\magnification 1900 \parskip 10pt \parindent 0pt \hoffset -0.3truein \hsize 7truein \voffset -0.4truein \vsize 10truein \def\emph{} \def\S{\Sigma_{i=1}^n} \def\Sm{\Sigma_{i=1}^m} \def\s{{\sigma}} \def\ep{\epsilon} \def\f{\vskip 10pt} \def\h{\hskip 10pt} \def\u{\vskip -8pt} \def\bh{\hfil \break} \def\N{{\bf N}} \def\S{\Sigma_{i=1}^n} \def\0{{\bf 0}} \def\Z{{\bf Z}} \def\R{{\bf R}} \def\C{{\bf C}} \def\x{{\bf x}} \def\a{{\bf a}} \def\y{{\bf y}} \def\e{{\bf e}} \def\i{{\bf i}} \def\j{{\bf j}} \def\k{{\bf k}} \def\p{{\bf p}} \def\f{\phi} \def\s{\sigma} \def\ep{\epsilon} \def\f{\vskip 10pt} \def\u{ \vskip -5pt } \def\h{\vskip -5pt \hskip 20pt} $C^{\infty}(M) = \{g ~|~ g^{smooth}: M \rightarrow \R \}$ $D$ is a derivation iff $ D: C^\infty (p) \rightarrow \R$ and $D$ is linear and satisfies the Leibniz rule. That is $D$ is a derivation if $D(f) \in \R$,\bh $D(cf) = cD(f)$, $D(f+g) = D(f) + D(g)$, \bh $D(fg) = f(p)Dg + g(p)Df$ Defn: A {\it vector field} or {\it section of the tangent bundle} $TM$ is a smooth function \bh $s{:~}M\rightarrow TM$ so that $\pi\circ s={id}$ [i.e., $s(p) = (p, v_p)]$. Ex: If $M = \R$, let $s(p) = (p, ({d\over dx})_p)$ Sometimes we will drop the p and write $s(p) = ({d\over d x})_p$ Let $f \in C^{\infty}(\R)$. For all $p \in \R$, $s(p)(f) = ({df \over dx})_p = {df \over dx}(p) $ Define $s_f: \R \rightarrow \R$, $s_f(p) = {df \over dx}(p)$. ~~~~~ I.e., $s_f = {df \over dx}$ Note $s_f$ is smooth. Lemma 3.4.1: For any vector field $s$ and smooth functions $f$ and $g$ on $M$, we have \centerline{$s_{fg}(p)=f(p)\cdot s_{g}(p)+s_{f}(p)\cdot g(p)$} Proof: ${d(fg) \over dx}(p) = f(p){dg \over d x}(p) + {df \over dx}(p)g(p)$ We can think of a vector field as a function \bh $S: C^{\infty}(M) \rightarrow C^{\infty}(M)$, $S(f) = s_f$ Ex: $S: C^{\infty}(\R) \rightarrow C^{\infty}(\R)$, $S(f) = {df \over dx}$.~~~~~~ I.e., $S = {d \over dx}$ Ex: If $M = \R$, then $s(p) = a(p)({d\over dx})_p$ where $a: \R \rightarrow \R$ is a smooth function. Let $f \in C^{\infty}(\R)$. \bh For all $p \in \R$, $s(p)(f) = a(p)({df \over dx})_p = a(p){df \over dx}(p) $ Define $s_f: \R \rightarrow \R$, $s_f(p) = a(p){df \over dx}(p)$. ~~~~~ I.e., $s_f = a {df \over dx}$ Note $s_f$ is smooth. Lemma 3.4.1: For any vector field $s$ and smooth functions $f$ and $g$ on $M$, we have \centerline{$s_{fg}(p)=f(p)\cdot s_{g}(p)+s_{f}(p)\cdot g(p)$} Proof: $a(p){d(fg) \over dx}(p) = a(p)f(p){dg \over d x}(p) + a(p){df \over dx}(p)g(p)$ We can think of a vector field as a function \bh $S: C^{\infty}(M) \rightarrow C^{\infty}(M)$, $S(f) = s_f$ Ex: $S: C^{\infty}(\R) \rightarrow C^{\infty}(\R)$, $S(f) = a{df \over dx}$ I.e., $S = a{d \over dx}$ \eject In the above we used the charts $\phi_p: \R \rightarrow \R, \phi_p(x) = x - p$. Thus ${d(g (\phi_p^{-1}(x)) ) \over dx}|_{x = 0} $ $ = {d(g (x + p) ) \over dx}|_{x = 0} = {dg \over dx}(p) $ Note $\phi_0(x) = \phi_p (x + p)$. Thus $ {d(\phi_p (\phi_0^{-1}(x) ) \over dx}|_{x = 0} = {d(\phi_p( \phi_p^{-1}(x+p) )) \over dx}|_{x = 0} = {d(x+p) \over dx}|_{x = 0} = 1$ \vfill If we use the chart $\psi_q: \R \rightarrow \R, \psi_q(x) = q - x$. Then ${d(g (\psi_p^{-1}(x)) ) \over dx}|_{x = 0} $ $ = {d(g (p-x) ) \over dx}|_{x = 0} = {-dg \over dx}(p) $ Note ${d(\psi_q (x + p) ) \over dx}|_{x = 0} = {d\psi_q \over dx}|_p= {d(q-x) \over dx}|_p = -1 $ Example of a non-smooth vector field on $\R$: If $p \geq 0$, let $s(p) = (p, ({d\over dx})_p)$ \bh [i.e., the basis element of $T_p(\R)$ from $\phi_p$] If $p < 0$, let $s(p) = (p, (-{d\over dx})_p)$ \bh [i.e., the basis element of $T_p(\R)$ from $\psi_p$] \eject Ex: If $M = \R^2$, then $s(\p) = a(\p) ({\partial \over \partial x})_\p + b(\p) ({\partial \over \partial y})_\p$ where $a, b: \R^2 \rightarrow \R$ are smooth functions. Ex: Let $\{( {\partial \over \partial x_1})_p, ..., ( {\partial \over \partial x_m})_p\}$ be a basis for $T_p(M)$. Let $s: M \rightarrow TM$, $s(p) = (p, \Sm a_i(p) ({\partial \over \partial x_i})_p)$ \end Ex: $M = S^1$. Let $\p \in S^1$, take a chart $(U, \phi)$ such that $\p \in U$. Let $\{ ( {\partial \over \partial x})_\p \}$ be a basis for $T_p(S^1)$. Let $s: S^1 \rightarrow TS^1$, Let $U = \{\theta ~|~ \theta \in (0, 2\pi) \}$ $s(p) = (p, ( {\partial \over \partial x})_\p )$. \eject Lemma 3.4.2: Let $S:C^{\infty}(M)\rightarrow C^{\infty}(M)$ be linear, and suppose $S(fg)(p)=f(p)\cdot S(g)(p)+S(f)(p)\cdot g(p)$. Then $S$ is a vector field. Proof: Define $S(p): C^{\infty}(M) \rightarrow \R$ to be the function which sends $f \in C^{\infty}(M)$ to $S(f)(p)$, i.e, the function $S(f)$ evaluated at $p$. Note that the hypothesis implies that $S(p)$ is linear and satisfies the Leibniz rule and hence is a derivation. Defn: If $A, B$ are vector fields, let $AB = A \circ B$ Defn: The {\it Lie Bracket} of vector fields $A$ and $B$ is $[A, B] = AB - BA: C^{\infty}(M)\rightarrow C^{\infty}(M)$. Thm: The Lie bracket of vector fields is a vector field. Randell 3.3 Defn: A {\it flow} on $M$ is a smooth action of the Lie group ${\R}^{1}$ on $M$, $\sigma$:~${\R}% ^{1}\times M\rightarrow M$. A flow is also called a {\it dynamical system}. $ \sigma(t, m) = \sigma_t(m) $ $\sigma_0(m) = m $, \hfil $\sigma_t \circ \sigma_s (m) = \sigma_{t+s} (m) = \sigma_s \circ \sigma_t (m)$ $ $\sigma_{-t} = \sigma_t^{-1}$ $\sigma_t: M \rightarrow M$ is a diffeomorphism. Ex: $\sigma: \R \times \R^2 \rightarrow \R^2$, $\sigma_t(x, y) = (x, y) + t(1, 2)$ \vfill Defn: The {\it orbit} of $x \in M$ = \hfil \break $\R(x) = \{ y \in M ~|~ \exists t \in \R$ such that $y = tx\}$ \vskip 10pt \line{A {\it flow line} is the smooth path $\alpha_p: \R \rightarrow M$, $\alpha_p(t) = \sigma(t, p)$.} Prop: each $q \in M$ lies on a unique flow line. \eject ``differentiating along the flow'': Given a flow $\sigma$ on $M$, define $s_{\sigma}$:~$M\rightarrow TM$ by $s_{\sigma}(p)=\left( p,d\alpha_{p}/dt|~_{t=0}\right) $ \vfill Proposition 3.3.3: $s_{\sigma}$ is a section of $TM$. \end $G(p) = \{[g] ~|~ g^{smooth}: U \rightarrow \R$, for some $U^{open}$ such that $p \in U \subset M \}$ $C^{\infty}(M) = \{g ~|~ g^{smooth}: M \rightarrow \R \}$ $Z_a = \{g ~|~ g^{smooth}: M \rightarrow \R, Dg_a = 0 \}$ The cotangent space $T^*_a = C^{\infty}(M)/ Z_a$. $T^*_a$ is an $m-$dimensional vector space: $(df)_a