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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf  Topologia das Variedades \\[15mm]
\small Cohomology, lecture 4: de Rham cohomology of a sphere}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]

{\small\bf Misha Verbitsky  
\\[2mm]
May 3, 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 Pullback of a differential form (reminder)}

\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 (reminder)}

\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.}

\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 $v$.}


\newpage

{\bf \blue Flow of diffeomorphisms (reminder)}


\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).

\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 $\frac d{dt}V_t^*=X_t$.}


\newpage

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

\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 $\Lie_{v_t}(\alpha) = (V_t^{-1})^*\frac{d}{dt}(V_t^*\alpha)$.}

\newpage

{\bf \blue Lie derivative and cohomology}

\claim
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 the pullback map 
$V_b^*$ acts on cohomology the same way as $V_a^*$.}

\proof
Since $(V_t^{-1})^*\frac {d V^*_t}{dt}(\eta)= \Lie_{X_t}(\eta)$, 
this map it acts trivially on cohomology. Then 
$V_b^*-V_a^*(\eta)=\int_{a}^b V^*_t\Lie_{X_t}(\eta)$ is exact for any closed $\eta$.
Therefore, $V_b^*(\eta)-V_a^*(\eta)$ is 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 $i>0$.}

\proof
Consider the map $V_t:\; U \arrow U $ mapping $x$ to $tx$,
where $t\in [0,1]$. Then the map 
$\eta\arrow (V_t^{-1})^*\frac {dV_t}{dt} (\eta)$ 
commutes with $d$ and maps closed forms to exact,
and hence acts trivially on cohomology. {\bf \purple Therefore, 
$V_1^*-V_0^*= \int_0^1 (V_t^{-1})^*\frac {dV_t}{dt}$
also acts trivially on cohomology.} However,
$V_0^*$ maps any $\eta \in \Lambda^i(M)$, $i>0$ to 0.
\endproof

\newpage

{\bf \blue Homogeneous spaces}

\definition
{\bf \blue A Lie group} is a smooth manifold equipped with a group
structure such that the group operations are smooth.
Lie group $G$ {\bf \blue acts on a manifold $M$}
if the group action is given by the smooth map
$G \times M \arrow M$. 

\definition
Let $G$ be a Lie group acting on a manifold $M$ transitively.
Then $M$ is called {\bf \blue a homogeneous space}.
For any $x\in M$ the subgroup $\St_x(G)=\{g\in G\ \ |\ \ g(x)=x\}$
is called {\bf \blue stabilizer of a point $x$}, or
{\bf \blue isotropy subgroup}.

\claim
For any homogeneous manifold $M$ with transitive
action of $G$, {\bf \purple one has $M=G/H$,} 
where $H=\St_x(G)$ is an isotropy subgroup.

\proof The natural surjective map $G\arrow M$ putting $g$ to $g(x)$
identifies $M$ with the space of conjugacy classes $G/H$.
\endproof

\remark Let $g(x)=y$. Then $\St_x(G)^g=\St_y(G)$:
{\bf \purple all the isotropy groups are conjugate.}

\newpage

{\bf \blue Isotropy representation}

\definition 
Let $M=G/H$ be a homogeneous space, $x\in M$ and
$\St_x(G)$ the corresponding stabilizer group.
The {\bf \blue isotropy representation}
is the natural action of $\St_x(G)$
on $T_x M$.

\definition
A tensor $\Phi$ on a homogeneous manifold $M=G/H$ 
is called {\bf \blue invariant} if it is mapped
to itself by all diffeomorphisms which come from
$g\in G$.

\remark
Let $\Phi_x$ be an isotropy invariant tensor
on $\St_x(G)$. For any $y\in M$ obtained as $y=g(x)$,
consider the tensor $\Phi_y$ on $T_y M$ obtained as
$\Phi_y:=g(\Phi)$. The choice of $g$ is not unique,
however, for another $g'\in G$ which satisfies $g'(x)=y$,
we have $g=g'h$ where $h\in \St_x(G)$. Since
$\Phi$ is $h$-invariant, {\bf \purple the tensor $\Phi_y$
is independent from the choice of $g$.}

We proved

\theorem
{\bf \red Homogeneous tensors on $M=G/H$ are in bijective
correspondence with isotropy invariant tensors on
$T_x M$,} for any $x\in M$.
\endproof

\newpage

{\bf \blue Cohomology of homogeneous spaces}

\claim
Let $M=G/H$ be a homogeneous space, with $G$ connected and compact.
Denote by $\Lambda^*(M)^G$ the space of $G$-invariant
differential forms. Then {\bf \red the natural morphism of complexes
\[
\begin{CD}
\Lambda^0(M)^G @>d>> \Lambda^1(M)^G @>d>> \Lambda^2(M)^G @>d>> ... \\
@V\iota VV @V\iota VV @V\iota VV \\
\Lambda^0(M) @>d>> \Lambda^1(M) @>d>> \Lambda^2(M) @>d>> ...
\end{CD}
\]
induces an isomorphism on cohomology.}

\proof
Consider the averaging map 
$\eta \stackrel \Av \arrow \frac {\int_{g\in G} g^*(\eta) dg}{\Vol(G)}$
where the volume and the integral is 
taken with respect to a $G$-invariant measure $dG$ on $G$.
Since $g^*$ commutes with the de Rham differential $d$,
the map $\Av$ also commutes with $d$.
Since $g^*(\eta)$ is cohomologous to $\eta$
as shown above, the form $\Av(\eta)$ is cohomologous to $\eta$.
Therefore, $\iota$ is invertible on cohomology:
$\iota \circ \Av =\Id, \Av\circ \iota=\Id$.
\endproof


\newpage

{\bf \blue Special orthogonal group $SO(n)$}

\definition
{\bf \blue Orthogonal group} $O(n)$ 
is the group of linear isometries of $\R^n$.
{\bf \blue Special orthogonal group},
denoted $SO(n)$, is the
group of orthogonal matrices
$A\in \End(\R^n)$ of determinant 1.

