diff --git a/content/Math4202/Math4202_L20.md b/content/Math4202/Math4202_L20.md new file mode 100644 index 0000000..01e0fdb --- /dev/null +++ b/content/Math4202/Math4202_L20.md @@ -0,0 +1,72 @@ +# Math4202 Topology II (Lecture 20) + +## Algebraic Topology + +### Retraction and fixed point + +#### Lemma of retraction + +Let $h:S^1\to X$ be a continuous map. The following are equivalent: + +- $h$ is null-homotopic ($h$ is homotopic to a constant map). +- $h$ extends to a continuous map $k:B_1(0)\to X$. ($k|_{\partial B_1(0)}=h$) + +> For this course, we use closed disk $B_1(0)=\{(x,y)|d((x,y),(0,0))\leq 1\}$. + +- $h_*$ is the trivial group homomorphism of fundamental groups (Image of $\pi_1(S^1,x_0)\to \pi_1(X,x_0)$ is trivial group, identity). + +
+Proof + +First we will show that (1) implies (2). + +By the null homotopic condition, there exists $H:S\times I\to X$ that $H(s,1)=h(s),H(s,0)=x_0$. + +Define the equivalence relation for all the point in the set $H(s,0)$. Then there exists $\tilde{H}:S\times I/\sim \to X$ is a continuous map. (by quotient map $q:S\times I\to S\times I/\sim$ and $H$ is continuous.) + +Note that the cone shape is homotopic equivalent to the disk using polar coordinates. $k|_{\partial B_1(0)}=H|_{S^1\times\{1\}}=h$. + +--- + +Then we will prove that (2) implies (3). + +Let $i: S^1\to B_1(0)$ be the inclusion map. Then $k\circ i=h$, $k_*\circ i_*=h_*:\pi_1(S^1,1)\to \pi_1(X,h(1))$. + +Recall that $k:B_1(0)\to X$, $k_*:\pi_1(B_1(0),0)\to \pi_1(X,k(0))$ is trivial, since $B_1(0)$ is contractible. + +Therefore $k_*\circ i_*=h_*$ is the trivial group homomorphism. + +--- + +Now we will show that (3) implies (1). + +Consider the map from $\alpha:[0,1]\to S^1$ by $\alpha:x\mapsto e^{2\pi ix}$. $[\alpha]$ is a generator of $\pi_1(S^1,1)$. (The lifting of $\alpha$ to $\mathbb{R}$ is $1$, which is a generator of $\mathbb{Z}$.) + +$h\circ \alpha:[0,1]\to X$ is a loop in $X$ representing $h_*([\alpha])$. + +As $h_*([\alpha])$ is trivial, $h\circ \alpha$ is homotopic to a constant loop. + +Therefore, there exist a homotopy $H:I\times I\to X$, where $H(s,0)=$ constant map, $H(s,1)=h\circ \alpha(s)$. + +Take $\tilde{H}:S\times I/\sim \to X$ by $\tilde{H}(\exp(2\pi is),0)=H(s,0)$, $\tilde{H}(\exp(2\pi is),t)=H(s,t)$. $x\in [0,1]$. + +(From the perspective of quotient map, we can see that $\alpha$ is the quotient map from $I\times I$ to $I\times I/\sim=S\times I$. Then $\tilde{H}$ is the induced continuous map from $S\times I$ to $X$.) + +
+ +#### Corollary of punctured plane + +$i:S^1\to \mathbb{R}^2-\{0\}$ is not null homotopic. + +
+Proof + +Recall from last lecture, $r:\mathbb{R}^2-\{0\}\to S^1$ is the retraction $x\mapsto \frac{x}{|x|}$. + +Therefore, we have $i_*:\pi_1(S^1,1)\to \pi_1(\mathbb{R}^2-\{0\},r(0))$: $\mathbb{Z}\to \pi_1(\mathbb{R}^2-\{0\},r(0))$ is injective. + +Therefore $i_*$ is non trivial. + +Therefore $i$ is not null homotopic. + +
\ No newline at end of file diff --git a/content/Math4202/_meta.js b/content/Math4202/_meta.js index be7b91d..a80187b 100644 --- a/content/Math4202/_meta.js +++ b/content/Math4202/_meta.js @@ -22,4 +22,5 @@ export default { Math4202_L17: "Topology II (Lecture 17)", Math4202_L18: "Topology II (Lecture 18)", Math4202_L19: "Topology II (Lecture 19)", + Math4202_L20: "Topology II (Lecture 20)", } diff --git a/content/Math4302/Math4302_L19.md b/content/Math4302/Math4302_L19.md new file mode 100644 index 0000000..94d49e3 --- /dev/null +++ b/content/Math4302/Math4302_L19.md @@ -0,0 +1,121 @@ +# Math4302 Modern Algebra (Lecture 19) + +## Solution for Midterm I + +Not applicable + +## Group + +Side notes: From previous lecture + +What is the group $\mathbb{Z}\times \mathbb{Z}/\langle (a,b)\rangle$, where $a,b\in \mathbb{Z}^+$? + +This should be isomorphic to $\mathbb{Z}\times \mathbb{Z}_{\operatorname{gcd}(a,b)}$ + +
