几何学 - 作业 3

T1

Choose any $\vec{v}\in {\overrightarrow{\mathbb{R}}}^2$ , (a) Show that given any $f\in \mathrm{Isom}{\mathbb{R}}^2$ , the composition $f\circ T_{\vec{v}}\circ f^{-1}$ is a translation. Find the translation vector associated to this translation.

(b) Draw one example and use it to explain the statement in the first question.

Proof

(a) Without loss of generality, let $f$ be an isometry that fix $O\in {\mathbb{R}}^2$ . (If not we can let $f=T_{\vec{v}}\circ f^{\prime }$ and $f^{\prime }$ will be the isometry that fix $O$). Then $f$ will be a linear map.

Thus,

$$f[(T_{\vec{v}}\circ f^{-1})(x)]=f[f^{-1}(x)+\vec{v}]=x+\vec{v}$$

Then the composition is a translation and translation vector will be $\vec{v}$ .

(b) See the example below. Consider the $30^{\circ }$ rotation by clockwise, trainslation $\vec{v}=(0,1{)}^{\mathrm{T}}$ .

T2

Show that all isometries of ${\mathbb{R}}^2$ can be written as compositions of reflections.

Proof

It is sufficient to prove the cases when isometries are translations and rotations.

When $f\in \mathrm{Isom}{\mathbb{R}}^2$ is a translation, consider $f(x)=x+v$ . Let $A$ be the reflection matrix which satisfies $Av=-v$ . Then

$$f_1(x)=Ax+v,f_2(x)=Ax$$

are all reflections. Then

$$f_1\circ f_2(x)=A(Ax)+v=x+v=f$$

When $f\in \mathrm{Isom}{\mathbb{R}}^2$ is a rotation, say

$$r(x)=B(x-p)+p$$

where $B$ is a rotation matrix and $p\in {\mathbb{R}}^2$ . Then we have

$$r(x)=t\circ h\circ t^{-1}$$

where $t(x)=x+p$ is a translation and $h$ is a rotation around $O$ . Then we only need to consider $h$ can be written as compositions of reflections. Let $B^{\prime }$ be any reflection matrix and we have

$$g_1(x)=B{B}^{\prime }x,g_2(x)=B^{\prime }x$$

Both of them are reflections. And $g_1\circ g_2(x)=Bx$ , this means $h$ can be written as composition of two reflections. $\square$

T3

Let $\underline{x}\in {\mathbb{R}}^2$ and $r\in {\mathbb{R}}_{>0}$ . Let $\underline{y}\in {\mathbb{R}}^2$ be such that $\underline{x}\ne \underline{y}$ . (a) Show that the distance between $\underline{y}$ and $C_{\underline{x},r}$ the circle centered at $\underline{x}$ of radius $r$ can be realized by a unique point $\underline{z}$ in $C_{\underline{x},r}$ i.e. there is a unique point $\underline{z}\in C_{\underline{x},r}$ such that

$$d_{\mathbb{E}}(\underline{y},\underline{z})=d_{\mathbb{E}}(\underline{y},C_{\underline{x},r})$$

(b) Show that the points $\underline{x},\underline{y},\underline{z}$ are collinear, i.e. on a same line in ${\mathbb{R}}^2$ .

Proof

(a) Without generality, let $\underline{x}=0$ (By translation).

Consider

$$f(z)=d_{\mathbb{E}}(\underline{y},z),z\in C_{\underline{x},r}$$

This is a continuous function and $C_{\underline{x},r}$ is a closed set. Then $f(z)$ is bounded and has its minimum value on $C_{\underline{x},r}$ . We can denote that

$$f(\underline{z})=\underset{z\in C_{\underline{x},r}}{\min }f(z)$$

About uniqueness we can consider

$$\begin{array}{rl}{f}^2(z)& =d_{\mathbb{E}}^2(\underline{y},z)=(y_1-z_1{)}^2+(y_2-z_2{)}^2\\ & =(y_1-r\cos \theta {)}^2+(y_2-r\sin \theta {)}^2\\ & =r^2+y_1^2+y_2^2-2r{y}_1\cos \theta -2r{y}_2\sin \theta \\ & =C-2r\frac{1}{\sqrt{y_1^2+y_2^2}}\sin (\theta +\gamma ),\theta \in [0,2\pi ]\end{array}$$

where $z=(z_1,z_2)=(r\cos \theta ,r\sin \theta )$ . We can see $f(z)$ reaches its minimum if and only if $\theta +\gamma =\frac{\pi }{2}$ , which make $\theta$ unique.

(b) By triangle inequality we can get

$$d_{\mathbb{E}}(\underline{y},\underline{z})⩽d_{\mathbb{E}}(\underline{y},\underline{z})+d_{\mathbb{E}}(\underline{z},\underline{x})$$

when equality holds we can get they are collinear. $\square$

T4

Let $f,g\in \mathrm{Isom}{\mathbb{R}}^2$ . Let $\underline{x},\underline{y}$ and $\underline{z}$ be three pairwise distinct points in ${\mathbb{R}}^2$ which are not collinear. Show the equivalence between the following two statements:

(a) We have $f=g$ .

(b) We have $f(\underline{x})=g(\underline{x}),f(\underline{y})=g(\underline{y})$ and $f(\underline{z})=g(\underline{z})$ at the same time.

Proof

It is sufficient to prove (b) $\Rightarrow$ (a) .

We have $f(\underline{x})=g(\underline{x}),f(\underline{y})=g(\underline{y})$ then we can say $f$ and $g$ has the same effect to the basis in ${\mathbb{R}}^2$ . If $f$ and $g$ fix the origin $O$ we say they are identical.

If they don’t fix the origin $O$ then we can apply same translation $T_{\vec{\underline{z}}}$ to make them fix $O$ since they have $f(\underline{z})=g(\underline{z})$ . Then we reduce the problem to the case above. $\square$