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\dateposted{April 1, 1999}
\PII{S 1079-6762(99)00056-6}
\copyrightinfo{1999}{American Mathematical Society}







\title{On quantum de Rham cohomology theory}

\author{Huai-Dong Cao}
\address{Department of Mathematics,
Texas A\&M University,
College Station, TX 77843}
\thanks{Authors' research was supported in part by NSF grants DMS-96-32028
and DMS-95-04925}

\author{Jian Zhou}
\address{Department of Mathematics,
Texas A\&M University,
College Station, TX 77843}

\subjclass{Primary 53C15, 58A12, 81R05}
\date{May 07, 1998}

\commby{Richard Schoen}

We define the quantum exterior product $\wedge_h$ 
and quantum exterior differential $d_h$
on  Poisson manifolds. The
quantum de Rham cohomology,
 which is a deformation quantization of the de Rham cohomology,
is defined as the cohomology of $d_h$.
We also define the quantum Dolbeault cohomology. 
A version of quantum integral on symplectic manifolds is considered and the 
corresponding quantum Stokes theorem is stated. We also derive the quantum hard 
Lefschetz theorem.
By replacing $d$ by $d_h$ and 
$\wedge$ by $\wh$ in the usual definitions, 
we define many quantum analogues of
important objects in differential geometry, e.g. 
quantum curvature. 
The quantum characteristic classes are then studied 
along the lines of the classical Chern-Weil theory. The
quantum equivariant de Rham cohomology is defined 
in the similar fashion.


In this note we announce a construction of a deformation of
the de Rham complex for any Poisson manifold.
In the case of a closed symplectic manifold,
its cohomology provides a deformation of the ring structure   
 on the de Rham cohomology.  
More precisely, on any Poisson manifold, we define a quantum
exterior product $\wh$ of exterior forms,
and quantum exterior differential $d_h$, such that $d_h ^ 2 = 0$,
and $d_h$  is a derivation for $\wh$. Here $h$ is an indeterminate.
We define the {\em quantum de Rham cohomology} as the cohomology of $d_h$.
Since $d_h$ is a derivation with respect to $\wh$,
there is an induced quantum multiplication on the quantum de Rham

This work grew out of our attempt to find a new way
to define quantum cohomology, which has recently attracted much
attention (see Tian \cite{Tia} for a survey on this topic,
and the introduction of Li-Tian
\cite{Li-Tia} for more recent development).
Intuitively, quantum cohomology  provides a deformation of the  ring
structure on the vector space
which underlies the de Rham cohomology by counting
pseudo-holomorphic curves in symplectic manifolds
or stable curves in  algebraic manifolds.
Unlike many cohomology theories in algebraic topology,
it is not defined as the cohomology of a graded differential algebra.
For the sake of being consistent with
other cohomology theories,
it would be  desirable to be able to do so,
even though the applications of quantum cohomology do not really require
this property. This is the first motivation for our construction of quantum 
de Rham cohomology. On the other hand,
from the point of view of finding deformations of
the ring structure on the de Rham cohomology {\em per se},
it is also interesting to see whether deformations of the de Rham complex
can provide new deformations of the de Rham cohomology. This is the 
second motivation to our construction.

Simple examples, such as flat tori and complex projective
spaces, show that the quantum de Rham cohomology does give nontrivial
deformation of the ring structure on the vector space underlying the de
Rham cohomology. However, such examples also show that it is 
different from the quantum cohomology.
Nevertheless, they show that the quantum de Rham cohomology
provides new deformations of
the de Rham cohomology: e.g.  for a symplectic flat torus, the
quantum cohomology does not give non-trivial deformations,
while the quantum de Rham cohomology does.
It is conceivable that the quantum de Rham cohomology turns out to be
different from the quantum cohomology, since it does not have ``strings" in it.
We speculate that a version of quantum de Rham cohomology on the
loop space might yield the quantum cohomology,
based on the description of quantum cohomology by Vafa \cite{Vaf}.

While it is disappointing that we cannot obtain an
easier definition of quantum cohomology 
along the lines of ordinary de Rham cohomology this way,
in retrospect,
the significance of our work is that it provides a very
simple version of quantum differential geometry,
in the sense that our objects contain a parameter $h$,  
which give us the classical objects in differential geometry when
$h = 0$.
For example, by replacing $d$ by $d_h$ and
$\wedge$ by $\wh$ in the usual definitions,
we define many quantum analogues of
important objects, such as quantum curvature
and quantum characteristic classes,
in differential geometry.
We believe this kind of quantum differential geometry should be useful
in formulating  quantum theories in physics.
Notice that this is quite different from Connes' non-commutative geometry.
We are in the process of investigating a relationship between the two.

The definition of the quantum exterior product $\wh$
is motivated by the Moyal-Weyl multiplication and Clifford multiplication.
For any finite dimensional vector space $V$, 
let  $\{ e_1, \cdots, e_m \}$ be a basis 
of $V$ and $\{ e^1, \cdots, e^m \}$ the dual basis.
Assume that $w = w^{ij} e_i \otimes e_j \in V \otimes V$.
Then $w$ defines a multiplication $\wedge_w$ on $\Lambda(V^*)$,
and a multiplication $*_w$ on $S(V^*)$,
such that $e^i \wedge_w e^j =   e^i \wedge e^j + w^{ij}$,
$e^i *_w e^j = e^i \odot  e^j + w^{ij}$.
If $w \in S^2(V)$, then $\wedge_w$ is the Clifford multiplication.
If $w \in \Lambda^2(V)$ is nondegenerate, $*_w$ is 
the Moyal-Weyl multiplication.
If $w \in \Lambda^2(V)$, then $\wedge_w$ is what we call the  {\em quantum 
exterior product}.
It is elementary to show that this multiplication is associative,
and we get a {\em quantum exterior algebra}. 
We use it to obtain a  quantum calculus on  any Poisson manifold.
In a way, quantum exterior algebra plays a role in Poisson or symplectic
geometry similar to that of Clifford algebra in Riemannian geometry.
Based on the success of Clifford algebra in Riemannian geometry,
we expect that the quantum exterior algebra should be useful in 
Poisson or symplectic geometry.

