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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf  Topologia das Variedades \\[15mm]
\small Cohomology, lecture 3: the Lie derivative}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]

{\small\bf Misha Verbitsky  
\\[2mm]
April 26, 2018\\[4mm]

{\tiny\bf IMPA, Tuesdays and Thursdays, 10:30, Sala 224}
}
\end{center}

\newpage

{\bf \blue The Grassmann algebra (reminder)}

\definition
Let $V$ be a vector space, and $W\subset V\otimes V$
a subspace generated by vectors $x\otimes
y + y\otimes x$ and $x\otimes x$, for all $x, y\in V$.
A graded algebra defined by the generator space $V$ and
the relation space $W$ is called {\bf\blue Grassmann algebra},
or {\bf\blue exterior algebra}, and denoted
$\Lambda^*(V)$. The space $\Lambda^i(V)$
is called {\bf\blue $i$-th exterior power} of $V$, and
the multiplication in $\Lambda^*(V)$ -- 
{\bf\blue exterior multiplication}. Exterior multiplication
is denoted $\wedge$.


\exercise
{\bf \purple Prove that $\Lambda^1 V$ is isomorphic to $V$.}


\definition
An element of Grassmann algebra is called {\bf\blue even}
if it lies in $\bigoplus_{i\in \Z} \Lambda^{2i}(V)$
and {\bf \blue odd} if it lies in $\bigoplus_{i\in \Z} \Lambda^{2i+1}(V)$.
For an even or odd $x\in \Lambda^*(V)$, we define a number
$\tilde x$ called {\bf\blue parity} of $x$.
The parity of $x$ is 0 for even $x$ and 1 for odd.

\claim
{\bf \purple In Grassmann algebra, 
$x\wedge y = (-1)^{\tilde x \tilde y}y\wedge x$.}

\newpage

{\bf \blue De Rham algebra (reminder)}


\definition
Let $\alpha\in (V^*)^{\otimes i}$ and $\alpha\in (V^*)^{\otimes j}$
be polylinear forms on $V$. Define the {\bf\blue tensor multiplication}
$\alpha \otimes \beta$ as\\
\phantom{xxxxxxx} $\alpha \otimes \beta(x_1, ..., x_{i+j}):= 
  \alpha(x_1, ..., x_j) \beta(x_{i+1}, ..., x_{i+j}).$

\definition
Let $\bigotimes_k T^*M \stackrel \Pi \arrow \Lambda^k M$ 
be the antisymmetrization map,
\[ \Pi(\alpha)(x_1, ..., x_n):= \frac 1{n!} \sum_{\sigma\in \Sym_n} 
(-1)^\sigma \alpha(x_{\sigma_1}, x_{\sigma_2}, ..., x_{\sigma_n}).
\]
Define {\bf \blue the exterior multiplication
$\wedge:\; \Lambda^i M\times \Lambda^j M \arrow \Lambda^{i+j} M$} as
$\alpha \wedge \beta := \Pi (\alpha \otimes \beta)$,
where $\alpha \otimes \beta$ is a section
$\Lambda^i M\otimes \Lambda^j M \subset \bigotimes_{i+j}
T^*M$ obtained as their tensor multiplication.

\remark The fiber of the bundle $\Lambda^* M$ at $x\in M$
{\bf \purple is identified with the Grassmann algebra $\Lambda^* T^*_x M$.}
This identification is compatible with the Grassmann product.


\definition
Let $t_1, ..., t_n$ be coordinate functions
on $\R^n$, and $\alpha\in \Lambda^* \R^n$ a monomial
obtained as a product of several $dt_i$:
$\alpha = dt_{i_1}\wedge dt_{i_2} \wedge ... \wedge dt_{i_k}$
$i_1 < i_2 < ...< i_k$. Then $\alpha$ is called {\bf \blue
a coordinate monomial}. 

\newpage

{\bf \blue De Rham differential (reminder)}

\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}\\
\phantom{XXX} 1. On functions, $d$ is equal to the differential.\\
\phantom{XXX} 2. $d^2=0$ \\
\phantom{XXX} 3. {\bf \blue (Graded Leibnitz identity)} $d(\eta \wedge \xi) = 
d(\eta) \wedge \xi + (-1)^{\tilde \eta}\eta \wedge
d(\xi)$,
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}.

\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 Graded algebras (reminder)}

\definition
An algebra $A$ is called {\bf\blue graded} if $A$ is 
represented as $A= \bigoplus A^i$, where $i\in \Z$,
and the product satisfies $A^i \cdot A^j
\subset A^{i+j}$. Instead of $\bigoplus A^i$
one often writes  $A^*$, 
where $*$ denotes all indices together.
Some of the spaces $A^i$ can be zero, but the 
ground field is always in $A^0$, so that it is non-empty. 

