\begin{document}

\title{Riemann-Roch Polynomials of the known Hyperkähler Manifolds}
\author{Ángel David Ríos Ortiz}
\address{Sapienza Universita di Roma, Dipartimento di Matematica, Piazzale Aldo Moro 5, 00185 Roma}
\email{rios@mat.uniroma1.it}
\date{}
\contrib[with an appendix by]{Yalong Cao and Chen Jiang}
\address{Kavli Institute for the Physics and Mathematics of the Universe (WPI),The University of Tokyo Institutes for Advanced Study, The University of Tokyo, Kashiwa, Chiba 277-8583, Japan}

\email{yalong.cao@ipmu.jp}
\address{Shanghai Center for Mathematical Sciences, Fudan University, 
Shanghai 200438, China}
\email{chenjiang@fudan.edu.cn}

\maketitle

\begin{abstract}
    We compute explicit formulas for the Euler characteristic of line bundles in the two exceptional examples of Hyperkähler Manifolds introduced by O'Grady. In an Appendix, Chen Jiang and Yalong Cao use our formulas to compute the Chern numbers of the example of O'Grady in dimension $10$.
\end{abstract}

\section{Introduction}

A compact Kähler manifold is called \emph{Hyperkähler} (HK) if it is simply connected and carries a holomorphic symplectic form that spans $H^{2,0}$. HK manifolds can be thought of as the higher dimensional analogues of $K3$ surfaces, and they constitute one of the three fundamental classes of varieties with vanishing first Chern class \cite{Beau83}.

Although any two $K3$ surfaces are deformation equivalent, this fact no longer holds in higher dimensions. The first two series of examples of deformation types in each (necesarily even) dimension were described by Beauville \cite{Beau83}; the first series, denoted by $K3^{[n]}$, is given by the Hilbert scheme of $n$ points in a K3 surface. The other one is a submanifold in the Hilbert scheme of $n$ points in an abelian surface. Generalizing the construction of a Kummer surface, this ($2n$-dimensional) deformation type is denoted by $\text{Kum}_n$.


Later, O'Grady introduced two new deformation types in dimensions $6$ and $10$  (\cite{OG99},\cite{OG03}), now denoted by OG6 and OG10 respectively. The construction of both exceptional examples is done by resolving a singular moduli space of sheaves on a K3 surface for OG10 and an abelian surface for OG6. In view of this analogies it is expected that the projective geometry of HK manifolds of $K3^{[5]}$-type (respectively $\text{Kum}_3$-type) should be related with that of OG10-type (respectively OG6-type).

The main result of this paper (cf. Theorem \ref{thm:RR}) gives, for the HK manifolds described by O'Grady, closed formulas that compute the Euler characteristic of any line bundle in terms of numerical polynomials that only depend in the Beauville-Bogomolov form --a canonical quadratic form in the second cohomology group of any HK. Surprisingly the formulas turn out to be exactly the same as those of the series described by Beauville.

In order to compute these polynomials we use two different methods. The first one exploits a recent description in \cite{LSV17} of OG10 as a compactification of a fibration associated with a cubic $4$-fold. The second one is based on the explicit descriptions of some uniruled divisors in two different models of OG6 given in \cite{MRS2018} and \cite{Nagai2014}.

Observe that in \cite{CaoChen19} the authors give a closed formula for the Riemann-Roch polynomial of OG6 in terms of the so-called $\lambda$-invariant, in our work the closed formula is obtained directly.

