}L^{[d]})$ given by Theorem 1 can be
computed via wall-crossing, say in rank $2$ for simplicity. In principle, we
are led to finding the Euler characteristics of certain higher rank twists of
the theta line bundles over the moduli space of semistable vector bundles,
whose intersection theory is known. However, obtaining exact formulas in this
fashion seems combinatorially difficult.

###  1.6. Acknowledgements

We thank Noah Arbesfeld, Patrick Girardet, Drew Johnson, Woonam Lim, Alina
Marian, Rahul Pandharipande and Ming Zhang for discussions related to
$K$-theoretic invariants of Quot and Hilbert schemes. The authors are
supported by the NSF through grant DMS 1802228.

##  2\. Rank zero quotients and exterior powers

In this section we prove Theorems 1 and 2. The proofs of the two theorems are
similar. The calculations for Theorem 1 are however simpler and already
illustrate the main points.

###  2.1. Torus action

We first establish Theorem 1 when $C=\mathbb{P}^{1}$ and
$E=\operatorname{\mathcal{O}}(a_{1})\oplus\cdots\oplus\operatorname{\mathcal{O}}(a_{N})$,
and for $L$ such that

$\deg L+a_{i}+1\geq 0$

for all $i$. The arbitrary genus case follows from here by universality
arguments, see Section 2.6 below.

Under the above assumptions, we seek to show that

(1)
$\sum_{d=0}^{\infty}q^{d}\chi(\operatorname{\mathsf{Quot}}_{d},\wedge_{y}L^{[d]})=(1-q)^{-1}(1+qy)^{\chi(E\otimes
L)}.$

Here, for simplicity, we wrote $\operatorname{\mathsf{Quot}}_{d}$ instead of
$\operatorname{\mathsf{Quot}}_{d}(E)$.

We evaluate expression (1) via Hirzebruch-Riemann-Roch

$\chi\left(\operatorname{\mathsf{Quot}}_{d},\wedge_{y}L^{[d]}\right)=\int_{\operatorname{\mathsf{Quot}}_{d}}\textnormal{ch}(\wedge_{y}L^{[d]})\,\textnormal{Td}\left(\operatorname{\mathsf{Quot}}_{d}\right),$

and we use $\mathbb{C}^{*}$-equivariant localization to compute the integral.
333It is natural to attempt localization directly in $K$-theory, but we were
unable to establish the result in this fashion.

To this end, we let $\mathbb{C}^{*}$ act on $E$ with weight $-w_{i}$ on the
summand $\operatorname{\mathcal{O}}(a_{i})$. This induces a
$\mathbb{C}^{\star}$-action on $\operatorname{\mathsf{Quot}}_{d}$. The fixed
subbundles correspond to split inclusions

$S=\bigoplus_{i=1}^{N}K_{i}(a_{i})\hookrightarrow
E=\bigoplus_{i=1}^{N}\mathcal{O}(a_{i}).$

Thus, the fixed loci are products of projective spaces

$\operatorname{\mathrm{F}}_{\vec{d}}=\mathbb{P}^{d_{1}}\times\cdots\times\mathbb{P}^{d_{N}}$

for vectors $\vec{d}=(d_{1},\ldots,d_{N})$ such that $d_{1}+\cdots+d_{N}=d$.
The factor $\mathbb{P}^{d_{i}}$ corresponds to the Hilbert scheme of $d_{i}$
points of $\mathbb{P}^{1}$ parameterizing short exact sequences

$0\to K_{i}\to\operatorname{\mathcal{O}}\to T_{i}\to 0$

such that $T_{i}$ is a torsion sheaf of length $d_{i}$.

There is a universal exact sequence

$0\to\mathcal{K}_{i}\to\operatorname{\mathcal{O}}\to\mathcal{T}_{i}\to 0$

over the product $\mathbb{P}^{1}\times\mathbb{P}^{d_{i}}$, with the universal
kernel given by

(2)
$\mathcal{K}_{i}=\operatorname{\mathcal{O}}_{\mathbb{P}^{1}}(-d_{i})\boxtimes\operatorname{\mathcal{O}}_{\mathbb{P}^{d_{i}}}(-1).$

For future reference, we note that the universal exact sequence
$0\to\mathcal{S}\to p^{*}E\to\mathcal{Q}\to 0$ over
$\mathbb{P}^{1}\times\operatorname{\mathsf{Quot}}_{d}$ restricts to
$\mathbb{P}^{1}\times\operatorname{\mathrm{F}}_{\vec{d}}$ as

(3)
$0\to\bigoplus_{i\in[N]}\mathcal{K}_{i}(a_{i})\to\operatorname{\mathcal{O}}(a_{1})\oplus\cdots\oplus\operatorname{\mathcal{O}}(a_{N})\to\bigoplus_{i\in[N]}\mathcal{T}_{i}(a_{i})\to
0,$

where pullbacks from the factors are understood above. We also set
$[N]=\\{1,2,\ldots,N\\}.$

By Atiyah-Bott localization, we have

(4)
$\chi(\operatorname{\mathsf{Quot}}_{d},\wedge_{y}L^{[d]})=\sum_{|\vec{d}|=d}\int_{\operatorname{\mathrm{F}}_{\vec{d}}}\textnormal{ch}(\wedge_{y}L^{[d]})\frac{\textnormal{Td}(\operatorname{\mathsf{Quot}}_{d})}{e_{\mathbb{C}^{*}}(\textrm{N}_{\vec{d}})}\bigg{|}_{\operatorname{\mathrm{F}}_{\vec{d}}}.$

Here $\textrm{N}_{\vec{d}}$ denotes the normal bundle of the fixed locus
$\textrm{F}_{\vec{d}}$.

###  2.2. Explicit calculations

We proceed to calculate the expressions appearing in the localization sum (4).
In the next subsections, we record the Todd genera, the normal bundle
contributions and the Chern characters of the tautological bundles.

####  2.2.1. Todd Classes

By (3), the tangent bundle
$T{\operatorname{\mathsf{Quot}}_{d}}=\operatorname{Hom}_{\pi}(\mathcal{S},\mathcal{Q})$
restricts to

$\bigoplus_{i,j\in[N]}\pi_{*}\left(\mathcal{K}^{\vee}_{i}(-a_{i})\otimes\mathcal{T}_{j}(a_{j})\right)$

over the fixed locus
$\operatorname{\mathrm{F}}_{\vec{d}}=\mathbb{P}^{d_{1}}\times\cdots\times\mathbb{P}^{d_{N}}$.
Here
$\pi:\operatorname{\mathsf{Quot}}_{d}\times\mathbb{P}^{1}\to\operatorname{\mathsf{Quot}}_{d}$
denotes the projection. In $K$-theory, the above expression equals

$\bigoplus_{i,j\in[N]}\pi_{*}\left(\mathcal{K}_{i}^{\vee}(a_{j}-a_{i})\right)-\bigoplus_{i,j\in[N]}\pi_{*}\left(\mathcal{K}^{\vee}_{i}\otimes\mathcal{K}_{j}(a_{j}-a_{i})\right).$

In this discussion $\pi_{\star}=R^{0}\pi_{\star}-R^{1}\pi_{\star}.$ Therefore
the Todd class of ${\operatorname{\mathsf{Quot}}_{d}}$ restricted to each
fixed locus is

$\prod_{i,j\in[N]}\textnormal{Td}\left(\pi_{*}\left(\mathcal{K}^{\vee}_{i}(a_{j}-a_{i})\right)\right)\left(\prod_{i,j\in[N],\,\,i\neq
j}\textnormal{Td}\left(\pi_{*}\left(\mathcal{K}^{\vee}_{i}\otimes\mathcal{K}_{j}(a_{j}-a_{i})\right)\right)\right)^{-1}.$

The above $(i,j)$-terms carry the weight $w_{i}-w_{j}$. The assumption $i\neq
j$ in the second product can be made since the term $i=j$ is trivial in genus
$0$.

####  2.2.2. Equivariant normal bundles

Over each fixed locus, the normal bundle is given by the moving part of the
tangent bundle:

(5)
$\displaystyle\textrm{N}_{\vec{d}}=T^{\text{mov}}\bigg{|}_{\operatorname{\mathrm{F}}_{\vec{d}}}$
$\displaystyle=\bigoplus_{i\neq
j}\pi_{*}\left(\mathcal{K}^{\vee}_{i}(-a_{i})\otimes\mathcal{T}_{j}(a_{j})\right)$
$\displaystyle=\bigoplus_{i\neq
j}\pi_{*}\left(\mathcal{K}_{i}^{\vee}(a_{j}-a_{i})\right)-\bigoplus_{i\neq
j}\pi_{*}\left(\mathcal{K}^{\vee}_{i}\otimes\mathcal{K}_{j}(a_{j}-a_{i})\right),$

where we continue to keep track of the weights $w_{i}-w_{j}$. Therefore, we
find the Euler classes

$\displaystyle\frac{1}{e_{\mathbb{C}^{*}}(\textrm{N}_{\vec{d}})}=\prod_{i,j\in[N],\,\,i\neq
j}\bigg{(}e_{\mathbb{C}^{*}}\left(\pi_{\star}\left(\mathcal{K}_{i}^{\vee}(a_{j}-a_{i})\right)\right)\bigg{)}^{-1}\prod_{i,j\in[N],\,i\neq
j}e_{\mathbb{C}^{*}}\left(\pi_{\star}\left(\mathcal{K}^{\vee}_{i}\otimes\mathcal{K}_{j}(a_{j}-a_{i})\right)\right).$

Collecting all expressions above, we obtain that over the fixed locus
$\operatorname{\mathrm{F}}_{\vec{d}}$, the factor
$\frac{\textnormal{Td}(\operatorname{\mathsf{Quot}}_{d})}{e_{\mathbb{C}^{*}}(\textrm{N}_{\vec{d}})}$
in the localization expression (4) restricts to

(6)
$\prod_{i\in[N]}\textnormal{Td}\left(\pi_{*}\left(\mathcal{K}_{i}^{\vee}\right)\right)\prod_{i,j\in[N],\,\,i\neq
j}\frac{\textnormal{Td}}{e_{\mathbb{C}^{*}}}\bigg{(}\pi_{*}\left(\mathcal{K}^{\vee}_{i}(a_{j}-a_{i})\right)\bigg{)}\prod_{i,j\in[N],\,i\neq
j}\frac{e_{\mathbb{C}^{*}}}{\textnormal{Td}}\bigg{(}\pi_{\star}\left(\mathcal{K}^{\vee}_{i}\otimes\mathcal{K}_{j}(a_{j}-a_{i})\right)\bigg{)}.$

####  2.2.3. Explicit contributions

The terms in (6) can be made explicit. For the first term, recalling (2), we
immediately compute

$\pi_{\star}(\mathcal{K}_{i}^{\vee})=\mathbb{C}^{d_{i}+1}\otimes\mathcal{O}_{{\mathbb{P}}^{d_{i}}}(1)\implies\textnormal{Td}(\pi_{*}(\mathcal{K}^{\vee}_{i}))=\left(\frac{h_{i}}{1-e^{-h_{i}}}\right)^{d_{i}+1},$

where $h_{i}$ is the hyperplane class on $\mathbb{P}^{d_{i}}$ (by abuse of
notation also pulled back to
$\mathbb{P}^{d_{1}}\times\cdots\times\mathbb{P}^{d_{N}}$). The equivariant
weights vanish for this term. (This is the Todd genus of the projective space,
as it should.)

Turning to the remaining terms, more generally, equation (2) straightforwardly
yields

$\displaystyle c(\pi_{*}(\mathcal{K}^{\vee}_{i}(a_{j}-a_{i})))$
$\displaystyle=(1+h_{i})^{d_{i}+a_{j}-a_{i}+1}$ $\displaystyle
c(\pi_{*}(\mathcal{K}^{\vee}_{i}\otimes\mathcal{K}_{j}(a_{j}-a_{i})))$
$\displaystyle=(1+(h_{i}-h_{j}))^{d_{i}-d_{j}+a_{j}-a_{i}+1}.$

In the equivariant cohomology, recalling that the above sheaves carry the
weight $w_{i}-w_{j}$, we obtain

$\displaystyle c(\pi_{*}(\mathcal{K}^{\vee}_{i}(a_{j}-a_{i})))$
$\displaystyle=(1+(h_{i}+w_{i}\epsilon-w_{j}\epsilon))^{d_{i}+a_{j}-a_{i}+1}$
$\displaystyle
c(\pi_{*}(\mathcal{K}^{\vee}_{i}\otimes\mathcal{K}_{j}(a_{j}-a_{i})))$
$\displaystyle=(1+(h_{i}+w_{i}\epsilon-
h_{j}-w_{j}\epsilon))^{d_{i}-d_{j}+a_{j}-a_{i}+1}.$

Here, $\epsilon$ denotes the equivariant parameter. This implies the following
expressions for the equivariant Todd genera

$\displaystyle\textnormal{Td}(\pi_{*}(\mathcal{K}^{\vee}_{i}(a_{j}-a_{i})))$
$\displaystyle=\bigg{(}\frac{h_{i}+w_{i}\epsilon-
w_{j}\epsilon}{1-e^{-(h_{i}+w_{i}\epsilon-
w_{j}\epsilon)}}\bigg{)}^{d_{i}+a_{j}-a_{i}+1}$
$\displaystyle\textnormal{Td}(\pi_{*}(\mathcal{K}^{\vee}_{i}\otimes\mathcal{K}_{j}(a_{j}-a_{i})))$
$\displaystyle=\bigg{(}\frac{h_{i}+w_{i}\epsilon-
h_{j}-w_{j}\epsilon}{1-e^{-(h_{i}+w_{i}\epsilon-
h_{j}-w_{j}\epsilon)}}\bigg{)}^{d_{i}-d_{j}+a_{j}-a_{i}+1}.$

Similarly we obtain the Euler classes

$\displaystyle
e_{\mathbb{C}^{*}}(\pi_{*}(\mathcal{K}^{\vee}_{i}(a_{j}-a_{i})))$
$\displaystyle=(h_{i}+w_{i}\epsilon-w_{j}\epsilon)^{d_{i}+a_{j}-a_{i}+1}$
$\displaystyle
e_{\mathbb{C}^{*}}(\pi_{*}(\mathcal{K}^{\vee}_{i}\otimes\mathcal{K}_{j}(a_{j}-a_{i})))$
$\displaystyle=(h_{i}+w_{i}\epsilon-
h_{j}-w_{j}\epsilon)^{d_{i}-d_{j}+a_{j}-a_{i}+1}.$

Simplification. All told, substituting the above expressions into (6) and
cancelling terms, we obtain

(7)
$\frac{\textnormal{Td}(\operatorname{\mathsf{Quot}}_{d})}{e_{\mathbb{C}^{*}}(\textrm{N}_{\vec{d}})}\bigg{|}_{\operatorname{\mathrm{F}}_{\vec{d}}}=\prod_{i\in[N]}h_{i}^{d_{i}+1}\prod_{i,j\in[N]}\bigg{(}\frac{z_{i}}{z_{i}-\alpha_{j}}\bigg{)}^{d_{i}+a_{j}-a_{i}+1}\prod_{i,j\in[N],\,i\neq
j}\bigg{(}\frac{z_{i}-z_{j}}{z_{i}}\bigg{)}^{d_{i}-d_{j}+a_{j}-a_{i}+1}$

where we set for notational convenience

$z_{i}=e^{h_{i}+w_{i}\epsilon},\quad\alpha_{i}=e^{w_{i}\epsilon}.$

We rewrite this in a slightly more convenient form in terms of the polynomial

$R(z)=\prod_{j\in[N]}(z-\alpha_{j}).$

Combining the $(i,j)$ and $(j,i)$-factors in the last product appearing in
(7), and judiciously accounting for the remaining terms, we eventually obtain

