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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf K\"ahler manifolds and holonomy \\[15mm]
\small lecture 4}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]

{\small Misha Verbitsky } 
\\[20mm]

{\tiny\bf Tel-Aviv University
\\[2mm]  December 23, 2010
}
\end{center}

\newpage

{\bf \blue K\"ahler manifolds}

{\bf\green DEFINITION:} An Riemannian metric $g$ on
an almost complex manifiold $M$ is called 
{\bf \blue Hermitian} if $g(Ix, Iy)= g(x,y)$.
In this case, $g(x, Iy)= g(Ix, I^2y) = - g(y, Ix)$,
hence $\omega(x,y):= g(x, Iy)$ is skew-symmetric.

{\bf\green DEFINITION:} The differential 
form $\omega\in \Lambda^{1,1}(M)$ is called
{\bf \blue the Hermitian form} of $(M,I,g)$.

\remark It is $U(1)$-invariant, hence {\bf \purple of Hodge type (1,1)}.

{\bf\green DEFINITION:} A complex Hermitian manifold $(M,I,\omega)$
is called {\bf \blue K\"ahler} if $d\omega=0$. 
The cohomology class $[\omega]\in H^2(M)$ of a form $\omega$ 
is called {\bf \blue the K\"ahler class} of $M$, and
$\omega$ {\bf \blue the K\"ahler form}. 

\newpage

\newpage

{\bf \blue Graded vector spaces and algebras}

\definition
A {\bf \blue graded vector space} is a space $V^* =\bigoplus_{i\in \Z} V^i$.

\remark If $V^*$ is graded, the endomorphisms space
$\End(V^*)=\bigoplus_{i\in \Z} \End^i(V^*)$ is also graded, with
$\End^i(V^*)= \bigoplus_{j\in \Z} \Hom(V^j, V^{i+j})$

\definition
A {\bf \blue graded algebra}(or ``graded associative algebra'')
is an associative algebra $A^*=\bigoplus_{i\in \Z} A^i$, with the product 
compatible with the grading: $A^i \cdot A^j \subset A^{i+j}$.

\remark A bilinear map of graded paces which satisfies
$A^i \cdot A^j \subset A^{i+j}$ is called {\bf \blue graded},
or {\bf \blue compatible with grading}.

\remark 
The category of graded spaces can be defined as a {\bf\purple category
of vector spaces with $U(1)$-action,} with the weight decomposition
providing the grading. Then {\bf\purple a graded algebra is an 
associative algebra in the category of spaces with $U(1)$-action}.


\definition
An operator on a graded vector space is called {\bf \blue
even} ({\bf \blue odd}) if it shifts the grading by even 
(odd) number. The {\bf\blue parity} $\tilde a$ of an operator
$a$ is 0 if it is even, 1 if it is odd. We say that
an operator is {\bf \blue pure} if it is even or odd.

\newpage


{\bf \blue Supercommutator}

\definition
A {\bf \blue supercommutator}  of pure operators
on a graded vector space is defined by a formula
$\{a,b\}= ab - (-1)^{\tilde a \tilde b}ba$.

\definition
A graded associative algebra is called {\bf \blue
graded commutative} (or ``supercommutative'')
if its supercommutator vanishes.

\example {\bf \purple The Grassmann algebra is supercommutative.}

\definition
{\bf \blue A graded Lie algebra} (Lie superalgebra)
is a graded vector space $\g^*$
equipped with a bilinear graded map 
$\{\cdot,\cdot\}:\; \g^*\times \g^* \arrow \g^*$
which is graded anticommutative:
$\{a,b\} = - (-1)^{\tilde a \tilde b}\{b,a\}$
and satisfies {\bf \blue the super Jacobi identity}
$\{c, \{a,b\}\} = \{\{c, a\},b\}+ (-1)^{\tilde a \tilde c}\{a,\{c, b\}\}$


