[LeeManifolds](Projects/Advanced%20Qual%20Projects/Geometry%20and%20Topology/Lee%20Manifolds%20Notes%20Project/LeeManifolds.md) Tags: #manifolds #reading_notes # Chapter 1: Point-Set Properties of Topological Manifolds Pages 1- 29. ## Notes Part 1 Definition (Topological Manifold) : A topological space $M$ that satisfies 1. $M$ is Hausdorff, i.e. points can be separated by open sets 2. $M$ is second-countable, i.e. has a countable basis 3. $M$ is locally Euclidean, i.e. every point has a neighborhood homeomorphic to an open subset $\hat U$ of $\RR^n$ for some fixed $n$. The last property says $p\in M \implies \exists U$ with $p\in U \subseteq M$, $\hat U\subseteq \RR^n$, and a homeomorphism $\phi: U \to \hat U$. > Note that second countability is primarily needed for existence of partitions of unity. Exercise : Show that the in the last condition, $\hat U$ can equivalently be required to be an open ball or $\RR^n$ itself. Theorem (Topological Invariance of Dimension) : Two nonempty topological manifolds of different dimensions can not be homeomorphic. Exercise : Show that in a Hausdorff space, finite subsets are closed and limits of convergent sequences are unique. Exercise : Show that subspaces and finite products of Hausdorff (resp. second countable) spaces are again Hausdorff (resp. second countable).\label{ex:subspaces_and_products_of_hausdorff} Thus any open subset of a topological manifold with the subspace topology is again a topological manifold. Exercise : Give an example of a connected, locally Euclidean Hausdorff space that is not second countable. Definition (Charts) : A chart on $M$ is a pair $(U, \phi)$ where $U\subseteq M$ is open and $\phi: U \to \hat U$ is a homeormohpsim from $U$ to $\hat U = \phi(U) \subseteq \RR^n$. If $p\in M$ and $\phi(p) = 0 \in \bar U$, then the chart is said to be *centered* at $p$. Note that any chart about $p$ can be modified to a chart $(\phi_1, \hat U_1)$ that is centered at $p$ by defining $\phi_1(x) = x - \phi(v)$. ![](Projects/Advanced%20Qual%20Projects/Geometry%20and%20Topology/Lee%20Manifolds%20Notes%20Project/sections/figures/image_2020-06-15-00-22-05.png) $U$ is the *coordinate domain* and $\phi$ is the *coordinate map*. Note that we can write $\phi$ in components as $\phi(p) = \thevector{x^1(p), \cdots, x^n(p)}$ where each $x^i$ is a map $x^i: U \to \RR$. The component functions $x^i$ are the *local coordinates* on $U$. Shorthand notation: $\thevector{x^i} \definedas \thevector{x^1, \cdots, x^n}$. Example (Graphs of Continuous Functions) : Define \begin{align*} \Gamma(f) = \theset{(x, y) \in \RR^{n} \cross \RR^{k} \suchthat x\in U,~ y = f(x) \in \hat U } .\end{align*} This is a topological manifold since we can take $\phi: \Gamma(f) \to U$ by restricting $\pi_1: \RR^{n}\cross \RR^k \to \RR^n$ to the subspace $\Gamma(f)$. Projections are continuous, restrictions of continuous functions are continuous.\todo{Thus graphs of continuous functions $f: \RR^n \rightarrow \RR^k$ are locally Euclidean?} This is a homeomorphism because the map $g: x \mapsto (x, f(x))$ is continuous and $g\circ \pi_1 = \id_{\RR^n}$ is continuous with $\pi_1 \circ g = \id_{\Gamma(f)}$. Note that $U \cong \Gamma(f)$, and thus $(U, \phi) = (\Gamma(f), \phi)$ is a single *global* coordinate chart, called the *graph coordinates* of $f$. Note that this works in greater generality:: \todo{Coordinates as numbers vs functions?} "The same observation applies to any subset of $\RR^{n+k}$ by setting *any* $k$ of the coordinates equal to some continuous function of the other $n$." Example (Spheres) : $S^n$ is a subspace of $\RR^{n+1}$ and is thus Hausdorff and second-countable by exercise \ref{ex:subspaces_and_products_of_hausdorff}.