Finally, let us point out some difference of our work with deformation quantizations
or star product, of symplectic or Poisson manifolds.
Let $M$ be a smooth manifold.
It is well-known that ${\cal A} = C^{\infty}(M)$ cannot be
deformed non-trivially by commutative algebras.
But now it is known through the work of Kontsevich  \cite{Kon} that it
is always possible to deform it by non-commutative algebras
if a Poisson bivector field $w$ on $M$ is given.
More precisely, there is a noncommutative multiplication $*$ on ${\cal A}[[h]]$
such that $({\cal A}[[h]], *)$ is an  associative algebra with unit,
\begin{equation*}f*g = fg + \sum_{n \geq 1} h^n B_n(f, g),\end{equation*}
such that
\lim_{h \rightarrow 0} \frac{f*g -g*f}{h} = \{f, g\},\end{equation*}
where $\{\cdot, \cdot\}$ is the Poisson bracket defined by $w$.
For some earlier results on star products on symplectic manifolds, see  
Bayen {\em et al} \cite{Bay}, De Wilde-Lecomte \cite{DeW-Lec},
Fedosov \cite{Fed}.
In general a $\bZ$-graded commutative algebra may have a deformation by
$\bZ_2$-graded commutative algebras, e.g. the quantum cohomology of 
Our quantum exterior product $\wedge_h$ defines
a $\bZ_2$-graded commutative, associative 
multiplication on $\Omega^*(M)[h]$,
and is trivial on $\Omega^0(M) = C^{\infty}(M)$. 
(It becomes $\bZ$-graded commutatively if we regard $h$ as
an element of grade $2$.)

We will present the  main results of our work below. Their proofs will appear
elsewhere \cite{Cao-Zho}.

%\noindent {\bf 1. Quantum exterior algebra}. 
\section*{1. Quantum exterior algebra}
Let $V$ be a finite dimensional vector space over a 
field $\bk$ of characteristic zero,
and $\Lambda(V^*)$ the exterior algebra generated by the dual vector space
For any $v \in V$ and $\alpha \in \Lambda^k(V^*)$, denote
(v \vdash \alpha ) (v_1, \cdots, v_{k-1}) & = &
	 \alpha (v, v_1, \cdots, v_{k-1}), \\
(\alpha \dashv v) (v_1, \cdots, v_{k-1}) & = & 
	\alpha (v_1, \cdots, v_{k-1}, v),
for $v_1, \cdots, v_{k-1} \in V$.
Let $\Lambda_h(V^*) = \Lambda(V^*)[h] = \Lambda(V^*) \otimes_{\bk} \bk[h]$.
For any  $w \in \Lambda^2(V)$, with $w =\sum_{i,j}w^{ij}e_i \wedge e_j$ with respect to a basis
$\{ e_1, \cdots, e_m \}$ of $V$, we define the {\em quantum exterior product} 
$\wedge_{h,w}: \Lambda(V^*) \otimes \Lambda(V^*) \rightarrow 
	\Lambda(V^*)[h]$ by
    \alpha \wedge_{h, w} \beta 
 =  \sum_{n \geq 0} 
	\frac{h^n}{n!} w^{i_1j_1} \cdots w^{i_nj_n}
	(\alpha \dashv e_{i_1} \dashv \cdots \dashv e_{i_n})
	\wedge (e_{j_n} \vdash \cdots \vdash e_{j_1} \vdash \beta), 
for $\alpha, \beta \in \Lambda(V^*)$.
This definition is evidently independent of the choice of the basis $\{ e_1, \cdots, e_m \}$.
We extend $\wedge_{h, w}$ as a $\bk[h]$-module map to 
$\Lambda_h(V^*) \otimes_{\bk[h]} \Lambda_h(V^*)$.
When there is no confusion about $w$, we will simply write
$\alpha \wh \beta$ for $\alpha \wedge_{h, w} \beta$.
We are interested in $\alpha \wedge_w \beta$, which is just 
$\alpha \wedge_{1, w} \beta$.
We assign to $h$ the degree $2$. Then $\Lambda_h(V^*)$ 
has a natural $\mathbb Z$-grading.
Denote by $\Lambda_h^{[n]}(V^*)$ the subspace of homogeneous elements
of degree $n$; then it is clear that
\begin{equation*}\Lambda_h^{[m]}(V^*) \wedge_h \Lambda_h^{[n]}(V^*) \subset

\begin{theorem}[Cao-Zhou \cite{Cao-Zho}] \label{Thm:algebraic}
The quantum exterior product satisfies the following properties:
& & \alpha \wh \beta = (-1)^{|\alpha||\beta|} \beta \wh \alpha, 
	\label{com} \\
& & (\alpha \wh \beta) \wh \gamma = \alpha \wh (\beta \wh \gamma), 
for  all $\alpha, \beta, \gamma \in \Lambda_h(V^*)$.
Therefore, $(\Lambda_h(V^*), \wedge_h)$ is a deformation quantization
of the exterior algebra $(\Lambda(V^*), \wedge)$.

