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% \keywords{Erd\H{o}s-Ko-Rado Theorem; Polar Space; Association Scheme; Cross-intersecting Family}
% 
% \subjclass{51E20; 05B25}
\title{\bf Cross-Intersecting Erd\H{o}s-Ko-Rado Sets\\ in Finite Classical Polar Spaces}

\author{Ferdinand Ihringer\\
\small Mathematisches Institut\\[-0.8ex]
\small Justus Liebig University Giessen\\[-0.8ex]
\small Arndtstra\ss{}e 2\\[-0.8ex]
\small 35392 Giessen, Germany\\
\small\tt Ferdinand.Ihringer@math.uni-giessen.de
}

\date{\dateline{Oct 10, 2014}{May 29, 2015}{Jun 15, 2015}\\
\small Mathematics Subject Classifications: 51E20, 05B25, 52C10}

\begin{document}

\maketitle

\begin{abstract}
  A cross-intersecting Erd\H{o}s-Ko-Rado set of generators of a finite classical polar space is a pair $(Y, Z)$ of sets of generators such that all $y \in Y$ and $z \in Z$ intersect in at least a point. We provide upper bounds on $|Y| \cdot |Z|$ and classify the cross-intersecting Erd\H{o}s-Ko-Rado sets of maximum size with respect to $|Y| \cdot |Z|$ for all polar spaces except some Hermitian polar spaces.
  
  \bigskip\noindent \textbf{Keywords:} Erd\H{o}s-Ko-Rado Theorem; Polar Space; Association Scheme; Cross-intersecting Family
\end{abstract}

\section{Introduction}

Erd\H{o}s-Ko-Rado sets (\emph{EKR sets}) were introduced by Erd\H{o}s, Ko, and Rado \cite{MR0140419} as a family of $k$-element subsets of $\{ 1, \ldots, n\}$ such that the elements of the family pairwise intersect non-trivially. In particular, Erd\H{o}s, Ko, and Rado partially classified all such $Y$ of maximum size.

\begin{theorem}[Theorem of Erd\H{o}s, Ko, and Rado]
  Let be $n \geq 2k$. Let $Y$ be an EKR set of $k$-element subsets of $\{ 1, \ldots, n\}$. Then
  \begin{align*}
   |Y| \leq \binom{n-1}{k-1}
  \end{align*}
  with equality for $n>2k$ if and only if $Y$ is set of all $k$-elemental sets containing a fixed element.
\end{theorem}

Stronger versions of this theorem were later proven by several authors including the 
famous works by Wilson \cite{wilson_ekr_1984}, and by Ahlswede and Khachatrian \cite{ahlswede_ekr_1997}.

This theorem for EKR sets was generalized to many structures, including subspaces of projective spaces \cite{maarten_ekr_planes,MR867648,MR0382015} and generators (maximal totally isotropic, respectively, singular subspaces) of polar spaces \cite{maarten_ekr_planes,maarten_ekr_hyperbolic,MR2755082,MR578319}. In polar spaces the problem is partially open, since the maximum size of EKR sets of generators of $\h(2d-1, q^2)$, $d > 3$ odd, is still unknown. To the knowledge of the author the best known upper bound is given in \cite{fi_km_ekr_in_hermitians}.

There exists the following modification of the original problem which generated a lot of interest: a \emph{cross-intersecting EKR set} is a pair $(Y, Z)$ of sets of subsets with $k$ elements of $\{ 1, \ldots, n\}$ such that all $y \in Y$ and $z \in Z$ intersect non-trivially. If one wants to generalize the theorem of Erd\H{o}s, Ko, and Rado to this structure, then the following question arises: how do we measure the size of $(Y, Z)$? There are at least two natural choices. Either one goes for an upper bound for $|Y|+|Z|$ or one tries to find an upper bound for $|Y| \cdot |Z|$. In the set case the first project was pursued in \cite{MR0444483}, while the second one was completed in \cite{MR1008163}. 
Results for vectors spaces are due to Tokushige \cite{tokushige_ev_method}. Again this problem can be generalized to polar spaces, where a \emph{cross-intersecting EKR set of generators} is a pair $(Y, Z)$ of sets of generators such that all $y \in Y$ and $z \in Z$ intersect in at least a point. In this setting this paper is only concerned with an upper bound for $|Y| \cdot |Z|$ and a classification of all cross-intersecting EKR sets reaching this bound.

One additional motivation for this problem is the following: as mentioned before the problem of EKR sets of maximum size in $\h(2d-1, q^2)$ is still open for $d > 3$ odd. Let $p$ be a point of $\h(2d-1, q^2)$ and let $X$ be an EKR set of $\h(2d-1, q^2)$. Furthermore, let $Y$ be the set of generators of $X$ on $p$ and $Z$ the set of generators of $X$ not on $p$. Now in the quotient geometry of $p$ isomorphic to $\h(2d-3, q^2)$ the projection of the generators of $Y$ and $Z$ onto the quotient geometry is a cross-intersecting EKR set. So both problems are related.

One last thing to point out is that this work does not provide tight upper bounds for cross-intersecting EKR sets in $\h(2d-1, q^2)$ for all $d > 2$. The problem is very similar to the open problem of the maximum size of EKR sets in $\h(9, q^2)$. Therefore, it could be reasonable to first solve the problem of the maximum size of cross-intersecting EKR sets in $\h(7, q^2)$ and then generalize the technique to EKR sets in $\h(9, q^2)$.

\section{Projective Spaces \& Polar Spaces}

We refer to \cite{hirschfeld1998projective} for details on \emph{projective spaces}. A projective space $\PG(n-1, q)$ of \emph{projective dimension} $n-1$ (respectively \emph{vector space dimension} $n$) over the field with $q$ elements has exactly
\begin{align*}
  \gauss{n}{k}_{q} := \prod_{i=1}^{k} \frac{q^{n-i+1}-1}{q^i-1}
\end{align*}
subspaces of (vector space) dimension $k$. 

\begin{Remark}
 All the used eigenvalue formulas are more convenient if we use vector space dimensions
 and not projective dimensions. Consequently, the word \emph{dimension} will always refer to
 the vector space dimension of a subspace.
\end{Remark}

We call $1$-dimensional subspaces \emph{points},
$2$-dimensional subspaces \emph{lines}, $3$-dimensional subspaces \emph{planes}, 
and $(n-1)$-dimensional subspaces  of an $n$-dimensional space \emph{hyperplanes}.
We denote the number of points in $\PG(n-1, q)$ by
\begin{align*}
  \gaussm{n}_{q} := \gauss{n}{1}_{q}.
\end{align*}
So we have
\begin{align*}
  \gauss{n}{k}_{q} = \prod_{i=1}^{k} \frac{\gaussm{n-i+1}_{q}}{\gaussm{i}_{q}}.
\end{align*}
We shall write $\gauss{n}{k}$ instead of $\gauss{n}{k}_q$ whenever the choice for $q$ is clear. We will often use
the following analog of the recursive definition of binomial coefficients.
\begin{align}
  \gauss{n+1}{k+1} = \gauss{n}{k+1} + q^{(n-k)} \gauss{n}{k} \label{eq_gaussian_rec}
\end{align}

A \emph{polar space of rank $d$} is an incidence geometry with subspaces of dimension from $0$ to $d$ defined by a non-degenerate reflexive sesquilinear form or a non-degenerate quadratic form. 
The \emph{finite classical polar spaces} are $\q^+(2d-1, q)$, $\q(2d, q)$, $\q^-(2d+1, q)$, $\w(2d-1, q)$, where $q$ is a prime power, $\h(2d-1, q)$, and $\h(2d, q)$, where $q$ is the square of a prime power.
We refer to \cite{hirschfeld1991general} for details. 
Denote the maximal totally isotropic, respectively, singular subspaces of (vector space) dimension $d$ as \emph{generators}.
We denote the tangent space of a totally isotropic subspace $s$ of a polar space by $s^\perp$.
Each subspace of (vector space) dimension $d-1$ of a polar space is incident with exactly $q^\pstype+1$ generators, where $\pstype = 0$ for $\q^+(2d-1, q)$, $\pstype = 1/2$ for $\h(2d-1, q)$, $\pstype = 1$ for $\q(2d, q)$ and $\w(2d-1, q)$, $\pstype = 3/2$ for $\h(2d, q)$, and $\pstype = 2$ for $\q^-(2d+1, q)$.
It is well known that a polar space possesses exactly 
\begin{align*}
 \prod_{i=0}^{d-1} (q^{i+\pstype}+1)
\end{align*}
generators and $(q^{d+e-1} + 1) \gaussm{d}$ points.

\section{The Association Scheme of a Polar Space}\label{sec_assoc}

We need some basic properties of an association scheme of generators on a dual polar space of rank $d$ and type $\pstype$. 
A complete introduction to association schemes can be found in \cite[Ch. 2]{brouwer1989distance}.

\begin{defi}
  Let $X$ be a finite set. A $d$-class association scheme is a pair $(X, \scrR)$, where $\scrR = \{R_0, \ldots, R_d\}$ is a set of symmetric binary relations on $X$ with the following properties:
  \begin{enumerate}
    \item $\{ R_0, \ldots, R_d\}$ is a partition of $X \times X$.
    \item $R_{0}$ is the identity relation.
    \item There are numbers $p_{ij}^k$ such that for $x, y \in X$ with $x R_k y$ there are exactly $p_{ij}^k$ elements $z$ with $x R_i z$ and $z R_j y$.
  \end{enumerate}
\end{defi}

The number $n_i := p_{ii}^{0}$ is called the $i$-valency of $R_i$. The total number of elements of $X$ is
\begin{align*}
  n := |X| = \sum_{i=0}^d n_i.
\end{align*}

The relations $R_i$ are described by their adjacency matrices $A_i \in \bbC^{n \times n}$ defined by
\begin{align*}
  (A_i)_{xy} = \begin{cases}
                 1 & \text{ if } x R_i y\\
                 0 & \text{ otherwise.}
               \end{cases}
\end{align*}
There exist (e.g. in \cite[p. 45]{brouwer1989distance}) idempotent Hermitian matrices $E_j \in \bbC^{n \times n}$ (hence they are positive semidefinite) with the properties
\begin{align*}
\begin{array}{lll}\displaystyle
\sum_{j=0}^d E_j = \vI, & \hspace*{2cm} &\displaystyle E_{0} = n^{-1} \vJ,\\
\displaystyle A_j = \sum_{i=0}^d P_{ij} E_i, &  &\displaystyle E_j = \frac{1}{n} \sum_{i=0}^d Q_{ij} A_i,
  \end{array}
\end{align*}
where $P = (P_{ij}) \in \bbC^{(d+1) \times (d+1)}$ and $Q = (Q_{ij}) \in \bbC^{(d+1) \times (d+1)}$ are the so-called eigenmatrices of the association scheme.
Notice that this implies that the matrices $A_j$ have $d+1$ common eigenspaces $V_i$,
where $V_i$ is the eigenspace of $A_j$ with eigenvalue $P_{ij}$. We will use the fact
that the matrices $A_j$ have common eigenspaces without further notice.