(8)
$\frac{\textnormal{Td}(\operatorname{\mathsf{Quot}}_{d})}{e_{\mathbb{C}^{*}}(\textrm{N}_{\vec{d}})}\bigg{|}_{\operatorname{\mathrm{F}}_{\vec{d}}}=\mathsf{u}\cdot\prod_{i\in[N]}\bigg{(}\frac{h_{i}}{R(z_{i})}\bigg{)}^{d_{i}+1}z_{i}^{d+1}\bigg{(}\frac{R(z_{i})}{\prod_{j\in[N]}(z_{j}-\alpha_{i})}\bigg{)}^{a_{i}+\ell+1}\cdot\prod_{i,j\in[N],\,\,i<j}(z_{i}-z_{j})^{2}$

for the sign

$\mathsf{u}=(-1)^{(N-1)(d+\sum(a_{i}+\ell+1))+\binom{N}{2}}.$

The integer $\ell$ included in the above expression will be useful later on.
For now, the value of $\ell$ plays no role. Any $\ell$ will work since

$\prod_{i}\frac{R(z_{i})}{\prod_{j}(z_{j}-\alpha_{i})}=1.$

####  2.2.4. Chern classes

For the remaining term in (4), we record the following

######  Lemma 6.

The equivariant restrictions of the Chern characters of the tautological
bundles to the fixed loci are given by

(9)
$\displaystyle\textnormal{ch}(\wedge_{y}L^{[d]})\bigg{|}_{\operatorname{\mathrm{F}}_{\vec{d}}}$
$\displaystyle=\prod_{i}\bigg{(}\frac{z_{i}(\alpha_{i}+y)}{\alpha_{i}(z_{i}+y)}\bigg{)}^{a_{i}+\ell+1}\bigg{(}\frac{z_{i}+y}{z_{i}}\bigg{)}^{d_{i}}$
(10)
$\displaystyle\textnormal{ch}\left((\wedge_{x}M^{[d]})^{\vee}\right)\bigg{|}_{\operatorname{\mathrm{F}}_{\vec{d}}}$
$\displaystyle=\prod_{i}\bigg{(}\frac{1+\alpha_{i}x}{1+z_{i}x}\bigg{)}^{a_{i}+m+1}(1+z_{i}x)^{d_{i}}$

where $L$ and $M$ are line bundles of degree $\ell$ and $m$ respectively.

###### Proof.

We only explain the first formula, the second assertion being entirely
similar. We note that over
$\textrm{F}_{\vec{d}}=\mathbb{P}^{d_{1}}\times\cdots\times\mathbb{P}^{d_{N}}$,
the bundle $L^{[d]}$ splits as contributions coming from each factor

$L^{[d]}=\pi_{\star}(\mathcal{Q}\otimes
p^{\star}L)=\bigoplus_{i\in[N]}\pi_{\star}(\mathcal{T}_{i}\otimes
p^{\star}L(a_{i})),$

with each summand acted on with $\mathbb{C}^{\star}$-weight $-w_{i}$. In
$K$-theory, we have by (2) that

$\mathcal{T}_{i}=\mathcal{O}-\mathcal{K}_{i}=\mathcal{O}_{\mathbb{P}^{1}\times\mathbb{P}^{d_{i}}}-\mathcal{O}_{\mathbb{P}^{1}}(-d_{i})\boxtimes\mathcal{O}_{\mathbb{P}^{d_{i}}}(-1).$

This yields

$\pi_{\star}(\mathcal{T}_{i}\otimes
p^{\star}L(a_{i}))=\mathbb{C}^{a_{i}+\ell+1}\otimes\mathcal{O}_{\mathbb{P}^{d_{i}}}-\mathbb{C}^{a_{i}-d_{i}+\ell+1}\otimes\mathcal{O}_{\mathbb{P}^{d_{i}}}(-1).$

The result follows immediately from here, using that
$\wedge_{y}(V+W)=\wedge_{y}V\cdot\wedge_{y}W$ and accounting for all terms. ∎

###  2.3. Proof of Theorem 1

With the above ingredients in place, the key steps of the argument are as
follows:

  * (i)

after judiciously accounting for all localization terms, the fixed point
contributions are summed using the Lagrange-Bürmann formula;

  * (ii)

next, the answer is recast as a quotient of suitable determinants. Schur
polynomials evaluated at the roots of a certain algebraic equation arise at
this step;

  * (iii)

finally, an application of the Jacobi-Trudi formula to the Schur polynomials
greatly simplifies the answer and gives the result.

To begin, we substitute equations (8) and (9) into the localization expression
(4). We obtain that $\chi(\operatorname{\mathsf{Quot}}_{d},\wedge_{y}L^{[d]})$
equals

(11) $\displaystyle\mathsf{u}\sum_{|\vec{d}|=d}\left[h_{1}^{d_{1}}\ldots
h_{N}^{d_{N}}\right]$
$\displaystyle\bigg{\\{}\prod_{i}\bigg{(}\frac{z_{i}(\alpha_{i}+y)}{\alpha_{i}(z_{i}+y)}\bigg{)}^{b_{i}}\bigg{(}\frac{z_{i}+y}{z_{i}}\bigg{)}^{d_{i}}\bigg{(}\frac{h_{i}}{R(z_{i})}\bigg{)}^{d_{i}+1}z_{i}^{d+1}\bigg{(}\frac{R(z_{i})}{\prod_{j}(z_{j}-\alpha_{i})}\bigg{)}^{b_{i}}$
$\displaystyle\cdot\prod_{i<j}(z_{i}-z_{j})^{2}\bigg{\\}}\bigg{|}_{\epsilon=0}.$

Here, we wrote for simplicity

$b_{i}=a_{i}+\ell+1.$

The brackets indicate extracting the coefficient of $h_{1}^{d_{1}}\ldots
h_{N}^{d_{N}}$ in the relevant expression; this corresponds to integration
over the product of projective spaces
$\textrm{F}_{\vec{d}}=\mathbb{P}^{d_{1}}\times\ldots\times\mathbb{P}^{d_{N}}.$
The equivariant parameter $\epsilon$ is set to $0$ at the end.

The rest of this section is dedicated to the explicit combinatorial
manipulations (i)-(iii) which bring the above expression (11) into the form
stated in Theorem 1. We first apply the multivariable Lagrange-Bürmann formula
[Ge]. The formulation we need in this case is as follows. Consider formal
power series $\Phi_{1}(h_{1}),\ldots,\Phi_{N}(h_{N})$ with $\Phi_{i}(0)\neq
0$, and consider a power series $\Psi(h_{1},\ldots,h_{N})$. We have

(12) $\sum_{(d_{1},\ldots,d_{N})}q_{1}^{d_{1}}\cdots
q_{N}^{d_{N}}\left[h_{1}^{d_{1}}\ldots
h_{N}^{d_{N}}\right]\left(\Phi_{1}(h_{1})^{d_{1}+1}\cdots\Phi_{N}^{d_{N}+1}(h_{N})\cdot\Psi(h_{1},\ldots,h_{N})\right)=\frac{\Psi}{J}$

for the change of variables

$q_{i}=\frac{h_{i}}{\Phi_{i}(h_{i})}$

with Jacobian

$J=\frac{dq_{1}}{dh_{1}}\cdots\frac{dq_{N}}{dh_{N}}.$

This formula will be used to derive equation (17) below. The intermediate
calculations are straightforward; nonetheless, we record the details for
completeness.

Set

$\Phi_{i}(h_{i})=\frac{h_{i}}{R(z_{i})}\frac{z_{i}+y}{z_{i}},$

and let

$\Psi=\mathsf{u}\,\prod_{i}\bigg{(}\frac{z_{i}(\alpha_{i}+y)}{\alpha_{i}(z_{i}+y)}\bigg{)}^{b_{i}}\bigg{(}\frac{z_{i}+y}{z_{i}}\bigg{)}^{-1}z_{i}^{d+1}\bigg{(}\frac{R(z_{i})}{\prod_{j}(z_{j}-\alpha_{i})}\bigg{)}^{b_{i}}\cdot\prod_{i<j}(z_{i}-z_{j})^{2}$

be determined by the remaining terms in (11). 444In the expression $\Psi$, we
regard the exponent $d$ in the term $z_{i}^{d+1}$ as an independent parameter,
foregoing for the moment the requirement that $d=d_{1}+\ldots+d_{N}$. The
careful reader may wish to replace the term $z_{i}^{d+1}$ by a more general
$z_{i}^{e+1}$ for $e\geq d$ in the proof below. This leads to (20) written
instead for the partition $\lambda_{k}=(e^{N},k)$ or $(k-N,(e+1)^{N})$. One
then specializes back to the case of interest $e=d$ and continues by applying
Lemma 7. Due to the factor $z_{i}-\alpha_{i}$ in $R(z_{i})$ which has a simple
zero at $h_{i}=0$, we have $\Phi_{i}(0)\neq 0$. We apply (12) with

$q_{1}=\ldots=q_{N}=q.$

Thus, letting $z_{i}$ be the root of the equation

$\displaystyle q(z_{i}+y)=z_{i}R(z_{i}),\quad z_{i}\big{|}_{q=0}=\alpha_{i},$

and letting $h_{i}$ be determined by $z_{i}=\alpha_{i}e^{h_{i}}$, we have
$q=\frac{h_{i}}{\Phi_{i}(h_{i})}.$ It follows from (11) and (12) that

$\chi\left(\operatorname{\mathsf{Quot}}_{d},\wedge_{y}L^{[d]}\right)=\left[q^{d}\right]\frac{\Psi}{J}\left(h_{1}(q),\ldots,h_{N}(q)\right).$

Equivalently,
$\chi\left(\operatorname{\mathsf{Quot}}_{d},\wedge_{y}L^{[d]}\right)$ equals

(13)
$\displaystyle\left[q^{d}\right]\mathsf{u}\prod_{i}\bigg{(}\frac{z_{i}(\alpha_{i}+y)}{\alpha_{i}(z_{i}+y)}\bigg{)}^{b_{i}}\bigg{(}\frac{z_{i}+y}{z_{i}}\bigg{)}^{-1}\frac{dh_{i}}{dq}z_{i}^{d+1}\bigg{(}\frac{R(z_{i})}{\prod_{j}(z_{j}-\alpha_{i})}\bigg{)}^{b_{i}}\cdot\prod_{i<j}(z_{i}-z_{j})^{2}\bigg{|}_{\epsilon=0}.$

Consider the polynomial

(14) $P(z)=zR(z)-q(z+y).$

Note that $P$ has degree $N+1$, so it admits $N+1$ roots, with
$z_{1},\ldots,z_{N}$ being $N$ of them. Let $z_{N+1}$ be the additional root
of $P$ which satisfies

$z_{N+1}\big{|}_{q=0}=0.$

We will greatly simplify (13) using the additional root $z_{N+1}$.

To this end, write $P(z)=(z-z_{1})\cdots(z-z_{N+1})$. A simple calculation
gives

(15)
$\displaystyle\frac{dq}{dh_{i}}=\frac{dq}{dz_{i}}\cdot\frac{dz_{i}}{dh_{i}}=z_{i}\frac{dq}{dz_{i}}=\frac{z_{i}}{z_{i}+y}P^{\prime}(z_{i}).$

Here, we used that

$q=\frac{z_{i}R(z_{i})}{z_{i}+y}\implies\frac{dq}{dz_{i}}=\frac{R(z_{i})}{z_{i}+y}+\frac{z_{i}R^{\prime}(z_{i})}{z_{i}+y}-\frac{z_{i}R(z_{i})}{(z_{i}+y)^{2}}=\frac{P^{\prime}(z_{i})}{z_{i}+y},$

where the definition of $P$ was necessary in the last equality. The terms in
(13) with exponent $b_{i}$ further simplify since

(16)
$\displaystyle\frac{z_{i}(\alpha_{i}+y)R(z_{i})}{\alpha_{i}(z_{i}+y)\prod_{j=1}^{N}(z_{j}-\alpha_{i})}=(-1)^{N}\frac{(z_{N+1}-\alpha_{i})}{\alpha_{i}}\iff(\alpha_{i}+y)z_{i}R(z_{i})=-(z_{i}+y)P(\alpha_{i})$

where the last identity holds true using (14) and the fact that $P(z_{i})=0,$
$R(\alpha_{i})=0$. Substituting identities (15) and (16) into (13), we obtain
the expression

(17)
$\left[q^{d}\right](-1)^{(N-1)d}\prod_{i=1}^{N}\bigg{(}\frac{\alpha_{i}-z_{N+1}}{\alpha_{i}}\bigg{)}^{b_{i}}\frac{z_{i}^{d+1}}{P^{\prime}(z_{i})}\cdot\prod_{1\leq
i\neq j\leq N}(z_{i}-z_{j})\bigg{|}_{\epsilon=0}.$

Having arrived at (17), a new idea is needed to go further. Note that

$\prod_{i=1}^{N}P^{\prime}(z_{i})=\prod_{i=1}^{N}\prod_{\stackrel{{\scriptstyle
j=1}}{{j\neq
i}}}^{N+1}(z_{i}-z_{j})\implies\prod_{i=1}^{N}\frac{1}{P^{\prime}(z_{i})}\cdot\prod_{1\leq
i\neq j\leq N}(z_{i}-z_{j})=\frac{\mathsf{V}_{N}}{\mathsf{V}_{N+1}}.$

Here, we introduced the two Vandermonde determinants

$\mathsf{V}_{N}=\begin{vmatrix}z_{1}^{N-1}&\cdots&z_{1}&1\\\
z_{2}^{N-1}&\cdots&z_{2}&1\\\ \vdots&\cdots&\vdots&\vdots\\\
z_{N}^{N-1}&\cdots&z_{N}&1\end{vmatrix},\quad\mathsf{V}_{N+1}=\begin{vmatrix}z_{1}^{N}&\cdots&z_{1}&1\\\
z_{2}^{N}&\cdots&z_{2}&1\\\ \vdots&\cdots&\vdots&\vdots\\\
z_{N+1}^{N}&\cdots&z_{N+1}&1\end{vmatrix}.$

We thus rewrite expression (17) in the form

(18)
$\left[q^{d}\right]\frac{(-1)^{(N-1)d}}{\mathsf{V}_{N+1}}\,\begin{vmatrix}z_{1}^{d+N}&z_{1}^{d+N-1}&\cdots&z_{1}^{d+1}\\\
z_{2}^{d+N}&z_{2}^{d+N-1}&\cdots&z_{2}^{d+1}\\\ \vdots&\vdots&\cdots&\vdots\\\
z_{N}^{d+N}&z_{N}^{d+N-1}&\cdots&z_{N}^{d+1}\end{vmatrix}\prod_{i=1}^{N}\bigg{(}\frac{\alpha_{i}-z_{N+1}}{\alpha_{i}}\bigg{)}^{b_{i}}\bigg{|}_{\epsilon=0}.$

Next, recall that $[q^{0}]z_{N+1}=0$, hence we may add terms which are
multiples of $z_{N+1}^{d+1}$ without changing the coefficient of $q^{d}$.
Using this observation, we enlarge the determinant in the numerator by adding
one more row and column. The answer is recast as the quotient of two
determinants of size $(N+1)\times(N+1)$:

(19)
$\displaystyle\left[\epsilon^{0}q^{d}\right]\frac{(-1)^{(N-1)d}}{\mathsf{V}_{N+1}}\,\begin{vmatrix}z_{1}^{d+N}&z_{1}^{d+N-1}&\cdots&z_{1}^{d+1}&\prod_{i=1}^{N}(\frac{\alpha_{i}-z_{1}}{\alpha_{i}})^{b_{i}}\\\
z_{2}^{d+N}&z_{2}^{d+N-1}&\cdots&z_{2}^{d+1}&\prod_{i=1}^{N}(\frac{\alpha_{i}-z_{2}}{\alpha_{i}})^{b_{i}}\\\
\vdots&\vdots&\cdots&\vdots&\vdots\\\
z_{N+1}^{d+N}&z_{N+1}^{d+N-1}&\cdots&z_{N+1}^{d+1}&\prod_{i=1}^{N}(\frac{\alpha_{i}-z_{N+1}}{\alpha_{i}})^{b_{i}}\end{vmatrix}.$

This does not change expression (18). Indeed, expanding along the last row,
the first entries do not contribute by the above reasoning, while the
rightmost corner contribution matches (18). The assumption $\deg L+a_{i}+1\geq
0$ made in the beginning of this section is also used here. This condition
rewrites as $b_{i}\geq 0$, so the terms we added on the last column do not
contribute poles at $q=0$.

