CONFIDENTIAL EXAM PAPER
This paper is not to be removed from the exam venue
Mathematics and Statistics
EXAMINATION
Semester 2 - Main, 2018
MATH3061 Geometry and Topology
EXAM WRITING TIME: 2 hours
READING TIME: 10 minutes
EXAM CONDITIONS:
This is a CLOSED book examination - no material permitted
During reading time - writing is not permitted at all
MATERIALS PERMITTED IN THE EXAM VENUE:
(No electronic aids are permitted e.g. laptops, phones)
Calculator - non-programmable
MATERIALS TO BE SUPPLIED TO STUDENTS:
2 x 12-page answer book
INSTRUCTIONS TO STUDENTS:
Please answer the questions for the Geometry and Topology components of the
course in separate booklets.
Please tick the box to confirm that your examination paper is complete.
Room Number ________
Seat Number ________
Student Number |__|__|__|__|__|__|__|__|__|
ANONYMOUSLY MARKED
(Please do not write your name on this exam paper)
For Examiner Use Only
Q Mark
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
Total ________
Page 1 of 6
Page 2 of 6 MATH3061 EXAM, SEMESTER 2 Semester 2, 2018
I. GEOMETRY — please answer in a separate booklet
1. a) Let 퐴퐵퐶 be a right triangle with 퐴퐵 perpendicular to 퐵퐶 , and let 휃 denote the angle 퐶퐴퐵. Draw
the diagram, and show that 푐표푠(휃) = |⃖⃖⃖⃗퐴퐵||⃖⃖⃖⃗퐴퐶| .b) Show that for any two points 푃 ,푄 in  and any number 휃, 휌푃 ,휃휌푄,−휃 is a translation.c) Given a vector 퐯 show that 휏퐯 is the composition of two rotations.
Solution
a) Since |⃖⃖⃖⃖⃖⃗퐵퐴|2 = ⃖⃖⃖⃖⃖⃗퐵퐴 ⋅ ⃖⃖⃖⃖⃖⃗퐵퐴 = ⃖⃖⃖⃖⃖⃗퐵퐴 ⋅ ( ⃖⃖⃖⃖⃖⃗퐵퐶 + ⃖⃖⃖⃖⃖⃗퐶퐴) = ⃖⃖⃖⃖⃖⃗퐵퐴 ⋅ ⃖⃖⃖⃖⃖⃗퐶퐴 = ⃖⃖⃖⃖⃖⃗퐴퐵 ⋅ ⃖⃖⃖⃖⃖⃗퐴퐶 = |⃖⃖⃖⃖⃖⃗퐴퐵|| ⃖⃖⃖⃖⃖⃗퐴퐶| cos(휃),
푐표푠(휃) = |⃖⃖⃖⃗퐴퐵||⃖⃖⃖⃗퐴퐶| .
b) The derivative of 휌푃 ,휃휌푄,−휃 is equal to 푅휃푅−휃 = 푅0 =
[
1 0
0 1
]
. Therefore, the composition is a
translation.
c) From class we know that 휏퐯 = 휎푙휎푚 for a pair of suitable parallel lines 푙, 푚. Let 푛 be a line notparallel to 푙, 푚, then from class we know that 휎푙휎푛 and 휎푛휎푚 are rotations. Therefore 휏퐯 = 휎푙휎푛휎푛휎푚is the composition of two rotations.
2. a) Let Φ = {(푥 ∶ 푦 ∶ 푧)|(2푥 − 푦 + 3푧)(푥 + 푧) = 0}.
i) Classify Φ as one of the 5 types of conics we studied.
ii) Find Φ ∩ 푙∞.b) Let Φ = {(푥 ∶ 푦 ∶ 푧)|푥2 + 2푦푧 + 푦2 = 0}, Φ0 = {(푥 ∶ 푦 ∶ 푧)|푥푦 = 푧2}. Let 휓푀 be a collineationwith 휓푀 (1 ∶ 0 ∶ 0) = (0 ∶ 0 ∶ 1), 휓푀 (0 ∶ 1 ∶ 0) = (1 ∶ −1 ∶ 1), 휓푀 (1 ∶ 1 ∶ 1) = (0 ∶ −2 ∶ 1).i) Find a matrix푀 such that 휓푀 (Φ0) = Φ.ii) Find Φ ∩ 푙∞ and classify Φ according to its intersection with  ′ (i.e. is it an ellipse orparabola or hyperbola).