The generators of a polar space define an association scheme if we say that two generators $a$ and $b$ are in relation $R_i$ if and only if $\codim(a \cap b) = i$.
Hence a cross-intersecting EKR set $(Y, Z)$ is a set of vertices such that there are no edges between $Y$ and $Z$ in the (distance-regular) graph associated with $A_d$.
This scheme is \emph{cometric}, so there exists a natural ordering of its $E_j$'s and its associated eigenspaces $V_j$ \cite[Sec. 2.7, Sec. 9.4]{brouwer1989distance}.
The matrix $P$ can be found in the literature (for example in \cite{MR618532} or \cite[Theorem 4.3.6]{vanhove_phd}).
In particular, the eigenvalues of $A_d$ are
\begin{align}
  (-1)^{r} q^{\binom{d-r}{2} + \binom{r}{2} + \pstype(d-r)}\label{eq_ev_disj}
\end{align}
for $r \in \{ 0, 1, \ldots, d \}$.
Here $V_0 = \langle \vj \rangle$, where $\vj$ is the all-ones vector. Furthermore, notice that
all eigenspaces $V_i$ of an association scheme are pairwise orthogonal (see \cite[Ch. 2]{brouwer1989distance}).

\section{An Algebraic Bound}

We shall apply a technique that was, to the knowledge of the author, first used by Haemers in \cite{haermers_ci_ekr_sets_interlacing_bound}.
The author learned about this technique from a paper by Tokushige \cite{tokushige_ev_method}, where he uses a variant of the result based on the work of
Ellis, Friedgut, and Pipel \cite{MR2784326}.
In this section we use the names $d$ and $E_i$ slightly differently compared to
the previous section on association schemes. We do so for two reasons:
\begin{itemize}
 \item For all the cases, which we consider, the $d$ and $E_i$ of the previous section is the same as in this section.
 \item We want to provide a complete version of Lemma \ref{lem_tok}, which is best given for regular graphs.
\end{itemize}

Let $G$ be a graph with $n$ vertices $\{ 1, \ldots, n\}$. A matrix $A = (a_{xy}) \in \bbC^{n \times n}$ is called \emph{extended weight adjacency matrix} of $G$ if $A$ is symmetric, and
\begin{enumerate}
 \item $a_{xy} \leq 0$ if $x$ and $y$ are non-adjacent,
 \item $a_{xx} = 0$,
 \item the all-ones vector $\vj$ is an eigenvector of $A$.
 \item $a_{xy} > 0$ for one $(x, y)$ with $x$ and $y$ adjacent.
\end{enumerate}

Let $V_0, V_1, \ldots, V_d$ be eigenspaces or subspaces of eigenspaces of $A$ such that
\begin{enumerate}
 \item $V_0 = \langle \vj \rangle$,
 \item $\bbC^n = V_0 \perp V_1 \perp \ldots \perp V_d$.
\end{enumerate}
Let $\lambda_i$ be the eigenvalue corresponding to $V_i$.
Denote the eigenvalue of $\vj$ by $k$. 
Denote the smallest eigenvalue with eigenvectors in $V_1 \perp \ldots \perp V_d$ by $\lambda_{-}$.
Denote the intersection of the eigenspace of $\lambda_-$ with $V_1 \perp \ldots \perp V_d$ by $V_-$.
Denote the largest eigenvalue with eigenvectors in $V_1 \perp \ldots \perp V_d$ by $\lambda_{+}$.
Denote the intersection of the eigenspace of $\lambda_+$ with $V_1 \perp \ldots \perp V_d$ by $V_+$.
Denote $\max\{ -\lambda_{-}, \lambda_{+}\}$ by $\lambda_{b}$.
We say that $\lambda_b$ is the \emph{second largest absolute eigenvalue} of $A$.
Notice that this actually does not mean that $\lambda_b$ is smaller than $k$.
The last condition ($a_{xy} > 0$) on extended weight adjacency matrices is merely technical and simplifies some of the stated results
as it guarantees $\lambda_- < 0 < \lambda_+$.

A \emph{characteristic vector} $\chi_Y$ of a subset $Y \subseteq \{ 1, \ldots, n\}$ is defined by
\begin{align*}
 \chi_{i} = \begin{cases}
    1 & \text{ if } i \in Y\\
    0 & \text{ if } i \notin Y
 \end{cases}.
\end{align*}

Let $E_0$ be the projection operator onto $\langle \vj \rangle$, i.e. $E_0 = \vJ/n$, where
$\vJ$ is the \emph{all-ones matrix}.\label{page_Eis}
More generally, let $E_i$ be the projection operator onto the eigenspace of $\lambda_i$ for $0 < i \leq d$.
Let $I_-$ be the subset of $\{ 1, \ldots, d\}$ with $\lambda_i = \lambda_-$
and let $I_+$ be the subset of $\{ 1, \ldots, d\}$ with $\lambda_i = \lambda_+$.
Then $V_- = \perp_{i \in I_-} V_i$ and $V_+ = \perp_{i \in I_+} V_i$, so we can
define the projections $E_-$, receptively, $E_+$ onto $V_-$, receptively, $V_+$ by
\begin{align*}
  E_- = \sum_{i \in I_-} E_i, && \text{ respectively,} && E_+ = \sum_{i \in I_+} E_i.
\end{align*}

We can write any vector of
$\chi \in V$ as
\begin{align*}
  \chi = \frac{\chi^\T \vj}{n} \vj + \sum_{i=1}^d E_i \chi.
\end{align*}
As the $\lambda_i$ are the eigenvalues of $A$, we can write 
\begin{align*}
  A = \frac{k}{n} \vJ + \sum_{i=1}^d \lambda_i E_i.
\end{align*}

An \emph{independent set} is a set of pairwise non-adjacent vertices of a graph.
Ellis, Friedgut, and Pipel \cite{MR2784326} used the following result, a generalization of the Hoffman bound for independent sets in graphs.
In this paper we say that a \emph{cross-intersecting independent set} is a pair $(Y, Z)$ such that
there are no edges between $Y$ and $Z$. In particular cross-intersecting sets are cross-intersecting independent sets for 
the appropriate graphs.
\begin{lemma}\label{lem_efp}
  Let $A$ be an extended weight adjacency matrix of a regular graph $G$ with $n$ vertices.
  Let $k$ be the eigenvalue of the all-ones vector $\vj$, and let $\lambda_b$ the second
  largest absolute eigenvalue of $A$.
  Let $(Y, Z)$ be a cross-intersecting independent set. 
  If $k + \lambda_b > 0$, then
  \begin{align*}
    \sqrt{|Y| \cdot |Z|} \leq \frac{\lambda_b}{k + \lambda_b} n.
  \end{align*}
\end{lemma}

\begin{Remark}
  In \cite{MR2784326} Lemma \ref{lem_efp} is restricted to \emph{weight adjacency matrices} or
  \emph{pseudo adjacency matrices} where $a_{ij}$ is zero if $i$ and $j$ are non-adjacent
  and $a_{ij} > 0$ if $i$ and $j$ are adjacent. Our more general definition 
  does not change anything and will turn out to be more convenient
  in Section \ref{sec_hermitian} where we shall not bother to calculate the exact 
  minimum of the Hoffman bound.
\end{Remark}

Tokushige reformulates Lemma \ref{lem_efp} in a more detailed way in \cite[Lemma 2]{tokushige_ev_method}. 
Unfortunately, his reformulation does not point out some details which are necessary for the special case handled in this paper. 
Furthermore, our notation differs significantly, so we shall prove Lemma \ref{lem_efp} again as
well as our version of Tokoshige's result, Lemma \ref{lem_tok}. First recall the inequality between the geometric and the arithmetic mean.

\begin{lemma}[Inequality of Arithmetic and Geometric Means]\label{lem_geo_mean}
  Let $0 \leq \alpha, \beta \leq 1$. Then we have
  \begin{align*}
    \sqrt{1-\alpha} \sqrt{1-\beta} \leq 1 - \sqrt{\alpha \beta}
  \end{align*}
  with equality if and only if $\alpha = \beta$.
\end{lemma}

Our variant of Tokoshige's lemma is as follows.

\begin{lemma}\label{lem_tok}
  If equality holds in Lemma \ref{lem_efp},
  then $|Y| = |Z| = \alpha n$ for some $\alpha \in \bbR$.
  Furthermore, one of the following cases occurs.
  \begin{enumerate}[(a)]
   \item We have $\lambda_{+} = \lambda_b > -\lambda_{-}$, $\chi_Y = \alpha \vj + v_+$, 
   and $\chi_Z = \alpha \vj - v_+$ for some vector $v_+ \in V_+$.
   \item We have $\lambda_{+} < \lambda_b = -\lambda_{-}$, $\chi_Y = \alpha \vj + v_-$, 
   and $\chi_Z = \alpha \vj + v_-$ for some vector $v_- \in V_-$. In this case $Y=Z$, and $Y$ is an independent set of maximum size.
   \item We have $\lambda_{+} = \lambda_b = -\lambda_{-}$, $\chi_Y = \alpha \vj + v_- + v_+$, 
   and $\chi_Z = \alpha \vj + v_- - v_+$ for some vectors $v_- \in V_-$ and $v_+ \in V_+$.
  \end{enumerate}
\end{lemma}
\begin{proof}
  Let $\chi = \chi_Y$ be the characteristic vector of $Y$.
  Let $\psi = \chi_Z$ be the characteristic vector of $Z$.
  We suppose that $(Y, Z)$ is a cross-intersecting independent set. 
  Since $A$ is an extended weight adjacency matrix, we have $\chi^\T A \psi \leq 0$
  
