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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf Complex geometry\\[15mm]
\small lecture 2: Cauchy formula }
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]

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

{\tiny\bf HSE, room 306, 16:20,
\\[2mm] 
September 26, 2020
}
\end{center}


\newpage



{\bf \blue The Grassmann algebra}

\definition
Let $V$ be a vector space. Denote by $\Lambda^i V$ the
space of antisymmetric polylinear $i$-forms on $V^*$, and
let $\Lambda^*V:=\bigoplus \Lambda^i V$.
Denote by $T^{\otimes i}V$ the algebra of {\bf \green all}
polylinear $i$-forms on $V^*$ (``tensor algebra''), and let 
$\Alt:\; T^{\otimes i}V\arrow \Lambda^i V$
be {\bf \blue the antisymmetrization},
\[\Alt(\eta)(x_1, ..., x_i):= 
\frac 1{i!}\sum_{\sigma\in 
\Sigma_i}(-1)^{\tilde \sigma}\eta(x_{\sigma_1}, ..., x_{\sigma_i})
\]
where $\Sigma_i$ is the group of permutations, and
$\tilde \sigma=1$ for odd permutations, and 0 for even.
Consider the multiplicative operation (``wedge-product'') on $\Lambda^*V$,
denoted by $\eta\wedge \nu:= \Alt (\eta \otimes \nu)$.
The space $\Lambda^*V$ with this operation is called
{\bf \blue the Grassmann algebra}.

\remark
{\bf \red It is an algebra of anti-commutative polynomials.}

{\bf \green Properties of Grassmann algebra:}

1. $\dim \Lambda^i V:= \binom{\dim V}{i}$, $\dim \Lambda^* V=2^{\dim V}$.

2. $\Lambda^*(V \oplus W)= \Lambda^*(V) \otimes \Lambda^*(W)$.

\newpage

{\bf \blue The Hodge decomposition in linear algebra}

\definition
Let $(V,I)$ be a space equipped with a complex structure.
{\bf \blue The Hodge decomposition} 
$V\otimes_\R \C:= V^{1,0}\oplus V^{0,1}$ is defined in such a way that
$V^{1,0}$ is a $\1$-eigenspace of $I$, and $V^{0,1}$ a $-\1$-eigenspace.

\remark 
Let $V_\C:= V \otimes_\R \C$.
The Grassmann algebra of skew-symmetric forms
$\Lambda^n V_\C:=\Lambda^n_\R V \otimes _\R C$
admits a decomposition
\[ 
  \Lambda^n V_\C= \bigoplus_{p+q=n} \Lambda^p V^{1,0} 
  \otimes \Lambda^q V^{0,1}
\]
We denote $\Lambda^p V^{1,0} \otimes \Lambda^q V^{0,1}$ by $\Lambda^{p,q}V$.
The resulting decomposition $\Lambda^n V_\C= \bigoplus_{p+q=n}\Lambda^{p,q}V$
is called {\bf \blue the Hodge decomposition of the Grassmann algebra}.

\remark The operator $I$ induces $U(1)$-action on $V$ by the formula
$\rho(t)(v) = \cos t\cdot v + \sin t \cdot I(v)$. We extend this
action on the tensor spaces by muptiplicativity.

\newpage

{\bf \blue $U(1)$-representations and the weight decomposition }

\remark {\bf \red Any complex representation $W$ of $U(1)$ is written
as a sum of 1-dimensional representations $W_i(p)$}, with $U(1)$
acting on each $W_i(p)$ as $\rho(t)(v) = e^{\1 pt}(v)$. The 1-dimensional
representations are called {\bf \blue weight $p$ 
representations of $U(1)$.}

\definition 
A {\bf \blue weight decomposition} of a $U(1)$-representation
$W$ is a decomposition $W= \oplus W^p$, where each $W^p=\oplus_i W_i(p)$ is
a sum of  1-dimensional representations of weight $p$.

