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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf Hodge theory \\[15mm]
\small lecture 15: K\"ahler relations (2)}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]
{\scriptsize NRU HSE, Moscow} \\[15mm]
{\small Misha Verbitsky, March 17, 2018 } 
\end{center}


\newpage

{\bf \blue Graded vector spaces and algebras (reminder)}

\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 (reminder)}

\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 Supersymmetry in K\"ahler geometry (reminder)}

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 $\goth{sl}(2)$-triple (reminder) }

\claim {\bf \blue (Lefschetz triples)}
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.}

\corollary 
The triple $L, \Lambda, H$ satisfies the relations for the $\goth{sl}(2)$
Lie algebra: $[L,\Lambda]=H$, $[H,L]=2L$, $[H,\Lambda]=2\Lambda$.

\newpage

{\bf \blue  Hodge components of $d$ (reminder)}

\claim
Let $(M,I)$ be an almost complex manifold, and
$d= \oplus d^{i, 1-i}$ be the Hodge components of $d$,
with $d^{a,b}:\; \Lambda^{p,q}(M) \arrow \Lambda^{p+a,
  q+b}(M)$.
{\bf \red Then there are only 4 components, $d= d^{2, -1}+ d^{1,0},
+ d^{0,1} + d^{-1, 2}$, with $d^{2, -1}$ and $d^{-1, 2}$
$C^\infty$-linear.} Moreover, {\bf \red 
the operators $d^{-1, 2}$ and $d^{2, -1}$ 
vanish when $I$ is (formally) integrable.}

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

\claim Let $(M,I)$ be a complex
manifold. {\bf \red 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:} The Hodge components of $d$ are expressed as
$d^{1,0}=\frac{d + \1 d^c}2$, $d^{0,1}=\frac{d - \1
  d^c}2$. Indeed,
$I(\frac{d + \1 d^c}2)I^{-1}=\1\frac{d + \1 d^c}2$, hence 
{\bf \purple $\frac{d + \1 d^c}2$ has Hodge type (1,0);} the same
argument works for $\bar\6$.
\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
$[{\cal W}, d^{0,1}] = - \1 d^{0,1}$. Then 
$[{\cal W}, d]= \1 d^{1,0}-\1 d^{0,1}=IdI^{-1}$. \endproof

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


\newpage

{\bf \blue Plurilaplacian (reminder)}

\theorem Let $M,I$ be a complex
manifold. {\bf \red Then 
1.  $\6^2=0$.\\
2. $\bar\6^2=0$.\\ 
3. $dd^c =- d^c d$\\ 
4. $dd^c= 2 \1 \6\bar\6$.}

\proof
The first is vanishing of (2,0)-part of $d^2$, and
the second is vanishing of its (0,2)-part.
Now, $\{d, d^c\}= -\{d, \{d, {\cal W}\}\}=0$
(Lemma 1), this gives $dd^c =- d^c d$.
Finally, $2 \1 \6\bar\6= \frac 1 2 (d+ \1 d^c)(d-\1 d^c)=
\frac 1 2 (dd^c - d^c d)= dd^c$.
\endproof

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

\exercise
Prove that {\bf \purple on a Riemannian surface $(M,I, \omega)$,
one has $dd^c(f) =\Delta(f) \omega$.}

\newpage

{\bf \blue Kodaira identities}

\theorem 
Let $M$ be a Kaehler manifold. One has the following
identities {\bf \red (``K\"ahler idenitities'', ``Kodaira idenities'').}
\[
 [\Lambda, \6] = \1 \bar\6^*,  \ \ \ 
 [L, \bar\6] = - \1 \6^*,  \ \ \ 
  [\Lambda, \bar\6^*] = - \1 \6, \ \ \ 
 [L, \6^*] = \1 \bar\6.
\]
Equivalently,
\[ 
  [\Lambda, d] = (d^c)^*,\ \ \  
\ \ [ L, d^*] = - d^c,\ \ \ \ \ [\Lambda, d^c] = - d^*,
\ \ \ \ \ [ L, (d^c)^*] = d.
\]
\pstep The first set of identities implies 
the second set. Indeed, by adding up appropriate
identities in the top set of their complex conjugate, 
we obtain ones in the bottom set; 
for example, adding $[\Lambda, \6] = \1 \bar\6^*$
and $[\Lambda, \bar\6] = - \1 \6^*,$, we obtain $[\Lambda, d] = (d^c)^*$.
Each of top identities is related to the other three 
by complex conjugation or by Hermitian conjugation, hence
they are all equivalent. Each of the bottom identities
implies the rest by Hermitian conjugation and conjugating
with $I$. Finally, $[\Lambda, \6] = \1 \bar\6^*$
can be obtained as a sum of $[\Lambda, d] = (d^c)^*$
and $[\Lambda, d^c] = - d^*$ with appropriate coefficients.
{\bf \red We obtained that all Kodaira identities are
implied by just one, say, $[L, d^*] =- d^c$}.

