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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf  Topologia das Variedades \\[15mm]
\small Cohomology, lecture 7: K\"unneth formula}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]

{\small\bf Misha Verbitsky  
\\[2mm]
May 29, 2018\\[4mm]

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

\newpage

{\bf \blue Tensor product of complexes}

\definition
{\bf \blue Bicomplex} of vector spaces
is a sequence of vector spaces $C^{p,q}$ equipped 
with the differentials $d^{1,0}:\; C^{p, q} \arrow C^{p+1, q}$
and $d^{0,1}:\; C^{p, q} \arrow C^{p, q+1}$, such that
$d^{0,1}\circ d^{0,1}= d^{1,0}\circ d^{1,0}=0$ and
$d^{0,1}\circ d^{1,0}= d^{1,0}\circ d^{0,1}$.

\example
Let $... \stackrel {d_C} \arrow C^i \stackrel {d_C} \arrow C^{i+1}  \stackrel {d_C} \arrow...$
and $... \stackrel  {d_D}\arrow D^i \stackrel {d_D} \arrow D^{i+1}  \stackrel {d_D} \arrow...$
be complexes of vector spaces. Their {\bf \blue tensor product} is
the bicomplex $C^p \otimes D^q$ with $d^{1,0} = d_C \otimes \Id_{D^*}$
and $d^{0,1}=\Id_{C^*} \otimes d_D$.

\definition
{\bf \blue Totalization} of a bicomplex $(C^{*,*}, d^{1,0}, d^{0,1})$
is the complex 
\[
... \stackrel {d}\arrow \bigoplus_{p+q=i} C^{p,q} \stackrel {d}\arrow \bigoplus_{p+q=i+1}  C^{p,q}
\stackrel {d}\arrow ... 
\]
where the differential $d$ is defined as $d\restrict{C^{p,q}}= d^{1,0} + (-1)^p d^{0,1}$.

\remark To see that $d^2=0$, we use $(d^{1,0})^2=0$, $(d^{0,1})^2=0$
and $d^{1,0} (-1)^p d^{0,1} + (-1)^{p+1} d^{0,1}d^{1,0}=0$.

\newpage

{\bf \blue Cohomology and the tensor product}

\theorem
Let $... \stackrel {d_C} \arrow C^i \stackrel {d_C} \arrow C^{i+1}  \stackrel {d_C} \arrow...$
and $... \stackrel  {d_D}\arrow D^i \stackrel {d_D} \arrow D^{i+1}  \stackrel {d_D} \arrow...$
be complexes of vector spaces, and $((C\otimes D)^*, d)$ be the totalization of
their tensor product bicomplex. {\bf \red Then 
\[ H^i((C\otimes D)^*)= 
\bigoplus_{p+q=i}(H^p(C)\otimes H^q(D)).\ \ \ \ (***)
\]}
\pstep
Consider the direct sum decomposition $C^*= C_1^* \oplus C_2^*$. Then 
$H^i((C\otimes D)^*)= H^i((C_1\otimes D)^*)\oplus H^i((C_2\otimes D)^*)$,
and $\bigoplus_{p+q=i}(H^p(C)\otimes H^q(D))=\bigoplus_{p+q=i}(H^p(C_1)\otimes H^q(D)) \oplus \bigoplus_{p+q=i}(H^p(C_2)\otimes H^q(D))$.
Therefore, whenever we represent $C^*$ as a direct sum $\bigoplus_i C^*_i$,
{\bf \purple it suffices to prove the isomorphism (***)
for each subcomplex $C_i^* \otimes D^*$.}

{\bf \green Step 2:} Every complex of vector spaces is a direct sum of two types of complexes,
\[... \stackrel d \arrow 0 \stackrel d \arrow V \stackrel d \arrow 0 \stackrel d \arrow...
\]
and 
\[ ... \stackrel d \arrow 0 \stackrel d \arrow V \stackrel {d=\Id_V} \arrow V \stackrel d \arrow
0 \stackrel d \arrow...
\]
where $V$ is a one-dimensional vector space.
It would suffice to prove (***)
when $C^*$ and $D^*$ is one of these two types of complexes.