\remark {\bf \red The Hodge decomposition 
$\Lambda^n V_\C= \bigoplus_{p+q=n}\Lambda^{p,q}V$
is a weight decomposition}, with $\Lambda^{p,q}V$
being a weight $p-q$-component of $\Lambda^n V_\C$.

\remark $V^{p,p}$ is the space of $U(1)$-invariant 
vectors in $\Lambda^{2p}V$.

Further on, {\bf \blue $TM$ is the tangent bundle on a manifold,
and $\Lambda^iM$ the space of differential $i$-forms.}
It is a Grassmann algebra on $TM$. 


\newpage

{\bf \blue Vector fields}

\definition
Let $X$ be the vector field on a manifold $M$, and $f$ a
function. Denote by $\Lie_X f$ {\bf \blue the derivatiive} of $f$ along
$X$. 

\definition
{\bf \blue A derivation} on a commutative ring is a map
$R \stackrel d \arrow R$ satisfying {\bf \blue the Leibniz
  identity} $d(xy) = d(x) y + x d(y)$.

\theorem
Each derivation of the ring $C^\infty M$ of smooth 
functions on $M$ is given by a vector field $X$;
{\bf \red this correspondence is bijective.}

\remark This can be used as a definition of a vector field.

\exercise
Prove that {\bf \red a commutator of two derivations is again a
derivation.}

\remark Vector fields are the same as derivations of
$C^\infty M$.
This allows us to define {\bf \blue the commutator of two
vector fields} as the commutator of the corresponding derivations. 

\definition
Denote by $TM$ the bundle of vector fields, and by
$\Lambda^1 M$ or $T^*$ the dual bundle, called
{\bf \blue the bundle of 1-forms}.
For any $f\in C^\infty M$,
the operation $X \arrow \Lie_X f$ is linear as a function
of $X$, hence it defines a section of $T^*M$. We denote 
this section $df$, and call it {\bf \blue the 
differential} of $f$.

\newpage

{\bf \blue De Rham algebra}

\definition
Let $\Lambda^* M$ denote the vector bundle
with the fiber $\Lambda^*T^*_xM$ at $x\in M$
($\Lambda^*T^*M$ is the Grassman algebra of the cotangent
space $T^*_x M$). The sections of $\Lambda^i M$
are called {\bf \blue differential $i$-forms}.
 The algebraic operation ``wedge product'' defined
on differential forms is $C^\infty M$-linear; 
the space $\Lambda^* M$ of all differential forms
is called {\bf \blue the de Rham algebra}.

\remark $\Lambda^0 M = C^\infty M$.

\theorem
{\bf \red There exists a unique operator 
$C^\infty M\stackrel d \arrow \Lambda^1 M\stackrel d
\arrow \Lambda^2 M \stackrel d \arrow \Lambda^3 M \stackrel d \arrow ...$
satisfying the following properties}


1. On functions, $d$ is equal to the differential.\\
2. $d^2=0$ \\
3. $d(\eta \wedge \xi) = 
d(\eta) \wedge \xi + (-1)^{\tilde \eta}\eta \wedge
d(\xi)$,
where $\tilde \eta=0$ where $\eta\in \lambda^{2i}M$ is
{\bf \blue an
even form,} and $\eta\in \lambda^{2i+1}M$ is {\bf \blue odd.}


\definition
The operator $d$ is called {\bf \blue de Rham differential}.

\exercise
{\bf \purple Prove it}.

\definition
A form $\eta$ is called {\bf \blue closed} if $d\eta=0$,
{\bf \blue exact} if $\eta \in \im d$. The group $\frac{\ker d}{\im d}$
is called {\bf \blue de Rham cohomology} of $M$.



\newpage

{\bf \blue Cauchy formula in dimension 1 (statement)}

\definition
Let $U\subset \C^n$ be an open subset, and
$f:\; U \arrow \C$ a function of class $C^1$ 
(differentiable at least once). We say that $f$
is {\bf \blue holomorphic} if the differential
$df:\; T_x U \arrow \C$ is complex linear at each
$x\in U$.

