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            lecture 11 \hfil
  \tiny M. Verbitsky }}
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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf Geometry of manifolds \\[15mm]
\small Lecture 11: the Lie derivative}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]

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

{\tiny\bf Math in Moscow and HSE
\\[2mm]  April 29, 2013
}
\end{center}

\newpage

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


\definition
Let $M$ be a smooth manifold. {\bf \blue A bundle of 
differential $i$-forms on $M$} is the bundle
$\Lambda^i T^* M$ of antisymmetric $i$-forms on $TM$.
It is denoted $\Lambda^i M$.

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

\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.


\newpage

{\bf \blue Coordinate monomials (reminder)}


\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}. 

\claim
{\bf \purple $\Lambda^*\R^n$ is a trivial bundle, 
and coordinate monomials are free generators of $\Lambda^*\R^n$.}



\definition
An associative algebra $A^* = \oplus_{i\in \Z}A^i$
is called {\bf \blue a graded algebra} if for all $a\in A^i$, $b\in A^j$, 
the product $ab$ lies in $A^{i+j}$.

\example De Rham algebra is a graded algebra.


\newpage

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


\definition
{\bf\blue De Rham differential}  $d:\; \Lambda^*M \arrow \Lambda^{*+1}M$
is an $\R$-linear map satisfying the following conditions.\\
\phantom{xxx} * For each $f \in \Lambda^0M=C^\infty M$,
$d(f)\in \Lambda^1 M$ is equal to the image
of the K\"ahler differential $df\in \Omega^1 M$
in $\Lambda^1 M = \Omega^1 M/K$.\\
\phantom{xxx} * {\bf \blue (Leibnitz rule)}
$d(a\wedge b) = da \wedge b + (-1)^j a\wedge
db$ for any $a\in \Lambda^i M, b \in \Lambda^j M$.\\
\phantom{xxx} * $d^2=0$.


\theorem\\
{\bf \red 
De Rham differential is uniquely determined by these axioms.}

\remark The proof of uniqueness is based on the following lemmas.

\lemma
Let $A=\bigoplus A^i$ be a graded algebra, $B\subset A$ a set of multiplicative 
generators, and $D_1, D_2:\; A \arrow A$ two odd derivations
which are equal on $B$. {\bf \purple Then $D_1=D_2$.} \endproof


\lemma 
{\bf \purple
$\Lambda^* M$ is generated by $C^\infty M$ and $d(C^\infty M)$.}

\remark
Let $t_1, ..., t_n$ be coordinate functions
on $\R^n$, $\alpha_i$ coordinate monomials, and $\alpha:= \sum f_i \alpha_i$.
Define $d(\alpha):=  \sum_i \sum_j\frac{df_i}{dt_j}dt_j \wedge \alpha_i$.
{\bf \red Then $d$ satisfies axioms of de Rham differential.} This proves 
its existence.


\newpage

{\bf \blue May 6: no lecture}

{\bf \Huge \red May 6, 2013:\\
NO LECTURE!}

Next lecture - May 13, 2013.



\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 
which shifts a grading by $i$ is called {\bf \blue even}
if $i$ is even, and {\bf \blue odd} otherwise.

\exercise
Prove that a 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.
Each time when in commutative case
two letters $A$, $F$ 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 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.}

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

\proof
By the super Jacobi identity, 
$\{d, \{d, E\}\}=-\{d, \{d, E\}\}+\{\{d, d,\} E\}\}$,
however, $\{d, d\} =2d^2 =0$.
\endproof

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

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


\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, commute 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 Leibnitz identity, they are equal
on $C^\infty M$ and commute with the de Rham differential.}
\endproof


\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
$\frac d{dt}V_t\restrict{t=c}$ is a derivation} (that is, a vector field).

\definition
The vector field $\frac d{dt}V_t\restrict{t=c}$
is called {\bf\blue a vector field tangent to a flow of
diffeomorphisms $V_t$ at $t=c$.}



\newpage

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


\definition
Let $v$ be a vector field on $M$, and
$V:\;M\times[a,b]\arrow M$ a flow of diffeomorphisms which satisfies
$\frac d{dt}V_t\restrict{t=c}=v$ for each $c$, and $V_0=\Id$.
Then $V_t$ is called {\bf\blue an exponent of $v$}.

\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 $v$ be a vector field, 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 $\Lie_v\alpha = \frac{d}{dt}(V_t^*\alpha)$.}

\proof
By definition, $\Lie_v=\frac{d}{dt}V_t$ on functions.
$\Lie_v$ 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 
$\Lie_v\alpha = \frac{d}{dt}(V_t^*\alpha)$.} \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


\definition
Let $d$ be de Rham differential.
A form in $\ker d$ is called
{\bf\blue closed}, a form in $\im d$ is called {\bf\blue exact}.
Since $d^2=0$, any exact form is closed.
{\bf\blue The group of $i$-th de Rham cohomology of $M$},
denoted $H^i(M)$, is a quotient of a space of closed
$i$-forms by the exact: $H^*(M)=\frac{\ker d}{\im d}$.