\label{ex:sphere_is_a_manifold} ![](figures/image_2020-06-15-21-28-56.png) To see that it's locally Euclidean, take \begin{align*} U_i^+ &\definedas \theset{\thevector{x^1, \cdots, x^n} \in \RR^{n+1} \suchthat x^i > 0} \qtext{for} 1 \leq i \leq n+1 \\ U_i^- &\definedas \theset{\thevector{x^1, \cdots, x^n} \in \RR^{n+1} \suchthat x^i < 0} \qtext{for} 1 \leq i \leq n+1 .\end{align*} Define \begin{align*} f: \RR^{n} &\to \RR^{\geq 0} \\ \vector x &\mapsto \sqrt{1 - \norm{\vector x}^2} .\end{align*} Note that we immediately need to restrict the domain to $\DD^n \subset \RR^n$, where $\norm{x}^2 \leq 1\implies 1 - \norm{x}^2 \geq 0$, to have a well-defined real function $f: \DD^n \to \RR^{\geq 0}$. Then (claim) \begin{align*} U_i^+ \intersect S^n \qtext{is the graph of} & x^i = f(x^1, \cdots, \hat{x^i}, \cdots, x^{n+1}) \\ U_i^- \intersect S^n \qtext{is the graph of} &x^i = -f(x^1, \cdots, \hat{x^i}, \cdots, x^{n+1}) .\end{align*} This is because \begin{align*} \Gamma(x^i) &\definedas \theset{(\vector x, f(\vector x)) \subseteq \RR^n \cross \RR} \\ &= \theset{ \thevector{x_1, \cdots, \hat{x^i}, \cdots, x^{n+1}}, f\qty{\thevector{x_1, \cdots, \hat{x^i}, \cdots, x^{n+1} }}\subseteq \RR^n \cross \RR } \\ &= \theset{ \thevector{x_1, \cdots, \hat{x^i}, \cdots, x^{n+1} }, \qty{1 - \sum_{\substack{j=1 \\ j\neq i}}^{n+1} (x^j)^2}^{1\over 2} \subseteq \RR^n \cross \RR } \\ \end{align*} and any vector in this set has norm satisfying \begin{align*} \norm{(\vector x, y)}^2 = \sum_{\substack{j=1 \\ j\neq i}}^{n+1} (x^j)^2 + \qty{1 - \sum_{\substack{j=1 \\ j\neq i}}^{n+1} (x^j)^2} = 1 \end{align*} and is thus in $S^n$. To see that any such point also has positive $i$ coordinate and is thus in $U_i^+$, we can rearrange (?) coordinates to put the value of $f$ in the $i$th coordinate to obtain \todo[fancyline]{Seems like $f$ is always the *last* coordinate in the graph} \begin{align*} \Gamma(x_i) = \theset{\thevector{x^1, \cdots, f(x^1, \cdots, \hat{x^i}, \cdots, x^n), \cdots, x^n }} \end{align*} and note that the square root only takes on positive values. Thus each $U_i^{\pm} \intersect S^n$ is the graph of a continuous function and thus locally Euclidean, and we can define chart maps \begin{align*} \phi_i^{\pm}: U_i^{\pm} \intersect S^n &\to \DD^n \\ \thevector{x^1, \cdots, x^n} &\mapsto [x^1, \cdots, \hat{x^i}, \cdots, x^{n+1}] \end{align*} yield $2(n+1)$ charts that are graph coordinates for $S^n$. Example (Projective Space) : Define $\RP^n$ as the space of 1-dimensional subspaces of $\RR^{n+1}$ with the quotient topology determined by the map \hspace{10em} \todo[fancyline]{How is this map a quotient map?} \begin{align*} \pi: \RR^{n+1}\smz &\to \RP^n\\ \vector x &\mapsto \spanof_\RR\theset{\vector x} .\end{align*} Notation: for $\vector x \in \RR^{n+1}\smz$ write $[\vector x] \definedas \pi(\vector x)$, the line spanned by $\vector x$. Define charts: \begin{align*} \tilde U_i \definedas \theset{\vector x \in \RR^{n+1}\smz \suchthat x^i \neq 0}, \quad U_i = \pi(\tilde U_i) \subseteq \RP^n \\ .\end{align*} and chart maps \begin{align*} \tilde \phi_i: \tilde U_i &\to \RR^n \\ \thevector{x^1, \cdots, x^{n+1}} &\mapsto \thevector{{x^1 \over x^i}, \cdots \hat{x^i}, \cdots {x^{n+1} \over x^i} } .