\remark Clearly, $f$ is holomorphic if and only if $df\in \Lambda^{1,0}(U)$,
where $\Lambda^{1,0}(U)$ is the Hodge (1,0)-component of the de Rham algebra.


{\bf \purple Taylor series decomposition for holomorphic
functions in 1 variable is implied by 
the Cauchy formula:} for any folomorphic function $f$ 
in disk $\Delta\subset \C$,
\[
\int_{\6 \Delta} \frac{f(z)dz}{z-a}= 2\pi\1 f(a),
\]
where $a\in \Delta$ any point,
and $z$ coordinate on $\C$. Indeed, in this case,
\[
2\pi\1 f(a) = \sum_{i\geq 0} a^i\int_{\6 \Delta} f(z) (z^{-1})^{i+1},
\]
because $\frac{1}{z-a}=z^{-1}\sum_{i\geq 0} (az^{-1})^i$.


\newpage

{\bf \blue Cauchy formula in dimension 1 (proof)}

Let's prove Cauchy formula, using Stokes' theorem.
Since the space $\Lambda^{1,0}\C$ is 1-dimensional,
$df\wedge dz=0$ for any holomorphic function on $\C$.
This gives

\claim
A function on a disk $\Delta\subset \C$ 
{\bf \red is holomorphic if and only if the form $\eta:=f dz$ is closed}
(that is, satisfies $d\eta=0$).
\endproof

Now, let $S_\epsilon$ be a radius $\epsilon$ circle around a point
$a\in \Delta$, $\Delta_\epsilon$ its interior, and 
$\Delta_0:=\Delta\backslash \Delta_\epsilon$. Stokes' theorem gives
\[
0= \int_{\Delta_0} d\left(\frac{f(z)dz}{z-a}\right)=
-\int_{S_\epsilon}\frac{f(z)dz}{z-a} + 
\int_{\6 \Delta} \frac{f(z)dz}{z-a},
\]
hence Cauchy formula would follow if we show that
$\lim\limits_{\epsilon\rightarrow 0}\int_{S_\epsilon}\frac{f(z)dz}{z-a}=2\pi\1 f(a).$
Assuming for simplicity $a=0$ and parametrizing the circle $S_\epsilon$
by $\epsilon e^{\1 t}$, we obtain 
\begin{multline*}
\int_{S_\epsilon}\frac{f(z)dz}{z}=
\int_0^{2\pi} \frac{f(\epsilon e^{\1t})}{\epsilon e^{\1t}} d(\epsilon e^{\1t})=\\
=\int_0^{2\pi} \frac{f(\epsilon e^{\1t})}{\epsilon e^{\1t}} \1 \epsilon e^{\1t}dt 
=\int_0^{2\pi}f(\epsilon e^{\1t})\1 dt
\end{multline*}
as $\epsilon$ tends to $0$, $f(\epsilon e^{\1t})$ tends to $f(0)$,
and this integral goes to $2\pi\1f(0)$.


\newpage

{\bf \blue Holomorphic functions on $\C^n$ (reminder)}


\theorem
Let $f:\; U \arrow \C$ be a differentiable function on 
an open subset $U\subset \C^n$. {\bf \red Then the following are equivalent.}\\
\ \ (1) {\bf \purple $f$ is holomorphic}.\\
\ \ (2) For any complex affine line $L\in \C^n$,
the restriction $f\restrict L=\C$ is {\bf \purple holomorphic  
as a function of one complex variable.}\\
\ \ (3) $f$ is expressed as a sum of Taylor series 
around any point $(z_1, ..., z_n)\in U$: for all
sufficiently small $t_1, ..., t_n$, one has
$f(z_1+t_1, z_2+ t_2, ..., z_n +t_n)= 
\sum_{i_1, ..., i_n}a_{i_1, ..., i_n} t_1^{i_1}t_2^{i_2}...t_n^{i_n}.$

\proof 
Equivalence of (1) and (2) is clear, because a restriction of
 $\theta \in \Lambda^{1,0}(M)$ to a line is a $(1,0)$-form on a line,
and, conversely, if $df$ is of type (1,0) on each complex line, it is
of type (1,0) on $TM$, which is implied by the following linear-algebraic 
observation.

