# Wednesday, August 24

:::{.remark}
Recall the notion of a smooth atlas, whose transition functions are $\phi_\beta \phi_\alpha\inv \in C^\infty( \varphi_ \alpha U_{\alpha \beta}, \varphi_{ \beta} U_{ \alpha \beta}) \subseteq C^\infty(\RR^d, \RR^d)$.
Example: for $S^2$, a smooth atlas is $\mca = \ts{\varphi_i, \varphi_+}$, the stereographic projections from the north and south poles respectively.
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:::{.definition title="Compatible charts"}
Given an atlas $\mca = \ts{ \varphi_ {\alpha}}$, a **compatible** chart is a homeomorphism $\phi: U\to \phi(U)$ such that $\phi \phi_{ \alpha}\inv, \varphi_{\alpha} \varphi\inv$ are smooth on their domains.
Equivalently, $\phi$ is compatible with $\mca$ iff $\mca'\da \mca \union \ts{ \varphi}$ is still a smooth atlas.
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:::{.definition title="Smooth structures and smooth manifolds"}
A $d\dash$dimensional **smooth structure** on a space $M$ is a maximal smooth $d\dash$dimensional atlas on $M$, i.e. a smooth atlas which is not properly contained in any other smooth atlas.
A $d\dash$dimensional **smooth manifold** is a second-countable Hausdorff space together with a $d\dash$dimensional smooth structure $\mca$.
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:::{.remark}
Note that a maximal atlas is quite large: e.g. for $S^2$, one can add uncountably many translations, rotations, and their compositions.
It suffices instead to give a single (potentially non-maximal) atlas $\mca_0$, which uniquely determines a maximal atlas $\mca$ by taking all possible charts which are compatible with $\mca_0$.
Important fact: the multivariable chain rule guarantees that the composition of smooth functions is smooth!
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:::{.remark}
$\RR$ as a bare topological space has many smooth structures.
Take $\mca_0 = \ts{ \id}: \RR\to \RR^1$, then clearly $\id\circ\id\inv$ is smooth so this forms a smooth atlas $\mca_0$ and thus is contained in a unique smooth structure $\hat \mca_0$ of all $\phi: U\to V$ on open subsets of $\RR$ which are homeomorphisms such that $\phi, \phi\inv$ are both smooth (since $\id\circ \phi\inv, \phi\circ\id\inv$ must be smooth to be compatible).
One can write an atlas which yields a distinct smooth structure by taking any $f\in \Top^\cong(\RR, \RR^1)$ is a homeomorphism but either $f$ or $f\inv$ isn't smooth.
For example, take $\mca_1\da\ts{f(x) = x^3}$, whose inverse isn't smooth.
The maximal atlas is $\hat\mca_1$ comprised of all $\phi\in \Top^\cong(U, V)$ such that $f \phi\inv, \phi f\inv$ are smooth.
Note that $f\not\in \hat\mca_0$, so $\hat\mca_0 \not\cong \hat\mca_1$.
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:::{.warnings}
One often hears that $\RR^4$ has infinitely many smooth structures, which means something slightly different!
In that case, we consider smooth structures up to diffeomorphism:
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:::{.definition title="Diffeomorphism"}
Let $M, N$ be smooth manifolds with atlases $\ts{ \varphi_ \alpha}, \ts{ \psi_ \beta}$ and $f\in\Top(M, N)$, and let $x\in M, f(x)\in N$.
Then $f$ is **smooth at $x$** if for one (and hence any) $\alpha, \beta$ with $x\in U_\alpha, f(x)\in V_\beta$, the map $\phi_\alpha\inv f \phi_\beta$ is smooth at $\phi_\alpha(x)$.

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The map $f$ is **smooth** iff it is smooth at every point, and is a **diffeomorphism** if $f$ is a bijection and $f,f\inv$ are smooth.
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:::{.exercise title="?"}
Show that $\hat\mca_0, \hat\mca_1$ from the previous example are diffeomorphic.
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:::{.remark}
A key feature of smooth manifolds: they admit many smooth functions, which can be used to probe different parts of the manifold separately.
An example is the smooth analog of the Urysohn lemma: if $A, B \subseteq M$ are disjoint and closed, there is a smooth function which is 1 on $A$ and 0 on $B$:

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