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%\Setcounter{section}{-1}

\begin{document}

\title{On the Chow groups of Quadratic Grassmannians}
\author{A.Vishik}
\maketitle

%\begin{center}
%{\small (preliminary version)}\\
%\end{center}
%
%\vspace{0.5cm}

\tableofcontents


\section{Introduction}

The current article is devoted to the computation of certain invariants
of smooth projective quadrics. Among the invariants of quadrics one can
distinguish those which could be called {\it discrete}. These are invariants
whose values are (roughly speaking) collections of integers. For a quadric
of given dimension such an invariant takes only finitely many values.
The first example is the usual dimension of anisotropic part of $q$.  
More sophisticated example is given by the {\it splitting pattern} of $Q$,
or the collection of {\it higher Witt indices} - see \cite{HR} and \cite{Kn1}.
The question of describing the set of possible values of this invariant 
is still open. Some progress in this direction was achieved by considering
the interplay of the splitting pattern invariant with another discrete
invariant, called, {\it motivic decomposition type} - see  
\cite{V}. 
The latter invariant measures in what pieces the Chow-motive of a quadric 
$Q$ could be decomposed. The splitting pattern invariant can be interpreted 
in terms of the existence of certain cycles on various flag varieties 
associated to $Q$, and the motivic decomposition type can be interpreted
in terms of the existence of certain cycles on $Q\times Q$. So, both
these invariants are faces of the following invariant $GDI(Q)$, which we
will call {\it (quite) generic discrete invariant}.   
Let $Q$ be a quadric of dimension $d$, and, for any $1\leq m\leq [d/2]+1$,
let $G(m,Q)$ be the grassmanian of projective subspaces of dimension
$(m-1)$ on $Q$. Then $GDI(Q)$ is the collection of the subalgebras
$$
C^*(G(m,Q)):=image(\op{CH}^*(G(m,Q))/2\row\op{CH}^*(G(m,Q)|_{\kbar})/2).
$$ 
It should be noticed that this invariant has a ''noncompact form'',
where one uses powers of quadrics $Q^{\times r}$ instead of $G(m,Q)$.
The equivalence of both forms follows from the fact that the Chow-motive of
$Q^{\times r}$ can be decomposed into the direct sum of the Tate-shifts of
the Chow-motives of $G(m,Q)$. The varieties $G(m,Q)|_{\kbar}$ have
natural cellular structure, so Chow-ring for them is a
finite-dimensional $\zzz$-algebra with the fixed basis 
parametrized by the Young diagrams of some kind. This way, $GDI(Q)$ appears as
a rather combinatorial object. 

The idea is to try to describe  the possible values
of $GDI(Q)$, rather than that of the certain faces of it.
In the present article we will address the computation of
$GDI(m,Q)$ for the biggest possible $m=[d/2]+1$. This case corresponds
to the Grassmannian of middle-dimensional planes on $Q$.  
It should be noticed, that it is sufficient to consider the case of
odd-dimensional quadrics. This follows from the fact that
for the quadric $P$ of even dimension $2n$ and arbitrary codimension 1
subquadric $Q$ in it, 
$G(n+1,P)=G(n,Q)\times_{\op{Spec}(k)}\op{Spec}(k\sqrt{det_{\pm}(P)})$.

Below we will show that, for $m=[d/2]+1$, the $GDI(m,Q)$ can be described
in a rather simple terms - see Main Theorem \ref{mt} and Definition \ref{j}.
The restriction on the possible values here is given by the Steenrod operations
- see Proposition \ref{sj}. And at the moment there is no other restrictions
known - see Question \ref{qj} (the author would expect that there is none).
Finally, in the last section we show that in the case of a {\it generic}
quadric, the Grassmannian of middle-dimensional planes is $2$-incompressible,
which gives a new proof of the conjecture of G.Berhuy and Z.Reihstein
(see \cite[Conjecture 12.4]{BR}). Also, we formulate a conjecture describing
the canonical dimension of arbitrary quadric - see Conjecture \ref{cdq}. 

Most of these results were announced at the conference on ``Quadratic forms''
in Oberwolfach in May 2002. This text was written while I was 
a member at the Institute for Advanced Study at Princeton, and I would like to
express my gratitude to this institution for the support, excellent
working conditions and very stimulating atmosphere. The support of the 
the Weyl Fund is deeply appreciated. Finally, I'm very gratefull to G.Berhuy 
for the numerous discussions concerning canonical dimension, which made it
possible for the final section of this article to appear. 


\section{The Chow ring of the last grassmannian}

Let $k$ be a field of characteristic different from $2$, and 
$q$ be a nondegenerate quadratic form on a $(2n+1)$-dimensional $k$-vector 
space $W_q$. Denote as $G(n,Q)$ the grassmannian of
$n$-dimensional totally isotropic subspaces in $W_q$. 
If $q$ is completely split, then
the corresponding grassmanian will be denoted as $G(n)$, and the underlying
space of the form $q$ will be denoted as $W_n$.
For small $n$, examples are: $G(1)\cong\pp^1$, $G(2)\cong\pp^3$, and 
$G(3)\cong Q_6$ - the 6-dimensional hyperbolic quadric.

The Chow ring $\op{CH}^*(G(n))$ has $\zz$-basis, consisting of the
elements of the type $z_I$, where $I$ runs over all 
subsets of $\{1,\ldots,n\}$ (including the empty one). In particular, 
$\op{rank}(\op{CH}^*(G(n)))=2^n$. The degree (codimension) of $z_I$ is 
$|I|=\sum_{i\in I}i$, and this cycle can be defined as the collection 
of such $n$-dimensional totally isotropic subspaces 
$A\subset W_q$, that 
$$
\ddim(A\cap\pi_{n+1-j})\geq \#(i\in I, i\geq j),\,\,\,\,\, 
\text{for all}\,\,\,1\leq j\leq n,
$$ 
where
$\pi_1\subset\ldots\subset\pi_{n}$ is the fixed flag of totally
isotropic subspaces in 
$W_q$. The element $z_{\emptyset}$ is the ring unit $1=[G(n)]$.

Other parts of the landscape are: the tautological $n$-dimensional bundle $V_n$
on $G(n)$, and the embedding $G(n-1)\stackrel{j_{n-1}}{\row} G(n)$ given
by the choice of a rational point $x\in Q$.

Fixing such a point $x$, let $M_n\subset G(n)\times G(n)$ 
be the closed subvariety of pairs
$(A,B)$,
satisfying the conditions:
$$ 
x\in B,\,\,\,\,\,\text{and}\,\,\,\,\, 
\op{codim}(A\cap B\subset A)\leq 1.
$$
The projection on the first factor $(A,B)\mapsto A$ defines a birational map
$g_n:M_n\row G(n)$.
In particular, by the projection formula, the map 
$g_n^*:\op{CH}^*(G(n))\row\op{CH}^*(M_n)$ is injective.
On the other hand, the rule $(A,B)\mapsto (B/x)$ defines the map 
$\pi:M_n\row G(n-1)$. Tautological bundle
$V_n$ is naturally a subbundle in the trivial $2n+1$-dimensional bundle
$pr^*(W_n)$, which we will denote still by $W_n$.
The variety $M_n$ can be also described as the variety of pairs
$B\subset C\subset W_n$, where $B$ is totally isotropic, $dim(B)=n$, 
$dim(C)=n+1$, and $x\in B$.
In other words, $M_n\cong \pp_{G(n-1)}(j_{n-1}^*(W_n/V_n))$, where
the identification is given by the rule: 
$$
(A,B)\mapsto (A+B)/B\,\,\,\,\,(\text{respectively},\,\,\,\,\,
(B,B)\mapsto B^{\perp}/B\,\,\,).
$$
Clearly, $j_{n-1}^*(W_n/V_n)=(W_{n-1}/V_{n-1})\oplus\oc$,
and $W_{n-1}/V_{n-1}\cong (V^{\perp}_{n-1})^{\vee}$. 
$V_{n-1}$ is a subbundle of $V_{n-1}^{\perp}$, and
$V^{\perp}_{n-1}/V_{n-1}=\Lambda^{2n-1}W_{n-1}\cong\oc$.
Thus, $M_n\cong\pp_{G(n-1)}(Y_{n-1})$, where 
$[Y_{n-1}]=[V_{n-1}^{\vee}]+2[\oc]\in\op{K}_0(G(n-1))$. 
We get a diagram 
$$
G(n)\stackrel{g_n}{\low}\pp_{G(n-1)}(Y_{n-1})
\stackrel{\pi_{n-1}}{\row}G(n-1).
$$
Using the exact sequences
$0\row A\row (A+B)\row (A+B)/A\row 0$ and 
$0\row B\row (A+B)\row (A+B)/B\row 0$,
and the fact that $q$ defines a nondegenerate pairing between the spaces
$(A+B)/B\cong A/(A\cap B)$ and $(A+B)/A\cong B/(A\cap B)$ for all pairs
$(A,B)$ aside from the codimension $>1$ subvariety 
$(\Delta(G(n))\cap M_n)\subset M_n$,
we get the exact sequences:
$$
0\row g_n^*(V_n)\row X_{n-1}\row\oc(1)\row 0,\,\,\,\,\,\text{and}
$$
$$
0\row\pi_{n-1}^*(V_{n-1})\oplus\oc\row X_{n-1}\row\oc(-1)\row 0,
$$
where $X_{n-1}$ is the bundle with the fiber $C$.
In particular, 
$$
[g_n^*(V_n)]=[\pi_{n-1}^*(V_{n-1})]+[\oc]+[\oc(-1)]-[\oc(1)].\,\,\,\,\,
\text{Also},
$$
$$
\op{CH}^*(\pp_{G(n-1)}(Y_{n-1}))=
\op{CH}^*(G(n-1))[\rho]/(\rho^2\cdot c(V_{n-1}^{\vee})(\rho)),
$$
where $\rho=c_1(\oc(1))$.