Finally we would like to point out the very recent paper \cite{Jiang20} where Chen Jiang proves the positivity of the coefficients of the Riemann-Roch polynomial for HK manifolds in general.

\begin{thankyou}
I am grateful to my PhD advisor Kieran O'Grady for his patience and deep insight. Thanks also goes to Federico Caucci, Antonio Rapagnetta and Domenico Fiorenza for many stimulating conversations and their mathematical suggestions. Finally I would like to thank heartily Chen Jiang and Yalong Cao for the interest they took in this work and for writting the Appendix.
\end{thankyou}

\section{Preliminaries}

Let $X$ be a HK of dimension $2n$ and $q_X$ its Beauville-Bogomolov form \cite{Beau83}. Recall that the Fujiki constant $c_X$ is defined as the rational  number such that for all $\alpha\in H^2(X)$ we have the so-called Fujiki relation:
\begin{equation}\label{eq:fujiki}
    \int_X \alpha^{2n} = c_X q_X(\alpha)^n
\end{equation}
\begin{rem}
The polarized form of Fujiki's relation is
\begin{equation}
    \int_X \alpha_1\smile\dots\smile\alpha_{2n} = \frac{c_X}{(2n)!}\sum_{\sigma\in S_{2n}} q_X(\alpha_{\sigma(1)},\alpha_{\sigma(2)})\cdot \cdot \cdot q_X(\alpha_{\sigma(2n-1)},\alpha_{\sigma(2n)})
\end{equation}
\end{rem}


Huybrechts further generalized this relation to all polynomials in the Chern classes, more specifically he proved the following

\begin{thm}[\cite{GHJ2003}, Corollary 23.17]\label{thm:constants}
Assume $\alpha\in H^{4j}(X,\Q)$ is of type $(2k,2k)$ for all small deformations of $X$. Then there exists a constant $C(\alpha)\in\Q$ such that
\begin{equation}\label{eq:todd}
    \int_X \alpha \smile \beta^{2n-2k} = C(\alpha)\cdot q_X(\beta)^{n-k}
\end{equation}
for all $\beta\in H^2(X,\Q)$.
\end{thm}


The odd Chern classes (hence the odd Todd classes) of $X$ vanish since the symplectic form on $X$ induces an isomorphism between $T_X$ and its dual. The Todd classes are topological invariants of $X$, so for any line bundle $L$ in $X$ we combine Theorem \ref{thm:constants} with Hirzebruch-Riemann-Roch Theorem to get 
\begin{equation}\label{eq:chi}
   \chi(X,L) =  \sum_{i=0}^n\frac{1}{(2i)!}\int_X Td_{2n-2i}(X)\smile c_1(L)^{2i} = \sum_{i=0}^n \frac{a_i}{(2i)!} \cdot q_X(L)^{i}
\end{equation}
where $a_i := C(Td_{2n-2i}(X))$.
\begin{defn}[\emph{Huybrechts, Nieper, Riess}]
The Riemann-Roch polynomial of $X$, denoted by $RR_X(t)$, is the polynomial 
\[
RR_X(t) = \sum_{i=0}^n \frac{a_i}{(2i)!}t^i.
\]
\end{defn}

Let us list a few well-known properties of this polynomial.

\begin{lem}\label{lem:RRproperties}
Let $X$ be a HK variety of dimension $2n$. The following properties hold:
\begin{enumerate}
    \item $RR_X$ depends only on the deformation class of $X$.
    \item The constant term is $a_0 = n+1$.
    \item The coefficient of the highest-order term  is $a_n = c_X$ and is positive.
    
\end{enumerate}
\end{lem}
\begin{proof}
We have already observed that the Todd classes are a deformation invariant of $X$. Hence each $a_i$ (and therefore $RR_X$) is also a deformation invariant of $X$. The constant term of $RR_X$ is the holomorphic Euler characteristic of $X$, this was computed \cite{Beau83} to be $n+1$. The constant $a_n = C(Td_0(X))$ is given by \eqref{eq:fujiki} so it is equal to $c_X$. Observe that $c_X$ is positive because the left hand side of the Fujiki relation \eqref{eq:fujiki} is a volume form.


