diff --git a/II_M/algebraic_topology.tex b/II_M/algebraic_topology.tex index 500c74a..043c51b 100644 --- a/II_M/algebraic_topology.tex +++ b/II_M/algebraic_topology.tex @@ -128,7 +128,7 @@ \subsection{Cell complexes} \] where the equivalence relation $\sim$ is the equivalence relation generated by $x\sim f(x)$ for all $x\in S^{n - 1}\subseteq D^n$ (and $\amalg$ is the disjoint union). - Intuitively, a map $f: S^{n - 1}\to X$ just picks out a subset $X$ that looks like the sphere. So we are just sticking a disk onto $X$ by attaching the boundary of the disk onto a sphere within $X$. + Intuitively, a map $f: S^{n - 1}\to X$ just picks out a subset of $X$ that looks like the sphere. So we are just sticking a disk onto $X$ by attaching the boundary of the disk onto a sphere within $X$. \begin{center} \begin{tikzpicture} \draw ellipse (0.5 and 1); @@ -701,7 +701,7 @@ \subsection{The fundamental group} Immediate from our previous lemmas. \end{proof} -Often, in mathematics, after defining a term, we give lots of examples of it. Unfortunately, it is rather difficult to prove that a space has a non-trivial fundamental group, until we have developed some relevant machinery. Hence we will have to wait for a while before we have some concrete examples. Instead, we will look at some properties of the fundamental group first. +Often in mathematics, after defining a term, we give lots of examples of it. Unfortunately, it is rather difficult to prove that a space has a non-trivial fundamental group, until we have developed some relevant machinery. Hence we will have to wait for a while before we have some concrete examples. Instead, we will look at some properties of the fundamental group first. \begin{defi}[Based space] A \emph{based space} is a pair $(X, x_0)$ of a space $X$ and a point $x_0\in X$, the \emph{basepoint}. A \emph{map of based spaces} @@ -1399,9 +1399,9 @@ \subsection{Covering space} \item The stabilizer of $\tilde{x}_0 \in p^{-1}(x_0)$ is $p_*(\pi_1(\tilde{X}, \tilde{x}_0)) \subseteq \pi_1(X, x_0)$. \item If $\tilde{X}$ is path connected, then there is a bijection \[ - p_* \pi_1(\tilde{X}, \tilde{x}_0)\backslash \pi_1(X, x_0) \to p^{-1}(x_0). + p_* (\pi_1(\tilde{X}, \tilde{x}_0))\backslash \pi_1(X, x_0) \to p^{-1}(x_0). \] - Note that $p_* \pi_1(\tilde{X}, \tilde{x}_0)\backslash \pi_1(X, x_0)$ is not a quotient, but simply the set of cosets. We write it the ``wrong way round'' because we have right cosets instead of left cosets. + Note that $p_*(\pi_1(\tilde{X}, \tilde{x}_0))\backslash \pi_1(X, x_0)$ is not a quotient, but simply the set of cosets. We write it the ``wrong way round'' because we have right cosets instead of left cosets. \end{enumerate} \end{lemma} Note that this is great! If we can find a covering space $p$ and a point $x_0$ such that $p^{-1}(x_0)$ is non-trivial, then we immediately know that $\pi_1(X, x_0)$ is non-trivial! @@ -1716,7 +1716,7 @@ \subsection{The Galois correspondence} Now we prove that every subgroup of $\pi_1$ comes from exactly one covering space. What this statement properly means is made precise in the following proposition: \begin{prop} - Let $(X, x_0)$, $(\tilde{X}_1, \tilde{x}_1)$, $(\tilde{X}_2, \tilde{X}_2)$ be path-connected based spaced, and $p_i: (\tilde{X}_i, \tilde{x}_i) \to (X, x_0)$ be covering maps. Then we have + Let $(X, x_0)$, $(\tilde{X}_1, \tilde{x}_1)$, $(\tilde{X}_2, \tilde{x}_2)$ be path-connected based spaced, and $p_i: (\tilde{X}_i, \tilde{x}_i) \to (X, x_0)$ be covering maps. Then we have \[ p_{1*}\pi_1(\tilde{X}_1, \tilde{x}_1) = p_{2*} \pi_1(\tilde{X}_2, \tilde{x_2}) \] @@ -2209,7 +2209,7 @@ \subsection{Seifert-van Kampen theorem} \[ \pi_1(S^n) \cong \pi_1(\R^n) \underset{\pi_1(S^{n - 1})}{*} \pi_1(\R^n) \cong 1 \underset{\pi_1(S^{n - 1})}{*} 1 \] - It is easy to see this is the trivial group. We can see this directly form the universal property of the amalgamated free product, or note that it is the quotient of $1 * 1$, which is $1$. +It is easy to see this is the trivial group. We can see this directly from the universal property of the amalgamated free product, or note that it is the quotient of $1 * 1$, which is $1$. So for $n \geq 2$, $\pi_1(S^n) \cong 1$. \end{eg} @@ -2428,7 +2428,7 @@ \subsection{The effect on \texorpdfstring{$\pi_1$}{pi1} of attaching cells} \end{proof} The more interesting case is when we have smaller dimensions. \begin{thm} - If $n = 2$, then the natural map $\pi_1(X, X_0) \to \pi_1(X\cup_f D^n, x_0)$ is \emph{surjective}, and the kernel is $\bra\bra [f] \ket \ket$. Note that this statement makes sense, since $S^{n - 1}$ is a circle, and $f: S^{n - 1} \to X$ is a loop in $X$. + If $n = 2$, then the natural map $\pi_1(X, x_0) \to \pi_1(X\cup_f D^n, x_0)$ is \emph{surjective}, and the kernel is $\bra\bra [f] \ket \ket$. Note that this statement makes sense, since $S^{n - 1}$ is a circle, and $f: S^{n - 1} \to X$ is a loop in $X$. \end{thm} This is what we would expect, since if we attach a disk onto the loop given by $f$, this loop just dies. @@ -2856,7 +2856,7 @@ \subsection{Simplicial complexes} The boundary of $\sigma$ is usually denoted by $\partial \sigma$, while the interior is denoted by $\mathring{\sigma}$, and we write $\tau \leq \sigma$ when $\tau$ is a face of $\sigma$. \end{defi} -In particular, the interior of a vertex is the vertex itself. Note that this notions of interior and boundary are distinct from the topological notions of interior and boundary. +In particular, the interior of a vertex is the vertex itself. Note that these notions of interior and boundary are distinct from the topological notions of interior and boundary. \begin{eg} The \emph{standard $n$-simplex} is spanned by the basis vectors $\{\mathbf{e}_0, \cdots, \mathbf{e}_n\}$ in $\R^{n + 1}$. For example, when $n = 2$, we get the following: @@ -3177,7 +3177,7 @@ \subsection{Simplicial approximation} \end{defi} \begin{prop} - $|K| = |K|'$ and $K'$ really is a simplicial complex. + $|K| = |K'|$ and $K'$ really is a simplicial complex. \end{prop} \begin{proof} @@ -3186,12 +3186,12 @@ \subsection{Simplicial approximation} We now have a slight problem. Even though $|K'|$ and $|K|$ are equal, the identity map from $|K'|$ to $|K|$ is not a simplicial map. -To solve this problem, we can choose any function $K \to V_K$ by $\sigma \mapsto v_\sigma$ with $v_\sigma \in \sigma$, i.e.\ a function that sends any simplex to any of its vertices. Then we can define $g: K' \to K$ by sending $\hat{\sigma} \mapsto v_\sigma$. Then this is a simplicial map, and indeed a simplicial approximation to the identity map $|K'| \to |K|$. +To solve this problem, we can choose any function $K \to V_K$ by $\sigma \mapsto v_\sigma$ with $v_\sigma \in \sigma$, i.e.