# 问题及解答

## $\inf_{f\in\alpha}\mathrm{dil}(f)$ 为代表 $\alpha$ 的最短闭曲线 $\gamma_0$ 的长度.

Posted by haifeng on 2020-07-05 23:45:34 last update 2020-07-06 22:18:23 | Edit | Answers (2)

$\mathrm{dil}(f)=\sup_{t\neq t'\\ t,t'\in I}\frac{d(f(t),f(t'))}{d(t,t')}$

$f(t)=\begin{cases} t, & t\in[0,1]\\ 2t-1, & t\in(1,2]\\ 3t-3, & t\in(2,3] \end{cases}$

$f(t)=\begin{cases} e^{i\frac{\pi}{2}t}, & t\in[0,1]\\ e^{i(\pi t-\frac{\pi}{2})}, & t\in(1,2]\\ e^{i\frac{\pi}{2}(t+1)}, & t\in(2,3] \end{cases}$

$\mathrm{dil}(f)=\sup_{t\neq t'\\ t,t'\in I}\frac{d(f(t),f(t'))}{d(t,t')}=\pi$

$h(s)=\begin{cases} e^{i\frac{\pi}{2}3s}, & s\in[0,\frac{1}{3}]\\ e^{i(\pi 3s-\frac{\pi}{2})}, & s\in(\frac{1}{3},\frac{2}{3}]\\ e^{i\frac{\pi}{2}(3s+1)}, & t\in(\frac{2}{3},1] \end{cases}$

$\mathrm{dil}(h)=\sup_{s\neq s'\\ s,s'\in I}\frac{d(h(s),h(s'))}{d(s,s')}=3\pi$

$\mathrm{dil}(h)=\mathrm{dil}(f)\cdot\mathrm{dil}(g).$

$\|\alpha\|=\inf_{f\in\alpha}\mathrm{dil}(f)\cdot\mathrm{vol}(B^1)$

$\mathrm{length}(\gamma)=\int_I |\dot{\gamma}|ds$

$\pi_n(X,x_0)$ 上的群结构

1

Posted by haifeng on 2020-07-06 21:54:59

(1) 若 $t,t'\in[0,1]$, 则 $f(t)=t$, $f(t')=t'$. 于是

$\frac{d(f(t),f(t'))}{d(t,t')}=\frac{|t-t'|}{|t-t'|}=1.$

(2) 若 $t,t'\in[1,2]$, 则

$\frac{d(f(t),f(t'))}{d(t,t')}=\frac{|(2t-1)-(2t'-1)|}{|t-t'|}=2.$

(3) 若 $t,t'\in[2,3]$, 则

$\frac{d(f(t),f(t'))}{d(t,t')}=\frac{|(3t-3)-(3t'-3)|}{|t-t'|}=3.$

(4) 若 $0 < t < 1 < t' < 2$, 则

$\begin{split} \frac{d(f(t),f(t'))}{d(t,t')}&=\frac{|t-(2t'-1)|}{|t-t'|}=\frac{|t-t'+1-t'|}{|t-t'|}\\ &\leqslant\frac{|t-t'|+|1-t'|}{|t-t'|}=1+\frac{|1-t'|}{|t-t'|}\\ &\leqslant 1+\frac{|1-t'|}{|1-t'|}\\ &=2 \end{split}$

$\sup_{t\neq t'\\ t,t'\in[0,3]}\frac{d(f(t),f(t'))}{d(t,t')}=3.$

2

Posted by haifeng on 2020-07-08 08:51:21

(1) 若 $t,t'\in[0,1]$,

$\frac{d(f(t),f(t'))}{d(t,t')}=\frac{|\frac{\pi}{2}t-\frac{\pi}{2}t'|}{|t-t'|}=\frac{\pi}{2}$

(2) 若 $t,t'\in[1,2]$,

$\frac{d(f(t),f(t'))}{d(t,t')}=\frac{|(\pi t-\frac{\pi}{2})-(\pi t'-\frac{\pi}{2})|}{|t-t'|}=\pi$

(3) 若 $t,t'\in[2,3]$,

$\frac{d(f(t),f(t'))}{d(t,t')}=\frac{|\frac{\pi}{2}(t+1)-\frac{\pi}{2}(t'+1)|}{|t-t'|}=\frac{\pi}{2}$

(4) 若 $t\in[0,1]$, $t'\in[1,2]$, 则

$\begin{split} \frac{d(f(t),f(t'))}{d(t,t')}&=\frac{|(\pi t'-\frac{\pi}{2})-\frac{\pi}{2}t|}{|t-t'|}\\ &=\frac{|\frac{\pi}{2}(t'-t)+\frac{\pi}{2}t'-\frac{\pi}{2}|}{|t-t'|}\\ &=\frac{\frac{\pi}{2}(t'-t)+\frac{\pi}{2}(t'-1)}{t'-t}\\ &=\frac{\pi}{2}+\frac{\frac{\pi}{2}(t'-1)}{t'-t}\\ &< \frac{\pi}{2}+\frac{\frac{\pi}{2}(t'-1)}{t'-1}\\ &=\frac{\pi}{2}+\frac{\pi}{2}\\ &=\pi \end{split}$

$\sup_{t,t'\in[0,3]\\ t\neq t'}\frac{d(f(t),f(t'))}{d(t,t')}=\pi.$