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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 591 Lecture 10}
\date{9/23/20} % Can also use \today

\begin{document}
\maketitle
\setlength\RaggedRightParindent{\parindent}
\RaggedRight

\subsection*{Review: Partial Derivatives}

\par\noindent
Given $p\in{}U\subseteq{}M$ and $\phi:U\to\R^{n}$, a coordinate chart, we can write $\phi=(x^{1},\ldots,x^{n})$ where each $x^{i}:U\to\R$. Then we defined
\[
	\pd{f}{x^{i}}(p)=\pd{f_{\phi}}{r^{i}}(\phi(p))
\]
where $f_{\phi}=f\of\phi\inv$, and $r^{i}$ is simply the coordinate in $\R^{n}$.\n

\ex{
	Take $M\subseteq\R^{3}$ defined as the graph of $G:A\to\R$ where $A$ is an open subset of $\R^{2}$ and $G$ is $C^{\infty}$. Then we can write $M=\set{(r^{1},r^{2},G(r^{1},r^{2}))\mid{}(r^{1},r^{2})\in{}A}$. There is one chart: $U=M$, $\phi$ is projection onto the $(r^{1},r^{2})$ plane. $\phi\inv(r^{1},r^{2})=(r^{1},r^{2},G(r^{1},r^{2}))$.\nn
	Let $f:M\to\R$ be the restriction of some $\tilde{f}:\R^{3}\to\R$ $C^{\infty}$. Then
	\[
		f_{\phi}(^{1},r^{2})=(f\of\phi\inv)(r^{1},r^{2})=\tilde{f}(\underbrace{r^{1},r^{2},G(r^{1},r^{2})}_{p\in{}M})
	\]
	Compute:
	\[
		\pd{f}{x^{1}}(p)=\pd{\tilde{f}}{r^{1}}(p)+\pd{G}{r^{1}}(r^{1},r^{2})\pd{\tilde{f}}{r^{3}}(p)
	\]
	Observe: $\pd{f}{x^{1}}(p)=\del\tilde{f}\cdot\dot\gamma$, for $\gamma(t)=(r^{1}+t,r^{2},G(r^{1}+t,r^{2}))\in{}M$, so $\restr{\dot\gamma(t)}{t=0}=(1,0,\pd{g}{r^{1}}(r^{1},r^{2}))$.\n
}

\par\noindent
Now, back to the theorem from last time:\n

\thm{
	If $p\in{}U\osubseteq{}M$, $\phi:U\to\R^{n}$ is a chart, and $\phi=(x^{1},\ldots,x^{n})$, then $\forall{}D\in{}T_{p}M$, one has
	\[
		D=\sum_{j=1}^{n}D([x^{i}])\restr{\pd{}{x^{i}}}{p},\;\;[x^{i}]\in{}C^{\infty}_{p}(M)
	\]
	Proof: It's based on the following observations:
	\begin{itemize}[leftmargin=3.5\parindent]
		\item The set of derivations, $T_{p}M$, is an $\R$-vector space.
		\item For any constant function $k$, $\forall{}D\in{}T_{p}M$, $D[k]=0$.\n
		Proof: $D([1])=D([1^{2}])=2D([1])$ by the product rule, so $D([1])=0$. Then linearity implies $D[k]=0$.
		\item $\forall{}D\in{}T_{p}M$, if $[f](p)=[g](p)=0$, then $D([f][g])=0$.
	\end{itemize}\up\n
	We'll start with what we had last time.
	\[
		f_{\phi}(r)=f(p)+\sum_{i=1}^{n}(r^{i}-r_{0}^{i})\pd{f_{\phi}}{r^{i}}(r_{0})+\frac{1}{2}\sum_{i,j=1}^{n}(r^{i}-r_{0}^{i})(r^{j}-r_{0}^{j})\cdot{}g_{ij}(r)
	\]
	Composing with $\phi$ yields
	\[
		f(r)=f(p)+\sum_{i=1}^{n}(x^{i}-x_{0}^{i})\pd{f}{x^{i}}(p)+\frac{1}{2}\sum_{i,j=1}^{n}(x^{i}-x_{0}^{i})(x^{j}-x_{0}^{j})\cdot{}g_{ij}(x)
	\]
	Apply $D$. Well, $f(p)$ is constant, so it vanishes. And the last term is second order, so based on the above observation, it vanishes as well. We're left with
	\[
		D([f])=\sum_{i=1}^{n}D([x^{i}-x_{0}^{i}])\pd{f}x^{i}(p)=\sum_{i=1}^{n}D([x^{i}])\pd{f}{x^{i}}(p)
	\]
	Thus,
	\[
		D=\sum_{i=1}^{n}D([x^{i}])\restr{\pd{}{x^{i}}}{p}
	\]
	\proven
}