\ a function that sends any simplex to any of its vertices. Then we can define $g: K' \to K$ by sending $\hat{\sigma} \mapsto v_\sigma$. Then this is a simplicial map, and indeed a simplicial approximation to the identity map $|K'| \to |K|$. We will revisit this idea later when we discuss homotopy invariance. The key theorem is that as long as we are willing to perform barycentric subdivisions, then we can always find a simplicial approximation. \begin{thm}[Simplicial approximation theorem] - Le $K$ and $L$ be simplicial complexes, and $f: |K| \to |L|$ a continuous map. Then there exists an $r$ and a simplicial map $g: K^{(r)} \to L$ such that $g$ is a simplicial approximation of $f$. Furthermore, if $f$ is already simplicial on $M\subseteq K$, then we can choose $g$ such that $|g||_M = f|_M$. + Let $K$ and $L$ be simplicial complexes, and $f: |K| \to |L|$ a continuous map. Then there exists an $r$ and a simplicial map $g: K^{(r)} \to L$ such that $g$ is a simplicial approximation of $f$. Furthermore, if $f$ is already simplicial on $M\subseteq K$, then we can choose $g$ such that $|g||_M = f|_M$. \end{thm} The first thing we have to figure out is how far we are going to subdivide. To do this, we want to quantify how ``fine'' our subdivisions are. @@ -3204,9 +3204,9 @@ \subsection{Simplicial approximation} We have the following lemma that tells us how large our mesh is: \begin{lemma} - Let $\dim K \leq n$, then + Let $\dim K = n$, then \[ - \mu(K^{(r)}) = \left(\frac{n}{n + 1}\right)^r \mu(K). + \mu(K^{(r)}) \leq \left(\frac{n}{n + 1}\right)^r \mu(K). \] \end{lemma} % insert proof The key point is that as $r \to \infty$, the mesh goes to zero. So indeed we can make our barycentric subdivisions finer and finer. The proof is purely technical and omitted. @@ -3647,7 +3647,7 @@ \subsection{Some homological algebra} & & D_n & D_{n + 1} \ar [l, "d_{n + 1}"] \end{tikzcd} \] -The intuition behind this definition is as follows: suppose $C_{\Cdot} = C_{\Cdot} (K)$ and $D_{\Cdot} = C_{\Cdot}(L)$ for $K, L$ simplicial complexes, and $f_{\Cdot}$ and $g_{\Cdot}$ are ``induced'' by simplicial maps $f, g: K \to L$. How can we detect if $f$ and $g$ are homotopic via the homotopy groups? +The intuition behind this definition is as follows: suppose $C_{\Cdot} = C_{\Cdot} (K)$ and $D_{\Cdot} = C_{\Cdot}(L)$ for $K, L$ simplicial complexes, and $f_{\Cdot}$ and $g_{\Cdot}$ are ``induced'' by simplicial maps $f, g: K \to L$, where by ``induced'' we mean where $n$-simplices are mapped to $k$-simplices for $k 0$ such that if $f, g: |K| \to |L|$ satisfy $\|f(x) - g(x)\| < \varepsilon$, then $f_* = g_*: H_n(K) \to H_n(L)$ for all $n$. \end{lemma} -The idea of the proof is that if $\|f(x) - g(x)\|$ is small enough, we can barycentrically subdivide $L$ such that we get a simplicial approximation to both $f$ and $g$. +The idea of the proof is that if $\|f(x) - g(x)\|$ is small enough, we can barycentrically subdivide $K$ such that we get a simplicial approximation to both $f$ and $g$. \begin{proof} By the Lebesgue number lemma, there is an $\varepsilon > 0$ such that each ball of radius $2\varepsilon$ in $|L|$ lies in some star $\St_L(w)$. @@ -4326,9 +4330,9 @@ \subsection{Continuous maps and homotopy invariance} \[ g(B_\delta(x)) \subseteq B_{2\varepsilon}(y) \subseteq \St_L(w). \] - Now subdivide $r$ times so that $\mu(K^{(r)}) < \frac{1}{2} \delta$. So for all $v \in V_K(R)$, we know + Now subdivide $r$ times so that $\mu(K^{(r)}) < \frac{1}{2} \delta$. So for all $v \in V_{K^(r)}$, we know \[ - \St_{K(r)} (v) \subseteq B_\delta(V). + \St_{K^{(r)}} (v) \subseteq B_\delta(v). \] This gets mapped by \emph{both} $f$ and $g$ to $\St_L(w)$ for the same $w \in V_L$. We define $s: V_{K^{(r)}} \to V_L$ sending $v \mapsto w$. \end{proof} @@ -4443,7 +4447,7 @@ \subsection{Homology of spheres and applications} \node at (-1.6, -1.2) [anchor = north east] {$g(x)$}; \end{tikzpicture} \end{center} - So we now show no such continuous retraction can exist. Suppose $r: D^n \to \partial D^n$ is a retraction, i.e.