\end{proof}


In view of the previous Lemma we can speak of the Riemann-Roch polynomial for a deformation \emph{type}. This has been done for the two series of examples introduced by Beauville.


\begin{exmp}[\cite{EGM01}, Lemma 5.1]
Let $X$ be a HK of $K3^{[n]}$-type, then the Riemann-Roch polynomial is given by
\[
RR_X(t) = \binom{t/2 + n + 1}{n}.
\]
\end{exmp}

\begin{exmp}[\cite{Nieper03}, Lemma 5.2]\label{exmp:Kum}
Let $X$ be a HK of $\text{Kum}_n$-type, then the Hilbert polynomial takes the form
\[
RR_X(t) = (n+1)\binom{t/2 + n}{n}.
\]
\end{exmp}

We will say that the Riemann Roch polynomial is of $K3^{[n]}$-type or $\text{Kum}_n$-type if it corresponds to one of the two examples above. Now we can state the main result of this paper.


\begin{thm}\label{thm:RR}
The Riemann-Roch polynomials for the deformation class of OG6 and OG10 are of $Kum_3$-type and $K3^{[5]}$-type respectively.
\end{thm}

The Theorem will be proved in Propositions \ref{prop:OG10} and \ref{prop:OG6} in the two Sections bellow.

\section{Riemann-Roch polynomial for OG10}

We will use the realization of OG10 constructed by Laza-Saccà-Voisin in \cite{LSV17}, denoted by $J$. This has a Lagrangian fibration $\pi:J\to \P^5$. Let $\Theta$ be the relative theta divisor. This is $\pi$-ample by \cite{LSV17}, Section 5. Denote by $X_t$ a smooth fiber of $\pi$ and by $\Theta_t$ the restriction of $\Theta$ to the fiber. Recall the following result by Sawon.
 
\begin{thm}[\cite{Saw16}, Claim 12]\label{thm:sawon}
Let $H$ be a $\pi$-ample divisor on $X$. Then the sheaf $\pi_*\O_X(H)$ is locally free of rank $h^0(X_t,H_t)$.
\end{thm}
\begin{rem}
In \cite{Saw16} it is assumed that $H$ is also ample, but tensoring with a sufficiently high multiple of the fiber yields the same result.
\end{rem}
The relative theta divisor is a principal polarization when restricted to any smooth fiber, hence $h^0(J_t,\Theta_t) = 1$. We obtain

\begin{prop}
$\pi_*(\O_J(\Theta))\cong \O_{\P^5}(k)$ for some $k\geq 0$. In particular
\begin{equation}\label{eq:binomialRR}
    h^0(J,\Theta +mF) = \binom{k+m+5}{5}
\end{equation}
for all $m\gg 0$.
\end{prop}
\begin{rem}
This computation also yields that the Fujiki costant of $J$ is $9!! = 945$, a result previously found by Rapagnetta.
\end{rem}

The class $F$ is isotropic with respect to $q_J$, we use the polarized version of the Fujiki formula to get
\[
\int_X \Theta^5\smile F^5 = \frac{c_X}{10!}\cdot (5!)^2\cdot 2^5\cdot q_X(F,\omega)^5.
\]

On the other hand, we can compute the left hand side to be

\[
\int_X \Theta^5\smile F^5  = \int_{X_t}\Theta^5|_{X_t} = \int_{X_t}\Theta_t^5 = 5!
\]
Hence $q_J(\Theta,F) = 1$. In particular $q_J(\Theta + mF) = q_J(\Theta) + 2m$.

\begin{prop}\label{prop:OG10}
The Riemann-Roch polynomial of OG10 is of $K3^{[5]}$-type.
\end{prop}
\begin{proof}
We substitute $t = q(\Theta + mF) = q(\Theta) +2m$ to get
\[
RR_X(t) = \binom{\frac{t+k-q_J(\Theta)}{2} + 5}{5}
\]
Evaluating at zero we get the following equation
\[
6 = \binom{\frac{k-q_J(\Theta)}{2} + 5}{5}
\]
whose only rational solution is $\frac{k-q_J(\Theta)}{2} =1$, the result follows.
\end{proof}

\begin{rem}
The same strategy can also be used to compute the Riemann-Roch polynomial for HK manifolds of $K3^{[n]}$-type.
\end{rem}

\section{Riemann-Roch polynomial for OG6}

Although there does exist a HK manifold of OG6-type with a Lagrangian fibration, we cannot use the same strategy as we did for OG10 because the abelian varieties appearing as smooth fibers are not principally polarized --i.e. there is not an ample divisor restricting to a principal polarization on every smooth fiber. Hence, we use an alternative method based on the explicit description of some divisors. 