  Put $y = |Y|$ and $z = |Z|$. Suppose without loss of generality $y \geq z > 0$.
  Recall the definitions of $E_i$, $E_+$ and $E_-$ at the beginning of the chapter on page~\pageref{page_Eis}.
  Write $\chi$ and $\psi$ as
  \begin{align*}
   &\chi = \frac{y}{n} \vj  + \sum_{i=1}^d E_i \chi,\\
   &\psi = \frac{z}{n} \vj  + \sum_{i=1}^d E_i \psi.
  \end{align*}
  Recall that
  \begin{align*}
    A = \frac{k}{n} \vJ + \sum_{i=1}^d \lambda_i E_i.
  \end{align*}
  Let $E_+$ the projection operator onto $V_+$ and $E_-$ the projection operator onto $V_-$.
  We have
  \begin{align}
    0 &\geq \chi^\T A \psi\notag\\
    &\geq \frac{k}{n} \chi^\T \vJ \psi - \sum_{i=1}^d \lambda_i \chi^\T E_i \psi \label{eq_ci_hoffman_coclique_eq0}\\
    &\geq \frac{kyz}{n} - \sum_{i=1}^d \left|\lambda_i \chi^\T E_i \psi \right| \label{eq_ci_hoffman_coclique_eq1}\\
    &\geq \frac{k yz}{n} - \lambda_b \sum_{i=1}^d |(E_i \chi)^\T E_i \psi| \label{eq_ci_hoffman_coclique_eq2}\\
    &\geq \frac{k yz}{n} - \lambda_b \sum_{i=1}^d \sqrt{(E_i \chi)^\T E_i \chi} \sqrt{(E_i \psi)^\T E_i \psi} \label{eq_ci_hoffman_coclique_eq2aa}\\
    &\geq \frac{k yz}{n} - \lambda_b \sqrt{\sum_{i=1}^d (E_i \chi)^\T(E_i \chi)} \cdot \sqrt{\sum_{i=1}^d (E_i \psi)^\T(E_i \psi)} \label{eq_ci_hoffman_coclique_eq2a}
%     &\geq \frac{k yz}{n} - \lambda_b |(E_++E_-) \chi| |(E_++E_-) \psi| \label{eq_ci_hoffman_coclique_eq2}
%   \\&\geq \frac{k yz}{n} - \lambda_b |(E_++E_-) \chi|^2 \label{eq_ci_hoffman_coclique_eq3}\\
  \end{align}
  Notice that we apply the Cauchy-Schwartz inequality twice when going from \eqref{eq_ci_hoffman_coclique_eq2} to \eqref{eq_ci_hoffman_coclique_eq2a}.
  Furthermore $\chi$ and $\psi$ are $0$-$1$-vectors, so
  \begin{align}
    y &= \chi^\T \chi = \frac{y^2}{n} + \sum_{i=1}^d (E_{i} \chi)^\T(E_{i} \chi),
    \intertext{and}
    z &= \frac{z^2}{n} + \sum_{i=1}^d (E_{i} \psi)^\T(E_{i} \psi).
  \end{align}
  If we put this back into our previous inequality \eqref{eq_ci_hoffman_coclique_eq2a}, 
  then we have by Lemma \ref{lem_geo_mean}
  \begin{align}
    0 &\geq \frac{k yz}{n} - \lambda_b \sqrt{yz} \sqrt{1-\frac{y}{n}} \sqrt{1-\frac{z}{n}} \label{eq_ci_hoffman_coclique_eq4a}\\
    &\geq \frac{k yz}{n} - \lambda_b \sqrt{yz} (1-\sqrt{yz}/n).\label{eq_ci_hoffman_coclique_eq4}
  \end{align}
  Rearranging yields Lemma \ref{lem_efp}.
  
  If this bound is tight, then we have equality from \eqref{eq_ci_hoffman_coclique_eq0}
  to \eqref{eq_ci_hoffman_coclique_eq4}. 
  Equality between \eqref{eq_ci_hoffman_coclique_eq1} and \eqref{eq_ci_hoffman_coclique_eq2} shows
  that we have $(E_i \chi)^\T E_i \psi = 0$ for all $i$ such that $|\lambda_i| \neq \lambda_b$.
  Equality between \eqref{eq_ci_hoffman_coclique_eq2} and \eqref{eq_ci_hoffman_coclique_eq2aa} shows
  that there exists a factor $\alpha_i$ with $E_i \chi = \alpha_i E_i \psi$ for all $i \in \{1, \ldots, d\}$.
  Equality between \eqref{eq_ci_hoffman_coclique_eq2aa} and \eqref{eq_ci_hoffman_coclique_eq2a} shows
  that there exist constants $\beta$ with $\sqrt{(E_i \chi)^\T E_i \chi} = \beta \sqrt{(E_i \psi)^\T E_i \psi}$ for all $i \in \{1, \ldots, d\}$.
  Hence, either $E_i \chi = E_i \psi = 0$ or $\beta = |\alpha_i|$.
  These two observations show that there $E_i \chi = E_i \psi = 0$ if $|\lambda_i | \neq \lambda_b$. Hence,
  \begin{align*}
   &\chi = \frac{y}{n} \vj  + E_- \chi + E_+ \chi\\
   &\psi = \frac{z}{n} \vj  + E_- \psi + E_+ \psi.
  \end{align*}
  Equality between \eqref{eq_ci_hoffman_coclique_eq4a} and \eqref{eq_ci_hoffman_coclique_eq4} shows $y = z$ (see Lemma \ref{lem_geo_mean}).
  By
  \begin{align*}
    \sum_{i=1}^d (E_{i} \psi)^\T(E_{i} \psi) = z - \frac{z^2}{n} = 
    y - \frac{y^2}{n} = \sum_{i=1}^d (E_{i} \chi)^\T(E_{i} \chi) = \beta \sum_{i=1}^d (E_{i} \psi)^\T(E_{i} \psi)
  \end{align*}
  we derive $\beta = 1$. Then equality between \eqref{eq_ci_hoffman_coclique_eq0} and \eqref{eq_ci_hoffman_coclique_eq1}
  shows that $\alpha_i = 1$ if $\lambda_i < 0$ and that $\alpha_i = -1$ if $\lambda_i > 0$.
  Hence, $E_- \chi = E_- \psi$ and $E_+ \chi = -E_+ \psi$.
  This proves Lemma \ref{lem_tok}.
\end{proof}
 
\section{Cross-intersecting EKR Sets of Maximum Size}

In this section we shall calculate tight upper bounds for all polar spaces except
$\h(2d-1, q^2)$, and classify all examples in case of equality.
For all polar spaces except $\h(2d-1, q^2)$ we can imitate the approach of Pepe, Storme, and Vanhove \cite{MR2755082}.
Recall from Section \ref{sec_assoc} that we have a natural ordering of the 
eigenspaces $V_0 (= \langle \vj \rangle), V_1, \ldots, V_d$ of the association scheme which
we defined on generators of a polar space of rank $d$.

\begin{lemma}\label{lem_thm_es_pre}
  Let $\scrP$ be a polar space over $\bbF_q$ with parameter $e$ and let $A_d$ be the 
  adjacency matrix matrix of the disjointness graph of generators of $\scrP$.
  Let $V_0, V_1, \ldots, V_d$ the common eigenspaces of the corresponding
  association scheme. Then $k = q^{\binom{d}{2}+de}$. Additionally, we have the following.
  \begin{itemize}
   \item If $\scrP \in \{ \q^+(2d-1, q), \h(2d-1, q)\}$, then $\lambda_b = q^{\binom{d}{2}}$. Moreover, if $d$ is
   even, then $\lambda_b = \lambda_+ > -\lambda_-$ and $V_+ = V_d$; if $d$ is odd, then $\lambda_b = -\lambda_- > \lambda_+$
   and $V_- = V_d$.
   \item If $\scrP \in \{ \q(2d, q), \w(2d-1, q) \}$, then $\lambda_b = q^{\binom{d}{2}}$.
   Moreover, if $d$ is even, then $\lambda_b = -\lambda_- = \lambda_+$, $V_- = V_1$ and $V_+ = V_d$;
   if $d$ is odd, then $\lambda_b = -\lambda_- > \lambda_+$ and $V_- = V_1 \perp V_d$.
   \item If $\scrP \in \{ \h(2d, q), \q^-(2d+1, q)\}$, then $\lambda_b = q^{\binom{d-1}{2}+de}$. Moreover,
   $\lambda_b = -\lambda_- > \lambda_+$ and $V_- = V_1$.
  \end{itemize}
\end{lemma}
\begin{proof}
  The eigenvalues of $A_d$ are given in \eqref{eq_ev_disj} as
  \begin{align*}
   (-1)^{r} q^{\binom{d-r}{2} + \binom{r}{2} + e(d-r)}.
  \end{align*}
  For $r=0$ this is the eigenvalue that belongs to the all-ones vector $\vj$, so
  with $k$ defined as in Lemma \ref{lem_efp} we have
  \begin{align*}
    k = q^{\binom{d}{2} + de}.
  \end{align*}
  For $e = 0$ note that the absolute eigenvalues for $r=0$ and $r=d$ are equal. Therefore,
  the eigenspace belonging to $k$ has dimension at least $2$ which make $k$ also the second
  largest absolute eigenvalue. Hence, we have the following for the different polar spaces.
  For $e \in \{ 0, 1/2 \}$ (i.e. $\scrP \in \{ \q^+(2d-1, q), \h(2d-1, q) \}$) the second largest absolute 
  eigenvalue occurs if and only if $r = d$, 
  for $e = 1$ (i.e.  $\scrP \in \{ \q(2d, q), \w(2d-1, q)\}$)
  the second largest absolute eigenvalue occurs if and only if $r \in \{ 1, d\}$, for
  $e \in \{ 3/2, 2\}$ (i.e.  $\scrP \in \{ \h(2d, q), \q^-(2d+1, q) \}$) the
  second largest absolute eigenvalue occurs if and only if $r = 1$.
\end{proof}

Recall that a cross-intersecting EKR set $(Y, Z)$ is a set of vertices of the disjointness graph
such that there are no edges between $Y$ and $Z$.
Hence, using Lemma \ref{lem_tok} and the classification of EKR sets of generators given in \cite{MR2755082} we get the following result.

\begin{cor}\label{cor_ci_ekr_gen_case}
  Let $(Y, Z)$ be a cross-intersecting EKR set of maximum size of a finite classical polar space $\scrP$ 
  not isomorphic to $\q(2d, q)$ with $d$ even, $\w(2d-1, q)$ with $d$ even, $\q^+(2d-1, q)$ 
  with $d$ even, or $\h(2d-1, q^2)$, where $|Y| \cdot |Z|$ reaches the bound in Lemma \ref{lem_efp}. 
  Then $Y = Z$, and $Y$ is an EKR set of maximum size.
\end{cor}
\begin{proof}
  By Lemma \ref{lem_tok} and Lemma \ref{lem_thm_es_pre}, all cross-intersecting EKR, which reach the bound,
  are EKR sets of maximum size. These EKR sets exist as shown in \cite{MR2755082}.
\end{proof}

The following cases besides $\h(2d-1, q^2)$ remain.

\begin{theorem}\label{thm_es}
 Let $(Y, Z)$ be a cross-intersecting EKR set of generators of maximum size of a polar space $\scrP$,
 with $\scrP$ isomorphic to $\q(2d, q)$, $d$ even, $\w(2d-1, q)$, $d$ even, or $\q^+(2d-1, q)$, $d$ even.
 Let $n$ be the number of generators of $\scrP$.
 Then we have the following:
 \begin{itemize}
  \item If $\scrP = \q^+(2d-1, q)$, then $\sqrt{|Y| \cdot |Z|}$ is at most $n/2$, and if this bound is reached, then 
  there is a $v_+ \in V_d$ such that $\chi_Y = \vj/2+ v_+$
  and $\chi_Z = \vj/2 - v_+$.
  \item If $\scrP \in \{ \q(2d, q), \w(2d-1, q)\}$, then $\sqrt{|Y| \cdot |Z|}$ is at most the number of generators on a fixed point, and if this bound is reached, then 
  there are $v_- \in V_1$ and $v_+ \in V_d$ such that $\chi_Y = \alpha \vj + v_- + v_+$
  and $\chi_Z = \alpha \vj + v_- - v_+$, with $\alpha = \frac{1}{q^d+1}$.
 \end{itemize}
\end{theorem}
\begin{proof}
  Apply Lemma \ref{lem_tok} and Lemma \ref{lem_thm_es_pre}.
\end{proof}

Similar to \cite{MR2755082} we shall continue by classifying the more complicated cases.