The proof of $(\ref{com})$ is trivial. The proof of $(\ref{ass})$ is of 
elementary nature but non-trivial.  It is proved first by brute force in the 
case of $\deg (\alpha) = 1$, and
then by induction on $\deg (\alpha)$ (see \cite{Cao-Zho} for details).

We can  also extend $\wh$ to $\Lambda_{h, h^{-1}}(V^*) 
= \Lambda(V^*)[h, h^{-1}] = \Lambda(V^*) \otimes_{\bk} \bk[h, h^{-1}]$. 

An algebra $A$ with unit $e \in A$ 
over a field $\bk$ is called a {\em $\bk$-Frobenius algebra} 
if there is a nondegenerate symmetric 
$\bk$-bilinear function
$\langle\cdot, \cdot\rangle: A \times A \rightarrow \bk$,
such that
\begin{equation*}\langle\alpha \beta, \gamma\rangle = \langle\alpha, \beta \gamma\rangle,
for all $\alpha, \beta, \gamma \in A$.
There is a simple way to construct a structure of Frobenius 
algebra on any $\bk$-algebra $A$ with unit. 
Let $\phi: A \rightarrow \bk$ be a nonzero $\bk$-functional 
on $A$.
Set $\langle\alpha, \beta\rangle_{\phi} = \phi(\alpha\beta)$ 
for $\alpha, \beta \in A$.
If it is nondegerate, then $(A, \langle\cdot, \cdot\rangle_{\phi})$ is
a $\bk$-Frobenius algebra.
Conversely, given any Frobenius algebra 
$(A, \langle\cdot, \cdot\rangle)$,
let $\phi(\alpha) = \langle\alpha, e\rangle$, for $\alpha \in A$;
then $\langle\cdot, \cdot\rangle = \langle\cdot, \cdot\rangle_{\phi}$.
Now on $\Lambda(V^*)$, consider a Berezin integral 
$\int: \Lambda(V^*) \rightarrow \bk$ ( i.e., a $\bk$-linear
functional which is only nonzero on $\Lambda^{top}(V^*)$).
Then it is clear that $\langle\alpha, \beta\rangle = \int \alpha \wedge_w \beta$
defines a structure of Frobenius algebra on $(\Lambda(V^*), \wedge_w)$.

\begin{example} \label{ex:2D}
Let $V$ be a two dimensional vector space with a basis $\{e_1, e_2\}$,
and dual basis $\{ e^1, e^2\}$.
Let $w = e_1 \wedge e_2$.
Then we have
&& e^1 \wh e^2 = e^1 \wedge e^2 + h, \\
&& e^1 \wh (e^1 \wedge e^2) = - h e^1, \\
&& e^2 \wh (e^1 \wedge e^2) = - h e^2, \\
&& (e^1 \wedge e^2) \wh (e^1 \wedge e^2) = - 2h e^1 \wedge e^2 - h^2.
It is a tedious but straightforward exercise to check the associativity.

\begin{example} \label{ex:symplectic}
Let $V$ be a $2n$-dimensional vector space with a basis 
$\{e_1, \cdots, e_{2n}\}$
and dual basis $\{e^1, \cdots, e^{2n}\}$.
Let $w = e_1 \wedge e_2 + \cdots + e_{2n-1} \wedge e_{2n}$,
and $\omega = e^1 \wedge e^2 + \cdots + e^{2n-1} \wedge e^{2n}$.
\begin{equation*}\omega_h = e^1 \wh e^2 + \cdots + e^{2n-1} \wh e^{2n}
= e^1 \wedge e^2 + \cdots + e^{2n-1} \wedge e^{2n} + n h.\end{equation*}
Then clearly we have
\begin{equation*}(\omega_h)_h^{n+1} : 
= \underbrace{\omega_h \wh \cdots \wh \omega_h}_{\textup{$n$ times}} = 0.\end{equation*}
In particular, when $n = 1$,
we get $(\omega + h) \wh (\omega + h) = 0$,
\begin{equation*}\omega \wh \omega = - 2h\omega - h^2,\end{equation*}
which recovers the last formula in Example 1.

%\noindent {\bf 2. Quantum de Rham complex}. 
\section*{2. Quantum de Rham complex}
Let $(P, w)$ be a  Poisson manifold, with bivector field $w$, 
whose Schouten-Nijenhuis bracket vanishes. 
(See Vaisman \cite{Vai} for definitions.) 
The fiberwise quantum exterior multiplication defines 
& \wh: \Omega_h(M) \otimes \Omega_h(M) \rightarrow \Omega_h(M), \\
& \wh: \Omega_{h, h^{-1}}(M) \otimes \Omega_{h, h^{-1}}(M) 
	\rightarrow \Omega_{h, h^{-1}}(M),
where $\Omega_{h}(M) 
= \Omega(M)[h] = \Omega(M) \otimes_{\bk} \bk[h]$
and $\Omega_{h, h^{-1}}(M) 
= \Omega(M)[h, h^{-1}] = \Omega(M) \otimes_{\bk} \bk[h, h^{-1}]$.
Koszul \cite{Kos} defined an operator 
$\delta: \Omega^{k}(M) \rightarrow \Omega^{k-1}(M)$ by
\begin{equation*}\delta \alpha = w \vdash d \alpha - d (w \vdash \alpha),\end{equation*}
for $\alpha \in \Omega^k(M)$.
He also showed that $\delta^2 = 0$, 
$d\delta + \delta d = 0$.
We define the quantum exterior differential $d_h = d - (h/2) \delta: \Omega(M)[h] 
\Omega(M)[h]$, and similarly on $\Omega(M)[h,h^{-1}]$.
Then it is easy to see that $d_h^2 = 0$.
One of the technical results in Cao-Zhou \cite{Cao-Zho} is the following

