4823eefff6
Some checks failed
Sync from Gitea (main→main, keep workflow) / mirror (push) Has been cancelled
46 lines
1.4 KiB
Markdown
46 lines
1.4 KiB
Markdown
# Math4202 Topology II (Lecture 22)
|
|
|
|
## Final reading, report, presentation
|
|
|
|
- Mar 30: Reading topic send email or discuss in OH.
|
|
- Apr 3: Finalize the plan.
|
|
- Apr 22,24: Last two lectures: 10 minutes to present.
|
|
- Final: type a short report, 2-5 pages.
|
|
|
|
## Algebraic topology
|
|
|
|
### Fundamental theorem of Algebra
|
|
|
|
For arbitrary polynomial $f(z)=\sum_{i=0}^n a_i x^i$. Are there roots in $\mathbb{C}$?
|
|
|
|
Consider $f(z)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_0$ is a continuous map from $\mathbb{C}\to\mathbb{C}$.
|
|
|
|
If $f(z_0)=0$, then $z_0$ is a root.
|
|
|
|
By contradiction, Then $f:\mathbb{C}\to\mathbb{C}-\{0\}\cong \mathbb{R}^2-\{(0,0)\}$.
|
|
|
|
#### Theorem for existence of n roots
|
|
|
|
A polynomial equation $x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_0=0$ of degree $>0$ with complex coefficients has at least one complex root.
|
|
|
|
There are $n$ roots by induction.
|
|
|
|
#### Lemma
|
|
|
|
If $g:S^1\to \mathbb{R}^2-\{(0,0)\}$ is the map $g(z)=z^n$, then $g$ is not nulhomotopic. $n\neq 0$, $n\in \mathbb{Z}$.
|
|
|
|
> Recall that we proved that $g(z)=z$ is not nulhomotopic.
|
|
|
|
Consider $k:S^1\to S^1$ by $k(z)=z^n$. $k$ is continuous, $k_*:\pi_1(S^1,1)\to \pi_1(S^1,1)$.
|
|
|
|
Where $\pi_1(S^1,1)\cong \mathbb{Z}$.
|
|
|
|
$k_*(n)=nk_*(1)$.
|
|
|
|
Recall that the path in the loop $p:I\to S^1$ where $p:t\mapsto e^{2\pi it}$.
|
|
|
|
$k_*(p)=[k(p(t))]$, where $n=\tilde{k\circ p}(1)$.
|
|
|
|
$k_*$ is injective.
|
|
|