\subsection{The Hyperbolic Quadric of Even Rank}

The generators of $\q^+(2d-1, q)$ can be partitioned into two sets $X_1$ and $X_2$ of generators (commonly known as \emph{latins} and \emph{greeks}) with $|X_1| = |X_2| = n/2$. 
For $x_1 \in X_1$ and $x_2 \in X_2$ the codimension of the intersection of $x_1 \cap x_2$ is odd. 
For $x_1, x_2 \in X_1$ the codimension of the intersection of $x_1 \cap x_2$ is even. 
This implies for $d$ even that $(X_1, X_2)$ is a cross-intersecting EKR set of maximum size according to Theorem \ref{thm_es}. 
There exist $x_1, x_2 \in X_1$ with $\dim(x_1 \cap x_2) = 0$ if $d$ is even, so $(X_1, X_1)$ is not a cross-intersecting EKR set.

\begin{theorem}\label{thm_ci_ekr_qp}
  Let $(Y, Z)$ be a cross-intersecting EKR set of maximum size of $\q^+(2d-1, q)$ with $d$ even. Then $Y = X_i$ and $Z=X_j$ for $\{ i, j\} = \{1, 2\}$.
\end{theorem}
\begin{proof}
  By Theorem \ref{thm_es}, we have $\chi_Y = \vj/2 + v_+$ and $\chi_Z = \vj/2 - v_+$.
  As in Theorem 16 of \cite{MR2755082} $V_{0} = \langle \vj \rangle$ is spanned by $\chi_{X_1}+\chi_{X_2}$, and $V_d$ is spanned by $\chi_{X_1}-\chi_{X_2}$. 
  Recall $\chi_Y + \chi_Z = \vj$ and $\chi_Y - \chi_Z = 2v_+ \in V_d$. Hence, $\chi_Y = \chi_{X_i}$ and $\chi_Z = \chi_{X_j}$ for some $\{i, j\} = \{1, 2\}$.
  Hence $\{ Y, Z \} = \{ X_1, X_2 \}$.
\end{proof}

\subsection{The Parabolic Quadric and the Symplectic Polar Space of Even Rank}

Before we classify cross-intersecting EKR sets of generators of maximum size in parabolic
quadrics and symplectic polar spaces, we mention a few simple, most likely well-known lemmas. We include their
proofs to make this article more self-contained.

\begin{lemma}\label{lem_nbr}
  Let $\chi \in \langle \vj \rangle \perp V$ for some eigenspace $V$ of an (extended weight) adjacency matrix $A$ of a $k$-regular graph with $n$ vertices associated with eigenvalue $\lambda$. 
  Then the characteristic vector $e_i$ of the $i$-th vertex satisfies
  \begin{align*}
    e_i^\T A \chi = \frac{\chi^\T \vj}{n} (k - \lambda) + \lambda e_i^\T \chi.
  \end{align*}
\end{lemma}
\begin{proof}~
  As $\chi \in \langle \vj \rangle \perp V$, we can write $\chi = \alpha \vj + v$ for some $v \in V$ and $\alpha = \frac{\chi^\T \vj}{n}$. Then
  \begin{align*}
   \hspace*{50mm} e_i^\T A \chi &= e_i^\T A (\alpha \vj + v)\\
   &= e_i^\T (\alpha k \vj + \lambda v)\\
   &= e_i^\T (\alpha (k-\lambda) \vj + \lambda \chi)\\
   &= \frac{\chi^\T \vj}{n} (k - \lambda) + \lambda e_i^\T \chi. && \hspace*{50mm} \qedhere
  \end{align*}
\end{proof}

\begin{cor}\label{lem_inmbs}
  Let $\chi, \psi \in \langle \vj \rangle \perp V_- \perp V_+$ for some eigenspaces $V_-$ and $V_+$ of a (weighted) adjacency matrix $A$ of a $k$-regular graph with $n$ vertices, where $\lambda_-$ is the eigenvalue associated with $V_-$ and $\lambda_+$ is the eigenvalue associated with $V_+$.
  If $\chi = \alpha \vj + v_- + v_+$ and $\psi = \alpha \vj + v_- - v_+$ for some $\alpha \in \bbR$, $v_- \in V_-$, and $v_+ \in V_+$, then
  \begin{align*}
    &e_i^\T A \chi = \alpha (k - \lambda_-) + \frac{(\lambda_-+\lambda_+)}{2} e_i^\T \chi + \frac{(\lambda_- - \lambda_+)}{2} e_i^\T \psi,\\
    &e_i^\T A \psi = \alpha (k - \lambda_-) + \frac{(\lambda_- + \lambda_+)}{2} e_i^\T \psi + \frac{(\lambda_--\lambda_+)}{2} e_i^\T \chi.
  \end{align*}
\end{cor}
\begin{proof}
  We have $\chi + \psi \in \langle \vj \rangle \perp V_-$ and $\chi - \psi \in V_+$. By Lemma \ref{lem_nbr} and $\vj^\T v_- = 0 = \vj^\T v_+$,
  \begin{align*}
    &e_i^\T A (\chi + \psi) = \frac{(\chi+\psi)^\T \vj}{n} (k - \lambda_-) + \lambda_- e_i^\T (\chi+\psi)\\
    &e_i^\T A (\chi - \psi) = \lambda_+ e_i^\T (\chi-\psi).
  \end{align*}
  Now the equations $2 e_i^\T A \chi = e_i^\T A (\chi + \psi) + e_i^\T A (\chi - \psi)$,
  $2 e_i^\T A \psi = e_i^\T A (\chi + \psi) - e_i^\T A (\chi - \psi)$ and $(\chi + \psi)^\T \vj = 2\alpha n$ yield the assertion.
\end{proof}

Recall for this section that all matrices $A_i$ of a given 
association scheme have the common eigenspaces $V_j$.

\begin{lemma}\label{lem_ev_q0}
  For the adjacency matrix $A_{d-s}$, $0 < s < d$, of the dual polar graph of $\q(2d, q)$ 
  or $\w(2d-1, q)$ the eigenspace $V_1$ is associated with eigenvalue
  \begin{align*}
    \lambda_{-,s} := -\gauss{d-1}{s} q^{\binom{d-s}{2}} + \gauss{d-1}{s-1} q^{\binom{d-s+1}{2}},
  \end{align*}
  the eigenspace $V_d$ is associated with eigenvalue
  \begin{align*}
    \lambda_{+,s} := (-1)^{d-s} \gauss{d}{s} q^{\binom{d-s}{2}},
  \end{align*}
  and the eigenspace $\langle \vj \rangle$ is associated with eigenvalue
  \begin{align*}
   k_s := \gauss{d}{s} q^{\binom{d-s+1}{2}} = \left( \gauss{d-1}{s} + \gauss{d-1}{s-1} q^{d-s} \right) q^{\binom{d-s+1}{2}}.
  \end{align*}
\end{lemma}
\begin{proof}
  See \cite[Theorem 4.3.6]{vanhove_phd}, where
  the eigenvalue of $V_j$ for $A_{i}$ is given by
  \begin{align*}
    \sum_{0, j-i \leq u \leq  d-i,j} (-1)^{j+u} \gauss{d-j}{d-i-u} \gauss{j}{u} q^{(u+i-j)(u+i-j+2\pstype -1)/2 + \binom{j-u}{2}}.
  \end{align*}
  Note that $e=1$ in the considered cases.
  For $j=0, 1, d$, and $i=d-s$, this formula yields the assertion.
  The last equality is an application of \eqref{eq_gaussian_rec}.
\end{proof}

If a cross-intersecting EKR set $(Y, Z)$ of $\q(2d, q)$ satisfies $\chi_Y, \chi_Z \in V_{0} \perp V_1$, then $Y = Z$, and $Y$ is an EKR set as before.
So only the case $\chi_Y, \chi_Z \in V_{0} \perp V_1 \perp V_d$ remains. In the following denote $V_1$ by $V_-$ and $V_d$ by $V_+$.
Furthermore, as in Lemma \ref{lem_tok} we write 
%\begin{align*}
 %\chi_Y &= \alpha \vj + v_- + v_+ \\
 %&= \frac{|Y|}{n} \vj + v_- + v_+ \\
 %&= \frac{\lambda_b}{k+\lambda_b} \vj + v_- + v_+
 %\intertext{ and }
 %\chi_Z &= \alpha \vj + v_- - v_+ \\
 %&= \frac{|Z|}{n} \vj + v_- - v_+ \\
 %&= \frac{\lambda_b}{k+\lambda_b} \vj + v_- - v_+
%\end{align*}
\begin{align*}
 \chi_Y &= \alpha \vj + v_- + v_+ 
 = \frac{|Y|}{n} \vj + v_- + v_+ 
 = \frac{\lambda_b}{k+\lambda_b} \vj + v_- + v_+
 \intertext{ and }
 \chi_Z &= \alpha \vj + v_- - v_+ 
 = \frac{|Z|}{n} \vj + v_- - v_+ 
 = \frac{\lambda_b}{k+\lambda_b} \vj + v_- - v_+
\end{align*}
with $v_- \in V_-$ and $v_+ \in V_+$.