\end{align*} Then (claim) this descends to a continuous map $\phi_i: U_i \to \RR^n$ by the universal property of the quotient: \begin{center} \begin{tikzcd} \tilde U_i \ar[d, "\pi_U"'] \ar[rd, "\tilde \phi_i"] & \\ U_i \ar[r, "\phi_i", dotted] & \RR^n \end{tikzcd} \end{center} - The restriction $\pi_U: \tilde U_i \to U_i$ of $\pi$ is still a quotient map because $\tilde U_i = \pi_U\inv(U_i)$ where $U_i\subseteq \RP^n$ is open in the quotient topology and thus $\tilde U_i$ is saturated. Thus $\pi_U$ sends saturated sets to open sets and is thus a quotient map. - $\tilde \phi_i$ is constant on preimages under $\pi_U$: fix $y\in U_i$, then $\pi_U\inv(\theset{y}) = \theset{\lambda \vector y \suchthat \lambda \in \RR\smz}$, i.e. the point $y \in \RP^n$ pulls back to every nonzero point on the line spanned by $\vector y\in \RR^n$. But \begin{align*} \tilde \phi_i(\lambda \vector y) &= \phi_i \qty{ \thevector{\lambda y^1, \cdots, \lambda y^i, \cdots, \lambda y^n} } \\ &= \thevector{{\lambda y^1 \over \lambda y^i}, \cdots, \hat{\lambda y^i}, \cdots, {\lambda y^{n+1} \over \lambda y^i}} \\ &= \thevector{{y^1 \over y^i}, \cdots, \hat{y^i}, \cdots, {y^{n+1} \over y^i}} \\ &= \tilde \phi_i(\vector y) .\end{align*} So this yields a continuous map \begin{align*} \phi_i: U_i \to \RR^n .\end{align*} We can now verify that $\phi$ is a homeomorphism since it has a continuous inverse given by \begin{align*} \phi_i\inv: \RR^n &\to U_i \subseteq \RP^n \\ \vector u \definedas \thevector{u^1, \cdots, u^n } &\mapsto \thevector{u^1, \cdots, u^{i-1}, {\color{red}1}, u^{i+1}, \cdots, u^n} .\end{align*} It remains to check: \todo{Exercise} 1. The $n+1$ sets $U_1, \cdots, U_{n+1}$ cover $\RP^n$. 2. $\RP^n$ is Hausdorff 3. $\RP^n$ is second-countable. Exercise (1.6) : Show that $\RP^n$ is Hausdorff and second countable. Exercise (1.7) : Show that $\RP^n$ is compact. (Hint: show that $\pi$ restricted to $S^n$ is surjective.) Definition (Topological Embedding) : A continuous map $f:X\to Y$ is a *topological embedding* iff it is injective and $\tilde f:X\to f(X)$ is a homeomorphism. Example (Product Manifolds) : Let $M \definedas M_1 \times \cdots \times M_k$ be a product of manifolds of dimensions $n_1, \cdots, n_k$ respectively. A product of Hausdorff/second-countable spaces is still Hausdorff/second-countable, so just need to check that it's locally Euclidean. - Let $\vector p \in \prod_{i=1}^N M_i$, so $p_i \in M_i$ - Choose a chart $(U_i, \phi_i)$ with $p_i\in U_i$ and assymble a product map: \begin{align*} \Phi \definedas \prod \phi_i: \prod U_i \to \prod R^{n_i} \cong \RR^{\Sigma n_i} \definedas \RR^N .\end{align*} - Claim: $\Phi$ is a homeomorphism onto its image in $R^N$. - Each $\phi_i$ is a homeomorphism onto $\phi_i(U_i)$ (by the definition of a chart on $M_i$) - It suffices to show that that $\Phi\inv$ exists and is continuous, where \begin{align*} \Phi\inv(V) \definedas \qty{\prod \phi_i}\inv \qty{\prod V_i} .\end{align*} - $\Phi$ is a product of continuous functions and thus continuous. - $\Phi\inv \definedas \qty{\prod \phi_i}\inv = \prod \phi_i\inv$, which are all assumed continuous since $\phi_i$ were homeomorphisms. Example (Torii) : $T^n \definedas \prod_{i=1}^n S^1$ is a topological $n\dash$manifold. Definition (Precompact) : A subset $A\subseteq X$ is *precompact* iff its closure $\cl_X(A)$ is compact in $X$. Proposition : Every topological manifold has a countable basis of precompact coordinate balls. Proposition : Let $M$ be a topological manifold. - $M$ is locally path-connected. - $M$ is connected $\iff M$ is path-connected - The connected components and path components of $M$ coincide. - $\pi_0(M)$ is countable and each component is open and a connected topological manifold. Proposition : Every topological manifold $M$ is locally compact. Proof : $M$ has a basis of precompact open sets. Theorem (Manifolds are Paracompact) : Given any open cover $\mcu \covers M$ of a topological manifold and any basis $\mcb$ for the topology on $M$, there exists a countable locally finite open refinement of $\mcu$ consisting of elements of $\mcb$. Proposition : $\pi_1(M)$ is countable.