Consider the open subvariety 
$\tilde{M}_n:=g_n^{-1}(G(n)\backslash j_{n-1}(G(n-1)))\subset M_n$
The map $\tilde{g}_n:\tilde{M}_n\row G(n)\backslash j_{n-1}(G(n-1))$ is an
isomorphism, and $\tilde{\pi}_{n-1}:\tilde{M}_n\row G(n-1)$ is an 
$n$-dimensional affine bundle over $G(n-1)$. 


\begin{proposition}
\label{es}
There is split exact sequence
\begin{equation*}
0\row\op{CH}^{*-n}(G(n-1))\stackrel{{j_{n-1}}_*}{\row}\op{CH}^*(G(n))
\stackrel{{j_{n-1}}^*}{\row}\op{CH}^*(G(n-1))\row 0.
\end{equation*}
\end{proposition}

\begin{proof}
Consider commutative diagram:
$$
\begin{CD}
G(n)&@<{g_n}<<&M_n&@>{\pi_{n-1}}>>&G(n-1)\\
@A{\ffi}AA&&@A{\psi}AA&&@|\\
G(n)\backslash j_{n-1}(G(n-1))&@<<{\tilde{g}_n}<&\tilde{M}_n&
@>>{\tilde{\pi}_{n-1}}>&G(n-1).
\end{CD}
$$



Notice that the choice of a point $y\in Q\backslash T_{x,Q}$ gives a section
$s:G(n-1)\row\tilde{M}_n$ of the affine bundle 
$\tilde{\pi}_{n-1}:\tilde{M}_n\row G(n-1)$. And the composition
$\ffi\circ\tilde{g}_n\circ s$ is equal to $j'_{n-1}$, where $j'_{n-1}$ is 
constructed
from the point $y\in Q$ in the same way as $j_{n-1}$ was constructed from
the point $x$. Thus, the isomorphism
$
\begin{CD}\op{CH}^*(G(n)\backslash j_{n-1}(G(n-1)))&
@>{(\tilde{\pi}_{n-1}^*)^{-1}\tilde{g}_n^*}>>&
\op{CH}^*(G(n-1))
\end{CD}
$
together with the localization at 
$G(n-1)\stackrel{j_{n-1}}{\row}G(n)\stackrel{\ffi}{\low} 
(G(n)\backslash G(n-1))$ gives us exact sequence
$$
\op{CH}^{*-n}(G(n-1))\stackrel{{j_{n-1}}_*}{\row}\op{CH}^*(G(n))
\stackrel{{j'_{n-1}}^*}{\row}\op{CH}^*(G(n-1))\row 0.
$$ 
Thus, $ker({j'_{n-1}}^*)=im({j_{n-1}}_*)$. Since it is true for arbitrary pair 
of points $x,y\in Q$ satisfying the condition that the line passing through 
them does not belong to a quadric, we get: $ker({j_{n-1}}^*)=im({j_{n-1}}_*)$.
On the other hand, the map 
${j'_{n-1}}_*:\op{CH}^{*-n}(G(n-1))\row\op{CH}^*(G(n))$ is split injective,
since 
$(\tilde{\pi}_{n-1})_*\circ\tilde{g}_n^*\circ\ffi^*\circ {j'_{n-1}}_*=id$.
Then the same is true for ${j_{n-1}}_*$. 
And we get the desired split exact sequence.
\Qed
\end{proof}

\begin{lemma}
\label{por}
The ring $\op{CH}^*(G(n))$ is generated by the elements of degree $\leq n$.
\end{lemma}

\begin{proof}
It easily follows by induction with the help of Proposition \ref{es}, and
projection formula.
\Qed
\end{proof}


\begin{proposition}
\label{canon}
Let $q$ be $2n+1$-dimensional split quadratic form. Then
\begin{itemize}
\item[$(1)$\ ] The group $O(q)$ acts trivially on $\op{CH}^*(G(n))$.
\item[$(2)$\ ] The maps $j_{n-1}^*$ and ${j_{n-1}}_*$ do not depend on
the choice of a point $x\in Q$.
\end{itemize}
\end{proposition}

\begin{proof}
Use induction on $n$. For $n=1$ the statement is trivial.
Suppose it is true for $(n-1)$. Let $j_{n-1,x}:G(n-1)_x\row G(n)$ be the map
corresponding to the point $x\in Q$.
For any $\ffi\in O(q)$ such that $\ffi(x)=y$, we have the map
$\ffi_{x,y}:G(n-1)_x\row G(n-1)_y$ such that $j_{n-1,y}\circ\ffi_{x,y}=
\ffi\circ j_{n-1,x}$. By the inductive assumption, the maps $\ffi_{x,y}^*$
and $(\ffi_{x,y})_*=((\ffi_{x,y})^*)^{-1}$ define canonical identification
of $CH^*(G(n-1)_x)$ and $CH^*(G(n-1)_y)$ which does not depend on the
choice of $\ffi$. And under this identification, 
$$
j_{n-1,x}^*\circ\ffi^*=j_{n-1,y}^*\,\,\,\text{and}\,\,\, 
\ffi_*\circ (j_{n-1,x})_*=(j_{n-1,y})_*.
$$ 

Let $\ffi\in O(q)$ be arbitrary
element, and $x,y\in Q$ be a such (rational) points that $\ffi(x)=y$. 
Let $z$ be arbitrary point on $Q$ such that neither of lines $l(x,z)$,
$l(y,z)$ lives on $Q$. Consider reflections $\tau_{x,z}$ and $\tau_{y,z}$.
They are rationally connected in $O(q)$. Consequently, 
for $\psi:=\tau_{y,z}\circ\tau_{x,y}$,
$\psi^*=id=\psi_*$. Thus, $(j_{n-1,x})^*=(j_{n-1,y})^*$ and  
$(j_{n-1,x})_*=(j_{n-1,y})_*$. 

From Proposition \ref{es} we get
the commutative diagram with exact rows:
{\footnotesize
$$
\begin{CD}
0&@>>>&\op{CH}^{*-n}(G(n-1)_y)&@>{(j_{n-1,y})_*}>>&\op{CH}^*(G(n))&
@>{j_{n-1,x}^*}>>&\op{CH}^*(G(n-1)_x)&@>>>&0\\
@.&&@|&&@AA{\ffi^*=(\ffi_*)^{-1}}A&&@|&&@.\\
0&@>>>&\op{CH}^{*-n}(G(n-1)_x)&@>{(j_{n-1,x})_*}>>&\op{CH}^*(G(n))&
@>{j_{n-1,y}^*}>>&\op{CH}^*(G(n-1)_y)&@>>>&0.
\end{CD}
$$
}
It implies that $\ffi^*$ is identity on elements of degree $\leq n$.
Since, by the Lemma \ref{por}, such elements generate $\op{CH}^*(G(n))$
as a ring, $\ffi^*=id=\ffi_*$, and $j_{n-1}^*,{j_{n-1}}_*$ are
well-defined.
\Qed
\end{proof}




\begin{proposition}
\label{oprz}
There is unique set of elements $z_i\in\op{CH}^i(G(n))$ defined for all 
$n\geq 1$ and satisfying the
properties:
\begin{itemize}
\item[$(0)$\ ] For $G(1)\cong\pp^1$, $z_1$ is the class of a point.
\item[$(1)$\ ] As a $\zzz$-module, 
$\op{CH}^*(G(n))=\operatornamewithlimits{\oplus}_{I\subset\{1,\ldots,n\}}
\zzz\cdot\operatornamewithlimits{\prod}_{i\in I}z_i$.
\item[$(2)$\ ] ${j_{n-1}}_*(1)=z_n$.
\item[$(3)$\ ] $j_{n-1}^*:\op{CH}^*(G(n))\row\op{CH}^*(G(n-1))$ is
given by the following rule on the additive generators above:
\begin{equation*}
\operatornamewithlimits{\prod}_{i\in I}z_i\mapsto
\begin{cases}
0,\,\,\text{if}\,\,n\in I;\\
\operatornamewithlimits{\prod}_{i\in I}z_i,\,\,\text{if}\,\,n\notin I.
\end{cases}
\end{equation*}
\end{itemize}
\end{proposition}

\begin{proof}
Let us introduce the elementary cycles $z_i\in\op{CH}^i(G(n))$
inductively as follows:
For $n=1$, $G(1)\cong\pp^1$, and $z_1$ is just the class of a point.
Let $z_i\in\op{CH}^i(G(n-1))$, for $1\leq i\leq n-1$ are defined and
satisfy the condition $(1)-(3)$. Let us define similar cycles on $G(n)$.