\ $r \circ i \simeq \id: \partial D^n \to D^n$. + So we now show no such continuous retraction can exist. Suppose $r: D^n \to \partial D^n$ is a retraction, i.e.\ $r \circ i \simeq \id: \partial D^n \to \partial D^n$. \[ \begin{tikzcd} S^{n - 1} \ar[r, "i"] & D^n \ar[r, "r"] & S^{n - 1} @@ -4772,7 +4776,19 @@ \subsection{Rational homology, Euler and Lefschetz numbers} \] In this case, we have not lost any information because there was no torsion part of the homology groups. - However, for the non-orientable surfaces, we have + However, for the non-orientable surfaces, since + + \[ + H_k(E_n) = + \begin{cases} + \Z & k = 0\\ + \Z^{n - 1} \times \Z / 2 & k = 1\\ + 0 & \text{otherwise} + \end{cases}, + \] + + (exercise) we have that + \[ H_k(E_n, \Q) = \begin{cases} @@ -4781,6 +4797,7 @@ \subsection{Rational homology, Euler and Lefschetz numbers} 0 & \text{otherwise} \end{cases}, \] + This time, this is different from the integral coefficient case, where we have an extra $\Z_2$ term in $H_1$. \end{eg} @@ -4866,13 +4883,13 @@ \subsection{Rational homology, Euler and Lefschetz numbers} There is an exact sequence \[ \begin{tikzcd} - 0 \ar[r] & B_i(K; Q) \ar[r] & Z_i(K; \Q) \ar[r] & H_i(K; \Q) \ar[r] & 0 + 0 \ar[r] & B_i(K; \Q) \ar[r] & Z_i(K; \Q) \ar[r] & H_i(K; \Q) \ar[r] & 0 \end{tikzcd} \] This is since $H_i(K, \Q)$ is defined as the quotient of $Z_i$ over $B_i$. We also have the exact sequence \[ \begin{tikzcd} - 0 \ar[r] & Z_i(K; Q) \ar[r] & C_i(K; \Q) \ar[r, "d_i"] & B_{i - 1}(K; \Q) \ar[r] & 0 + 0 \ar[r] & Z_i(K; \Q) \ar[r] & C_i(K; \Q) \ar[r, "d_i"] & B_{i - 1}(K; \Q) \ar[r] & 0 \end{tikzcd} \] This is true by definition of $B_{i - 1}$ and $Z_i$. Let $f_i^H, f_i^B, f_i^Z, f_i^C$ be the various maps induced by $f$ on the corresponding groups. Then we have @@ -4937,7 +4954,7 @@ \subsection{Rational homology, Euler and Lefschetz numbers} \end{eg} \begin{eg} - Suppose $G$ is a path-connected topological group, i.e.\ $X$ is a group and a topological space, and inverse and multiplication are continuous maps. + Suppose $G$ is a path-connected topological group, i.e.\ $G$ is a group and a topological space, and inverse and multiplication are continuous maps. If $g \not= 1$, then the map \begin{align*} diff --git a/II_M/probability_and_measure.tex b/II_M/probability_and_measure.tex index aefdd88..a9dcdc9 100644 --- a/II_M/probability_and_measure.tex +++ b/II_M/probability_and_measure.tex @@ -703,7 +703,7 @@ \subsection{Probability measures} \liminf A_n &= \bigcup_n \bigcap_{m \geq n} A_m. \end{align*} \end{defi} -To parse these definitions more easily, we can read $\cap$ as ``for all'', and $\cup$ as ``there exits''. For example, we can write +To parse these definitions more easily, we can read $\cap$ as ``for all'', and $\cup$ as ``there exist''. For example, we can write \begin{align*} \limsup A_n &= \forall n,\exists m \geq n\text{ such that }A_m\text{ occurs}\\ &= \{x: \forall n, \exists m \geq n, x \in A_m\}\\ @@ -1030,9 +1030,9 @@ \subsection{Constructing new measures} \draw [->] (0, 0) -- (4, 0); \draw [->] (0, 0) -- (0, 4); - \draw [thick, mblue] (0, 0) -- (1.5, 2) -- (2.5, 2) node [draw, fill=white, circle, inner sep = 0, minimum size = 3] {}; + \draw [thick, mblue] (0, 0) -- (1.5, 2) -- (2.5, 2) node [circ] {}; - \draw [thick, mblue] (2.5, 3) node [circ] {} -- (4, 4); + \draw [thick, mblue] (2.5, 3) node [draw, fill=white, circle, inner sep = 0, minimum size = 3] {} -- (4, 4); \end{tikzpicture} \end{center} \end{eg}