\begin{theorem} \label{Thm:main}
For any Poisson manifold $(M, w)$, 
$d_h$ satisfies 
\begin{eqnarray} \label{derivation2}
d_h(\alpha \wh \beta) = 
	(d_h \alpha) \wh \beta + 
	(-1)^{|\alpha|}\alpha \wh (d_h \beta) 
for $\alpha, \beta$ in $\Omega(M)[h]$,  or 
 $\alpha, \beta$ in $\Omega(M)[h, h^{-1}]$. 

This is proved first for $\deg (\alpha) = 1$ by brute force,
then by induction  on $\deg (\alpha)$.
For regular Poisson manifolds (e.g., symplectic manifolds),
there is an easier proof.
On such Poisson manifolds, there always exists a torsionless connection
$\nabla$ which preserves $w$.
Then for any local frame $\{ e^1, \cdots, e^n \}$, 
and $\alpha \in \Omega(M)$, we have 
\begin{eqnarray} \label{eqn:connection}
d_h \alpha = e^i \wh \nabla_{e_i} \alpha.
This is the analogue of a similar expression for $d + d^*$ in
Riemannian geometry (Lawson-Michelsohn \cite{Law-Mic}, Lemma II.5.13).
Using the analogue of normal coordinates, 
the proof of Theorem \ref{Thm:main} reduces to 
the associativity of the quantum exterior multiplication.

%\noindent {\bf 3. Quantum de Rham cohomology}.
\section*{3. Quantum de Rham cohomology}

For any Poisson manifold $(M, w)$, 
the {\em (polynomial) quantum de Rham cohomology} is defined
\begin{equation*}Q_hH_{dR}^*(M) = \Ker d_h / \Img d_h,\end{equation*}
for the quantum exterior differential $d_h: \Omega(M)[h] \rightarrow \Omega(M)[h]$.
The {\em Laurent quantum de Rham cohomology} is
\begin{equation*}Q_{h, h^{-1}}H_{dR}^*(M) = \Ker d_h / \Img d_h,\end{equation*}
for  $d_h: \Omega(M)[h, h^{-1}] \rightarrow 
\Omega(M)[h, h^{-1}]$.
As a consequence of Theorem \ref{Thm:algebraic} and Theorem \ref{Thm:main}, 
we have
\begin{theorem} The quantum de Rham cohomology 
$Q_hH_{dR}^*(M)$  of a Poison manifold  
has the following properties:
\alpha \wh \beta & = & (-1)^{|\alpha||\beta|}\beta \wh \alpha, \\
(\alpha  \wh \beta) \wh \gamma & = & \alpha \wh (\beta \wh \gamma),
for $\alpha, \beta, \gamma \in Q_hH^*_{dR}(M)$.
Similar results hold for the Laurent quantum de Rham cohomology.

The complex $(\Omega(M)[h], d_h)$ can be regarded as a double complex
$(C^{p, q}, -h\delta/2, d)$, where
$C^{p,q} = h^p\Omega^{q-p}(M)$, $p \geq 0$.
This is the analogue of Brylinski's double complex ${\cal C}_{..}(M)$ 
(\cite{Bry}, $\S 1.3$).
By the standard theory for a double complex
(Bott-Tu \cite{Bot-Tu}, $\S 14$), 
there are two spectral sequences $E$ and $E'$ abutting to 
$H^*(\Omega[h], d_h) = Q_hH^*_{dR}(M)$,
with $E_1^{p, q} = h^pH^q(C^{p, *}, d) = h^pH^{q-p}_{dR}(M)$,
$(E_1')^{p, q} = h^pH^*(C^{*, q}, \delta) = h^pPH_{q-p}(M)$, $p \geq 0$. 
Since nontrivial $E_1^{p, q}$ all have $p+ q$ even,
and the differential $d_r$ changes the parity of $p+q$, 
it is routine to prove the following

\begin{theorem} \label{Thm:spectral1}
For a Poisson manifold with odd Betti numbers all vanishing,
the spectral sequence $E$ degenerates at $E_1$, i.e. $d_r = 0$ for
all $r \geq 0$. Hence
$Q_hH^*_{dR}(M)$ is a deformation quantization of $H^*_{dR}(M)$.

Brylinski \cite{Bry} proved that on a closed K\"{a}hler manifold $(M, \omega)$,
every de Rham cohomology class has a representative $\alpha$ such that
$d \alpha = 0$, $\delta \alpha = 0$.
This implies the following

For a closed K\"{a}hler manifold $M$,
the spectral sequence $E$ degenerates at $E_1$, i.e. $d_r = 0$ for
all $r \geq 0$. Hence
$Q_hH^*_{dR}(M)$ is a deformation quantization of $H^*_{dR}(M)$.