From the Proposition \ref{es} we get: ${j_{n-1}}^*$ is an 
isomorphism on $\op{CH}^i$, for $i<n$.
Now, for $1\leq i\leq n-1$, we define $z_i\in\op{CH}^i(G(n))$ as unique 
element corresponding under this isomorphism to $z_i\in\op{CH}^i(G(n-1))$.
And put: $z_n:={j_{n-1}}_*(1)$.
We automatically get $(2)$ satisfied. 

Let $J\subset\{1,\ldots,n-1\}$. 
From the projection formula we get: 
$$
{j_{n-1}}_*(\prod_{j\in J}z_j)=
z_n\cdot\prod_{j\in J}z_j.
$$

Applying once more Proposition \ref{es}, we get condition (1) and (3).
\Qed
\end{proof}

{\bf Remark:} The cycle $z_i$
we constructed is given by the set of $n$-dimensional 
totally isotropic subspaces $A\subset W_n$ satisfying the condition:
$A\cap\pi_{n+1-i}\neq 0$ for fixed  totally
isotropic subspace $\pi_{n+1-i}$ of dimension $(n+1-i)$.
%\\   
%
%\smallskip

Consider the commutative diagram:
\begin{equation*}
\begin{CD}
G(n)&@<<g_n<&\pp_{G(n-1)}(Y_{n-1})&@>>\pi_{n-1}>&G(n-1)\\
@Aj_{n-1}AA&&@AjAA&&@AAj_{n-2}A\\
G(n-1)&@<<g_{n-1}<&\pp_{G(n-2)}(Y_{n-2})&@>>\pi_{n-2}>&G(n-2).
\end{CD}
\end{equation*}


By Proposition \ref{oprz}, the ring homomorphism
$$
j^*:\op{CH}^*(\pp_{G(n-1)}(Y_{n-1}))\row\op{CH}^*(\pp_{G(n-2)}(Y_{n-2}))
$$
is a surjection, and it's kernel is generated as an ideal by the elements
$\pi_{n-1}^*(z_{n-1})$ and $\rho^2\cdot c(V_{n-2}^{\vee})(\rho)$.
In particular, $(j^*)^k$ is an isomorphism for all $k<n-1$, and the kernel of 
$(j^*)^{n-1}$ is additively generated by $\pi_{n-1}^*(z_{n-1})$.

\begin{theorem}
\label{o}
Let $V_n$ be tautological bundle on $G(n)$, $z_i$ be elements defined in
Proposition $\ref{oprz}$, and $\rho=c_1(\oc(1))$. Then:
\begin{itemize}
\item[$(1)$\ ] $g_n^*(z_k)=\rho^k+2\sum_{0<i<k}\rho^{k-i}\pi_{n-1}^*(z_i)+
\pi_{n-1}^*(z_k)$, for all $0<k<n$. 
\item[$(2)$\ ] $g_n^*(z_n)=\rho^n+2\sum_{0<i<n}\rho^{n-i}\pi_{n-1}^*(z_i)$.
\item[$(3)$\ ] 
$c(V_n)(t)=t^n+2\operatornamewithlimits{\sum}_{1\leq i\leq n}(-1)^iz_it^{n-i}$.
\end{itemize}
\end{theorem}

\begin{proof}
$G(1)$ is a conic and $V_1\cong\oc(-1)$.
Hence, $c(V_1)(t)=t-2z_1$. Let now $(1)_{m,k}$ is proven for all
$m<n$ and all $0<k<m$, and $(2)_m$ and $(3)_m$ are proven for all $m<n$.

Since 
$(j^*)^k:\op{CH}^k(\pp_{G(n-1)}(Y_{n-1}))\row\op{CH}^k(\pp_{G(n-2)}(Y_{n-2}))$
is an isomorphism for $k<n-1$, 
the condition $(1)_{n,k}$ follows from $(1)_{n-1,k}$ for all such $k$
in view of 
$j^*(\rho)=\rho$ and 
$j^*(\pi_{n-1}^*(z_k))=\pi_{n-2}^*(z_k)$
(Proposition \ref{oprz}(3)).
Analogously, since the kernel $(j^*)^{n-1}$ is additively generated by  
$\pi_{n-1}^*(z_{n-1})$, the condition $(2)_{n-1}$ implies that 
$g_n^*(z_{n-1})=\rho^{n-1}+2\sum_{1\leq i< n-1}\rho^{n-1-i}\pi_{n-1}^*(z_i)+
\lambda\cdot\pi_{n-1}^*(z_{n-1})$, where 
$\lambda\in\zz$.
Since $Y_{n-1}=\oc\oplus(V_{n-1}^{\perp})^{\vee}$, the projection 
$\pi_{n-1}:\pp_{G(n-1)}(Y_{n-1})\row G(n-1)$ has the section $s$
(given by the rule: $(B/x)\mapsto (B,B)$). It satisfies: $g_n\circ s=j_{n-1}$.
Since $s^*(\pi_{n-1}^*(z_{n-1}))=z_{n-1}$, $s^*(\rho)=0$ and
$j_{n-1}^*(z_{n-1})=z_{n-1}$, we get $\lambda=1$, which implies
$(1)_{n,n-1}$.

Choose some rational point $y\in Q\backslash T_{x,Q}$.
By Propositions \ref{canon}(2) and \ref{oprz}(2), 
the cycle $z_n$ is defined as the set of such planes $A$, that $y\in A$. Then 
the cycle $g_n^*(z_n)$
is the set of such pairs $(A,B)$, that $y\in A,\,x\in B$ and 
$\ddim(A+B/A)\leq 1$. Thus $A+B=y+B$, and $g_n^*(z_n)$ is given by the section
$\pp_{G(n-1)}(\oc)\subset\pp_{G(n-1)}(Y_{n-1})$. Since
$c(Y_{n-1})(t)=t^2\cdot c(\pi_{n-1}^*(V_{n-1}^{\vee}))(t)$, this class can be 
expressed as $\rho\cdot c(\pi_{n-1}^*(V_{n-1}^{\vee}))(\rho)$. 
The last expression is equal to
$\rho^n+2\rho^{n-1}\pi_{n-1}^*(z_1)+\ldots+2\rho\pi_{n-1}^*(z_{n-1})$ because
of $(3)_{n-1}$. The statement $(2)_n$ is proven.

Finally,  since $[g_n^*(V_n)]=[\pi_{n-1}^*(V_{n-1})]+[\oc]+[\oc(-1)]-[\oc(1)]$,
\\
$g_n^*(c(V_n)(t))=
\pi_{n-1}^*(c(V_{n-1})(t))\cdot\frac{t\cdot(t-\rho)}{t+\rho}$.
In the light of $(3)_{n-1}$, this is equal to
$$
\left(t^{n-1}+2\sum_{1\leq i\leq n-1}(-1)^i\pi_{n-1}^*(z_i)t^{n-1-i}\right)
\cdot\frac{t\cdot (t-\rho)}{t+\rho}.
$$ 
Using the equality 
$\rho^2(\rho^{n-1}+2\rho^{n-2}\pi_{n-1}^*(z_1)+\ldots+
2\pi_{n-1}^*(z_{n-1}))=0$, as well as the conditions $(1)_{n,k}$ and $(2)_n$,
we can rewrite the last expression as:
$$
t^n+2\sum_{1\leq i\leq n}(-1)^ig_n^*(z_i)t^{n-i}.
$$
Since $g_n^*$ is injective (the map $g_n$ is birational), we get:
$$
c(V_n)(t)=t^n+2\sum_{1\leq i\leq n}(-1)^iz_it^{n-i}.
$$ 
The statement $(3)_n$ is proven.
\Qed
\end{proof}





\section{Multiplicative structure}

The multiplicative structure of $\op{CH}^*(G(n))$ was studied extensively
by H.Hiller, B.Boe, J.Stembridge, P.Pragacz and J.Ratajski - see \cite{HB},
\cite{PR}.

We can compute this ring structure from Theorem \ref{o}. 
Although, we restrict our consideration only to $(\,mod\,2)$ case, 
it should be pointed out that the integral case can be obtained in
exactly the same way.