\example
Consider the algebra $\End(A^*)$ of operators on a 
graded vector space, with supercommutator as above.
{\bf \purple Then $\End(A^*), \{\cdot,\cdot\}$ is a graded Lie algebra.}

{\bf \green Lemma 1:}
Let $d$ be an odd element of a Lie superalgebra, satisfying
$\{d,d\}=0$, and $L$ an even or odd element. {\bf \red Then $\{\{L, d\}, d\}=0$.}

{\bf \green Proof:} 
$0=\{L,\{d,d\}\}= \{\{L, d\}, d\}+(-1)^{\tilde L}\{d,\{L, d\}\}=2\{\{L, d\}, d\}.$ \endproof

\newpage

{\bf \blue Hodge $*$ operator}

Let $V$ be a vector space. {\bf \blue A metric $g$ on $V$ induces
a natural metric on each of its tensor spaces:}
$g(x_1\otimes x_2 \otimes ... \otimes x_k, x_1'\otimes x_2' \otimes ... \otimes x_k') = g(
x_1, x'_1)g(x_2, x'_2) ... g(x_k, x'_k)$.

{\bf \purple This gives a natural positive definite scalar product
on differential forms over a Riemannian manifold $(M,g)$:}
$g(\alpha, \beta) := \int_M g(\alpha, \beta) \Vol_M$

Another non-degenerate 
form is provided by the {\bf \blue Poincare pairing:}\\
$\alpha, \beta \arrow \int_M \alpha \wedge \beta$.

\definition
Let $M$ be a Riemannian $n$-manifold.
Define {\bf \blue the Hodge $*$ operator}
$*:\; \Lambda^k M \arrow \Lambda^{n-k} M$ by the following relation:
{\bf \red $g(\alpha, \beta) = \int_M \alpha \wedge *\beta.$}

\remark {\bf \red The Hodge $*$ operator always exists.} It is defined
explicitly in an orthonormal basis $\xi_1, ..., \xi_n \in \Lambda^1 M$:
\[ * (\xi_{i_1}\wedge\xi_{i_2} \wedge ... \wedge\xi_{i_k}) =
(-1)^s \xi_{j_1}\wedge\xi_{j_2} \wedge ... \wedge\xi_{j_{n-k}},
\]
where $\xi_{j_1},\xi_{j_2}, ...,  \xi_{j_{n-k}}$ is a complementary
set of vectors to $\xi_{i_1}, \xi_{i_2}, ..., \xi_{i_k}$, and
$s$ the signature of a permutation $(i_1, ..., i_k, j_1, ..., j_{n-k})$.

\remark $*^2\restrict{\Lambda^k(M)}=(-1)^{k(n-k)}\Id_{\Lambda^k(M)}$

\newpage

{\bf \blue Hodge theory}

\claim
On a compact Riemannian $n$-manifold, one has
$d^*\restrict{\Lambda^k M} = (-1)^{nk} *d*$,
where $d^*$ denotes {\bf \blue the adjoint operator}, which
is defined by the equation $(d\alpha, \gamma) = (\alpha, d^*\gamma)$.

{\bf \green Proof:} Since 
\[ 
  0=\int_M d(\alpha\wedge \beta) = \int_M d(\alpha)\wedge \beta +
  (-1)^{\tilde \alpha} \alpha \wedge d(\beta),
\]
one has $(d\alpha, *\beta) = (-1)^{\tilde \alpha} (\alpha, *d\beta)$.
Setting $\gamma:= *\beta$, we obtain
\[
  (d\alpha, \gamma) = (-1)^{\tilde \alpha} (\alpha, *d(*)^{-1}\gamma)=
  (-1)^{\tilde \alpha} (-1)^{\tilde \alpha(\tilde n-\tilde\alpha)}
  (\alpha, *d*\gamma)= (-1)^{\tilde \alpha\tilde n}(\alpha, *d*\gamma).
\]
\endproof


\definition
The anticommutator $\Delta:=\{d, d^*\}= dd^* + d^* d$
is called {\bf \blue the Laplacian} of $M$. It is self-adjoint
and positive definite: $(\Delta x, x)= (dx, dx) + (d^*x, d^*x).$
Also, $\Delta$ commutes with $d$ and $d^*$ (Lemma 1).