Similarly, we regard $(\Omega(M)[h, h^{-1}], d - h \delta/2)$ 
 as a double complex 
$(\widetilde{C}^{p, q}, -h\delta/2, d)$,
where $\widetilde{C}^{p, q} = h^p\Omega^{q-p}(M)$, 
$p, q \in {\mathbb Z}$. 
This is essentially Brylinski's double complex 
${\cal C}_{..}^{per}$, but with a different bigrading.
We get two spectral sequences $\tilde{E}$ and  $\tilde{E}'$
abutting to $Q_{h, h^{-1}}H^*_{dR}(M)$, 
with $\tilde{E}_1^{p, q} = h^pH^{q-p}_{dR}(M)$, 
$(\tilde{E}_1')^{p, q} = h^pH^*(C^{*, q}, \delta) = h^pH_{q-p}(M)$,
$p, q \in {\mathbb Z}$. 
It is clear that an analogue of Theorem \ref{Thm:spectral1}
holds for $\tilde{E}$.
On the other hand, by a method of Brylinski \cite{Bry}, 
one can prove the following

\begin{theorem} \label{thm:degeneracy}
For any compact symplectic manifold without boundary,
the spectral sequences $\tilde{E}$ and $\tilde{E}'$ degenerate 
at $\tilde{E}_1$ and $\tilde{E}_1'$, respectively.
$Q_{h, h^{-1}}H^*_{dR}(M)$ is a Laurent 
deformation quantization of $H^*_{dR}(M)$.

Fixing an isomorphism $H^{2n}_{dR}(M) \cong {\mathbb R}$ then defines
a structure of Frobenius algebra on $(H^*_{dR}(M), \wedge_w)$.

It would be interesting to know whether Theorems 
\ref{Thm:spectral1}--\ref{thm:degeneracy} hold
for regular Poisson manifolds, or even general Poisson manifolds.

%\noindent {\bf 4. Some examples}.
\section*{4. Some examples}
Our first example is the flat symplectic torus $(T, \omega)$.
By virtue of Theorem \ref{thm:degeneracy}, 
$Q_{h, h^{-1}}H^*_{dR}(T)$ is isomorphic to $H^*(T) \otimes \bR[h, h^{-1}]$.
Picking up a flat Riemannian metric on $T$ which is compatible with
we can represent de Rham cohomology classes by harmonic forms on $T$.
But all these forms are parallel, hence,
by (\ref{eqn:connection}), $d_h$-closed. 
The quantum exterior product of any two such forms can be found
by restricting to a point on the torus. 
Therefore, the ring structure on quantum de Rham cohomology
is the quantum exterior algebra for the tangent space of any point on the torus.
The case of a $2$-torus can be explicitly described by Example \ref{ex:2D}.
Notice that since the second homotopy group of a torus is trivial,
there is no nontrivial pseudo-holomorphic $S^2$ in a torus.
Hence the quantum cohomology of a symplectic torus is trivial.
But the quantum de Rham cohomology is obviously nontrivial.

Similarly, for a complex projective space $\bC\bP_n$ with standard 
K\"{a}hler structure,
both ordinary and quantum de Rham cohomology is generated by the symplectic 
form $\omega$.
Again since harmonic forms are parallel, we can reduce the calculation to
the tangent space of a point, which is 
a symplectic vector space.
Then the result of Example \ref{ex:symplectic} can be used.
It is clear now 
that the quantum cohomology and quantum de Rham cohomology produce
different deformations:
e.g. on $\bC\bP_1$,
the quantum cohomology is 
\begin{equation*}\bR[\omega][q]/(\omega^2 - q),\end{equation*}
while the quantum de Rham cohomology is
\begin{equation*}\bR[\omega][h]/(\omega^2 + 2h \omega + h^2).\end{equation*}

%\noindent {\bf 5. Quantum Hard Lefschetz Theorem}.
\section*{5. Quantum Hard Lefschetz Theorem}
For a closed symplectic manifold, the analogue of the Hard Lefschetz
Theorem (see \cite{Gri-Har}) holds for $Q_{h, h^{-1}}H^*_{dR}(M)$.
We begin with a $2n$-dimensional symplectic vector space $(V, \omega)$.
Brylinski \cite{Bry} defined a symplectic star operator 
$*: \Lambda^k(V^*) \rightarrow \Lambda^{2n-k}(V^*)$.
We can extend it to $\Lambda_{h, h^{-1}}$ by setting $*h = h^{-1}$,
and $*h^{-1} = h$.
Define $L_h: \Lambda_{h, h^{-1}}(V^*) \rightarrow \Lambda_{h, h^{-1}}(V^*)$
by $L_h(\alpha) = \omega \wh \alpha$.
Define $L_h^* = -*L*$, and $A_h: \Lambda_{h, h^{-1}}(V^*) \rightarrow 
	\Lambda_{h, h^{-1}}(V^*)$
by $A_h(\alpha) = (n - k) \alpha$, 
for $\alpha \in \Lambda^{[k]}_{h, h^{-1}}(V^*)$.
Then we have

The following identities hold:
[L_h, L_h^*] = 0, & [L_h, A_h] = 2 L_h, & [L^*_h, A_h] = - 2 L^*_h.
Furthermore, if we regard multiplications by $h$ and $h^{-1}$
as operators, then we have 
[h, h^{-1}] = 0, & [L_h, h^{\pm 1}] = [L_h^*, h^{\pm 1}] = 0, &
[A_h, h^{\pm}] = \pm 2 h^{\pm 1}.