Expression (19) is symmetric in the roots $z_{1},\ldots,z_{N+1}$ of $P(z)$.
The answer can be rewritten in terms of the elementary symmetric functions in
the $z_{i}$’s which depend polynomially on the $\alpha_{i}$’s. Thus (19) is a
rational fraction in the $\alpha$’s, with denominator
$\prod_{i=1}^{N}\alpha_{i}^{b_{i}}$ (coming from the last column). Since we
are interested in the coefficient of $\epsilon^{0}$, since the elementary
symmetric functions depend continuously on the $\alpha$’s, we may substitute
$\alpha_{i}=1$, noting that there are no poles in (19) at these values.

After the substitution, the $z_{i}$’s solve $z(z-1)^{N}-q(z+y)=0$.
Furthermore, since

$\sum_{i=1}^{N}b_{i}=\chi(E\otimes L):=\chi,$

the entries in the last column of the determinant (19) become

$(1-z_{i})^{\chi}=\sum_{k\geq 0}(-1)^{k}\binom{\chi}{k}z_{i}^{k}.$

Expanding the determinant along the last column yields sums over Schur
polynomials. Specifically, we obtain

(20) $\left[q^{d}\right]\sum_{k\geq
0}(-1)^{(N-1)d+k}\binom{\chi}{k}s_{\lambda_{k}}(z_{1},\ldots,z_{N+1}).$

Here, we set

$\lambda_{k}=(d^{N},k)=(d,\ldots,d,k),$

and $s_{\lambda_{k}}(z_{1},\dots z_{N+1})$ denotes the corresponding Schur
polynomial, when $k\leq d$. The terms for $d<k\leq d+N$ have vanishing
contribution due to repeating columns in the determinant. To account for the
ordering of the exponents, the shape of the partition changes when $k>d+N$. In
all cases, we find

$\lambda_{k}=\begin{cases}(d^{N},k)&\text{if }k\leq d\\\
(k-N,(d+1)^{N})&\text{if }k>d+N\end{cases}.$

The lemma below identifies the coefficient of $q^{d}$ in
$s_{\lambda_{k}}(z_{1},\ldots,z_{N+1})$. We obtain

$\displaystyle\chi(\operatorname{\mathsf{Quot}}_{d},\wedge_{y}L^{[d]})=\sum_{k=0}^{d}\binom{\chi}{k}y^{k}=\left[q^{d}\right](1+qy)^{\chi}(1-q)^{-1}.$

This completes the proof of Theorem 1 in genus $0$ under the assumption
$b_{i}\geq 0$ for all $1\leq i\leq N$.

######  Lemma 7.

We have

$\left[q^{d}\right]s_{\lambda_{k}}(z_{1},\dots,z_{N+1})=\begin{cases}(-1)^{d(N-1)}(-y)^{k}&\text{if
}k\leq d\\\ 0&\text{if }k>d+N\end{cases}.$

###### Proof.

Since the $z_{i}$’s are the roots of the polynomial $P(z)=z(z-1)^{N}-q(z+y)$,
the elementary symmetric functions in $z_{1},\ldots,z_{N+1}$ are

$\displaystyle e_{j}=\begin{cases}\binom{N}{j}&\text{if }j\neq N,N+1\\\
1+(-1)^{N-1}q&\text{if }j=N\\\ (-1)^{N}qy&\text{if }j=N+1.\\\ \end{cases}$

Assume $k\leq d$ so that $\lambda_{k}=(d^{N},k)$. The Jacobi-Trudi formula
expresses the Schur polynomial as a $d\times d$ determinant in the elementary
symmetric functions. The entries are dictated by the conjugate partition
$\lambda_{k}^{\prime}=((N+1)^{k},N^{d-k}),$ so that

(31) $\displaystyle
s_{\lambda_{k}}=\left|\begin{array}[]{ccccccccccc}e_{N+1}&0&0&\cdots&0&\vline&\quad
0&\cdots&0&0\\\ e_{N}&e_{N+1}&0&\cdots&0&\vline&\quad 0&\cdots&0&0\\\
e_{N-1}&e_{N}&e_{N+1}&\cdots&0&\vline&\quad 0&\cdots&0&0\\\
\vdots&\vdots&\vdots&\cdots&\vdots&\vline&\,\,\,\,\;\;\vdots&\cdots&\vdots&\vdots&\\\
e_{N-k+2}&e_{N-k+3}&e_{N-k+4}&\cdots&e_{N+1}&\vline&\quad 0&\cdots&0&0\\\
&&&&&\vline&&&&\\\ \hline\cr
e_{N-k}&e_{N-k+1}&e_{N-k+2}&\cdots&e_{N-1}&\vline&\quad e_{N}&\cdots&0&0\\\
\vdots&\vdots&\vdots&\cdots&\vdots&\vline&\vdots&\cdots&\vdots&\vdots\\\
e_{N-d+2}&e_{N-d+3}&e_{N-d+4}&\cdots&e_{N-d+k+1}&\vline&\quad
e_{N-d+k+2}&\cdots&e_{N}&e_{N+1}\\\
e_{N-d+1}&e_{N-d+2}&e_{N-d+3}&\cdots&e_{N-d+k}&\vline&\quad
e_{N-d+k+1}&\cdots&e_{N-1}&e_{N}\\\ \end{array}\right|.$

Each of the $e_{j}$’s is at most linear in $q$. Since the determinant has size
$d$, extracting the $q^{d}$ coefficient is immediate. In fact, we can replace
the $e_{j}$’s by their linear terms in $q$; these are zero unless $j=N$ or
$j=N+1$. We obtain that

$\displaystyle\left[q^{d}\right]s_{\lambda_{k}}=\left[q^{d}\right]\left|\begin{array}[]{cccccc}\begin{matrix}e_{N+1}&0&0&\cdots&0&0\\\
e_{N}&e_{N+1}&0&\cdots&0&0\\\ \vdots&\vdots&\vdots&\cdots&\vdots&\vdots\\\
0&0&0&\cdots&e_{N}&e_{N+1}\end{matrix}&\vline&\mbox{\Large 0}\\\
\hline\cr\mbox{\Large 0}&\vline&\begin{matrix}e_{N}&e_{N+1}&0&\cdots&0&0\\\
0&e_{N}&e_{N+1}&\cdots&0&0\\\ \vdots&\vdots&\vdots&\cdots&\vdots&\vdots\\\
0&0&0&\cdots&0&e_{N}\end{matrix}\end{array}\right|.$

Thus,

$\displaystyle[q^{d}]s_{\lambda_{k}}=\left[q^{d}\right]e_{N+1}^{k}e_{N}^{d-k}=(-1)^{kN+(d-k)(N-1)}y^{k}.$

The case $k>d+N$ changes the conjugate partition $\lambda^{\prime}_{k}$, but
the reasoning is identical. ∎

###  2.4. Proof of Theorem 2

A similar but slightly more involved argument yields Theorem 2 in genus $0$
when $b_{i}\geq 0$ for all $1\leq i\leq N$. Specifically, we prove that

(32)
$\chi\left(\operatorname{\mathsf{Quot}}_{d},\wedge_{y}L^{[d]}\otimes_{p=1}^{r}\left(\wedge_{x_{p}}M_{p}^{[d]}\right)^{\vee}\right)=\left[q^{d}\right](1-q)^{-1}(1+qy)^{\chi(E\otimes
L)}\prod_{p=1}^{r}(1-x_{p}yq)^{-\chi(L\otimes M_{p}^{\vee})}.$

We indicate some of the steps.

Just as in Theorem 1, we begin by applying Hirzebruch-Riemann-Roch followed by
Atiyah-Bott localization:

(33)
$\chi\left(\operatorname{\mathsf{Quot}}_{d},\wedge_{y}L^{[d]}\otimes_{p=1}^{r}\left(\wedge_{x_{p}}M_{p}^{[d]}\right)^{\vee}\right)=\sum_{\vec{d}}\int_{\operatorname{\mathrm{F}}_{\vec{d}}}\textnormal{ch}(\wedge_{y}L^{[d]})\prod_{p=1}^{r}\textnormal{ch}\left(\left(\wedge_{x_{p}}M_{p}^{[d]}\right)^{\vee}\right)\frac{\textnormal{Td}(\operatorname{\mathsf{Quot}}_{d})}{e_{\mathbb{C}^{*}}(\textrm{N}_{\vec{d}})}\bigg{|}_{\operatorname{\mathrm{F}}_{\vec{d}}}.$

All terms that appear here have been computed in the previous subsections.
Using (8), (9) and (10), we rewrite (33) as

$\displaystyle\mathsf{u}\sum_{|\vec{d}|=d}\left[h_{1}^{d_{1}}\cdots
h_{N}^{d_{N}}\right]$
$\displaystyle\,\bigg{\\{}\,\prod_{i=1}^{N}\,\,\bigg{(}\bigg{(}\frac{z_{i}(\alpha_{i}+y)}{\alpha_{i}(z_{i}+y)}\bigg{)}^{b_{i}}\bigg{(}\frac{z_{i}+y}{z_{i}}\bigg{)}^{d_{i}}\prod_{p=1}^{r}\bigg{(}\frac{1+\alpha_{i}x_{p}}{1+z_{i}x_{p}}\bigg{)}^{a_{i}+m_{p}+1}(1+z_{i}x_{p})^{d_{i}}\bigg{(}\frac{h_{i}}{R(z_{i})}\bigg{)}^{d_{i}+1}$
$\displaystyle
z_{i}^{d+1}\bigg{(}\frac{R(z_{i})}{\prod_{j=1}^{N}(z_{j}-\alpha_{i})}\bigg{)}^{b_{i}}\bigg{)}\cdot\prod_{1\leq
i<j\leq N}(z_{i}-z_{j})^{2}\bigg{\\}}\,\bigg{|}_{\epsilon=0}$

where $b_{i}=a_{i}+\ell+1$ and $\mathsf{u}=(-1)^{(N-1)(d+\sum
b_{i})+\binom{N}{2}}$. Here, we set $m_{p}=\deg M_{p}$.

Next, the Lagrange-Bürmann formula with the change of variables

(34) $\displaystyle q(z_{i}+y)\prod_{p=1}^{r}(1+z_{i}x_{p})=z_{i}R(z_{i})$

turns (33) into the following unwieldy expression

$\displaystyle\left[q^{d}\right]\,\mathsf{u}\,\,\prod_{i=1}^{N}$
$\displaystyle\left[\bigg{(}\frac{z_{i}(\alpha_{i}+y)}{\alpha_{i}(z_{i}+y)}\bigg{)}^{b_{i}}\prod_{p=1}^{r}\bigg{(}\frac{1+\alpha_{i}x_{p}}{1+z_{i}x_{p}}\bigg{)}^{a_{i}+m_{p}+1}\bigg{(}\frac{z_{i}+y}{z_{i}}\prod_{p=1}^{r}(1+z_{i}x_{p})\bigg{)}^{-1}\frac{dh_{i}}{dq}z_{i}^{d+1}\bigg{(}\frac{R(z_{i})}{\prod_{j}(z_{j}-\alpha_{i})}\bigg{)}^{b_{i}}\right]$
$\displaystyle\cdot\prod_{1\leq i<j\leq
N}(z_{i}-z_{j})^{2}\bigg{|}_{\epsilon=0}.$

However, there are further simplifications. To this end, we define the
polynomial

$P(z)=zR(z)-q(z+y)\prod_{p=1}^{r}(1+zx_{p}).$

Since $r\leq N-1$, the degree of $P$ is $N+1$, so there is an additional root
$z_{N+1}$ for $P$. Following the same steps that led to (17), we simplify the
above expression to

(35)
$\left[q^{d}\right](-1)^{(N-1)d}f(z_{N+1})\,\prod_{i=1}^{N}\,\frac{z_{i}^{d+1}}{P^{\prime}(z_{i})}\prod_{1\leq
i\neq j\leq N}(z_{i}-z_{j})\bigg{|}_{\epsilon=0}$

where

$f(z)=\prod_{p=1}^{r}(1+zx_{p})^{m_{p}-\ell}\prod_{i=1}^{N}\left(\frac{\alpha_{i}-z}{\alpha_{i}}\right)^{b_{i}}.$

We record the details of the simplification in the lemma below; the reader can
also skip directly to (38).