Solution
a) i) It’s the union of two lines. The lines ⟨2 ∶ −1 ∶ 3⟩ and ⟨1 ∶ 0 ∶ 1⟩.
ii) We substitute 푧 = 0 in both equations to obtain points (1 ∶ 2 ∶ 0) and (0 ∶ 1 ∶ 0).
b) i) The point (1 ∶ 0 ∶ 1) is the intersection of the tangent lines toΦ at (0 ∶ 0 ∶ 1) and (1 ∶ −1 ∶
1) and therefore is equal to 휓푀 (0 ∶ 0 ∶ 1). Therefore we can assume푀 =
⎡⎢⎢⎣
0 휇 휈
0 −휇 0
휆 휇 휈
⎤⎥⎥⎦
and
⎡⎢⎢⎣
0 휇 휈
0 −휇 0
휆 휇 휈
⎤⎥⎥⎦
⎡⎢⎢⎣
1
1
1
⎤⎥⎥⎦ =
⎡⎢⎢⎣
0
−2푘
푘
⎤⎥⎥⎦. We can choose 휇 = 2, 휇 = −휈, 휆 + 휇 + 휈 = 1. We obtain
푀 =
⎡⎢⎢⎣
0 2 −2
0 −2 0
1 2 −2
⎤⎥⎥⎦.
ii) Substituting 푧 = 0 produces 푥2 + 푦2 = 0, no solution in  . Therefore it’s an ellipse.
3. a) Let 훼 = 휏퐯훼푃 ,푀 , where 퐯 =
[
−4
3
]
, 푃 = (1, 2),푀 =
[
1 1
−2 3
]
. (Recall, 훼푃 ,푀 is the unique affine
trasnformation fixing 푃 with derivative푀 .)
i) Find the formula for 훼(푥, 푦).
ii) Is 훼 an isometry?
b) Let Φ1 and Φ2 be two conics which are hyperbolas in  ′. Is there always an affine transformationof  ′ mapping  ′ ∩ Φ1 to  ′ ∩ Φ2? Explain.
Solution
a) i) 훼(푥, 푦) = (푀((푥, 푦)−푃 )+푃 )+퐯 =
[
1 1
−2 3
] [
푥 − 1
푦 − 2
]
+(1, 2)+
[
−4
3
]
=
[
푥 + 푦 − 3
−2푥 + 3푦 − 4
]
+
(1, 2) +
[
−4
3
]
= (푥 + 푦 − 2,−2푥 + 3푦 − 2) +
[
−4
3
]
= (푥 + 푦 − 6,−2푥 + 3푦 + 1).
ii) No, because푀푇푀 is not the identity matrix.
b) Yes. A conic is a hyperbola means it is non-degenerate and its intersection with 푙∞ comprises twopoints. We can find a collineation mapping Φ1 to Φ0 (from previous question) sending points atinfinity to (1 ∶ 0 ∶ 0) and (0 ∶ 1 ∶ 0). Similarly for Φ2. Therefore both of these collineations fix
푙∞ setwise. And from class we know such collineations restrict to affine transformations of  ′.
Page 4 of 6 MATH3061 EXAM, SEMESTER 2 Semester 2, 2018
II. TOPOLOGY — please answer in a separate booklet
4. Give examples of the following. You do not need to justify your answers.
a) A graph with Euler characteristic 5
b) List all of the trees with 4 vertices
c) A surface with Euler characteristic −5
d) A non-planar connected graph with Euler characteristic −7
e) A Eulerian graph
f) A closed surface that is not orientable
g) A graph that does not have a spanning tree
h) A map that cannot be 3-coloured
i) A knot that is 3-colourable
j) The polygonal decomposition of a surface with one boundary circle and Euler characteristic −1.
Solution
a) Let 퐺 be the (disconnected) graph with 5 vertices and no edges.
b) There are two trees with four vertices:
and
c) There are many, such as #7픻2 or #7ℙ
d) There are many examples. By lectures we know that퐾5 is not planar. Since 휒(퐾5) = 5−
(5
2
)
= −5
we can obtain a non-planar graph of Euler characteristic −7 by adding two edges to 퐾5:
e) There are many examples, such as the cycle graphs 퐶푛 and the complete graphs 퐾푛.f) There are many examples such as ℙ and 핂.
g) Take any disconnected graph.
h) Again there are many, but probably the simplest example is:
i) By lectures the (reverse) trefoil is 3-colourable
j) The surfaces with one boundary circle are of the form 픻2 ##푡핋 and 픻2 ##푝ℙ, which has Euler
characteristic 1−2푡 and 1−푝, respectively. Hence, the possible surfaces are 픻2 # 핋 and 픻2 ##2ℙ.
5. a) Draw polygons for the standard surfaces 픸, 핂, ℙ and 핋 .
b) Let 푆 be the surface given by the word 푎푏푐푎푏.
i) Draw a polygonal decomposition of 푆.
ii) Use surgery to write 푆 as a standard surface.