\newpage

{\bf \blue Cohomology and the tensor product (2)}

\theorem
Let $... \stackrel {d_C} \arrow C^i \stackrel {d_C} \arrow C^{i+1}  \stackrel {d_C} \arrow...$
and $... \stackrel  {d_D}\arrow D^i \stackrel {d_D} \arrow D^{i+1}  \stackrel {d_D} \arrow...$
be complexes of vector spaces, and $((C\otimes D)^*, d)$ be the totalization of
their tensor product bicomplex. {\bf \red Then 
\[ H^i((C\otimes D)^*)= 
\bigoplus_{p+q=i}(H^p(C)\otimes H^q(D)).\ \ \ \ (***)
\]}
{\bf \green Step 2:} Every complex of vector spaces is a direct sum of two types of complexes,
$... \stackrel d \arrow 0 \stackrel d \arrow V \stackrel d \arrow 0 \stackrel d \arrow...
$
and 
$ ... \stackrel d \arrow 0 \stackrel d \arrow V  \xlongrightarrow{d=\Id_V} V \stackrel d \arrow
0 \stackrel d \arrow...
$
where $V$ is a one-dimensional vector space.
It would suffice to prove (***)
when $C^*$ and $D^*$ is one of these two types of complexes.

{\bf \green Step 3:}
When $C^* = ... \stackrel d \arrow 0 \stackrel d \arrow V \stackrel d \arrow 0 \stackrel d \arrow...$,
with $V$ in degree 0, we 
have $(C\otimes D)^* \cong D^*$ and $\bigoplus_{p+q=i}(H^p(C)\otimes H^q(D))\cong H^i(D)$
hence  (***) holds tautologically. 

{\bf \green Step 4:}
When both $C^*$ and $D^*$ have form
$... \stackrel d \arrow 0 \stackrel d \arrow V  \xlongrightarrow{d=\Id_V} V \stackrel d \arrow
0 \stackrel d \arrow...$, the totalization of $C^*\otimes D^*$ takes form
\[... \stackrel d \arrow 0 \stackrel d \arrow V^{\otimes 2}
\xlongrightarrow {d=(\Id_{V^{\otimes 2}}, \Id_{V^{\otimes 2}})} V^{\otimes 2}\oplus V^{\otimes 2} 
\xlongrightarrow  {d=(\Id_{V^{\otimes 2}}, -\Id_{V^{\otimes 2}})} V ^{\otimes 2} \stackrel d \arrow 0 \stackrel d \arrow ...
\]
therefore $H^*((C\otimes D)^*)=0$. Also, $H^*(C)\otimes H^*(D)=0$,
hence (***) also holds.
\endproof

\newpage 

{\bf \blue Tensor product and functions on a product}

\theorem
Let $C(M)$ be the space of functions $f:\; M \arrow \R$,
and $C(N)$ the spave of functions $f:\; N \arrow \R$.
Consider the natural map $\Psi:\; C(M)\otimes C(N)\arrow C(M\times N)$.
{\bf \red Then $\Psi$ is injective.}


\pstep
{\bf \purple For $N, M$ finite $\Psi$ is an isomorphism.} Indeed,
for any $m\in M$ and $n\in N$, the tensor product 
product $\chi_m\otimes \chi_n$
of atomic functions $\chi_m$ and $\chi_n$
is mapped to $\chi_{(m,n)}$, hence $\Psi$ is surjective,
and it is injective because $\dim C(M)\otimes C(N)= |M||N|=
\dim C(M\times N)$.