\lemma
Let $\eta\in V^*\otimes \C$ be a complex-valued linear form
on a real vector space $(V,I)$ equipped with a complex
structure $I$.
{\bf \purple Then $\eta\in \Lambda^{1,0}(V)$ if and only if its restriction
to any $I$-invariant 2-dimensional subspace $L$ belongs to 
$\Lambda^{1,0}(L)$.} \\
\exercise {\bf \red Prove it.}

(3) clearly implies (1). 
(1) implies (3) by Cauchy formula (many variables), 
proven below.

\newpage

{\bf \blue Cauchy formula (many variables)}

\remark Let $U\subset \C^n$ be an open subset,
and $z_1, ..., z_n$ complex coordinates.
Holomorphicity of $f:\; U \arrow \C$ is equivalent
to $df\in \Lambda ^{1,0}(M)$, which is equivalent
to $df\wedge dz_1 \wedge dz_1 \wedge ... \wedge dz_n=0$.
Denote the form $dz_1 \wedge dz_1 \wedge ... \wedge dz_n$ by $\Phi$.
We obtain that {\bf \purple $f$ is holomorphic
if and only if the form $f\Phi$ is closed}

\theorem {\bf \blue (Cauchy formula in dimension $n$)}\\
Let $\Delta\subset \C^n$ be a polydisk (product of disks)
of radius 1, and   $\alpha_1, ..., \alpha_n\in \Delta$
complex numbers. Denote by $S\subset \C^n$ the product of circles
of radius 1 in variables $z_1, ..., z_n$:, 
$S= S_1(z_1) \times S_1(z_2) \times ... \times S_1(z_n)$.
Let $f$ be a holomorphic function in a polydisk. {\bf \red
  Then $\int_S V=
(2\pi\1)^n f(\alpha_1, ... \alpha_n)$, where
\[ V=\frac{f\Phi}{(z_1-\alpha_1) (z_2-\alpha_2) \times ... \times (z_n-\alpha_n)}.
\]}
\phantom{.\hspace{-20pt}} \pstep
Denote by $Z$ the set $\bigcup_{i=1}^n \{(z_1, ...,
z_n)\ \ |\ \ z_i=\alpha_i\}$. The complement of $Z$ is the
set of definition of the closed differential form $V$.
Let $S_\epsilon$ be the product of circles of radius
$\epsilon$ with center in  $\alpha_1, ..., \alpha_n$.
Then $S, S_\epsilon\subset \C^n \backslash Z$,
and {\bf \purple the tori $S$, $S_\epsilon$ are homotopy equivalent
in the domain $\C^n \backslash Z$, where $V$ is closed.}
{\bf \red It remains to show that
 $\lim_{\epsilon \rightarrow 0}\int_{S_\epsilon} V=
(2\pi\1)^n f(\alpha_1, ... \alpha_n)$.}


\newpage

{\bf \blue Cauchy formula (many variables), part 2}


\theorem {\bf \blue (Cauchy formula in dimension $n$)}\\
Let $\Delta\subset \C^n$ be a polydisk (product of disks)
of radius 1, and   $\alpha_1, ..., \alpha_n\in \Delta$
complex numbers. Denote by $S\subset \C^n$ the product of circles
of radius 1 in variables $z_1, ..., z_n$:, 
$S= S_1(z_1) \times S_1(z_2) \times ... \times S_1(z_n)$.
Let $f$ be a holomorphic function in a polydisk. {\bf \red
  Then $\int_S V=
(2\pi\1)^n f(\alpha_1, ... \alpha_n)$, where
\[ V=\frac{f\Phi}{(z_1-\alpha_1) (z_2-\alpha_2)  ...  (z_n-\alpha_n)}.
\]}
\phantom{.\hspace{-30pt}} \pstep
Let $S_\epsilon$ be a product of circles of radius
$\epsilon$
with center in $\alpha_1, ..., \alpha_n$.
{\bf \purple It remains to show that
 $\lim_{\epsilon \rightarrow 0}\int_{S_\epsilon} V=
(2\pi\1)^n f(\alpha_1, ... \alpha_n)$.}

