Download Geometry in S2: Lines, Reflections, and Isometries - Prof. Darryl Cullough and more Exams Mathematics in PDF only on Docsity!
Mathematics 4513-
Final Examination
December 19, 2008
Name (please print)
Instructions: Give brief, clear answers. If asked for a definition, give the definition that we have used in this
course. In some of the problems, you will need to calculate using the formula ΩℓX = X − 2 〈X, P 〉P for reflection
of S
2 across the line with pole P.
I.
(i) Define a line in R^2.
A line is a set of the form P + [v], where P and v are vectors with v 6 = 0, and [v] = {λv | λ ∈ R}.
(ii) Using the notation of the definition, write an expression for the line thorough (1, 0) and (3, 5).
For the line through (1, 0) and (3, 5), we can take P = (1, 0) and v = (3, 5) − (1, 0) = (2, 5), so the line
can be written as (1, 0) + [(2, 5)].
(iii) Find a unit normal to the line in (ii).
The direction vector is (2, 5), so a normal vector is (2, 5)
⊥ = (− 5 , 2), and a unit normal is
II.
Let X be a set with a distance function d : X × X → X. Let f and g be functions from X to X.
(i) Define what it means to say that f is an isometry of X.
f is an isometry when d(f (x), f (y)) = d(x, y) for all x, y ∈ X.
(ii) Prove that if f and g are isometries, then their composition f g is also an isometry.
For all x, y ∈ X, d(f g(x), f g(y)) = d(g(x), g(y)) = d(x, y).
III.
Recall that for a, b ∈ R
3 we defined the cross product a × b to be the unique vector in R
3 such that
〈a × b, x〉 = det(x, a, b), where det(x, a, b) is the determinant of the matrix whose rows are x, a, and b. Use
this definition to verify the following facts about the cross product. You may use standard properties of
the determinant, and may use the fact that if 〈a, x〉 = 〈b, x〉 for all x, then a = b.
(i) a × b = −b × a
For all x, 〈a × b, x〉 = det(x, a, b) = − det(x, b, a) = −〈b × a, x〉 = 〈−b × a, x〉, so a × b = −b × a.
(ii) 〈a × b, c〉 = 〈a, b × c〉
〈a × b, c〉 = det(c, a, b) = − det(a, c, b) = det(a, b, c) = 〈b × c, a〉 = 〈a, b × c〉.
(iii) If a = (a 1 , a 2 , a 3 ) and b = (b 1 , b 2 , b 3 ), then a × b = (a 2 b 3 − a 3 b 2 , a 3 b 1 − a 1 b 3 , a 1 b 2 − a 2 b 1 ).
For all x = (x 1 , x 2 , x 3 ),
〈a × b, x〉 = det
x 1 x 2 x 3
a 1 a 2 a 3
b 1 b 2 b 3
= x 1 (a 2 b 3 − a 3 b 2 ) − x 2 (a 1 b 3 − a 3 b 1 ) + x 3 (a 1 b 2 − a 2 b 1 )
= 〈(a 2 b 3 − a 3 b 2 , a 3 b 1 − a 1 b 3 , a 1 b 2 − a 2 b 1 ), (x 1 , x 2 , x 3 )〉
for all x, so a × b = (a 2 b 3 − a 3 b 2 , a 3 b 1 − a 1 b 3 , a 1 b 2 − a 2 b 1 ).
Page 2
IV.
Recall that a line in S
2 with pole P is defined to be the set {X ∈ S
2 | 〈X, P 〉 = 0}. Let P and Q be
distinct points in S
2 with P 6 = −Q.
(i) Give an expression (in terms of P and Q) for a pole of the line ℓ that contains P and Q.
The pole would be a unit vector perpendicular to P and Q, so we can take P × Q/|P × Q|.
(ii) Give an expression (in terms of P and Q) for a pole of the line perpendicular to ℓ that contains Q.
The pole would be perpendicular to the pole P × Q/|P × Q| of ℓ, and hence perpendicular to P × Q,
and would also be perpendicular to Q since the line contains Q. So a pole would be
(P × Q) × Q
|(P × Q) × Q|
V.
