# Friday, January 15

:::{.remark}
Let \[
V &\da \ts{a^2 + b^2 + c^2 + d^3 + e^{6k-1} = 0} \subseteq \CC^5 \\
S_\eps &\da \ts{ \abs{a}^2 + \abs{b}^2 + \abs{c}^2 + \abs{d}^2 + \abs{e}^2 = 1}
.\]
Then $V_k \intersect S_\eps \cong S^7$ is a homeomorphism, and taking $k=1,2,\cdots, 28$ yields the 28 smooth structures on $S^7$.
Note that $V_k$ is the cone over $V_k \intersect S_\eps$.

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> ? Admits a smooth structure, and $\bar{V}_k \subseteq \CP^5$ admits no smooth structure.
:::

:::{.question}
Is every triangulable manifold PL, i.e. homeomorphic to a simplicial complex?
:::

:::{.answer}
No!
Given a simplicial complex, there is a notion of the **combinatorial link** $L_V$ of a vertex $V$:

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It turns out that there exist simplicial manifolds such that the link is not homeomorphic to a sphere, whereas every PL manifold admits a "PL triangulation" where the links are spheres.
:::

:::{.remark}
What's special in dimension 4?
Recall the **Kirby-Siebenmann** invariant $\ks(x) \in H^4(X; \ZZ_2)$ for $X$ a topological manifold where $\ks(X) = 0 \iff X$ admits a PL structure, with the caveat that $\dim X \geq 5$.
We can use this to cook up an invariant of 4-manifolds.
:::

:::{.definition title="Kirby-Siebenmann Invariant of a 4-manifold"}
Let $X$ be a topological 4-manifold, then 
\[
\ks(X) \da \ks(X \cross \RR)
.\]
:::

:::{.remark}
Recall that in $\dim X\geq 7$, every PL manifold admits a smooth structure, and we can note that 
\[
H^4(X; \ZZ_2) = H^4(X \cross \RR; \ZZ_2) = \ZZ_2,
.\]
since every oriented 4-manifold admits a fundamental class.
Thus
\[
\ks(X) = 
\begin{cases}
0 & X \cross \RR \text{ admits a PL and smooth structure} 
\\
1 & X \cross \RR \text{ admits no PL or smooth structures }.
\end{cases}
\]
:::

:::{.remark}
$\ks(X) \neq 0$ implies that $X$ has no smooth structure, since $X \cross \RR$ doesn't.
Note that it was not known if this invariant was ever nonzero for a while!
:::

:::{.remark}
Note that $H^2(X; \ZZ)$ admits a symmetric bilinear form $Q_X$ defined by 
\[
Q_X: H^2(X; \ZZ)^{\tensor 2} &\to \ZZ \\
\alpha \tensor \beta &\mapsto \int_X \alpha\wedge \beta \da (\alpha \cupp \beta)([X])
.\]
where $[X]$ is the fundamental class.
:::

# Main Theorems for the Course

:::{.remark}
Proving the following theorems is the main goal of this course:
:::

:::{.theorem title="Freedman"}
If $X, Y$ are compact oriented topological 4-manifolds, then $X\cong Y$ are homeomorphic if and only if $\ks(X) = \ks(Y)$ and $Q_X \cong Q_Y$ are isometric, i.e. there exists an isometry 
\[
\varphi: H^2(X; \ZZ) \to H^2(Y; \ZZ)
.\]
that preserves the two bilinear forms in the sense that $\inner {\varphi \alpha}{ \varphi \beta} = \inner{ \alpha}{ \beta}$.


Conversely, every **unimodular** bilinear form appears as $H^2(X; \ZZ)$ for some $X$, i.e. the pairing induces a map 
\[
H^2(X; \ZZ) &\to H^2(X; \ZZ)\dual \\
\alpha &\mapsto \inner{ \alpha }{ \wait}
.\]
which is an isomorphism.
This is essentially a classification of simply-connected 4-manifolds.
:::

:::{.remark}
Note that preservation of a bilinear form is a stand-in for "being an element of the orthogonal group", where we only have a lattice instead of a full vector space.
:::

:::{.remark}
There is a map \( H^2(X; \ZZ) \mapsvia{PD} H_2(X; \ZZ) \) from Poincaré , where we can think of elements in the latter as closed surfaces $[\Sigma]$, and
\[
\inner{ \Sigma_1 }{ \Sigma_2 } = \text{signed number of intersections points of } \Sigma_1 \transverse \Sigma_2
.\]
Note that Freedman's theorem is only about homeomorphism, and is not true smoothly.
This gives a way to show that two 4-manifolds are homeomorphic, but this is hard to prove!
So we'll black-box this, and focus on ways to show that two *smooth* 4-manifolds are *not* diffeomorphic, since we want homeomorphic but non-diffeomorphic manifolds.
:::