{\bf \green Step 2:} 
For any linearly independent set of $k$ functions 
$f_1, ..., f_k \in C(M)$,
consider restriction of $f_1, ..., f_k$ to a  finite subset
$M_0\subset M$. If there is a linear relation 
$\sum_i \lambda_i f_i \restrict{{M_0}}$ for each
finite subset, this linear relation is true on $M$.
Therefore, {\bf \purple linearly independent functions remain
linearly independent if restricted on a sufficiently
big finite subset.}

{\bf \green Step 3:} Let $\{f_\alpha\}$ be a basis in
$C(M)$, $\{g_\alpha\}$ a basis in $C(N)$.
Then $\{f_\alpha\otimes g_\alpha\}$ is a basis in 
$C(M)\otimes C(N)$, indexed by $\alpha \in A$,
$\beta\in B$. Any vector $x\in C(M)\otimes C(N)$ 
takes form $x= \sum_{i\in A_0, j \in B_0} x_{ij}f_i\otimes g_j$,
where $A_0\subset A$, $B_0\subset B$ are finite subsets.
Then $x\restrict{M_0\times N_0}$ is non-zero for some
finite subsets $M_0\subset M$, $N_0\subset N$
(Step 2). This implies that $\Psi(x)\restrict{M_0\times N_0}$
is also non-zero (Step 1). \endproof


\newpage

{\bf \blue Stone-Weierstrass theorem}


\definition
Let  $A\subset C^0 M$ be a subspace in the space of
continuous functions. We say that $A$ {\bf\blue separates the points} of $M$
if for all distinct points $x, y\in M$, there exists $f\in A$ such that
$f(x) \neq f(y)$.

\definition
Let $M$ be a topological space, and
$\|f\|:= \sup_M |f|$ {\bf\blue  the sup-norm on functions}.
The {\bf \blue $C^0$-topology}, or {\bf \blue uniform
  topology}  on the space $C^0(M)$ of continuous functions
is topology defined by the sup-norm.


\theorem {\bf\blue (Stone-Weierstrass)}\\
Let $A\subset  C^0 M$ be a subring separating points,
and $\bar A$ its closure in the $C^0$-topology. {\bf \red Then
$\bar A= C^0(M)$.}

\corollary For any two manifolds $M$ and $N$,
{\bf \red the ring $C^\infty(M) \otimes C^\infty(N)$ is dense in $C^0(M\times N)$.}

\proof
$C^\infty(M) \otimes C^\infty(N)$ is a subring in $C^0(M\times N)$,
as shown above, and it is dense because it separates the points.
\endproof


\newpage

{\bf \blue Differential forms on a product}

\claim Let $V, W$ be vector spaces.
{\bf \red Then $\Lambda^i(V \oplus W) = \bigoplus_{p+q=i} \Lambda^p(V) \otimes \Lambda^q(W)$.}

\proof The multiplication map 
$\bigoplus_{p+q=i} \Lambda^p(V) \otimes \Lambda^q(W)
\arrow \Lambda^i(V \oplus W)$ takes the monomial basis
in $\bigoplus_{p+q=i} \Lambda^p(V) \otimes \Lambda^q(W)$
to the monomial basis in $\Lambda^i(V \oplus W)$.
\endproof

\corollary
For any two manifolds $M_1$ and $M_2$,
consider the projection maps $\pi_i:\; M_1 \times M_2
\arrow M_i$. Then {\bf \red the natural multiplication map
$\pi^*_1 \Lambda^*(M_1) \otimes \pi^*_2
\Lambda^*(M_2)\arrow  \Lambda^*(M_1\times M_2)$
is an isomorphism. }

\proof Indeed, $\Lambda^1(M_1\times M_2) = \pi^*_1
\Lambda^1(M_1)\oplus \pi^*_2
\Lambda^1(M_2)$, hence 
\begin{multline*}
\Lambda^i(M_1 \times M_2) = \Lambda^i(\Lambda^1(M_1\times
M_2))=\\ = \Lambda^i\bigg( \pi^*_1
\Lambda^1(M_1)\oplus \pi^*_2
\Lambda^1(M_2)\bigg) = \bigoplus_{p+q=i}  \pi^*_1
\Lambda^p(M_1) \otimes \pi^*_2\Lambda^q(M_2).
\end{multline*}
\endproof