\begin{prop}\label{prop_q0_inmbs}
  Let $(Y, Z)$ be a cross-intersecting EKR set of $\q(2d, q)$ or $\w(2d-1, q)$, $d$ even, of maximum size such that $Y \nsubseteq Z$. Let $g \in Y \setminus Z$.
  \begin{enumerate}[(a)]
   \item If $d-s$ is even, then $g$ meets $0$ elements of $Z$ in dimension $s$.
   \item If $d-s$ is odd, then $g$ meets $0$ elements of $Y$ in dimension $s$.
   \item If $d-s$ is even, then $g$ meets 
   \begin{align*}
      \gauss{d}{s} q^{\binom{d-s}{2}}
   \end{align*}
   elements of $Y$ in dimension $s$.
   \item If $d-s$ is odd, then $g$ meets
   \begin{align*}
      \gauss{d}{s} q^{\binom{d-s}{2}}
   \end{align*}
   elements of $Z$ in dimension $s$.
   \end{enumerate}
\end{prop}
\begin{proof}
  We can calculate these numbers with Lemma \ref{lem_ev_q0} and Corollary \ref{lem_inmbs} by choosing $\chi_{\{g\}}$ as $e_i$. 
  For $A_{d-s}$ the parameters are given by
  \begin{align*}
   k_s - \lambda_{-,s} &= \left( \gauss{d-1}{s} + \gauss{d-1}{s-1} q^{d-s} \right) q^{\binom{d-s+1}{2}} \\
   &+ \gauss{d-1}{s} q^{\binom{d-s}{2}} - \gauss{d-1}{s-1} q^{\binom{d-s+1}{2}}\\
   &= q^{\binom{d-s}{2}} \gauss{d-1}{s} \left( q^{d-s} +1 \right) + q^{\binom{d-s+1}{2}} \gauss{d-1}{s-1} \left( q^{d-s} -1 \right) \\
   &\stackrel{\text{Def.}}{=} q^{\binom{d-s}{2}} \gauss{d-1}{s} \left( q^{d-s} +1 \right) + q^{d-s} \cdot q^{\binom{d-s}{2}} \gauss{d-1}{s} \left( q^s -1 \right) \\
   &=q^{\binom{d-s}{2}} \gauss{d-1}{s} \left( q^{d}+1 \right),
   \intertext{for $d-s$ even}
   & \lambda_{-,s} + \lambda_{+,s} \stackrel{\eqref{eq_gaussian_rec}}{=} 2 \gauss{d-1}{s-1} q^{\binom{d-s+1}{2}},\\
   & \lambda_{-,s} - \lambda_{+,s} \stackrel{\eqref{eq_gaussian_rec}}{=} -2 \gauss{d-1}{s} q^{\binom{d-s}{2}},
   \intertext{for $d-s$ odd}
   & \lambda_{-,s} + \lambda_{+,s} \stackrel{\eqref{eq_gaussian_rec}}{=} -2 \gauss{d-1}{s} q^{\binom{d-s}{2}},\\
   & \lambda_{-,s} - \lambda_{+,s} \stackrel{\eqref{eq_gaussian_rec}}{=} 2 \gauss{d-1}{s-1} q^{\binom{d-s+1}{2}}.
   \intertext{Recall that}
   \frac{(\chi_Y+\chi_Z)^\T \vj}{2n} &= \frac{\lambda_+}{k + \lambda_+}
   = \frac{q^{\binom{d}{2}}}{q^{\binom{d+1}{2}} + q^{\binom{d}{2}}}
   = \frac{1}{q^{d}+1}.
   \intertext{Hence by Corollary \ref{lem_inmbs} and $g \notin Z$ (i.e. $e_i^T \psi = 0$),}
   e_i^\T A_{d-s} \chi_Y &= \frac{k_s - \lambda_{-,s}}{q^d+1} + \frac{(\lambda_-+\lambda_+)}{2} e_i^\T \chi\\
   &= q^{\binom{d-s}{2}} \gauss{d-1}{s} + \frac{(\lambda_-+\lambda_+)}{2} e_i^\T \chi.
   \intertext{If $d-s$ is even, then by \eqref{eq_gaussian_rec}  and $e_i^\T \chi_Y = 1$}
   e_i^\T A_{d-s} \chi_Y &= q^{\binom{d-s}{2}} \gauss{d-1}{s} + q^{\binom{d-s+1}{2}} \gauss{d-1}{s-1} = q^{\binom{d-s}{2}} \gauss{d}{s}.
   \intertext{If $d-s$ is odd, then by $e_i^\T \chi_Y = 1$}
   e_i^\T A_{d-s} \chi_Y &= q^{\binom{d-s}{2}} \gauss{d-1}{s} - q^{\binom{d-s}{2}} \gauss{d-1}{s} = 0.
  \end{align*}
  All the remaining cases are either similar or trivial.
\end{proof}

\begin{cor}\label{cor_Y_Z_diff_imp_Y_Z_disj}
  Let $(Y, Z)$ be a cross-intersecting EKR set of $\q(2d, q)$ or $\w(2d-1, q)$, $d$ even, of maximum size such that $Y \nsubseteq Z$.
  Then $Y \cap Z = \emptyset$.
\end{cor}
\begin{proof}
  Let $g$ be as in Proposition \ref{prop_q0_inmbs}. 
  By Proposition \ref{prop_q0_inmbs}, no element of $Y$ meets $g$ in the same dimension as an element of $Z$, hence $Y \cap Z = \emptyset$.
\end{proof}

Now we have a strong combinatorial information about cross-intersecting EKR sets $(Y, Z)$ of maximum size which are not EKR sets.
By adding some geometrical arguments this leads to a complete classification of cross-intersecting EKR sets
of maximum size in these parabolic and symplectic polar spaces
as we shall see in the following.

\begin{lemma}\label{lem_intersection_lemma}
  Let $g_1, g_2, g_3$ be pairwise different generators of a polar space of rank $d$
  such that 
  \begin{align*}
   \dim(g_1 \cap g_2 \cap g_3) = d-2 && \text{and} && g_1 \cap g_2 \neq g_1 \cap g_3\\
   && \text{and} && \dim(g_1 \cap g_2) = \dim(g_1 \cap g_3) = d-1.
  \end{align*}
  Then $g_2 \cap g_3$ is a $(d-2)$-dimensional subspace of $g_1$.
\end{lemma}
\begin{proof}
  We have 
  \begin{align*}
    \dim(g_1 \cap g_2), \dim(g_1 \cap g_3) > \dim(g_1 \cap g_2 \cap g_3).
  \end{align*}
  By $\dim(g_1 \cap g_2) = \dim(g_1 \cap g_3) = d-1$,
  \begin{align*}
    \dim(\langle g_1 \cap g_2, g_1 \cap g_3 \rangle) = \dim(g_1 \cap g_2) + \dim(g_1 \cap g_3) - \dim(g_1 \cap g_2 \cap g_3) = d = \dim(g_1).
  \end{align*}
  Hence, 
  \begin{align}
   \langle g_1 \cap g_2, g_1 \cap g_3 \rangle = g_1. \label{eq_g1g2_g1g3_dm1}
  \end{align}
  Suppose contrary to the first part of the assertion that $g_2 \cap g_3 \nsubseteq g_1$.
  By \eqref{eq_g1g2_g1g3_dm1}, we have
  \begin{align*}
    &\dim(\langle g_1 \cap g_2, g_1 \cap g_3, g_2 \cap g_3 \rangle) \\
   = &\dim(g_2 \cap g_3) + \dim(\langle g_1 \cap g_2, g_1 \cap g_3 \rangle) - \dim(g_2 \cap g_3 \cap \langle g_1 \cap g_2, g_1 \cap g_3 \rangle)\\
   = &\dim(g_2 \cap g_3) + \dim(g_1) - \dim(g_1 \cap g_2 \cap g_3)\\
   > &(d-2) + d - (d-2) = d.
  \end{align*}
  As $\langle g_1 \cap g_2, g_1 \cap g_3, g_2 \cap g_3 \rangle$ is a totally isotropic subspace
  of a polar space rank $d$, 
  this is a contradiction. Hence, $g_2 \cap g_3 \subseteq g_1$.
  Furthermore, $\dim(g_2 \cap g_3) = \dim(g_1 \cap g_2 \cap g_3) = d-2$.
\end{proof}

\begin{lemma}\label{lem_nmb_dm1z}
  Let $(Y, Z)$ be a cross-intersecting EKR set of $\q(2d, q)$ or $\w(2d-1, q)$, $d$ even, of maximum size with $Y \neq Z$. 
  Let $g, h \in Y$ disjoint. Let $\pi_1, \ldots, \pi_{\gaussm{d}} \subseteq g$ be the $\gaussm{d}$ subspaces of dimension $d-1$ of $g$.
  Then the following holds:
  \begin{enumerate}[(a)]
   \item Exactly $\gaussm{d}$ elements $z_1, \ldots, z_{\gaussm{d}}$ of $Z$ meet $g$ in dimension $d-1$.\label{lem_nmb_dm1z_a}
   \item We have $\{ z_i ~|~ i \in \{ 1, \ldots, \gaussm{d}\} \} = \{ \langle \pi_i, \pi_i^\perp \cap h \rangle ~|~ i \in \{ 1, \ldots, \gaussm{d}\} \}$.
   \item We have that $z_i \cap z_j$ is a $(d-2)$-dimensional subspace of $g$.
  \end{enumerate}
\end{lemma}
\begin{proof}
  By Proposition \ref{prop_q0_inmbs} (d) and $Y \neq Z$, exactly $\gaussm{d}$ elements of $Z$ meet $g$ in dimension $d-1$. This shows \eqref{lem_nmb_dm1z_a}.
  By Proposition \ref{prop_q0_inmbs} (b), $\dim(z_i \cap z_j) < d-1$ for $i \neq j$. 
  Hence, each hyperplane $\pi_i$ of $g$ lies in exactly one element $z_i$, and $z_i$ satisfies $z_i \subseteq \pi_i^\perp$.
  Since $(Y, Z)$ is a cross-intersecting EKR set, all $z_i$ meet $h$ in at least a point.
  Since $\pi_i \subseteq g$, $\dim(\pi_i) = d-1$, and $g \cap h = \emptyset$, we see that $\pi_i^\perp \cap h$ is a point.
  Hence,
  \begin{align*}
   \{ z_i ~|~ i \in \{ 1, \ldots, \gaussm{d}\} \} = \{ \langle \pi_i, \pi_i^\perp \cap h \rangle ~|~ i \in \{ 1, \ldots, \gaussm{d}\} \}.
  \end{align*}
  This shows (b).
  Lemma \ref{lem_intersection_lemma} (with $g_1 = g, g_2 = z_i, g_3 = z_j$)
  shows (c).
\end{proof}

\subsubsection{The Parabolic Quadric \texorpdfstring{$\q(2d, q)$}{Q(2d, q)}, \texorpdfstring{$d$}{d} even}

Let $s$ be a subspace of $\PG(2d, q)$. We write $Y \subseteq s$ if all elements of 
$Y$ are subspaces of $s$, $Y \cap s$ for all elements of $Y$ in $s$, and $Y \setminus s$ for all elements of $Y$ not in $s$.
\begin{lemma}\label{lem_q0inhyp_pre2}
  Let $g$ and $h$ be disjoint generators of $\q(2d, q)$.
  Then $\scrQ := \langle g, h \rangle \cap \q(2d, q)$ is isomorphic to $\q^+(2d-1, q)$.
\end{lemma}
\begin{proof}
  The generators $g$ and $h$ are disjoint, hence $\scrQ$ is not degenerate.
  The hyperplane $\langle g, h \rangle$ obviously contains generators, hence $\scrQ$ does not have type $\q^-(2(d-1)+1, q)$.
  Therefore, the intersection $\langle g, h \rangle \cap \q(2d, q)$ is isomorphic to $\q^+(2d-1, q)$.
\end{proof}

\begin{lemma}\label{lem_q0inhyp_pre1}
  Let $(Y, Z)$ be a cross-intersecting EKR set of $\q(2d, q)$, $d$ even, of maximum size such that $Y \cap Z = \emptyset$. 
    Let $g \in Y$. Let $\tilde{Y}$ be the set of generators of $Y$ disjoint to $g$.
    \begin{enumerate}[(a)]
     \item There exists a hyperplane $\eta$ of type $\q^+(2d-1, q)$ such that $g, \tilde{Y} \subseteq \eta$.
     \item Let $z \in Z$. If $z$ meets an element of $\{ g \} \cup \tilde{Y}$ in a subspace of dimension $d-1$,
     then $z \in \eta$ and all elements of $Z$ disjoint to $z$ are in $\eta$.
%      \item If $\tilde{G} \in Y \setminus h$, then $\tilde{G}$ meets all elements of $\tilde{Y}$ non-trivially.
     \item If $\tilde{g} \in Y$ and $\dim(\tilde{g} \cap h) = d-2$ for an $h \in \tilde{Y}$, 
     then $\tilde{g}$ and the $q^{\binom{d}{2}}$ generators disjoint to $\tilde{g}$ are in $\eta$.
    \end{enumerate}
\end{lemma}
\begin{proof}
  By Proposition \ref{prop_q0_inmbs} (c), a generator $g \in Y$ is disjoint to $q^{\binom{d}{2}}$ generators of $Y$. 
  Let $h \in \tilde{Y}$.
  By Lemma \ref{lem_q0inhyp_pre2}, $\eta := \langle g, h \rangle$ has type $\q^+(2d-1, q)$. 
  We shall show $\tilde{Y} \subseteq \eta$.
  