Thus we cannot use the representation theory of $sl(2, {\mathbb C})$
as in the classical theory.
Notice that multiplication by $h$ is an isomorphism, whose inverse is
multiplication by $h^{-1}$.
Let $M_h = h^{-1}L_h$, $M_h^* = hL_h^*$; then $M_h, M_h^*, A_h$
form an abelian algebra. 
Notice that it now suffices to find the eigenvalues of $M_h$ 
on $\Lambda^{[0]}_{h, h^{-1}}(V^*)$ and $\Lambda^{[1]}_{h, h^{-1}}(V^*)$.
In Cao-Zhou \cite{Cao-Zho},
this is done by induction on the dimension of $V$.
In suitable bases,
the matrix $M_n$ of $M_h$ for $V$ (of dimension $2n$) 
can be expressed in terms of
the matrix $M_{n-1}$ of $M_h$ for a symplectic subspace of dimension $2n - 2$.
Essentially, it is of the form
M_{n} = \left( \begin{array}{cccc} 
M_{n-1} & -I & 0 & 0 \\
\\I & M_{n-1} + 2I & 0 & 0 \\
0 & 0 & M_{n-1} + I & 0 \\
0 &0 & 0 & M_{n-1} + I
\end{array} \right).
For details, see Cao-Zhou \cite{Cao-Zho}.
We have the following 

 Let $\{ M_n \}$ be a sequence of square 
matrices with coefficient in $\bk$,
obtained in the following way:
M_{n+1} = \left( \begin{array}{cc} 
M_n & -I \\ I & M_n + 2I \end{array} \right)
for $n \geq 1$,
where $I$ is the identity matrix of the same size as $M_n$.
For any $\lambda \in \bk$, and $n \geq 1$, we have
\begin{equation*}\det (M_{n+1} + \lambda I)  
=  \det [M_n + (\lambda +1)I]^2.\end{equation*}
Therefore, the eigenvalues of $M_{n+1}$ can be 
obtained by adding $1$ to those of $M_n$,
with the multiplicities doubled.

As a consequence,
we find that the eigenvalues of $M_h$ on $\Lambda_{h, h^{-1}}(V^*)$
are $n$ and $n \pm \sqrt{5}/2$, when $\dim (V) = 2n$.
For $M_h^*$, they are $-n$ and $-n \pm \sqrt{5}/2$. Therefore, we have

For a symplectic vector space $V$, the operators $L_h$ and
$L_h^*$ are  isomorphisms.
Furthermore, $\Lambda_{h, h^{-1}}(V^*)$ decomposes 
into one dimensional eigenspaces of $h^{-1}L_h$ 
(or $hL_h^*$) with nonzero eigenvalues.

On a symplectic manifold $(M, \omega)$, we have
[L_h, d_h] = 0, & [L_h^*, d_h] = 0, & [A_h, d_h] = -d_h.

As a consequence, we get:

\begin{theorem}[Quantum Hard Lefschetz Theorem]
For any symplectic manifold \linebreak $(M^{2n},  \omega)$,
its Laurent quantum de Rham cohomology 
$Q_{h, h^{-1}}H^*_{dR}(M)$ decomposes into 
one-dimensional eigenspaces of  the operator $h^{-1}L_h$
(or $hL_h^*$) with nonzero eigenvalues. 
In particular, $L_h$ and $L_h^*$ are isomorphisms.

In classical algebraic geometry, the Hard Lefschetz Theorem can be proved
by considering the Lie algebra generated by 
the operator $L$ given by multiplication 
with the symplectic form $\omega$ and its adjoint $\Lambda$ 
by commutators. 
The above theorem get its name since 
we consider $L_h$ given by the quantum exterior
product with $\omega$.
It is not related to the symplectic version of the
Hard Lefschetz Theorem proved by Mathieu \cite{Mat} and Yan \cite{Yan},
which does not hold for all closed symplectic manifolds.

%\noindent {\bf 6. Complexified quantum exterior algebra}.
\section*{6. Complexified quantum exterior algebra}
We also consider real vector space $V$ with an
almost complex structure $J\in \End (V)$
 such that $J^2 = - Id$.
There is an induced linear transformation 
$\Lambda^2J: \Lambda^2(V) \rightarrow \Lambda^2(V)$.
For any bivector $w \in \Lambda^2(V)$,
$J$ is said to preserve $w$ if 
$\Lambda^2J (w) = w$.
Given any bivector $w$ which is preserved by $J$, 
we can define the quantum exterior product on 
$\Lambda_h(V^*)$ as in the last section.
Now if we tensor everything by $\mathbb C$,
we get a complex algebra ${\mathbb C}\Lambda_h(V^*)$, which is
a deformation quantization of ${\mathbb C}\Lambda(V^*) :=
\Lambda(V^*) \otimes_{\mathbb R} {\mathbb C} 
= \Lambda_{\mathbb C}(V^* \otimes_{\mathbb R} {\mathbb C})$.  As in complex geometry, 
we can exploit a natural decomposition as follows. $J$ can be uniquely extended to a 
complex linear endomorphism, denoted also by $J$, of  ${\mathbb C}V$ also satisfying 
$J^2 = - Id$. There is a natural identification of complex vector
spaces  ${\mathbb C}V \cong V^{1, 0} \oplus V^{0, 1}$ , where $V^{1, 0}$ and $V^{0, 1}$
are eigenspaces of $J$ with eigenvalues $\sqrt {-1}$ and $-\sqrt {-1}$ respectively.
As a consequence, there are decompositions
{\mathbb C}\Lambda(V) & = & \bigoplus_{p, q} \Lambda^{p, q}(V), \\
{\mathbb C}\Lambda(V^*) & = & \bigoplus_{p, q} \Lambda^{p, q}(V^*),
where $\Lambda^{p, q}(V) \cong \Lambda_{\mathbb C}^p(V^{1, 0})
\otimes_{\mathbb C} \Lambda^q_{\mathbb C}(V^{0, 1})$,
and $\Lambda^{p, q}(V^*) \cong \Lambda_{\mathbb C}^p((V^{1, 0})^*)
\otimes_{\mathbb C} \Lambda^q_{\mathbb C}((V^{0, 1})^*)$.
We give ${\mathbb C}\Lambda_h(V^*)$ the following 
${\mathbb Z} \times {\mathbb Z}$-bigrading: 
elements in $\Lambda^{p, q}(V^*)$
has bidegree $(p, q)$,  and $h$ has bidegree $(1, 1)$.
Since $w$ is preserved by $J$, it belongs to $\Lambda^{1, 1}(V)$
after complexification.
Denote by $\Lambda_h^{[p, q]}(V^*)$ the space of homogeneous
elements of bidegree $(p, q)$.
It is then straightforward to see that
\begin{equation*}\Lambda_h^{[p, q]}(V^*) \wh \Lambda_h^{[r, s]}(V^*)
	\subset \Lambda_h^{[p+r, q+t]}(V^*).\end{equation*}