\newpage

{\bf \blue Differential forms on a product (2)}

\corollary
Consider the natural multiplicative map
$\Psi:\; \bigoplus_{p+q=i}\Lambda^p(M_1) \otimes_\R \Lambda^q(M_2)\arrow
\Lambda^i(M_1 \times M_2)$ obtained by taking pullbacks
and multiplying. {\bf \red Then $\Psi$ is injective, and its image is dense in $C^0$-topology.}

\proof
The image is generated multiplicatively by forms
which are lifted from $M_1$ and $M_2$. It would
suffice to prove the statement locally in coordinates. Let
$V_i$ be the vector spaces
generated by $dx_1, ..., dx_n$, $dy_1, ..., dy_m$, where $x_1, ..., x_n$
and $y_1, ..., y_m$  are coordinate functions on $M_i$, defined in a neighbourhood
of a point. Then $\Lambda^*(M_1)$ is $C^\infty M_1 \otimes_\R \Lambda^*(V_2)$, and
$\Lambda^*(M_2)$ is $C^\infty M_2 \otimes_\R
\Lambda^*(V_2)$. Similarly,
$\Lambda^*(M_1\times M_2)= C^\infty (M_1\times M_2)
\otimes_\R \Lambda^*(V_1 \oplus V_2)$.
Since $\Lambda^*(M_1)\otimes \Lambda^*(M_2)$
is equal to $C^\infty (M_1) \otimes C^\infty (M_2)\otimes
\Lambda^*(V_1 \oplus V_2)$, the second is dense in the
first as we have already proven.
\endproof


\newpage

{\bf \blue K\"unneth theorem}

\theorem
Consider the map $\Psi:\; \bigoplus_{p+q=i}\Lambda^p(M_1) \otimes_\R \Lambda^q(M_2)\arrow
\Lambda^i(M_1 \times M_2)$ obtained by taking pullbacks
and multiplying. Define the differential 
on the totalization of $\Lambda^*(M_1) \otimes_\R \Lambda^*(M_2)$
in the usual way. {\bf \red Then $\Psi$ commutes with the
differential and induces identity on the cohomology.}

{\em \green We prove this result later this lecture.}

\corollary {\bf \blue (K\"unneth theorem)}\\
For any two manifolds $M_1, M_2$, {\bf \red there is an isomorphism
$H^i(M_1\times M_2) \cong \bigoplus_{p+q=i} H^p(M_1)
\otimes H^q(M_2)$.}

\proof The differential in the totalization of the bicomplex
$\Lambda^*(M_1) \otimes_\R \Lambda^*(M_2)$ is compatible with the de Rham
differential on $\pi_1^*(\Lambda^*(M_1)) \otimes
\pi_2^*(\Lambda^*(M_2))= \Lambda^*(M_1 \otimes M_2)$. However, cohomology of the
latter is $H^*(M_1\times M_2)$, cohomology of the former is
$H^*(M_1)\otimes H^*(M_2)$. Since $\Psi$ induces an
isomorphism on cohomology, one has $H^*(M_1\times M_2)
\cong H^*(M_1)\otimes H^*(M_2)$.
\endproof


\newpage

{\bf \blue Hermann K\"unneth}

{\small 
Hermann K\"unneth (1892-1975) was the son of the high
school ("Gymnasium", the highest form of high school)
teacher Christian K\"unneth. Beginning with 1910, he
studied mathematics at the Universit"at Erlangen and the
Ludwig-Maximilians-Universit"at M\"unchen with a "break"
from 1914-1919 where he served in the German army; he was
injured twice and was prisoner of war with the
British. K\"unneth was member of the AMV Fridericiana
Erlangen, a musically oriented fraternity. His professors
in Erlangen were Ernst Sigismund Fischer, Paul Gordan, Max
Noether, Richard Baldus and Erhard Schmidt.