Let G be the group whose elements are points in the plane R
2 , and whose operation is the usual vector
addition, (x 1 , y 1 ) + (x 2 , y 2 ) = (x 1 + y 1 , x 2 + y 2 ). Define Φ : G → G by Φ(X) = (3x, − 2 y), where X = (x, y).
Verify that Φ is an isomorphism (that is, verify that Φ is injective, surjective, and satisfies Φ(X 1 + X 2 ) =
Φ(X 1 ) + Φ(X 2 )).
Injectivity: Suppose that Φ((x 1 , y 1 )) = Φ((x 2 , y 2 )). Then (3x 1 , − 2 y 1 ) = (3x 2 , − 2 y 2 ). So 3x 1 = 3x 2
and − 2 y 1 = − 2 y 2 , giving x 1 = x 2 and y 1 = y 2. So (x 1 , y 1 ) = (x 2 , y 2 ).
Surjectivity: Let (x, y) ∈ G. Then Ψ(x/ 3 , −y/2) = (x, y).
Homomorphism: Φ(X 1 + X 2 ) = Φ((x 1 , y 1 ) + (x 2 , y 2 )) = Φ((x 1 + x 2 , y 1 + y 2 ))
= (3(x 1 + x 2 ), −2(y 1 + y 2 )) = (3x 1 , − 2 y 1 ) + (3x 2 , − 2 y 2 ) = Φ(X 1 ) + Φ(X 2 ).
VI.
(i) Calculate the determinants of the matrices rot(θ) and ref(θ).
det(rot(θ)) = det
cos(θ) − sin(θ)
sin(θ) cos(θ)
= cos
2 (θ) − (− sin
2 (θ)) = 1
det(ref(θ)) = det
cos(2θ) sin(2θ)
sin(2θ) − cos(2θ)
= − cos
2 (2θ) − sin
2 (2θ)) = − 1
(ii) Explain geometrically why these are the values one would expect for these determinants.
Both of these linear transformations are isometries, so they preserve area, and therefore their determi-
nants should have absolute value 1. Since rot(θ) preserves the sense but ref(θ) reverses it, we expect
det(rot(θ)) = 1 and det(ref(θ)) = −1.
Page 4
X.
Let α, β, and γ be lines in S
2 with α and γ both perpendicular to β.
(i) Sketch a picture of these three lines in S^2 , oriented so that the intersection points of α and γ appear as the
north and south poles.
β
α γ
(ii) Suppose that {e 1 , e 2 , e 3 } is an orthonormal basis for which e 3 is an intersection point of α and γ, say the
north pole in part (i). Write the general form of the matrices of Ωα, Ωβ , and Ωγ with respect to this basis
(you do not need to do any calculations, just use your knowledge of how the matrices of these reflections
look for this kind of basis). It is a good idea to write them in block form, so that even though they are 3 × 3
matrices, they appear visually as 2 × 2 matrices.
Ωα =
ref(θ)
, Ωβ =
I
, and Ωγ =
ref(φ)
where I is the 2 × 2 identity matrix.
(iii) By considering the product of the three matrices in part (ii), show that the composition ΩαΩβ Ωγ is a glide
reflection.
Multiplying the three matrices gives
ref(θ) ref(φ)
rot(2(θ − φ))
which is the matrix of a glide reflection. Specifically, it is the product
rot(2(θ − φ))
I
of a rotation about e 3 through the angle 2(θ − φ) and a reflection fixing e 1 and e 2 and sending e 3 to
−e 3 , i. e. the reflection Ωβ.
XI.
Let {E 1 , E 2 , E 3 } be the standard basis of R
3 , and let J be an isometry of S
2
. Use the Orthonormal Basis
Theorem to express J(X) in terms of E 1 , E 2 , and E 3.
