Merge branch 'main' of https://git.trance-0.com/Trance-0/NoteNextra-origin

content/Math4202/Math4202_L21.md

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+# Math4202 Topology II (Lecture 21)
+
+## Algebraic Topology
+
+### Application of fundamental groups
+
+Recall from last Friday, $j:S^1\to \mathbb{R}^2-\{0\}$ is not null homotopic
+
+#### Hairy ball theorem
+
+Given a non-vanishing vector field on $B^2=\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq 1\}$, ($v:B^2\to \mathbb{R}^2$ continuous and $v(x,y)\neq 0$ for all $(x,y)\in B^2$) there exists a point of $S^1$ where the vector field points directly outward, and a point of $S^1$ where the vector field points directly inward.
+
+<details>
+<summary>Proof</summary>
+
+By our assumption, then $v:B^2\to \mathbb{R}^2-\{0\}$ is a continuous vector field on $B^2$.
+
+$v|_{S^1}:S^1\to \mathbb{R}^2-\{0\}$ is null homotopic.
+
+We prove by contradiction.
+
+Suppose $v:B^2\to \mathbb{R}^2-\{0\}$ and $v|_{S^1}:S^1\to \mathbb{R}^2-\{0\}$ is everywhere outward. (for everywhere inward, consider $-v$ must be everywhere outward)
+
+Because $v|_{S^1}$ extends continuously to $B^2$, then $v|_{S^1}:B^2\to \mathbb{R}^2-\{0\}$ is null homotopic.
+
+We construct a homotopy for functions between $v|_{S^1}$ and $j$. (Recall $j:S^1\to \mathbb{R}^2-\{0\}$ is not null homotopic)
+
+Define $H:S^1\times I\to \mathbb{R}^2-\{0\}$ by affine combination
+
+$$
+H((x,y),t)=(1-t)v(x,y)+tj(x,y)
+$$
+
+we also need to show that $H$ is non zero.
+
+Since $v$ is everywhere outward, $v(x,y)\cdot j(x,y)$ is positive for all $(x,y)\in S^1$.
+
+$H((x,y),t)\cdot j(x,y)=(1-t)v(x,y)\cdot j(x,y)+tj(x,y)\cdot j(x,y)=(1-t)(v(x,y)\cdot j(x,y))+t$
+
+which is positive for all $t\in I$, therefore $H$ is non zero.
+
+So $H$ is a homotopy between $v|_{S^1}$ and $j$.
+</details>
+
+#### Corollary of the hairy ball theorem
+
+$\forall v:B^2\to \mathbb{R}^2$, if on $S^1$, $v$ is everywhere outward/inward, there is $(x,y)\in B^2$ such that $v(x,y)=0$.
+
+#### Brouwer's fixed point theorem
+
+If $f:B^2\to B^2$ is continuous, then there exists a point $x\in B^2$ such that $f(x)=x$.
+
+<details>
+<summary>Proof</summary>
+
+We proceed by contradiction again.
+
+Suppose $f$ has no fixed point, $f(x)-x\neq 0$ for all $x\in B^2$.
+
+Now we consider the map $v:B^2\to \mathbb{R}^2$ defined by $v(x,y)=f(x)-x$, this function is continuous since $f$ is continuous.
+
+$forall x\in S^1$, $v(x)\cdot x=f(x)\cdot x-x\cdot x=f(x)\cdot x-1$.
+
+Recall the cauchy schwartz theorem, $|f(x)\cdot x|\leq \|f(x)\|\cdot\|x\|\leq 1$, note that $f(x)\neq 0$ for all $x\in B^2$, $v(x)\cdot x<0$. This means that all $v(x)$ points inward.
+
+This is a contradiction to the hairy ball theorem, so $f$ has a fixed point.
+</details>

content/Math4202/_meta.js

@@ -24,4 +24,5 @@ export default {
     Math4202_L18: "Topology II (Lecture 18)",
     Math4202_L19: "Topology II (Lecture 19)",
     Math4202_L20: "Topology II (Lecture 20)",
+    Math4202_L21: "Topology II (Lecture 21)",
 }

