Answers to Exercises

Solutions Manual to accompany

Geometry:

from Isometries to Special Relativity Nam-Hoon Lee 1 / 4

Answers to Exercises Chapter 1 1.1 (a) Letp1=0andp2= (1;0). Then,d(p1;p2) =1; however,

d(f(p1);f(p2)) =26=d(p1;p2):

Hence,fis not an isometry.(b) Letp1= (0;1)andp2= (1;0). Then,d(p1;p2) =2; however,

d(f(p1);f(p2)) =06=d(p1;p2):

Hence,fis not an isometry.(c) For two pointsp1= (x1;y1);p2= (x2;y2)2R 2 , d(f(p1);f(p2)) = 1 p 2 q ((x1y1)(x2y2)) 2

• ((x1+y1+1)(x2+y2+1))

2 = 1 p 2 q 2((x1x2) 2

• (y1y2)

2 ) = q (x1x2) 2

• (y1y2)

2

=d(p1;p2):

Therefore,fis an isometry.(d) For two pointsp1= (x1;y1);p2= (x2;y2)2R 2 ,

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• Answers to Exercises

d(f(p1);f(p2)) = 1 5 q ((3x1+4y1)(3x2+4y2)) 2

• ((4x13y1)(4x23y2))

2 = 1 5 q 25((x1x2) 2

• (y1y2)

2 ) = q (x1x2) 2

• (y1y2)

2

=d(p1;p2):

Therefore, f is an isometry.

1.2 Note that a 6= 0 or b 6= 0. First assume that a 6= 0.Let p1=

a c a ;b

and p2=

a c a ;b

:

Then, forp= (x;y)2R 2 , p2Lp1;p2 ,d(p1;p) =d(p2;p) ,d(p1;p) 2 =d(p2;p) 2 ,

xa+ c a

2

• (yb)

2 =

x+a+ c a

2

• (y+b)

2 ,2

a+ c a

x+

a+ c a

2 2by=2

a+ c a

x+

a+ c a

2 +2by ,0=4ax+4by+4c

,0=ax+by+c:

Therefore,L=Lp1;p2 .Now assumea=0. Thenb6=0 and the lineLis dened by the equation y= c b

:

Let p1=

0; c b 1

and p2=

0; c b +1

:

Then we have L = Lp1;p2

also in this case.

1.3 For an isometry f and a circle C=fp2R 2 jd(p ;q) =rg ; whereqis the center andris the radius, let 3 / 4

Answers to Exercises3 C

=fp2R 2 jd(p;f(q)) =rg; a circle of radiusr, centered atf(q).We will show that C

=f(C):

p2C

,d(p;f(q)) =r ,d(f 1 (p);f 1 (f(q))) =r ,d(f 1 (p);q)) =r ,f 1 (p)2C

,p2f(C):

Hence,C

=f(C):

1.4Iff(p) =f(q)andfis an isometry, then d(f(p);f(q)) =0; and thus,d(p;q) =0, which impliesp=q.

1.5Ifa 2 +b 2 =1, thena=cosqandb=sinqfor someq. Hence,fis a rotation around the origin.Conversely, if f(x;y) = (axby;bx+ay) is a rotation around the origin, thena=cosqandb=sinqfor someq. Therefore, a 2 +b 2 =1.

1.6 Note that r (a;b);q=t (a;b)rqt

(a;b):

Hence, r (a;b);q(x;y) = (t (a;b)rqt (a;b))(x;y) = (t (a;b)rq)(xa;yb) =t (a;b)((xa)cosq(yb)sinq;(xa)sinq+ (yb)cosq)

= (a+ (xa)cosq(yb)sinq;b+ (xa)sinq+ (yb)cosq):

1.7

• For allp;q2R

2 , d(id R 2(p);id R

2(q)) =d(p;q):

Hence, the identity map is an isometry.

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