\example
{\bf \purple The tensor algebra 
$T(V)$ and the polynomial algebra $\Sym^*(V)$
are obviously graded.}


\definition
A subspace $W\subset A^*$ of a graded algebra
is called {\bf\blue graded} if $W$ is a direct sum of components
$W^i \subset A^i$. 

\exercise
Let $W\subset  T(V)$ be a graded subspace.
Prove that then {\bf \purple the algebra generated by $V$ with relation
space $W$ is also graded.}

\newpage

{\bf \blue Superalgebras}

\definition
Let  $A^* = \oplus_{i\in \Z}A^i$ be a graded algebra over a field.
It is called
{\bf\blue graded commutative}, or {\bf\blue supercommutative},
if $ab = (-1)^{ij} ba$
for all $a\in A^i, b \in A^j$. 

\example
{\bf \purple Grassmann algebra $\Lambda^* V$
is clearly supercommutative.}


\definition
Let $A^*$ be a graded commutative algebra, 
and $D:\; A^* \arrow A^{*+i}$ be a map which
shifts grading by $i$. 
It is called a
{\bf\blue  graded derivation}, if
$D(ab) = D(a) b + (-1)^{ij} a D(b)$,
for each $a \in A^j$. 

\remark
If $i$ is even, graded derivation is a usual derivation.
If it is even, it an odd derivation.

\definition
Let $M$ be a smooth manifold, and  $X\in TM$ 
a vector field. Consider an operation of {\bf \blue 
convolution with a vector field} 
$i_X:\; \Lambda^i M \arrow \Lambda^{i-1}M$, 
mapping an $i$-form $\alpha$ to an $(i-1)$-form
$v_1, ..., v_{i-1} \arrow \alpha(X, v_1, ..., v_{i-1})$


\exercise
{\bf \purple Prove that $i_X$ is an odd derivation.}



\newpage

{\bf \blue Supercommutator}


\definition
Let $A^*$ be a graded vector space, and 
$E:\; A^*\arrow A^{*+i}$, $F:\; A^*\arrow A^{*+j}$
operators shifting the grading by $i, j$.
Define  {\bf\blue the supercommutator} $\{E, F\}:= EF - (-1)^{ij} FE$.

\definition
An endomorphism of a graded vector space
which shifts grading by $i$ is called {\bf \blue even}
if $i$ is even, and {\bf \blue odd} otherwise.

\exercise
Prove that the supercommutator satisfies 
{\bf\blue graded Jacobi identity,}
\[
\{ E, \{F, G\}\} = \{\{ E, F\}, G\} + (-1)^{\tilde E
\tilde F}  \{ F, \{E, G\}\}
\]
where  $\tilde E$ and $\tilde F$ are 0 if
$E, F$ are even, and 1 otherwise.


\remark
There is a simple mnemonic rule
which allows one to remember a superidentity,
if you know the commutative analogue.
{\bf\red Each time when in commutative case
two letters $E$, $F$ are exchanged, in supercommutative
case one needs to multiply by $(-1)^{\tilde E\tilde F}$.}


\exercise
{\bf \purple Prove that a supercommutator of superderivations
is again a superderivation.}


\newpage


{\bf \blue Pullback of a differential form}

\definition
Let  $M \stackrel \phi\arrow N$ be a morphism of smooth manifolds,
and $\alpha \in \Lambda^i N$ be a differential form.
Consider an $i$-form $\phi^* \alpha$ taking value
\[ \alpha\restrict{\phi(m)}(D_\phi(x_1), ... D_\phi(x_i))\] on
$x_1, ..., x_i\in T_m M$. It
is called {\bf\blue the pullback of $\alpha$}. 
If $M\stackrel \phi \arrow N$ is a closed embedding,
the form $\phi^* \alpha$ is called
{\bf \blue the restriction} of $\alpha$ to 
$M \hookrightarrow N$.

