12_18.tex

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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 591 Lecture 40}
\date{12/18/20} % Can also use \today

\begin{document}
\maketitle
\setlength\RaggedRightParindent{\parindent}
\RaggedRight

\par\noindent
We'll begin with a very brief look at the algebra behind cohomology.\n

\defn{
	A \u{cochain complex} $\mathcal{A}$ of vector spaces is a sequence of linear maps
	\[
		\begin{tikzcd}[ampersand replacement=\&, column sep=20pt]
			0 \arrow[r] \& A^{0} \arrow[r,"d"] \& A^{1} \arrow[r,"d"] \& A^{2} \arrow[r,"d"] \& \cdots
		\end{tikzcd}
	\]
	\st{} $d\of{}d=0$ (whenever defined).\n
}

\defn{
	The \u{cohomology} of a cochain complex $\mathcal{A}$ is, $\forall{}k\in\N$, $H^{k}(\mathcal{A})=Z^{k}(\mathcal{A})/B^{k}(\mathcal{A})$, where $Z^{k}(\mathcal{A})=\ker(d)$ (with $d:A^{k}\to{}A^{k+1}$) and $B^{k}(\mathcal{A})=\im(d)$ (with $d:A^{k-1}\to{}A^{k}$).\n
}

\defn{
	If $\mathcal{A}$ and $\mathcal{B}$ are cochain complexes, a map $f:\mathcal{A}\to\mathcal{B}$ between them is a sequence: $\forall{}k\in\N$, we have $f_{k}:A^{k}\to{}B^{k}$ \st{} $d\of{}f^{k}=f^{k}\of{}d$. I.e., the following diagram commutes:
	\[
		\begin{tikzcd}[ampersand replacement=\&]
			\cdots \arrow[r,"d"] \& A^{k} \arrow[r,"d"] \arrow[d,"f^{k}"] \& A^{k+1} \arrow[r,"d"] \arrow[d,"f^{k+1}"] \& \cdots\\
			\cdots \arrow[r,"d"] \& B^{k} \arrow[r,"d"] \& B^{k+1} \arrow[r,"d"] \& \cdots
		\end{tikzcd}
	\]
}

\lemma{
	Such an $f:\mathcal{A}\to\mathcal{B}$ induces $f^{\sharp}:H^{k}(\mathcal{A})\to{}H^{k}(\mathcal{B})$ by $f^{\sharp}[a]=[f(a)]$ for any $a\in{}Z^{k}(\mathcal{A})$, and $f^{\sharp}$ is well-defined.\n
}

\par\noindent
Observe:
\begin{enumerate}[label=\alph*), itemsep=0pt, topsep=0pt]
	\item $(f\of{}g)^{\sharp}=f^{\sharp}\of{}g^{\sharp}$.
	\item For de Rham theory, if $F:M\to{}N$ is $C^{\infty}$, then we get $f:\Omega^{*}(N)\to\Omega^{*}(M)$ ($\Omega^{*}(N)$ is the de Rham complex of $N$), where $\forall\alpha\in\Omega^{k}(N)$, $f(\alpha)=F^{*}\alpha$.
\end{enumerate}

\section*{Homotopies between Maps of Cochain Complexes}

\defn{
	Say $f,g:\mathcal{A}\to\mathcal{B}$. A (\u{chain\vphantom{py}}) \u{homotopy} (\u{operator}) between them is a sequence of maps: $\forall{}k$, $h:A^{k}\to{}B^{k-1}$ \st{} the following diagram commutes:
	\[
		\begin{tikzcd}[ampersand replacement=\&]
			\& \cdots \arrow[r,"d"] \& A^{k} \arrow[dl,"h"] \arrow[d,"f-g"] \arrow[r,"d"] \& A^{k+1} \arrow[dl,"h"] \arrow[r,"d"] \& \cdots\\
			\cdots \arrow[r,"d"] \& B^{k-1} \arrow[r,"d"] \& B^{k} \arrow[r,"d"] \& \cdots
		\end{tikzcd}
	\]
	That is, $h\of{}d+d\of{}h=f-g$.\n
}

\lemma{
	If there exists a homotopy between $f$ and $g$, then $f^{\sharp}=g^{\sharp}$.\n
}

