\classheader{2013-01-09}

\subsection*{Review of Previous Class}

For any topological space $Y$, we define $C^0(Y)$ to be the set of continuous real-valued functions on $Y$. For any open subsets $U\subseteq V$ of $Y$, there is a restriction map $\Res(U,V):C^0(V)\to C^0(U)$. Recall that a presheaf $R$ of subsets of $C_Y^0$ consists of, for all open $U\subseteq Y$, a choice of $R(U)\subseteq C^0(U)$ such that whenever $U\subseteq V$, we have $\Res(U,V)(f)\in R(U)$ for any $f\in R(V)$.

Given a presheaf $R$ of subsets of $C_Y^0$, we say that it is a sheaf of subsets when the following condition is satisfied: given any open $V\subseteq Y$ and $f\in C^0(V)$, if for all $p\in V$ there is a neighborhood $V_p\subseteq V$ of $p$ such that $\Res(V_p,V)(f)\in R(V_p)$, then $f\in R(V)$.

We say that a sheaf $R$ of subsets of $C_Y^0$ is a sheaf of $\R$-subalgebras if $R(U)\subseteq C^0(U)$ is an $\R$-subalgebra for all open $U$.

For us, a ringed space will consist of a pair $(Y,R)$ of a topological space $Y$ and a sheaf $R$ of $\R$-subalgebras of $C_Y^0$. A morphism $(Y,R)\to(Y',R')$ of ringed spaces is a continuous map $f:Y\to Y'$ such that for all open $V'\subseteq Y'$ and $g\in R'(V')$, the pullback $f^*g$, i.e. the composition
\[f^{-1}(V')\xrightarrow{\;\;f|_{f^{-1}(V')}\;\;} V'\xrightarrow{\;\;\;g\;\;\;}\R,\]
is an element of $R(f^{-1}(V))$. An isomorphism $(Y,R)\to (Y',R')$ is a homeomorphism $f:Y\to Y'$ such that $f^{-1}:(Y',R')\to (Y,R)$ is also a morphism of ringed spaces.

Fix some $k\in\{0,1,2,\ldots,\infty,\omega\}$. A standard example of a ringed space is $(\Omega,C_\Omega^k)$ where $\Omega$ is an open subset of $\R^n$, and $C_\Omega^k$ is the sheaf defined by $C_\Omega^k(V)=C^k(V)$ for any open $V\subseteq\Omega$.

\begin{definition}
A $C^k$ manifold is a ringed space $(M,R)$ such that $M$ is covered by open sets $U_i$ where each $(U_i,R_{U_i})$ is isomorphic as a ringed space to some $(\Omega_i,C_{\Omega_i}^k)$, where $\Omega_i$ is an open subset of $\R^{n_i}$. We then refer to $R$ as $C_M^k$.

Given $C^k$ manifolds $M$ and $N$, a map $f:M\to N$ is a $C^k$ map when it is a morphism of ringed spaces $(M,C_M^k)\to (N,C_N^k)$.
\end{definition}

\subsection*{Tangent Vectors}

A natural way we think about tangent vectors arising is as the tangent vector to a curve. We can define an equivalence relation $\sim$ on curves through a point $p\in M$, according to when they have the same tangent vector at $p$.

Given a $C^k$ manifold $M$ and a point $p\in M$, we consider two $C^k$ maps $\gamma:(-a,a)\to M$ and $\delta:(-b,b)\to M$ where $\gamma(0)=\delta(0)=p$. When do we want to say that $\gamma\sim\delta$?

First, let's define germs. If $U,V\subseteq M$ are open subsets containing $p$ and $f\in C_M^k(U)$ and $g\in C_M^k(V)$, we say that $f\sim g$ if there is an open $W\subseteq U\cap V$ containing $p$ such that $f|_W=g|_W$. The collection
\[C_{M,p}^k:=\coprod_{\text{open }U\ni p}C_M^k(U)\bigg\slash\sim\]
is the set of germs of functions on $M$ at $p$. In fact, it is an $\R$-algebra.

Given a germ $f$ at $p$, say defined on $U$ (so that $f:U\to\R$ is a $C^k$ function), there is some $0<a'<a$ such that the image of $\gamma|_{(-a',a')}$ is contained in $U$, so we can consider the composition
\[(-a',a')\xrightarrow{\;\;\gamma|_{(-a',a')}\;\;} U\xrightarrow{\;\;f\;\;}\R.\]
The composition is defined on a (smaller) neighborhood of 0, so we can still consider $(f\circ \gamma)'(0)$.

The above procedure gives a linear functional $D_\gamma:C_{M,p}^k\to\R$, assigning to a germ $f$ the value of $(f\circ \gamma)'(0)$. We say that $\gamma\sim\delta$ precisely when $D_\gamma=D_\delta$. A tangent vector of $M$ at $p$ is then an equivalence class of curves under this relation.