Now let  $\omega$ be a symplectic 
form on $V$ which is compatible with an almost complex structure
on $V$. Namely, rank of $\omega$ is $2n = \dim (V)$, 
$w(J\cdot, J\cdot) = \omega(\cdot, \cdot)$,
and $g(\cdot, \cdot) := \omega(\cdot, J\cdot)$ is 
a positive definite element of $S^2(V^*)$.
Then $\omega$ induces a natural Hermitian metric $H$ on 
${\mathbb C}\Lambda(V^*)$, such that
\begin{equation*} H(\alpha \ww \beta, \gamma)  
= H(\alpha, \beta \ww \gamma)\end{equation*}
for any $\alpha, \beta, \gamma \in {\mathbb C}\Lambda(V^*)$.
Here $w \in \Lambda^2(V)$ is obtained from $\omega$ by 
``raising the indices" (for details, see Cao-Zhou \cite{Cao-Zho}, $\S 1.5$). 
This shows that the algebra $({\mathbb C}\Lambda(V^*), \wedge_w)$ 
has a structure of Hermitian Frobenius algebra.

%\noindent {\bf 7. Quantum Dolbeault cohomology}.
\section*{7. Quantum Dolbeault cohomology}
On a complex manifold $(M, J)$ with a Poisson structure $w$, 
such that $J$ preserves $w$,
we define $\delta^{-1, 0}: \Omega^{p, q}(M) 
	\rightarrow \Omega^{p-1, q}(M)$
and $\delta^{0, -1}(M): \Omega^{p, q}(M) 
	\rightarrow \Omega^{p, q-1}(M)$ by
\delta^{0, -1} \alpha  & = & w\vdash (\partial \alpha) 
	- \partial (w \vdash \alpha), \\
\delta^{-1, 0} \alpha  & = & 
	w\vdash (\overline{\partial} \alpha) 
	- \overline{\partial} (w \vdash \alpha),
for $\alpha \in \Omega^{p, q}(M)$.
Set $\partial_h = \partial - (h/2) \delta^{0, -1}$,
and $\overline{\partial}_h = \overline{\partial} - (h/2) \delta^{-1, 0}$.
Then $d_h = \partial_h + \overline{\partial}_h$.
Now $0 = d_h^2 = \partial_h^2 + 
	+ \overline{\partial}_h\partial_h)
	+ \overline{\partial}_h^2$,
since they have bidegrees $(2, 0)$, 
$(1, 1)$ and $(0, 2)$ respectively.
Hence, we have
\begin{eqnarray*} \label{double}
\partial_h^2 = 0, & 
	+ \overline{\partial}_h\partial_h = 0, &
\overline{\partial}_h^2 = 0.
We then define
Q_hH^{p, *}(M) & = & H(\Omega_h^{[p, *]}(M), \overline{\partial}_h), \\
Q_{h, h^{-1}}H^{p, *}(M) & = & 
	H(\Omega_{h, h^{-1}}^{[p, *]}(M), \overline{\partial}_h).
They will be called the {\em quantum Dolbeault cohomology} and
{\em Laurent quantum Dolbeault cohomology} respectively.
Several relevant spectral sequences and their degeneracy are considered
in Cao-Zhou \cite{Cao-Zho}.

%\noindent {\bf 8. Quantum integral and quantum Stokes Theorem}
\section*{8. Quantum integral and quantum Stokes Theorem}
Let $(M, \omega)$ be a closed $2n$-dimensional symplectic manifold.
Define an integral $\int_h: \Omega_h(M) \rightarrow {\mathbb R}[h]$
as follows.
For any $\alpha \in \Omega^{j}(M)$,
if $j$ is odd, set $\int_h \alpha = 0$;
if $j = 2n - 2k $ for some integer $k$,
\begin{equation*}\int_h \alpha = \int_M \alpha \wedge \frac{\omega^k}{k!}.\end{equation*}
Extend $\int_h$ to $\Omega_h(M)$ as a ${\mathbb R}[h]$-module map.
We call $\int_h$ the {\em quantum integral}.
Then we have

\begin{theorem}[Quantum Stokes Theorem]
For any $\alpha \in \Omega^{j}(M)$, we have $\int_h d \alpha = 0$, 
        $\int_h h\delta \alpha = 0$. Therefore
\begin{equation*}\int_h d_h \alpha = 0.\end{equation*}

