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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf  Topologia das Variedades \\[15mm]
\small Cohomology, lecture 9: Poincar\'e duality }
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]

{\small\bf Misha Verbitsky  
\\[2mm]
June 7, 2018\\[4mm]

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


\newpage

{\bf \blue Cohomology  with compact support}

\definition
Let $\omega$ be a differential form on $M$ {\bf \blue Support}
$\Supp(\omega)$ of $\omega$ is the closure of the set of all points where $\omega\neq 0$.
We say that $\omega$ {\bf \blue has compact support}
if its support is compact.

\remark
Clearly, {\bf \purple a differential of a form with compact support 
has compact support.}

\definition
{\bf \blue Cohomology with compact support} of a manifold $M$
is cohomology of the complex $(\Lambda^*_c(M), d)$ 
of differential forms with compact support.

\remark Let $U\subset M$ be an open subset. {\bf \red Then
$\Lambda^*_c(U) \subset \Lambda^*_c(M)$.} Indeed, 
any two non-intersecting closed subsets 
of a manifold can be separated by non-intersecting open subsets.
Applying this to $\Supp(\alpha)$ and $M \backslash U$,
we obtain an open set $V\supset M \backslash U$ such that
$\alpha\restrict {V\cap U}=0$. Extend $\alpha$ to $V$
by setting $\alpha=0$ on the whole $V$, and we obtain 
a smooth extension of $\alpha$ to $M$.

\remark
This defines a map $H^*_c(U) \arrow H^*_c(M)$.
{\bf \purple Cohomology is contravariant functor} (the functor
which inverts the arrows), {\bf \purple cohomology with compact
support is covariant functor}.

\newpage

{\bf \blue Long exact sequence (reminder)}

\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} $(C^*, d)\arrow (C_1^*, d)$ of complexes 
is a sequence of homomorphism $C^i \arrow C_1^i$
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 $\frac{\ker d}{\im d}$.
One denotes the $i$-th group of cohomology 
of a complex by $H^i(C^*)$

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

\definition
{\bf \blue Short exact sequence of complexes}\\
$0 \arrow A^* \arrow B^* \arrow C^* \arrow 0$
is a sequence of morphisms of complexes
$A^* \stackrel x \arrow B^* \stackrel y\arrow C^*$
such that $x:\; A^i \arrow B^i$ is injective,
$y:\; B^i \arrow C^i$ is surjective (for all $i$),
and $\ker y= \im x$.

\theorem
Let  $0 \arrow A^* \arrow B^* \arrow C^* \arrow 0$
be an exact sequence of complexes. Then 
there exists a {\bf \blue long exact sequence of cohomology}
\[... \arrow H^{i-1}(C^*) \arrow H^i(A^*) \arrow  H^i(B^*) 
\arrow H^{i}(C^*) \arrow H^{i+1}(A^*) \arrow ...
\]

\newpage

{\bf \blue Differential forms on closed subsets (reminder)}

\definition
Let $M$ be a manifold, and $Z\subset M$
a closed subset. Let $\alpha, \beta$
be two differential forms defined in 
open sets $U_\alpha$ and $U_\beta$ containing $Z$.
We say that $\alpha$ and $\beta$ are equivalent
if $\alpha=\beta$ on $U_\alpha\cap U_\beta$.
The space of equivalence classes is denoted
$\Lambda^*(Z)$ and called {\bf \blue the space of
differential forms on $Z$} or {\bf \blue
the space of germs of differentials forms in $Z$}.

\remark
{\bf \purple If $Z$ is a manifold with boundary, 
$\Lambda^*(Z)$ is the space of differential
forms on $Z$,} by definition of differential
forms on manifolds with boundary.

\exercise
Using partition of unity, {\bf \red prove that
the natural restriction map $\Lambda^*(M)\arrow \Lambda^*(Z)$
is surjective for any closed $Z\subset M$.}

\remark
The de Rham differential is well defined on $\Lambda^*(Z)$,
allowing us to {\bf \purple define the de Rham cohomology of $Z$ as usual.}
Poincar\'e lemma holds in this situation, too.

\claim
Let $X\subset \R^n$ be a closed starlike subset.
{\bf \red Then $H^i(X)=0$ for all $i>0$.}

\proof Same as for open $X$. \endproof

\newpage

{\bf \blue Cohomology of a closed subset}


\remark 
Let $Z\subset M$ be a closed subset of a manifold.
{\bf \red The notation $\Lambda^*_c(Z)$ is slightly misleading,} because
the forms in this space are defined in a neighbourhood of $Z$ in $M$.
To avoid confusion, we denote the cohomology of this complex by $H^*_Z(M)$.

