Chapter 5: Solutions
Exercise 5.1 Choose any distinct points p and q on the n-sphere Sn (n > 0). So p and q are unit vectors in Rn+1 .
We’ll use a great circle on Sn to connect these points. There are certainly other approaches, e.g., someone from class
had the good idea of using longitudes to connect p to q through the “north pole”.
The method outlined here works nicely because we won’t need to refer to coordinates for p and q.
(a) It may be unlikely, but if it does happens that p ⊥ q (i.e., p · q = 0) then define, for θ ∈ R, the points
α(θ) = (cos θ)p + (sin θ)q.
2
2
2
(1)
n
Since kα(θ)k = α(θ) · α(θ) = cos θ + sin θ = 1 then α(θ) ∈ S for all θ ∈ R. The smooth curve α traces out
a great circle on Sn as θ ranges through R with α(0) = p and α(π/2) = q.
(b) If q = −p, that is, if p and q are both on a line through the origin (they’re linearly dependent and called
antipodal points) then intuitively any great circle containing either p or q should contain the other. Let’s check
it: first choose a point e ∈ Sn orthogonal to p and proceed1 as in part (a) above, except with e in place of q:
α(θ) = (cos θ)p + (sin θ)e for θ ∈ R. Then α(θ) ∈ Sn for each θ ∈ R and that α(0) = p and α(π) = −p = q.
(c) If p and q are not on the same line (i.e., if p and q are linearly independent) then try this approach: replace q
with another point e ∈ Sn such that e ⊥ p and e is in the same plane as the points p, q and 0. Then, again as in
(a) put α(θ) = (cos θ)p + (sin θ)e. The great circle traced out by α passes through p (θ = 0), e (θ = π/2) and,
hopefully, q (θ =?). How do you find the correct θ for q? Well, p · q = cos γ for some angle γ with 0 < γ < π
(remember p and q are unit vectors). Setting θ = γ could work.
To do this first find the unit vector e and then compute α(γ) to see if α(γ) = q. Construct e by projecting q
along p, then subtracting this projection from q, and then normalizing the result to unit length – so that it’s
on Sn (this basic construction should be quite familiar to you by now):
e=
q − (cos γ)p
q − (q · p)p
q − (cos γ)p
q − (q · p)p
=
=p
=p
.
2
2
kq − (q · p)pk
sin γ
1 − (q · p)
1 − cos γ
Now let’s see if α(γ) equals q:
q − (cos γ)p
α(γ) = cos(γ)p + sin(γ)e = cos(γ)p + sin(γ)
= cos(γ)p + q − (cos γ)p = q.
sin γ
Exercise 5.2 Let p, q ∈ S and α : [a, b] → S a continuous function with α(a) = p and α(b) = q. The composition
g ◦ α : I → {−1, 1} is continuous because both α and g are. Suppose g(p) = 1, i.e., (g ◦ α)(a) = 1. If −1 = g(q) =
(g ◦ α)(b) then the intermediate value theorem from calculus tells us that, for some t0 ∈ (a, b), (g ◦ α)(t0 ) = 0. Of
course, this contradicts the assumption that g assumes only the values −1 and 1. So it must be that g(q) = 1 = g(p).
A similar arguments handles the case of g(p) = −1 and we conclude that, in either case, g(q) = g(p) and conclude
that either g(q) = 1 for all q ∈ S or g(q) = −1 for all q ∈ S.
Exercise 5.4 If f (x1 , . . . , xn+1 ) = x21 + · · · + x2n+1 and if p = (x1 , . . . , xn+1 ) with kpk = r we have ∇f (p) =
2(x1 , . . . , xn+1 ) = 2p with k∇f (p) = 2(x1 , . . . , xn+1 )k = 2r.
If N1 (p) = ∇f (p)/k∇f (p)k then N1 (p) = p/r which means, N1 (p) = (p, p/r) and N2 (p) = −N1 (p) = −(p, p/r).
Exercise 5.5
(a) n = 1: Rotate N(p) = (p, 0, 1) clockwise 90◦ gives (p, 1, 0).
R2
N(p)
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•
p
R
1 You could construct such a point e from p by first choosing a point r ∈ Sn not on the line through p and q, setting u = r − (r · p)p,
and then e = u/kuk.
(b) n = 2: In the picture below the two vectors v and Rθ v are in the surface (and tangent space) Sp ' R2p , as
is the vector N(p) × v. And I’ve taken θ < 0. The ordered orthogonal basis {v, N(p) × v, N(p)} for R3 is a
right-handed system because of the convention we take (in calculus, physics, engineering, etc.) in defining the
cross product of two 3-dimensional vectors, in this case, v × [N(p) × v] = kvk2 N(p).
R3
N(p)
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N(p) × v
R2
p•
Rθ v
θ
v
(c) n = 3:


1
e1
0
 e2 


det 
 e3  = det 0
0
N (p)

