\magnification 1400 \parskip 10pt \parindent 0pt \hoffset -0.3truein \hsize 7truein \voffset -0.3truein \vsize 10truein \def\R{{\bf R}} \def\C{{\bf C}} \def\N{{\bf N}} \def\r{{\bf r}} \def\c{{\bf c}} \def\p{{\bf p}} \def\x{{\bf x}} \def\y{{\bf y}} \def\e{{\bf e}} \def\i{{\bf i}} \def\j{{\bf j}} \def\k{{\bf k}} \def\0{{\bf 0}} \def\a{{\bf a}} \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\l{\vskip 5pt \hrule \vskip -5pt} Suppose $f: N \rightarrow M$ is smooth. Then $d_pf: T_p(N) \rightarrow T_p(M)$. If $df_{p}$ is $1-1$, for all $p\in N$, then $f$ is called an \emph{\it immersion}. \centerline{I.e., $f$ is an immersion iff $f$ has rank $n$} If $df_{p}$ is onto for all $p\in N$, then $f$ is called a \emph{\it submersion}. \centerline{I.e., $f$ is an submersion iff $f$ has rank $m$} Defn. Suppose $f: M \rightarrow N$ is smooth. $p \in M$ is a {\it critical point} and $f(p)$ is a {\it critical value} if rank $df_p < n$. If $p \in M$ is not a critical point, then it is a {\it regular point}. If $q \in N$ is not a {\it critical value}, then it is a {\it regular value}. Note: $q \in N$ is a regular value iff $f^{-1}(q) = \emptyset$ or $\forall p \in f^{-1}(q)$, $df_p = n$. Defn: If $K$ is a submanifold of $M$, then $K$ has the subspace topology. \hfil \break Also, for all $p \in K$, there exists a chart $(\phi, U)$ for $M$ such that $\phi(U) = \Pi_1^m(-\ep, \ep)$ and $\phi(U\cap K) = \Pi_1^k(-\ep, \ep) \times \Pi_{k+1}^m\{0\}$ , then \hfil \break Also, $(\phi_{U \cap K }, U \cap K )$, $\phi|_{U \cap K }: U \cap K \rightarrow \Pi_1^k(-\ep, \ep)$ is a chart for $K$.\bh The collection of such charts form a pre-atlas for $K$. \vfill Thm 2.3.13: Let $q$ be a regular value of $f$:~$M\rightarrow N$. Then either $f^{-1}(q)=\emptyset $ or $f^{-1}(q)$ is an $(m-n)$-submanifold of $% M$. Defn: Suppose $f$\emph{:}~$M\rightarrow N$ is a $1-1$ immersion, and suppose $f$\emph{:}~$M\rightarrow f(M)$ is a homeomorphism, where $% f(M)\subset N$ has the relative topology. Then $f$ is an \emph{\it embedding}, and $f(M)$ is an embedded submanifold. Thm. $f: M \rightarrow N$ embedding implies $f(M)$ is a submanifold of $N$. \end \l Thm. $f: M \rightarrow N$ embedding implies $f(M)$ is a submanifold of $N$. Recall $K$ is a submanifold of $N$ if $\forall q \in K \subset N$, $\exists g^{smooth}: V^{open} \subset N \rightarrow \R^{n-m}$, $q \in V$ such that $K \cap V = g^{-1}(0)$ and rank $d_p g = n-m$ Proof. Since $f: M \rightarrow N$ embedding, $f: M \rightarrow N$ is a 1-1 immersion and $f:M \rightarrow f(M)$ is a homeomorphism where $f(M)$ is a subspace of $N$ Take $q \in f(M)$. Since $f$ is 1:1, $\exists ! p \in M$ such that $f(p) = q$. $f: M \rightarrow N$ an immersion implies $f$ has rank $m \leq n$. Thus by the rank theorem, \l Randell's Submanifolds (2.3) = Boothby's Regular submanifold (III.5): Prop: If $U^{open} \subset M^m$, then $U$ is an $m$-dimensional submanifold of $M$. Prop: If $K$ is a submanifold of $M$, then $i: K \rightarrow M$, $i(k) = k$, the inclusion map is smooth. Ex: Find a counterexample to the above if we replace the hypothesis $K$ is a submanifold of $M$ with $K \subset M$. Prop: If $f: N \rightarrow M$ is smooth and if $H$ is a submanifold of $N$, then $f: H \rightarrow M$ is smooth Ex: Find a counterexample to the above if we replace the hypothesis $H$ is a submanifold of $N$ with $H \subset N$. Prop: If $f: N \rightarrow M$ is smooth and if $K$ is a submanifold of $M$ and if $f(N) \subset K$, then $f: N \rightarrow K$ is smooth. Ex: Find a counterexample to the above if we replace the hypothesis $K$ is a submanifold of $M$ with $K \subset M$. \eject Boothy III.6 = Randell Chapter 1.3 \vskip 5pt \hrule \vskip -5pt Defn: $G$ is a {\it topological group} if \vskip -5pt \h 1.) $(G, *)$ is a group \h 2.) $G$ is a topological space. \h 3.) $*: G \times G \rightarrow G$, $*(g_1, g_2) = g_1*g_2$, and \leftline{~~~~~~~~$In: G \rightarrow G$, $In(g) = g^{-1}$ are both continuous functions.} Defn: $G$ is a {\it Lie group} if \h 1.) $G$ is a group \h 2.) $G$ is a smooth manifold. \h 3.) $*$ and $In$ are smooth functions. Ex: $Gl(n, \R) = $ set of all invertible $n \times n$ matrices is a Lie group: \h 1.) $(Gl(n, \R), matrix ~multiplication)$ is a group \h 2.) $(Gl(n, \R) $ is a smooth manifold. \h 3.) $* (Gl(n, \R) \times (Gl(n, \R) \rightarrow (Gl(n, \R)$, \h ~~~~~~~~~~~~~~~ $*(A, B) = AB$ and \h ~~~~$In: (Gl(n, \R) \rightarrow (Gl(n, \R)$ \h ~~~~~$In(A) = A^{-1}$ are smooth functions. \l Ex: $(\C -\{\0\}, \cdot )$, is a Lie group. Thm: If $G$ is a Lie group and $H$ is a submanifold, then $H$ is a Lie group. Ex: $(S^1, \cdot )$ Ex: $G_1, G_2$ lie groups implies $G_1 \times G_2$ is a lie group. Ex: $T^n = S^1 \times ... \times S^1$ is a Lie group. \vskip 5pt \hrule \vskip -5pt The following maps are diffeomorphisms: $In: G \rightarrow G$, $In(g) = g^{-1}$. For $a \in G$, $L_a: G \rightarrow G$, $L_a(g) = ag$ $R_a: G \rightarrow G$, $R_a(g) = ga$ Ex: $O(n) = \{M \in GL(n, \R) ~|~ M^tM = I \} $ is a Lie group. Ex: $Sl(n, \R) = \{M \in GL(n, \R) ~|~ det(M) = 1 \} $ is a Lie group. Defn: $F$ is a {\it homomorphism} of Lie groups if $F$ is an algebraic homomorphism of Lie groups and $F$ is smooth. Ex: $F: GL(n, \R) \rightarrow \R - \{\0\}$, $F(M) = det(M)$ is a homomorphism. \l Randell's Submanifolds (2.3) = Boothby's Regular submanifold (III.5): $K \subset N$ is a $k$-submanifold of $N$ if $\forall p \in K$, there exists, \l Suppose $f: N \rightarrow M$ is smooth and has constant rank $k$. If $q \in M$, then $f^{-1}(q)$ is a submanifold of $N$ of dimension $n - k$. Proof: Let $p \in f^{-1}(q)$. By the rank theorem, \l Ex: $F: (\R, +) \rightarrow (S^1, \cdot)$, $F(t) = e^{2\pi it}$ is a