Let us denote as $\ov{u}$ the image of $u$ under the map 
$\op{CH}^*\row\op{CH}^*/2$.

\begin{proposition}
\label{giperb}
$$
\op{CH}^*(G(n))/2=
\operatornamewithlimits{\otimes}_{1\leq d\leq n;\,d-\text{odd}}
(\zzz/2[\ov{z}_d]/(\ov{z}_d^{2^{m_d}})),
$$
where $m_d=[log_2(n/d)]+1$.
\end{proposition}

\begin{proof}
Consider the diagram:
\begin{equation*}
\begin{CD}
G(n)&@<g_n<<&\pp_{G(n-1)}(Y_{n-1})&@>\pi_{n-1}>>&G(n-1).
\end{CD}
\end{equation*}
From Theorem \ref{o}, $g_n^*(\ov{z}_k)=\ov{\rho}^k+\pi_{n-1}^*(\ov{z}_k)$,
for $k<n$, and $g_n^*(\ov{z}_n)=\ov{\rho}^n$. 
Then it easily follows by the induction on $n$, that 
$\ov{z}_k^2=\ov{z}_{2k}$ (where we assume $\ov{z}_r=0$ if $r>n$).

Thus, we have surjective ring homomorphism
\begin{equation*}
\operatornamewithlimits{\otimes}_{1\leq d\leq n;\,d-\text{odd}}
(\zzz/2[\ov{z}_d]/(\ov{z}_d^{2^{m_d}}))\row\op{CH}^*(G(n))/2.
\end{equation*}
Since the dimensions of both rings are equal to $2^n$, it is an isomorphism.
\Qed
\end{proof}


Let $J$ be a set. Let us call a {\it multisubset} the collection
$\Lambda=\coprod_{\beta\in B}\Lambda_{\beta}$ of disjoint subsets of $J$.
For a subset $I$ of $J$, we will denote by the same symbol $I$ the
multisubset $\coprod_{i\in I}\{i\}$.
Let $B=\coprod_{\gamma\in C}B_{\gamma}$, and 
$\Lambda'_{\gamma}=\coprod_{\beta\in B_{\gamma}}\Lambda_{\beta}$.
Then the multisubset $\Lambda':=\coprod_{\gamma\in C}\Lambda'_{\gamma}$
is called the {\it specialization} of $\Lambda$.
We call the specialization {\it simple} if $\#(B_{\gamma})\leq 2$,
for all $\gamma\in C$. 

Let $J$ now be some set of natural numbers (it may contain multiple entries).
Then to any finite multisubset 
$\Lambda=\coprod_{\beta\in B}\Lambda_{\beta}$ of $J$ we 
can assign the set of natural numbers 
$\ov{\Lambda}:=\{\sum_{i\in\Lambda_{\beta}}i\}_{\beta\in B}$. 
We call the specialization $\Lambda$ {\it good} if 
$\ov{\Lambda}\subset\{1,\ldots,n\}$.
 
Suppose $I$ be some finite set of natural numbers. Let us define 
the element $\ov{z}_I\in\op{CH}^*(G(n))/2$ by the formula:
$$
\ov{z}_I=
\sum_{\Lambda}\prod_{j\in\ov{\Lambda}}\ov{z}_{j},
$$ 
where we assume $\ov{z}_r=0$, if $r>n$, and the sum is taken over all
simple specializations $\Lambda$ of the multisubset $I=\coprod_{i\in I}\{i\}$.


\begin{lemma}
\label{om}
If $I\not\subset\{1,\ldots,n\}$, then $\ov{z}_I=0$.
\end{lemma}

\begin{proof}
If $I$ contains an element $r>n$, then $\ov{z}_I$ is clearly zero.
Suppose now that $I$ contains some element $i$ twice, say as $i_1$ and $i_2$.
Consider the subgroup $Z_2\subset S_s$ interchanging $i_1$ and $i_2$ and
keeping all other elements in place. We get $\zzz/2$-action on our 
specializations.
The terms which are not stable under this action will appear with multiplicity
$2$, so, we can restrict our attention to the stable terms.
But such specializations have the property that $\{i_1,i_2\}$ is disjoint from
the rest of $i$'s, and the corresponding sum looks as:
$\sum_{M}\prod_{j\in\ov{M}}\ov{z}_{j}
\cdot(\ov{z}_i^2+\ov{z}_{2i})$,
where the sum is taken over all simple specializations of the multisubset 
$I\backslash\{i_1,i_2\}$.
Since $\ov{z}_i^2=\ov{z}_{2i}$, this expression is zero.
\Qed
\end{proof}

We immediately get the (modulo 2) version of the Pieri formula
proved by H.Hiller and B.Boe:

\begin{proposition}{\rm (\cite{PR})}
\label{pieri}
$$
\ov{z}_I\cdot \ov{z}_j=
\ov{z}_{I\cup j}+\sum_{i\in I}\ov{z}_{(I\backslash i)\cup (i+j)},
$$
where we omit terms $\ov{z}_J$ with $J\not\subset\{1,\ldots,n\}$
(in particular, if $J$ contains some element with multiplicity $>1$).
\end{proposition}

\begin{proof}
$\ov{z}_{I\cup j}=
\sum_{\Lambda}\prod_{l\in\ov{\Lambda}}\ov{z}_{l}$, where
the sum is taken over all simple specializations of the multisubset 
$I\cup j$.
We can distinguish two types of specializations: 1) $j$ is separated from $I$;
2) $j$ is not separated from $I$, that is, there is $\beta$ such that
$\Lambda_{\beta}=\{i,j\}$, for some $i\in I$. Let us call the latter 
specializations to be of type $(2,i)$. 
Clearly, the sum over specializations of the first kind is
equal to $\ov{z}_I\cdot\ov{z}_j$, and the sum over the specializations of the
type $(2,i)$ is equal to $\ov{z}_{(I\backslash i)\cup (i+j)}$.
Finally, the terms with $J\not\subset\{1,\ldots,n\}$ could be ommited by
Lemma \ref{om}.
\Qed
\end{proof}

We also get the expression of monomials on $z_i$'s in terms of $z_I$'s.

\begin{proposition}
\label{razl}
\begin{itemize}
\item[$(1)$\ ] The set $\{\ov{z}_I\}_{I\subset\{1,\ldots,n\}}$ is a
basis of $\op{CH}^*(G(n))/2$.
\item[$(2)$\ ] $\prod_{i\in I}\ov{z}_i=\sum_{\Lambda}\ov{z}_{\ov{\Lambda}}$, 
where sum is taken over all good specializations of $I$.
\end{itemize}
\end{proposition}

\begin{proof}

(1) On the $\zzz/2$-vector space 
$\op{CH}^*(G(n))/2=
\oplus_{I\subset\{1,\ldots,n\}}\zzz/2\cdot\prod_{i\in I}\ov{z}_i$ we have
lexicographical filtration. Consider the linear map 
$\eps:\op{CH}^*(G(n))/2\row\op{CH}^*(G(n))/2$ sending $\prod_{i\in I}\ov{z}_i$
to $\ov{z}_I$. Then the associated graded map: $gr(\eps)$ is the identity.
Thus, $\eps$ is invertible, and the set  
$\{\ov{z}_I\}_{I\subset\{1,\ldots,n\}}$ form a basis.

(2) Consider the $\zzz/2$-vector spaces
$W_1:=\oplus_{\Lambda}\zzz/2\cdot x_{\Lambda}$, and
$W_2:=\oplus_{\Lambda}\zzz/2\cdot y_{\Lambda}$,
where $\Lambda$ runs over all finite multisubsets of $\nnn$.

Consider the linear maps $\psi:W_2\row W_1$ which sends
$y_{\Lambda}$ to the $\sum_{\Lambda'}x_{\Lambda'}$, where
the sum is taken over all specializations of $\Lambda$,
and $\ffi:W_1\row W_2$ which sends $x_{\Lambda}$ to the
$\sum_{\Lambda'}y_{\Lambda'}$, where the sum is taken over all
simple specializations $\Lambda'$ of $\Lambda$.
It is an easy exercise to show that $\ffi$ and $\psi$
are mutually inverse.


Consider the linear surjective maps: $w_1:W_1\row\op{CH}^*(G(n))/2$ and
$w_2:W_2\row\op{CH}^*(G(n))/2$ given by the rule:
$w_1(x_{\Lambda}):=\ov{z}_{\ov{\Lambda}}$, and
$w_2(y_{\Lambda}):=\prod_{j\in\ov{\Lambda}}\ov{z}_{j}$.


Then, by the definition of $\ov{z}_I$, $w_1=w_2\circ\ffi$.
Then $w_2=w_1\circ\psi$, which implies that
$\prod_{i\in I}\ov{z}_i=\sum_{\Lambda}\ov{z}_{\ov{\Lambda}}$, where the
sum is taken over all specializations of $I$.
It remains to notice, that nongood specializations do not contribute to the sum
(by Lemma \ref{om}).
\Qed
\end{proof}


{\bf Examples:} 1) $\ov{z}_i\cdot \ov{z}_j=\ov{z}_{i,j}+\ov{z}_{i+j}$,
where the first term is
ommited if $i=j$ and the second if $i+j>n$.
2) $\ov{z}_{i,j,k}=\ov{z}_i\cdot \ov{z}_j\cdot \ov{z}_k+
\ov{z}_{i+j}\cdot \ov{z}_k+\ov{z}_{j+k}\cdot \ov{z}_i+
\ov{z}_{i+k}\cdot \ov{z}_j$.