content/Math4302/Math4302_L21.md

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+# Math4302 Modern Algebra (Lecture 21)
+
+## Groups
+
+### Group acting on a set
+
+#### Definition of orbits
+
+We define the equivalence relation on $X$
+
+$$
+x\sim y\iff y=g\cdot x\text{ for some }g
+$$
+
+So we get a partition of $X$ into equivalence classes: orbits
+
+$$
+Gx\coloneqq \{g\cdot x|g\in G\}=\{y\in X|x\sim y\}
+$$
+
+is the orbit of $X$.
+
+$x,y\in X$ either $Gx=Gy$ or $Gx\cap Gy=\emptyset$.
+
+$X=\bigcup_{x\in X}Gx$.
+
+<details>
+<summary>Example</summary>
+
+Let $D_4$ acting on $X=\{1,2,3,4\}$. Let $D_4=\{e,\rho,\rho^2,\rho^3,\mu,\mu\rho,\mu\rho^2,\mu\rho^3\}$.
+
+define $\phi\in D_4$, $i\in X$, $\phi\cdot i=\phi(i)$
+
+The orbits are:
+
+orbit of 1: $D_4\cdot 1=\{1,2,3,4\}$. This is equal to orbit of 2,3,4.
+
+---
+
+Let $G=S_3$ acting on $X=S_3$ via conjugation, let $\sigma\in X$ and $\phi\in G$, we define $\phi\cdot\sigma\coloneqq \phi\sigma\phi^{-1}$.
+
+$S_3=\{e,(1,2,3),(1,3,2),(1,2),(1,3),(2,3)\}$.
+
+The orbits are:
+
+orbit of $e$: $G e=\{e\}$. since $geg^{-1}=e$ for all $g\in S_3$.
+
+orbit of $(1,2,3)$:
+
+- $e(1,2,3)e^{-1}=(1,2,3)$
+- $(1,3,2)(1,2,3)(1,3,2)^{-1}=(1,2,3)$
+- $(1,2,3)(1,2,3)(1,2,3)^{-1}=(1,2,3)$
+- $(1,2)(1,2,3)(1,2)^{-1}=(2,3)(1,2)=(1,3,2)$
+- $(1,3)(1,2,3)(1,3)^{-1}=(1,2)(1,3)=(1,3,2)$
+- $(2,3)(1,2,3)(2,3)^{-1}=(1,3)(2,3)=(1,3,2)$
+
+So the orbit of $(1,2,3)$ is equal to orbit of $(1,3,2)$. $=\{(1,2,3),(2,3,1)\}$.
+
+orbit of $(1,2)$:
+
+- $(1,2,3)(1,2)(1,2,3)^{-1}=(1,3)(1,3,2)=(2,3)$
+- $(1,3,2)(1,2)(1,3,2)^{-1}=(2,3)(1,2,3)=(1,3)$
+
+Therefore orbit of $(1,2)$ is equal to orbit of $(2,3)$, $(1,3)$. $=\{(1,2),(2,3),(1,3)\}$
+
+The orbits may not have the same size.
+
+</details>
+
+#### Definition of isotropy subgroup
+
+Let $X$ be a $G$-set, the stabilizer (or isotropy subgroup) corresponding to $x\in X$ is
+
+$$
+G_x=\{g\in G|g\cdot x=x\}
+$$
+
+$G_x$ is a subgroup of $G$. $G_x\leq G$.
+
+- $e\cdot x=x$, so $e\in G_x$
+- If $g_1,g_2\in G_x$, then $(g_1g_2)\cdot x=g_1\cdot(g_2\cdot x)=g_1 \cdot x$, so $g_1g_2\in G_x$
+- If $g\in G_x$, then $g^{-1}\cdot g=x=g^{-1}\cdot x$, so $g^{-1}\in G_x$
+
+<details>
+<summary>Examples of isotropy subgroups</summary>
+
+Let $D_4$ acting on $X=\{1,2,3,4\}$, find $G_1$, $G_2$, $G_3$, $G_4$.
+
+$G_1=G_3=\{e,\mu\}$, $G_2=G_4=\{e,\mu\rho^2\}$.
+
+---
+
+Let $S_3$ acting on $X=S_3$. Find $G_{e}$, $G_{(1,2,3)}$, $G_{(1,2)}$.
+
+$G_{e}=S_3$, $G_{(1,2,3)}=G_{(1,3,2)}=\{e,(1,2,3),(1,3,2)\}$, $G_{(1,2)}=\{e,(1,2)\}$, ($G_{(1,3)}=\{e,(1,3)\}$, $G_{(2,3)}=\{e,(2,3)\}$)
+
+> The larger the orbit, the smaller the stabilizer.
+
+</details>
+
+#### Orbit-stabilizer theorem
+
+If $X$ is a $G$-set and $x\in X$, then
+
+$$
+|Gx|=(G:G_x)=\text{ number of left cosets of }G_x=\frac{|G|}{|G_x|}
+$$
+
+<details>
+<summary>Proof</summary>
+
+Define $\alpha$ be the function that maps the set of left cosets of $G_x$ to orbit of $x$. $gG_X\mapsto g\cdot x$.
+
+This function is well defined. And $\alpha$ is a bijection.
+
+Continue next lecture.
+
+</details>