{\bf \green Step 2:} To simplify notation we set
$\alpha_i=0$. Parametrize $S_\epsilon$ by the cube
 $[0, 2\pi]^n$ using the map 
$t_1, ..., t_n\arrow \epsilon e^{\1 t_1}, ..., \epsilon e^{\1 t_n}$.
This gives
\begin{multline*}
\int_{S_\epsilon} V= 
\int_{S_\epsilon}f(z) \frac{dz_1}{z_1}\wedge ... \wedge\frac{dz_n}{z_n}  =\\
=\int_0^{2\pi} ... \int_0^{2\pi} 
\frac{f(\epsilon e^{\1t_1 }, \epsilon e^{\1t_2 }, ...,\epsilon e^{\1t_n })}{\epsilon e^{\1t_1}\epsilon e^{\1t_2}... \epsilon e^{\1t_n}}\epsilon^n d\left(e^{\1t_1}\right)d\left(e^{\1t_2}\right) ...d\left(e^{\1t_n}\right) =\\
=(\1)^n\int_0^{2\pi}...\int_0^{2\pi}f(\epsilon e^{\1t_1}, ..., \epsilon e^{\1t_n })dt_1dt_2...dt_n,
\end{multline*}
which converges to $(2\pi\1)^n f(0,
..., 0)$. 
\endproof

\newpage

{\bf \blue Cauchy formula and Taylor expansion}

\remark Cauchy formula implies that
{\bf \red holomorphic functions defined in a polydisk
have Taylor expansion in this polydisk}.
Indeed, 
\[
f(\alpha_1, ... \alpha_n)= \frac1{(2\pi\1)^n} \int_S \frac{fdz_1 \wedge ... \wedge dz_n}{(
z_1-\alpha_1) (z_2-\alpha_2) \times ... \times (z_n-\alpha_n)}
\]
Take the Taylor expansion of $(z_i-\alpha_i)^{-1}$ using
\[
\frac 1 {(z_i-\alpha_i)}= \frac {z_i^{-1}} {(1-\alpha_i z_i^{-1})}
= \sum_{l=0}^\infty \alpha_i^l z_i^{-l-1}.
\]
Then
\[
f(\alpha_1, ... \alpha_n)= \sum_{i_1=0}^\infty ... \sum_{i_n=0}^\infty 
\alpha_1^{i_1} .... \alpha_{i_n}^{i_n} \int_{S} f(z_1,..., z_n)z_1^{-i_1-1} ...
z_n^{-i_n-1} dz_1 \wedge ... \wedge dz_n.
\]




\newpage

{\bf \blue Complex manifolds (reminder)}

\definition
{\bf \blue A holomorphic function} on $\C^n$ is a function
$f:\; \C^n \arrow \C$ such that $df$ is complex linear, that
is $df\in \Lambda^{1,0}(M)$.

\remark
Holomorphic functions form a sheaf.

\definition
{\bf \blue A complex manifold} $M$ is a ringed space which is locally isomorphic
to an open ball in $\C^n$ with a sheaf of holomorphic functions.

\remark 
In other words, {\bf \purple $M$ is covered with open balls embedded to $\C^n$}
and transition functions (being coordinate functions for one
ball considered in other coordinate system) {\bf \purple are holomorphic.}

\newpage


{\bf \blue Complex manifolds (reminder)}

\definition
{\bf \blue A holomorphic function} on $\C^n$ is a function
$f:\; \C^n \arrow \C$ such that $df$ is complex linear, that
is $df\in \Lambda^{1,0}(M)$.

\remark
Holomorphic functions form a sheaf.