0
1
0
0
0
0
1
0

0
0
 = 1 > 0.
0
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Exercise 5.6 Follow the text’s hint: Fix p ∈ C and set N (p) = (cos θ, sin θ). Then N(p) = (p, N (p)) =
(p, (cos θ, sin θ)) and the positive tangent direction at p is (by definition) the unit vector
u = p, cos(θ − π/2), sin(θ − π/2) .
Because the tangent
space Cp is one dimensional
then v = (p, v) must be a scale of this direction, that is, v =
±kvku = ±kvk p, cos(θ − π/2), sin(θ − π/2) . In the figure below v is drawn in the negative tangent direction
at p.
{v} is inconsistent with orientation N at p.
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N(p)
•
p
C
v
The basis {v} for Cp will be consistent with the orientation N on Cp provided
v
det
> 0.
N (p)
But
v
cos(θ − π/2) sin(θ − π/2)
sin θ
det
= ±kvk det
= ±kvk det
N (p)
cos θ
sin θ
cos θ
v
v
So det
> 0 if and only if v = kvku, that is, if and only if u =
.
N (p)
kvk
2
− cos θ
sin θ
= ±kvk.
Exercise 5.7
(a) For the first part compute the values of v · (v × w) (and w · (v × w)). You should get 0 (in both cases).
To show that kv × wk = kvkkwk sin θ It’s be easier to square both sides first.
Here is an identity that may help:
2
v·w
v·w
If θ ∈ [0, π] is such that cos θ =
. This means
then sin θ ≥ 0 and sin2 θ = 1 − cos2 θ = 1 −
kvkkwk
kvkkwk
"
2 #
v
·
w
kvk2 kwk2 sin2 θ =kvk2 kwk2 1 −
kvkkwk
=kvk2 kwk2 − (v · w)2 .
So it just needs to be checked that:
kv × wk2 = kvk2 kwk2 − (v · w)2 .
(b) Expand both u · (v × w) and the determinant. The expressions will match. This shows:


u
u · (v × w) = det  v 
w
Swapping different rows in a matrix will change the sign of its determinant. So






w
v
u
det  v  = det  w  = det  u 
v
u
w
that is,
u · (v × w) = v · (w × u) = w · (u × v).
(c) If, for any vector u ∈ R3p , u · x = u · (v × w) then u · [x − (v × w)] = 0. Setting u = x − (v × w) gives
kx − (v × w)k2 = 0 i.e., x = v × w.
Exercise 5.8
(a) By Exercise 5.7(b) and by properties of determinants:




N (p)
v
N(p) · (v × w) = det  v  = det  w  > 0
w
N (p)
if and only if {v, w} is consistent with N(p).
(b) Assume, without loss of generality, that both v = (p, v) and w = (p, w) are unit vectors, that is, kvk = kwk = 1.
Then v and N(p) × v are perpendicular unit vectors (problem 5.7(a) or because of properties of the cross
product you learned in calculus) and so {v, N(p) × w} is an orthonormal basis for the tangent space Sp .
Since w is in Sp then there are unique scalars a and b so that
w = av + b[N(p) × v].
Take the dot product of w, first with v and then with N(p) × v, to see that a = w · v and b = w · [N(p) × v].
The vectors w and v are on the unit circle in Sp . The image of the rotation Rθ v sweeps through all points on
the unit circle exactly once as θ ranges through the angles −π < θ ≤ π. That is, there must be a unique angle
θ with −π < θ ≤ π and w = Rθ v.
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So w = cos θv + sin θ[N(p) × v] for some angle θ with −π < θ ≤ π. This implies that a = cos θ and, more
importantly, b = sin θ, so that




w
v
sin θ = b = w · [N(p) × v] = det  N (p)  = det  w  .
v
N (p)


v
This last expression shows that {v, w} is consistent with the orientation N iff det  w  > 0 iff sin θ > 0
N (p)
iff 0 < θ < π.
Exercise 5.9
(a) The vectors v = (p, v), w = (p, w) are given and N(p) = (p, N (p)) depends on the surface S at the point p.
u
 v 
4

Regard the expression det 
 w  as a function of its first row u = (u1 , u2 , u3 , u4 ) ∈ R . That is,
N (p)

u
 v 

det 
 w  = au1 + bu2 + cu3 + du4 = u · x = u · x
N (p)

where x = (a, b, c, d) ∈ R4 , x = (p, x), u = (p, u) and where the coefficients a, b, c, and d come from the first
row determinant expansion and, as such, are uniquely determined by vectors v, w, and N(p) (the other rows).
But is x = (p, x) tangent to S at p? We just need to check that x ⊥ N(p). But this is easy now:


N (p)
 v 

N(p) · x = det 
 w  = 0.
N (p)
So x ∈ Sp and, once the surface S is oriented, depends uniquely on v, w ∈ Sp . Denote by v × w the vector x.
(b) Routine computations, many of which follow from properties of determinants.
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