# Thursday, December 02

:::{.remark}
Today: the Hirzebruch signature theorem and exotic $S^7$.
We saw that $\Omega_* \tensor \QQ$ is the polynomial algebra on $\QQ[v_1, v_2\cdots]$ where $v_2$ corresponds to $\CP^{2n}$.
This was because $\ker\qty{\Omega_n \mapsvia{\ts{p_{I_n}}} \ZZ\cartpower{n} }$ can only be torsion. 
:::

:::{.definition title="Signature"}
For $B$ a symmetric bilinear form, $\signature(B) = \dim V^+ - \dim V^-$, the dimensions of positive/negative definite subspaces respectively, or equivalently the difference in the number of positive and negative eigenvalues.
For $M\in \smooth\Mfd^{4n}$, there is a middle-dimensional pairing
\[
H^{2n}(M; \ZZ)\tensorpower{\ZZ}{2} &\to \ZZ \\
a\tensor b &\mapsto \inner{a \cupprod b}{[M]}
,\]
and we write $\sigma(M)$ for the signature of this form.
:::

:::{.theorem title="?"}
The signature induces a ring morphism
\[
\sigma: (\Omega_*, \disjoint, \times) \to (\ZZ, +, \cdot)
,\]
so

- $\sigma(M \disjoint N) = \sigma(M) + \sigma(N)$,
- $\sigma(M\times N) = \sigma(M) \sigma(N)$,
- $\sigma(M) = 0$ if $M = \bd M'$ for some $M'$.

:::

:::{.lemma title="?"}
If there is a half-dimensional subspace $L \subseteq H^{2n}(M; \QQ)$ such that $\inner{a \cupprod}{[M]} = 0$ for all $a,b\in L$ then $\sigma(M) = 0$.
:::

:::{.proof title="?"}
Use Poincare duality to get an isomorphism:

\begin{tikzcd}
	{H^{2n}(N; \QQ)} && {H^{2n}(M; \QQ)} && {H^{2n+1}(N, M; \QQ)} \\
	\\
	&& {H_{2n}(M; \QQ)} && {H_{2n}(N; \QQ)}
	\arrow["{i_*}", from=3-3, to=3-5]
	\arrow["{\cong, \PD}"', from=1-5, to=3-5]
	\arrow["\cong", from=1-3, to=3-3]
	\arrow["{f}", from=1-3, to=1-5]
	\arrow["{i^*}", from=1-1, to=1-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbNCwwLCJIXnsybisxfShOLCBNOyBcXFFRKSJdLFs0LDIsIkhfezJufShOOyBcXFFRKSJdLFsyLDAsIkheezJufShNOyBcXFFRKSJdLFsyLDIsIkhfezJufShNOyBcXFFRKSJdLFswLDAsIkheezJufShOOyBcXFFRKSJdLFszLDEsImlfKiJdLFswLDEsIlxcY29uZywgXFxQRCIsMl0sWzIsMywiXFxjb25nIl0sWzIsMF0sWzQsMiwiaV4qIl1d)

Note that $\rank i^* = \rank i_*$ since these spaces are dual, and this has the same rank as $f$ since the sides are isomorphisms.
Then $\dim H^{2n}(M) = \dim \im i^* + \dim \coker i^*$ and $\dim \coker i^* = \dim \im i_*$, so $\dim H^{2n}(M) = 2\dim \im i^* \leq 2\dim \im i^*$.
Then writing $i^* \alpha = a, i^* \beta = b\in \im i^*$,
\[
\inner{ i^*(a\cupprod b)}{[M]} = \inner{ \alpha\cupprod \beta}{i_*[M]} =0
,\]
since $i_*[M] = 0$.
Thus $\sigma(M) = 0$.
:::

:::{.corollary title="?"}
One can compute the signature in terms of Pontryagin numbers, i.e. there is a formula for $\sigma(M)$ in terms of $\ts{p_{I_n}(M)}$.
The explicit formula is the **Hirzebruch signature theorem**.
:::

:::{.example title="?"}
Write $\Omega_4 \tensor \QQ = \gens{\CP^2}$ and consider $M \in \smooth\Mfd^4$.
Note $\CP^2$ has only 1 Pontryagin number and $c(\CP^n) = (1+x)^{n+1}$, so we have a formula
\[
1 - p_1(\CP^2) 
&= c(\CP^2) \cdot c(\bar{\CP^2}) \\
&= (1+x)^3(1-x)^3 \\
&= (1-x^2)^3 \\
&= 1 + -3x^2 + 3x^4 - x^6 \\
&= 1 - 3x^2
,\]
where we've used that $H^{\geq 4} = 0$.
So $p_1(\CP^2)= 3x^2$, which implies
\[
\inner{ p_1 \CP^2}{[\CP^2]} = \inner{3x^2}{[\CP^2]} = 3
.\]
Using that $\sigma(\CP^2) = 1$, we can deduce $\sigma(M) = p_1(M)/3$ by considering eigenvalues of the intersection pairing.
:::