\definition
{\bf \blue A complex manifold} $M$ is a ringed space which is locally isomorphic
to an open ball in $\C^n$ with a sheaf of holomorphic functions.

\remark 
In other words, {\bf \purple $M$ is covered with open balls embedded to $\C^n$}
and transition functions (being coordinate functions for one
ball considered in other coordinate system) {\bf \purple are holomorphic.}

\newpage

{\bf \blue Integrability of almost complex structures}


\definition
An almost complex structure $I$ on a manifold is called {\bf\blue integrable}
if any point of $M$ has a neighbourhood $U$ diffeomorphic 
to an open subset of $\C^n$, in such a way that 
the almost complex structure $I$ is induced by the standard
one on $U\subset \C^n$.

\claim 
{\bf \red  Complex structure on a manifold $M$ uniquely determines
an integrable almost complex structure, and is determined by 
it.}

\proof
Complex structure on a manifold $M$ is determined by the sheaf of holomorphic
functions $\calo_M$, and $\calo_M$ is determined by $I$ as explained above.
Therefore, an integrable almost complex structure defines a complex structure.
Conversely, every complex structure gives a sub-bundle
in $\Lambda^{1,0}(M)=d\calo_M\subset \Lambda^1(M,\C)$, and
{\bf \purple  such a sub-bundle defines an almost complex structure 
by Remark 1.}
\endproof

\newpage

{\bf \blue Frobenius form}

\claim
Let $B\subset TM$ be a sub-bundle of a tangent bundle of a 
smooth manifold. Given vector fiels $X,Y\in B$, consider their
commutator $[X,Y]$, and lets $\Psi(X,Y)\in TM/B$
be the projection of $[X,Y]$ to $TM/B$. {\bf \red Then $\Psi(X,Y)$
is $C^\infty(M)$-linear in $X$, $Y$: }
\[ \Psi(fX,Y)= \Psi(X,fY)=f\Psi(X,Y).
\]
\proof Leibnitz identity gives $[X,fY]=f[X,Y]+ X(f) Y$,
and the second term belongs to $B$, hence does not influence the
projection to $TM/B$. \endproof

\definition
This form is called {\bf \blue the Frobenius form} of the
sub-bundle $B\subset TM$. This bundle is called {\bf\blue involutive},
or {\bf \blue integrable}, or {\bf \blue holonomic} if $\Psi=0$.

\exercise {\bf \purple Give an example of a non-integrable sub-bundle.}

\newpage

{\bf \blue Formal integrability}


\definition
An almost complex structure $I$ on $(M,I)$
is called {\bf\blue formally integrable}
if $[T^{1,0}M, T^{1,0}]\subset T^{1,0}$, that is,
if $T^{1,0}M$ is involutive.

\definition The Frobenius form $\Psi\in \Lambda^2(\Lambda^{1,0}M)\otimes T^{0,1}M$
is called {\bf \blue the Nijenhuis tensor}.

\claim 
{\bf \purple 
If a complex structure $I$ on $M$ is integrable, it is formally integrable.}

\proof Locally, the bundle $T^{1,0}(M)$ is generated by $d/dz_i$, where $z_i$ are
complex coordinates. These vector fields commute, hence satisfy
$[d/dz_i, d/dz_j]\in T^{1,0}(M)$. This means that the Frobenius
form vanishes. \endproof

\theorem
{\bf \blue (Newlander-Nirenberg)}\\
{\bf\red A complex structure $I$ on $M$ is integrable if and only if it is
formally integrable.}

\proof (real analytic case) next lecture, probably.

\remark {\bf \purple In dimension 1, formal integrability is automatic.}
Indeed, $T^{1,0}M$ is 1-dimensional, hence all skew-symmetric
2-forms on $T^{1,0}M$ vanish.




\newpage

{\bf \blue Possible topics for the next lectures }

1. Proof of Frobenius theorem.

2. Newlander-Nirenberg theorem for real analytic almost complex manifolds.

\end{document}
