\classheader{2013-02-01}


Today we'll introduce a purely sheaf-theoretic definition of constructions such as the tensor product of bundles, which is the one that is used in practice.

Everything we talk about today will be $\Cinfty$.

Let $p:V\to B$ be a vector bundle on $B$. Given an open set $U\subseteq B$, a section $s$ on $U$ is a map $s:U\to V$ making the diagram commute:
\begin{center}
\begin{tikzcd}
{}& V\ar{d}{p}\\
U \ar{ru}{s} \ar[hook]{r} & B
\end{tikzcd}
\end{center}
The set of $\Cinfty$ sections $s:U\to V$ is denoted by $\Gamma(U,V)$. We'll use the notation $\Cinfty(V)$ for $\Gamma(B,V)$.

For any open set $U\subseteq B$, $\Gamma(U,V)$ is a module over the ring $\Cinfty_B(U)=\{\Cinfty\text{ maps }f:U\to\R\}$, and for any open sets $U'\subseteq U$ of $B$, the restriction map $\Gamma(U,V)\to\Gamma(U',V)$ is a homomorphism of $\Cinfty_B(U)$-modules, where $\Gamma(U',V)$ is given the structure of a $\Cinfty_B(U)$-module by restriction of scalars along the map $\Cinfty_B(U)\to\Cinfty_B(U')$. We say that $U\mapsto \Gamma(U,V)$ is a sheaf of modules over $C_B^\infty$. Here is a diagram of the situation:
\begin{center}
\begin{tikzcd}
\Gamma(U,V) \ar{d}[swap]{\Cinfty(U)\text{ module hom. }} & \text{ is a module over } & \Cinfty(U)\ar{d}{\text{ ring hom.}}\\
\Gamma(U',V) & \text{ is a module over }& \Cinfty(U')
\end{tikzcd}
\end{center}
An example of a sheaf of $C_B^\infty$-modules is
\[(C_B^\infty)^k=\underbrace{C_B^\infty\oplus\cdots\oplus C_B^\infty}_{k\text{ times}}.\]
If $p:V\to B$ is the trivial bundle of rank $k$, so that $V=B\times\R^k$ and $p$ is the projection map, then clearly we can identify $\Cinfty(V)$ with $(C_B^\infty)^k$.

\begin{definition}
A sheaf $\cal{F}$ of modules over $\Cinfty_B$ is locally free of rank $k$ if there is an open cover $B=\bigcup_{i\in I} U_i$ such that $\cal{F}|_{U_i}\cong(C_{U_i}^\infty)^k$ for all $i\in I$.
\end{definition}


\begin{theorem}
The functor
\[\text{category of vector bundles on }B\to \text{category of locally free, finite rank sheaves of }C_B^\infty\text{-modules}\]
which sends $V\mapsto \Cinfty(V)$ is an equivalence of categories.
\end{theorem}
\begin{proof}
To see that it is fully faithful, 

(I wasn't able to get down this part of the argument in class. The next page is a supplement provided to me by Professor Nori to fill in this gap.)

\includepdf[pages={-}]{math318lecture11supplement.pdf}

We also need to check that this functor is essentially surjective. Let $\cal{F}$ be locally free. There exists an open cover of $U_i$'s such that  $\psi_i:\cal{F}_{U_i}\xrightarrow{\;\cong\;}(\Cinfty_{U_i})^k$. Restricted to $U_i\cap U_j$, we get an isomorphism
\begin{center}
\begin{tikzcd}[column sep=0.6in]
(\Cinfty_{U_i\cap U_j})^k\ar{r}{\psi_j\circ\psi_i^{-1}}[swap]{\cong} &(\Cinfty_{U_i\cap U_j})^k
\end{tikzcd}
\end{center}
which as we saw before corresponds to a $\Cinfty$ map $\varphi_{ij}:U_i\cap U_j\to \GL_k(\R)$. Check that $\varphi_{ij}$ satisfy the cocycle condition, which will then let us construct the required $V$.
\end{proof}


Let $V_1,V_2$ be vector bundles, and let $s_1:U\to V_1$ and $s_2:U\to V_2$ be $\Cinfty$ sections of $V_1$ and $V_2$ respectively on $U$. Then whatever the definitions are, we would like to be able to say that
\begin{center}
``$s_1\tensor s_2$ is a $\Cinfty$ section of $V_1\tensor V_2$ on $U$''.
\end{center}
Recall that we can take the tensor product of two sheaves. The sheaf $\Cinfty(V_1)\tensor_{\Cinfty_B}\Cinfty(V_2)$ is a locally free sheaf of $\Cinfty_B$-modules. By the equivalence we just proved, we get a vector bundle $Z$ with an isomorphism
\begin{center}
\begin{tikzcd}
\Cinfty(Z)\ar{r}{T}[swap]{\cong}& \Cinfty(V_1)\otimes\Cinfty(V_2),
\end{tikzcd}
\end{center}
and we define $Z$ to be $V_1\otimes V_2$. We have that $s_1\otimes s_2$ is a section of the bundle on the right on $U$, and $T^{-1}(s_1\otimes s_2)$ is a section of $Z$ on $U$.

Let $V$ be a vector bundle on $B$ of rank $k$. We define $\det(V)=\Lambda^k(V)$, the top exterior power of $V$, which is a line bundle on $B$ (i.e., a vector bundle of rank 1).