%\noindent {\bf 9. Quantum Chern-Weil theory}.
\section*{9. Quantum Chern-Weil theory}
Given a real or complex vector bundle $E \rightarrow M$  over a 
Poisson manifold $M$, and a connection $\nabla^E$ on it,
we define the quantum covariant derivative
\begin{equation*}d_h^{\nabla^E}: \Omega_h^*(E) \rightarrow \Omega_h^*(E)\end{equation*}
 as follows.
Let $\bs$ be a local frame of $E$ and $\theta$ the 
connection $1$-form in this frame. Then we have 
$\nabla \bs = \bs \otimes \theta$, i.e.,
\begin{equation*}\nabla \bs_j = \sum_{k=1}^{n} \bs_k \otimes \theta_j^k.\end{equation*}
For $\alpha =\bs \otimes \phi$, where $\phi$ is a vector-valued form,
we define
\begin{equation*}d_h^{\nabla^E} \alpha = \bs \otimes (\theta \wh \phi + d_h \phi) 
	= \sum  \bs_k \otimes ( \theta_j^k \wh \phi^j + d_h \phi^k).\end{equation*}
It is straightforward to check 
that the definition of $d_h^{\nabla^E}$ is independent of 
the choice of the local frames (Cao-Zhou \cite{Cao-Zho}, Lemma 7.1).
Furthermore, there is an element $R^E_h \in
\Omega^2_h(\End(E))$ such that
for each $k \geq 0$, $(d_h^{\nabla^E})^2$ on $\Omega^k_h(M)$ is
given by $(d_h^{\nabla^E})^2 \Phi = \Phi \wh R_h^E$, for
any $\Phi \in \Omega^*_h(E)$.
$R^E_h$ is called the quantum curvature of $\nabla^E$.
In a local frame, $R^E_h$ is given by
\begin{equation*}F_h = d_h \theta + \theta \wh \theta,\end{equation*}
where $\theta$ is the connection $1$-form matrix in the local frame. 

If $p$ is a polynomial on the space of $n\times n$-matrices, 
such that $p(G^{-1}AG) = p(A)$, 
for any $n \times n$-matrix $A$, and
invertible $n \times n$-matrix $G$,
then $p(F_h)$ for different frames patch up to 
a well-defined element $p(R^E) \in \Omega^*(M)[h]$.
Similarly to the ordinary Chern-Weil theory, 
it is easy to see that $d_h p(R^E) = 0$.
So it defines a class in $Q_hH^*_{dR}(M)$.
The usual  construction of transgression operator  
carries over to show that
this class is independent of the choice of the connection $\nabla^E$.
In this way,  one can define quantum Chern classes, 
quantum Euler class, etc.
We will call them quantum characteristic classes.
It is clear that we can repeat the same story in the Laurent case.

%\noindent {\bf 10. Quantum equivariant de Rham cohomology}.
\section*{10. Quantum equivariant de Rham cohomology}
Let $(M, w)$ be a Poisson manifold that 
admits an action by a compact connected Lie group $G$,
such that the $G$-action preserves the Poisson bivector field $w$.
Let $\fg$ be the Lie algebra of $G$, $\{ \xi_a \}$ a basis of $\fg$ and 
$\{ \Theta^a \}$ the dual basis in $S^1(\fg^*)$. 
Denote by $\iota_a$ the contraction by the vector field 
generated by the one parameter group corresponding to $\xi_a$,
and $L_a$ the Lie derivative by the same vector field.
Imitating the Cartan model for equivariant cohomology, 
we consider the operator $D_{hG} = d_h + \Theta^a \iota_a 
= d - h \delta /2+ \Theta^a \iota_a$
acting on $(S(\fg^*) \otimes \Omega(M))^G[h]$.  
It is well known that $d + \Theta^a \iota_a$ maps 
$(S(\fg^*) \otimes \Omega(M))^G$ to itself.
Since the $G$-action preserves $w$, it is easy to check that
$\delta$ also preserves $(S(\fg^*) \otimes \Omega(M))^G$.
Therefore, $D_{hG}$ is an operator from 
$(S(\fg^*) \otimes \Omega(M))^G[h]$
to itself. 
Now on $(S(\fg^*) \otimes \Omega(M))^G[h]$, we have 
& = & d_h^2 + (\Theta^a\iota_a)^2 + \Theta^a (d \iota_a + \iota_a d)
	- h \Theta^a (\delta \iota_a + \iota_a \delta) \\
& = &	- h \Theta^a (\delta \iota_a + \iota_a \delta).
Since $\delta = \iota_{w} d - d \iota_{w}$, it is straightforward to verify
that $\delta \iota_a + \iota_a \delta =0$.
%\delta \iota_a + \iota_a \delta 
%& = & \iota_{w}d \iota_a - d \iota_{w} \iota_a 
%	+ \iota_a \iota_{w} d - \iota_a d \iota_{w} \\
%& = & \iota_{w}d \iota_a - d \iota_a \iota_{w}
%	+ \iota_{w} \iota_a d - \iota_a d \iota_{w} \\
%& = & \iota_{w}L_a - L_a \iota_{w} = -\iota_{L_aw} = 0.
Hence, $D_{hG}^2 =0$.
We call the cohomology 
\begin{equation*}Q_hH^*_{G}(M) := H^*((S(\fg^*) \otimes \Omega(M))^G[h], D_{hG})\end{equation*}
the quantum equivariant de Rham cohomology. 
Similar definitions can be made by using Laurent deformation. 
We will study the quantum equivariant de Rham cohomology in a forthcoming


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