\par\noindent
Last time, we showed that for $X\in\mathfrak{X}(M)$, with $\varphi$ the flow of $X$ (which we assume to be complete), then $\forall\omega\in\Omega^{k}(M)$, $\frac{d}{dt}\varphi_{t}^{*}\omega=\varphi_{t}^{*}\mathcal{L}_{X}\omega=\varphi_{t}^{*}(\iota_{X}d\omega+d\iota_{X}\omega)$. Thus, $\varphi_{1}^{*}\omega-\omega=\int_{0}^{1}\varphi_{t}^{*}(\iota_{X}d\omega+d\iota_{X}\omega)\,dt$.\n

\par\noindent
Check: If we define $h(\omega)=\int_{0}^{1}\varphi_{t}^{*}(\iota_{X}\omega)\,dt\in\Omega^{k-1}(M)$, hen the above formula shows that $h$ is a chain homotopy between $\varphi_{1}^{*}$ and the identity map.\n

\newpage
\section*{Mayer-Vietoris Sequence}

\par\noindent
Motivation: How can we compute $H^{*}(S^{2})$?\n

\par\noindent
Well, we can describe $S^{2}$ as the union of $U$ and $V$, where $U$ and $V$ are diffeomorphic to the open disk, and their intersection is diffeomoprhic to the cylinder $S^{1}\times(-1,1)$. Can we say anything about $H^{*}(S^{2})$ in terms of $H^{*}(U)$, $H^{*}(V)$, and $H^{*}(U\cap{}V)$?\n

\par\noindent
Hypothesis: In general, for $U,V$ open with $M=U\cup{}V$, we have
\[
	\begin{tikzcd}[row sep=tiny]
		& U \arrow[dr, hookrightarrow]\\
		U\cap{}V \arrow[ur, hookrightarrow] \arrow[dr, hookrightarrow] & & M\\
		& V \arrow[ur, hookrightarrow]
	\end{tikzcd}
\]
We can then form, $\forall{}k\in\N$,
\[
	\begin{tikzcd}[column sep=20pt]
		0 \arrow[r] & \Omega^{k}(M) \arrow[r,"f"] & \Omega^{k}(U)\oplus\Omega^{k}(V) \arrow[r,"g"] & \Omega^{k}(U\cap{}V) \arrow[r] & 0 \\[-20pt]
		& & (\alpha,\beta) \arrow[r,mapsto] & \restr{(\alpha-\beta)}{U\cap{}V}
	\end{tikzcd}
\]
where $f$ is the pullback/restriction.\n

\lemma{
	$\forall{}k\in\N$, this is an exact sequence, i.e., the image of each map is the kernel of the next one. (This is true iff it's a complex with zero cohomology).\nn
	Proof: We have exactness at $\Omega^{k}(M)$ iff $f$ is injecive. This is true because $M=U\cup{}V$, and $U$ and $V$ are both open.\nn
	We have exactness at $\Omega^{k}(U)\oplus\Omega^{k}(V)$ iff $\im(f)=\ker(g)$. Well, $\im(f)$ is the set of restrictions of globally-defined forms, so we're still okay.\nn
	We have exactness at $\Omega^{k}(U\cap{}V)$ iff $g$ is surjective. Let $\omega\in\Omega^{k}(U\cap{}V)$. We need to show $\exists\alpha\in\Omega^{k}(U),\beta\in\Omega^{k}(V)$ \st{} $\restr{(\alpha-\beta)}{U\cap{}V}=\omega$. Let $\set{\chi_{U},\chi_{V}}$ be a subordinate partition of unity to $\set{U,V}$. Define
	\[
		\alpha(p)\eqdef\left\{\begin{array}{ll}
			\chi_{V}\omega & p\in{}U\cap{}V\\
			0 & p\in{}U\cut{}V
		\end{array}\right.
		\qquad
		\beta(p)\eqdef\left\{\begin{array}{ll}
			-\chi_{U}\omega & p\in{}U\cap{}V\\
			0 & p\in{}V\cut{}U
		\end{array}\right.
	\]
	Then $\restr{(\alpha-\beta)}{U\cap{}V}=\restr{\chi_{V}\omega+\chi_{I}\omega}{U\cap{}V}=\omega$.\nn
	\proven
}