\theorem {\bf \blue (The main theorem of Hodge theory)}\\
{\bf \red There is a basis in the Hilbert space $L^2(\Lambda^*(M))$
consisting of eigenvectors of $\Delta$.}

\theorem {\bf \blue (``Elliptic regularity for $\Delta$'')}
Let $\alpha\in L^2(\Lambda^k(M))$ be an eigenvector of $\Delta$.
{\bf \red Then $\alpha$ is a smooth $k$-form.}

\newpage

{\bf \blue De Rham cohomology}

\definition 
The space 
$H^i(M):= \frac {\ker d\restrict{\Lambda^iM}}{d\left(\Lambda^{i-1}M\right)}$
is called {\bf \blue the de Rham cohomology of $M$}.

\definition
A form $\alpha$ is called {\bf\blue harmonic} if $\Delta(\alpha)=0$.

\remark Let $\alpha$ be a harmonic form. {\bf \red Then 
$(\Delta x, x)= (dx, dx) + (d^*x, d^*x),$} hence 
$\alpha \in \ker d \cap \ker d^*$

\remark {\bf \purple The projection ${\cal H}^i(M) \arrow H^i(M)$
from harmonic forms to cohomology is injective.} Indeed,
a form $\alpha$ lies in the kernel of such projection if
$\alpha=d\beta$, but then 
$(\alpha,\alpha)=(\alpha, d\beta) = (d^* \alpha, \beta) =0$.

\theorem
{\bf \red The natural map ${\cal H}^i(M) \arrow H^i(M)$ is an isomorphism}\\
(see the next page).

\remark {\bf \purple 
Poincare duality immediately follows from this theorem.}



\newpage

{\bf \blue Hodge theory and the cohomology}

\theorem
{\bf \red The natural map ${\cal H}^i(M) \arrow H^i(M)$ is an isomorphism.}

{\bf \green Proof. Step 1:}
Since $d^2=0$ and $(d^*)^2=0$, one has 
$\{d, \Delta\}=0$.
This means that {\bf \red $\Delta$ commutes with the de Rham differential.}

{\bf \green Step 2:} Consider the eigenspace decomposition
$\Lambda^*(M) \tilde = \bigoplus_\alpha {\cal H}^*_\alpha(M)$,
where $\alpha$ runs through all eigenvalues of $\Delta$,
and ${\cal H}^*_\alpha(M)$ is the corresponding eigenspace.
{\bf \purple For each $\alpha$, de Rham differential defines a complex
\[
{\cal H}^0_\alpha(M) \stackrel d \arrow 
{\cal H}^1_\alpha(M) \stackrel d \arrow 
{\cal H}^2_\alpha(M) \stackrel d \arrow ...
\]
}

{\bf \green Step 3:} On ${\cal H}^*_\alpha(M)$, one has
$dd^* + d^* d= \alpha$. When $\alpha \neq 0$, and $\eta$ closed,
this implies $dd^*(\eta) + d^* d(\eta)= dd^* \eta = \alpha\eta$,
hence $\eta= d\xi$, with $\xi:= \alpha^{-1}d^* \eta$.
This implies that {\bf \purple the complexes $({\cal H}^*_\alpha(M), d)$
don't contribute to cohomology.}

{\bf \green Step 4:} We have proven that
\[
H^*(\Lambda^* M, d) = \bigoplus_\alpha H^* ({\cal H}^*_\alpha(M),d)=
H^* ({\cal H}^*_0(M),d)={\cal H}^*(M).
\]
\endproof

\newpage

{\bf \blue Supersymmetry in K\"ahler geometry}

Let $(M, I, g)$ be a Kaehler manifold, $\omega$ its Kaehler form.
{\bf \blue On $\Lambda^*(M)$, the following operators are defined.}

0. $d$, $d^*$, $\Delta$, because it is Riemannian.