:::{.definition title="Signature"}
The **signature** of a topological 4- manifold is the signature of $Q_X$, where we note that $Q_X$ is a symmetric nondegenerate bilinear form on $H^2(X; \RR)$ and for some $a, b$
\[
(H^2(X; \RR), Q_x) \mapsvia{\text{isometric}} \RR^{a, b}
.\]
where $a$ is the number of $+1$s appearing in the matrix and $b$ is the number of $-1$s.
This is $\RR^{ab}$ where $e_i^2 = 1, i=1\cdots a$ and $e_i^2 = -1, i=a+1, \cdots b$, and is thus equipped with a specific bilinear form corresponding to the Gram matrix of this basis.
\[
\begin{bmatrix}
1 & 0 & 0 & 0 & 0
\\
0 & 1 & 0 & 0 & 0
\\
0 & 0 & \ddots & 0 & 0
\\
0 & 0 & 0 & -1 & 0
\\
0 & 0 & 0 & 0 & -1
\end{bmatrix}
= I_{a\times a} \oplus -I_{b \cross b}
.\]
Then the signature is $a-b$, the dimension of the positive-definite space minus the dimension of the negative-definite space.

:::

:::{.theorem title="Rokhlin's Theorem"}
Suppose $\inner{ \alpha}{\alpha} \in 2\ZZ$ and \( \alpha\in H^2(X; \ZZ) \) and $X$ a simply connected **smooth** 4-manifold. Then 16 divides $\sig(X)$.
:::

:::{.remark}
Note that Freedman's theorem implies that there exists topological 4-manifolds with no smooth structure.
:::

:::{.theorem title="Donaldson"}
Let $X$ be a smooth simply-connected 4-manifold.
If $a=0$ or $b=0$, then $Q_X$ is diagonalizable and there exists an orthonormal basis of $H^2(X; \ZZ)$.
:::

:::{.remark}
This comes from Gram-Schmidt, and restricts what types of intersection forms can occur.
:::

## Warm Up: $\RR^2$ Has a Unique Smooth Structure

:::{.remark}
Last time we showed $\RR^1$ had a unique smooth structure, so now we'll do this for $\RR^2$.
The strategy of solving a differential equation, we'll now sketch the proof.


:::

:::{.definition title="Riemannian Metrics"}
A **Riemannian metric** $g\in \Gamma( \Sym^2 T\dual X)$ for $X$ a smooth manifold is a metric on every $T_p X$, so $g_p \in (T_p X^{\tensor 2})\dual$, such that
\[
g_p: T_pX \tensor T_p X &\to \RR && g(v, v) \geq 0, \quad g(v,v) = 0 \iff v=0
.\]

:::

:::{.definition title="Almost complex structure"}
An **almost complex structure** is a morphism $J\in \Endo_{\Vect(X)}(TX)$ of vector bundles over $X$ such that $J^2 = -\id_{TX}$.
:::

:::{.definition title="Integrable"}
An almost-complex structure is **integrable** $J$ if it comes from a complex structure in the following sense:
for a complex manifold $M\in \Mfd(\CC)$, take holomorphic coordinates $z = x+iy$ and set $J \dd{}{x} \da \dd{}{y}$ and $J\dd{}{y} \da -\dd{}{x}$.
:::

:::{.remark}
A manifold $M\in \smooth\Mfd(\RR)$ admits an almost-complex structure iff $TM$ admits a reduction of structure group $\GL_{2n}(\RR) \to \GL_n(\CC)$.
:::

:::{.remark}
Let $e\in T_p X$ and $e\neq 0$, then if $X$ is a surface then $\ts{e, J_p e}$ is a basis of $T_p X$, where $J_p$ is the restriction of $J$ to $T_p X$:

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:::

:::{.exercise title="?"}
Show that \( \ts{ e, J_p e } \) are linearly independent in $T_p X$.
In particular, $J_p$ is determined by a point in $\RR^2\smts{\text{the }x\dash\text{axis}}$

:::

:::{.proof title="That R2 admits a unique smooth structure (sketch)"}

Let $\tilde \RR^2$ be an exotic $\RR^2$.

### Step 1

Choose a metric on $\tilde \RR^2$, say $g \da \sum f_I g_i$ with $g_i$ metrics on coordinate charts $U_i$ and $f_i$ a partition of unity.

### Step 2

Find an almost complex structure on $\tilde \RR^2$.
Choosing an orientation of $\tilde \RR^2$, the metric $g$ defines a unique almost complex structure $J_p e \da f\in T_p \tilde \RR^2$ such that 

- $g(e, e) = g(f, f)$ 
- $g(e, f) = 0$.
- $\ts{e, f}$ is an oriented basis of $T_p \tilde \RR^2$ 

This is because after choosing $e$, there are two orthogonal vectors, but only one choice yields an *oriented* basis.

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  };
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### Step 3

We then apply a theorem:

:::{.theorem title="Almost-complex structures on surfaces come from complex structures"}
Any almost complex structure on a surface comes from a complex structure, in the sense that there exist charts \( \varphi_i: U_i \to \CC \) such that $J$ is multiplication by $i$.
:::

So 
\[ d \varphi(J \cdot e) = i \cdot d \varphi_i (e)
,\] 
and $(\tilde \RR^2, J)$ is a complex manifold. 
Since it's simply connected, the Riemann Mapping Theorem shows that it's biholomorphic to $\DD$ or $\CC$, both of which are diffeomorphic to $\RR^2$.
:::

:::{.remark}
See the Newlander-Nirenberg theorem, a result in complex geometry. 
:::