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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf Hodge theory \\[15mm]
\small lecture 9: Complex manifolds}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]
{\scriptsize NRU HSE, Moscow} \\[15mm]
{\small Misha Verbitsky, February 21, 2018 } 
\end{center}

\newpage

{\bf \blue Complex structure on vector spaces}

\definition
Let $V$ be a vector space over $\R$, and $I:\; V \arrow V$
an automorphism which satisfies $I^2 = - \Id_V$. 
Such an automorphism is called {\bf\blue a complex structure
operator} on $V$.

{\bf \blue We extend the
action of $I$ on the tensor spaces 
$V\otimes V \otimes ... \otimes V \otimes 
V^*\otimes V^* \otimes ... \otimes V^*$ by multiplicativity:}
$I(v_1 \otimes ... \otimes w_1 \otimes ... \otimes w_n)=
I(v_1) \otimes ... \otimes I(w_1)\otimes ... \otimes I(w_n)$.


{\bf \red Trivial observations:}\\
1. {\bf \purple The eigenvalues $\alpha_i$ of $I$ are $\pm \1$.}
Indeed, $\alpha_i^2=-1$.

2. {\bf \purple $V$ admits an $I$-invariant, positive definite
scalar product (``metric'') $g$.}
Take any metric $g_0$, and let $g:= g_0 + I(g_0)$.

3. {\bf \purple $I$ is orthogonal for such $g$.}\\
Indeed, $g(Ix,Iy)=g_0(x,y)+ g_0(Ix,Iy)=g(x,y)$.


4. {\bf \purple $I$ diagonalizable over $\C$.} 
Indeed, any orthogonal matrix is diagonalizable.

5. {\bf \purple There are as many $\1$-eigenvalues as there are
$-\1$-eigenvalues.} 

\newpage

{\bf \blue Comples structure operator in coordinates}

This implies that in an appropriate basis in $V \otimes_\R \C$, the almost
complex structure operator is diagonal, as follows:

\[\left[
\begin{array}{c|c}
\begin{array}{cccc}
\1 &  & &\\
  &  \1 & & \\
  &    & \ddots &  \\
  &    &  &  \1 
\end{array}
 & 0 \\ \hline
0 & \begin{array}{cccc}
-\1 &  & &\\
  &  -\1 & & \\
  &    & \ddots &  \\
  &    &  &  -\1 
\end{array}
\end{array}\right]
\]
We also obtain its normal form in a real basis:
\[ \begin{bmatrix}
0 & -1 \\
1 &  0 \\
  &    & 0 & -1 \\
  &    & 1 &  0 \\
  &    &   &   & \ddots   \\
  &    &   &   & & \ddots \\
  &    &   &   & &       & 0 & -1 \\
  &    &   &   & &       & 1 &  0
\end{bmatrix}
\]

\newpage

{\bf \blue Hermitian structures}


% Denote by $\nu$ the {\bf \blue real structure operator,}
%$\nu(\sum \lambda_i w_i)=\sum \bar \lambda_i w_i$, where $w_i\in V$
%is a basis. Then $\nu(I(z))=I(\nu(z))$, that is, {\bf \red $I$ is real}. 
%For any $\1$-eigenvector
%$w$, one has $I(\nu(w))=\nu(I(w))=\nu(\1 w)=-\1 w$,
%hence {\bf \purple 
%$\nu$ exchanges $\1$-eigenvectors and $-\1$-eigenvectors.}


\definition An $I$-invariant positive definite
scalar product on $(V,I)$ is called
{\bf \blue an Hermitian metric}, and $(V,I,g)$ -- 
an Hermitian space.



\remark
Let $I$ be a complex structure operator on a real 
vector space $V$, and $g$ -- a Hermitian metric. Then
{\bf \red
the bilinear form $\omega(x,y) := g(x, Iy)$ is skew-symmetric.}
Indeed, $\omega(x,y) = g(x, Iy) = g(Ix, I^2y) = -g(Ix, y) = -\omega(y, x)$.

\definition
A skew-symmetric form
 $\omega(x,y)$ is called {\bf \blue an Hermitian form on $(V,I)$}.

\remark 
In the triple $I, g, \omega$, {\bf \purple each element can 
recovered from the other two.}


\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}.