From now on, assume that all our topological spaces $Z$ to be connected, and satsify the following property ($\ast$): the collection of contractible open subsets of $Z$ forms a basis for the topology of $Z$.

Recall the definition of a constant presheaf: given an abelian group $A$, we define $P_A(U)=A$ for all $U$ open on $Y$. The sheafification of $P_A$ is denoted by $A_Y$.

\begin{definition}
The higher direct image is defined as follows. Given a continuous map $f:X\to Y$, fix an integer $q\geq 0$. For any open $U\subseteq Y$, we define $P^q(U):=H^q(f^{-1}(U))$. Of course, for $U'\subseteq U$, we have $f^{-1}(U')\subseteq f^{-1}(U)$, and cohomology is contravariant, which is the right direction. Thus, $P^q$ is a presheaf on $Y$.
\end{definition}

The $q$th higher direct image of $f$ is the sheafification of $P^q$, and we denote it by $R^qf_*\Z_X$. As the sheafification, it satisfies a certain universal property: for any sheaf $\cal{G}$ on $Y$ and $h:P^q\to\cal{G}$, there is a unique $\widetilde{h}$ such that
\begin{center}
\begin{tikzcd}
P^q\ar{d} \ar{r}{h} & \cal{G}\\
R^qf_*\Z_X \ar[dashed]{ru}[swap]{\widetilde{h}}
\end{tikzcd}
\end{center}
What if $f:X\to Y$ is a fiber bunde with fiber $F$? In particular, we might be interested in the case when $X=Y\times F$ and $f=p_1$.

Because cohomology is contravariant, we get a map $H^q(F)\xrightarrow{\;p_2^*\;}H^q(Y\times F)$. By definition, we have $H^q(Y\times F)=P^q(Y)$, which comes with a restriction map to $P^q(U)=H^q(U\times F)$.

We claim that there is a natural map $H^q(F)_Y\xrightarrow{\;\phi\;}R^qf_*\Z_X$, and that this is an isomorphism.

The natural map $\phi$ is defined by
\begin{center}
\begin{tikzcd}
P_{H^q(F)}(U)=H^q(F) \ar{rd} \ar{r} & P^q(U) \ar{r} & (R^qf_*\Z_X)(U)\\
 & H^q(F)_Y \ar{ru}[swap]{\phi}
\end{tikzcd}
\end{center}
To prove $\phi$ is an isomorphism of sheaves, it suffices to show it induces isomorphisms on all stalks.

Because the stalk at a point $y\in Y$ is the limit over all open sets containing $y$, and because $Y$ satisfies the property ($\ast$), we have
\begin{center}
\begin{tikzcd}
H^q(F) \ar{r} \ar[bend left]{rr}{\id} & R^qf_*\Z_Y \ar{r}{\cong} & H^q(F)
\end{tikzcd}
\end{center}

\begin{corollary}
The sheaf $R^qf_*\Z_Y$ is locally constant, with stalks $\cong$ $H^q(F)$.
\end{corollary}

\begin{definition}
Let $p:V\to B$ be a real vector bundle of rank $k$. Let $0_V:B\to V$ be the zero section, and let $V'=V-0_V(B)$. Then we can make a new bundle $p:(V,V')\to B$, with fibers $(\R^k,\R^k-\{0\})$.  For any open set $U\subseteq B$, the map $U\mapsto \cal{P}(U):=H^k(p^{-1}(U),p^{-1}(U)\cap V')$ is a presheaf; take the associated sheaf. This sheaf is $\Or_V$, the orientation sheaf of $V$.
\end{definition}

Note that $\Or_V$ is a locally constant sheaf with all stalks $\cong\Z$. An orientation of $V$ is a global section $s$ of $\Or_V$ such that $s(b)\in H^k(p^{-1}(b),p^{-1}(b)-\{0\})$ is a generator for all $b\in B$.

\begin{theorem}[Thom isomorphism theorem]
There is a natural isomorphism
\[H^i(B,\Or_V)\xrightarrow{\;\cong\;} H^{i+k}(V,V')\]
where the left denotes cohomology with values in an abelian group (the sheaf $\Or_V$ is locally constant, so this makes sense), and the right is cohomology of the pair $(V,V')$.
\end{theorem}

\begin{theorem}[Thom isomorphism theorem, for us]
$\text{}$

\begin{enumerate}
\item $H^i(V,V')=0$ for all $i<k$, when $V$ is a vector bundle of rank $k$.
\item The natural map $H^k(V,V')\to \Gamma(B,\Or_V)$ is an isomorphism.
\item Let $s$ be an orientation of $V$, i.e. $s\in\Gamma(B,\Or_V)$. Then $s$ has a preimage $\theta\in H^k(V,V')$ by part 2. The map
\begin{center}
\begin{tikzcd}[row sep=0.1in]
H^i(B) \ar{r} & H^{i+k}(V,V')\\
\alpha\ar[mapsto]{r} & p^*\alpha\cupprod\theta
\end{tikzcd}
\end{center}
is an isomorphism.
\end{enumerate}
\end{theorem}
Recall that we can form the cohomology ring $\bigoplus_{m=0}^\infty H^m(X)$, where cup product is the multiplication. For any $Y\subseteq X$, the group $\bigoplus_{m=0}^\infty H^m(X,Y)$ is a module over the cohomology ring.