(To receive marks for this part you must use surgery.)
c) Let 푇 be the surface given by the word 푎푏푐푑푓푏푑푒푐 푒.
i) Is 푇 orientable? Justify your answer.
ii) Draw a polygonal decomposition of 푇
iii) How many vertices and rim circles does your polygonal decomposition of 푇 have?
iv) Compute the Euler characteristic of 푇
v) Describe 푇 as a standard surface
Solution
a) There are many ways to do this. Here is one of them:
푎
푏
푐
푏
푎
푏
푎
푏
푎
푏
푎
푏
푎
푏
푎
푏
픸 핂 ℙ 핋
b) i) A polygonal decomposition of 푎푏푐푎푏 is
푎
푏푐
푎
푏ii) We want to use surgery to glue along the unoriented edge 푎:
푎
푏푐
푎
푏
푑
≅
푑
푏푐
푏
푑
푎
≅ 픻2 #
푏
푏
푑
푑 ≅ 픻
2 ##2ℙ.
c) i) The surface 푇 is not orientable because 푏 is an unoriented edge.
ii) A polygonal decomposition of 푇 is
푎
푏
푐
푑
푓
푏
푑
푒
푐
푒푥
푥
푥
푦 푥
푥
푥
푥
푥푦
We have labelled the vertices in the polygon for part (iii).
iii) The labelling of the vertices in (ii) shows that the polygon has 2 vertices. The only unpaired
edges are 푎 and 푓 and, taken together, these give one boundary circle from 푥 → 푦 → 푥.
Hence, 푇 has only one boundary circle.
iv) By part (iii), 푇 has 2 vertices, 6 edges and one face, so 휒(푇 ) = 2 − 6 + 1 = −3.
v) Since 푇 is unoriented with one boundary circle 푇 = 픻2 ##푝ℙ, for some 푝. Therefore,
−3 = 휒(푇 ) = 휒(픻2 ##푝ℙ) = 1 − 푝,
so that 푝 = 4. Hence, 푇 = 픻2 ##4ℙ is a punctured sphere with 4 cross caps.
6. a) A polygonal decomposition of the double torus #2핋 is made by gluing together squares so that
푝 edges meet at each vertex. (Note that 휒(#2핋 ) = −2.) Suppose that this decomposition has 푣
vertices, 푒 edges and 푓 faces.
i) Explain why 푝푣 = 2푒 = 4푓 .
ii) Show that 푝 must be even and that 푝 > 4.
b) Compute the knot determinant of the following knot:
Page 6 of 6 MATH3061 EXAM, SEMESTER 2 Semester 2, 2018
푐4 푐5
푐1
푐2
푐3
using the given labelling of the arcs by 푐1, 푐2, 푐3, 푐4 and 푐5. Hence, determine whether or not theknot is 3-colourable.
Solution
a) i) Consider the degree sum ∑푥∈푉 deg 푥, where 푉 is the set of vertices. As each vertex hasdegree 푝, this sum is 푝푣. Similarly, since each edge has 2 vertices the sum is 2푒 and, since
as each face has 4 vertices, the sum is also equal to 4푓 . Hence,∑
푥∈푉
deg 푥 = 푝푣 = 2푒 = 4푓.
ii) Since this surface is a polygonal decomposition of #2핋 and 휒(#2핋 ) = −2. Therefore, by
part (a),
−2 = 푣 − 푒 + 푓 = 푣 − 푝푣
2
+ 푝푣
4
= 푣
(
1 − 푝
4
)
.
Rearranging, 푝
4
− 푣
2
= 1, so that 푝 = 2푣 + 4 Consequently, 푝 is even. Moreover, since 푣 ⩾ 1,
푝 > 4. for this
b) We should first observe that this knot is alternating. Therefore, using the given labelling of the
arcs, the knot matrix for this knot is [
2 0 −1 −1 0
0 2 0 −1 −1
−1 0 2 0 −1
−1 −1 0 2 0
0 −1 −1 0 2
]
.
Therefore, taking the (5, 5)-minor the knot determinant is|||| 2 0 −1 −10 2 0 −1−1 0 2 0−1 −1 0 2 |||| 퐶2+2퐶4= |||| 2 −2 −1 −10 0 0 −1−1 0 2 0−1 3 0 2 |||| = − ||| 2 −2 −1−1 0 2−1 3 0 ||| 퐶3+2퐶1= − ||| 2 −2 3−1 0 0−1 3 −2 ||| = 4 − 9 = −5.
Hence, the knot determinant is 5, so this knot is not 3-colourable.
This is the last page of the exam paper.

欢迎咨询51作业君