\newpage

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

\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.


\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 Holomorphic functions}

\definition
Let $I:\; TM \arrow TM$ be an endomorphism of a tangent bundle
satisfying $I^2=-\Id$. Then $I$ is called {\bf \blue
almost complex structure operator}, and the pair $(M,I)$
{\bf \blue an almost complex manifold}.

\example
$M=\C^n$, with complex coordinates $z_i=x_i + \1 y_i$,
and $I(d/dx_i)=d/dy_i$, $I(d/dy_i)=-d/dx_i$.

\example 
In complex dimension 1, {\bf \purple almost complex structure
is the same as conformal structure with orientation}
{\bf \red (prove it).}

\definition
A function $f:\; M \arrow \C$ on an almost
complex manifold is called {\bf \blue holomorphic}
if $df \in \Lambda^{1,0}(M)$.

\remark
For some almost complex manifolds, {\bf \red there are no holomorphic
functions at all}, even locally. 

Example: $S^6$ with a certain
canonical ($G_2$-invariant) almost complex structure.


\newpage

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

\theorem
Let $f:\; M \arrow \C$ be a differentiable function on 
an open subset $M\subset \C^n$, with the natural almost
complex structure. {\bf \red Then the following are equivalent.}\\
\ \ (1) {\bf \purple $f$ is holomorphic}.\\
\ \ (2) The differential $df:\; TM \arrow \C$,
considered as a form on the vector space $T_x M=T_x\C^n=\C^n$
{\bf \purple is $\C$-linear.} \\
\ \ (3) For any complex affine line $L\in \C^n$,
the restriction $f\restrict L=\C$ is {\bf \purple holomorphic (complex analytic) 
as a function of one complex variable.}\\
\ \ (4) $f$ is expressed as a sum of Taylor series around any point $(z_1, ..., z_n)\in M$:
\[
f(z_1+t_1, z_2+ t_2, ..., z_n +t_n)= 
\sum_{i_1, ..., i_n}a_{i_1, ..., i_n} t_1^{i_1}t_2^{i_2}...t_n^{i_n}.
\]
(here we assume that the complex numbers $t_i$ satisfy
$|t_i|<\epsilon$, where $\epsilon$ depends on $f$ and $M$). 

\proof (1) and (2) are tautologically equivalent.
Equivalence of (1) and (3) is also 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.

\newpage

{\bf \blue Holomorphic functions on $\C^n$ (2)}

\theorem
Let $f:\; M \arrow \C$ be a differentiable function on 
an open subset $M\subset \C^n$, with the natural almost
complex structure. {\bf \red Then the following are equivalent.}\\
\ \ (1) {\bf \purple $f$ is holomorphic}.\\
\ \ (2) The differential $df:\; TM \arrow \C$,
considered as a form on the vector space $T_x M=T_x\C^n=\C^n$
{\bf \purple is $\C$-linear.} \\
\ \ (3) For any complex affine line $L\in \C^n$,
the restriction $f\restrict L=\C$ is {\bf \purple holomorphic (complex analytic) 
as a function of one complex variable.}\\
\ \ (4) $f$ is expressed as a sum of Taylor series 
around any point $(z_1, ..., z_n)\in M$.

\lemma
Let $\eta\in V^*\otimes \C$ be a complex-valued linear form
on a vector space $(V,I)$ equipped with a complex structure.
{\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.}

(4) clearly implies (2). 
(1) implies (4) by Cauchy formula.


\newpage

{\bf \blue Taylor decomposition from Cauchy formula}

{\bf \purple Taylor series decomposition on a line is implied by 
the Cauchy formula:}
\[
\int_{\6 \Delta} \frac{f(z)dz}{z-a}= 2\pi\1 f(a),
\]
where $\Delta\subset \C$ is a disk, $a\in \Delta$ 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}

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 Sheaves}

\definition
A {\bf\blue presheaf of functions} on a
topological space $M$ is a collection of subrings 
${\cal F}(U)\subset C(U)$ in
the ring $C(U)$ of all functions on $U$, for each open 
subset $U\subset M$, such that the restriction of
every $\gamma\in{\cal F}(U)$ to an open subset $U_1\subset U$ belongs to
${\cal F}(U_1)$. 