Assume that $M$ is an open set $\Omega\subseteq\R^n$. Then the chain rule tells us that $(f\circ \gamma)'(0)=f'(p)\gamma'(0)$, where $f'(p):\R^n\to\R$ is a linear functional. Then in this case, $\gamma_1\sim\gamma_2$ $\iff$ $\gamma_1'(0)=\gamma_2'(0)$, so this matches our intuition about how tangent vectors should behave. We see that the space of tangent vectors of $\Omega$ at $p$ can be canonically identified with $\R^n$.

Note that there is no natural notion of adding curves, but when we think about the corresponding linear functionals $D_\gamma$, the vector space structure becomes clear. According to our definition, $D_\gamma$ is an element of the dual space of $C_{M,p}^k$. Thus, by definition, the tangent space of $M$ at $p$ can be considered as a subset of the dual space, i.e. $T_pM\hookrightarrow(C_{M,p}^k)^*$.

\begin{theorem}
$T_pM$ is a linear subspace of $(C_{M,p}^k)^*$.
\end{theorem}
\begin{proof}
It suffices to check in the case that $M=$ an open subset $\Omega\subseteq\R^n$. In this case, we see that
\[T_pM=\{\textstyle\sum_{i=1}^na_i\partial_i(p)\},\]
where for any germ $f\in C_{M,p}^k$, the functional $\partial_i|_p$ sends $f\mapsto \partial_i f|_p:=\frac{\partial f}{\partial x_i}|_p$. 
\end{proof}

Note that for any curve $\gamma$, the corresponding functional $D=D_\gamma$ satisfies the Leibniz condition:
\[D(fg)=f(p)Dg+g(p)Df.\]
Thus, it is reasonable to include this condition in our definition of tangent vector.
\begin{proposition}
For any $p\in\R^n$, the only linear functionals on $C_{\R^n,p}^\infty$ that satisfy the Leibniz condition are those of the form $f\mapsto \sum_{i=1}^na_i\partial_i f|_p$.
\end{proposition}

{\color{myred}{\rule{\textwidth}{0.02in}}\vspace{-0.1in}
\begin{center}\bf
$\blacktriangledown$\quad What is a counterexample?\quad $\blacktriangledown$
\end{center}}

You might find it interesting to show that the proposition is false even for $C^1_{\R}$.

\vspace{-0.1in}{\color{myred}{\rule{\textwidth}{0.02in}}}

\begin{proof}[Proof of Proposition]
WLOG, let's say that $p=0$. First, we make two observations:
\begin{enumerate}
\item If we put $f=1$ in the Leibniz condition, we get that $D(g)=D(g)+g(p)Df$ for all $g$, so we must have $D(1)=0$, and hence by linearity, $D(\lambda)=0$ for any $\lambda\in\R$.
\item  $\partial_i(x_j)=\delta_{ij}$.
\end{enumerate}
Let $D$ be a functional that satisfies the Leibniz condition, and put $D(x_i)=a_i$. Let
\[E=D-\sum_{i=1}^na_i\partial_i.\]
We see that $E(x_i)=0$ for all $i$, and we want to show that this forces $E=0$, i.e. $E(f)=0$ for all $f\in C_{\Omega,p}^\infty$. Equivalently, we can show that $E(f-f(p))=0$, since $E(f(p))=0$.

\begin{lemma}
If $f$ is a $C^\infty$ function defined in a neighborhood of 0 in $\R^n$ and $f(0)=0$, then there exist $C^\infty$ functions $g_1,\ldots,g_n$ (defined in a possibly smaller neighborhood of 0) such that $f=\sum_{i=1}^nx_ig_i$.
\end{lemma}

Intuitively, this lemma makes sense, because if $f$ vanishes along the hyperplane $x_1=0$, this suggests that $x_1$ ``divides'' $f$, i.e. that $f(x_1,x_2)=x_1g(x_1,x_2)$ for some function $g$; and if $f$ vanishes on the space where $x_1=x_2=0$, this suggests that $f$ is a sum of a multiple of $x_1$ and a multiple of $x_2$. Proving this lemma will be a part of the first homework assignment.

Using the lemma, we can conclude the argument. Because $E$ satisfies the Leibniz condition, we have
\[E(f)=\sum_{i=1}^n\underbrace{x_i(0)}_{=\,0}E(g_i)+g_i(0)\underbrace{E(x_i)}_{=\,0}=0.\qedhere\]
\end{proof}

We define the contangent space $T_p^*M$ of $M$ at $p$ to be the dual space $(T_pM)^*$. If $v_1,\ldots,v_n$ are a basis for $T_pM\subset(C_{M,p}^k)^*$, then letting $D=\bigcap_{i=1}^n\ker(v_i)$, we have $T_p^*M=C_{M,p}^k/D$.

Our proposition implies that when $k=\infty$, we have $D=\R+\frak{m}^2$ where $\frak{m}=\{f\in C^\infty_{M,p}\mid f(p)=0\}$.