  Suppose to the contrary that there exists a generator $\tilde{h} \in \tilde{Y}$ not in $\eta$.
  In the notation of Lemma \ref{lem_nmb_dm1z} we write
  \begin{align*}
    Z_a & = \{ z_i ~|~ i \in \{1, \ldots, \gaussm{d}\}\}\\
    &= \{ z \in Z ~|~ \dim(z \cap a) = d-1 \} \subseteq Z
  \end{align*}
  for $a \in Y$.

  Define the set of points
  \begin{align*}
   \scrP := \{ \tilde{h} \cap z ~|~ z \in Z_g \}.
  \end{align*}
  By Lemma \ref{lem_nmb_dm1z} (c), $|\scrP| = |Z_g| = \gaussm{d}$.
  Furthermore, $Z_g \subseteq \eta$ as the elements of $Z_g$ meet $h$ non-trivially, 
  so $\scrP \subseteq \tilde{h} \cap \eta$.
  Hence,
  \begin{align*}
    \gaussm{d} = |\scrP| \leq |\tilde{h} \cap \eta| = \gaussm{d-1}.
  \end{align*}
  This is a contradiction.
  Thus, $\tilde{Y} \subseteq \eta$.
  This proves (a).
  
  To prove (b), suppose without loss of generality that $z$ meets $g$ in a subspace of dimension $d-1$.
  Then $z \subseteq \eta$ as $z \in Z_{g}$ and $Z_g \subseteq \eta$.
  Let $\pi$ be a $(d-1)$-dimensional space of $h$ disjoint to $z$. Define $\tilde{z}$
  as $\langle \pi, \pi^\perp \cap g \rangle$. Then $\tilde{z} \in Z_{h}$ by Lemma \ref{lem_nmb_dm1z} (b)
  and $\tilde{z}$ is disjoint to $z$ by Proposition \ref{prop_q0_inmbs} (b). 
  Then $\tilde{z} \subseteq \eta$.
  Hence, $\langle z, \tilde{z} \rangle = \eta$. By (a), all elements of $Z$ disjoint to 
  $z$ are in $\eta$. This shows (b).
  
  Let $\tilde{g} \in Y$ with $\dim(h \cap \tilde{g})=d-2$.
  By Lemma \ref{lem_nmb_dm1z} (c), there exists a $z \in Z_{h}$ with $h \cap \tilde{g} \subseteq z$. By (b), $\tilde{g} \subseteq \eta$.
  Let $\tilde{h} \in Y$ be disjoint to $\tilde{g}$. As $(Y, Z)$ is a cross-intersecting EKR set,
  $z$ meets $\tilde{h}$ in a point. Let $\bar{z} \in Z_{\tilde{h}}$
  with $z$ disjoint to $\bar{z}$ ($\bar{z}$ exists for the same reasons as $\tilde{z}$ exists).
  By (b) and $\dim(h \cap z)=d-1$, $\langle z, \bar{z} \rangle = \eta$.
  Hence, $\tilde{h} \subseteq \eta$. By (a), all elements of $Y$ disjoint to $\tilde{g}$
  are in $\eta$. This shows (c).
\end{proof}

\begin{cor}\label{cor_disj_and_dm2_trans}
  Let $(Y, Z)$ be a cross-intersecting EKR set of $\q(2d, q)$, $d$ even, of maximum size such that $Y \cap Z = \emptyset$. 
  Let $g_1, g_2, g_3 \in Y$. Then
  \begin{align*}
    \dim(g_1 \cap g_2) \in \{ 0, d-2 \} && \text{ and } && \dim(g_2 \cap g_3) = 0
  \end{align*}
  implies that $g_1 \subseteq \langle g_2, g_3 \rangle$.
\end{cor}
\begin{proof}
  As $g_2$ and $g_3$ are disjoint, $\langle g_2, g_3 \rangle$ is a hyperplane $\eta$.
  By Lemma \ref{lem_q0inhyp_pre1} (a), $g_1 \subseteq \langle g_2, g_3 \rangle$ if $\dim(g_1 \cap g_2) = 0$.
  By Lemma \ref{lem_q0inhyp_pre1} (c), $g_1 \subseteq \langle g_2, g_3 \rangle$ if $\dim(g_1 \cap g_2) = d-2$.
\end{proof}

\begin{cor}\label{cor_two_dm2_gens}
  Let $(Y, Z)$ be a cross-intersecting EKR set of $\q(2d, q)$, $d$ even, of maximum size such that $Y \cap Z = \emptyset$.
  Suppose that there are $g, h, \tilde{g} \in Y$ with 
  \begin{align*}
    \dim(g \cap h) = d-2 \text{ and } \dim(g \cap \tilde{g}) > 0.
  \end{align*}
  Let $\eta$ be the hyperplane, which contains $g$ and all elements of $Y$ disjoint to $g$ (see Proposition \ref{lem_q0inhyp_pre1} (a)).
  If $\tilde{g}$ and $h$ are disjoint, then $\tilde{g} \in \eta$ or all generators $\tilde{h} \in Y$
  with $\dim(\tilde{g} \cap \tilde{h}) = d-2$ meet $g$ non-trivially.
\end{cor}
\begin{proof}
  Suppose that $\tilde{g}$ and $h$ are disjoint. By Proposition \ref{prop_q0_inmbs} (b),
  we have that $\dim(g \cap \tilde{g})$ is even.
  By $\dim(g \cap h) = d-2$ and $\dim(g \cap \tilde{g}) > 0$, we have that
  $\dim(g \cap \tilde{g}) = 2$. Let $\tilde{h} \in Y$ with $\dim(\tilde{g} \cap \tilde{h}) = d-2$.
  If $g$ and $\tilde{h}$ are disjoint, then $\tilde{g} \subseteq \eta = \langle g, \tilde{h} \rangle$.
  Hence, if $\tilde{g} \nsubseteq \eta$, then $\dim(g \cap \tilde{h}) > 0$.
\end{proof}

We need the following bound.

\begin{lemma}\label{lem_bound_gens}
  Let $q \geq 2$. Let $d \geq 1$. Then
  \begin{align*}
    \prod_{i=1}^{d-1} (q^i+1) \leq \frac{2q^d}{q^d+1} \left( q^{\binom{d}{2}} - q^{\binom{d-1}{2}} + 1\right) + q^{\binom{d-2}{2} + 2(d-2)}.
  \end{align*}
\end{lemma}
\begin{proof}
  We will prove the assertion by induction over $d$.
  It can be easily checked that the assertion is true for $d \leq 4$.
  If the assertion is true for $d \geq 4$, then
  \begin{align*}
    \prod_{i=1}^{d} (q^i+1) &\leq (q^d+1) \left(\frac{2q^d}{q^d+1} \left( q^{\binom{d}{2}} - q^{\binom{d-1}{2}} + 1\right) + q^{\binom{d-2}{2} + 2(d-2)} \right) \\
    &\stackrel{(*)}{\leq} \frac{2q^{d+1}}{q^{d+1}+1} \left(q^{\binom{d+1}{2}} - q^{\binom{d}{2}} + 1\right) + q^{\binom{d-1}{2} + 2(d-1)}.
  \end{align*}
  The difference between the right side of $(*)$ and the left side of $(*)$ equals
  \begin{align*}
   \frac{q^{d}}{q^{d+1}+1}  \left( 2q^{\binom{d}{2}+2} - 2q^{\binom{d}{2}+1} - 3 q^{\binom{d}{2}} + 2 q^{\binom{d-1}{2}} - q^{\binom{d-1}{2}-2} - 2q^{d+1} + 2q-2 \right)
  \end{align*}
  which is a positive expression for $q \geq 2$.
\end{proof}

\begin{prop}\label{lem_q0inhyp}
  Let $(Y, Z)$ be a cross-intersecting EKR set of $\q(2d, q)$ of maximum size such that $Y \cap Z = \emptyset$. Then there exists a hyperplane $\eta$ such that $Y, Z \subseteq \eta$.
\end{prop}
\begin{proof}
  Let $g \in Y$.
  In the view of Lemma \ref{lem_q0inhyp_pre1}, we find a hyperplane $\eta$ that contains $g$, 
  the set $Y_1$ of $q^{\binom{d}{2}}$ generators of $Y$ disjoint to $g$, and, by Proposition \ref{prop_q0_inmbs} and 
  Lemma \ref{lem_q0inhyp_pre1} (c), the set $Y_2$ of $\gauss{d}{2} q^{\binom{d-2}{2}}$ generators of $Y$ which meet $g$ in dimension $d-2$.
  % Make Y_2 to \dim(h \cap \tilde{g}) >= 2 or \dim(h \cap g) >= 2.
  
  Suppose contrary to the assertion that there exists an element $\tilde{g} \in Y$ that is not in $\eta$.
  Then $\tilde{g}$ and the set $Y_3$ of $q^{\binom{d}{2}}$ generators of $Y$ disjoint to $\tilde{g}$ lie in a second hyperplane $\eta' \neq \eta$ by Lemma \ref{lem_q0inhyp_pre1}.
  Let $Y_4$ be the set of $\gauss{d}{2} q^{\binom{d-2}{2}}$ generators of $Y$ which meet $\tilde{g}$ in dimension $d-2$.
  
  First we show $|(Y_2 \cup Y_4) \setminus (Y_1 \cup Y_3)| \geq \gauss{d}{2} q^{\binom{d-2}{2}}$.
  If $Y_2 \cap Y_3$ is empty, then the claim is clear.
  Suppose that there exists a generator $h \in Y_2 \cap Y_3$.
  By Corollary \ref{cor_two_dm2_gens}, then $Y_4$ is disjoint to $Y_1$.
  The claim follows.
  
  Now we want to show that $Y_1 \cap Y_3$ is empty.
  Suppose that there exists a generator $h \in Y_1 \cap Y_3$.
  By Corollary \ref{cor_disj_and_dm2_trans} with $g_1  = \tilde{g}$, $g_2 = h$, and $g_3 = g$,
  we obtain the contradiction $\tilde{g} \subseteq \langle g, h \rangle = \eta$.
  Hence, 
  \begin{align}
    |Y| &\geq |Y_1| + |Y_3| + |(Y_2 \cup Y_4) \setminus (Y_1 \cup Y_3)| + 2 \notag \\
    & = 2\left(q^{\binom{d}{2}} + 1\right)  + \gauss{d}{2} q^{\binom{d-2}{2}} \geq 2\left(q^{\binom{d}{2}} + 1\right)  + q^{\binom{d-2}{2} + 2(d-2)} .\label{eq_Y_bnd_1}
  \end{align}
  According to Theorem \ref{thm_es},
  \begin{align*}
    |Y| = \prod_{i=1}^{d-1} (q^i+1).
  \end{align*}
  This contradicts \eqref{eq_Y_bnd_1} by Lemma \ref{lem_bound_gens}.
\end{proof}

\begin{theorem}\label{satz_ci_ekr_q0}
  Let $(Y, Z)$ be a cross-intersecting EKR set of $\q(2d, q)$, or $\w(2d-1, q)$, $q$ even, of maximum size such that $Y \neq Z$.
  Then $d$ even and $(Y, Z)$ is a cross-intersecting EKR set of generators of a subgeometry $\scrQ$ isomorphic to $\q^+(2d-1, q)$
  (i.e. $Y$ are latins and $Z$ are greeks).
\end{theorem}
\begin{proof}
  First consider $\q(2d, q)$.  By Corollary \ref{cor_Y_Z_diff_imp_Y_Z_disj}, then $Y \cap Z = \emptyset$.
  By Proposition \ref{lem_q0inhyp}, $Y, Z \subseteq \eta$ for some hyperplane $\eta$ isomorphic to $\q^+(2d-1, q)$ if not $Y=Z$.
  Hence, $(Y, Z)$ is a cross-intersecting set of $\q^+(2d-1, q)$ of maximum size.
  These sets were classified in Theorem \ref{thm_ci_ekr_qp}.
  