1. $L(\alpha):= \omega\wedge \alpha$

2. $\Lambda(\alpha) := * L * \alpha$. 
It is easily seen that $\Lambda= L^*$.

3. The Weil operator $W\restrict{\Lambda^{p,q}(M)}=\1(p-q)$

\theorem
{\bf \red These operators generate a Lie superalgebra
$\goth a$ of dimension $(5|4)$,} 
acting on $\Lambda^*(M)$. Moreover, the Laplacian $\Delta$ is
central in $\goth a$, hence {\bf \purple $\goth a$ also acts on the
cohomology of $M$. }

\remark This is a convenient way to summarize 
the K\"ahler relations and the Lefschetz' $\goth{sl}(2)$-action.


\newpage

{\bf \blue Reference:}


JM Figueroa-O'Farrill, C Koehl, B Spence,
{\em \blue \bf Supersymmetry and the cohomology of (hyper)Kaehler 
manifolds,} arXiv:hep-th/9705161,
Nucl.Phys. B503 (1997) 614-626

M. Verbitsky,
{\em \blue\bf  Hyperkaehler manifolds with torsion, 
supersymmetry and Hodge theory,} arXiv:math/0112215,
Asian J. Math. Vol. 6, No. 4, pp. 679-712 (2002)

Elena Poletaeva,
{\em \blue\bf  Superconformal algebras and 
Lie superalgebras of the Hodge theory}, arXiv:hep-th/0209168,
        J.Nonlin.Math.Phys. 10 (2003) 141-147

\newpage

{\bf \blue The coordinate operators }

Let $V$ be an even-dimensional real vector space equipped with
a scalar product, and $v_1, ..., v_{2n}$ an orthonormal basis.
Denote by $e_{v_i}:\; \Lambda^k V  \arrow \Lambda^{k+1} V$
an operator of multiplication, $e_{v_i}(\eta) = e_i \wedge \eta$.
Let $i_{v_i}:\; \Lambda^k V  \arrow \Lambda^{k-1} V$
be an adjoint operator, $i_{v_i}= * e_{v_i} *$.

\claim
The operators $e_{v_i}$, $i_{v_i}$, $\Id$ are a basis of an {\blue \bf 
odd Heisenberg Lie superalgebra $\goth H$}, 
with {\bf \purple the only non-trivial 
supercommutator given by the formula $\{ e_{v_i}, i_{v_j}\} = \delta_{i,j}\Id$.}

Now, consider the tensor $\omega= \sum_{i=1}^n v_{2i-1} \wedge v_{2i}$,
and let $L(\alpha) = \omega \wedge \alpha$, and $\Lambda:= L^*$
be the corresponding {\bf \blue Hodge operators}.

\claim From the commutator relations in $\goth H$,
one obtains immediately that
\[
H:=[L, \Lambda] = 
\left [\sum e_{v_{2i-1}} e_{v_{2i}}, \sum i_{v_{2i-1}} i_{v_{2i}}\right]
= \sum_{i=1}^{2n} e_{v_i} i_{v_i} - \sum_{i=1}^{2n} i_{v_i} e_{v_i},
\]
{\bf \red is a scalar operator acting as $k-n$ on $k$-forms.}

\newpage

{\bf \blue Integrability of the complex structure }

\claim {\bf \blue (``Cartan's formula'')}
The de Rham differential of can be expressed
through the commutator of vector fields:
\begin{multline*}
d\eta(X_1, ... X_{d+1}) = \sum (-1)^{i+1} D_{X_i}(\eta(X_1, ...,\check X_i,
...,  X_{d+1}) \\- \sum_{i<j} (-1)^{i+j+1}\eta([X_i,X_j], X_1, ...,\check X_i,
..., \check X_j, ..., X_{d+1}).
\end{multline*}
{\bf \purple For a 1-form $\eta$, this gives 
$d\eta(X_1,X_2) = D_{X_1}\eta(X_2)-D_{X_2}\eta(X_1)-\eta([X_1, X_2])$.}

\corollary Let $(M,I)$ be an almost complex manifold. Then
the following assertions are equivalent.