\definition
A presheaf of functions
${\cal F}$ is called {\bf\blue a sheaf of functions} 
if these subrings
satisfy the following condition. Let $\{U_i\}$ be a cover of an open
subset $U\subset M$ (possibly infinite)
and $f_i\in{\cal F}(U_i)$ a family of
functions defined on the open sets of the cover and 
compatible on the pairwise intersections:
$$f_i|_{U_i\cap U_j}=f_j|_{U_i\cap U_j}$$
for every pair of members of the cover. {\bf \purple Then there exists
$f\in{\cal F}(U)$ such that $f_i$ is the restriction of $f$ to $U_i$ for
all $i$.}


\newpage

{\bf \blue Sheaves and presheaves: examples}

{\bf \green Examples of sheaves:}

* Space of continuous functions

* Space of smooth functions, any differentiability class

* Space of real analytic functions


{\bf \green Examples of presheaves which are not sheaves:}

* Space of constant functions {\bf \purple (why?)}

* Space of bounded functions {\bf \purple (why?)}

\newpage

{\bf \blue Ringed spaces}

A {\bf\blue ringed space} $(M,{\cal F})$ is a
topological space equipped with a sheaf of
functions. A~{\bf \blue morphism}
$(M,{\cal F})\stackrel\Psi\longrightarrow(N,{\cal F}')$ of ringed spaces
is a continuous map $M\stackrel\Psi\longrightarrow N$ such that, for
every open subset $U\subset N$ and every function $f\in{\cal F}'(U)$, the
function $\psi^* f:=f\circ\Psi$ belongs to the ring
${\cal F}\big(\Psi^{-1}(U)\big)$. An {\bf\blue  isomorphism} of ringed spaces
is a homeomorphism $\Psi$ such that $\Psi$ and $\Psi^{-1}$ 
are morphisms of ringed spaces.

\example
Let $M$ be a manifold of class $C^i$ and let
$C^i(U)$ be the space of functions of this class. 
{\bf \purple Then $C^i$ is a sheaf of functions, and 
$(M, C^i)$ is a ringed space.}


\remark
Let $f:\; X \arrow Y$ be a smooth map of smooth manifolds.
Since a pullback $f^*\mu$ of a smooth function $\mu\in C^\infty(M)$
is smooth, {\bf \purple a smooth map of 
smooth manifolds defines a morphism of ringed spaces.}


\newpage

{\bf \blue Complex manifolds}

\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 and almost complex manifolds}

\definition {\bf \blue Standard almost complex structure}
is $I(d/dx_i)=d/dy_i$, $I(d/dy_i)=-d/dx_i$ on $\C^n$ with
complex coordinates $z_i=x_i+\1 y_i$.

\definition
A map $\Psi:\; (M,I)\arrow (N,J)$ 
from an almost complex manifold to an almost complex
manifold is called {\bf \blue holomorphic} if 
$\Psi^*(\Lambda^{1,0}(N))\subset \Lambda^{1,0}(M)$.

\remark
This is the same as $d\Psi$ being complex linear; 
for standard almost complex structures, {\bf \purple this is the same
as the coordinate components of $\Psi$ being holomorphic functions.}


\definition
{\bf \blue A complex manifold} is a manifold equipped with an atlas
with charts identified with open subsets of $\C^n$ and transition
functions 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 \purple 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.
\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 $\Psi\in \Lambda^{2,0}M\otimes TM$
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.

\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 Distributions}

\definition
{\bf \blue Distribution} on a manifold is 
a sub-bundle $B\subset TM$

\remark
Let  $\Pi:\; TM \arrow TM/ B$ be the projection, and
$x, y \in B$ some vector fields. Then 
$[fx, y]= f[x,y] - D_y (f) x$. This implies that
{\bf \purple  $\Pi([x,y])$ 
is $C^\infty(M)$-linear as a function of $x$ and $y$.}

\definition
The map $[B,B]\arrow TM/B$ we have constructed is
called {\bf \blue Frobenius bracket} (or {\bf \blue Frobenius form}); 
it is a skew-symmetric $C^\infty(M)$-linear form on $B$ with values in $TM/B$.

\definition
A distribution is called {\bf \blue integrable},
or {\bf \blue holonomic}, or {\bf \blue involutive}, if
its Frobenuus form vanishes.