\cor{
	It's easy to check that $\displaystyle\pd{x^{i}}{x^{j}}=\delta_{ij}$, so $\displaystyle\set{\pdat{}{x^{i}}{p}}_{i=1}^{k}$ is a basis of $T_{p}M$ over $\R$.\n
}

\par\noindent
In summary,
\begin{itemize}
	\item We are defining tangent vectors by $T_{p}M$, which is the set of derivations at $p$. We'll be changing our notation: $u,v,w,\ldots\in{}T_{p}M$.
	\item Representation of vectors in coordinates:
	\[
		v=\sum_{i=1}^{n}v_{i}\pdat{}{x^{i}}{p}\qquad(v_{1},\ldots,v_{n})\in\R^{n}
	\]
	\item Also, $v=\dot\gamma$ for some $\gamma:(t_{0}-\varepsilon,t_{0}+\varepsilon)\to{}M$ $C^{\infty}$, with $\gamma(t_{0})=p$.
	\item Note for later: $p\ne{}q\Rightarrow{}T_{p}M\cap{}T_{q}M=\emptyset$.
\end{itemize}

\lemma{
	$\displaystyle\dot\gamma(t_{0})=\sum_{i=1}^{n}\pd{x^{i}}{t}(t_{0})\pdat{}{x}{p}$ if $\gamma(t)=(x^{1}(t),\ldots,x^{n}(t))\in\R^{n}$.\n
}

\subsection*{Differentials of Functions}

\defn{
	Let $p\in{}U\osubseteq{}M$, $f:U\to\R$ $C^{\infty}$. Then we define
	\[
		\map{df_{p}:T_{p}M}{\R}{v}{v[f]}
	\]
	Notation: we say $T_{p}U\eqdef{}T_{p}M$.\n
}

\par\noindent
Note: $df_{p}\in(T_{p}M)^{*}$, the dual of the tangent space.\n

\defn{
	$T_{p}^{*}M=(T_{p}M)^{*}$ is called the \u{cotangent space} of $M$ at $p$. $df_{p}\in{}T_{p}^{*}M$.\n
}

\par\noindent
Note: $df_{p}(v)=v[f]$.\n

\par\noindent
In coordinates, we saw that if $\phi=(x^{1},\ldots,x^{n})$ and $\phi(p)=(x_{0}^{1},\ldots,x_{0}^{n})\in\R^{n}$, then
\[
	f(x)=f(p)+\sum_{i=1}^{n}(x^{i}-x_{0}^{i})+O(2)
\]
(With $O(2)$ denoting something that vanishes in the second order at $p$.) Then
\[
	v[f]=\sum_{i=1}^{n}\pd{f}{x^{i}}v([x^{i}])
\]
By definition, $v[x^{i}]=dx^{i}(v)$. We conclude that
\[
	df_{p}=\sum_{i=1}^{n}\pd{f}{x^{i}}(p)dx^{i}_{p}
\]
This is just like it was in Calc III!\n

\par\noindent
Note that, in some ways, it makes more sense to first define $T_{p}^{*}M$, and then obtain $T_{p}M$ as its dual.\n

\defn{
	$I_{p}=\set{[f]\in{}C^{\infty}_{p}(M)\mid{}[f](p)=0}$, an ideal in the ring of germs.
}

\[
	I_{p}^{2}=\set{\sum_{i,j}[f_{i}][g_{j}]:[f_{i}],[g_{j}]\in{}I_{p}}
\]
is the set of ``$O(2)$ germs''. We claim that $T_{p}^{*}M\cong{}I_{p}/I_{p}^{2}$. Then $df$ is the class of $[f-f(p)]\in{}I_{p}/I_{p}^{2}$.

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