(i) {\bf \red $d\eta \subset \Lambda^{0,2}(M) \oplus \Lambda^{1,1}(M)$}
 for any $\eta \in \Lambda^{0,1}(M)$.

(ii) {\bf \red $I$ is integrable.}

\remark {\blue This is equivalent to $d\restrict {\Lambda^1M}$
having only two Hodge components: $d=d^{1,0}+d^{0,1}$}
(for a non-integrable complex structure, there are 4:
 $d= d^{2,-1}+d^{1,0}+d^{0,1}+d^{-1,2}$). 

\remark Since $\Lambda^* M$ is multiplicatively generated by
$\Lambda^1(M)$, {\bf \purple
the decomposition $d= d^{2,-1}+d^{1,0}+d^{0,1}+d^{-1,2}$
holds for any almost complex manifold.}


\newpage

{\bf \blue  Integrability and the Hodge decomposition}

\claim
{\bf \red A manifold $(M,I)$ is integrable
if and only if $(d^{0,1})^2|_{C^\infty M}=0$.}

{\bf \green Proof. Step 1:}
{\bf \purple The bundle $\Lambda^{1,0}(M)$ is generated over $C^{\infty}M$
by \\ $d^{1,0}(C^\infty M)$.} Indeed, it is $n$-dimensional, $n=\dim_\C M$
and to prove this one needs to find $n$ functions $f_1, ..., f_n$
with $d^{1,0}f_i$ linearly independent at a point. This is done
by taking $2n$ functions $f_1, ..., f_{2n}$ with $df_i$
linearly independent, and finding an appropriate subset.

{\bf \green Step 2:} Then, the integrability condition 
$d(\Lambda^{1,0}(M)) \subset \Lambda^{2,0}(M) \oplus \Lambda^{1,1}(M)$
is equivalent to 
$d d^{1,0}(C^\infty M)\subset \Lambda^{2,0}(M) \oplus \Lambda^{1,1}(M)$
$\Leftrightarrow$  $d^{-1,2} (d^{1,0}(C^\infty M))=0$. 

{\bf \green Step 3:} The $(0,2)$ component of $d^2=0$
gives $\{d^{-1,2}, d^{1,0}\} = \{d^{0,1}, d^{0,1}\} = 2 (d^{0,1})^2=0$.
From Step 2, we obtain that
{\bf \blue
$(d^{0,1})^2\restrict{C^\infty M}=0$ is equivalent to integrability.}
\endproof

\remark {\bf \purple The above claim provides an equivalence
$d^{2,-1}=0$ $\Leftrightarrow$ $\{d^{-1,2}, d^{1,0}\}=0$
$\Leftrightarrow$ $(d^{0,1})^2=0$.}

\newpage

{\bf \blue  The twisted differential $d^c$}


\definition The {\bf \blue twisted differential}
is defined as $d^c:=I d I^{-1}$.

\claim Let $(M,I)$ be a complex
manifold. {\bf \blue Then 
$\6:= \frac{d + \1 d^c}2$, $\bar \6:= \frac{d - \1 d^c}2$
are the Hodge components of $d$}, $\6= d^{1,0}$, 
$\bar\6= d^{0,1}$. 