######  Lemma 8.

We have

(36)
$\displaystyle\bigg{(}\frac{z_{i}+y}{z_{i}}\prod_{p=1}^{r}(1+z_{i}x_{p})\bigg{)}\frac{dq}{dh_{i}}=P^{\prime}(z_{i})$

and

(37)
$\prod_{i=1}^{N}\left(\bigg{(}\frac{z_{i}(\alpha_{i}+y)}{\alpha_{i}(z_{i}+y)}\,\frac{R(z_{i})}{\prod_{j=1}^{N}(z_{j}-\alpha_{i})}\bigg{)}^{b_{i}}\prod_{p=1}^{r}\bigg{(}\frac{1+\alpha_{i}x_{p}}{1+z_{i}x_{p}}\bigg{)}^{a_{i}+m_{p}+1}\right)=(-1)^{(N-1)\sum
b_{i}}f(z_{N+1}).$

###### Proof.

Equation (36) follows by differentiating the expression for $q$ given in (34).
For (37), recall $b_{i}=a_{i}+\ell+1$, and use the following identities

$\displaystyle z_{i}R(z_{i})$
$\displaystyle=q(z_{i}+y)\prod_{p=1}^{r}(1+z_{i}x_{p}),$
$\displaystyle\prod_{j=1}^{N}(z_{j}-\alpha_{i})$
$\displaystyle=(-1)^{N+1}\frac{P(\alpha_{i})}{z_{N+1}-\alpha_{i}}=(-1)^{N}q\frac{(\alpha_{i}+y)\prod_{p=1}^{r}(1+\alpha_{i}x_{p})}{z_{N+1}-\alpha_{i}}.$

In the last line we used the definition of $P$ and the fact that
$R(\alpha_{i})=0.$ Then (37) becomes

$\prod_{i=1}^{N}\left(\bigg{(}(-1)^{N}\frac{(z_{N+1}-\alpha_{i})}{\alpha_{i}}\bigg{)}^{b_{i}}\prod_{p=1}^{r}\bigg{(}\frac{1+\alpha_{i}x_{p}}{1+z_{i}x_{p}}\bigg{)}^{m_{p}-\ell}\right).$

Finally, recalling that $\alpha_{i}$ and $z_{i}$ are roots of $R$ and $P$, for
each fixed $p$ we have

$\displaystyle\prod_{i=1}^{N}\frac{1+\alpha_{i}x_{p}}{1+z_{i}x_{p}}=\frac{R\left(-1/x_{p}\right)}{P\left(-1/x_{p}\right)}\left(-\frac{1}{x_{p}}-z_{N+1}\right)=(1+z_{N+1}x_{p}).$

In the last equality, we used again the definition of $P$ in terms of $R$. The
lemma follows from here. ∎Having arrived at (35), by the same reasoning as in
(18) we rewrite the answer as the quotient of two determinants

(38)
$\left[\epsilon^{0}q^{d}\right]\frac{(-1)^{(N-1)d}}{\det(z_{i}^{N-j+1})}\,\begin{vmatrix}z_{1}^{d+N}&z_{1}^{d+N-1}&\cdots&z_{1}^{d+1}\\\
z_{2}^{d+N}&z_{2}^{d+N-1}&\cdots&z_{2}^{d+1}\\\ \vdots&\vdots&\cdots&\vdots\\\
z_{N}^{d+N}&z_{N}^{d+N-1}&\cdots&z_{N}^{d+1}\end{vmatrix}f(z_{N+1}).$

The denominator is the Vandermonde determinant of size $(N+1)\times(N+1)$,
while the numerator has size $N\times N$. Using the previous arguments, in
particular that $[q^{0}]z_{N+1}=0$, we enlarge the determinant appearing in
the numerator of (38) by adding one row and one column:

(39)
$\displaystyle[\epsilon^{0}q^{d}]\frac{(-1)^{(N-1)d}}{\det(z_{i}^{N-j+1})}\,\begin{vmatrix}z_{1}^{d+N}&z_{1}^{d+N-1}&\cdots&z_{1}^{d+1}&f(z_{1})\\\
z_{2}^{d+N}&z_{2}^{d+N-1}&\cdots&z_{2}^{d+1}&f(z_{2})\\\
\vdots&\vdots&\cdots&\vdots&\vdots\\\
z_{N+1}^{d+N}&z_{N+1}^{d+N-1}&\cdots&z_{N+1}^{d+1}&f(z_{N+1})\end{vmatrix}.$

Since (39) is symmetric in $z_{i}^{\prime}s$, it can be written as a rational
function in the $\alpha_{i}$’s whose denominator equals
$\prod_{i=1}^{N}\alpha_{i}^{b_{i}}$ coming from the denominator of $f$. The
substitution $\alpha_{i}=1$ therefore makes sense. After this substitution,
the last column can be rewritten in terms of

$f(t)|_{\alpha_{i}=1}=\prod_{p=1}^{r}(1+x_{p}t)^{m_{p}-\ell}\cdot\left(1-t\right)^{\chi}$

for the values $t=z_{1},z_{2},\ldots,z_{N+1}$. Here $\chi=\chi(E\otimes L).$
We expand $f(t)$ into powers $t^{k}$, and then we expand the determinant (39)
along the last column yielding

(40) $(-1)^{(N-1)d}\sum_{k\geq
0}\left[t^{k}\right]f(t)\cdot\left[q^{d}\right]s_{\lambda_{k}}(z_{1},\ldots,z_{N+1}),$

for the partition

$\lambda_{k}=\begin{cases}(d^{N},k)&\text{if }k\leq d\\\
(k-N,(d+1)^{N})&\text{if }k>d+N.\end{cases}$

By Lemma 9 below, for $k\leq d$ we have

$\left[q^{d}\right]s_{\lambda_{k}}(z_{1},\ldots,z_{N+1})=(-1)^{(N-1)d}(-y)^{k}\left[t^{d-k}\right]\frac{1}{(1-t)(1-x_{1}yt)\cdots(1-x_{r}yt)}.$

Substituting the last formula into (40), we obtain that

$\displaystyle\chi$
$\displaystyle\left(\operatorname{\mathsf{Quot}}_{d},\wedge_{y}L^{[d]}\otimes_{p=1}^{r}\big{(}\wedge_{x_{p}}M_{p}^{[d]}\big{)}^{\vee}\right)$
$\displaystyle=\sum_{k=0}^{d}\left[t^{k}\right]\left((1-t)^{\chi(E\otimes
L)}\prod_{p=1}^{r}(1+x_{p}t)^{m_{p}-\ell}\right)\cdot(-y)^{k}\left[t^{d-k}\right]\frac{1}{(1-t)\prod_{p=1}^{r}(1-x_{p}yt)}$
$\displaystyle=\sum_{k=0}^{d}\left[t^{k}\right]\left((1+yt)^{\chi(E\otimes
L)}\prod_{p=1}^{r}(1-x_{p}yt)^{m_{p}-\ell}\right)\cdot\left[t^{d-k}\right]\frac{1}{(1-t)\prod_{p=1}^{r}(1-x_{p}yt)}$
$\displaystyle=\left[t^{d}\right]\frac{(1+yt)^{\chi(E\otimes
L)}}{(1-t)\prod_{p=1}^{r}(1-x_{p}yt)^{\chi(L\otimes M_{p}^{\vee})}}.$

This completes the proof of (32) and of Theorem 2 in genus $0$ when $b_{i}\geq
0$ for all $i$.

###  2.5. Schur polynomials

Let $z_{1},\dots z_{N+1}$ be $N+1$ roots of

$P(z)=z(z-1)^{N}-q(z+y)(1+zx_{1})\cdots(1+zx_{r}),$

where $0\leq r\leq N-1$. We show

######  Lemma 9.

For the partition $\lambda_{k}$ above, and $k\leq d$, we have

$\left[q^{d}\right]s_{\lambda_{k}}(z_{1},\dots,z_{N+1})=(-1)^{(N-1)d}(-y)^{k}\left[t^{d-k}\right]\frac{1}{(1-t)(1-x_{1}yt)\cdots(1-x_{r}yt)}.$

If $k>d+N$, the coefficient vanishes.

###### Proof.

The proof is similar to that of Lemma 7. Assume $k\leq d$, the other case
being analogous. Since the $z_{i}$’s are the roots of the polynomial $P(z)$,
the elementary symmetric functions are

$\displaystyle
e_{j}=\binom{N}{j}+(-1)^{j-1}q\left[z^{N+1-j}\right](y+z)(1+zx_{1})\cdots(1+zx_{r}).$

We examine again the Jacobi-Trudi determinant (31)

$\displaystyle
s_{\lambda_{k}}=\left|\begin{array}[]{cccccccccc}e_{N+1}&0&\cdots&0&\vline&\quad
0&\cdots&0&0\\\ e_{N}&e_{N+1}&\cdots&0&\vline&\quad 0&\cdots&0&0\\\
e_{N-1}&e_{N}&\cdots&0&\vline&\quad 0&\cdots&0&0\\\
\vdots&\vdots&\cdots&\vdots&\vline&\quad\vdots&\cdots&\vdots&\vdots&\\\
e_{N-k+2}&e_{N-k+3}&\cdots&e_{N+1}&\vline&\quad 0&\cdots&0&0\\\
&&&&\vline&&&&&\\\ \hline\cr e_{N-k}&e_{N-k+1}&\cdots&e_{N-1}&\vline&\quad
e_{N}&\cdots&0&0\\\
\vdots&\vdots&\cdots&\vdots&\vline&\vdots&\cdots&\vdots&\vdots\\\
e_{N-d+2}&e_{N-d+3}&\cdots&e_{N-d+k+1}&\vline&\quad
e_{N-d+k+2}&\cdots&e_{N}&e_{N+1}\\\
e_{N-d+1}&e_{N-d+2}&\cdots&e_{N-d+k}&\vline&\quad
e_{N-d+k+1}&\cdots&e_{N-1}&e_{N}\\\ \end{array}\right|.$

The $e_{j}$’s are at most linear in $q$. To find the coefficient of $q^{d}$ in
the above $d\times d$ determinant, we may thus replace $e_{j}$ with the
coefficient of the linear term in $q$. Thus, we may take

(41) $e_{j}=(-1)^{j-1}\left[z^{N+1-j}\right](y+z)(1+zx_{1})\cdots(1+zx_{r}).$

In particular $e_{N+1}=(-1)^{N}y.$ Furthermore, note that the first $k\times
k$ block of the determinant is lower triangular, hence

$\left[q^{d}\right]s_{\lambda_{k}}=e_{N+1}^{k}\cdot
T_{d-k}=(-1)^{Nk}y^{k}\cdot T_{d-k}$

where $T_{m}$ is the $m\times m$ determinant

$T_{m}=\begin{vmatrix}e_{N}&e_{N+1}&0&\cdots&0&0\\\
e_{N-1}&e_{N}&e_{N+1}&\cdots&0&0\\\ e_{N-2}&e_{N-1}&e_{N}&\cdots&0&0\\\
\vdots&\vdots&\vdots&\cdots&\vdots\\\
e_{N-m+2}&e_{N-m+3}&e_{N-m+2}&\cdots&e_{N}&e_{N+1}\\\
e_{N-m+1}&e_{N-m+2}&e_{N-m+1}&\cdots&e_{N-1}&e_{N}\end{vmatrix}.$

The argument is completed using the Lemma below. ∎

######  Lemma 10.

Assume $e_{1},\ldots,e_{N+1}$ are given by (41). For any $m\geq 0$, we have

(42)
$T_{m}=(-1)^{(N-1)m}\left[t^{m}\right]\frac{1}{(1-t)(1-x_{1}yt)\cdots(1-x_{r}yt)}.$

###### Proof.

We set $T_{0}=1$ and $T_{\ell}=0$ for $\ell<0$. By expanding the determinant
$T_{m}$ along the first column and then successively along the rows, we obtain
the recursion

$T_{m}=\sum_{j=0}^{r}(-1)^{j}e_{N+1}^{j}e_{N-j}T_{m-j-1}\quad\text{ for all
}m>0.$

Note that by (41), for degree reasons we have $e_{N-j}=0$ if $j>r$. This
explains the upper bound of the index $j$ in the sum. Forming the generating
series

$T=\sum_{m=0}^{\infty}T_{m}t^{m},$

the above recursion immediately yields

$T=\bigg{(}1-\sum_{j=0}^{r}(-1)^{j}e_{N+1}^{j}e_{N-j}t^{j+1}\bigg{)}^{-1}.$

Substituting the values of $e_{j}$ from (41), we obtain for all $0\leq j\leq
r$ that

$\displaystyle(-1)^{j+1}e_{N+1}^{j}e_{N-j}$
$\displaystyle=(-1)^{N(j+1)}y^{j+1}\left[z^{j+1}\right]\left(\left(1+\frac{z}{y}\right)(1+zx_{1})\cdots(1+zx_{r})\right)$
$\displaystyle=\left[t^{j+1}\right]\left((1-(-1)^{N-1}t)(1-(-1)^{N-1}x_{1}yt)\cdots(1-(-1)^{N-1}x_{r}yt)\right),$

where the substitution $z=(-1)^{N}yt$ was carried out in the last step.
Therefore

$\displaystyle
T=\bigg{(}1-\sum_{j=0}^{r}(-1)^{j}e_{N+1}^{j}e_{N-j}t^{j+1}\bigg{)}^{-1}=\frac{1}{(1-(-1)^{N-1}t)\prod_{p=1}^{r}(1-(-1)^{N-1}x_{p}yt)}.$

Taking the coefficient of $t^{m}$ gives the required identity. ∎

###  2.6. Arbitrary genus

Relying on the ideas of [EGL], we show how the calculations for
$C=\mathbb{P}^{1}$ imply Theorems 1 and 2 for arbitrary genus. We explain this
for Theorem 1, the case of Theorem 2 being entirely similar. The argument is
also noted and used in [OP] over surfaces for punctual quotients of trivial
bundles, and extended to quotients of arbitrary vector bundles in [St2]. The
case of curves is analogous, but we record the details for the benefit of the
readers who seek a self-contained account.

Step1. The first goal is to show that the Euler characteristic

(43) $\chi(\operatorname{\mathsf{Quot}}_{d}(E),\wedge_{y}L^{[d]})$

is a polynomial (that may depend on $N$) in $\deg E$, $\deg L$ and
$\chi(\mathcal{O}_{C})$ (with coefficients in $\mathbb{Q}[y]$), for all smooth
projective possibly disconnected curves $C$. In fact, the statement holds for
all tautological integrals of the form

(44) $\int_{\operatorname{\mathsf{Quot}}_{d}(E)}\mathsf{P}$

where $\mathsf{P}$ is a polynomial in the Chern classes of the tangent bundle
of $\operatorname{\mathsf{Quot}}_{d}(E)$ and $L^{[d]}.$

We first analyze the case of split vector bundles

$E=\bigoplus_{i=1}^{N}F_{i},\quad\text{rk }F_{i}=1.$

For such a vector bundle, we can use the action of $\mathbb{C}^{\star}$ on the
summands of $E$ to evaluate (44), just as we have done for genus $0$ above.
This way, we are led to considering integrals of the form

(45) $\int_{C^{[d_{1}]}\times\cdots\times C^{[d_{N}]}}\mathsf{Q}$

where $\mathsf{Q}$ is a polynomial involving the Chern classes of

$\pi_{\star}\left(\mathcal{K}_{i}\otimes
M\right),\quad\pi_{\star}\left(\mathcal{K}_{i}^{\vee}\otimes
M\right),\quad\pi_{\star}\left(\mathcal{K}_{i}^{\vee}\otimes\mathcal{K}_{j}\otimes
M\right)$

for various $M\to C$, including $M=F_{i}^{\vee}\otimes F_{j}$ or $M=L\otimes
F_{i}$. We can evaluate these with the aid of Grothendieck-Riemann-Roch. The
integrals (45) can be pulled back via the finite map

$p:C^{d}\to C^{[d_{1}]}\times\cdots\times C^{[d_{N}]}.$

The pullbacks of $\mathcal{K}_{i}^{\vee}$ over $C^{d}\times C$ correspond to
sums of diagonals $\Delta_{\bullet,d+1}$, and thus $\eqref{locpro}$ takes the
form

$\frac{d_{1}!\cdots d_{N}!}{d!}\int_{C^{d}}\widetilde{\mathsf{Q}}$

where $\widetilde{\mathsf{Q}}$ is a universal expression in the diagonals and
classes from $C$. In general, monomials in diagonals and classes from $C$ can
be evaluated explicitly using that for $\Delta\hookrightarrow C\times C$ we
have

$\Delta^{2}=2\chi(\mathcal{O}_{C}),\quad\Delta\cdot M=\deg M,$

for all smooth projective possibly disconnected curves $C$, $M\to C$.
Therefore (44) is a polynomial in $\deg F_{i}$, $\deg L$ and
$\chi(\mathcal{O}_{C})$.