:::{.example title="?"}
Consider $M^8$ and $\Omega_8 \tensor \QQ = \gens{\CP^4, \CP^2\times \CP^2}$.
Then $\Sigma(\CP^4) = \sigma(\CP^2\cartpower{2}) = 1$, what are the Pontryagin numbers?
Write $v_2 = \CP^2, v_2^2 = \CP^2 \times \CP^2, v_4 = \CP^4$,then
\[
1 - p_1(v_4) + p_2(v_4) 
&= c(v_4)c(\bar v_4) \\
&= (1+x)^5(1-x)^5 \\
&= (1-x^2)^5 \\
&= 1-5x^2 + 10x^4
,\]
 so $p_1(v_4) = 5x^2$ and $p_2(v_4) = 10x^4$.
Then
\[
\inner{p_1^2 v_4}{[v_4]} &= \inner{25x^4}{[v_4]} = 25 \\
\inner{p_2 v_4}{[v_4]} &= \inner{10x^4}{[v_4]} = 10 
.\]
Similarly,
\[
1 + p_1(v_4^2) + p_2(v_4^2) = (1+ 3x^2)(1+3y^2) = 1 + 3x^2 + 3y^2 + 9x^2y^2
,\]
so
\[
\inner{p_1^2 v_2^2}{[v_2^2]} 
= \inner{(3x^2 + 3y^2)^2 }{[v_2^2]}
= \inner{9x^4 + 9y^4 + 18x^2 y^2 }{[v_2^2]}
= 18
,\]
since the cohomology ring is $\FF[x]/\gens{x^2} \tensor \FF[y]/\gens{y^2}$, and a similar calculation shows
\[
\inner{p_2(v_2^2)}{[v_2^2]} = 9
.\]

Summarizing, where we abuse notation and identity classes with numbers,

- $p_1^2(v_4) = 25$,
- $p_2(v_4) = 10$,
- $p_1^2(v_2^2) = 18$,
- $p_2(v_2^2) = 9$.

Writing $\sigma = ap_1^2 + b p_2$, 
\[
1 = \sigma(v_4) &= 25 a + 10 b \\
1 = \sigma(v_2^2) &= 18a + 9b
.\]

This yields $b = {1\over 9}-2a$ and $1 = 25a + 10\qty{{1\over 9} - 2a}$ yields $a=-1/45$ and $b=7/45$.
Thus
\[
\sigma = {1\over 45}\qty{2p_2 - p_1^2}
.\]
:::

:::{.theorem title="Thom"}
\[
\Omega_7 = 0
.\]
:::

:::{.definition title="?"}
Given $M^7$ with $H^3(M) = H^4(M)$ and suppose $M = \bd B^*$.
Let $i: H^4(B; M) \to H^4(B)$ and define
\[
\lambda(M) \da 2 \inner{ (i\inv p_1 \T B)^2 }{[B]} - \sigma(B)
.\]
Here $\sigma(B)$ is the signature of
\[
H^4(B, M ; \QQ)\tensorpower{\ZZ}{2} &\to \ZZ \\
a\tensor b &\mapsto \inner{a \cupprod b}{[B]} 
.\]
:::

:::{.proposition title="?"}
$\lambda(M)$ is independent of $B$.
:::

:::{.remark}
Let $C \da B_1 \glue{M} B_2$ where $\bd B_i = M$, then there is a relation expressing $\inner{p_1(C)^2}{[C]}$ in terms of pairings of $p(B_i)$ against $[B_i]$, and $\sigma(C) = \sigma(B_1) - \sigma(B_2)$.
We have 
\[
45\sigma(C) + \inner{p_1(C)^2}{[C]} &\equiv 0 \mod 7 \\
\implies -40\sigma(C) + \inner{p_1(C)^2}{[C]} &\equiv 0 \mod 7 \\
\implies \sigma(C) + 2\inner{p_1(C)^2}{[C]} &\equiv 0 \mod 7 \\
\implies (\sigma(B_1) - \sigma(B_2) ) - 2 \qty{ \inner{(i\inv p_1 B_1)^2 }{[B_1]} \cdots }
&= \lambda_{B_1}(M) - \lambda_{B_2}(M)
.\]
:::

:::{.remark}
If the $\lambda$s differ, the manifolds can not be diffeomorphic.
Constructing $S^7$s: take $S^3$ bundles over $S^4$.
Pick any map $S^3\injects \SO_4(\RR) \subseteq \Homeo(S^3)$ to get clutching data, which are elements of $\pi_3(\SO_4) = \ZZ\cartpower{2}$.
Label these bundles with parameters $(m, n) \in \ZZ\cartpower{2}$ so that $\xi_{1, 0}$ corresponds to lifting $[S^3, \SO_3]$ to $[S^3, \SO_4]$ fixing the $x\dash$axis, and $\xi_{1, 1}$ is the canonical for $\HH\PP^1$.
If $M_k$ is the total space of $\xi_{k, 1}$, then $\lambda(M_k) \equiv k^2-1 \mod 7$, so aren't diffeomorphic.
Then one can show that the $M_k$ are homeomorphic by finding a Morse function with exactly 2 critical points.
:::