{\bf \green Proof:} Let $V$ be a space generated by $d, I d I$.
The natural action of $U(1)$ generated by $e^{\cal W}$
preserves $V$. {\bf \purple Since $d$ has only two Hodge components.
$U(1)$ acts with weights $\1$ and
$-\1$,} and its Hodge components are expressed as
above. \endproof

\claim
On a complex manifold, one has 
{\bf \red $d^c = [{\cal W}, d]$. }

{\bf \green Proof:} Clearly, $[{\cal W}, d^{1,0}]= \1 d^{1,0}$ and
$[{\cal W}, d^{0,1}] = - \1 d^{0,1}$. Adding these equations,
obtain $d^c = [{\cal W}, d]$.

\corollary $\{d, d^c\} = \{d, \{d, {\cal W}\}\}=0$ (Lemma 1).


\newpage

{\bf \blue De Rham differential on Kaehler manifolds}


\theorem {\red \bf The following statements are equivalent.}

1. $I$ is integrable.\ \ 2. $\6^2=0$.\ \ 
3. $\bar\6^2=0$.\ \ 
4. $dd^c =- d^c d$\ \ 
5. $dd^c= 2 \1 \6\bar\6$.

\definition The
operator $dd^c$ is called {\bf \blue the pluri-Laplacian}.

\theorem 
Let $M$ be a Kaehler manifold. One has the following
identities {\bf \red (``K\"ahler idenitities'').}
\[
 [\Lambda, \6] = \1 \bar\6^*,  \ \ \ 
 [L, \bar\6] = - \1 \6^*,  \ \ \ 
  [\Lambda, \bar\6^*] = - \1 \6, \ \ \ 
 [L, \6^*] = \1 \bar\6.
\]
Equivalently,
\[ 
  [\Lambda, d] = (d^c)^*,\ \ \  
\ \ [ L, d^*] = - d^c,\ \ \ \ \ [\Lambda, d^c] = - d^*,
\ \ \ \ \ [ L, (d^c)^*] = d.
\]


\newpage

{\bf \blue Laplacians and supercommutators}

\theorem
Let 
\[ \Delta_d:= \{d, d^*\}, \ \ \Delta_{d^c}:= \{d^c, {d^c}^*\}, \ \ 
   \Delta_\6:= \{\6,\6^*\}, \Delta_{\bar \6}:= \{\bar\6,\bar\6^*\}.
\]
{\bf \red Then $\Delta_d=\Delta_{d^c}=2\Delta_\6=2\Delta_{\bar \6}.$}
In particular, {\bf \purple $\Delta_d$ preserves the Hodge decomposition.}

{\bf \green Proof:} By Kodaira relations, $\{d,d^c\}=0$.
Graded Jacobi identity gives
\[ \{d, d^*\}= -\{d,\{\Lambda, d^c\}\}=
\{ \{\Lambda, d\}, d^c\}= \{d^c, {d^c}^*\}.
\]
Same calculation with $\6, \bar\6$ gives $\Delta_\6=\Delta_{\bar \6}.$.
Also, $\{\6,\bar \6^*\}= \1 \{\6,\{ \Lambda, \6\}\}=0$, 
(Lemma 1), and the same argument implies that {\bf \purple all anticommutators
$\6, \bar\6^*$, etc. all vanish except $\{\6,\6^*\}$ and
$\{\bar\6,\bar\6^*\}$.} This gives $\Delta_d=\Delta_\6+\Delta_{\bar \6}.$
\endproof


\definition The operator  
$\Delta:=\Delta_d$ is called {\bf\blue the Laplacian}.

\remark We have proved that {\bf \red operators 
$L, \Lambda,  d,  {\cal W}$ generate
a Lie superalgebra of dimension $(5|4)$ (5 even, 4 odd), 
with a 1-dimensional center $\R\Delta$.}

\newpage

{\bf \blue The Lefschetz  ${\goth sl}(2)$-action}

\corollary
The operators $L, \Lambda, H$ form a basis of a Lie algebra
isomorphic to ${\goth sl}(2)$, with relations
\[
[L, \Lambda] =H, \ \ [H, L]= 2 L, \ \ [H, \Lambda] = -2 \Lambda.
\]

\definition
$L, \Lambda, H$ is 
called {\bf \blue the Lefschetz $\goth{sl}(2)$-triple}.