Step 2. We next argue that the above polynomial only depends on $\deg
E=\sum_{i}\deg F_{i}$, $\deg L$ and $\chi(\mathcal{O}_{C})$. This requires
additional considerations. We write

$x_{i}=\deg F_{i},\quad y=\deg L,\quad z=\chi(\mathcal{O}_{C}),$

and $\mathsf{R}(x_{1},\ldots,x_{N},y,z)$ for the universal polynomial found
above. The polynomial $\mathsf{R}$ is certainly symmetric in
$x_{1},\ldots,x_{N}$.

We claim that if $x_{i}$ are sufficiently large,
$\mathsf{R}(x_{1},\ldots,x_{N},y,z)$ is in fact a polynomial in
$\sum_{i=1}^{N}x_{i}.$ Indeed, for large degrees, the line bundles $F_{i}$ are
globally generated (over connected curves $C$). Thus we can write $E$ as a
quotient

$0\to K\to W\to E\to 0,$

where $W$ is a trivial bundle (whose rank depends on $\deg E$). By [St1,
Theorem 5], modified from the original setting of surfaces to the case of
curves, there is an embedding

(46)
$\operatorname{\mathsf{Quot}}_{d}(E)\hookrightarrow\operatorname{\mathsf{Quot}}_{d}(W)$

cut out by a canonical section of the bundle $\left(K^{\vee}\right)^{[d]}$.
With this observation, the integral (44) rewrites as

(47)
$\int_{\operatorname{\mathsf{Quot}}_{d}(E)}\mathsf{P}=\int_{\operatorname{\mathsf{Quot}}_{d}(W)}\widetilde{\mathsf{P}}$

where $\mathsf{P}$ is a polynomial in the Chern classes of the tangent bundle
of $\operatorname{\mathsf{Quot}}_{d}(W)$ and the tautological bundles
$\left(K^{\vee}\right)^{[d]}$ and $L^{[d]}$. Applying the localization
argument in Step 1 once again, this time to
$\operatorname{\mathsf{Quot}}_{d}(W),$ we see that (47) only depends on

$\deg K^{\vee}=\deg E,\quad\deg L,\quad\chi(\mathcal{O}_{C}).$

Thus, the polynomial $\mathsf{R}(x_{1},\ldots,x_{N},y,z)$ is a function of
$\sum_{i=1}^{N}x_{i},y,z,$ when $x_{i}$ are large. Hence

$\mathsf{R}(x_{1},\ldots,x_{N},y,z)=\mathsf{S}(x_{1}+\ldots+x_{N},y,z),$

for a new universal polynomial $\mathsf{S}$. This proves the statement we need
about (44) when the bundle $E$ splits.

Step 3. The general case follows from the following observation. Assume $E$
sits in an extension

$0\to E_{1}\to E\to E_{2}\to 0.$

Considering the universal extension

$0\to p^{\star}E_{1}\to\mathcal{E}\to p^{\star}E_{2}\to 0$

over $p:C\times\text{Ext}^{1}(E_{2},E_{1})\to C$, and constructing the
relative Quot scheme $\operatorname{\mathsf{Quot}}_{d}(\mathcal{E})$ over the
extension space, we see that

$\int_{\operatorname{\mathsf{Quot}}_{d}(E)}\mathsf{P}=\int_{\operatorname{\mathsf{Quot}}_{d}(E_{1}\oplus
E_{2})}\mathsf{P}.$

To reduce to the case of split $E$, consider $M$ a line bundle such that

$0\to M\to E\to F\to 0$

is exact and $F$ is a vector bundle of smaller rank. By the above observation
we can replace $E$ by $M\oplus F$, and then continue inductively.

Step 4. We return to the series appearing in Theorem 1, namely

$\mathsf{Z}_{C,L,E}=\sum_{d=0}^{\infty}q^{d}\chi\left(\operatorname{\mathsf{Quot}}_{d}(E),\wedge_{y}L^{[d]}\right).$

Consider a disconnected curve $C=C_{1}\sqcup C_{2}$, $E=E_{1}\sqcup E_{2}$ and
$L=L_{1}\sqcup L_{2}$. We compare the Quot schemes of $C,C_{1},C_{2}$ and the
tautological bundles over them:

$\mathsf{Quot}_{d}(E)=\bigsqcup_{d_{1}+d_{2}=d}\mathsf{Quot}_{d_{1}}(E_{1})\times\mathsf{Quot}_{d_{2}}(E_{2}),\quad
L^{[d]}=\bigsqcup_{d_{1}+d_{2}=d}L_{1}^{[d_{1}]}\boxtimes L_{2}^{[d_{2}]}.$

This implies

(48)
${\mathsf{Z}}_{C,L,E}={\mathsf{Z}}_{C_{1},L_{1},E_{1}}\cdot{\mathsf{Z}}_{C_{2},L_{2},E_{2}}.$

By the arguments of [EGL, Theorem 4.2], the factorization (48) shows that

(49)
$\mathsf{Z}_{C,L,E}=\mathsf{A}^{\chi(C,\mathcal{O}_{C})}\cdot{\mathsf{B}}^{\deg
L}\cdot\mathsf{C}^{\deg E},$

for universal series $\mathsf{A},\mathsf{B},\mathsf{C}\in\mathbb{Q}(y)[[q]]$
that depend only on $N$. We specialize to
$(C,L)=(\mathbb{P}^{1},\mathcal{O}_{\mathbb{P}^{1}}(\ell))$ with $\ell$
sufficiently large, and
$E=\mathcal{O}(a_{1})\oplus\ldots\oplus\mathcal{O}(a_{N})$. Comparing (1) and
(49), we obtain

${\mathsf{A}}=(1-q)^{-1}\cdot(1+qy)^{N},\quad{\mathsf{B}}=(1+qy)^{N},\quad{\mathsf{C}}=1+qy.$

Substituting these expressions back into (49), we obtain Theorem 1 for all
genera:

$\mathsf{Z}_{C,L,E}=\mathsf{A}^{\chi(C,\mathcal{O}_{C})}\cdot{\mathsf{B}}^{\deg
L}\cdot\mathsf{C}^{\deg
E}=(1-q)^{-\chi(\mathcal{O}_{C})}\cdot(1+qy)^{\chi(E\otimes L)}.$

This completes the argument. ∎

######  Example 11.

Theorem 1 in higher genus immediately implies

$\chi\left(\operatorname{\mathsf{Quot}}_{d}(E),\wedge^{k}L^{[d]}\right)=0\,\,\text{
if }d\geq k+g,\,\,g\geq 1.$

This follows by examining the coefficient of $q^{d}y^{k}$ in the expression
$(1-q)^{-\chi(\mathcal{O}_{C})}(1+qy)^{\chi(E\otimes L)}.$

##  3\. Symmetric Powers

###  3.1. Genus zero.

Theorem 4 concerns the symmetric powers of the tautological bundles
$\mathsf{Sym}_{y}L^{[d]}$ in genus $0$ and is proven in a similar fashion as
Theorem 1. The calculations are however more involved. The higher genus case
and Theorem 5 will be considered in Section 3.2.

By Section 2.6, for each $d$ and $k$, the Euler characteristic of

$\chi\left(\operatorname{\mathsf{Quot}}_{d},\mathsf{Sym}^{k}L^{[d]}\right)$

depends polynomially on $\ell$. To prove Theorem 4, it suffices to assume
$b_{i}=\ell+a_{i}+1\geq d+1$ for all $i$.

By Hirzebruch-Riemann-Roch followed by Atiyah-Bott localization, we calculate

(50)
$\chi\left(\operatorname{\mathsf{Quot}}_{d},\mathsf{Sym}_{y}L^{[d]}\right)=\int_{\operatorname{\mathsf{Quot}}_{d}}\textnormal{ch}(\mathsf{Sym}_{y}L^{[d]})\,\textnormal{Td}\left(\operatorname{\mathsf{Quot}}_{d}\right)=\sum_{\vec{d}}\int_{\operatorname{\mathrm{F}}_{\vec{d}}}\textnormal{ch}(\mathsf{Sym}_{y}L^{[d]})\frac{\textnormal{Td}(\operatorname{\mathsf{Quot}}_{d})}{e_{\mathbb{C}^{*}}(\textrm{N}_{\vec{d}})}\bigg{|}_{\operatorname{\mathrm{F}}_{\vec{d}}}.$

Instead of Lemma 6, for the current computation we use the expression

(51)
$\textnormal{ch}(\mathsf{Sym}_{y}L^{[d]})\bigg{|}_{\textrm{F}_{\vec{d}}}=\prod_{i\in[N]}\bigg{(}\frac{\alpha_{i}(z_{i}-y)}{z_{i}(\alpha_{i}-y)}\bigg{)}^{a_{i}+\ell+1}\bigg{(}\frac{z_{i}}{z_{i}-y}\bigg{)}^{d_{i}}.$

The Todd genera and the normal bundle contributions are found in (8). We
substitute (8) and (51) into (50) and apply Lagrange-Bürmann. Carrying out
these steps carefully, we arrive at the following. Consider the polynomial

$P(z)=(z-y)R(z)-qz,$

and let $z_{1},\ldots,z_{N+1}$ be its roots with $z_{i}(q=0)=\alpha_{i}$ for
$1\leq i\leq N.$ Then, just as in the derivation leading up to (17) for
exterior powers, (50) turns into

$\displaystyle(-1)^{(N-1)d}\left[q^{d}\right]\prod_{i\in[N]}\bigg{(}\frac{\alpha_{i}-z_{N+1}}{\alpha_{i}-y}\bigg{)}^{b_{i}}\bigg{(}\frac{z_{i}}{z_{i}-y}\bigg{)}^{-1}\frac{z_{i}^{d+1}}{P^{\prime}(z_{i})}\prod_{i,j\in[N],\,\,\,i\neq
j}(z_{i}-z_{j})\bigg{|}_{\epsilon=0}.$

This simplification makes use of the fact that

$\frac{dq}{dh_{i}}=P^{\prime}(z_{i}).$

As in (18), the above expression can be recast as the quotient of determinants

$\displaystyle\left[\epsilon^{0}q^{d}\right]\frac{(-1)^{(N-1)d}}{\det(z_{i}^{N-j+1})}\begin{vmatrix}(z_{1}-y)z_{1}^{d+N-1}&\cdots&(z_{1}-y)z_{1}^{d}\\\
(z_{2}-y)z_{2}^{d+N-1}&\cdots&(z_{2}-y)z_{2}^{d}\\\ \vdots&\cdots&\vdots\\\
(z_{N}-y)z_{N}^{d+N-1}&\cdots&(z_{N}-y)z_{N}^{d}\end{vmatrix}\prod_{i\in[N]}\bigg{(}\frac{\alpha_{i}-z_{N+1}}{\alpha_{i}-y}\bigg{)}^{b_{i}}.$

The same derivation that led to (19) yields the enlarged $(N+1)\times(N+1)$
determinant

$\displaystyle[\epsilon^{0}q^{d}]\frac{(-1)^{(N-1)d}}{\det(z_{i}^{N-j+1})}\begin{vmatrix}(z_{1}-y)z_{1}^{d+N-1}&(z_{1}-y)z_{1}^{d+N-1}&\cdots&(z_{1}-y)z_{1}^{d}&\prod_{i=1}^{N}(\frac{\alpha_{i}-z_{1}}{\alpha_{i}-y})^{b_{i}}\\\
(z_{2}-y)z_{2}^{d+N-1}&(z_{2}-y)z_{2}^{d+N-1}&\cdots&(z_{2}-y)z_{2}^{d}&\prod_{i=1}^{N}(\frac{\alpha_{i}-z_{2}}{\alpha_{i}-y})^{b_{i}}\\\
\vdots&\vdots&\cdots&\vdots&\vdots\\\
(z_{N+1}-y)z_{N+1}^{d+N-1}&(z_{N+1}-y)z_{N+1}^{d+N-1}&\cdots&(z_{N+1}-y)z_{N+1}^{d}&\prod_{i=1}^{N}(\frac{\alpha_{i}-z_{N+1}}{\alpha_{i}-y})^{b_{i}}\end{vmatrix}.$

This uses $b_{i}\geq d+1$ for all $i$, and the fact that $\alpha_{i}-z_{i}$
has no free $q$-term, so in particular the first $N$ entries of the last
column do not contribute to the $q^{d}$-coefficient.

The expression above is symmetric in the roots of $P$, and as previously
remarked the substitution $\alpha_{i}=1$ is allowed to obtain the coefficient
of $\epsilon^{0}$. Thus $z_{1},\ldots,z_{N+1}$ become roots of

$P(z)=(z-1)^{N}(z-y)-qz.$

This also turns the last column into the vector with entries

$\frac{(1-z_{i})^{\chi}}{(1-y)^{\chi}}=\frac{1}{(1-y)^{\chi}}\sum_{\ell=0}^{\chi}\binom{\chi}{\ell}(-1)^{\ell}z_{i}^{\ell}.$

Here $\chi=\sum_{i}b_{i}=\chi(E\otimes L)$.

Using the additivity of the determinant with respect to the first $N$ columns,
we split the last determinant into a sum

$\displaystyle\left[q^{d}\right]\sum_{\ell=0}^{\chi}\sum_{m=0}^{N}\frac{(-1)^{(N-1)d+\ell}}{(1-y)^{\chi}}\binom{\chi}{\ell}(-y)^{m}\frac{1}{\det(z_{i}^{N-j+1})}\begin{vmatrix}z_{1}^{d+N}&\cdots&z_{1}^{d+m+1}&z_{1}^{d+m-1}&\cdots&z_{1}^{d}&z_{1}^{\ell}\\\
z_{2}^{d+N}&\cdots&z_{2}^{d+m+1}&z_{2}^{d+m-1}&\cdots&z_{2}^{d}&z_{2}^{\ell}\\\
\vdots&\cdots&\vdots&\vdots&\cdots&\vdots&\vdots\\\
z_{N+1}^{d+N}&\cdots&z_{N+1}^{d+m+1}&z_{N+1}^{d+m-1}&\cdots&z_{N+1}^{d}&z_{N+1}^{\ell}\end{vmatrix}.$

Indeed, from each of the first $N$ columns we select $N$ powers of $z_{i}$
whose exponents range from $d$ to $d+N$. Exactly one value $d+m$ must be
skipped, giving a term in the sum. The contribution $(-y)^{m}$ comes from
terms with exponents between $d$ and $d+m-1$.

Regarding the last sum, we make the following three remarks.