+Proof + +If $\operatorname{gcd}(a,b)=1$, then $\mathbb{Z}\times \mathbb{Z}/\langle (a,b)\rangle\simeq \mathbb{Z}$. + +The general isomorphism $\phi:\mathbb{Z}\times \mathbb{Z}/\langle (a,b)\rangle\to \mathbb{Z}\times \mathbb{Z}_{\operatorname{gcd}(a,b)}$ is given by + +$$ +(a,b)\mapsto (\frac{ma-nb}{d},au+bv\mod d) +$$ + +where $u,v$ are two integers such that $mu+nv=d$. + +Geometrically, partition the lattice to grids then map each element in grid to continuous integer in $ab/\operatorname{gcd}(a,b)=\operatorname{lcm}(a,b)$. + +$\phi$ is surjective homomorphism, and $\ker(\phi)=\langle (a,b)\rangle$ + +
+ +### Center of a group + +Recall from previous lecture, the center of a group $G$ is the subgroup of $G$ that contains all elements that commute with all elements in $G$. + +$$ +Z(G)=\{a\in G\mid \forall g\in G, ag=ga\} +$$ + +this subgroup is normal and measure the "abelian" for a group. + +
+Example + +$Z(S_3)=\{e\}$, all the transpositions are not commutative, so $Z(S_3)=\{e\}$. + +--- + +$Z(GL_n(\mathbb{R}))=\{A\in GL_n(\mathbb{R})\mid AB=BA\text{ for all }B\in GL_n(\mathbb{R})\}$ + +This is all the multiples of the identity matrix. + +Ideas of proof: + +1. Consider all matrix that is have non-zero element other than diagonal, we can always find a matrix that don't commute with it. +2. Consider if the diagonal entry are different, the row flip matrix will make your life worse. + +Therefore the only choice left the multiple of the identity matrix. + +
+ +#### Definition of the commutator of a group + +Let $G$ be a group and $a,b\in G$, the commutator $[a,b]$ is defined as $aba^{-1}b^{-1}$. + +$[a,b]=e$ if and only if $a$ and $b$ commute. + +Some additional properties: + +- $[a,b]^{-1}=[b,a]$ + +#### Definition of commutator subgroup + +Let $G'$ be the subgroup of $G$ generated by all commutators of $G$. + +$$ +G'=\{[a_1,b_1][a_2,b_2]\ldots[a_n,b_n]\mid a_1,a_2,\ldots,a_n,b_1,b_2,\ldots,b_n\in G\} +$$ + +Then $G'$ is the subgroup of $G$. + +- Identity: $[e,e]=e$ +- Inverse: $([a_1,b_1],\ldots,[a_n,b_n])^{-1}=[b_n,a_n],\ldots,[b_1,a_1]$ + +Some additional properties: + +- $G$ is abelian if and only if $G'=\{e\}$ +- $G'\trianglelefteq G$ +- $G/G'$ is abelian +- If $N$ is a normal subgroup of $G$, and $G/N$ is abelian, then $G'\leq N$. + +> These implies that $G'$ is the smallest abelian normal subgroup of $G$. + +
+Proof + +First we will show that $G'$ is normal + +Let $g\in G$ and $h=[a_1,b_1],\ldots,[a_n,b_n]\in G'$, then it is sufficient to show that $ghg^{-1}\in G'$. + +Consider arbitrary $[a,b]$ + +$$ +\begin{aligned} +g[a,b]g^{-1}&=ga_1 b_1 a_1^{-1}b_1^{-1}g^{-1}\\ +&=ga(g^{-1}g)b(g^{-1} g)a^{-1}(g^{-1}g)b^{-1}g^{-1}\\ +&=(gag^{-1})(gbg^{-1})(ga^{-1}g^{-1})(gb^{-1}g^{-1})\\ +&=(gag^{-1})(gbg^{-1})(gag^{-1})^{-1}(gbg^{-1})^{-1}\\ +&=[gag^{-1},gbg^{-1}] +\end{aligned} +$$ + +--- + +Them we will show that $G/G'$ is abelian. + +For all $a,b\in G$, we have $[a,b]\in G'$, so $[a,b]^{-1}\in G'$. Therefore $abG'=baG'\iff [b^{-1},a^{-1}]\in G'$. Therefore $G/G'$ is abelian. + +
diff --git a/content/Math4302/_meta.js b/content/Math4302/_meta.js index baf2b7b..728669f 100644 --- a/content/Math4302/_meta.js +++ b/content/Math4302/_meta.js @@ -21,4 +21,5 @@ export default { Math4302_L16: "Modern Algebra (Lecture 16)", Math4302_L17: "Modern Algebra (Lecture 17)", Math4302_L18: "Modern Algebra (Lecture 18)", + Math4302_L19: "Modern Algebra (Lecture 19)", }