\remark {\bf \purple Finite-dimensional 
representations of ${\goth sl}(2)$ are semisimple}.

\remark A simple finite-dimensional 
representation $V$ of ${\goth sl}(2)$
is generated by $v\in V$ which satisfies 
$\Lambda (v) =0$, $H (v) = p v$ {\bf \blue (``lowest weight vector'')},
where $p\in \Z^{\geq 0}$. Then
$v, L(v), L^2(v), ..., L^p(v)$
form a basis of $V_p:=V$. {\bf \red This
representation is determined uniquely by $p$.}

\remark In this 
basis, {\bf \purple 
$H$ acts diagonally:} $H(L^i(v))= (2i-p)L^i(v)$.


\remark One has $V_p= \Sym^p V_1$, where $V_1$ is
a 2-dimensional tautological representation.
It is called {\bf \blue a weight $p$ representation of
${\goth sl}(2)$}.

\corollary For a finite-dimensional representation $V$
of ${\goth sl}(2)$, denote by $V^{(i)}$ the eigenspaces of
$H$, with $H\restrict{V^{(i)}}=i$. {\bf \red Then $L^i$ induces
an isomorphism $V^{(-i)} \stackrel{L^i}\arrow V^{(i)}$
for any $i>0$.}


\newpage

{\bf \blue Lefschetz action on cohomology.}

From the supersymmetry theorem, the following result
follows.

\corollary
The ${\goth sl}(2)$-action $\langle 
L, \Lambda, H\rangle$ and the action of Weil operator
commute with Laplacian, hence {\bf \red preserve the harmonic
forms on a K\"ahler manifold}.

\corollary 
Any cohomology class can be represented as
a sum of closed $(p,q)$-forms, giving a decomposition
$H^i(M) = \bigoplus_{p+q=i}H^{p,q}(M)$, with
$\overline{H^{p,q}(M)} = H^{q,p}(M)$. 

\corollary {\bf\purple odd cohomology of a compact
K\"ahler manifold are even-dimensional.}

\corollary
Let $M$ be a compact, K\"ahler manifold of complex dimension
$n$, and $i+p+q=n$. Then $L^i$ defines {\bf \blue
the Lefschetz isomorphism} $H^{p,q} \stackrel{L^i}\arrow H^{p+2i,q+2i}(M)$

\newpage

{\bf \blue The Hodge diamond:}
{\small \[
\begin{array}{ccccccccc}
&&&&H^{n,n}&&&& \\[8mm]
&&&H^{n,n-1}&&H^{n-1,n}&&& \\[8mm]
\ \ \ \ &&H^{n,n-2}&&H^{n-1,n-1}&&H^{n-2,n}&& \ \ \ \ \\[8mm]
&H^{n,n-3}(M)& &H^{n-1,n-2}(M) && H^{n-2,n-1}(M) &&H^{n-3,n}(M) &\\[8mm]
& \vdots &&\vdots &&\vdots&&\vdots &\\[8mm]
&H^{3,0}(M)& &H^{2,1}(M) && H^{1,2}(M) &&H^{0,3}(M) &\\[8mm]
&&H^{2,0}&&H^{1,1}&&H^{0,2}&& \\[8mm]
&&&H^{1,0}&&H^{0,1}&& &\\[8mm]
&&&&H^{0,0}&&&& \\[8mm]
\end{array}
\]}

\newpage

{\bf \blue Hyperk\"ahler manifolds}

\definition (E. Calabi, 1978)\\ Let $(M, g)$ be a Riemannian
manifold equipped with three complex structure operators
$I, J, K:\; TM\arrow TM$, satisfying the quaternionic relation
\[ I^2=J^2=K^2=IJK=-\Id.\]  Suppose that $I$, $J$, $K$ are
K\"ahler. Then $(M, I, J, K, g)$ is called {\bf \blue hyperk\"ahler}.