\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

\newpage

{\bf \blue Poincar\'e lemma}

\definition
An open subset $U\subset \R^n$
is called {\bf\blue starlike} if for any
$x\in U$ the interval $[0,x]$ belongs to $U$.


\theorem {\bf \blue (Poicar\'e lemma)}
Let $U\subset \R^n$ be a starlike subset.
{\bf \red Then $H^i(U)=0$ for $i>0$.}

\remark
{\bf \purple The proof would follow if we construct a vector field
$\vec{r}$ such that $\Lie_{\vec r}$ is invertible on
$\Lambda^*(M)$:} $\Lie_{\vec r}R= \Id$. Indeed, for any
closed form $\alpha$ we would have 
$\alpha = \Lie_{\vec r} R\alpha= di_{\vec r} R\alpha+ 
i_{\vec r} Rd\alpha=di_{\vec r} R\alpha,$
hence any closed form is exact.

Then Poincar\'e lemma is implied by the following 
statement.

\proposition
Let $U\subset \R^n$ be a starlike subset, $t_1, ..., t_n$ 
coordinate functions,
and $\vec r:= \sum t_i \frac d {dt_i}$ the radial
vector field. {\bf \red Then $\Lie_{\vec r}$ is invertible on
$\Lambda^i(U)$ for $i>0$.}

\newpage

{\bf \blue Radial vector field on starlike sets}

\proposition
Let $U\subset \R^n$ be a starlike subset, 
 $t_1, ..., t_n$ coordinate functions,
and $\vec r:= \sum t_i \frac d {dt_i}$ the radial
vector field. {\bf \red Then $\Lie_{\vec r}$ is invertible on
$\Lambda^i(U)$ for $i>0$.}

{\bf \green Proof. Step 1:}
Let $t$ be the coordinate function on a real line,
$f(t)\in C^\infty \R$ a smooth function,
and $v:= t \frac d {dt}$ a vector field. Define
$R(f)(t):=\int^1_0 \frac{f(\lambda t)}{\lambda} d\lambda$.
Then this integral converges whenever $f(0)=0$, 
and satisfies $\Lie_v R(f)=f$. Indeed, 
\[ \int^1_0 \frac{f(\lambda t)}{\lambda} d\lambda=
\int^t_0 \frac{f(\lambda t)}{t\lambda} d(t\lambda)=
\int^t_0 \frac{f(z)}{z} d(z),\] hence
$\Lie_v R(f)= t \frac{f(t)}{t}= f(t)$.

{\bf \green Step 2:}
Consider a function $f\in C^\infty \R^n$ satisfying
$f(0)=0$, and  $x=(x_1, ..., x_n)\in \R^n$. {\bf \purple Then
\[
R(f)(x):=\int_0^1 \frac{f(\lambda x)}{\lambda} d\lambda
\]
converges, and satisfies $\Lie_{\vec{r}} R(f)=f$.}


\newpage

{\bf \blue Radial vector field on starlike sets (cont.)}

{\bf \green Step 3:}
Consider a differential form  $\alpha\in\Lambda^i$,
and let $h_\lambda x\arrow \lambda x$ be the homothety with coefficient
$\lambda\in [0,1]$. Define
\[ 
R(\alpha):= \int^1_0 \lambda^{-1} h_\lambda^*(\alpha) d\lambda.
\]
Since $h_\lambda^*(\alpha)=0$ for $\lambda=0$, this integral
converges. {\bf \red It remains to prove that $\Lie_{\vec r}R=\Id$.}

{\bf \green Step 4:} Let $\alpha$ be a coordinate monomial, $\alpha=dt_{i_1}
\wedge dt_{i_2} \wedge ... \wedge dt_{i_k}$.
Clearly, $\Lie_{\vec r} (T^{-1}\alpha)=0$, where
$T=t_{i_1}t_{i_2}... t_{i_k}$. 
{\bf \purple Since $h_\lambda^* (f\alpha)= h_\lambda^*(tf) T^{-1}\alpha$, we have
$R (f\alpha)= R(Tf) T^{-1}\alpha$ for any function $f\in C^\infty M$.}
This gives 
\[ \Lie_{\vec r} R(f\alpha) = \Lie_{\vec r}R(Tf) T^{-1}\alpha=
TfT^{-1}\alpha = f\alpha.
\]
\endproof






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