\claim
Let $Z\subset M$ be a smooth subvariety. Assume that 
there exists a neighbourhood $U\supset Z$ and
a family of smooth maps $\phi_t:\; U\arrow U$
smoothly depending on $t\in [0, 1]$
such that $\phi_0$ is identity and
$\phi_1:\; U \arrow Z$ is identity on $Z$.
{\bf \red Then $H^*_Z(M)= H^*(Z)= H^*(U)$.}

\proof
The maps $\phi_t$ induce the same map on cohomology
of $U$ and on $H^*_Z(M)$ as follows from the homotopy
invariance of de Rham cohomology (same argument as used
in the proof of Poincar\'e lemma). Then $\phi_1:\; U \arrow U$
induces an identity map on $H^*_Z(M)$ and on $H^*(U)$.
Therefore, the composition $U \stackrel {\phi_1}\arrow Z \hookrightarrow U$
induces identity on cohomology, both $H^*(U)$ and $H^*_Z(U)$. The composition 
$Z \hookrightarrow U \stackrel {\phi_1}\arrow Z$ is the identity map.
We obtain that $H^*(U)= H^*_Z(U)= H^*(Z)$.
\endproof


\newpage

{\bf \blue Excision for cohomology with compact support}

\proposition
Let $Z\subset M$ be a closed subset of a manifold.
{\bf \red Then the following sequence of complexes is exact.
\[
0 \arrow \Lambda^*_c(M \backslash Z) \arrow \Lambda^*_c(M) \stackrel R \arrow 
\Lambda^*_c(Z)  \arrow 0.
\]}
\proof Surjectivity in the last term and injectivity in the first is clear.
Now, let $R(\alpha)=0$, where $\alpha$ is a form with compact
support. Then $\alpha=0$ in a neighbourhood of $Z$.
This means that $\Supp(\alpha)\subset M \backslash Z$.
\endproof

\corollary {\bf \blue (Excision exact sequence)}\\
 Let $Z\subset M$ be a compact subset of a manifold.
{\bf \red There is a long exact sequence
\[ ...\arrow H^i_Z(M)\arrow H^{i+1}_c(M \backslash Z)\arrow 
H^{i+1}_c(M)\arrow  H^{i+1}_Z(M) \arrow H^{i+2}_c(M \backslash Z)\arrow ...
\]}
\endproof

\newpage

{\bf \blue Cohomology of $\R^n$ with compact support}


\corollary {\bf \blue (Excision exact sequence)}\\
 Let $Z\subset M$ be a compact subset of a manifold.
{\bf \red There is a long exact sequence
\[ ...\arrow H^i_Z(M)\arrow H^{i+1}_c(M \backslash Z)\arrow 
H^{i+1}_c(M)\arrow  H^{i+1}_Z(M) \arrow H^{i+2}_c(M \backslash Z)\arrow ...
\]}
\endproof


Let us apply this to $M=S^n$ and $Z$ a point. Using the previous
claim, we can assume that $H^*_Z(M) = H^*(pt)$, where $H^*(pt)$
denotes cohomology of a point.
Then
\[ ...\arrow H^i(pt)\arrow H^{i+1}_c(\R^n)\arrow 
H^{i+1}_c(S^n)\arrow  H^{i+1}(pt) \arrow H^{i+2}_c(\R^n)\arrow ...
\]
Positive-dimensional cohomology 
of a point vanish, and the only non-zero
cohomology of $S^n$ are in dimension $n$. Therefore, all
positive-dimensional 
terms in this exact sequence vanish, except in dimension n, where we 
get $H^{n-1}(pt)= \arrow H^{n}_c(\R^n)\arrow  H^{n}(S^n)=\R \arrow H^{n}(pt)=0$, hence
$H^{n}_c(\R^n)=\R$.

\corollary {\bf \red $H^i_c(\R^n)=\R$ for $i=n$ and vanishes otherwise.}

\proof
Since the closed 0-forms are constant functions, and they
don't have compact support on non-compact manifold, $H^0_c(\R^n)=0$.
The rest of cohomology vanish except $H^{n}_c(\R^n)=\R$ as 
shown above.
\endproof

\newpage

{\bf \blue Poincar\'e lemma with compact support}

\lemma Let $M \subset \R^n$ be an open starlike set. Then 
{\bf \red Then $H^i_c(M)=\R$ for $i=n$ and vanishes otherwise.}