  * (i)

When $\ell<d$, the above quotient of determinants is the Schur polynomial for
the partition $\lambda=(d^{N-m},(d-1)^{m},\ell)$. Using Jacobi-Trudi as in
Lemma 7, we obtain that

$\left[q^{d}\right]s_{\lambda}(z_{1},\ldots,z_{N+1})=\begin{cases}(-1)^{(N-1)d}&\text{
if }\ell=m=0\\\ 0&\text{ otherwise }\end{cases}.$

  * (ii)

When $\ell>d+N$, the shape of the partition changes to
$\lambda=(\ell-N,(d+1)^{N-m},d^{m})$, and we also acquire an additional
$(-1)^{N}$ coming from permuting the columns to bring the last one to the
front. Note that $\lambda$ contains the rectangular partition $(d^{N+1})$ and
a hook partition $\mu:=(\ell-N-d,1^{N-m})$. Examining the determinant, we can
factor $z_{i}^{d}$ from each column. Thus

$s_{\lambda}=e_{N+1}^{d}\cdot s_{\mu}=y^{d}\cdot s_{\mu}.$

Here we used that $e_{N+1}=y$ which can be seen from the expression
$P(z)=(z-1)^{N}(z-y)-qz$.

  * (iii)

Finally, for $d\leq\ell\leq d+N$, the only value that can contribute is
$\ell=d+m$, in which case we can directly evaluate the corresponding quotient
of determinants to be $(-1)^{m}y^{d}$. The coefficient of $q^{d}$ vanishes in
this case (for $d\neq 0$).

Putting everything together we conclude

$\chi\left(\operatorname{\mathsf{Quot}}_{d},\mathsf{Sym}_{y}L^{[d]}\right)=\frac{1}{(1-y)^{\chi}}+O(y^{d}).$

Consequently, for $d>k$, we have

$\chi\left(\operatorname{\mathsf{Quot}}_{d},\mathsf{Sym}^{k}L^{[d]}\right)=\left[y^{k}\right]\frac{1}{(1-y)^{\chi}}=\binom{\chi+k-1}{k}.$

The result is also correct for $d=k$; this can be seen for instance from the
result below. ∎With a bit more effort, the same ideas (combined with a residue
calculation) yield a general expression in genus $0$. We need this result in
order to prove Theorem 5 in all genera in Section 3.2.

######  Theorem 12.

When $C=\mathbb{P}^{1}$ and $\chi=\chi(E\otimes L)$, we have

$\chi(\operatorname{\mathsf{Quot}}_{d}(E),\mathsf{Sym}_{y}L^{[d]})=\sum_{k=0}^{d}\binom{-\chi+d(N+1)}{k}\frac{(-y)^{k}}{(1-y)^{d(N+1)}}.$

###### Proof.

Since both sides depend polynomially on $\ell$, see for instance the arguments
in Section 2.6 for the left hand side, we may assume $\ell$ is sufficiently
large. In this case, we have seen above that

$\chi(\operatorname{\mathsf{Quot}}_{d}(E),\mathsf{Sym}_{y}L^{[d]})=\frac{1}{(1-y)^{\chi}}+\sum_{\ell>d+N}^{\chi}\frac{1}{(1-y)^{\chi}}(-1)^{(N-1)d+N+\ell}\binom{\chi}{\ell}y^{d}\left[q^{d}\right]\sum_{m=0}^{N}(-y)^{m}s_{\mu(\ell,m)},$

for the partition $\mu(\ell,m)=(\ell-N-d,1^{N-m}).$

Lemma 13 below evaluates the sum over $m$. We obtain

$\displaystyle\chi(\mathsf{Sym}_{y}L^{[d]})$
$\displaystyle=\frac{1}{(1-y)^{\chi}}\bigg{[}1+\sum_{\ell>d+N}^{\chi}(-1)^{(N-1)d+\ell}\binom{\chi}{\ell}y^{d+1}\left[t^{\ell-
N-d}\right]\frac{t^{N(d-1)+1}}{(1-t)^{Nd}(1-yt)^{d+1}}\bigg{]}$
$\displaystyle=\frac{1}{(1-y)^{\chi}}\bigg{[}1+\sum_{\ell>d+N}^{\chi}(-1)^{(N-1)d+\ell}\binom{\chi}{\ell}y^{d+1}\text{Res}_{t=0}\,\,\frac{t^{(N+1)d-\ell}}{(1-t)^{Nd}(1-yt)^{d+1}}\,dt\bigg{]}.$

We can allow all values $\ell\geq 0$ in the sum above since the residue
vanishes in the range $\ell\leq N+d$. The binomial theorem evaluates the sum
over $\ell$. Letting

$\omega=\frac{t^{(N+1)d-\chi}(1-t)^{\chi-Nd}}{(1-yt)^{d+1}}\,dt,$

we conclude that

$\displaystyle\chi(\mathsf{Sym}_{y}L^{[d]})=\frac{1}{(1-y)^{\chi}}\bigg{[}1+(-1)^{(N-1)d+\chi}y^{d+1}\text{Res}_{t=0}\,\,\omega\bigg{]}.$

Lemma 14 finishes the proof. ∎

######  Lemma 13.

Let $z_{1},\ldots,z_{N+1}$ be the roots of $P(z)=(z-1)^{N}(z-y)-qz.$ For
$\ell>0$, we have

$\displaystyle\left[q^{d}\right]\sum_{m=0}^{N}(-y)^{m}s_{(\ell,1^{N-m})}(z_{1},\ldots,z_{N+1})=$
$\displaystyle(-1)^{N}\left[t^{\ell}\right]\frac{yt^{N(d-1)+1}}{(1-t)^{Nd}(1-yt)^{d+1}}.$

###### Proof.

Using Jacobi-Trudi, the left hand side of the expression in the lemma equals
the $\ell\times\ell$ determinant

$\displaystyle\sum_{m=0}^{N}(-y)^{m}\begin{vmatrix}e_{N+1-m}&e_{N+2-m}&e_{N+3-m}&\cdots&e_{N+\ell-m}\\\
e_{0}&e_{1}&e_{2}&\cdots&e_{\ell-1}\\\ 0&e_{0}&e_{1}&\cdots&e_{\ell}\\\
\vdots&\vdots&\vdots&\cdots&\vdots\\\ 0&0&0&\cdots&e_{1}\end{vmatrix}.$

Summing with respect to $m$, we obtain that the $i$th term in first row
becomes

$\displaystyle A_{i}=\sum_{m=0}^{N}(-y)^{m}e_{N+i-m}$
$\displaystyle=\left[t^{N+i}\right](1-yt+\cdots+(-1)^{N}y^{N}t^{N})(1+e_{1}t+\cdots+e_{N+1}t^{N+1})$
$\displaystyle=(-1)^{N+i}\left[t^{N+i}\right]\frac{(1-(yt)^{N+1})}{1-yt}\cdot
t^{N+1}P(1/t).$

Expanding with respect to the first row, we obtain the required determinant
equals

$\displaystyle A_{1}h_{\ell-1}-A_{2}h_{\ell-2}+\cdots+(-1)^{\ell-1}A_{\ell}$

where $h_{j}=s_{(j)}$ is the homogeneous symmetric polynomial. We know that
the homogeneous symmetric polynomials are given by

$\displaystyle h_{i}$
$\displaystyle=\left[t^{i}\right]\frac{1}{(1-z_{1}t)\cdots(1-z_{N+1}t)}=\left[t^{i}\right]\frac{1}{t^{N+1}P(1/t)}.$

Thus the required sum equals

(52) $\displaystyle\sum_{i=1}^{\ell}(-1)^{i-1}A_{i}h_{\ell-i}$
$\displaystyle=\sum_{i=1}^{\ell}(-1)^{N+1}\bigg{[}\left[t^{N+i}\right]\frac{(1-(yt)^{N+1})}{1-yt}t^{N+1}P(1/t)\bigg{]}\bigg{[}\left[t^{\ell-i}\right]\frac{1}{t^{N+1}P(1/t)}\bigg{]}$
$\displaystyle\approx\sum_{j=0}^{N}(-1)^{N}\bigg{[}\left[t^{N-j}\right]\frac{(1-(yt)^{N+1})}{1-yt}t^{N+1}P(1/t)\bigg{]}\bigg{[}\left[t^{\ell+j}\right]\frac{1}{t^{N+1}P(1/t)}\bigg{]}$
$\displaystyle\approx\sum_{j=0}^{N}(-1)^{N}\bigg{[}\left[t^{N-j}\right]\frac{t^{N+1}P(1/t)}{1-yt}\bigg{]}\bigg{[}\left[t^{\ell+j}\right]\frac{1}{t^{N+1}P(1/t)}\bigg{]},$

where $\approx$ means equality of the $q^{d}$ coefficients. To justify the
second line, we note that the difference with the previous term equals

$\left[q^{d}\right](-1)^{N}\left[t^{N+\ell}\right]\bigg{(}\frac{1-(yt)^{N+1}}{1-yt}t^{N+1}P(1/t)\cdot\frac{1}{t^{N+1}P(1/t)}\bigg{)}=0$

for $d>0$. Moreover, since $j$ runs from $0$ to $N$, we may also ignore the
term $(yt)^{N+1}$ in the second line, thus yielding the third equality.

Note that

$t^{N+1}P(1/t)=(1-yt)(1-t)^{N}-qt^{N}.$

Thus

$\displaystyle\left[t^{N-j}\right]\frac{t^{N+1}P(1/t)}{1-yt}=\left[t^{N-j}\right]\left((1-t)^{N}-\frac{qt^{N}}{1-yt}\right)=\begin{cases}(-1)^{N-j}\binom{N}{N-j}&\text{if
}j>0\\\ (-1)^{N}-q&\text{if }j=0\end{cases}.$

Hence the $q^{d}$-coefficient in the sum (52) equals

$\displaystyle\left[q^{d}\right]\sum_{j=0}^{N}$
$\displaystyle(-1)^{j}\binom{N}{j}\left[t^{\ell+j}\right]\frac{1}{(1-yt)(1-t)^{N}-qt^{N}}+(-1)^{N+1}\left[q^{d-1}\right]\left[t^{\ell}\right]\frac{1}{(1-yt)(1-t)^{N}-qt^{N}}$
$\displaystyle=(-1)^{N}\left[q^{d}\right]\left[t^{\ell+N}\right]\frac{(1-t)^{N}}{(1-yt)(1-t)^{N}-qt^{N}}+(-1)^{N+1}\left[q^{d-1}\right]\left[t^{\ell}\right]\frac{1}{(1-yt)(1-t)^{N}-qt^{N}}.$

We note that the order in which we take the $q^{d}$ and
$t^{\ell+N}$-coefficients can be switched. This is allowed in our case since
we are considering expressions of the form
$\left(1-\mathsf{A}(q,t)\right)^{-1}$ expanded near $q=t=0$, where
$\mathsf{A}$ is a polynomial in $q,t$ (and $y$). Thus, taking the respective
coefficient of powers of $q$ in the above expression we obtain

$\displaystyle(-1)^{N}\left[t^{\ell+N}\right]\frac{t^{Nd}}{(1-yt)^{d+1}(1-t)^{Nd}}+(-1)^{N+1}\left[t^{\ell}\right]\frac{t^{N(d-1)}}{(1-yt)^{d}(1-t)^{Nd}}.$

This immediately implies the lemma. ∎

######  Lemma 14.

For $\chi\geq Nd$, set

$\omega=\frac{t^{(N+1)d-\chi}(1-t)^{\chi-Nd}}{(1-yt)^{d+1}}\,dt.$

We have

$1+(-1)^{(N-1)d+\chi}y^{d+1}\text{Res}_{t=0}\,\,\omega=\sum_{k=0}^{d}\binom{-\chi+(N+1)d}{k}\frac{(-y)^{k}}{(1-y)^{(N+1)d-\chi}}.$

###### Proof.

Since $\chi\geq Nd$, the form $\omega$ has poles at worst at $t=0$, $t=\infty$
and $t=\frac{1}{y}.$ By the residue theorem, we have

$\text{Res}_{t=0}\,\omega=-\text{Res}_{t=\infty}\,\omega-\text{Res}_{t=1/y}\,\omega.$

Changing variables $t=\frac{1}{s}$, we compute

$\text{Res}_{t=\infty}\,\omega=-\text{Res}_{s=0}\,(s-1)^{\chi-
Nd}(s-y)^{-d-1}\frac{ds}{s}=(-1)^{\chi-(N+1)d}y^{-d-1}.$

Similarly, changing variables $t=\frac{1-s}{y}$, we find

$\displaystyle\text{Res}_{t=\frac{1}{y}}\,\omega$
$\displaystyle=-\text{Res}_{s=0}\,(1-s)^{-\chi+(N+1)d}(s+y-1)^{\chi-
Nd}y^{-d-1}\frac{ds}{s^{d+1}}$
$\displaystyle=-y^{-d-1}\left[s^{d}\right](1-s)^{-\chi+(N+1)d}(s+y-1)^{\chi-
Nd}$
$\displaystyle=-y^{-d-1}\sum_{k=0}^{d}(-1)^{k}\binom{-\chi+(N+1)d}{k}\binom{\chi-
Nd}{d-k}(y-1)^{\chi-Nd-d+k}.$

Collecting terms, we obtain

$\displaystyle 1+(-1)^{(N-1)d+\chi}y^{d+1}\text{Res}_{t=0}\,\omega$
$\displaystyle=\sum_{k=0}^{d}\binom{-\chi+(N+1)d}{k}\binom{\chi-
Nd}{d-k}(1-y)^{\chi-Nd-d+k}$
$\displaystyle=\sum_{k=0}^{d}\binom{-\chi+(N+1)d}{k}\frac{(-y)^{k}}{(1-y)^{(N+1)d-\chi}}.$

To justify the last equality, we write $u=-\chi+(N+1)d$ and show more
generally

$\sum_{k=0}^{d}\binom{u}{k}\binom{-u+d}{d-k}(1-y)^{k}=\sum_{k=0}^{d}\binom{u}{k}{(-y)^{k}}.$

This follows by induction on $d$. Indeed, write $\mathsf{L}_{d}$ for the left
hand side. Using Pascal’s identity and then rewriting the binomials, we obtain

$\displaystyle\mathsf{L}_{d+1}-\mathsf{L}_{d}$
$\displaystyle=\sum_{k=0}^{d+1}\binom{u}{k}\left(\binom{-u+d+1}{d+1-k}-\binom{-u+d}{d-k}\right)(1-y)^{k}=\sum_{k=0}^{d+1}\binom{u}{k}\binom{-u+d}{d+1-k}(1-y)^{k}$
$\displaystyle=\sum_{k=0}^{d+1}\binom{u}{d+1}\binom{d+1}{k}(-1)^{d-k+1}(1-y)^{k}=\binom{u}{d+1}(-y)^{d+1}.$

The proof follows immediately from here. ∎

###  3.2. Universal functions

Over a smooth projective curve $C$ of arbitrary genus, let

$\mathsf{W}=\sum_{d=0}^{\infty}q^{d}\chi\left(\operatorname{\mathsf{Quot}}_{d}(E),\mathsf{Sym}_{y}L^{[d]}\right).$

The arguments in Section 2.6 exhibit $\mathsf{W}$ as a product of universal
series 555Strictly speaking, we only explained the factorization
$\mathsf{W}=\mathsf{A}_{1}^{\chi(\mathcal{O}_{C})}\cdot\mathsf{B}_{1}^{\deg
E}\cdot\mathsf{B}_{2}^{\deg L}$ in terms of $3$ universal series. An argument
of [St2] shows that only $2$ series are needed. Indeed, tensorization by a
line bundle $M\to C$ gives an isomorphism
$\operatorname{\mathsf{Quot}}_{d}(E)\simeq\operatorname{\mathsf{Quot}}_{d}(E\otimes
M)$ in such a fashion that $L^{[d]}$ gets identified with $\left(L\otimes
M^{-1}\right)^{[d]}$. On the level of generating series this implies
$\mathsf{B}_{1}^{N}=\mathsf{B}_{2}$, which then yields the result with
$\mathsf{A}=\mathsf{A}_{1}\cdot\mathsf{B}_{1}^{-N},$
$\mathsf{B}=\mathsf{B}_{1}$.