content/Math4302/Math4302_L22.md

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+# Math4302 Modern Algebra (Lecture 22)
+
+## Groups
+
+### Group acting on a set
+
+Let $X$ be a $G$-set, recall that the orbit of $x\in X$ is $\{g\cdot x|g\in G\}$.
+
+#### The orbit-stabilizer theorem
+
+For any $x\in X$, ,$G_x=\{g\in G|g\cdot x=x\}\leq G$.
+
+Let $(G:G_x)$ denote the index of $G_x$ in $G$, then $(G:G_x)=\frac{|G|}{|G_x|}$, which equals to the number of left cosets of $G_x$ in $G$.
+
+<details>
+<summary>Proof</summary>
+
+Define $\alpha:gG_x\mapsto g\cdot x$.
+
+$\alpha$ is well-defined and injective.
+
+$$
+gG_x=g'G_x\iff g^{-1}g'\in G_x\iff (g^{-1}g')\cdot x=x\iff g^{-1}\cdot(g'\cdot x)=x\iff g'\cdot x=g\cdot x
+$$
+
+$\alpha$ is surjective, therefore $\alpha$ is a bijection.
+
+</details>
+
+<details>
+<summary>Example</summary>
+
+Number of elements in the orbit of $x$ is $1$ if and only if $g\cdot x=x$ for all $g\in G$.
+
+if and only if $G_x=G$.
+
+</details>
+
+#### Theorem for orbit with prime power groups
+
+Suppose $X$ is a $G$-set, and $|G|=p^n$ for some prime $p$. Let $X_G$ be the set of all elements in $X$ whose orbit has size $1$. (Recall the orbit divides $X$ into disjoint partitions.) Then $|X|\equiv |X_G|\mod p$.
+
+<details>
+<summary>Examples</summary>
+
+Let $G=D_4$ acting on $\{1,2,3,4\}=X$.
+
+$X_G=\emptyset$ since there is no element whose orbit has size $1$.
+
+---
+
+Let $G=\mathbb{Z}_{11}$ acting on a set with $|X|=20$ if the action is not trivial, then what is $|X_G|$?
+
+Using the theorem we have $|X_G|\equiv 20\mod 11=9$. Therefore $|X_G|=9$ or $20$, but the action is not trivial, $|X_G|=9$.
+
+An instance for such $X=\mathbb{Z}_{11}\sqcup\{x_1,x_2,\ldots,x_9\}$, where $\mathbb{Z}_{11}$ acts on $\{x_1,x_2,\ldots,x_9\}$ trivially. and $\mathbb{Z}_{11}$ acts on $x_1$ with addition.
+
+</details>
+
+<details>
+<summary>Proof</summary>
+
+If $x\in X$ such that $|Gx|\geq 2$, then $\frac{|G|}{|G_x|}=|Gx|\geq 2$.
+
+So $|G|=|G_x||Gx|\implies |Gx|$ divides $|G|$.
+
+So $|Gx|=p^m$ for some $m\geq 1$.
+
+Note that $X$ is the union of subset of elements with orbit of size $1$, and distinct orbits of sizes $\geq 2$. (each of them has size positive power of $p$)
+
+So $p|(|X|-|X_G|)$.
+
+this implies that $|X_G|\equiv |X_G|\mod p$.
+
+</details>
+
+#### Corollary: Cauchy's theorem
+
+If $p$ is prime and $p|(|G|)$, then $G$ has a subgroup of order $p$.
+
+> This does not hold when $p$ is not prime.
+>
+> Consider $A_4$ with order $12$, and $A_4$ has no subgroup of order $6$.
+
+<details>
+<summary>Proof</summary>
+
+It is enough to show, there is $a\in G$ which has order $p$: $\{e,a,a^2,\ldots,a^{p-1}\}\leq G$.
+
+Let $X=\{(g_1,g_2,\ldots,g_p)|g_i\in G,g_1g_2g_3\ldots g_p=e\}$.
+
+Then $|X|=|G|^{p-1}$ since $g_p$ is determined uniquely by $g_p=(g_1,g_2,\ldots,g_{p-1})^{-1}$.
+
+Therefore we can define $\mathbb{Z}_p$ acts on $X$ by shifting.
+
+$i\in \mathbb{Z}_p$ $i\cdot (g_1,g_2,\ldots,g_p)=(g_{i+1},g_2,\ldots,g_p,g_1,\ldots,g_i)$.
+
+$X$ is a $\mathbb{Z}_p$-set.
+
+- $0\cdot (g_1,g_2,\ldots,g_p)=(g_1,g_2,\ldots,g_p)$.
+- $j\cdot (i\cdot (g_1,g_2,\ldots,g_p))=(i+j)\cdot (g_1,g_2,\ldots,g_p)$.
+
+By the previous theorem, $|X|\equiv |X_G|\mod p$.
+
+Since $p$ divides $|G|^{p-1}$, $p$ also divides $|X_G|$. Therefore $(e,e,e,\ldots,e)\in X_G$. Therefore $|X_G|\geq 1$.
+
+So $|X_G|\geq 2$, we have $(a,a,\ldots,a)\in X_G$, $a\neq e$, but $a^p=e$, so $ord(a)=p$.
+
+</details>

content/Math4302/_meta.js

@@ -23,4 +23,5 @@ export default {
     Math4302_L18: "Modern Algebra (Lecture 18)",
     Math4302_L19: "Modern Algebra (Lecture 19)",
     Math4302_L20: "Modern Algebra (Lecture 20)",
+    Math4302_L21: "Modern Algebra (Lecture 21)",
 }