\lemma {\bf \blue (*)}
Let $\Psi_1, \Psi_2:\; \Lambda^* N \arrow \Lambda^* M$
be two maps which satisfy graded Leibnitz identity, supercommutes with
de Rham differential, and satisfy \\ $\Psi_1\restrict{C^\infty M}=
\Psi_2\restrict{C^\infty M}$. {\bf \red Then $\Psi_1 = \Psi_2$.}

\proof
The algebra $\Lambda^* M$ is generated multiplicatively
by $C^\infty M$ and $d(C^\infty M)$; restrictions of $\Psi_i$
to these two spaces are equal. \endproof


\claim
{\bf \red Pullback commutes with the de Rham differential.}

\proof
Let $d_1, d_2:\; \Lambda^* N \arrow \Lambda^{*+1} M$
be the maps $d_1 = \phi^* \circ d$ and $d_2= d\circ \phi^*$.
{\bf \purple These maps satisfy the Leibnitz identity, and they are equal
on $C^\infty M$.} The super-commutator $\delta:= \{d_i, d\}$
is equal to $d \circ\phi^* \circ d$, it commutes 
with $d$, and equal 0 on functions. By Lemma (*), $\delta=0$.
Then $d_i$ supercommutes with $d$. Applying Lemma (*) again,
we obtain that $d_1=d_2$.
\endproof


\newpage

{\bf \blue Lie derivative}

\definition
Let $B$ be a smooth manifold, and $v\in TM$ a vector field.
An endomorphism $\Lie_v:\; \Lambda^* M \arrow \Lambda^* M$,
preserving the grading is called
{\bf\blue a Lie derivative along $v$} if it satisfies the 
following conditions.\\
\phantom{XX} (1) On functions $\Lie_v$ is 
equal to a derivative along $v$.  (2) $[\Lie_v, d]=0$.\\
 \phantom{XX} (3) $\Lie_v$ is a derivation of the de Rham algebra.

\remark
The algebra $\Lambda^*(M)$ is generated 
by $C^\infty M=\Lambda^0(M)$ 
and $d(C^\infty M)$. The restriction
$\Lie_v\restrict{C^\infty M}$ is determined by the first axiom.
On $d(C^\infty M)$ is also determined
because $\Lie_v(df) =d(\Lie_v f)$. {\bf \red Therefore,
$\Lie_v$ is uniquely defined by these axioms.}

\exercise Prove that
{\bf \purple $\{d, \{d, E\}\}=0$
for each $E\in \End(\Lambda^* M)$.}


\theorem
{\bf \blue (Cartan's formula)}
Let $i_v$ be a convolution with a vector field, $i_v(\eta) = \eta(v, \cdot, \cdot, ..., \cdot)$
{\bf \red Then $\{d, i_v\}$ is equal to the Lie derivative along $v$.}

\proof
$\{d, \{d, i_v\}\}=0$ by the lemma above.
A supercommutator of two graded derivations is a graded derivation.
Finally, $\{d, i_v\}$ acts on functions as $i_v(df)=\langle v, df\rangle.$
\endproof

\newpage

\ \ \\[2mm]

{\bf \blue Cartan's magic formula:} $d\circ i_x + i_x
\circ d =\Lie_x$. 

\centerline{\red \bf Which Cartan?} 

{\small
\centerline{\begin{tabular}{cc}
\'Elie Cartan (1869-1951) & Henri Cartan (1904-2008)\\ 
\epsfig{file=Cartan_big.jpg,height=0.4\linewidth} &
\epsfig{file=Henri-Cartan.jpg,height=0.4\linewidth}\\
 {\bf \green \'Elie Cartan?} & {\bf \green Henri Cartan?}\\
(Robert Bryant and Dick Palais
& (S.S. Chern: Lectures \\
Mathoverflow) & on differential geometry)
\end{tabular}}
}



\newpage

{\bf \blue Flow of diffeomorphisms}


\definition
Let $f:\; M \times [a,b]\arrow M$ be a smooth map
such that for all $t\in [a,b]$ the restriction
$f_t:= f\restrict{M\times\{t\}}:\; M \arrow M$
is a diffeomorphism. Then $f$ is called {\bf\blue a flow
of diffeomorphisms}. 

\claim
Let $V_t$ be a flow of diffeomorphisms, $f\in C^\infty M$,
and $V^*_t(f)(x):= f(V_t(x))$. Consider the map
$\frac d{dt}V_t\restrict{t=c}:\; C^\infty M \arrow C^\infty M$,
with $\frac d{dt}V_t\restrict{t=c}(f)=
(V_c^{-1})^*\frac{dV_t}{dt}\restrict{t=c}f$. 
{\bf \red Then
$f\arrow (V_t^{-1})^*\frac d{dt}V_t^*f$ is a derivation} 
(that is, a vector field).