  The part of the assertion for $\w(2d-1, q)$, $q$ even, follows, since $\q(2d, q)$ and $\w(2d-1, q)$ are isomorphic for $q$ even.
\end{proof}

\subsubsection{The Symplectic Polar Space \texorpdfstring{$\w(2d-1, q)$}{W(2d-1, q)}, \texorpdfstring{$d$}{d} even, \texorpdfstring{$q$}{q} odd}

Similar to \cite{MR2755082} we use the following property of $\w(2d-1, q)$, $d$ even (see \cite[1.3.6, 3.2.1, 3.3.1]{Payne2009}):

\begin{theorem}\label{thm_wq_odd_3lines}
  Let $\ell_1, \ell_2, \ell_3$ be three pairwise disjoint lines of $\w(3, q)$, $q$ odd. 
  Then the number of lines meeting $\ell_1, \ell_2, \ell_3$ is $0$ or $2$.
\end{theorem}

\begin{theorem}
  Let $(Y, Z)$ be a cross-intersecting EKR set of maximum size of $\w(2d-1, q)$, $d$ even, $q$ odd. Then $Y=Z$.
\end{theorem}
\begin{proof}
  Suppose to the contrary that $Y \neq Z$. By Corollary \ref{cor_Y_Z_diff_imp_Y_Z_disj}, then $Y \cap Z = \emptyset$.
  By Proposition \ref{prop_q0_inmbs} (c), we can find two disjoint generators $g$ and $h$ in $Y$.
  Again by Proposition \ref{prop_q0_inmbs} (c), there are exactly $q \gaussm{d} \gaussm{d-1}/(q+1)$ generators $Y' \subseteq Y$ which meet $g$ in a subspace of dimension $d-2$.
  The generator $g$ has $\gaussm{d} \gaussm{d-1}/(q+1)$ subspaces of dimension $d-2$.
  Hence, we find a subspace $\ell \subseteq g$ of dimension $d-2$ such that $\ell$ is contained in $q$ elements of $Y'$.
  Since $q$ is odd, there are at least three elements $y_1, y_2, y_3$ of $Y$ through $\ell$.
  
  Consider the quotient geometry $\scrW_3$ of $\ell$ isomorphic to $\w(3, q)$ and the projection of the elements of $Y$ and $Z$ onto $\scrW_3$ from $\ell$.
  Since elements of $Y$ do not meet each other in dimension $d-1$ by Proposition \ref{prop_q0_inmbs}, $y_1, y_2, y_3$ are three disjoint lines in $\scrW_3$ after projection.
  The subspace $\ell^\perp \cap h$ has dimension $2$, so we find a subspace $\tilde{\ell} \subseteq h$ of dimension $d-2$ disjoint to $\ell^\perp$.
  Let $\pi_1, \pi_2, \pi_3$ be subspaces of dimension $d-1$ in $h$ with $\tilde{\ell} \subseteq \pi_1, \pi_2, \pi_3$.
  By Lemma \ref{lem_nmb_dm1z} (b), we find $z_1, z_2, z_3 \in Z$ with $\pi_i \subseteq z_i$ for $i \in \{1, 2, 3\}$.
  By Lemma \ref{lem_nmb_dm1z} (c), the pairwise meets of $z_1, z_2, z_3$ are contained in $\tilde{\ell}$.
  As we have that $\tilde{\ell}$ meets $\ell^\perp$ trivially, we have that $z_1$, $z_2$, $z_3$ are projected onto three disjoint lines on $\scrW_3$.
  These three lines have to meet the projections of $y_1$, $y_2$, and $y_3$, since $(Y, Z)$ is a cross-intersecting EKR set.
  This contradicts Theorem \ref{thm_wq_odd_3lines}.
\end{proof}

\section{The Hermitian Polar Space \texorpdfstring{$\h(2d-1, q^2)$}{H(2d-1, q2)}}\label{sec_hermitian}

It is well-known (see for example \cite{MR2312327}) that the linear programming 
bound given in \cite{fi_km_ekr_in_hermitians} can be reformulated as a weighted 
Hoffman bound. Hence, Lemma \ref{lem_efp} is applicable if $d > 1$. 
The original bound on EKR sets on $\h(2d-1, q^2)$ is as follows.

\begin{theorem}[\cite{fi_km_ekr_in_hermitians}]\label{lem_h2dp1_normal}
  Let $Y$ be an EKR set of $\h(2d-1, q^2)$ with $d > 2$ odd. Then 
  \begin{align*}
    |Y| \le \frac{nq^{d-1} - f_1(q^{d-1}-1)\left(1 - c\right)}{q^{2d-1}+q^{d-1}+f_1(q^{d-1}-1)c}\approx q^{d^2-2d+2},
  \end{align*}
  where $n=\prod_{i=0}^{d-1} (q^{2i+1}+1)$, $f_1=q^2 \gaussm{d}_{q^2} \frac{q^{2d-3}+1}{q+1}$ and $c=\frac{q^2-q-1+q^{-2d+3}}{q^{2d}-1}$.
\end{theorem}

The result by Luz \cite{MR2312327} which shows
that the linear programming bound is a special case of the weighted Hoffman bound\footnote{This seems to be part of the mathematical folklore for a long time, but the author is not aware of any source older than \cite{MR2312327}.}
also holds for cross-intersecting EKR sets, but we feel that we should show this 
directly, since the transition from the linear programming technique used in 
\cite{fi_km_ekr_in_hermitians} to the weighted Hoffman bound is not obvious.
We shall prove a cross-intersecting result similar to \cite{fi_km_ekr_in_hermitians}
in the following.

\begin{theorem}\label{lem_h2dp1}
  Let $(Y, Z)$ be a cross-intersecting EKR set of $\h(2d-1, q^2)$ with $d > 2$. Then 
  \begin{align*}
    \sqrt{|Y|\cdot |Z|} \le \frac{n\lambda_b}{\lambda_b-k} \approx q^{d^2-2d+2},
  \end{align*}
  where $n=\prod_{i=0}^{d-1} (q^{2i+1}+1)$, $\lambda_b = -q^{(d-1)^2} - \alpha \left( 1 -  f_1 \frac{1 - c}{n}\right)$,
  $k = q^{d^2} + \alpha f_1 \left( c + \frac{1-c}{n} \right)$,
  $f_1=q^2 \gaussm{d}_{q^2} \frac{q^{2d-3}+1}{q+1}$, $c=\frac{q^2-q-1+q^{-2d+3}}{q^{2d}-1}$,
  and
\begin{align*}
  \alpha = \begin{cases}
            q^{d(d-1)} - q^{(d-1)^2} & \text{ if $d$ odd,}\\
            \frac{q^{d(d-1)} - q^{(d-1)^2}}{1 - 2(1-c)f_1/n} & \text{ if $d$ even.}
           \end{cases}
\end{align*}
\end{theorem}

\begin{proof}[Proof of Theorem \ref{lem_h2dp1}]
Let $d > 1$.
Let $A_d$ be the disjointness matrix as defined in Section \ref{sec_assoc}. Consider
the matrix $A$ defined by
\begin{align*}
  A = A_d - \alpha E_1 + \frac{\alpha f_1 c}{n} \vJ + \alpha f_1 \frac{1 - c}{n} \vI
\end{align*}

\textbf{Claim 1.}
Our first claim is that $\alpha$ is positive. For $d$ odd, this is obvious.
For $d$ even it is sufficient to show that $1 - 2f_1/n$ is positive as we have $c \in [0, 1]$
and $f_1 > 0$. By $d \geq 4$,
\begin{align*}
  1 &< \frac{(q+1)(q^2-1)}{2q^2}\\
  &\leq \frac{\left( \prod_{i=0}^{d-3} \left( q^{2i+1} + 1 \right) \right) (q^{2d-1}+1)}{2q^2 \gaussm{d}_{q^2}}\\
  &\leq \frac{\left( \prod_{i=0}^{d-1} \left( q^{2i+1} + 1 \right) \right) (q+1)}{2q^2 \gaussm{d}_{q^2} \left(q^{2d-3}+1\right)}
  = \frac{n}{2f_1}.
\end{align*}

\textbf{Claim 2.}
Our second claim is that $A$ is an extended weight adjacency matrix. By Section \ref{sec_assoc},
it is clear that the entry $(x, y)$ of $E_1$ equals $Q_{i,1}/n$ if $x$ and $y$ meet
in codimension $i$. It was shown in \cite[Equations (6)--(11)]{fi_km_ekr_in_hermitians} 
that the following\footnote{
  Note the following about \cite{fi_km_ekr_in_hermitians}:
\begin{enumerate}
 \item The equations (6)--(11) in \cite{fi_km_ekr_in_hermitians} do not depend on $d$ odd.
 \item The relations in \cite{fi_km_ekr_in_hermitians} are numbered $-1, 0, \ldots, d$ for a Hermitian polar space of rank $d+1$. This implies that
  one has to replace an index $i$ by $i+1$ (in particular, $d$ by $d+1$) if one goes from the notation of \cite{fi_km_ekr_in_hermitians} to the notation of this article.
  \item We use $\epsilon$ in \cite{fi_km_ekr_in_hermitians} instead of $e$ as in this article, so instead of type $\epsilon = -1/2$ we have type $e=1/2$.
\end{enumerate}} holds:
\begin{multicols}{2}
\begin{enumerate}[(a)]
 \item $Q_{0, 1} = f_1$, 
 \item $Q_{d-1, 1} = f_1 c$,
 \item $Q_{s, 1} \geq f_1 c$ if $s < d$,
 \item $Q_{d, 1} < 0$.
\end{enumerate}
\end{multicols}
Hence, the entry $(x, y)$ of the matrix $A$ is $0$ if $x = y$, it is less than or equal to
zero if $1 \leq \codim(x \cap y) \leq d-1$, and it is larger than $1$ if $x$ and $y$ are disjoint.
This shows that $A$ is an extended weight adjacency matrix of the disjointness graph of generators.

