# Monday, August 22


:::{.remark}
Recall the definition of an atlas for a topological space as homeomorphisms $\phi_\alpha: U_\alpha \to \RR^d$ onto open subsets of $\RR^d$, and a topological manifold  is a 2nd countable Hausdorff space which admits an atlas.
An example is $S^2$, one can stereographically project from the north pole:

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\end{tikzpicture}

One can use similar triangles to ascertain the formula for this projection.
Similarly one can project from the south pole, and these two charts form an atlas to $\RR^2$:
\[
\phi_-: S^2\smts{\tv{0,0,1}} &\to \RR^2 \\
\tv{x,y,z} &\mapsto \tv{{x\over 1-z}, {y\over 1-z}} \\ \\
\phi_+: S^2\smts{\tv{0,0,-1}} &\to \RR^2 \\
\tv{x,y,z} &\mapsto \tv{{x\over 1+z}, {y\over 1+z}}
.\]

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:::{.remark}
We want to define a **smooth** manifold, and in particular we want to say when a map $f: M\to \RR$ is "differentiable" at a point.
One can try to define this by saying $f$ is smooth if $f \circ \phi_\alpha\inv$ at $\phi_\alpha(x)$ as a function $\RR^d\to \RR$ -- however, there may be issues on overlaps.
We resolve this by imposing stronger conditions on the atlas:
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:::{.definition title="Smooth atlases"}
A **smooth** $d\dash$dimensional atlas on a topological space $M$ is $d\dash$dimensional atlas $\mca \da \ts{\varphi_\alpha : U_{\alpha} \to \RR^d}$ such that for every \( \alpha, \beta\in \mca \), the transition function \( \varphi_ \beta \circ \varphi_ \alpha: \varphi_{\alpha}(U_{ \alpha \beta}) \to \varphi_ \beta(U_{ \alpha \beta}) \) is in $C^\infty(\RR^d, \RR^d)$:

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\end{tikzpicture}

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:::{.remark}
Similarly one can require $C^1$ atlases, analytic atlases, complex atlases, etc by imposing similar conditions on transition functions.
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:::{.exercise title="?"}
Check that the stereographic projection atlas for $S^2$ is smooth.

> Hint: to get $\phi_-\inv$, solve $\tv{{x\over 1-z}, {y\over 1-z}} = \tv{u, v}$ using $x^2+y^2=1-z^2$ to get $z = {u^2+v^2-1\over u^2+v^2+1}$.
> Check $1-z = {2\over u^2+v^2-1}$ to get $x={2u\over u^2+v^2+1}, y= {2v\over u^2+v^2+1}$.
> Wind up with $\phi_+ \circ \phi_-\inv(u, v) = \tv{{u\over u^2+v^2}, {v\over u^2+v^2}}$, which has partial derivatives of all orders as a map $\RR^2\smz\to \RR^2\smz$.

Note that this yields the map $\vector w \mapsto {\vector w \over \norm{\vector w}^2}$, which in complex coordinates is $w \mapsto {w\over \abs{w}^2} = {1\over \bar{w}}$, which is not holomorphic. However, one could change one of the signs in $\phi_{\pm}$ to fix this and obtain a complex structure.
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