\newpage

{\bf \blue Smooth submersions}

\definition
Let $\pi:\; M \arrow M'$ be a smooth map of manifolds.
This map is called {\bf \blue submersion} if 
at each point of $M$ the differential $D\pi$ is surjective,
and {\bf \blue immersion} if it is injective.

\claim
Let $\pi:\; M \arrow M'$ be a submersion. Then each
$m\in M$ has a neighbourhood $U\cong V\times W$, where
$V, W$ are smooth and {\bf \red $\pi \restrict U$
is a projection of $V\times W=U\subset M$ to $W\subset M'$ along $V$.}

\exercise {\bf \purple Deduce this result from the inverse function theorem.}

\exercise
{\bf \blue (``Ehresmann's fibration theorem'')}\\
Let $\pi:\; M \arrow M'$ be a smooth submersion of compact
manifolds. Prove that $\pi$ is a locally trivial fibration.

\definition
{\bf \blue Vertical tangent space} $T_\pi M\subset TM$ 
of a submersion $\pi:\; M \arrow M'$ is  the kernel of $D\pi$. 

\claim
Let $\pi:\; M \arrow M'$ be a submersion and
$T_\pi M\subset TM$ the vertical tangent space.
{\bf \red Then $T_\pi M$ is an involutive subbundle.}

\proof $D_\pi([X, Y])= [D_\pi(X), D_\pi(Y)]=0$ for any $X, Y \in \ker D_\pi$.
\endproof



\newpage

{\bf \blue Frobenius theorem (statement)}


{\bf \green Frobenius Theorem:}
Let $B\subset TM$ be a sub-bundle. Then $B$ is involutive
if and only if each point $x\in M$ has a neighbourhood
$U\ni x$ and {\bf \red a smooth submersion $U\stackrel \pi \arrow V$ 
such that $B$ is its vertical tangent space: $B= T_\pi M$.}


\remark
The implication {\bf \blue ``$B=T_\pi M$'' $\Rightarrow$ ``Frobenius form vanishes''}
was proven above.


\definition
The fibers of $\pi$ are called {\bf \blue leaves},
or {\bf \blue integral submanifolds} of the distribution $B$.
Globally on $M$, {\bf \blue a leaf of $B$} is a maximal
connected manifold $Z\hookrightarrow M$ which is immersed to $M$ 
and tangent to $B$ at each point.
A distribution for which Frobenius theorem holds is called
{\bf \blue integrable}. If $B$ is integrable, the set of 
its leaves is called {\bf \blue a foliation}. The leaves
are manifolds which are immersed to $M$, but not necessarily closed.




\newpage

{\bf \blue Frobenius theorem: existence of integral submanifolds}

\remark
To prove the Frobenius theorem for $B\subset TM$, {\bf \purple it suffices to show that
each point is contained in an interal submanifold.} In this case,
the smooth submersion $U\stackrel \pi \arrow V$  is a projection
to the leaf space of the distribution.

\remark
{\bf \red When $B$ is 1-dimensional} (in this case one says that
{\bf \blue $B$ has rank 1}, denoted $\rk B=1$), {\bf \red Frobenius theorem
follows from existence of the diffeomorphism flow associated with
a vector field.} Indeed, locally we may assume that $B$ admits
a non-degenerate section $v$. Let $V_t:\; M \times \R \arrow M$ be the corresponding 
flow of diffeomorphisms. Then $Z_m:=V_t(\{m\}\times \R$ is tangent to
$v$ everywhere, hence it is a 1-dimensional manifold immersed in
$M$. Clearly, $Z_m$ is a leaf this distribution. {\bf \purple Since $B$
is a tangent to a foliation, it is integrable.}


Further on we shall need the following exercise.

\exercise
Let $V_t= e^{v_t}$ be a diffeomorphism flow on $M$, and
$F\subset TM$ a vector bundle. Assume that $[v_t, F]\subset F$.
{\bf \red Prove that then $V_t$ preserves $F\subset TM$.}

\newpage

{\bf \blue Basic sub-bundles (1)}

\definition
Let $B\subset TM$ be an involutive sub-bundle.
A sub-bundle $F\subset TM$ is called {\bf \blue basic} for $B$
if $F\supset B$ and for all $b\in B, b'\in F$, one has 
 $[b, b']\in F$.