\pstep Consider the map $\Lambda^n_c(M) \arrow \R$ mapping
$\alpha$ to $\int_M \alpha$. Clearly, it is non-zero on 
positively oriented $n$-forms with compact support
and zero on $d(\Lambda^{n-1}_c(M))$ by Stokes' theorem.
To prove the Poincar\'e lemma with compact support,
{\bf \purple it remains to show that all the $n$-forms $\alpha$
with $\int_M \alpha=0$ and all closed $i$-forms with $i<n$ are
exact.}

{\bf \green Step 2:} 
Denote by $\Gamma^t$ the homothety map $z \arrow tz$,
where $0 < t \leq 1$. Then $\Gamma^t$ maps forms
with compact support to forms with compact support;
denote this map by $\Gamma^t_*:\; \Lambda^i_c(M)\arrow \Lambda^i_c(M)$.
Clearly $\frac{\Gamma^t_*}{dt}= -\Lie_{\vec r}$, where $\vec
r$ is a radial vector field. Since $\Lie_x$ acts by zero
on cohomology, {\bf \purple $\Gamma_*^t$ acts on $H^*_c(M)$ as
identity.}

{\bf \green Step 3:} 
Let $\alpha\in \Lambda^i_c(M)$ be an
$n$-form with $\int_M \alpha=0$ or a closed $i$-form with
$i<n$. Since $\alpha$ is cohomologous to 0 in
$H^*_c(\R^n)$, one has $\alpha=d\beta$, where $\beta$
is a form on $\R^n$ with compact support.
 Choose $t$ in such a way
that support of $\Gamma_*^t\beta$ belongs to $M$.
{\bf \purple Then $\Gamma_*^t\alpha$ is cohomologous to 0 in
$H^*_c(M)$.} However, the cohomology class of $\alpha$
is equal to the cohomology class of $\Gamma_*^t\alpha$ 
(Step 2). \endproof

\newpage

{\bf \blue Mayer-Vietoris with compact support (1)}

\claim Consider open subsets $U, V$ of a manifold $M$,
Then 
{\bf \red the natural embedding maps define an exact sequence of complexes:}
\[ 
0 \arrow \Lambda^*_c(U\cup V)\stackrel \phi \arrow \Lambda^*_c(U)
\oplus  \Lambda^*_c(V) \stackrel \psi \arrow \Lambda^*_c(U\cup V)\arrow 0. \ \ (*)
\]
Here $\phi$ is the natural embedding on both components, and
$\psi$ is  the natural embedding 
on the first component and its opposite on the second component.

\pstep
The map $\phi$ is clearly injective. 
Also, the sequence is exact in the middle term, because
for any $\lambda \in \Lambda^*_c(U)$, $\mu\in\Lambda^*_c(V)$
such that $\lambda-\mu=0$ in $\Lambda^*_c(U\cup V)$, one has
$\Supp(\lambda) = \Supp(\mu) \subset U\cup V$.

{\bf \green Step 2:}
It remains to prove that $\psi$ is surjective.
Let $Z=\Supp(\alpha)$ be a compact subset of a form $\alpha \in \Lambda^*_c(U\cup V)$,
and let $Z_U:= Z\backslash V$, $Z_V:= Z \backslash U$.
These are non-intersecting compact subsets, hence there
exists a covering $\{U_i\}$ such that
none of $U_i$ intersects both $Z_V$ and $Z_U$.
Let $\psi_i$ be the corresponding partition of unity,
let $\Psi_U$ be the sum of all $\phi_i$ intersecting
$Z_U$ and $\Psi_V$ be the sum of the rest of $\phi_i$.
Then $\Psi_U \restrict {Z_V}=0$ and  $\Psi_V \restrict {Z_U}=0$,
hence $\Psi_U \alpha\in \Lambda^*_c(U)$ and 
$\Psi_V \alpha\in \Lambda^*_c(V)$.
We have decomposed $\alpha$ to a sum of two forms
which lie in $\Lambda^*_c(U)$ and $\Lambda^*_c(V)$.
\endproof

\newpage

{\bf \blue Mayer-Vietoris with compact support (2)}

\claim Consider open subsets $U, V$ of a manifold $M$,
Then 
{\bf \red the natural embedding maps define an exact sequence of complexes:}
\[ 
0 \arrow \Lambda^*_c(U\cap V)\stackrel \phi \arrow \Lambda^*_c(U)
\oplus  \Lambda^*_c(V) \stackrel \psi \arrow \Lambda^*_c(U\cup V)\arrow 0. \ \ (*)
\]
Here $\phi$ is the natural embedding on both components, and
$\psi$ is  the natural embedding 
on the first component and its opposite on the second component.