\section{Action of the Steenrod algebra}

On the Chow-groups modulo prime $l$ there is the action of the Steenrod 
algebra.
Such action was constructed by V.Voevodsky in the context of arbitrary motivic
cohomology - see \cite{VoO}, and then a simpler construction was
given by P.Brosnan for the case of usual Chow groups - see \cite{PB}. 
For quadratic grassmannians we will be interested only in the case $l=2$.

We can compute the action of the Steenrod squares on the cycles $\ov{z}_i$.
Let $\sti$ be the total Steenrod opeartion $id+\sti^1+\sti^2+\ldots$.

\begin{theorem}
$$
\sti(\ov{z}_i)=
\sum_{m=i}^{\op{min}(n,2i)}\binom{i}{m-i}\cdot\ov{z}_m
$$
\end{theorem}

\begin{proof}
Use induction on $n$. The base is trivial.
Suppose the statement is true for $(n-1)$.
By Theorem \ref{o}, $g_n^*(\ov{z}_i)=\ov{\rho}^i+\pi_{n-1}^*(\ov{z}_i)$,
where we assume $\ov{z}_n\in\op{CH}^*(G(n-1))/2$ to be zero.
Also, $\ov{c}(V_n)(t)=t^n$, which implies that $\ov{\rho}^{n+1}=0$.
Using the fact that $\sti$ commutes with the pull-back morphisms
(see \cite{PB}), the latter equality, and inductive assumption, 
we get:
\begin{equation*}
\begin{split}
g_n^*(\sti(\ov{z}_i))=\sti(g_n^*(\ov{z}_i))=\sti(\ov{\rho}^i)+
\pi_{n-1}^*\left(\sum_{m=i}^{\op{min}(n-1,2i)}\binom{i}{m-i}\cdot\ov{z}_m
\right)=\\
\sum_{m=i}^{2i}\binom{i}{m-i}\ov{\rho}^m+
\sum_{m=i}^{\op{min}(n-1,2i)}\binom{i}{m-i}
\cdot(\ov{\rho}^m+g_n^*(\ov{z}_m))=\\
\sum_{m=i}^{\op{min}(n,2i)}\binom{i}{m-i}g_n^*(\ov{z}_m).
\end{split}
\end{equation*}
Now, the statement follows from the injectivity of $g_n^*$.
\Qed
\end{proof}













\section{Main theorem}


Let $X$ be some variety over the field $k$. We will denote:
$$
C^*(X):=image(\op{CH}^*(X)/2\row\op{CH}^*(X|_{\kbar})/2).
$$

Let now $Q$ be a smooth projective quadric of dimension $2n-1$, and 
$X=G(n,Q)$ be the grassmanian of middle-dimensional projective planes on it.
Then $X|_{\kbar}=G(n)$.
In this section we will show that, as an algebra, $C^*(G(n,Q))$ is generated
by the elementary cycles $\ov{z}_i$ contained in it.


Let $F(n,Q)$ be the variety of complete flags
$(l_0\subset l_1\subset\ldots\subset l_{n-1})$ of projective
subspaces on $Q$. Then $F(n,Q)$ is naturally isomorphic to the
complete flag variety $F_{G(n,Q)}(V_n)$ of the tautological $n$-dimensional
bundle on $G(n,Q)$. On the variety $F(n,Q)$ there are natural (subquotient)
line bundles $\lc_1,\ldots,\lc_n$. 
The first chern classes $c_1(\lc_i),\, 1\leq i\leq n$ generate the
ring $\op{CH}^*(F(n,Q))$ as an algebra over $\op{CH}^*(G(n,Q))$, and
the relations among them depend only on the chern classes $c_j(V_n)$.
Let $F_n$ be the variety of complete
flags of subspaces of the $n$-dimensional vector space $V$. It also has
natural line bundles $\lc'_1,\ldots,\lc'_n$.
Again, the first chern classes $c_1(\lc'_i)$ generate the ring
$\op{CH}^*(F_n)$.
By Theorem \ref{o} (3), modulo $2$, all chern classes $c_j(V_n)$ are the 
same as
the chern classes of the trivial $n$-dimensional bundle $\oplus_{i=1}^n\oc$. 
Thus, modulo $2$, the Chow ring of $F_{G(n,Q)}(V_n)$ is isomorphic
to the Chow ring of $F_{G(n,Q)}(\oplus_{i=1}^n\oc)$. We get:

\begin{theorem}
\label{fig}
There is a ring isomorphism
$$
\op{CH}^*(F(n,Q))/2\cong\op{CH}^*(G(n,Q))/2\otimes_{\zzz/2}\op{CH}^*(F_n)/2,
$$
where the map $\op{CH}^*(G(n,Q))\row\op{CH}^*(F(n,Q))$ is induced by the
natural projection $F(n,Q)\row G(n,Q)$, and the map
$\op{CH}^*(F_n)/2\row\op{CH}^*(F(n,Q))/2$ is given on the generators by the
rule: $c_1(\lc'_i)\mapsto c_1(\lc_i)$.
\end{theorem}

\Qed

Notice, that the map $ac:\op{CH}^*(F_n)\row\op{CH}^*(F_n|_{\kbar})$ is
an isomorphism, and, 
$C^*(F(n,Q))=C^*(G(n,Q))\otimes_{\zzz/2}\op{CH}^*(F_n|_{\kbar})/2$.
Thus, we have:

\begin{statement}
\label{GF}
Let $v_1,\ldots,v_s$ be linearly independent elements of
$\op{CH}^*(F_n|_{\kbar})/2$, and $x_i\in\op{CH}^*(G(n,Q)|_{\kbar})/2$, then 
$x=\sum_{i=1}^sx_i\cdot v_i$ belongs to $C^*(F(n,Q))$ if and only if
all $x_i\in C^*(G(n,Q))$.
\end{statement}

\Qed

The ring $\op{CH}^*(F_n)$ can be described as follows.
Let us denote $c_1(\lc_j)$ as $h_j$, and the set $\{h_j,\ldots, h_n\}$ as 
$\un{h}(j)$ (and $\un{h}(1)$ as $\un{h}$).
For arbitrary set of variables $\un{u}=\{u_1,\ldots, u_r\}$ let us define 
the degree $i$ polynomials $\sigma_i(\un{u})$ and $\sigma_{-i}(\un{u})$ 
from the equation:
$$
\prod_{l}(1+u_l)=\sum_i\sigma_i(\un{u})=
\left(\sum_i\sigma_{-i}(\un{u})\right)^{-1}.
$$

It is well-known that:

\begin{statement}
\label{fn}
$$
\op{CH}^*(F_n)=\zzz[\un{h}]/(\sigma_i(\un{h}),\,\,1\leq i\leq n)=
\zzz[\un{h}]/(\sigma_{-i}(\un{h}(i)),\,\,1\leq i\leq n).
$$
\end{statement}

Since $\sigma_{-i}(\un{h}(i))$ is the $\pm$-monic polynomial in $h_i$ with
coefficients in the subring, generated by $\un{h}(i+1)$, we get:
$\op{CH}^*(F_n)$ is a free module over the subring 
$\zzz[\un{h}(n)]/(\sigma_{-n}(\un{h}(n)))=\zzz[h_n]/(h_n^n)$.


Let $\pi:F(n,Q)\row F(n-1,Q)$ be the natural projection between full
flag varieties. We will denote by the same symbol $\ov{z}_I$ the images
of $\ov{z}_I$ in $\op{CH}^*(F(n,Q))/2$.

The following statement is the key for the Main Theorem.

\begin{proposition}
$$
\pi^*\pi_*(\ov{z}_I)=\sum_{i\in I}\ov{z}_{(I\backslash i)}\cdot
\pi^*\pi_*(\ov{z}_i).
$$
\end{proposition}

\begin{proof}
$F(n,Q)$ is a conic bundle over $F(n-1,Q)$ inside the projective bundle
$\pp_{F(n-1,Q)}(V)$, where, in $K_0(F(n,Q))$,
$\pi^*[V]=[\lc_n]+[\lc_n^{-1}]+[\oc]$. Sheaf $\lc_n$ is nothing else
but the restriction of the sheaf $\oc(-1)$ from $\pp_{F(n-1,Q)}(V)$ to
$F(n,Q)$. 

\begin{lemma}
\label{fi}
Let $V$ be a $3$-dimensional bundle over some variety $X$ equipped with
the nondegenerate quadratic form $p$. Let $\pi:Y\row X$ be conic bundle of
$p$-isotropic lines in $U$. Then there is a $CH^*(X)$-algebra
automorphism $\phi:CH^*(Y)\row CH^*(Y)$ of exponent $2$ such that
\begin{itemize}
\item[$(1)$\ ]
$\phi(c_1(\oc(-1)|_Y))=c_1(\oc(1)|_Y)$. 
\item[$(2)$\ ]
$\pi^*\pi_*(x)\cdot c_1(\oc(1)) =x-\phi(x)$
\end{itemize}
\end{lemma}

\begin{proof}
Consider variety $Y\times_X Y$ with the natural projections
$\pi_1$ and $\pi_2$ on the first and second factor, respectively.
Then divisor $\Delta(Y)\subset Y\times_X Y$ defines an invertible
sheaf $\lc$ on $Y\times_X Y$ such that 
$\lc^2\cong\pi_1^*(\oc(1))\otimes\pi_2^*(\oc(1))$ and
$\Delta^*(\lc)=\oc(1)$. Consider the map 
$f:=\Delta\circ\pi_2:Y\times_X Y\row Y\times_X Y$.
Define $\phi:\op{CH}^*(Y)\row\op{CH}^*(Y)$ as 
$id-\Delta^*\circ f_*\circ\pi_1^*$.