(53)
$\mathsf{W}=\mathsf{A}^{\chi(\mathcal{O}_{C})}\cdot\mathsf{B}^{\chi(E\otimes
L)}.$

In principle Theorem 12 determines both series $\mathsf{A},\mathsf{B}$ from
the genus $0$ answer. Theorem 5 asserts that more precisely we have

$\mathsf{B}=f\left(\frac{qy}{(1-y)^{N+1}}\right)$

where $f(z)$ is the solution to the equation

$f(z)^{N}-f(z)^{N+1}+z=0,\,\quad f(0)=1.$

Proof of Theorem 5. The function $f$ is most conveniently expressed in terms
of a change of variables. We have

$f(z)=\frac{1}{1+t}\quad\text{ for }z=-\frac{t}{(1+t)^{N+1}}.$

We record the one-variable version of the general Lagrange-Bürmann formula
(12). Assuming $\Phi(0)\neq 0$, for the change of variables
$z=\frac{t}{\Phi(t)}$, the following general identity holds

(54)
$\sum_{d=0}^{\infty}z^{d}\cdot\left(\left[t^{d}\right]\Phi(t)^{d}\cdot\Psi(t)\right)=\frac{\Psi(t)}{\Phi(t)}\cdot\frac{dt}{dz}\,.$

We introduce two functions which will be useful in the argument. Write

(55)
$\mathsf{F}_{\chi}(z)=\sum_{d=0}^{\infty}z^{d}\binom{-\chi+(N+1)d}{d}\implies
F_{\chi}(z)=\sum_{d=0}^{\infty}z^{d}\left(\left[t^{d}\right](1+t)^{-\chi+(N+1)d}\right).$

An immediate application of (54) yields

(56) $\mathsf{F}_{\chi}(z)=\frac{(1+t)^{-\chi+1}}{1-Nt}\quad\text{for
}z=\frac{t}{(1+t)^{N+1}}.$

Setting $\chi=0$ and integrating, we also obtain the expression

(57)
$\mathsf{G}(z)=\sum_{d=1}^{\infty}z^{d}\cdot\frac{N}{d}\binom{(N+1)(d-1)}{d-1}=1-\frac{1}{(1+t)^{N}},$

for the same change of variables. With this understood, we note that for the
function $f$ in the theorem, we have

$f(-z)^{N}=1-\mathsf{G}(z).$

The statement to be proven thus becomes

$\mathsf{B}^{N}=1-\mathsf{G}\left(-\frac{qy}{(1-y)^{N+1}}\right)$

or equivalently

(58)
$\mathsf{B}^{N}=1+\sum_{d=1}^{\infty}(-1)^{d+1}\frac{N}{d}\cdot\binom{(N+1)(d-1)}{d-1}\cdot\left(\frac{qy}{(1-y)^{N+1}}\right)^{d}.$

Turning to the generating series (53), we specialize to genus $0$ and we keep
track on the dependence on $\deg L=\ell$ in the notation, so that

$\mathsf{W}_{\ell}=\sum_{d=0}^{\infty}q^{d}\chi\left(\operatorname{\mathsf{Quot}}_{d}(E),\mathsf{Sym}_{y}L^{[d]}\right)=\mathsf{A}^{-1}\cdot\mathsf{B}^{\chi}.$

As usual, $\chi=\chi(E\otimes L)$. This yields

(59) $\mathsf{W}_{\ell+1}=\mathsf{W}_{\ell}\cdot\mathsf{B}^{N}.$

By Theorem 12, we have

$\mathsf{W}_{\ell}=\sum_{d=0}^{\infty}\mathsf{c}_{d}(\chi)\cdot
q^{d},\quad\mathsf{W}_{\ell+1}=\sum_{d=0}^{\infty}\mathsf{c}_{d}(\chi+N)\cdot
q^{d},$

where for simplicity, we wrote

(60)
$\mathsf{c}_{d}(\chi)=\sum_{k=0}^{d}\binom{-\chi+d(N+1)}{k}\frac{(-y)^{k}}{(1-y)^{d(N+1)}}.$

Examining the coefficient of $q^{d}$ in the identity (59), it follows that in
order to confirm (58) it suffices to prove

$\mathsf{c}_{d}(\chi+N)=\mathsf{c}_{d}(\chi)+\sum_{\ell=1}^{d}\mathsf{c}_{d-\ell}(\chi)\cdot(-1)^{\ell+1}\frac{N}{\ell}\binom{(N+1)(\ell-1)}{\ell-1}\left(\frac{y}{(1-y)^{N+1}}\right)^{\ell}.$

We use the defining expressions (60) to verify this equality. After
multiplying by $(1-y)^{d(N+1)}$ and extracting the coefficient of $y^{k}$ on
both sides, we need to show that for $0\leq k\leq d$, we have

(61)
$\binom{-\chi-N+d(N+1)}{k}=\binom{-\chi+d(N+1)}{k}-\sum_{\ell=1}^{k}\frac{N}{\ell}\binom{(N+1)(\ell-1)}{\ell-1}\binom{-\chi+(d-\ell)(N+1)}{k-\ell}.$

Using Pascal’s identity, it is easy to see that if (61) holds for $k$ and all
$\chi$, then it also holds for $k-1$ and all $\chi$. Thus, by downward
induction it suffices to assume $k=d$. In this case, we seek to show

$\sum_{\ell=1}^{d}\frac{N}{\ell}\binom{(N+1)(\ell-1)}{\ell-1}\binom{-\chi+(d-\ell)(N+1)}{d-\ell}=\binom{-\chi+d(N+1)}{d}-\binom{-\chi-N+d(N+1)}{d}.$

This is indeed correct. Recalling (55) and (57), we see that the two sides
equal the $z^{d}$-coefficient in the identity

$\mathsf{G}(z)\cdot\mathsf{F}_{\chi}(z)=\mathsf{F}_{\chi}(z)-\mathsf{F}_{\chi+N}(z).$

The latter equality is immediately justified using the explicit formulas (56)
and (57) after changing variables from $z$ to $t$ as above.

##  4\. Higher rank quotients

It is natural to wonder whether the formulas proven in Section 2 extend to
Quot schemes parametrizing quotients of rank $r>0$. We write
$\operatorname{\mathsf{Quot}}_{d}(E,r)$ for the corresponding Quot scheme. We
restrict to the case $C=\mathbb{P}^{1}$, and $E$ a trivial rank $N$ vector
bundle. In this case, $\operatorname{\mathsf{Quot}}_{d}(E,r)$ is smooth.

Theorem 3 gives an expresion for the Euler characteristics

$\displaystyle\chi(\operatorname{\mathsf{Quot}}_{d}(E,r),\wedge_{y}L^{[d]})$

in terms of the roots $z_{1},\dots,z_{N}$ of the equation

$(z-1)^{N}-q(z+y)z^{r-1}=0.$

Unlike our previous computations, this result does not imply the answer in
higher genus, since the universality arguments in Section 2.6 fail in this
case. Proof of Theorem $3$. We follow the same steps as in the proof of
Theorem 1. However, some modifications are necessary. Write

$s=N-r$

for the rank of the subbundles. We use the torus action on
$\operatorname{\mathsf{Quot}}_{d}(E,r)$ coming from the torus action on
$E=\mathbb{C}^{N}\otimes\mathcal{O}_{\mathbb{P}^{1}}$ with weights
$-w_{1},\ldots,-w_{N}$. The fixed loci are parameterized by pairs
$(\vec{d},I)$, where $\vec{d}=(d_{1},\ldots,d_{s})$ with
$|\vec{d}|=d_{1}+\cdots+d_{s}=d$, and $I\subset[N]$ is a subset of cardinality
$s$. The fixed loci are products of symmetric powers of $\mathbb{P}^{1}$:

$\operatorname{\mathrm{F}}_{\vec{d},I}=\mathbb{P}^{d_{1}}\times\cdots\times\mathbb{P}^{d_{s}}.$

Each such product appears $\binom{N}{s}$ times corresponding to the choice $I$
of $s$ summands of the trivial bundle $E$ into which the kernel injects

$0\to S\to\bigoplus_{i\in
I}\mathcal{O}_{\mathbb{P}^{1}}\to\bigoplus_{i\in[N]}\mathcal{O}_{\mathbb{P}^{1}}=E.$

This changes slightly the expressions for the normal bundles
$\textrm{N}_{\vec{d},I}$. With the same notation as in Section 2.1, we find
that (5) becomes

$\displaystyle\textrm{N}_{\vec{d},I}=\bigoplus_{i\in I,\,\,j\in[N],\,\,i\neq
j}\pi_{*}\left(\mathcal{K}_{i}^{\vee}\right)-\bigoplus_{i,j\in I,\,\,i\neq
j}\pi_{*}\left(\mathcal{K}^{\vee}_{i}\otimes\mathcal{K}_{j}\right).$

Compared to prior expressions, the range of the indices $i,j$ has changed. The
above sheaves carry weights $w_{i}-w_{j}$. By direct calculation, we obtain
the analogue of equation (8)

$\frac{\textnormal{Td}(\operatorname{\mathsf{Quot}}_{d})}{{e}_{\mathbb{C}^{*}}(\textrm{N}_{\vec{d},I})}=(-1)^{(s-1)d}\prod_{i\in
I}\bigg{(}\frac{h_{i}z_{i}^{N-s}}{R(z_{i})}\bigg{)}^{d_{i}+1}z_{i}^{d+1}\prod_{i,j\in
I,\,i\neq j}(z_{i}-z_{j}),$

where $z_{i}$, $\alpha_{i}$ and $R(z)=\prod_{j=1}^{N}(z-\alpha_{j})$ were
defined before. Similarly, the analogue of Lemma 6 becomes

$\displaystyle\textnormal{ch}(\wedge_{y}L^{[d]})\bigg{|}_{\operatorname{\mathrm{F}}_{\vec{d},I}}=\prod_{j\in[N]}\bigg{(}1+\frac{y}{\alpha_{j}}\bigg{)}^{\ell+1}\prod_{i\in
I}\bigg{(}1+\frac{y}{z_{i}}\bigg{)}^{d_{i}-(\ell+1)}.$

We substitute these expressions into Hirzebruch-Riemann-Roch and Atiyah-Bott
localization:

$\displaystyle\chi(\operatorname{\mathsf{Quot}}_{d},\wedge_{y}L^{[d]})=\sum_{\vec{d},I}\int_{\operatorname{\mathrm{F}}_{\vec{d},I}}\textnormal{ch}(\wedge_{y}L^{[d]})\,\frac{\textnormal{Td}(\operatorname{\mathsf{Quot}}_{d})}{\mathsf{e}_{\mathbb{C}^{*}}(\textrm{N}_{\vec{d},I})}\,\bigg{|}_{\operatorname{\mathrm{F}}_{\vec{d},I}}.$

After substitution, we invoke Lagrange-Bürmann formula, but the change of
variables takes a slightly different form for $r>0$. If

$P(z)=R(z)-q(z+y)z^{r-1}$

and $z_{i}$ are the roots of the above equation, the reader can verify by
direct calculation that we arrive at the following expression

$\left[q^{d}\right](-1)^{(s-1)d}\,\frac{\prod_{j=1}^{N}(\alpha_{j}+y)^{\ell+1}}{\prod_{j=1}^{N}\alpha_{j}^{\ell+1}}\,\cdot\sum_{\vec{d},I}\,\prod_{i\in
I}(z_{i}+y)^{-(\ell+1)}\frac{z_{i}^{d+\ell+N-s+1}}{P^{\prime}(z_{i})}\prod_{i,j\in
I,\,i\neq j}(z_{i}-z_{j})\bigg{|}_{\epsilon=0}.$

Since $P(-y)=R(-y)$, it follows that

$\prod_{j=1}^{N}(\alpha_{j}+y)=\prod_{j=1}^{N}(z_{j}+y),$

and thus the previous expression can be further simplified to

(62)
$\left[q^{d}\right]\frac{(-1)^{(s-1)d}}{\prod_{j=1}^{N}\,\,\alpha_{j}^{\ell+1}}\,\,\sum_{\vec{d},I}\prod_{j\not\in
I}(z_{j}+y)^{\ell+1}\,\,\prod_{i\in
I}\frac{z_{i}^{d+\ell+N-s+1}}{P^{\prime}(z_{i})}\prod_{i,j\in I,\,i\neq
j}(z_{i}-z_{j})\bigg{|}_{\epsilon=0}.$

Compared to Theorem 1, for $r>0$ we do not have an additional root, so we
finish the argument in a different way. The key observation is that we can
rewrite (62) as the $q^{d}$-coefficient in the quotient of two $N\times N$
determinants

(63)
$\frac{(-1)^{(s-1)d}}{\prod_{j=1}^{N}\alpha_{j}^{\ell+1}}\frac{1}{\det(z_{i}^{N-j})}\left|\begin{array}[]{cccc}z_{1}^{d+\ell+N}&z_{2}^{d+\ell+N}&\cdots&z_{N}^{d+\ell+N}\\\
z_{1}^{d+\ell+N-1}&z_{2}^{d+\ell+N-1}&\cdots&z_{N}^{d+\ell+N-1}\\\
\vdots&\vdots&\cdots&\vdots\\\
z_{1}^{d+\ell+N-s+1}&z_{2}^{d+\ell+N-s+1}&\cdots&z_{N}^{d+\ell+N-s+1}\\\ \\\
z_{1}^{N-s-1}(z_{1}+y)^{\ell+1}&z_{2}^{N-s-1}(z_{2}+y)^{\ell+1}&\cdots&z_{N}^{N-s-1}(z_{N}+y)^{\ell+1}\\\
\vdots&\vdots&\cdots&\vdots\\\
(z_{1}+y)^{\ell+1}&(z_{2}+y)^{\ell+1}&\cdots&(z_{N}+y)^{\ell+1}\end{array}\right|.$

This agrees with the prior expression (62). To justify this assertion, we use
generalized Laplace expansion of the determinant in the numerator along the
last $N-s$ rows simultaneously. Picking $s$ columns labeled by the index set
$I$, the corresponding minors of (63) (evaluated using Vandermonde
determinants) contribute exactly the $I$-term of (62).