\par\noindent
Observe: $f$ and $g$ are cochain maps -- they commute with $d$!\n

\lemma{
	(Zig-Zag Lemma) Let $\mathcal{A},\mathcal{B},\mathcal{C}$ be cochain complexes, and $f:\mathcal{A}\to\mathcal{B}$, $g:\mathcal{B}\to\mathcal{C}$ cochain maps, \st{} $\forall{}k\in\N$,
	\[
		\begin{tikzcd}[ampersand replacement=\&, column sep=20pt]
			0 \arrow[r] \& A^{k} \arrow[r,"f"] \& B^{k} \arrow[r,"g"] \& C^{k} \arrow[r] \& 0
		\end{tikzcd}
	\]
	is exact. Then $\forall{}k$, $\exists\delta_{k}:H^{k}(\mathcal{C})\to{}H^{k+1}(\mathcal{A})$, a linear map referred to as the connecting morphism, \st{} the following sequence is exact:
	\[
		\begin{tikzcd}[ampersand replacement=\&, column sep=20pt, row sep=small]
			0 \arrow[r] \& H^{0}(\mathcal{A}) \arrow[r,"f^{\sharp}"] \& H^{0}(\mathcal{B}) \arrow[r,"g^{\sharp}"] \arrow[d, phantom, ""{coordinate, name=ref1}] \& H^{0}(\mathcal{C}) \arrow[dll, rounded corners, to path={
			-- ([xshift=10pt]\tikztostart.east)
			|- (ref1)
			-| ([xshift=-10pt]\tikztotarget.west)
			-- (\tikztotarget)
			}]\\
			\& \cdots \& \phantom{\cdots}\\
			\& \& \phantom{\cdots} \arrow[d,phantom,""{coordinate, name=ref2}] \& \cdots \arrow[dll, rounded corners, to path={
			-- ([xshift=10pt]\tikztostart.east)
			|- (ref2)
			-| ([xshift=-10pt]\tikztotarget.west)
			-- (\tikztotarget)
			}]\\
			\& H^{k}(\mathcal{A}) \arrow[r,"f^{\sharp}"] \& H^{k}(\mathcal{B}) \arrow[r,"g^{\sharp}"] \arrow[d,phantom,""{coordinate, name=ref3}] \& H^{k}(\mathcal{C}) \arrow[dll, rounded corners, to path={
			-- ([xshift=10pt]\tikztostart.east)
			|- (ref3)
			-| ([xshift=-10pt]\tikztotarget.west)
			-- (\tikztotarget)
			}]\\
			\& H^{k+1}(\mathcal{A}) \arrow[r,"f^{\sharp}"] \& H^{k+1}(\mathcal{B}) \arrow[r,"g^{\sharp}"] \arrow[d,phantom,""{coordinate, name=ref4}] \& H^{k+1}(\mathcal{C}) \arrow[dll, rounded corners, to path={
			-- ([xshift=10pt]\tikztostart.east)
			|- (ref4)
			-| ([xshift=-10pt]\tikztotarget.west)
			-- (\tikztotarget)
			}]\\
			\& \cdots \& \phantom{\cdots}
		\end{tikzcd}
	\]
}

\par\noindent
Observe: This applied to the case $M=U\cup{}V$ is precisely the Mayer-Vietoris sequence.\n