1912 he took his first Staatsexamen to become a teacher
and 1920 he took his second. In 1920 he became teacher in
Bavaria, in particular at high schools ("Gymnasien") in
Kronach and Erlangen. He remained in contact with the
University in Erlangen, where he got in PhD under the
direction of Tietze in 1922 (and was assistant (professor)
beginning with 1921). The title of his thesis was "\"Uber
die Bettischen Zahlen einer Produktmannigfaltigkeit" -
"About the Betti numbers of a product manifold" (where he
proved the K\"unneth formula).

1923 he became assistant (professor) in Berlin;
interestingly, he became 1923 also teacher ("Studienrat")
in Kronach. As already indicated, the switched to the high
school Fridericianum in Erlangen in 1925, where he became
Oberstudienrat in 1950 [this would not be a very high
  position at a high school these days, but I am not sure
  how it was then]. In 1942 he habilitated in Erlangen and
was Privatdozent (a kind of freelancing professor) after
that. After he retired in 1957 from his teaching job, he
became associate professor in Erlangen. Otto Haupt said
about this: "[he] developed an amazing and surprising
scientific activity." (at the age of 65)

1964 he got the Bundesverdienstkreuz am Band (the second
lowest order of the "Order of Merit of the Federal
Republic of Germany"). (See this newspaper article - it
says: "His chivalric personality, of clear judgement,
emanates human kindness and witty humour.")
}

\newpage

{\bf \blue Hermann K\"unneth (1892-1975)}

\begin{center}
\epsfig{file=Kunneth.jpg,width=0.30\linewidth}\\

Hermann K\"unneth (1892-1975)
\end{center}



\newpage

{\bf \blue The proof of K\"unneth theorem}

\theorem
Consider the map $\Psi:\; \bigoplus_{p+q=i}\Lambda^p(M_1) \otimes_\R \Lambda^q(M_2)\arrow
\Lambda^i(M_1 \times M_2)$ obtained by taking pullbacks
and multiplying. Define the differential 
on the totalization of $\Lambda^*(M_1) \otimes_\R \Lambda^*(M_2)$
in the usual way. {\bf \red Then $\Psi$ commutes with the
differential and induces identity on the cohomology.}

\pstep
For any open subset $W\subset M_1\times M_2$,
denote by $\Lambda^*_p(W)$ the algebra of differential
forms multiplicatively generated by products of $\pi_1^*(\alpha)$ and
$\pi_2^*(\beta)$. As shown above, $\Lambda^*_p(A\times B)
= \Lambda^*(A) \otimes_\R \Lambda^*(B)$. Denote by 
$H^*_p(W)$ the cohomology of $(\Lambda^*_p(W), d)$.
Since $H^*((C_1\otimes C_2)^*)= H^*(C_1)\otimes H^*(C_2)$,
one has $H^*_p(A\times B)=H^*(A) \otimes H^*(B)$.

{\bf \green Step 2:} 
We assume that $M_1, M_2$ are compact or have a finite polyhedral
structure.  Let $U_1, ..., U_n$ be a cover of $M_1$ 
and $V_1, ..., V_m$ a cover of $M_2$ 
with all $U_i$, $V_j$ and all their intersections
starlike and closed. As explained in Lecture 6,
to build such a cover, one can take small convex
neighbourhood of each polyhedra, or the Voronoi cells
for an appropriate Riemannian structure.
Then $U_i \times V_j$ is a cover of $M_1 \times M_2$
with the same properties. 
Since $H^*_p(A\times B)=H^*(A) \otimes H^*(B)$, the
natural map $\Psi:\; H^*_p(U_i \times V_j)\arrow H^*(U_i \times V_j)$
is an isomorphism.