JX = 〈JX, E 1 〉E 1 + 〈JX, E 2 〉E 2 + 〈JX, E 3 〉E 3
A few people used the fact that isometries of S
2 are linear to give another response, which was also
worth full credit:
JX = J(〈X, E 1 〉E 1 + 〈X, E 2 〉E 2 + 〈X, E 3 〉E 3 ) = 〈X, E 1 〉JE 1 + 〈X, E 2 〉JE 2 + 〈X, E 3 〉JE 3
Page 5 Name (please print)
XII.
Let P , Q, and R be points in S
2 , with P 6 = −Q. Explain why P , Q, and R are collinear exactly when
〈R, P × Q〉 = 0.
A pole of the line containing P and Q is
P × Q
|P × Q|
. R lies in the line if and only if it is orthogonal to
the pole, that is, if and only if 0 =
R,
P × Q
|P × Q|
|P × Q|
〈R, P × Q〉. Since |P × Q| is a nonzero
scalar, the latter is equivalent to 〈R, P × Q〉 = 0.
XIII.
Let G be the group of isometries of S
2 , and let H be the subgroup of G consisting of the isometries that
take the point (0, 0 , 1) to itself, that is, the isometries J of S
2 with J((0, 0 , 1)) = (0, 0 , 1).
(i) Tell what one would need to do to prove that H is not normal in G.
One would need to find an isometry J of S
2 with J ∈ H, and another isometry K of S
2 so that the
composition KJK
− 1 is not in H. That is, J(0, 0 , 1) = (0, 0 , 1) but KJK
− 1 (0, 0 , 1) 6 = (0, 0 , 1).
(ii) Prove that H is not normal in G. There are many ways to do this, here are three (the second and third are
due to students in our class who did this problem, good work!):
Solution 1: Let ℓ be the equator, so ℓ has pole (0, 0 , 1), and Ωℓ ∈ H. Let e be the line of points
equidistant from (1, 0 , 0) and (0, 0 , 1); a pole of e is
We know that Ωe interchanges (1, 0 , 0) and (0, 0 , 1), and Ωℓ fixes every point on the equator ℓ, so we
have
ΩeΩℓΩe(0, 0 , 1) = ΩeΩℓ(1, 0 , 0) = Ωe(1, 0 , 0) = (0, 0 , 1)
and therefore ΩeΩℓΩe ∈ H. But
Ωe(ΩeΩℓΩe)Ω
− 1 e = Ω
2 eΩℓ = Ωℓ ,
and Ωℓ(0, 0 , 1) = (0, 0 , −1) so Ωℓ ∈/ H. Therefore H is not a normal subgroup of G.
Solution 2: Let R be a rotation whose only fixed points are (0, 0 , ±1), and let m be any line such that
Ωm(0, 0 , 1) 6 = (0, 0 , ±1) (that is, any line except the equator). Then R ∈ H. To show that ΩmRΩ
− 1 m ∈/^ H,
suppose for contradiction that that ΩmRΩ
− 1 m ∈^ H. Using the fact that Ωm^ = Ω
− 1 m , we find
ΩmRΩ
− 1 m (0,^0 ,^ 1) = (0,^0 ,^ 1)
− 1 m RΩm(0,^0 ,^ 1) = (0,^0 ,^ 1)
R(Ωm(0, 0 , 1)) = Ωm(0, 0 , 1)
Therefore Ωm(0, 0 , 1) is a fixed point of R, so Ωm(0, 0 , 1) = (0, 0 , ±1), contradicting the choice of m.
This contradiction shows that ΩmRΩ
− 1 m ∈/^ H, so^ H^ is not a normal subgroup.
Solution 3: Let R be the rotation about (1, 0 , 0) that takes (0, 0 , 1) to (0, 1 , 0) (i. e. rotation by π/2).
Note that R
− 1 (0, 0 , 1) = (0, − 1 , 0). Let m be the line with pole (0, 1 , 0). Since (0, 0 , 1) ∈ m, we have
Ωm(0, 0 , 1) = (0, 0 , 1) and hence Ωm ∈ H. But
RΩmR
− 1 (0, 0 , 1) = RΩm(0, − 1 , 0) = R(0, 1 , 0) = (0, 0 , −1) ,
so RΩmR
− 1 ∈/ H. Therefore H is not a normal subgroup.