\newpage
\par\noindent
Sketch of the proof:
\begin{enumerate}
	\item Check exactness at $H^{k}(\mathcal{B})$:
	\[
		\begin{tikzcd}[ampersand replacement=\&, column sep=20pt]
			H^{k}(\mathcal{A}) \arrow[r,"f^{\sharp}"] \& H^{k}(\mathcal{B}) \arrow[r,"g^{\sharp}"] \& H^{k}(\mathcal{C})
		\end{tikzcd}
	\]
	We need to show $\im(f^{\sharp})=\ker(g^{\sharp})$. Well, we know $0=(g\of{}f)^{\sharp}=g^{\sharp}\of{}f^{\sharp}$, so $\im(f^{\sharp})\subseteq\ker(g^{\sharp})$. For the reverse inclusion, let $[\beta]\in\ker(g^{\sharp})$, so $\beta\in{}Z^{k}(\mathcal{B})$. We rely on the following commutative diagram:
	\[
		\begin{tikzcd}[ampersand replacement=\&, column sep=20pt]
			0 \arrow[r] \& A^{k} \arrow[r,"f"] \& B^{k} \arrow[r,"g"] \& C^{k} \arrow[r] \& 0\\
			0 \arrow[r] \& A^{k-1} \arrow[u,"d"] \arrow[r,"f"] \& B^{k-1} \arrow[u,"d"] \arrow[r,"g"] \& C^{k-1} \arrow[u,"d"] \arrow[r] \& 0
		\end{tikzcd}
	\]
	Assume that $g^{\sharp}[\beta]=0$, i.e., $\exists{}c\in{}C^{k-1}$ \st{} $g(\beta)=dc$. Then $\exists{}b\in{}B^{k-1}$ \st{} $g(b)=c$.\n
	Thus, $g(\beta)=dc=dg(b)=gd(b)$. This means $g(\beta-db)=0$, so $\exists{}a\in{}A^{k}$ \st{} $f(a)=\beta-db$, so $\beta=db+f(a)$.\n
	We need to show $[\beta]\in\im{}f^{\sharp}$, so we need to have $da=0$. Well, $0=df(a)=fda$. Because $f$ is injective, we must have $da=0$. We conclude that $\beta=db+f(a)$ and $da=0$, so $[\beta]=[f(a)]=f^{\sharp}[a]$.\n
	\item Check existence of $\delta$:
	\[
		\begin{tikzcd}[ampersand replacement=\&, column sep=20pt]
			0 \arrow[r] \& A^{k} \arrow[r] \arrow[d] \& B^{k} \arrow[r] \arrow[d] \& C^{k} \arrow[r] \arrow[d] \arrow[lld, dashed] \& 0\\
			0 \arrow[r] \& A^{k+1} \arrow[r] \& B^{k+1} \arrow[r] \& C^{k+1} \arrow[r] \& 0
		\end{tikzcd}
	\]
	Let $c\in{}Z^{k}(\mathcal{C})$, so $c\in{}C^{k}$, $dc=0$. Then $\exists{}b\in{}B^{k}$ \st{} $g(b)=c$. So $0=dc=dg(b)=g(db)$. Thus, $db\in\ker(g)=\im(f)$, so $\exists{}a\in{}A^{k+1}$ \st{} $f(a)=db$. In summary, $c=g(b)$ and $db=f(a)$. We claim:
	\begin{enumerate}[label=(\roman*)]
		\item $da=0$.
		\item $[a]\in{}H^{k+1}(\mathcal{A})$ depends only on $[c]$.
	\end{enumerate}
	So we define $\delta([c])=[a]$. Check:
	\begin{enumerate}[label=(\roman*)]
		\item $fda=df(a)=ddb=0$. $f$ is injective, so $da=0$.
		\item This just requires more diagram chasing.
	\end{enumerate}
\end{enumerate}

\cor{
	(Mayer-Vietoris Sequence) If $M=U\cup{}V$, there is an exact sequence
	\[
		\begin{tikzcd}[ampersand replacement=\&, column sep=20pt, row sep=tiny]
			0 \arrow[r] \& H^{0}(M) \arrow[r,"f^{\sharp}"] \& H^{0}(U)\oplus{}H^{0}(V) \arrow[r,"g^{\sharp}"] \arrow[d, phantom, ""{coordinate, name=ref1}] \& H^{0}(U\cap{}V) \arrow[dll, rounded corners, to path={
			-- ([xshift=10pt]\tikztostart.east)
			|- (ref1)
			-| ([xshift=-10pt]\tikztotarget.west)
			-- (\tikztotarget)
			}]\\
			\& H^{1}(M) \arrow[r,"f^{\sharp}"] \& H^{1}(U)\oplus{}H^{1}(V) \arrow[r,"g^{\sharp}"] \arrow[d, phantom, ""{coordinate, name=ref2}] \& H^{1}(U\cap{}V) \arrow[dll, rounded corners, to path={
			-- ([xshift=10pt]\tikztostart.east)
			|- (ref2)
			-| ([xshift=-10pt]\tikztotarget.west)
			-- (\tikztotarget)
			}]\\
			% \& \makebox[\widthof{$H^{1}(M)$}][c]{$\cdots$} \& \phantom{\cdots}
			\& \cdots \& \phantom{\cdots}
		\end{tikzcd}
	\]
	with $f^{\sharp}$ and $g^{\sharp}$ given as above.\n
}

