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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf  Topologia das Variedades \\[15mm]
\small Cohomology, lecture 2: de Rham differential}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[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 Antisymmetrization (reminder)}

The natural embedding from $\Lambda^n V$ to $T^{\otimes n} V$
is given by {\bf \blue antisymmetrization}
\[\Alt(x_1\wedge ...\wedge x_n):= 
\frac 1{n!}\sum_{\sigma\in \Sigma_n}(-1)^{\tilde \sigma}
x_{\sigma(1)}\otimes ...\otimes x_{\sigma(n)}.
\]
{\bf \green Its properties:}

1. Clearly, $\im \Alt$ lies in the space of antisymmetric 
tensors, and $\Alt(\eta - (-1)^{\tilde \sigma} \eta)=0$,
hence {\bf \purple $\Alt$ defines a map from Grassmann algebra
to the space of antisymmetric tensors.}

2. The natural {\bf \purple projection 
from antisymmetric tensors to the Grassmann algebra is
inverse to $\Alt$.}

3. This map {\bf \purple defines an isomorphism of the Grassmann algebra
and the space of antisymmetric tensors.}

\newpage

{\bf \blue Grassmann algebra: dimension of components (reminder)}

\remark
For linearly independent vectors $x_1, ..., x_k$,
the antisymmetrization 
$x_1 \wedge x_2 \wedge ... \wedge x_k :=\frac {1}{k!} \sum (-1)^{\tilde \sigma}
x_{\sigma(1)}\otimes ...\otimes x_{\sigma(k)}$
is non-trivial. Indeed, the monomials
$x_{\sigma(1)}\otimes ...\otimes x_{\sigma(k)}$
are linearly independent in $V^{\otimes k}$.
{\bf \purple This implies $\dim \Lambda^k V\geq \binom{\dim V}{k}$}.

\claim {\bf \red $\dim \Lambda^k V= 
\binom{\dim V}{k}$,  and
$\dim \Lambda^* V=2^{\dim V}$.}

\proof Let $x_1, ..., x_n$ be a basis in $V$.
Then {\bf \purple the space $\Lambda^k V$ is generated
by antisymmetric tensors $x_{i_1} \wedge x_{i_2}  \wedge ... \wedge x_{i_k}$,
$i_1 < i_2 < ... < i_k$,
which are all linearly independent.}
\endproof

\newpage

{\bf \blue Tensor product of vector bundles}


\definition
Let $V,V'$ be $R$-modules, $W$ a free abelian group
generated by $v\otimes v'$, with $v\in V, v'\in V'$,
and $W_1\subset W$ a subgroup generated by combinations
$rv \otimes v'-v\otimes rv'$, $(v_1+ v_2)\otimes v'-
v_1 \otimes v' - v_2 \otimes v'$ and $v\otimes (v'_1+ v'_2)-
v\otimes v'_1 - v\otimes v'_2$.
Define {\bf\blue the tensor product} $V \otimes_R V'$
as a quotient group $W/W_1$.

\exercise
Show that $r \cdot v\otimes v'\mapsto (rv)\otimes v'$
{\bf \purple defines an $R$-module structure on $V \otimes_R V'$.}


\definition
{\bf \blue Tensor product} of vector bundles
is a tensor product of the corresponding sheaves of modules.

\exercise 
 Let $B$ and $B'$ ve vector bundles on $M$,
$B\restrict x$, $B'\restrict x$ their fibers, and
$B\otimes_{C^\infty M} B'$ their tensor product.
{\bf \purple
Prove that $B\otimes_{C^\infty M} B'\restrict x = 
B\restrict x\otimes_\R B'\restrict x$.}

\newpage

{\bf \blue Cotangent bundle}

\definition
Let $M$ be a smooth manifold, 
$TM$ the tangent bundle, and $\Lambda^1 M = T^*M$ its dual bundle.
It is called {\bf \blue cotangent bundle} of $M$.
Sections of $T^*M$ are called {\bf \blue 1-forms}
or {\bf \blue covectors} on $M$.
For any $f\in C^\infty M$, consider a functional
$TM\arrow C^\infty M$ obtained by mapping $X\in TM$ to
a derivation of $f$: $X \arrow D_X(f)$. Since
this map is linear in $X$, it defines a section
$df\in T^*M$ called {\bf \blue the differential} of $f$.