Let $X$ be a HK manifold of OG6-type. The formula \eqref{eq:chi} for $X$ is:

\begin{equation}\label{eq:RRforOG6}
    \chi(X,L) = a_0 + \frac{a_1}{2!}q_X(L) + \frac{a_2}{4!}q_X(L)^2 + \frac{a_3}{6!}q_X(L)^3 = RR_X(q_X(L))
\end{equation}


By Lemma \ref{lem:RRproperties} we have $a_0 = 4$ and also $a_3 = c_X = 60$, by \cite{Rap07}. We will find divisors whose invariants reduce equation \ref{eq:RRforOG6} to a linear system of equations. In order to do this, we will introduce the divisors $\widetilde{\Sigma}$ and $\widetilde{B}$, its Beauville-Bogomolov form and Euler characteristic are outlined in the table bellow.

\begin{table}[ht]
\centering
\begin{tabular}{l|l|l|}
\cline{2-3}
                                           & \textbf{$\chi$} & \textbf{$q_X$} \\ \hline
\multicolumn{1}{|l|}{$\widetilde{\Sigma}$} &        -4         & -8                \\ \hline
\multicolumn{1}{|l|}{$\widetilde{B}$}      &         0        & -2                \\ \hline
\end{tabular}\caption{Invariants for $\widetilde{\Sigma}$ and $\widetilde{B}$.}\label{table}
\end{table}

\begin{prop}\label{prop:OG6}
The Riemann-Roch polynomial of OG6 is of $Kum_3$-type.
\end{prop}
\begin{proof}
If we plug into Equation \eqref{eq:RRforOG6} the known invariants of $\widetilde{\Sigma}$ and $\widetilde{B}$ given in Table \ref{table}, then we have a complete system of \emph{linear} equations because the Beauville-Bogomolov square of both divisors is different. Hence, constants $a_1$ and $a_2$ are uniquely determined. The coefficients $a_0,\dots,a_4$ are the same as those of the Riemann-Roch polynomial of $Kum_3$-type given in Example \ref{exmp:Kum}.
\end{proof}


In the rest of this Section we are going to compute the values of the Table \ref{table}. Let us recall the following easy

\begin{lem}
Let $E\subseteq X$ be an effective smooth divisor, then $\chi(X,\O(E)) = 4-\chi(E,\O_E)$.
\end{lem}
\begin{proof}
The exact sequence induced by a nonzero section of $E$ gives a formula for its Euler characteristic:
\begin{equation}\label{eq:eulercharacteristicdivisor}
    \chi(X,\O(E)) = \chi(X,\O_X) + \chi(E,\O_E(E)) = 4 + \chi(E,\O_E(E)).
\end{equation}
By adjunction $\O_E(E)\cong\omega_E$. If $E$ is smooth, then by Serre duality $\chi(E,\omega_E) = -\chi(E,\O_E)$.
\end{proof}

In order to describe the divisors we need to introduce some notation. Let $A$ be an abelian surface, and denote by $A^\vee$ its dual. Multiplication by $-1$ is an involution on $A\times A^\vee$. The first divisor $\widetilde{\Sigma}$ corresponds to the exceptional locus of a resolution of a singular moduli space of sheaves on $A$.


