Geometry: from Isometries to Special Relativity (Instructor Solution Manual, Solutions)

61 Pages·2021·0.389 MB·English

by  Nam-Hoon Lee

Instructor's Solutions
Manual
to accompany
Geometry:
from Isometries to Special Relativity
Nam-Hoon Lee
Undergraduate Texts in Mathematics
Springer, New York, 2020.
Preface
Thismanualcontainssolutionstoallexercisesinthetext,
Geometry:fromIsometriestoSpecialRelativity,
publishedbySpringerin2020.Eachexercisewhosesolutionisincludedinthetext
hasanunderlinednumbering.Pleasenote:forstudentstakingacoursebasedonthe
text, copying solutions from this Manual can place you in violation of academic
honesty.Theremaybestillsomeerrorsinthesolutions.Ifyouhaveanycomments,
corrections or feedback regarding the book and this solutions manual, feel free to
[email protected].
Nam-HoonLee
Contents
1 Euclidean Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
2 Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3 Stereographic Projection and Inversions . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4 Hyperbolic Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
5 Lorentz–Minkowski Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
6 Geometry of Special Relativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

Answers to Exercises
Chapter1
1.1
(a)Let p =0and p =(1,0).Then,d(p ,p )=1;however,
1 2 1 2
d(φ(p ),φ(p ))=2(cid:54)=d(p ,p ).
1 2 1 2
Hence,φ isnotanisometry.
(b)Let p =(0,1)and p =(−1,0).Then,d(p ,p )=2;however,
1 2 1 2
d(φ(p ),φ(p ))=0(cid:54)=d(p ,p ).
1 2 1 2
Hence,φ isnotanisometry.
(c)Fortwopoints p =(x ,y ),p =(x ,y )∈R2,
1 1 1 2 2 2
1 (cid:113)
d(φ(p ),φ(p ))= √ ((x −y )−(x −y ))2+((x +y +1)−(x +y +1))2
1 2 1 1 2 2 1 1 2 2
2
1 (cid:113)
= √ 2((x −x )2+(y −y )2)
1 2 1 2
2
(cid:113)
= (x −x )2+(y −y )2
1 2 1 2
=d(p ,p ).
1 2
Therefore,φ isanisometry.
(d)Fortwopoints p =(x ,y ),p =(x ,y )∈R2,
1 1 1 2 2 2
1
2 AnswerstoExercises
1(cid:113)
d(φ(p ),φ(p ))= ((3x +4y )−(3x +4y ))2+((4x −3y )−(4x −3y ))2
1 2 1 1 2 2 1 1 2 2
5
1(cid:113)
= 25((x −x )2+(y −y )2)
1 2 1 2
5
(cid:113)
= (x −x )2+(y −y )2
1 2 1 2
=d(p ,p ).
1 2
Therefore, φ is an isometry.
1.2 Note that a (cid:54)= 0 or b (cid:54)= 0. First assume that a (cid:54)= 0.
Let
(cid:16) c (cid:17)
p = a− ,b
1
a
and
(cid:16) c (cid:17)
p = −a− ,−b .
2
a
Then,for p=(x,y)∈R2,
p∈L ⇔d(p ,p)=d(p ,p)
p1,p2 1 2
⇔d(p ,p)2=d(p ,p)2
1 2
(cid:16) c(cid:17)2 (cid:16) c(cid:17)2
⇔ x−a+ +(y−b)2= x+a+ +(y+b)2
a a
(cid:16) c(cid:17) (cid:16) c(cid:17)2 (cid:16) c(cid:17) (cid:16) c(cid:17)2
⇔2 −a+ x+ −a+ −2by=2 a+ x+ a+ +2by
a a a a
⇔0=4ax+4by+4c
⇔0=ax+by+c.
Therefore,L=L .
p1,p2
Nowassumea=0.Thenb(cid:54)=0andthelineLisdefinedbytheequation
c
y=− .
b
Let
(cid:16) c (cid:17)
p = 0,− −1
1
b
and
(cid:16) c (cid:17)
p = 0,− +1 .