\remark
Clearly, $SO(n)$ {\bf \purple acts on $(n-1)$-dimensional
sphere $S^{n-1}$ transitively. }

\newpage

{\bf \blue The special orthogonal group $SO(n)$ is conected}

\theorem
{\bf \red The group $SO(n)$ is connected.}\\
\proof
Each element  $A\in SO(n)$ can be represented in a certain
basis by a block matrix,
\[\small
gAg^{-1}=\begin{pmatrix}
        A_{\alpha_1} &               &        & 0\\
                      & A_{\alpha_2} &        &  \\
                      &               & \ddots &  \\
           0          &               &        & A_{\alpha_k}
\end{pmatrix}
\]
where $A_{\alpha_1}\in \End(\R^2)$ are rotation matrices,
\[\small
A_{\alpha_i}=\begin{pmatrix}\cos\alpha_i & \sin \alpha_i \\ -\sin \alpha_i & \cos\alpha_i
\end{pmatrix}
\]
Then 
\[\small
A_t:=g^{-1}\begin{pmatrix}
        A_{t\alpha_1} &               &        & 0\\
                      & A_{t\alpha_2} &        &  \\
                      &               & \ddots &  \\
           0          &               &        & A_{t\alpha_k}
\end{pmatrix} g
\]
is a homotopy connecting $A_t=A$ to $A_0=\Id$.
\endproof

\corollary
{\bf \red Cohomology of a sphere $S^n$ are equal to the cohomology
of the algebra of $SO(n+1)$-invariant differential forms.}

\proof We have just shown that $SO(n+1)$ is connected;
its compactness is left as an exercise. \endproof

\newpage

{\bf \blue Special orthopgonal group $SO(n)$ acting on a sphere}

\theorem
Let $\alpha$ be an $SO(n+1)$-invariant $p$-form on
$S^n$. {\bf \red Then $p=0$ or $p=n$ and $\alpha$ is either 
a constant or a constant times the Riemannian volume form.}

\pstep Using induction on $n$, we can assume 
that the statement of the theorem is true for $S^k$, $k<n$
(it is clearly true for $S^1$).
Since $O(k+1)$ acts on $S^k$ in non-orientable way,
{\bf \purple all $O(k+1)$-invariant forms on $S^k$ are constant
0-forms.}

{\bf \green Step 2:}
Let $S^{k} \subsetneq S^n$ be a sphere associated
with a $k+1$-dimensional subspace $V= \R^{k+1}\subset W= \R^{n+1}$.
Clearly, the stabilizer of $S^{k}$ in $SO(n+1)$ acts on $S^k$
as a full orthogonal group $O(k+1)$. {\bf \purple Therefore, any
$SO(n+1)$-invariant form restricted to $S^k= S^n \cap V$ gives 
a constant 0-form,} for any $k+1$-dimensional subspace $V\subset W$.

{\bf \green Step 3:} Suppose that there exists a non-zero
$SO(n+1)$-invariant $k$-form $\alpha$ on $S^n$, with $0<k<n$.
Then $\alpha$ restricted to $S^k$ is 0, for any sphere
${\goth S}=S^k\subset S^n$ obtained as ${\goth S}= S^n \cap V$.
However, there is such a sphere tangent to any $k$-dimensional
subspace in $T_x S^n$ for each $x\in S^n$. Therefore,
$\alpha(v_1, ..., v_k)=0$ for any $v_1, ..., v_k\in T_x S^n$.
\endproof


\newpage

{\bf \blue Cohomology of $S^n$}

\theorem
{\bf \red De Rham cohomology $H^p(S^n)$ of the sphere $S^n$ are 
1-dimensional for $p=0, n$ and 0-dimensional otherwise.}

\pstep 
De Rham cohomology $H^0(X)=\R$ for any connected manifold $M$.
Indeed, $\ker(d) \cap \Lambda^0(X)$ is the space of constant 
functions.

{\bf \green Step 2:} 
De Rham cohomology $H^p(S^n)$  are represented by $SO(n+1)$-invariant forms,
hence they vanish for $0<p<n$. The space of $SO(n)$-invariant
$n$-forms is 1-dimensional, because all such forms are proportional
to the space of Riemannian volume forms. These forms are
not cohomologous to 0 because the space $\Lambda^{n-1}(S^n)^{SO(n+1)}$
is trivial, and one can compute cohomology of $S^n$ by taking the
cohomology of the complex of $SO(n+1)$-invariant forms.
\endproof


\end{document}