\claim {\bf \red $\Lambda^1 M$ is
generated as a $C^\infty M$-module by
$d(C^\infty M)$.}

\proof Locally in coordinates $x_1, ..., x_n$ this is clear,
because the covectors $dx_1, ..., dx_n$ give a basis
in $T^* M$ dual to the basis $\frac d{dx_1}, ..., \frac d{dx_n}$ in 
$TM$. To obtain the same statement globally, use the
partition of unity. 
\endproof


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

\newpage

{\bf \blue De Rham algebra}


\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}

\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. $d(\eta \wedge \xi) = 
d(\eta) \wedge \xi + (-1)^{\tilde \eta}\eta \wedge
d(\xi)$,
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 Derivations determined by their values on generators}

\remark
A map $d$ on a graded algebra which satisfies the Leibnitz rule
$d(\eta \wedge \xi) = 
d(\eta) \wedge \xi + (-1)^{\tilde \eta}\eta \wedge
d(\xi)$ is called {\bf \blue an odd derivation}.

\remark The following two lemmas are needed to prove uniqueness
of de Rham differential.

\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 \red 
$\Lambda^* M$ is generated by $C^\infty M$ and $d(C^\infty M)$.}

\proof
By definition, $\Lambda^* M$ is generated by $\Lambda^0 M = C^\infty M$
and $\Lambda^1 M$. However, $d(C^\infty M)$ generate $\Lambda^1 M$,
as shown above. \endproof


\newpage

{\bf \blue De Rham differential: uniqueness and existence}


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

\proof De Rham differential is an odd derivation.
Its value on $C^\infty M$ is defined by the first axiom.
On $d(C^\infty M)$ de Rham differential valishes, because $d^2=0$.
\endproof

\definition
Let $t_1, ..., t_n$ be coordinate functions
on $\R^n$, $\alpha_i$ coordinate monomials, and $\alpha:= \sum f_i \alpha_i$.
Define $d(\alpha):=  \sum_i \sum_j\frac{df_i}{dt_j}dt_j \wedge \alpha_i$.

\exercise\\
{\bf \purple Check that $d$ satisfies the properties of de Rham differential.}

\corollary {\bf \red De Rham differential exists on any smooth manifold.}

\proof Locally, de Rham differential $d$ exists, as follows from the
construction above. Cover $M$ by open sets where $d$ is defined.
Since $d$ is unique, it is compatible on intersections of these
open sets. {\bf \purple
Therefore, it is well defined globally on $M$.}
\endproof



\newpage

{\bf \blue Georges de Rham (1903-1990)}

\begin{center}\epsfig{file=derham8.jpg,width=0.62\linewidth}\\[4mm]

{\it\small
Georges de Rham (with Marcel Berger)}
\end{center}




\newpage

{\bf \blue Integration}

\definition
Let $M$ be an $n$-dimensional manifold.
Clearly, $\Lambda^n(M)$ is a rank 1 vector bundle.
This bundle is trivial if and only if it has
a non-vanishing section $\eta$. 
Let's call two such sections $\eta$, $\eta'$ {\bf\blue equivalent}
if $\eta = f \eta'$, where $f\in C^\infty M$ is positive everywhere.
A choice of equivalence class of such sections is called {\bf \blue
an orientation} on $M$, and $M$ is called {\bf\blue oriented}. 
A section of $\Lambda^n(M)$
which belongs to this class is called {\bf \blue
a differential $n$-form, positive everywhere}, or 
{\bf\blue  a volume form}.

\theorem
Let $M$ be an $n$-dimensional compact oriented
manifold. Denote by $\Lambda^n (M)$ the space of sections
of the vector bundle $\Lambda^n (M)$.
{\bf \red Then there exists a 
linear functional $\int_M:\; \Lambda^n (M)\arrow \R$}
which is positive on volume forms and invariant
under the  $\Diff(M)$ (diffeomorphism group of $M$) action.  Moreover,
{\bf \red  such a functional is unique up to a positive constant
multiplier.}

\remark To fix the normalization, it would suffice
to take a diffeomorphism from a unit cube to $U\subset M$
which can be extended to a small neighbourhood of the cube.
Consider the pushforward $\eta$ of the standard volume form 
$dx_1 \wedge ... \wedge dx_n$ from the cube to $M$,
and let $\int_M \eta=1$. It is not hard to see
that {\bf \purple all such maps from the cube to $U\subset M$
are diffeomorphism-conjugate}, hence {\bf \red this choice
defines the functional $\int_M$ uniquely.}

\newpage

{\bf \blue Stokes' theorem and Poincar\'e lemma}

\definition
The map $\int_M:\; \Lambda^n (M)\arrow \R$
defined above is called {\bf\blue the integral}
of the differential form.

{\bf \green Stokes' theorem:} {\bf \blue (Poincar\`e)}\\
Let $\eta$ be $n-1$-form on $n$-manifold $M$ with a boundary $\6 M$.
{\bf \red Then \[\int_M d\eta=\int_{\6 M}\eta.\]}

{\bf \green Existence and uniqueness of the 
integral and Stokes' theorems are left as exercises for now.}

{\bf \green Poincar\'e lemma:}\\
 {\bf \red De Rham cohomology $H^i(B)$ of an open
ball $B$ vanish for $i>0$.}



\end{document}