\begin{thm}[\cite{MRS2018}, Corollary 2.8]\label{thm:Sigma}
The divisor $\widetilde{\Sigma}$ is a $\P^1$-bundle over the nonsingular variety
\[
\overline{\Sigma} := Bl_{\text{Sing}((A\times A^\vee)/\pm1)}(A\times A^\vee)/\pm1.
\]
\end{thm}
The Euler characteristic is multiplicative for smooth fibrations, so $\chi(\O_{\widetilde{\Sigma}}) = \chi(\O_{\overline{\Sigma}})$. The Hodge numbers are given by
\[
H^{4,q}(\overline{\Sigma}) = H^{4,q}(A\times \widehat{A})^{inv}.
\]
Since the action in cohomology is given in differential forms by multiplication with $(-1)^{\deg}$, the cohomology groups are
\[
H^{4,q}(A\times \widehat{A}) = 
\begin{cases}
    0 & \text{if} \;\;\;\; q=2,4\\
    6 & \text{if} \;\;\;\; q=3\\
    1 & \text{if} \;\;\;\; q=1,5.
\end{cases}
\]
It follows that $\chi(\O_{\overline{\Sigma}}) = -8$ and hence $\chi(\O_X(\widetilde{\Sigma})) = -4$ by \eqref{eq:eulercharacteristicdivisor}. By Theorem 3.3.1 of \cite{Rap07} it is known that $q_X(\widetilde{\Sigma}) = -8$. 


The other divisor $\widetilde{B}$ corresponds to the locus of non-locally free sheaves in a moduli space of sheaves on $A$.

\begin{thm}[\cite{Nagai2014}]
$\widetilde{B}$ is a $\P^1$-bundle over $K(A)\times K(A^\vee)$, where $K(A)$ denotes the smooth Kummer surface associated to $A$.
\end{thm}

The previous Theorem yields $\chi(\O_{\widetilde{B}}) = \chi(\O_{K(A)\times K(A^\vee)}) = 4$, since both $K(A)$ and $K(A^\vee)$ are K3 surfaces. Finally by Formula \eqref{eq:eulercharacteristicdivisor} we have $\chi(\O_X(\widetilde{B})) =0$. By Theorem 3.5.1 of \cite{Rap07} we have that $q_X(\widetilde{B}) = -2$; this finishes the proof of the Proposition, and of the Main Theorem.


\bibliography{bib.bib}{}
\bibliographystyle{alpha}

\newpage
\appendix

\section{Chern numbers of OG10, by Yalong Cao and Chen Jiang}
An interesting question is to compute topological invariants (e.g.  topological Euler characteristic and Chern numbers) of hyperk\"ahler varieties. 

For a HK of $K3^{[n]}$-type, Ellingsrud--G\"{o}ttsche--Lehn \cite{egl} showed that the Chern numbers can be efficiently calculated,  in terms of the Chern numbers of the varieties $(\mathbb{P}^2)^{[k]}$ and $
(\mathbb{P}^1 \times \mathbb{P}^{1})^{[k]}$
(which  can be calculated by Bott's residue formula via \cite{ES87, ES96}), though no explicit formula is known.

For a HK of $\text{Kum}_{n}$-type, Nieper \cite{nieper2002} showed that the Chern numbers can be efficiently calculated,   in terms of the Chern numbers of the varieties $(\mathbb{P}^2)^{[k]}$ (see also \cite{BN, SawonPhd}).

For OG6, the Chern numbers are computed by Mongardi--Rapagnetta--Sacc\`{a}
\cite[Corollary 6.8]{mrs} (see also \cite{Rapa-thesis}).

For OG10, the topological Euler characteristic is computed by Mozgovyy \cite{Moz-thesis} (see also \cite{HLS}) and the Hodge numbers are computed by  de Cataldo--Rapagnetta--Sacc\`a \cite{CRS}. But the Chern numbers of OG10 have not been computed yet. In fact, Hodge numbers could not provide enough linear equations to solve Chern numbers (cf. \cite[Section 5.2]{SawonPhd}). 

As an application to the Riemann--Roch polynomial of OG10, combining with the Hodge numbers, we can compute $7$ Chern numbers of OG10.