\textbf{Claim 3.}
Our third claim is that the second largest absolute
eigenvalue of $A$ is
\begin{align*}
  q^{(d-1)^2} + \alpha \left( 1 -  f_1 \frac{1 - c}{n}\right),
  \intertext{ and that }
  k = q^{d^2} + \alpha f_1 \left( c + \frac{1-c}{n} \right).
\end{align*}
Recall that $A$ is a linear combination of the $A_i$s of the scheme,
so the common eigenspaces of the $A_i$s are the eigenspaces of $A$.
Notice for the following that $c \in [0, 1]$ and $\alpha, f_1 > 0$.
By \eqref{eq_ev_disj}, the eigenvalues of $A$ are 
\begin{align*}
  &q^{d^2} + \alpha f_1 \left( c + \frac{1-c}{n} \right) \text{ for } \langle \vj \rangle,\\
  &-q^{(d-1)^2} - \alpha \left( 1 -  f_1 \frac{1 - c}{n}\right) \text{ for } V_1,\\
  &(-1)^r q^{(d-r)^2 + r(r-1)} + \alpha f_1 \frac{1 - c}{n}\text{ for } V_r \text{ with } 1 < r < d,\\
  &(-1)^d q^{d(d-1)} + \alpha f_1 \frac{1 - c}{n} \text{ for } V_d.
\end{align*}
If $d$ even, then we have that 
\begin{align*}
 q^{(d-1)^2} + \alpha \left( 1 -  f_1 \frac{1 - c}{n}\right) = q^{d(d-1)} + \alpha f_1 \frac{1 - c}{n}
\end{align*}
is the second largest absolute eigenvalue.
If $d$ odd, then 
\begin{align*}
  q^{(d-1)^2} + \alpha \left( 1 -  f_1 \frac{1 - c}{n}\right) = q^{d(d-1)} - \alpha f_1 \frac{1 - c}{n}
\end{align*}
is the second largest absolute eigenvalue.
This proves our claim. 

Now we can apply Lemma \ref{lem_efp} with these values. Note that $k$ has approximately
size $q^{d^2+d-2}$, the second largest absolute eigenvalue $\lambda_b$ has approximately size
$q^{d(d-1)}$, and $n$ has approximately size $q^{d^2}$. Therefore,
\begin{align*}
  \frac{n\lambda_b}{\lambda_b+k}
\end{align*}
has approximately size $q^{d^2-2d+2}$. 
\end{proof}

Note that the normal adjacency matrix of the disjointness graph of generators of $\h(2d-1, q^2)$ only
yields $q^{d^2-d}$ as an upper bound, so this improves the bound significantly.

For the sake of completeness we want to mention the cross-intersecting EKR sets for $d=2$ as Lemma \ref{lem_efp} only considered the case $d > 2$.
We will do this after providing a general geometrical result on (maximal) cross-intersecting EKR sets, where we call a cross-intersecting EKR set $(Y, Z)$ \emph{maximal} if there exists no generator $x$ such that $(Y \cup \{ x \}, Z)$ or $(Y, Z \cup \{ x \})$ is a cross-intersecting EKR set.

\begin{lemma}\label{lem_zy_intersect}
  Let $(Y, Z)$ be a maximal cross-intersecting EKR set in a finite classical polar space of rank $d$. 
  If two distinct elements $y_1, y_2 \in Y$ meet in a subspace of dimension $d-1$, then all elements of $Z$ meet this subspace in at least a point.
\end{lemma}
\begin{proof}
 Assume that there exists a generator $z$ which meets $y_1$ and $y_2$ in points $p, r$ not in $y_1 \cap y_2$. 
   Then $\langle p, r, y_1 \cap y_2 \rangle$ is a totally isotropic subspace of dimension $d+1$. Contradiction.
\end{proof}

\begin{theorem}\label{thm_h3}
  Let $(Y, Z)$ be a maximal cross-intersecting EKR set of $\h(3, q^2)$ with $|Y| \geq |Z|$. Then one of the following cases occurs:
  \begin{enumerate}[(a)]
   \item The set $Y$ is the set of all lines of $\h(3, q^2)$, and $Z = \emptyset$. Here $|Y| \cdot |Z| = 0$.\label{thm_h3_a}
   \item The set $Y$ is the set of all lines meeting a fixed line $\ell$ in at least a point, and $Z = \{ \ell \}$. Here $|Y| \cdot |Z| = (q^2+1)q + 1$.\label{thm_h3_b}
   \item The set $Y$ is the set of all lines on a fixed point $p$, and $Y = Z$. Here $|Y| \cdot |Z| = (q+1)^2$.\label{thm_h3_c}
   \item The set $Y$ is the set of lines meeting two disjoint lines $\ell_1, \ell_2$, and $Z = \{ \ell_1, \ell_2\}$. Here $|Y| \cdot |Z| = 2(q^2+1)$.\label{thm_h3_d}
   \item The set $Y$ is the set of lines meeting three disjoint lines $\ell_1, \ell_2, \ell_3$, and $Z$ is the set of all $q+1$ lines meeting the lines of $Y$. Here $|Y| \cdot |Z| = (q+1)^2$.\label{thm_h3_e}
  \end{enumerate}
\end{theorem}
\begin{proof}
  Assume that \eqref{thm_h3_a} does not occur. 
  
  By Lemma \ref{lem_zy_intersect}, as soon as two elements of $Y$ meet in a point $p$, then all elements of $Z$ contain $p$. 
  Hence, \eqref{thm_h3_b} occurs or at least $2$ elements of $Z$ meet in $p$. Hence, all elements of $Y$ contain $p$ by Lemma \ref{lem_zy_intersect}.
  This is case \eqref{thm_h3_c}.
  
  So assume that $Y$ and vice-versa $Z$ only consist of disjoint lines. 
  If there are two lines $\ell_1, \ell_2 \subseteq Z$, then there are $q^2+1$ (disjoint) lines $L$ meeting $\ell_1$ and $\ell_2$ in a point (hence $|Y| \leq q^2+1$). 
  If more than $q+1$ of these lines meet $\ell_1$ (hence $|Y| > q+1$), then $Z$ contains at most two lines, since in $\h(3, q^2)$ exactly $q+1$ lines meet $3$ pairwise disjoint lines in a point.
  This yields \eqref{thm_h3_d}.
  If $|Z| \geq 3$, then $|Y| \leq q+1$ by the previous argument.
  We may assume $|Y| \geq 3$.
  Then it is well-known that there are exactly $q+1$ lines meeting the $q+1$ lines of $Y$. 
  Hence, we can add these lines and then $Z$ is maximal.
  This yields \eqref{thm_h3_e}.
\end{proof}

The author tried to prove that the maximum cross-intersecting EKR set of $\h(5, q^2)$ is 
the unique EKR of maximum size given in \cite{MR2755082}, but aborted this attempt after he got lost in too many case distinctions.
The EKR set $Y$ of all generators meeting a fixed generator in at least a line corresponds 
to the largest cross-intersecting EKR set $(Y, Y)$ known to the author and has size $q^5+q^3+q+1$.
The largest example known to the author for $\h(7, q^2)$ is the following.

\begin{Example}\label{ex_largst_h7}
  Let $g$ be a generator of $\h(7, q^2)$.
  Let $Y$ be the set of all generators that meet $g$ in at least a $2$-dimensional subspace.
  Let $Z$ be the set of all generators that meet $g$ in at least a $3$-dimensional subspace.
  Then $(Y, Z)$ is a cross-intersecting EKR set.
\end{Example}
\begin{proof}
 A generator of $\h(7, q^2)$ is a $4$-dimensional subspace.
 A plane and a line of a $4$-dimensional subspace meet pairwise in at least a point.
 Hence, $(Y, Z)$ is a cross-intersecting EKR set.
\end{proof}

In this example $Y$ has
\begin{align*}
  &1 + q + q^3 + q^4 + q^5 + q^6 + q^7 + 2q^8 + q^{10} + q^{12}
\intertext{elements, $Z$ has}
  &1 + q + q^3 + q^5 + q^7
\end{align*}
elements, so in total the cross-intersecting EKR set has size
\begin{align*}
  \sqrt{|Y|\cdot |Z|} \approx q^{19/2}.
\end{align*}
The bound given in Theorem \ref{lem_h2dp1} for this case is approximately $q^{10}$.
For $\h(2d-1, q^2)$, $d > 4$, the largest example known to the author is the EKR set 
of all generators on a fixed point. The author assumes that the largest known examples
are also the largest examples.

\section{Summary}

We summarize our results in the following table. 
We only list the cases, where cross-intersecting EKR sets of maximum size are not necessarily EKR sets.
The table includes the size of the largest known example if it is not known if 
the best known bound does not seem to be tight.

\begin{footnotesize}
\begin{center}
\begin{tabular}{p{3.2cm}lp{4.5cm}l}\toprule
\raggedright Polar Space & Maximum Size $\sqrt{|Y|\cdot |Z|}$ & \raggedright Largest (known) Examples & Reference \\\toprule \raggedright
$\q^+(2d-1, q)$, $d$ even & $n/2$ & \raggedright $Y$ latins, $Z$ greeks & Th. \ref{thm_ci_ekr_qp}\\ \raggedright
$\q(2d, q)$, $d$ even & $(q+1) \cdot \ldots \cdot (q^{d-1}+1)$ & \raggedright $Y$ latins and $Z$ greeks of a $\q^+(2d+1, q)$, or $Y=Z$ EKR set & Th. \ref{satz_ci_ekr_q0} \\ \raggedright
$\w(2d-1, q)$, $d$ even, $q$ even & \raggedright $(q+1) \cdot \ldots \cdot (q^{d-1}+1)$ & see $\q(2d, q)$ & Th. \ref{satz_ci_ekr_q0} \\ \raggedright
$\h(3, q^2)$ & $q^3+q+1$ & \raggedright $Z = \{\ell\}$, $Y$ all lines meeting $\ell$ & Th. \ref{thm_h3} \\ \raggedright
$\h(5, q^2)$ & $\lessapprox q^{5}$ & \raggedright largest EKR set, size $\approx q^{5}$ & Th. \ref{lem_h2dp1} \\ \raggedright
$\h(7, q^2)$ & $\lessapprox q^{10}$ & \raggedright Example \ref{ex_largst_h7}, size $\approx q^{19/2}$ & Th. \ref{lem_h2dp1} \\ \raggedright
$\h(2d-1, q^2)$, $d > 4$ & $\lessapprox q^{(d-1)^2+1}$ & \raggedright all generators on a point, size $\approx q^{(d-1)^2}$ & Th. \ref{lem_h2dp1}\\
\bottomrule
\end{tabular}
\end{center}
\end{footnotesize}

\section{Open Problems}

As mentioned before, the case $\h(2d-1, q^2)$ is mostly open. The next logical step would be
to solve the problem for $\h(5, q^2)$ and $\h(7, q^2)$. It might be also interesting to study 
the same problem for proper subspaces of generators. 
For classical EKR sets this was done recently by Klaus Metsch \cite{Metsch2014}
using mostly geometrical arguments.

\section*{Acknowledgments}

I would like to thank one anonymous referee, Linda Beukemann and Klaus Metsch for their helpful
remarks, which improved the presentation of our results very much.
% In particular the comments of the referee were very, very detailed and I am very grateful for them.

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\appendix

\end{document}