Since the above expression is a symmetric function in the $z_{i}$’s, it can
also be expressed in terms of the elementary symmetric functions, which are in
turn polynomials in $\alpha_{j}$. Setting $\epsilon=0$ corresponds to setting
$\alpha_{j}=1$ for all $j$, or in turn working with the roots of

$P(z)=(z-1)^{N}-q(z+y)z^{r-1}=0.$

This completes the argument. ∎

######  Corollary 15.

As in Theorem 3, let $C=\mathbb{P}^{1}$, $E$ is trivial of rank $N$, $\deg
L=\ell\geq-d-1$. Then

$\displaystyle\chi\left(\operatorname{\mathsf{Quot}}_{d}(E,r),\det
L^{[d]}\right)=(-1)^{(N-r-1)d}\left[q^{d}\right]s_{\lambda}(z_{1},z_{2},\dots
z_{N})$

for the partition $\lambda=((d+\ell+1)^{N-r})$. The $z_{i}$’s are the distinct
roots of the equation $(z-1)^{N}-qz^{r-1}=0.$

###### Proof.

The statement follows by running the argument above for the determinant $\det
L^{[d]}$ instead of $\wedge_{y}L^{[d]}$. (We prefer this route since
extracting the top $y$-coefficient in the determinant (63) requires some
care.) The reader can verify that this results in the following two changes:

  * (i)

the new localization sum requires a new change of variables, so in particular,
$z_{1},\ldots,z_{N}$ are roots of $R(z)-z^{r-1}q=0$;

  * (ii)

the analogue of expression (62) is

$\left[q^{d}\right]\frac{(-1)^{(s-1)d}}{\prod_{j=1}^{N}\alpha_{j}^{\ell+1}}\,\sum_{\vec{d},I}\,\,\prod_{i\in
I}\frac{z_{i}^{d+\ell+N-s+1}}{P^{\prime}(z_{i})}\,\,\prod_{i,j\in I,\,i\neq
j}(z_{i}-z_{j})\bigg{|}_{\epsilon=0}.$

As a result, the counterpart of (63) takes the form

$\left[\epsilon^{0}q^{d}\right]\frac{(-1)^{(s-1)d}}{\prod_{j=1}^{N}\alpha_{j}^{\ell+1}}\frac{1}{\det(z_{i}^{N-j})}\left|\begin{array}[]{cccc}z_{1}^{d+\ell+N}&z_{2}^{d+\ell+N}&\cdots&z_{N}^{d+\ell+N}\\\
z_{1}^{d+\ell+N-1}&z_{2}^{d+\ell+N-1}&\cdots&z_{N}^{d+\ell+N-1}\\\
\vdots&\vdots&\cdots&\vdots\\\
z_{1}^{d+\ell+N-s+1}&z_{2}^{d+\ell+N+1-s}&\cdots&z_{N}^{d+\ell+N-s+1}\\\ \\\
z_{1}^{N-s-1}&z_{2}^{N-s-1}&\cdots&z_{N}^{N-s-1}\\\
\vdots&\vdots&\cdots&\vdots\\\ 1&1&\cdots&1\end{array}\right|.$

The proof is completed setting $\alpha_{j}=1$, and noting that when
$d+\ell+1\geq 0$, the last expression is exactly the Schur polynomial of the
partition $\lambda=((d+\ell+1)^{s})$. ∎

######  Example 16.

In the simplest case $d=0$, the Quot scheme is the Grassmannian $G=G(s,N)$ and
$\det L^{[d]}=\mathcal{O}_{G}(\ell+1)$. In the corollary, since we are
extracting the coefficient of $q^{0}$, we can set $z_{1}=z_{2}=\ldots=z_{N}=1$
to obtain the correct identity

$\chi(G,\mathcal{O}_{G}(\ell+1))=s_{\lambda}(1,\ldots,1),\quad\lambda=((\ell+1)^{s}).$

######  Example 17.

An interesting specialization of Corollary 15 arises for $\ell=0$. We show

$\displaystyle\chi\left(\operatorname{\mathsf{Quot}}_{d}(E,r),\det\mathcal{O}^{[d]}\right)=\binom{N}{r+d}.$

We have

$\chi(\operatorname{\mathsf{Quot}}_{d},\det\mathcal{O}^{[d]})=\left[q^{d}\right](-1)^{(s-1)d}s_{\lambda}(z_{1},\dots,z_{N})$

where $\lambda=((d+1)^{s})$. The elementary symmetric functions in
$z_{1},\dots z_{N}$ are

$\displaystyle e_{j}=\begin{cases}\binom{N}{j}&j\neq s+1\\\
\binom{N}{j}+(-1)^{s}q&j=s+1\end{cases}.$

Using Jacobi-Trudi, we have

$\displaystyle
s_{\lambda}(z_{1},\dots,z_{N})=\begin{vmatrix}e_{s}&e_{s+1}&e_{s+2}&\cdots&e_{s+d-1}&e_{s+d}\\\
e_{s-1}&e_{s}&e_{s+1}&\cdots&e_{s+d-2}&e_{s+d-1}\\\
\vdots&\vdots&\vdots&\cdots&\vdots&\vdots\\\
e_{s-d+1}&e_{s-d+2}&e_{s-d+3}&\cdots&e_{s}&e_{s+1}\\\
e_{s-d}&e_{s-d+1}&e_{s-d+2}&\cdots&e_{s-1}&e_{s}\\\ \end{vmatrix}.$

In the $(d+1)\times(d+1)$ determinant, the only term yielding the power
$q^{d}$ is $(-1)^{d}e_{s+1}^{d}e_{s-d}$, coming from the lower left corner
$e_{s-d}$ and the terms $e_{s+1}$ above the diagonal. To conclude, it remains
to note that

$\left[q^{d}\right]e_{s+1}^{d}e_{s-d}=(-1)^{sd}\binom{N}{s-d}.$

######  Example 18.

Assume $d>s(\ell+1)$. The Schur polynomial $s_{\lambda}$ has weighted degree
$|\lambda|=s(d+\ell+1)<(s+1)d$ in the elementary symmetric functions $e_{i}$,
where we set $\deg e_{i}=i$. We noted in Example 17 that only $e_{s+1}$
contains a linear $q$-term. By degree reasons, $e_{s+1}$ appears in
$s_{\lambda}$ with exponent $<d$. Thus, in this case the $q^{d}$-coefficient
vanishes, and

$\chi\left(\operatorname{\mathsf{Quot}}_{d}(E,r),\det L^{[d]}\right)=0.$

######  Example 19.

Assume $d=s(\ell+1)$, so that $d+\ell+1=(s+1)(\ell+1)$ and $|\lambda|=d(s+1)$
for $\lambda=((d+\ell+1)^{s})$. With these numerics, we claim that

(64) $s_{\lambda}=(-1)^{sd}e_{s+1}^{d}+\text{ lower order terms in }e_{s+1}.$

Using that the only nonzero $q$-contribution in $e_{j}(z_{1},\ldots,z_{N})$ is
given by

$\left[q\right]e_{s+1}(z_{1},\ldots,z_{N})=(-1)^{s},$

we obtain

$\left[q^{d}\right]s_{\lambda}(z_{1},\ldots,z_{N})=1,\text{ and thus
}\chi\left(\operatorname{\mathsf{Quot}}_{d}(E,r),\det L^{[d]}\right)=1.$

To justify (64), we let

$(x_{1},\ldots,x_{N})=(1,\zeta,\zeta^{2},\ldots,\zeta^{s},0,\ldots,0),$

where $\zeta$ is a primitive $(s+1)$-root of $1$. In this case, we have

$e_{s+1}(x_{1},\ldots,x_{N})=(-1)^{s},\quad e_{j}(x_{1},\ldots,x_{N})=0\text{
for }j\neq 0,\,\,j\neq s+1.$

Thus, to confirm (64) it remains to show that

(65) $s_{\lambda}(x_{1},\ldots,x_{N})=1.$

This follows from the (first) Jacobi-Trudi identity

$s_{\lambda}=\begin{vmatrix}h_{(s+1)(\ell+1)}&h_{(s+1)(\ell+1)+1}&\cdots&h_{(s+1)(\ell+1)+(s-1)}\\\
h_{(s+1)(\ell+1)-1}&h_{(s+1)(\ell+1)}&\cdots&h_{(s+1)(\ell+1)+(s-2)}\\\
\vdots&\vdots&\cdots&\vdots\\\
h_{(s+1)(\ell+1)-(s-2)}&h_{(s+1)(\ell+1)-(s-1)}&\cdots&h_{(s+1)(\ell+1)+1}\\\
h_{(s+1)(\ell+1)-(s-1)}&h_{(s+1)(\ell+1)-(s-2)}&\cdots&h_{(s+1)(\ell+1)}\\\
\end{vmatrix},$

where $h_{j}$ are the homogeneous symmetric functions. In our case, we have

$h_{j}(x_{1},\ldots,x_{N})=1\text{ if }j\equiv 0\mod s+1,\quad
h_{j}(x_{1},\ldots,x_{N})=0\text{ otherwise}.$

Hence the above matrix evaluated at $(x_{1},\ldots,x_{N})$ is the identity,
yielding (65).

##  5\. Further questions

###  5.1. Cohomology groups

It is natural to inquire whether Theorem 1 can be refined to yield information
about all cohomology groups of the tautological bundles $\wedge^{k}L^{[d]}.$
We ask:

######  Question 20.

Is it true that

(66)
$H^{\bullet}\left(\mathsf{Quot}_{d}(E),\wedge^{k}L^{[d]}\right)=\wedge^{k}H^{\bullet}(E\otimes
L)\otimes\mathsf{Sym}^{d-k}H^{\bullet}(\mathcal{O}_{C})?$

To explain the notation, if $V^{\bullet}=V_{0}\oplus V_{1}$ is a
$\mathbb{Z}_{2}$-graded vector space, we define the graded vector spaces

$\wedge^{k}V^{\bullet}=\bigoplus_{i+j=k}\wedge^{i}V_{0}\otimes\mathsf{Sym}^{j}V_{1},\quad\mathsf{Sym}^{k}V^{\bullet}=\bigoplus_{i+j=k}\mathsf{Sym}^{i}V_{0}\otimes\wedge^{j}V_{1}$

where the summands have degree $j$. With the convention

$\dim\,W^{\bullet}=\sum(-1)^{j}\dim\,W^{j}$

for the superdimension of a graded vector space, the usual formulas hold true

$\dim\,\wedge^{k}V^{\bullet}=\binom{\dim
V^{\bullet}}{k},\quad\dim\,\mathsf{Sym}^{k}V^{\bullet}=(-1)^{k}\binom{-\dim
V^{\bullet}}{k}.$

Thus, taking dimensions in (66), we immediately match the expressions in
Theorem 1. There is also a natural analogue for Theorem 2. The study of these
questions may require understanding the derived category of
$\mathsf{Quot}_{d}(E)$.

Evidence. Formula (66) is true in the following cases

  * (i)

over the symmetric product of a curve, that is for $\text{rank }E=1$. This was
shown in [K2, Section 3] using the derived category;

  * (ii)

for $d=1$ so that $\mathsf{Quot}_{1}(E)=\mathbb{P}(E),$ the projective bundle
of length $1$ quotients of $E$;

  * (iii)

for $k=0$, the formula predicts the Hodge numbers
$h^{p,0}(\mathsf{Quot}_{d}(E))=\binom{g}{p}$ for $p\leq d$. This follows from
[BFP, R] which give the Hodge polynomials

$\sum
h^{p,q}(\mathsf{Quot}_{d}(E))(-u)^{p}(-v)^{q}t^{d}=\prod_{i=0}^{\text{rk}(E)-1}\frac{(1-u^{i}v^{i+1}t)^{g}(1-u^{i+1}v^{i}t)^{g}}{(1-u^{i}v^{i}t)(1-u^{i+1}v^{i+1}t)}.$

###  5.2. Rationality

For any line bundles $L_{1},\ldots,L_{\ell}\to C$ and integers
$k_{1},\ldots,k_{\ell}\geq 0$, the series of $K$-theoretic invariants

$\mathsf{Z}_{C,E}\left(L_{1},\ldots,L_{\ell}\,|\,k_{1},\ldots,k_{\ell}\right)=\sum_{d}q^{d}\chi\left(\operatorname{\mathsf{Quot}}_{d}(E),\wedge^{k_{1}}L_{1}^{[d]}\otimes\cdots\otimes\wedge^{k_{\ell}}L_{\ell}^{[d]}\right)$

are given by rational functions with pole at $q=1$. This assertion was proved
in [AJLOP] in the context of punctual Quot schemes of surfaces, endowed with
the virtual class, but the same argument applies here as well. (The argument
proceeds by localization when $E$ is split, but this is sufficient in light of
the universality statements of Section 2.6.) Keeping with the theme of Section
1.2, we note that a similar result also holds true for Hilbert schemes of
points on surfaces (without virtual classes). This was conjectured in [AJLOP]
and proved in [A2].

Theorem 1 gives the simple expression

$\mathsf{Z}_{C,E}(L\,|\,k)=\binom{\chi(E\otimes
L)}{k}q^{k}\cdot(1-q)^{-\chi(\mathcal{O}_{C})}.$

######  Question 21.

What is the structure of the rational functions
$\mathsf{Z}_{C,E}(L_{1},\ldots,L_{\ell}\,|\,k_{1},\ldots,k_{\ell})$? Do they
admit explicit formulas?

It is natural to inquire whether the results proved here carry over to the
Quot schemes $\mathsf{Quot}_{d}(E,r)$ parametrizing quotients of $E$ of any
rank $r$ and degree $d$. The latter possess $2$-term perfect obstruction
theories [CFK, MO1].

######  Question 22.

Are the series

$\mathsf{Z}_{C,E}^{(r)}\left(L_{1},\ldots,L_{\ell}\,|\,k_{1},\ldots,k_{\ell}\right)=\sum_{d=0}^{\infty}q^{d}\chi^{\textrm{vir}}\left(\mathsf{Quot}_{d}(E,r),\wedge^{k_{1}}L_{1}^{[d]}\otimes\ldots\otimes\wedge^{k_{\ell}}L_{\ell}^{[d]}\right)$

rational functions with pole only at $q=1$? Are there explicit expressions for
$\mathsf{Z}_{C,E}^{(r)}\left(L_{1},\ldots,L_{\ell}\,|\,k_{1},\ldots,k_{\ell}\right)$?

Here, for a scheme $Y$ with a $2$-term perfect obstruction theory and virtual
structure sheaf $\mathcal{O}_{Y}^{\text{vir}}$, and for $V\to Y$, we set

$\chi^{\textrm{vir}}(Y,V)=\chi(Y,V\otimes\mathcal{O}_{Y}^{\text{vir}}).$

###  5.3. Higher rank

Extending Theorems 1 to $K$-theory classes $V\to C$ of arbitrary rank is not
immediate. In general, a change of variables is likely needed.

######  Question 23.

Find a closed-form expression for the series

$\sum_{d=0}^{\infty}q^{d}\chi\left(\operatorname{\mathsf{Quot}}_{d}(E),\wedge_{y}V^{[d]}\right).$

This may potentially be used to address Question 21 as well. Theorem 5
partially addresses the case $V=-L$, for $L$ a line bundle.

Turning to higher rank quotients and Theorem 3, we could ask for the arbitrary
genus version:

######  Question 24.

For line bundles $L\to C$, find a closed-form expression for
$\chi^{\text{vir}}(\operatorname{\mathsf{Quot}}_{d}(E,r),\wedge_{y}L^{[d]}).$

For example, in genus $0$, for $E$ trivial of rank $N$, for rank $r>0$, $\deg
L=\ell$, numerical experiments suggest that the answer stabilizes to

$\chi\left(\operatorname{\mathsf{Quot}}_{d}(E,r),\wedge_{y}L^{[d]}\right)=(1+y)^{N(\ell+1)},$

as soon as $d\geq(N-r)(\ell+1)$.

Analogous questions can be asked about symmetric powers or other Schur
functors as well.

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