\subsection{Relations of $q_X$ and $\lambda$ via Riemann--Roch polynomials}
Let $X$ be a HK variety. For a line bundle $L$ on $X$, Nieper \cite[Definition 17]{nieper} defined the {\it characteristic value} of $L$,
$$
\lambda(L):=\begin{cases}\frac{24n\int_{X}\ch(L)}{\int_{X}c_{2}(X) \ch(L)} & \text{if well-defined;}\\ 0 & \text{otherwise.}\end{cases}
$$
Note that $\lambda(L)$ is a positive (topological constant) multiple of $q_X(c_1(L))$ (cf. \cite[Proposition 10]{nieper}), more precisely,
$$
\lambda(L)=\frac{12c_X}{(2n-1)C(c_2(X))}q_X(c_1(L)).
$$
Here we denote the ratio $$M_{\lambda, q_X}:=\lambda(L)/q_X(c_1(L))=\frac{12c_X}{(2n-1)C(c_2(X))}.$$
So for the Riemann--Roch polynomial, we can write it in terms of either $q_X$ (say $\text{\rm RR}_{X, q_X}(t)$) or $\lambda$ (say $\text{\rm RR}_{X, \lambda}(t)$). They coincide up to a multiple of invariable, that is, $\text{\rm RR}_{X, q_X}(t)=\text{\rm RR}_{X, \lambda}(M_{\lambda, q_X}t)$.

 An observation (maybe well-known to experts) is that, even though we do not know the value of  $M_{\lambda, q_X}=\frac{12c_X}{(2n-1)C(c_2(X))}$ priorly, we can see it at once we know the expression of  the Riemann--Roch polynomial, that is, $\text{\rm RR}_{X, q_X}(t)$  determines $M_{\lambda, q_X}$,  $(\text{\rm RR}_{X, \lambda}(t), c_X)$  determines $M_{\lambda, q_X}$.
 Hence as a consequence, 
$\text{\rm RR}_{X, q_X}(t)$ and $(\text{\rm RR}_{X, \lambda}(t), c_X)$ determine each other.

\begin{lem}
The  coefficients of the first 2 leading terms of $\text{\rm RR}_{X, q_X}(t)$ determines $M_{\lambda, q_X}$. The  coefficient of the leading term $\text{\rm RR}_{X, \lambda}(t)$ and  $c_X$ determines $M_{\lambda, q_X}$.
\end{lem}

\begin{proof}
Write $\text{\rm RR}_{X, q_X}(t)=at^n+bt^{n-1}+(\text{\rm lower terms}).$ 
Then we have
$$
\int_X \td(X)\exp(L)=\text{\rm RR}_{X, q_X}(q_X(L)).
$$
Comparing both sides, 
this implies that 
$$
\int_X \frac{1}{(2n)!}L^{2n}=aq_X(L)^n,
$$
and 
$$
\int_X\td_2(X) \frac{1}{(2n-2)!}L^{2n-2}=bq_X(L)^{n-1}.
$$
Using Fujiki's relation, these implies that 
$$
c_X=(2n)!a
$$
and 
$$
C(c_2(X))=12(2n-2)!b.
$$
Hence $M_{\lambda, q_X}=\frac{12c_X}{(2n-1)C(c_2(X))}=\frac{2na}{b}$.


Write $\text{\rm RR}_{X, \lambda}(t)=a't^n+(\text{\rm lower terms}).$ 
Then
$\text{\rm RR}_{X, q_X}(t)=\text{\rm RR}_{X, \lambda}(M_{\lambda, q_X}t)=a'M_{\lambda, q_X}^nt^n+(\text{\rm lower terms})$. Then $(2n)!a'M_{\lambda, q_X}^n=c_X$, which determines $M_{\lambda, q_X}$ as $M_{\lambda, q_X}$ is a positive real number. 
\end{proof}