\remark A hyperk\"ahler manifold $M$ is
equipped with 3 symplectic forms $\omega_I$, $\omega_J$, 
$\omega_K$. The form 
$\Omega:= \omega_J+\1\omega_K$ 
{\bf \purple is a holomorphic symplectic 2-form on
$(M,I)$.} \endproof


\theorem (Calabi-Yau)
Let $M$ be a compact, holomorphically symplectic K\"ahler
manifold. Then {\bf \red $M$ admits a hyperk\"ahler metric,} which is
uniquely determined by the cohomology class of its 
K\"ahler form $\omega_I$.

{\green \it Hyperk\"ahler geometry is essentially
the same as holomorphic symplectic geometry}

\newpage

{\bf \blue Supersymmetry in hyperk\"ahler geometry}

Let $(M, I, J,K, g)$ be a hyperkaehler manifold, $\omega_I$,
$\omega_J$, $\omega_K$ its Kaehler forms.
{\bf \blue On $\Lambda^*(M)$, the following operators are defined.}

0. $d$, $d^*$, $\Delta$, because it is Riemannian.

1. $L_I(\alpha):= \omega_I\wedge \alpha$

2. $\Lambda_I(\alpha) := * L_I * \alpha$. 
It is easily seen that $\Lambda_I= L^*_J$.

3. Three Weil operators
$W_{I}\restrict{\Lambda^{p,q}(M,I)}=\1(p-q)$,
$W_{ J}\restrict{\Lambda^{p,q}(M,J)}=\1(p-q)$,
$W_{K}\restrict{\Lambda^{p,q}(M,K)}=\1(p-q)$

\theorem
{\bf \red These operators generate a Lie superalgebra
$\goth a$} of dimension $(11|8)$,
 acting on $\Lambda^*(M)$. Moreover, the Laplacian $\Delta$ is
central in $\goth a$, hence {\bf \purple $\goth a$ also acts on the
cohomology of $M$. }


\remark The Weil operators form the Lie algebra
$\goth{su}(2)$ of unitary quaternions. This means that {\bf \blue 
the quaternionic action belongs to $\goth a$}. In particular,
$L_J, L_K, \Lambda_J$ and $\Lambda_K$.

\remark The twisted de Rham differentials 
$d_I, d_J, d_K$, associated to $I,J,K$ also belong to
$\goth a$: {\bf \purple $d_I= [ W_I, d]$, $d_J= [ W_J, d]$, 
$d_K= [ W_K, d]$}

\newpage

{\bf \blue Supersymmetry and the Hodge decomposition }

\remark 1.  $[L_I, \Lambda_J]=W_K$,
 $[L_J, \Lambda_K]=W_I$, $[L_I, \Lambda_K]=-W_J$.

2. The even part of $\goth a$
{\bf \red is isomorphic to $\goth{sp}(1,1, {\Bbb H})\oplus\R \cdot \Delta $.}

3. The odd part $\langle d, d_I, d_J, d_K, d,^* d_I^*, d_J^*, d_K^*\rangle$
{\bf \red generates the 9-dimensional odd Heisenberg algebra,} with the
only non-trivial supercommutators being 
$\{d, d^*\}=\{d_I, d^*_I\}=\{d_J, d^*_J\}=\{d_K, d^*_K\}=\Delta$

4. The action of $\goth a_{even}$ on $\goth a_{odd}$
{\bf \purple is the fundamental representation of $\goth{sp}(1,1, {\Bbb H})$ in 
${\Bbb H^2}$,} with the quaternionic Hermitian 
metric on $\goth a_{odd}$ provided
by the anticommutator.

\remark
The weight decomposition of the $\goth{sp}(1,1, {\Bbb H})=
\goth{so}(1,4)$-action on $H^*(M)$ 
{\bf \purple coincides with the Hodge decomposition.}






\end{document}