\newpage

{\bf \blue The proof of K\"unneth theorem (2)}

\theorem
Consider the map $\Psi:\; \bigoplus_{p+q=i}\Lambda^p(M_1) \otimes_\R \Lambda^q(M_2)\arrow
\Lambda^i(M_1 \times M_2)$ obtained by taking pullbacks
and multiplying. Define the differential 
on the totalization of $\Lambda^*(M_1) \otimes_\R \Lambda^*(M_2)$
in the usual way. {\bf \red Then $\Psi$ commutes with the
differential and induces identity on the cohomology.}

{\bf \green Step 3:} The Mayer-Vietoris long exact sequence
holds for $H^*_p(\cdot)$-cohomology of closed subsets in 
$M_1 \times M_2$; the argument is the same as for the
usual Mayer-Vietoris sequence. Assume that 
$\Psi:\; H^*_p(K)\arrow H^*(K)$ is an isomorphism
for all $K\subset M_1 \times M_2$ obtained as union
of at most $n-1$  sets obtained by intersections of 
some of $U_i\times V_j$, hence closed and starlike
(indeed, $U\times V \cap U' \times V'= (U\cap U')\times (V \cap V')$).

We prove that $\Psi:\; H^*_p(K)\arrow H^*(K)$  is an isomorphism for all $n$ using induction
by $n$. Let $X=U_i\times V_j$, and $Y$ be the union of $n-1$
such sets. Mayer-Vietoris exact sequence gives
\[\small
\begin{CD}
... @>>> H^{i-1}_{p}( X\cap  Y) @>>>  H^{i}_{p}( X\cup  Y) 
@>>>  H^{i}_{p}( X)\oplus  H^{i}_{p}( X)@>>> H^{i+1}_{p}( X\cap  Y)@>>>
...\\
&& @VVV @VVV @VVV @VVV \\
... @>>> H^{i-1}_{}(X\cap Y) @>>>  H^{i}_{}(X\cup Y) 
@>>>  H^{i}_{}(X)\oplus  H^{i}_{}(X)@>>> H^{i+1}_{}(X\cap Y)@>>>...
\end{CD}
\]
By induction assumption,
the vertical arrows in this diagram are isomorphisms for all terms
except for $H^*_p( X \cup  Y)\arrow H^*(X \cup Y)$.
{\bf \purple The 5-lemma implies that they are isomorphisms in all terms.}
\endproof.








%\newpage
%
%{\bf \blue Piecewise polynomial differential forms}
%
%\observation
%Let $V, W$ be polyhedra in $\R^n, \R^m$,
%and $C^\pol(V)$, $C^\pol(W)$ the ring of polynomial
%functions. Then $C^\pol(V\times W) \cong
%C^\pol(V)\otimes_\R C^\pol(W)$, because 
%$C^\pol(R^n)\otimes_\R C^\pol(R^m) \cong C^\pol(R^{n+m})$.
%
%\definition
%Let $M$ be a polyhedral $n$-manifold, 
%For each $n$-dimensional face $F$ of $M$ 
%we choose a homeomorphism of $F$ with a convex polyhedron in
%$\R^n$, such that the gluing maps between the codimension
%1 faces are linear. Consider the ring $C^\pol(M)$ 
%of {\bf \blue piecewise polynomial functions} (continuous functions which are
%polynomial on each $n$-face). 
%
%\definition
%Let $M$ be a smooth $n$-manifold with a polyhedral structure.
%Define the space of {\bf \blue piecewise polynomial differential
%forms on $M$} $\Lambda^*_\pol(M)$ as the space of
%continuous sections 
%$\alpha$ of the
%the bundle $\Lambda^*(M)$ which have polynomial
%coefficients on each $n$-dimensional face, and such that 
%$d\alpha$ (which is a priori continuous only on each 
%$n$-face) is continuous in $M$
%
%\exercise
%Prove that {\bf \purple the product and the differential of piecewise
%polynomial differential forms is again piecewise polynomial.}
%
%\theorem
%Cohomology of piecewise polynomial forms
%coincides the same as cohomology.

\end{document}