\corollary {\bf \blue (Mayer-Vietoris with compact support)}\\
Let $U, V\subset M$ be open subsets.
{\bf \red Then there is a long exact sequence associated with (*):}
\begin{multline*}... \arrow H^{i-1}_c(U)
\oplus H^{i-1}_c(U) \arrow H^{i-1}_c(U\cup V) \arrow H^i_c(U\cap V) \arrow \\ \arrow  H^i_c(U)
\oplus H^i_c(V)  
\arrow H^{i}_c(U\cup V) \arrow H^{i+1}_c(U\cap V) \arrow ...
\end{multline*}
\endproof

\newpage

{\bf \blue Poincar\'e duality}

\remark
Let $M$ be a manifold.
Consider the multiplication map
$\Lambda^*(M) \otimes\Lambda^*_c(M) \arrow \Lambda^*_c(M)$.
As we have already seen, a product of closed forms
is closed, and a product of a closed and exact forms is exact.

\theorem {\bf \blue (Poincar\'e duality theorem)}\\
Let $M$ be an $n$-dimensional oriented connected manifold.
Then the integration map $\alpha \arrow \int_M \alpha$
gives an isomorphism $H^n_c(M)=\R$. Moreover, {\bf \red the 
multiplication $H^i_c(M) \times H^{n-i}(M) \arrow H^n_c(M)=\R$
defines a non-degenerate pairing between $H^i_c(M)$
and $H^{n-i}(M)$.}

{\em \green It will be proven later today}

\remark
On a compact manifold, $H^i_c(M)= H^i(M)$, and then the 
{\bf \blue Poincar\'e pairing} $\alpha, \beta \arrow \int_M \alpha \wedge \beta$
defines a non-degenerate bilinear form {\bf \blue (the Poincar\'e form)}
on cohomology.

\remark Note that {\bf \purple 1-dimensionality of $H^n_c(M)$ is itself a consequence
of Poincare duality,} because $H^n_c(M)$  is dual to $H^0(M)=\R$.


\newpage

{\bf \blue Proof of Poincar\'e duality}

\theorem {\bf \blue (Poincar\'e duality theorem)}\\
Let $M$ be an $n$-dimensional oriented connected manifold.
Then the integration map $\alpha \arrow \int_M \alpha$
gives an isomorphism $H^n_c(M)=\R$. Moreover, {\bf \red the 
multiplication $H^i_c(M) \times H^{n-i}(M) \arrow H^n_c(M)=\R$
defines a non-degenerate pairing between $H^i_c(M)$
and $H^{n-i}(M)$.}

\proof
Let $M$ be a manifold and $\{U_i\}$ a finite open cover such that
all $U_i$, all the closures ${\overline U_i}$ 
and all their intersections are starlike (such a cover
can be obtained from a polyhedral structure or from
a convex cover as in Lecture 6).
Poincar\'e lemma implies that the 
Poincare pairing map $H^*(\bar K)\arrow H^{n-*}_c(\bar K)^*$
is an isomorphism for any $K$ which is starlike, 
where $\bar K$ is the closure of $K$.
Using induction in $n$, we may assume that
$H^*(\bar K)\arrow H^{n-*}_c(\bar K)^*$
is an isomorphism for any $K$ which is obtained as a union 
of $n-1$ or less of $U_i$. Let us prove it for $K=\bigcup_{i=1}^n U_i$.
Mayer-Vietoris exact sequences, applied to 
 $X=U_0$ and $Y=\bigcup_{i=2}^n U_i$, 
give the following diagram.
\[\small
\begin{CD}
... @>>> H^{i-1}(\bar X\cap \bar Y) @>>>  H^{i}(\bar X\cup \bar Y) 
@>>>  H^{i}(\bar X)\oplus  H^{i}(\bar X)@>>> H^{i+1}(\bar X\cap \bar Y)@>>>
...\\
&& @VVV @VVV @VVV @VVV \\
... @>>> H^{n-i+1}_{c}(X\cap Y)^* @>>>  H^{n-i}_{c}(X\cup Y)^* 
@>>>  H^{n-i}_{c}(X)^*\oplus  H^{n-i}_{c}(Y)^*@>>> H^{n-i-1}_{c}(X\cap Y)^*@>>>...
\end{CD}
\]
By induction assumption,
the vertical arrows in this diagram are isomorphisms for all terms
except possibly for $H^*(\bar X \cup \bar Y)\arrow H^{n-*}_c(X \cup Y)^*$.
{\bf \purple The 5-lemma implies that they are isomorphisms in all terms.}
\endproof.



\end{document}