\remark One should think of basic sub-bundles as of {\bf \purple sub-bundles
preserved by all diffeomorphisms obtaned from exponentiation 
of a vector field $v\in B$.}

\lemma Let $B\subset TM$ be an integrable distribution, 
$\pi:\; M \arrow M_1$ projection to the leaf space of $B$,
and $F\supset B$ a sub-bundle of $TM$ containing $B$.
Then the following conditions are equivalent: {\bf \red
(a) $F$ is basic for $B$.\\
(b) There exists a sub-bundle $F_1\subset TM_1$ 
such that $\pi^{-1}(F_1) = F$.}

\proof Next slide.

\newpage

{\bf \blue Basic sub-bundles (2)}

\lemma Let $B\subset TM$ be an integrable distribution, 
$\pi:\; M \arrow M_1$ projection to the leaf space of $B$,
and $F\supset B$ a sub-bundle of $TM$ containing $B$.
Then the following conditions are equivalent: {\bf \red
(a) $F$ is basic for $B$.\\
(b) There exists a sub-bundle $F_1\subset TM_1$ 
such that $\pi^{-1}F_1 = F$.}

\pstep 
Consider coordinates $x_1, ..., x_n$ on $M$ such that
$x_{k+1}=\pi^*(x_{k+1}', ..., x_n=\pi^*(x_n)$, where $x_i', i= k+1, k+2,..., n$ 
are coordinates on $M_1$, and $\frac {d}{dx_1}, ..., \frac {d}{dx_k}$
generate $B$. Locally such coordinates always exist, because 
$B$ is integrable. 
Denote by $G$ a subgroup of $\Diff(M)$ obtained by exponents
of $\frac {d}{dx_1}, ..., \frac {d}{dx_k}$. Since $[B, F]\subset F$,
the corresponding diffeomorphisms preserve $F$. {\bf \purple Therefore,
$F$ is a $G$-invariant sub-bundle of $TM$.}

{\bf \green Step 2:} {\bf \purple Any $G$-invariant sub-bundle
$F\supset B$ is obtained as $\pi^{-1}(F_1)$ for some
sub-bundle $F_1 \subset TM_1= M/G$.} Indeed, since the action
of $G_1$ is free, the bundle $F$ is generated over $C^\infty M$
by $G$-invariant sections. However, any $G$-invariant 
bundle $F$ containing $B$ is generated by $G$-invariant sections,
which can be lifted from $M/G$ {\bf \purple (check this)}.

{\bf \green Step 3:} Conversely, if $F$ is lifted from 
$M_1=M/G$, it is $G$-invariant, hence $e^{tb}(b')\subset F$,
and this gives $[b, b']\subset F$ {\bf \purple (check this)}.
\endproof

\newpage

{\bf \blue Frobenius theorem (proof)}


{\bf \green Frobenius Theorem:}
Let $B\subset TM$ be a sub-bundle. Then $B$ is involutive
if and only if each point $x\in M$ has a neighbourhood
$U\ni x$ and {\bf \red a smooth submersion $U\stackrel \pi \arrow V$ 
such that $B$ is its vertical tangent space: $B= T_\pi M$.}

\pstep
Consider a rank 1 sub-bundle $B_1 \subset B$.
Using the diffeomorphism flow as above, we prove
that $B_1$ is integrable. Since  $[B_1, B]\subset B$,
the bundle $B$ is basic with respect to $B_1$.
{\bf \purple Therefore, $B= \pi^{-1}(B')$ for some $B'\subset TM_1$,
where $M_1$ is the leaf space of $B_1$.}

{\bf \green Step 2:} Let 
$\pi:\; M \arrow M_1$ be the projection to the leaf space.
Then $B= \pi^{-1}(B')$, where $\rk B'=\rk B-1$. 
Using induction in $\rk B$, we can assume that $B'$
is integrable. Let $\pi_0:\; M_1 \arrow M_0$ be the
projection to the leaf space of $B'$, defined locally in $M$.
{\bf \purple Then $\pi\circ\pi_0:\; M \arrow M_0$  is the projection to the
leaf space of $B$.}
\endproof




\end{document}