\par\noindent
Application: $H^{k}(S^{n})=\left\{\begin{array}{ll}
	\R & k\in\set{0,n}\\
	0 & \ptxt{otherwise}
\end{array}\right.$ We can prove this using induction on $n$. For example, for $n=2$,
\[
	\begin{tikzcd}[column sep=20pt, row sep=tiny]
		& S^{2} & U\sqcup{}V & U\cap{}V\\
		H^{0} & \R \arrow[r] & \R\oplus\R \arrow[r] & \R\\[-10pt]
		& x \arrow[r,mapsto] & (x,x)\\[-10pt]
		& & (x,y) \arrow[r,mapsto] \arrow[d, phantom,""{coordinate, name=ref1}] & x-y\\
		H^{1} & H^{1}(S^{2}) \arrow[r] \arrow[uuurr, leftarrow, dashed, rounded corners, to path={
			-- ([xshift=-10pt]\tikztostart.west)
			|- (ref1)
			-| ([xshift=15pt]\tikztotarget.east)
			-- (\tikztotarget)
		}] & 0 \arrow[r] \arrow[d,phantom,""{coordinate, name=ref2}] & \R \arrow[dll, dashed, rounded corners, to path={
			-- ([xshift=10pt]\tikztostart.east)
			|- (ref2)
			-| ([xshift=-10pt]\tikztotarget.west)
			-- (\tikztotarget)
		}]\\
		H^{2} & H^{2}(S^{2}) \arrow[r] & 0 \arrow[r] & 0
	\end{tikzcd}
\]
We have the exact sequence \begin{tikzcd}[cramped, sep=small]
	0 \arrow[r] & \R \arrow[r] & H^{2}(S^{2}) \arrow[r] & 0
\end{tikzcd}, so the mapping from $\R$ to $H^{2}(S^{2})$ must be injective and surjective, so $H^{2}(S^{2})=\R$. As for $H^{1}(S^{2})$, the map $(x,y)\mapsto{}x-y$ is surjective, so the map into $H^{1}(S^{2})$ must be the zero map. By exactness at $H^{1}(S^{2})$, we must have the kernel of the map from $H^{1}(S^{2})$ to $0$ also be $0$, so we must have $H^{1}(S^{2})=0$. Then, our inductive step uses the fact that the ``equator'' $U\cap{}V$ is homotopy equivalent to $S^{n-1}$.\n

\par\noindent
Another example is the $2$-torus, $T^{2}$. We can cut the torus in half to get two components $U,V$, each of which is diffeomorphic to the cylinder, which in turn is homotopy equivalent to $S^{1}$. The $U\cap{}V$ is the disjoint union of $2$ cylinders. We then have
\[
	\begin{tikzcd}[column sep=20pt, row sep=tiny]
		& T^{2} & U\sqcup{}V & U\cap{}V\\
		H^{0} & \R \arrow[r] & \R\oplus\R \arrow[r] \arrow[d,phantom,""{coordinate, name=ref1}] & \R\oplus\R\\
		H^{1} & H^{1}(T^{2}) \arrow[r] \arrow[urr, leftarrow, dashed, rounded corners, to path={
			-- ([xshift=-10pt]\tikztostart.west)
			|- (ref1)
			-| ([xshift=15pt]\tikztotarget.east)
			-- (\tikztotarget)
		}] & \R\oplus\R \arrow[r] \arrow[d,phantom,""{coordinate, name=ref2}] & \R\oplus\R \arrow[dll, dashed, rounded corners, to path={
			-- ([xshift=10pt]\tikztostart.east)
			|- (ref2)
			-| ([xshift=-10pt]\tikztotarget.west)
			-- (\tikztotarget)
		}]\\
		H^{2} & H^{2}(T^{2}) \arrow[r] & 0 \arrow[r] & 0
	\end{tikzcd}
\]

\exer{
	Show that $H^{k}(T^{2})=\left\{\begin{array}{ll}
		\R & k\in\set{0,2}\\
		\R^{2} & k=1
	\end{array}\right.$, and that $H^{1}(T^{2})$ is generated by $[dx^{1}]$ and $[dx^{2}]$.\n
}

\thm{
	If $M$ is a compact, oriented, connected manifold (with $m=\dim{}M$), then
	\[
		\int_{M}:H^{k}(M)\to\R
	\]
	is an isomorphism, so $H^{m}(M)\cong\R$.\n
}

\end{document}