\proof
$\frac d{dt}V_t^*(fg)= V_t^*(f)\frac d{dt}V_t^* g+ \frac d{dt}V_t^* f  V_t^*(g)$
by the Leignitz rule, giving
\[ (V_t^{-1})^*\frac d{dt}V_t^*(fg)= f (V_t^{-1})^*\frac d{dt}V_t^*g + 
   g (V_t^{-1})^*\frac d{dt}V_t^*f.
\]
\endproof\\
\definition
The vector field $\frac d{dt}V_t\restrict{t=c}$
is called {\bf\blue the vector field tangent to a flow of
diffeomorphisms $V_t$ at $t=c$.}

\claim
Let $V_t$ be a flow of diffeomorphisms and $X_t$ the corresponding
vector field. {\bf \red Then for any $\eta\in \Lambda^* M$, one has
$\frac d{dt}V_t^*(\eta)= \Lie_{X_t}(\eta)$.}

\proof The operators $\frac d{dt}V_t^*$ and $\Lie_{X_t}$
are equal on functions, satisfy the Leibitz identity and
commute with $d$. \endproof

\newpage

{\bf \blue Flow of diffeomorphisms obtained from vector fields}

\exercise
Let $M$ be a compact manifold, and 
$\Psi:\; C^\infty M \arrow C^\infty M$
is a ring automorphism. Prove that {\bf \purple 
$\Psi$ is induced by an action of a diffeomorphism
of $M$.}


\theorem
Let $M$ be a compact manifold, and $X_t\in TM$ a family
of vector fields smoothly depending on $t\in [0,a]$.
{\bf \red Then there exists a unique diffeomorphism flow $V_t$,  $t\in [0,a]$,
such that $V_0=\Id$ and $\frac d{dt}V_t^*=X_t$.}

\pstep
Given $f\in C^\infty M$, we can solve an equation
$\frac d{dt}W_t(f)=\Lie_{X_t}(f)$ (here $\Lie_{X_t}(f)$
denotes the derivative along the vector field). 
The solution $W_t(f)$ exists for all $t\in [0,x]$ 
and is unique by Peano theorem on existence and
uniqueness of solutions of ODE,

{\bf \green Step 2:} Since 
\[ 
  \frac d{dt}W_t(fg)= \Lie_{X_t}(f) g + \Lie_{X_t}(g)f =
  \frac d{dt}(W_t(f)W_t(g)),
\]
$W_t$ is multiplicative. Also, it is invertible.
Applying the previous exercise, we obtain that
$W_t$ is a diffeomorphism.
\endproof

For another proof see Chapter 5 of
 	{\em Arnold V.I.,	 Ordinary Differential Equations}

\newpage

{\bf \blue Flow of diffeomorphisms 
from vector fields on maifolds with boundary}

\remark
The same result {\bf \purple holds for compact manifolds with boundary
if the vector field is tangent to the boundary.} The proof is the same.

\remark {\bf \purple The same result holds if $M$ is a complete
Riemannian manifold and $X_t$ is $C$-Lipschitz for all $t$}, that is,
satisfies $\bigg ||X_t|_x - |X_t|_y\bigg|\leq Cd(x,y)$.
It also follows from the Peano theorem.

Let $M$ be a compact manifold with boundary $\6 M$, and $X_t\in TM$ a family
of vector fields smoothly depending on $t\in [0,a]$. Suppose that 
$X_t \restrict {\6M}$ is pointed inside, and not outside.
We can embed $M$ to $M_1= M \cup \6M\times [0, 1]$
obtained by gluing $\6 M \subset M$ to 
$\6 M \times \{0\}\subset \6M\times [0, 1]$
and extend $X_t$ to a vector field (also denoted by $X_t$)
on $M_1$, tangent to $\6 M_1$ on the boundary
$\6 M_1$. Applying the previous theorem, and using
the fact that the phase trajectories of $X_t$
cross $\6 M \subset M_1$ in one direction, we obtain 
the following theorem.


\theorem
Let $M$ be a compact manifold with boundary, and $X_t\in TM$ a family
of vector fields smoothly depending on $t\in [0,a]$. Suppose that
$X_t \restrict {\6M}$ points inside $M$.
{\bf \red Then there exists a unique family of smooth embeddings
$V_t:\; M \arrow M$,  $t\in [0,a]$,
such that $\frac d{dt}V_t^*=X_t$ and $V_0=\Id$.}
\endproof


\newpage

{\bf \blue Lie derivative and a flow of diffeomorphisms}


\definition
Let $v_t$ be a vector field on $M$, smoothly depending on 
the ``time parameter'' $t\in[a,b]$, and
$V:\;M\times[a,b]\arrow M$ a flow of diffeomorphisms which satisfies
$\frac d{dt}V_t=v_t$ for each $t\in[a,b]$, and $V_0=\Id$.
Then $V_t$ is called {\bf\blue an exponent of $v_t$}.

