# Monday, September 12

:::{.remark}
Next: derivatives of maps between smooth manifolds.
For $F: U \subseteq \RR^n \to \RR^k$, recall that

- For all $p\in U$, there is a linear map $DF(p): \RR^n\to \RR^n$, a matrix satisfying $DF(p)v = \dd{}{t} \evalfrom{t=0} D(p + tv)$.
- The definition is local in the following sense: if $p\in U' \subseteq U$ is in a smaller open set, $D\ro{F}{U'}(p)$ is the same linear map as $DF(p)$.
- There is a chain rule: $D(G\circ F) (p) = DG( F(p) ) DF(p)$.
- $D\id_U (p) = \id_U$.

Extending this to map of smooth manifolds $f: M\to Q$ with $\dim M = n, \dim Q = k$, what kind of object is $df_p$ for $p\in M$?
This will be a linear map $df_p: \T_p M \to T_{f(p)} Q$, which articulates directions one can travel from $p$, which necessarily requires these tangent spaces (to be defined shortly) to be vector spaces.
Note that we don't want to define $df_p$ in terms of embeddings $M \injects \RR^N$ or charts, since we want this to be intrinsic and independent of such choices, although these notions should coincide e.g. for $T^2 \embeds \RR^3$ embedded mapping to $S^1 \embeds \RR^2$.

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:::{.remark}
Some desirable properties of tangent spaces:

- $\T_p M$ should be a vector space and $df_p$ a linear map between vector spaces.
- Locality: if $p\in U \subseteq M$ is open, then there should be a canonical isomorphism $\T_p U \iso \T_p M$.
- If $\phi: U\to \RR^n$ is a smooth chart then $d\phi_p$ should identify $\T_p M \iso \T_{\phi(p)} \RR^n$.

Idea: identify a vector $v$ with an operation on $C^\infty(M)$, the directional derivative in the direction $v$.
For $f: U\to \RR$, this is can be written as $(Df)_p(v) = \ip{ (\grad f)(p) }{v} = \qty{ \sum_{i=1}^n v_i \ro{ \dd{}{x_i} }{p} } f$.
So we will identify $\T_p U \cong \spanof_\RR\ts{ \ro{\dd{}{x_i} }{p} }^{1\leq i \leq n}$ in the vector space of linear functions $C^\infty(M) \to \RR$.
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:::{.definition title="Tangent spaces"}
Let $M\in\smooth\Mfd$ and $p\in M$.
A **derivation** at $p$ is a map $v: C^\infty(M) \to \RR$ which is $\RR\dash$linear and satisfies the Leibniz rule at $p$:
\[
v(fg)\evalfrom{p} = f(p) v(g) + v(f) g(p)
.\]
We then define $\T_p M$ to be the space of all derivations at $p$.
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:::{.remark}
Note that it is not obvious that $\T_p M$ is precisely spanned by $\ts{\dd{}{x_i}\evalfrom{p} }_{i=1}^n$, although one inclusion is clear.
This definition is chart-independent, $\T_p M$ is a vector space, and there are several properties that can immediately be read off.
Let $v\in \T_p M, f,g\in C^\infty(M)$, then

- If $f(x) = c$ is a constant function, $v(f) = 0$.
  By linearity it suffices to show this for $f(x) = 1$.
  Apply the Leibniz rule: $v(1) = v(1 \cdot 1) = 1v(1) + v(1) 1 \implies v(1) = 2v(1) \implies v(1) = 0$.
- If $f(p) = g(p) = 0$ then $v(fg) = 0$ by the Leibniz rule.

Idea: $v(f)$ only depends on first derivatives of $f$, since the constants vanish, and the 2nd and higher order terms involve $(\cdots)x_i x_j$ and thus vanish.
Note that the locality property is not obvious, since neither of $C^\infty(U)$ and $C^\infty(M)$ contain the other.
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