\newpage

{\bf \blue Kodaira identities (2)}

\pstep {\bf \red We reduced the Kodaira identities to just one, 
$[L, d^*] =- d^c$.}

{\bf \green Step 2:} Let ${\goth E}:\; \Lambda^iM 
\otimes \Lambda^1 M\arrow \Lambda^{i+1}(M)$ be the multiplication, 
and ${\goth I}:\; \Lambda^iM \otimes 
\Lambda^1 M\arrow \Lambda^{i-1}(M)$ the map that takes
$\alpha \wedge \theta$ and puts it to $*(*\alpha \wedge
\theta)$. In other words, ${\goth I}$ takes a tensor 
$\alpha \otimes \theta$, with $\alpha \in \Lambda^iM$
and $\theta \in \Lambda^1 M$, uses the metric $g$
to produce a vector field $X$ from $\theta$,
and maps $\alpha$ to $\alpha\cntrct X$ (convution
of $\alpha$ and $X$). 

{\bf \green Step 3:}
Let $\nabla$ be the Levi-Civita connection.
Then $d\alpha = {\goth E}(\nabla(\alpha))$, because $\nabla$ is
torsion-free. Since $d^*=*d*$, one has 
$d^*(\alpha)= {\goth I}(\nabla(\alpha))$.
Let $x_1, y_1, ..., x_n, y_n\in \Lambda^1_m M$
be an orthonormal basis such that 
$\omega= \sum x_i \wedge y_i$. 
Then ${\goth I}(\nabla(\alpha))= \sum_{i} i_{x_i}(\nabla_{x_i}
\alpha)+ i_{y_i}(\nabla_{y_i} \alpha)$.
Taking a commutator with $L= \sum e_{x_i}e_{y_i}$
and using the commutator relations between $e_{v}$ and
$i_w$ found earlier, we obtain
\[
[L, d^*]= \sum_i \nabla_{x_i}[e_{x_i}e_{y_i}, i_{x_i}]+
\nabla_{y_i}[e_{x_i}e_{y_i}, i_{y_i}]= \sum_i \nabla_{y_i}
e_{x_i}- \nabla_{x_i}e_{y_i}.
\]
(the operator $\nabla_{w}$ commutes with $L$, because
$\omega$ is parallel).
However, 
\[ \sum_i \nabla_{y_i}
e_{x_i}- \nabla_{x_i}e_{y_i}= 
-I\left(\sum_i\nabla_{x_i}e_{x_i} +
\nabla_{y_i}e_{y_i}\right) = - d^c
\]
which gives $[L, d^*] =- d^c$. \endproof


\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 $\R\Delta$ central.}

\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 \[
\begin{array}{ccccccccc}
&&&&H^{n,n}&&&& \\[8mm]
&&&H^{n,n-1}&&H^{n-1,n}&&& \\[8mm]
\ \ \ \ &&H^{n,n-2}&&H^{n-1,n-1}&&H^{n-2,n}&& \ \ \ \ \\[8mm]
&H^{n,n-3}(M)& &H^{n-1,n-2}(M) && H^{n-2,n-1}(M) &&H^{n-3,n}(M) &\\[8mm]
& \vdots &&\vdots &&\vdots&&\vdots &\\[8mm]
&H^{3,0}(M)& &H^{2,1}(M) && H^{1,2}(M) &&H^{0,3}(M) &\\[8mm]
&&H^{2,0}&&H^{1,1}&&H^{0,2}&& \\[8mm]
&&&H^{1,0}&&H^{0,1}&& &\\[8mm]
&&&&H^{0,0}&&&& \\[8mm]
\end{array}
\]}




\end{document}