The described maps fit into the diagram:
$$
\begin{CD}
Y&@<{\pi_1}<<&Y\times_X Y&&\\
@V{\pi}VV&&@VV{\pi_2}V&&\\
X&@<<{\pi}<&Y&@>>{\Delta}>&Y\times_X Y
\end{CD}
$$ 
with the transversal cartesian square ($\pi_1^*(T_{\pi})=T_{\pi_2}$). 
Consequently,
$\pi^*\circ\pi_*={\pi_2}_*\circ{\pi_1}^*$. Since
$\oc(\Delta(Y))|_{Y}=\oc(1)$, we have:
$$
\pi^*\pi_*(x)\cdot c_1(\oc(1))=\Delta^*\Delta_*{\pi_2}_*\pi_1^*(x)=
x-\phi(x).
$$

Consider the map 
$\psi:=id-f_*:\op{CH}^*(Y\times_X Y)\row \op{CH}^*(Y\times_X Y)$. 
We claim that $\psi$ is a ring homomorphism.
Really, $Y\times_X Y\cong\pp_Y(U)$, where the projection $\pp_Y(U)\row Y$
is given by $\pi_2$ and $c(U)(t)=t(t-c_1(\oc(1)))$.
Thus $\op{CH}^*(Y\times_X Y)=\op{CH}^*(Y)[\rho]/(\rho(\rho-c_1(\oc(1))))$,
where the map $\op{CH}^*(Y)\row\op{CH}^*(Y\times_X Y)$ is $\pi_2^*$.
Notice, that $f_*\pi_2^*=\Delta_*{\pi_2}_*\pi_2^*=0$, 
that is, $\psi|_{\op{CH}^*(Y)}$ is the identity. At the same time,
$\psi(\rho)=\rho-\rho=0$. Since 
$\op{CH}^*(Y\times_X Y)$ is free $\op{CH}^*(Y)$-module of rank $2$ with
the basis $1$, $\rho$, by the projection formula,
we get that $\psi$ is an endomorphism of
$\op{CH}^*(Y\times_X Y)$ considered as an $\op{CH}^*(Y)$-algebra.

Since, $\phi=\Delta^*\circ\psi\circ\pi_1^*$, it is a homomorphism of 
$\op{CH}^*(X)$-algebras.
Also, $\phi(c_1(\oc(1)))=-c_1(\oc(1))$. 
Finally, since the composition 
${\pi_2}_*\pi_1^*\Delta^*\Delta_*:\op{CH}^*(Y)\row\op{CH}^*(Y)$ is equal
$2\cdot id$, we get 
$$
(\Delta^*\circ f_*\circ\pi_1^*)^{\circ 2}=2(\Delta^*\circ f_*\circ\pi_1^*),
$$
which is equivalent to:  $\phi^{\circ 2}=id$. Thus, $\phi$ is an
automorphism of exponent $2$.
\Qed
\end{proof}

Let us compute the action of $\phi$ on basis elements $\ov{z}_I$.
Let $\sigma_i$ be elementary symmetric functions in $h_i$'s.
Since $h_i\in CH^*(F(n-1,Q))$, for $i<n$, we have
equality $\phi(h_i)=h_i$ for them, and $\phi(h_n)=-h_n$.
We know that $\sigma_i=(-1)^i2z_i$. We immediately conclude:

\begin{lemma}
\label{zi}
$\phi(z_i)=z_i+\sum_{0<l<i}2z_{i-l}h_n^l+h_n^i$.
\end{lemma}

\begin{lemma}
\label{zii}
$\phi(\ov{z}_I)=\ov{z}_I+\sum_{i\in I}\ov{z}_{(I\backslash i)}\ov{h}_n^i$. 
\end{lemma}

\begin{proof}
Let us define the {\it size} $s(I)$ of $I$ as the number of it's elements.
Use induction on the size of $I$. The case of size $=1$ is OK by 
the previous lemma. Suppose the statement is known for sizes $<s(I)$.

Let $i$ be some element of $I$. 
We know from Proposition \ref{pieri} that 
$\ov{z}_I=\ov{z}_{(I\backslash i)}\cdot \ov{z}_i+
\sum_{j\in I,j\neq i}\ov{z}_{(I\backslash\{i,j\})\cup (i+j)}$.
Since $\phi$ is a ring homomorphism, we get:
\begin{equation*}
\begin{split}
\phi(\ov{z}_I)=\phi(\ov{z}_{(I\backslash i)})\cdot\phi(\ov{z}_i)+
\sum_{j\in I\backslash i}\phi(\ov{z}_{(I\backslash\{i,j\})\cup (i+j)})=\\
\left(\ov{z}_{(I\backslash i)}+
\sum_{l\in I\backslash i}\ov{z}_{(I\backslash\{i,l\})}h_n^l\right)
\cdot(\ov{z}_i+\ov{h}_n^i)+\\
\sum_{j\in I\backslash i}\left(\ov{z}_{(I\backslash\{i,j\})\cup (i+j)}+
\ov{z}_{(I\backslash\{i,j\})}\ov{h}_n^{i+j}+
\sum_{m\in I\backslash\{i,j\}}\ov{z}_{(I\backslash\{i,j,m\})\cup(i+j)}
\ov{h}_n^m\right)=\\
\ov{z}_I+\ov{z}_{(I\backslash i)}\ov{h}_n^i+
\sum_{l\in I\backslash\{i\}}(\ov{z}_{(I\backslash\{i,l\})}\cdot \ov{z}_i)
\ov{h}_n^l+
\sum_{m\neq j\in I\backslash\{i\}}\ov{z}_{(I\backslash\{i,j,m\})\cup(i+j)}
\ov{h}_n^m=\\
\ov{z}_I+\ov{z}_{(I\backslash i)}\ov{h}_n^i+\sum_{j\in I\backslash i}
\ov{z}_{(I\backslash j)}\ov{h}_n^j+
2\cdot
\sum_{m\neq j\in I\backslash\{i\}}
\ov{z}_{(I\backslash\{i,j,m\})\cup(i+j)}\ov{h}_n^m=\\
\ov{z}_I+\sum_{j\in I}\ov{z}_{(I\backslash j)}\ov{h}_n^j
\end{split}
\end{equation*}
(as usually, one should omit $\ov{z}_J$ with $J\not\subset\{1,\ldots,n\}$).
\Qed
\end{proof}

Let $p=q\perp\hh$. Then $Q$ can be identified with the quadric of
projective lines on $P$ passing through fixed rational point $y$.
This identifies the complete flag variety
$F(r,Q)$ with the subvariety of  $F(r+1,P)$ consisting of flags
containing our point $y$. We get an embedding $i_r:F(r,Q)\row F(r+1,P)$.
It is easy to see that the diagram
$$
\begin{CD}
F(n,Q)&@>i_n>>&F(n+1,P)\\
@V{\pi}VV&&@VV{\pi'}V\\
F(n-1,Q)&@>>i_{n-1}>&F(n,P)
\end{CD}
$$
is cartesian, and since $\pi'$ is smooth, we have an equality:
$\pi^*\circ\pi_*\circ i_n^*=i_n^*\circ{\pi'}^*\circ\pi'_*$.

 
It follows from Lemmas \ref{fi} and \ref{zii} that
$$
{\pi'}^*\pi'_*(\ov{z}_I)\cdot \ov{h}_{n+1}=
\sum_{i\in I}\ov{z}_{(I\backslash i)}\ov{h}_{n+1}^i.
$$
Thus, modulo the kernel of multiplication by $\ov{h}_{n+1}$, 
${\pi'}^*\pi'_*(\ov{z}_I)\equiv 
\sum_{i\in I}\ov{z}_{(I\backslash i)}\ov{h}_{n+1}^{i-1}$.
But, by the Statement \ref{fn} and Theorem \ref{fig}, such kernel is generated 
by $\ov{h}_{n+1}^n$.

Since $i_n^*(\ov{z}_I)=\ov{z}_I$, $i_n^*(\ov{h}_{n+1})=\ov{h}_n$ and 
$\ov{h}_n^n=0$ on $F(n,Q)$, we get:

$$
\pi^*\pi_*(\ov{z}_I)=i_n^*{\pi'}^*\pi'_*(\ov{z}_I)=
\sum_{i\in I}\ov{z}_{(I\backslash i)}\ov{h}_n^{i-1}=
\sum_{i\in I}\ov{z}_{(I\backslash i)}\pi^*\pi_*(\ov{z}_i).
$$
\Qed
\end{proof}

Notice, that the elements $\pi^*\pi_*(z_i)$ belong to 
$\op{CH}^*(F_n)/2$, and they are
linearly independent (being nonzero and having different degrees).