\claim 
Exponent of a vector field is unique; it exists when $M$
is compact. This statement is called {\bf \blue ``Picard-Lindel\"of theorem''}
or {\bf \blue ``uniqueness and existence of solutions of ordinary differential
equations''}. 

\proposition
Let $v_t$ be a time-dependent vector field, $t\in[a,b]$,
and $V_t$ its exponent.
For any $\alpha \in \Lambda^*M$, consider $V_t^*\alpha$
as a $\Lambda^*M$-valued function of $t$.
{\bf \red Then $\Lie_{v_0}(\alpha) = \frac{d}{dt}\restrict{t=0}(V_t^*\alpha)$.}


\proof
By definition of differential, $\Lie_{v_0}f= \langle df, v_0\rangle$,
hence $\Lie_{v_0}f=\frac{d}{dt}\restrict{t=0}V_t^*(f)$.
The operator $\Lie_{v_0}$ commutes with de Rham differential, because
$\Lie_v=i_v d+di_v$. The map $\frac{d}{dt}V_t$ commutes
with de Rham differential, because it is a derivative of a 
pullback. Now {\bf \purple Lemma (*) is applied to show that 
$\Lie_{v_0}\alpha = \frac{d}{dt}\restrict{t=0}(V_t^*\alpha)$.} \endproof



\newpage

{\bf \blue Homotopy operators}

\definition
{\bf\blue A complex} is a sequence of vector spaces
and homomorphisms
$... \stackrel d\arrow C_{i-1} \stackrel d\arrow
C_i\stackrel d\arrow C_{i+1} \stackrel d\arrow...$\\
satisfying $d^2=0$.  {\bf \blue
Homomorphism} $(C_*, d)\arrow (C_*', d)$ of complexes 
is a sequence of homomorphism $C_i \arrow C'_i$
commuting with the differentials.

\definition
An element $c\in C_i$ is called {\bf\blue closed}
if $c\in \ker d$ and {\bf \blue exact} if $c\in \im d$.
{\bf \blue Cohomology} of a complex
is a quotient $\frac{\ker d}{\im d}$.

\remark A homomorphism of complexes induces a natural
homomorphism of cohomology groups.


\definition
Let $(C_*, d)$, $(C_*', d)$ be a complex.
{\bf \blue Homotopy} is a sequence of maps $h:\; C_*
\arrow C'_{*-1}$. Two homomorphisms $f, g:\; (C_*, d)\arrow (C_*', d)$ 
are called {\bf \blue homotopy equivalent} if $f-g =\{h,d\}$
for some homotopy operator $h$.

\claim
Let  $f, f':\; (C_*, d)\arrow (C_*', d)$ be homotopy
equivalent maps of complexes. {\bf \red Then $f$ and $f'$ induce
the same maps on cohomology.}

{\bf \green Proof. Step 1:}
Let $g:= f-f'$. It would suffice to prove that $g$
induces 0 on cohomology. 


\newpage

{\bf \blue Lie derivative and homotopy}

\claim
Let  $f, f':\; (C_*, d)\arrow (C_*', d)$ be homotopy
equivalent maps of complexes. {\bf \red Then $f$ and $f'$ induce
the same maps on cohomology.}

{\bf \green Proof. Step 1:}
Let $g:= f-f'$. It would suffice to prove that $g$
induces 0 on cohomology. 

{\bf \green Step 2:}
Let $c\in C_i$ be a closed element.
{\bf \purple Then $g(c)= dh(c)+ hd(c)=dh(c)$ exact.} \endproof

\remark
Let $v$ be a vector field, and $\Lie_v:\; \Lambda^* M\arrow \Lambda^* M$
be the corresponding Lie derivative. Then {\bf \red $\Lie_v$
commutes with the de Rham differential, and acts
trivially on the de Rham cohomology.}

\proof 
$\Lie_v=i_v d+di_v$ maps closed forms to exact. \endproof

\corollary
Let $V_t$, $t\in[a, b]$ be a flow of diffeomorphisms on a manifold $M$.
{\bf \red 
Then $V_b^*$ (the pullback) acts on cohomology the same way as $V_a^*$.}

\proof
Since $\frac {d V^*_t}{dt}(\eta)= \Lie_{X_t}(\eta)$, 
it acts trivially on cohomology. Then 
$V_b^*-V_a^*(\eta)=\int_{a}^b \Lie_{X_t}(\eta)$ is exact for any closed $\eta$.
Therefore, $V_b^*(\eta)-V_a^*(\eta)$ is exact.
\endproof


\end{document}
