Manifolds

TODO: description.

Day 1

Convention: always embed $X$ into a $\mathbb{R}^n$ for some very large $n > 0$.

Definition. Let $U\subseteq \mathbb{R}^n$ be open. A map $f:U \to \mathbb{R}^m$ is smooth if has continuous partial derivatives of all orders.

Example. Polynomials are clearly smooth.

Example. $\exp(x^2 + y^2 + z^2)$.

Definition. Let $X \subseteq \mathbb{R}^n$. Then $f:X \to \mathbb{R}^m$ is smooth if it can be locally extended to a smooth map $f:U \to \mathbb{R}^m$ where $X \subseteq U$ and $U$ is open in $\mathrm{R}^n$. By local, we mean that for any point $x \in X$, there is some open $U\subseteq \mathbb{R}^n$ such that $f:X \cap U \to \mathbb{R}^m$ can be extended to a smooth map $U \to \mathbb{R}^m$.

Example. $X = [0, 1]$. Then $f:X \to \mathbb{R}^2$ given by $x \mapsto (x, x)$ is smooth.

Example. $X = S^1$.

Definition. We say $f:X \to Y$ is diffeomorphic if $f$ is a smooth bijection and $f^{-1}$ is smooth.

Example. $\mathbb{R}$ is diffeomorphic to $S^1$ with the north pole removed by stereographic projection.

Remark. Milnor’s exotic spheres: there are 28 non-diffeomorphic structures on $S^7$.

Remark. Donaldson. $\mathbb{R}^4$ is the unique Euclidean space among all $\mathbb{R}^n$ such that it has uncountable many non-diffeomorphic structures.

Definition. $X$ is a $k$-dimensional smooth manifold if it is locally diffeomorphic to $\mathbb{R}^k$. That is, for any $x \in X$, there is an open neighborhood $V$ of $x$ such that there is a diffeomorphism $\psi:V \to U \subseteq \mathbb{R}^k$ with $U$ open.

Example. $S^1 = {x^2 + y^2 = 1}$. Implicit function theorem go brrr

Remark. We say that $\psi^{-1}$ is a parameterization of the neighborhood $V$ and $\psi$ is coordinate system. Spelling this all out, $\psi = (x_1, \ldots, x_k)$ with each $x_i$ a smooth function on $X$.

Definition. Let $U \subseteq \mathbb{R}^n$ and $f:U \to \mathbb{R}^m$ be smooth. Choose a vector $h \in \mathbb{R}^n$. The derivative of $f$ in the derection of $h$ at a point $x$ is given by \(df_x(h) = \lim_{t \to 0} \frac{f(x + th) - f(x)}{t}.\)

DO THE STARRED EXERCISES *-ex in section 1. DUE NEXT WEEK

Day 2

Let $U \subseteq \mathbb{R}^n$ be open and $f:U \to \mathbb{R}^m$ a smooth map. Let $h$ be a vector in $\mathbb{R}^n$. Recall that we define the directional derivative of $f$ at $x$ along $h$ as

Notice that this gives you a linear map $df_x : \mathbb{R}^n \to \mathbb{R}^m$. If $f = (f_1, \ldots, f_m)$, then

There is also a chain rule. Let $U \subseteq \mathbb{R}^n$ and $V \subseteq \mathbb{R}^m$ open subsets with $U$ a neighborhood of $x$. If $f:U \to V$ and $g:V \to \mathbb{R}^\ell$ with $U$, then

Suppose $X \subseteq \mathbb{R}^n$ is a $k$-dimensional smooth manifold. Let $\phi:U \to X$ be a parameterization with $U\subseteq \mathbb{R}^k$ open. In particular, let $\phi(0) = x$ with $x \in X$.

Definition. The tangent space of $X$ at point $x$ is defined to be the image of $d\phi_0 : \mathbb{R}^k \to \mathbb{R}^n$ written as $T_xX$.

Exercise. Check $T_xX$ is independent of choice of parameterization.

Suppose $f:X \to Y$ is a smooth map between manifolds and $f(x) = y$.

Definition. The derivative of $f$ is $df_x : T_xX \to T_yY$.

Construction of $df_x$:

• Choose parametrizations $\phi:U \to X$ with $0 \mapsto x$. Let $\psi:V \to Y$ with $0\mapsto y$.
• Let $h = \psi^{-1} \circ f \circ \phi$.
• Define

• Above, we have
  • $d \phi_0: T_0\mathbb{R}^k \to \mathbb{T}_x X$
  • $(d\phi_0)^{-1}: T_x X\to T_0 \mathbb{R}^k$
  • $dh_0 : T_0\mathbb{R}^k \to T_0 \mathbb{R}^\ell$
  • $d\psi_0 : T_0 \mathbb{R}^\ell \to T_y Y$

Example. Let $S^1 = {x^2 + y^2 = 1}$. Suppose $p = (1, 0) \in S^1$. If we parameterize $x(y) = \sqrt{1 - y^2}$. Then the parameterization $\phi:(-1, 1) \to S^1$ is given by $y \mapsto (\sqrt{1 - y^2}, y)$. Then $T_pS^1 := d\phi_0((-1, 1))$. Since

So then $T_x S^1 = \mathbb{R}\langle 0, 1\rangle$.

Note that this is totally cumbersome. So here’s a different method: Notice that

We can actually obtain this directly from $x^2 + y^2 = 1$ via the implicit function theorem. Let $x = x(y)$. Then

In other words, the tangent vector is actually tangent to $(x, y)$.

Exercise. If $f$ is a local diffeomorphism at $x$, then $df_x:T_x X \mapsto T_y Y$ is an isomorphism.

Theorem (Inverse function theorem). If $f:X \to Y$ is a smooth map such that $df_x$ is an isomorphism for some point $x \in X$, then $f$ is a local diffeomorphism at $x$.

Definition. Let $f:X\to Y$, $f’:X’ \to Y’$. We say that $f$ and $f’$ are equivalent if there exists diffeomorphisms $\alpha:X \to X’, \beta:Y \to Y’$ such that $\beta\circ f = f’\circ \alpha$.

Exercise. Inverse function theorem is equivalent to saying that if $df_x$ is an isomorphism, then $f:X \to Y$ is locally equivalent to the identity map $\mathrm{Id}:\mathbb{R}^{k} \to \mathbb{R}^k$ with $\dim X = \dim Y = k$.

The case where $df_x$ is an isomorphism is the simplest case which leads to diffeomorphism. The case where $df_x$ is injective leads to an immersion.

Theorem (Local immersion theorem). If $f:X \to Y$ is smooth, $df_x$ is injective for $x \in X$, then there exist local coordinates around $x$ and $f$ such that

Day 3

Definition. Recall that $f : X \to Y$ smooth is an immersion provided $df_x$ is injective for every $x \in X$.

Theorem (Local immersion theorem). If $f:X \to Y$ is an immersion, then there exist local coordinates on $X$ and $Y$ so that

$\dim X = k$, $\dim Y = \ell \ge k$.