As a corollary, we get:

\begin{maintheorem}
\label{mt}
As an algebra, $C^*(G(n,Q))$ is generated by the elementary classes $\ov{z}_i$ 
contained in it.
\end{maintheorem}

\begin{proof}
Let $\ov{z}$ be an element of $C^*(G(n,Q))$. It can be expressed as a linear
combination of the basis elements $\ov{z}_I$'s. 
Let us define the {\it size} $s(\ov{z})$ of the
element $\ov{z}=\sum \ov{z}_{I_a}$ as the maximum of sizes of $I_a$ involved. 
Let $m(\ov{z})$ be the main term of $\ov{z}$, that is, 
$\sum_{a:s(I_a)=s(\ov{z})}\ov{z}_{I_a}$.

\begin{lemma}
\label{mtzi}
Let $\ov{z}=\sum_{a}\ov{z}_{I_a}\in C^*(G(n,Q))$. Let $s(I_a)=s(\ov{z})$,
and $i\in I_a$.
Then the elementary cycle
$\ov{z}_i$ belongs to $C^*(G(n,Q))$.
\end{lemma}

\begin{proof}
Let $i\in I_a$, and $I_a\backslash i=\{j_2,\ldots,j_s\}$.
Denote the operation $\pi^*\pi_*$ as $D$. Then
$D(\ov{z})=\sum_{1\leq j\leq n} d_j(\ov{z})\cdot D(\ov{z}_j)$, where 
$d_j(\ov{z})\in\op{CH}^*(G(n,Q)|_{\kbar})/2$, and the elements
$D(\ov{z}_j)\in\op{CH}^*(F_n)/2$ are linearly independent.
Since $D$ is defined over the base field, $D(\ov{z})\in C^*(F(n,Q))$, and,
by the Statement \ref{GF}, $d_j(\ov{z})\in C^*(G(n,Q))$. Clearly, 
$m(d_j(\ov{z}))=d_j(m(\ov{z}))$. It is easy to see that
$d_{j_s}\ldots d_{j_2}(\ov{z})=\ov{z}_i$, since for arbitrary $I_b$ with 
$s(I_b)<s=s(\ov{z})$ we have: $d_{j_s}\ldots d_{j_2}(\ov{z}_{I_b})=0$, 
or $1$, and
for $I_c\neq I_a$ with $s(I_c)=s$, $d_{j_s}\ldots d_{j_2}(\ov{z}_{I_c})$ 
is either $0$, or has degree different from $i$.
Thus, $\ov{z}_i\in C^*(G(n,Q))$. 
\Qed
\end{proof}

Let us prove by induction on the size of $\ov{z}$, that $\ov{z}$ belongs to the
subring of $C^*(G(n,Q))$ generated by $\ov{z}_j$'s. The base of induction, 
$s=1$ is trivial.
Notice that $m(\prod_{i\in I}\ov{z}_i)=\ov{z}_I$. Thus,
the size of $\ov{z}'=\ov{z}-\sum_{a:s(I_a)=
s(\ov{z})}\prod_{i\in I_a}\ov{z}_i$ is smaller than 
that of $\ov{z}$. But by the Lemma \ref{mtzi}, all the $\ov{z}_i$'s 
appearing in 
this expression belong to $C^*(G(n,Q))$. By the inductive assumption, 
$\ov{z}'$ 
belongs to the subring of $C^*(G(n,Q))$ generated by $\ov{z}_j$'s. 
Then so is $\ov{z}$. 
\Qed
\end{proof}

{\bf Remark.} Actually, for the proof of the Main Theorem one just needs
the statement of the Lemma \ref{fi}(1).

\begin{cor}
\label{arb}
For arbitrary smooth projective quadric $Q$,
$$
C^*(G(n,Q))=
\operatornamewithlimits{\otimes}_{1\leq d\leq n;\,d-\text{odd}}
(\zzz/2[\ov{z}_{d2^{l_d}}]/(\ov{z}_{d2^{l_d}}^{2^{(m_d-l_d)}})),
$$
for certain $0\leq l_d\leq m_d=[log_2(n/d)]+1$.
\end{cor}

\begin{proof}
It immediately follows from the Main Theorem \ref{mt}, 
Proposition \ref{giperb}, and the fact that $\ov{z}_s^2=\ov{z}_{2s}$
(or $0$, if $2s>n$). 
\Qed
\end{proof}

Now we can introduce:

\begin{definition}
\label{j}
\begin{itemize}
\item[$(1)$\ ]
Let $Q$ be a quadric of dimension $2n-1$.
Denote as $J(Q)$ the subset of $\{1,\ldots,n\}$ consisting of those $i$, for 
which $\ov{z}_i\in C^*(G(n,Q))$.
\item[$(2)$\ ]
Let $P$ be a quadric of dimension $2n$.
Let $Q$ be arbitrary subquadric of codimension $1$ in $P$. Then
$J(P)$ is a subset of $\{0,1,\ldots,n\}$, where $0\in J(P)$ iff 
$det_{\pm}(P)=1$ and, for $i>0$, $i\in J(P)$ iff 
$i\in J(Q|_{k\sqrt{det_{\pm}(P)}})$.
\end{itemize}
\end{definition}

\noindent\hspace{2mm}
{\bf Remark:} The definition (2) above is motivated by the fact that
$G(n+1,P)$ is isomorphic to 
$G(n,Q)\times_{\op{Spec}(k)}\op{Spec}(k\sqrt{det_{\pm}(P)})$.\\

It follows from the Main Theorem \ref{mt} that $C^*(G(n,Q))$ is
exactly the subring of $\op{CH}^*(G(n,Q)|_{\kbar})/2$ generated by
$\ov{z}_i,\,i\in J(Q)$. In particular, $J(Q)$ carries all the information
about $C^*(G(n,Q))$. Notice, that the same information is contained in
the sequence $\{l_d\}_{d-\text{odd};1\leq d\leq n}$.

What restrictions do we have on the possible values of $J(Q)$?
Because of the action of the Steenrod operations, we get:

\begin{proposition}
\label{sj}
Let $i\in J(Q)$, and $r\in\nnn$ is such that 
$\binom{i}{r}\equiv 1\,(\,mod\,2)$, and $i+r\leq n$. Then
$(i+r)\in J(Q)$.
\end{proposition}

\begin{proof}
$j$ belongs to $J(Q)$ if and only if the cycle 
$\ov{z}_j\in\op{CH}^j(G(n,Q)|_{\kbar})/2$ is defined over the base field.
Since $\ov{z}_i$ has such a property, and the Steenrod operation $\sti^r$
is defined over the base field too, we get:
$\ov{z}_{(i+r)}=\sti^r(\ov{z}_{i})$ is also defined over the base field.
\Qed
\end{proof}

The natural question arises:

\begin{question}
\label{qj}
Do we have other restrictions on $J(Q)$? 
In other words, let $J\subset\{1,\ldots,n\}$ be a subset satisfying the
conditions of Proposition $\ref{sj}$. Does there exist a quadric  
such that $J(Q)=J$?
\end{question}

It is not difficult to check that, at least, for $n\leq 4$, there is no
other restrictions.


\section{On the canonical dimension of quadrics}

In this section we will show that in the case of a {\it generic} quadric
the variety $G(n,Q)$ is $2$-incompressible, and also will
formulate the conjecture describing the {\it canonical dimension}
of arbitrary quadric. I would like to point out that the current section
would not appear without the numerous discussions with G.Berhuy,
who brought this problem to my attention.

We start by computing the characteristic classes of the variety
$G(n,Q)$.

Let $W$ be $(2n+1)$-dimensional vector space over $k$ equipped with
the nondegenerate quadratic form $q$. Let $F(r)=F(r,Q)$ be the
variety of complete flags $(\pi_1\subset\ldots\subset\pi_r)$ of 
totally isotropic subspaces in $W$. Thus, $F(0)=\op{Spec}(k)$,
$F(1)=Q$, etc. ... . We get natural smooth projective maps:
$\eps_r:F(r+1)\row F(r)$ with fibers - quadrics of dimension $2n-2r+1$.