2
b
Then we have L = Lp1,p2 also in this case.
1.3 For an isometry φ and a circle
C={p∈R2|d(p,q)=r},
whereqisthecenterandristheradius,let
AnswerstoExercises 3
C(cid:48)={p∈R2|d(p,φ(q))=r},
acircleofradiusr,centeredatφ(q).
Wewillshowthat
C(cid:48)=φ(C).
p∈C(cid:48)⇔d(p,φ(q))=r
⇔d(φ−1(p),φ−1(φ(q)))=r
⇔d(φ−1(p),q))=r
⇔φ−1(p)∈C
⇔p∈φ(C).
Hence,C(cid:48)=φ(C).
1.4Ifφ(p)=φ(q)andφ isanisometry,then
d(φ(p),φ(q))=0,
andthus,d(p,q)=0,whichimplies p=q.
1.5 Ifa2+b2=1,thena=cosθ andb=sinθ forsomeθ.Hence,φ isarotation
aroundtheorigin.
Conversely,if
φ(x,y)=(ax−by,bx+ay)
isarotationaroundtheorigin,thena=cosθ andb=sinθ forsomeθ.Therefore,
a2+b2=1.
1.6 Note that
r =t ◦r ◦t .
(a,b),θ (a,b) θ −(a,b)
Hence,
r (x,y)=(t ◦r ◦t )(x,y)
(a,b),θ (a,b) θ −(a,b)
=(t ◦r )(x−a,y−b)
(a,b) θ
=t ((x−a)cosθ−(y−b)sinθ,(x−a)sinθ+(y−b)cosθ)
(a,b)
=(a+(x−a)cosθ−(y−b)sinθ,b+(x−a)sinθ+(y−b)cosθ).
1.7
1.Forall p,q∈R2,
d(id (p),id (q))=d(p,q).
R2 R2
Hence,theidentitymapisanisometry.
4 AnswerstoExercises
2.Letφ beanisometryofR2.Forall p,q∈R2,
d(φ−1(p),φ−1(q))=d(φ(φ−1(p)),φ(φ−1(q))))
=d(p,q).
Hence,φ−1isalsoanisometry.
3.Letφ andψ beisometriesofR2.Forall p,q∈R2,
d((φ◦ψ)(p),(φ◦ψ)(q))=d(φ(ψ(p)),φ(ψ(q)))
=d(ψ(p),ψ(q))
=d(p,q).
Hence,φ◦ψ isalsoanisometry.
1.8
(a)Foreach(x,y)∈R2,
(r¯◦r¯)(x,y)=r¯(x,−y)=(x,y).
Hence,r¯◦r¯=id ,whichmeansthat
R2
r¯−1=r¯.
(b)Foreach p∈R2,
(t ◦t )(p)=t (β+p)=α+β+p=t (p).
α β α α+β
Hence,
t ◦t =t .
α β α+β
Notethat
id =t =t =t ◦t .
R2 0 α+(−α) α −α
Therefore,
t−1=t .
α −α
(c)Foreach(x,y)∈R2,
(rθ◦rθ(cid:48))(x,y)=rθ(xcosθ(cid:48)−ysinθ(cid:48),xsinθ(cid:48)+ycosθ(cid:48))
=((xcosθ(cid:48)−ysinθ(cid:48))cosθ−(xsinθ(cid:48)+ycosθ(cid:48))sinθ,
(xcosθ(cid:48)−ysinθ(cid:48))sinθ+(xsinθ(cid:48)+ycosθ(cid:48))cosθ)
=(xcos(θ+θ(cid:48))−ysin(θ+θ(cid:48)),xsin(θ+θ(cid:48))+ycos(θ+θ(cid:48)))
=r (x,y).
θ+θ(cid:48)
Hence,rθ◦rθ(cid:48) =rθ+θ(cid:48).
AnswerstoExercises 5
Notethat
id =r =r =r ◦r .
R2 0 θ+(−θ) θ −θ
Therefore,r−1=r .