\subsection{Chern numbers of OG10}

\begin{thm}
Let $X$ be a HK of OG10-type. Then the Chern numbers of $X$ are the following:
\begin{align*}
&(c_2^5, c_2^3c_4, c_2^2c_6, c_2c_8, c_2c_4^2, c_4c_6, c_{10})\\
=&(127370880, 53071200, 12383280, 1791720, 22113000, 5159700, 176904)
\end{align*}
\end{thm}
\begin{proof}
Let $X$ be a HK of OG10-type.
Then the Riemann--Roch polynomial is
$$
\text{\rm RR}_{X, q_X}(t)=\binom{t/2+6}{5}.
$$
By the discussion of the first section, 
we know that 
\begin{equation}
\text{\rm RR}_{X, \lambda}(t)=\text{\rm RR}_{X, q_X}(4t)=\binom{2t+6}{5}.
\label{ortiz RR} \end{equation}
That is, for a line bundle $L$ on $X$, 
$$
\chi(X,L)=\binom{2\lambda(L)+6}{5}.
$$
On the other hand, Nieper \cite[Theorem 5.2]{nieper} proved that
\begin{align}\chi(X,L){}&=\int_{X}\exp\bigg(-2\sum_{k=1}^{\infty}\frac{B_{2k}}{4k}\ch_{2k}(X)T_{2k}\bigg(\sqrt{\frac{\lambda(L)}{4}+1}\bigg)\bigg)\notag\\
{}&=\int_{X}\exp\bigg(-2\sum_{k=1}^{\infty}b_{2k}s_{2k}(X)T_{2k}\bigg(\sqrt{\frac{\lambda(L)}{4}+1}\bigg)\bigg)
\label{nieper RR} \end{align}
where $B_{2k}$ are the Bernoulli numbers, 
$b_{2k}=\frac{B_{2k}}{4k(2k)!}$ are modified Bernoulli numbers with $b_0=1$, $s_{2k}=(2k)!\ch_{2k}(X)$, and $T_{2k}$ are even Chebyshev polynomials of the first kind.


So by \eqref{ortiz RR} and \eqref{nieper RR}, we get an identity between polynomials in terms of $y=\sqrt{{\lambda(L)}/{4}+1}$:
\begin{equation}
\binom{8y^2-2}{5}=\int_{X}\exp\bigg(-2\sum_{k=1}^{\infty}b_{2k}s_{2k}(X)T_{2k}(y)\bigg).\label{ortizRR=nieperRR}
\end{equation}
Comparing coefficients of \eqref{ortizRR=nieperRR}, we have $6$ linear equations of $7$ Chern numbers $c_2^5$, $c_2^3c_4$, $c_2^2c_6$, $c_2c_8$, $c_2c_4^2$, $c_4c_6$, $c_{10}$.
Of course this is not sufficient to get a unique solution. In fact, among $6$ equations, there are only $3$ linearly independent linear equations (from comparing coefficients of $y^8$, $y^6$, and $y^2$).


On the other hand, we have more equations form the Hodge numbers of $X$ computed by de Cataldo--Rapagnetta--Sacc\`a \cite{CRS}.
This gives us 
\begin{align*}

\chi^{1}(X){}&=-111;\\
\chi^{2}(X){}&=1062;\\
\chi^{3}(X){}&=-7151;\\
\chi^{4}(X){}&=33534.
\end{align*}
Here $\chi^p(X)=\sum_{q=0}^{\dim X}(-1)^{q}h^{p,q}(X)=\int_X\ch(\Omega_X^p)\td(X)$.
Again, expressing the left-hand sides by Chern numbers, we get $4$  linear equations  of $7$ Chern numbers. 
We can use Mathematica to solve the linear equations for 
$7$ Chern numbers of OG10 as the following:
\begin{align*}
&(c_2^5, c_2^3c_4, c_2^2c_6, c_2c_8, c_2c_4^2, c_4c_6, c_{10})\\
=&(127370880, 53071200, 12383280, 1791720, 22113000, 5159700, 176904).
\end{align*}
\end{proof}
{"config": "arxiv", "file": "2006.09307/article.tex"}