Let $\lc_i$ be the standard subquotient linear bundles
$\lc_i:=\pi_i/\pi_{i-1}$, and $h_i=c_1(\lc_i)$. 
The bundle $\lc_i$ is defined
on $F(r)$, for $r\geq i$. These divisors $h_i$ are the roots of the
tautological vector bundle $V_n$ studied above.  

\begin{proposition}
The Chern polynomial of the tangent bundle $T_{F(r)}$ is equal to:
\begin{equation*}
c(T_{F(r)})=\frac{\prod_{1\leq i\leq r}(1-h_i)^{2n+1}}
{\prod_{1\leq i\leq r}(1-2h_i)
\cdot\prod_{1\leq i<j\leq r}\left((1-h_j+h_i)\cdot(1-h_j-h_i)\right)}.
\end{equation*}
\end{proposition}

\begin{proof}
Let $V_r$ be a tautological vector bundle on $F(r)$. Then $V_r$ is an 
isotropic subbundle of $\eta^*(W)$ ($\eta:F(r)\row\op{Spec}(k)$ is the 
projection), and on the subquotient $W_r:=V_r^{\perp}/V_r$ we have a 
nondegenerate quadratic form $q_{\{r\}}$. Then the variety $F(r+1)$ is 
defined as zeroes of this quadratic form. Thus, $F(r+1)$ is the divisor
of the sheaf $\oc(2)$ on the projective bundle $\pp_{F(r)}(W_r)$, and
we have exact sequence:
$$
0\row \oc(2)|_{F(r+1)}\row T_{\pp_{F(r)}(W_r)}|_{F(r+1)}\row T_{F(r+1)}\row 0.
$$  
On the other hand, we have sequences:
$$
0\row T_{\eps_r}\row T_{\pp_{F(r)}(W_r)}\row\eps^*(T_{F(r)})\row 0\,\,\, 
\text{and}
$$
$$
0\row\oc\row W_r\otimes\lc_{r+1}^{-1}\row T_{\eps_r}\row 0.
$$
It remains to notice, that in $K_0$, 
$[W_r]=(2n+1)[\oc]-\sum_{i=1}^r([\lc_i]+[\lc_i^{-1}])$, to get the
equality:
\begin{equation*}
c(T_{F(r+1)})=\eps_r^*(c(T_{F(r)}))\cdot
\frac{(1-h_{r+1})^{2n+1}}
{(1-2h_{r+1})\cdot\prod_{i=1}^r(1-h_{r+1}+h_i)(1-h_{r+1}-h_i)}
\end{equation*}
The statement now easily follows by induction on $r$.
\Qed
\end{proof}


Now, it is easy to compute the characteristic classes of the 
quadratic grassmanians.

\begin{proposition}
\label{cg}
\begin{equation*}
c(T_{G(r)})=\frac{\prod_{1\leq i\leq r}(1-h_i)^{2n+1}}
{\prod_{1\leq i\leq r}(1-2h_i)
\cdot\prod_{1\leq i<j\leq r}\left((1-(h_j-h_i)^2)\cdot(1-h_j-h_i)\right)},
\end{equation*}
where $h_j$ are the roots of the tautological vector bundle $V_r$ on $G(r)$. 
\end{proposition}

\begin{proof}
Consider the (forgetting) projection $\delta_r:F(r)\row G(r)=G(r,Q)$.
We have natural identification of $F(r)$ with the variety of complete
flags corresponding to the tautological bundle $V_r$ on $G(r)$
(we will permit ourselves to use the same notation for the tautological 
bundles on $G(r)$ and $F(r)$ - this is justified by the fact that they
are related by the map $\delta_r^*$). 
One readily deduces:
\begin{equation*}
c(T_{\delta_r})=\prod_{1\leq i<j\leq r}(1+h_j-h_i),
\end{equation*} 
and the statement follows.
\Qed
\end{proof}

Now we can prove the following Conjecture of G.Berhuy (proven by him for 
$n\leq 4$):

\begin{theorem}
$degree(c_{dim(G(n))}(-T_{G(n)}))\equiv 2^n\,\,(\,mod\,2^{n+1})$. 
\end{theorem}

\begin{proof}
By Proposition \ref{cg}, Chern classes of $(-T_{G(n)})$ can be expressed
as polynomials in the Chern classes of the tautological vector bundle $V_n$.
From Theorem \ref{o} we know that $c_j(V_n)=\sigma_j=(-1)^j2z_j$, where 
$z_j$ are elementary cycles defined in Proposition \ref{oprz}. 


Since, in $K_0$, $[V_n]+[V_n^{\vee}]=2n[\oc]$, we get the relations
on $\sigma_j$:

\begin{lemma}
\label{sootn}
$\sigma_i^2=2(-1)^i(\sigma_{2i}+
\sum_{1\leq j<i}(-1)^j\sigma_j\cdot\sigma_{2i-j})$.
\end{lemma}

\begin{proof}
It is just the component of degree $2i$ of the relation
$$
\left(1+\sum_i\sigma_i\right)\cdot\left(1+\sum_i(-1)^i\sigma_i\right)=1.
$$
\Qed
\end{proof}


Let $A:=\zzz[\tilde{\sigma}_1,\ldots,\tilde{\sigma}_n]$. We have ring 
homomorphism $\psi:A\row\op{CH}^*(G(n))$ sending $\tilde{\sigma}_i$ to
$\sigma_i$. It follows from the Lemma \ref{sootn}, that for arbitrary 
$f\in A$ there exists some $g\in A$
such that $g$ does not contain squares, and 
$\psi(f-g)\in 2^{n+1}\op{CH}^*(G(n))$.
If $f$ has degree $=dim(G(n))$, then $g$ got to 
be monomial $\lambda\cdot\prod_{1\leq i\leq n}\sigma_i$. 
Moreover, if $f$ was a monomial divisible by $2$, or containing square,
then $\lambda$ will be divisible by $2$. 
Consider ideal $L\subset A$ generated by $2$ and squares of elements of
positive degree.
Let $R$ be a quotient ring, and $\ffi:A\row R$ be the projection.

Since $\prod_{1\leq i\leq n}\sigma_i=
(-1)^{n(n+1)/2}
2^n\prod_{1\leq i\leq n}z_i$, and $\prod_{1\leq i\leq n}z_i$ is the
class of a rational point (by Proposition \ref{oprz}), we get that for 
arbitrary $f\in A$, the 
$degree(\psi(f))$ is divisible by $2^n$, and  for $f\in L$ 
the degree is divisible by $2^{n+1}$.
Thus, modulo $2^{n+1}$, the degree of $\psi(f)$ depends only on $\ffi(f)$.

In $R$ we have the following equalities:
$$\ffi(\prod_{1\leq i<j\leq n}(1-(h_i-h_j)^2))=1,\,\,\,\text{and}$$
$$\ffi(\prod_{1\leq i<j\leq n}(1-h_i-h_j))=
\ffi(\prod_{j=1}^{n-1}(1+\sum_{1\leq s\leq j}\sigma_s)).$$
Also, clearly, $\ffi((1+f)^2)=1$ for any $f$ of positive degree.
Thus,
$$
\ffi(c(-T_{G(n)}))=\ffi(\prod_{j=1}^n(1+\sum_{1\leq s\leq j}\sigma_s)).
$$
And the component of this polynomial of dimension $=dim(G(n))$ is just
$\ffi(\prod_{1\leq i\leq n}\sigma_i)$. Consequently,
$degree(c_{dim(G(n))}(-T_{G(n)}))\equiv 2^{n}\,\,(\,mod\,2^{n+1})$.

\Qed
\end{proof}

We recall from \cite{Me} that a variety $X$ is $p$-compressible if
there is a rational map $X\dashrightarrow Y$ to some variety $Y$ such that
$dim(Y)<dim(X)$ and $v_p(n_X)\leq v_p(n_Y)$, where $n_Z$ is the
image of the degree map $deg:\op{CH}_0(Z)\row\zzz$.

From the Rost degree formula (\cite[Theorem 6.4]{Me}) for the characteristic 
number $c_{dim(G(n))}$ modulo $2$ 
(see \cite[Corollary 7.3, Proposition 7.1]{Me}), we get:

\begin{proposition}
\label{inc}
Let $Q$ be a smooth $2n+1$-dimensional quadric, all splitting fields of which 
have degree divisible by $2^n$ (we call such $Q$ - generic). 
Then the variety $G(n,Q)$ is $2$-incompressible.
\end{proposition}

\Qed


Call two smooth varieties $X$ and $Y$ equivalent if there are
rational maps $X\dashrightarrow Y$ and $Y\dashrightarrow X$.
Then let $d(X)$ be the minimal dimension of a variety equivalent to $X$.  
Recall from \cite{BR} that a {\it canonical dimension} $cd(q)$ of a 
quadratic form $q$ is defined as $d(G(n,Q))$, where $n=[dim(q)/2]+1$. 

Proposition \ref{inc} gives another proof of the fact
that the canonical dimension of a generic $(2n+1)$-dimensional form
is $n(n+1)/2$, which computes the canonical dimension of the
groups $SO_{2n+1}$ and $SO_{2n+2}$ (cf. \cite[Theorem 1.1, Remark 1.3]{Ka}). 

Our computations of the generic discrete invariant $GDI(m,Q)$ permit to
conjecture the answer in the case of arbitrary smooth quadric $Q$:

\begin{conjecture}
\label{cdq}
Let $Q$ be smooth projective quadric of dimension $d$. Then
$$
cd(Q)=\operatornamewithlimits{\sum}_{j\in\{1,\ldots,[d+1/2]\}\backslash J(Q)}
j, 
$$
where
$J(Q)$ is the invariant from the Definition \ref{j}.
\end{conjecture} 

If $Q$ is generic, then $J(Q)$ is empty, and $cd(Q)$ is indeed
equal $\sum_{1\leq i\leq n}i=n(n+1)/2$.


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\end{document}