θ −θ
1.9
Letβ beapointonthelinethrough pandqandR =d(p,β),R =d(q,β).Let
p q
C andC becirclesofradiiR andR ,centeredat pandq,respectively.Notethat
p q p q
C ∩C ={β}.
p q
From the solution of Exercise 1.3 and using the condition that p and q are fixed
pointsofanisometryφ,
φ(C )=C andφ(C )=C .
p p q q
Hence,
{φ(β)}=φ({β})=φ(C ∩C )=φ(C )∩φ(C )=C ∩C ={β},
p q p q p q
andtherefore,φ(β)=β.Weshowedthateverypointonthelinethrough pandqis
afixedpointofφ.
1.10
Thegivenconditionimpliesthatr¯−1=r¯ .Notethatr¯−1=r¯ .Hence,
L1 L2 L1 L1
r¯ =r¯ ,
L1 L2
whichimpliesthatL =L .
1 2
1.11 Choose non-collinear points p1, p2 and p3. The isometry φ maps the triangle
(cid:52)p1 p2 p3 to a congruent triangle
(cid:52)φ(p )φ(p )φ(p ),
1 2 3
whichimpliesthatthepointsφ(p ),φ(p )andφ(p )arealsonon-collinear.Asin
1 2 3
the proof of Theorem 1.6, by composing reflections, we can build an isometry ψ
suchthat
φ(p )=ψ(p ),φ(p )=ψ(p )andφ(p )=ψ(p ).
1 1 2 2 3 3
UsingargumentssimilartothoseintheproofofTheorem1.4,weshowthatφ =ψ.
Sinceψ isbijective,φ isalsobijective.
1.12
(a)
6 AnswerstoExercises
(φ◦ψ)ξ =(φ◦ψ)◦ξ◦(φ◦ψ)−1
=φ◦ψ◦ξ◦ψ−1◦φ−1
=φ◦(ψ◦ξ◦ψ−1)◦φ−1
=φ◦(ψξ)◦φ−1
= φ(ψξ)
and
φ(ψ◦ξ)=φ◦(ψ◦ξ)◦φ−1
=φ◦ψ◦φ−1◦φ◦ξ◦φ−1
=(cid:0)φ◦ψ◦φ−1(cid:1)◦(cid:0)φ◦ξ◦φ−1(cid:1)
=(cid:0)φψ(cid:1)◦(cid:0)φξ(cid:1).
(b)Choosetwopointsp1,p2∈L(cid:48)=r¯M(L).Then,r¯M(pi)∈L.Notethatr¯L(cid:48)(pi)=
p.However,
i
r¯Mr¯ (p)=(r¯ ◦r¯ ◦r¯−1)(p)
L i M L M i
=(r¯ ◦r¯ ◦r¯ )(p)
M L M i
=(r¯ ◦r¯ )(r¯ (p))
M L M i
=r¯ (r¯ (r¯ (p)))
M L M i
=r¯ (r¯ (p))
M M i
=p
i
fori=1,2.Chooseanotherpoint p3thatdoesnotbelongtoL(cid:48);then,L(cid:48)=Lp3,p(cid:48)3 for
somepoint p(cid:48).Notethat
3
L=(r¯M)−1(L(cid:48))=r¯M(L(cid:48))=r¯M(Lp3,p(cid:48)3)=Lr¯M(p3),r¯M(p(cid:48)3).
Clearly,r¯L(cid:48)(p3)=p(cid:48)3.Notethat
r¯Mr¯ (p )=(r¯ ◦r¯ ◦r¯ )(p )
L 3 M L M 3
=r¯ (r¯ (r¯ (p )))
M L M 3
=r¯ (r¯ (p(cid:48)))
M M 3
=p(cid:48).
3
Insum,
r¯L(cid:48)(pi)=r¯Mr¯L(pi)
fori=1,2,3,and p ,p and p arenon-collinear.AccordingtoTheorem1.4,
1 2 3
r¯L(cid:48) =r¯Mr¯L.