homomorphism. Ex: $F: (\R^n, +) \rightarrow (T^n, \cdot)$, $F(t_1, ..., t_n) = (e^{2\pi it_1}, ..., e^{2\pi it_n}$ is a homomorphism. Thm: If $F: G_1 \rightarrow G_2$ is a homomorphism of Lie groups, then \h 1.) rank(F) is constant. \h 2.) kernel of $F$ = $F^{-1}(e)$ is a closed submanifold \h 3.) $F^{-1}(e)$ is a Lie group. \h 4.) $dim(ker~F) = dim(G_1) - rank(F)$ Thm: If $H$ is a submanifold and an algebraic subgroup of $G$, then $H$ is closed in $G$. \vskip 5pt \hrule \vskip -5pt Defn: $G$ = group, $X =$ set. {\it $G$ acts on $X$} (on the left) if $\exists \s: G \times X \rightarrow X$ such that \h 1.) $\s(e, x) = x ~~\forall x \in X$ \h 2.) $\s(g_1, \s(g_2, x)) = \s(g_1g_2, x)$ Notation: $\s(g, x) = gx$. \hfil \break Thus 1) $ex = x$; 2) $g_1(g_2x) = (g_1g_2)(x)$. \u If $G$ is a Lie group and $X$ is a smooth manifold, then we require $\s$ to be smooth. Defn: The {\it orbit} of $x \in X$ = \hfil \break $G(x) = \{ y \in X ~|~ \exists g$ such that $y = gx\}$ Note: \hfil \break 1.) $x \in G(x)$ \hfil %%\break 2.) If $G(x) \cap G(y) \not= \emptyset$, then $G(x) = G(y)$ Thus we can use an action of $G$ to partition $X$ into disjoint subsets. Defn: If $G$ acts on $X$, then $X/G = X/\sim$ where $x \sim y$ iff $y \in G(x)$ iff $\exists g$ such that $y = gx$. If $X$ is a topological space, then $X/G = X/\sim$ is a topological space with the quotient topology. When is $X/G = X/\sim$ a manifold? Ex: $G= ({\bf Z}, +)$, $M = \R$, $\s(n, x) = n + x$. $M/G =$ Ex: $G= ({\bf Z} \times {\bf Z}, +)$, $M = \R^2$, \hfil \break $\s((n, m), (x,y)) = (n + x, m + y)$. $M/G =$ Ex: $G= ({\bf Z_2}, +)$, $M = S^n$, $\s(0, x) = x$, $\s(1, x) = -x$, . $M/G =$ \end \eject Defn: The action of $G$ on $X$ is {\it free } if $gx = x$ implies $g = e$. Thm 1.3.9: If $M$ is a smooth $n$-manifold, and $G$ is a finite Lie group acting freely on $M$, then $M/G$ is a smooth $n$-manifold. Also, $p: M \rightarrow M/G$ is smooth. Defn: $G$ is a {\it discrete group} if \vskip -5pt \h 0.) $G$ is a group. \vskip -5pt \h 1.) $G$ is countable \vskip -5pt \h 2.) $G$ has the discrete topology Note a discrete group is a Lie group. Defn: The action of $G$ on $M$ is {\it properly discontinuous} if $\forall x \in M$, $\exists U^{open}$ such that $x \in U$ and $U \cap gU = \emptyset ~~\forall g \in G$. \vskip 5pt \hrule \vskip -5pt Ex: $({\bf Z}, +)$ acting on $\R^1$ where $\s(n, x) = n + x$. Thm 1.3.2: $M$ smooth $n$-manifold, $G$ discrete group acting properly discontinuously on $M$ implies $M/G$ is a smooth $n$-manifold. Also, $p: M \rightarrow M/G$ is smooth. \end 3, 6, 7 \eject $\matrix{ & f && & \cr M & \longrightarrow & f(M) & \subset & N\cr \cup & &&& \cup \cr U & & && V \cr \downarrow \phi^{homeo}&&& & \psi^{homeo} \downarrow \cr & \psi \circ f \circ \phi = \pi_(1...m) && & \cr \Pi_1^m (-\ep, \ep) & \longrightarrow && & \Pi_1^n (-\ep, \ep) )\cr }$ \end