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Math 60370, Fall ‘ Assignment 8 Due Friday, November 6 (by noon)
No textbook problems this week. Note that I am using C rather thanˆ C to denote the Riemann sphere in these problems.
Problem 1. Conformality of the stereographic projection.
- In class, I discussed the stereographic projection π : S^2 → Cˆ from the unit sphere in R^3 onto the Riemann sphere. Specifically, π(u 1 , u 2 , u 3 ) = u^11 −+uiu 32 when u 3 6 = 1, and π(0, 0 , 1) = ∞. Show that π is conformal. That is, show that if p ∈ S^2 , then there is a constant r(p) > 0 such that ||Dpπ(v)|| = r(p)||v|| for every vector v ∈ R^3 tangent to S^2 at p. You needn’t treat the case p = (0, 0 , 1) here.
Solution. Let p = (u 1 , u 2 , u 3 ) We showed in class that π(p) = ( (^1) −u^1 u 3 , (^1) −u^2 u 3 ). Hence
Dpπ =
[ 1
1 −u 3 0
u 1 (1−u 3 )^2 (^0 1) −^1 u 3 (1−^ uu^22 ) 2
]
Now a vector v = (v 1 , v 2 , v 3 ) is tangent to S 2 at p, if and only if p·u = 0. Hence
Dpπ(v) =
(v 1 (1 − u 3 ) + u 1 v 3 , v 2 (1 − u 3 ) + u 2 v 3 ) (1 − u 3 )^2 The square of the length of the numerator in this expression is equal to L^2 := (v 12 + v 22 )(1 − u 3 )^2 + v^23 (u^21 + u^22 ) + 2v 3 (1 − u 3 )(v 1 u 1 + v 2 u 2 ) = (v 12 + v 22 )(1 − u 3 )^2 + v^23 (u^21 + u^22 ) − 2 v^23 u 3 (1 − u 3 ) = (v 12 + v 22 )(1 − u 3 )^2 + v^23 (− 2 u 3 + u^23 + 1) = ‖v‖^2 (1 − u 3 )^2. (The first equality holds because p · v = 0 and the second because ‖p‖^2 = 1.) Hence
‖Duπ(v)‖ =
L
(1 − u 3 )^2
‖v‖ 1 − u 3
Hence the assertion holds with r(p) = r(u 1 , u 2 , u 3 ) = (^1) −^1 u 3.
- Give a formula for r(p) in terms of the image z = π(p) instead of p. Given a C^1 curve ˜γ : [a, b] → S^2 and its image γ = π ◦ ˜γ : [a, b] → C, how would you express the length of ˜γ as an integral over γ? Solution. Letting p = (u 1 , u 2 , u 3 ) ∈ S^2 and z = π(p), I have from u^21 + u^22 + u^23 = 1 that |z|^2 = 1 −u
(^23) (1−u 3 )^2 =^
1+u 3 1 −u 3. Solving for^ u^3 in terms of^ z then gives
r(p) =
1 − u 3
1 + |z|^2 2
More specifically, if I take p = ˜γ(t) and z = π(˜γ(t)) = γ(t), then the chain rule and the previous formula for r(p) give me
|γ′(t)| = |Dpπ · ˜γ′(t)| = |r(p)| ‖γ˜′(t)‖ = ‖γ˜′(t)‖
1 + |γ(t)|^2 2
That is, ‖˜γ′(t)‖ = 2 |γ
′(t)| 1+|γ(t)|^2. This implies that
Length(˜γ) =
∫ (^) b
a
‖γ˜′(t)‖ dt =
∫ (^) b
a
2 |γ˜′(t)| 1 + |γ(t)|^2
dt =
γ
2 |dz| 1 + |z|^2
Problem 2. An interesting fact about the steregraphic projection π : S^2 → Cˆ is that it identifies circles in S^2 with circles and lines in C. In this direction, show that
- the preimage of a circle in C under π is a circle in S^2 ; Solution. A circle in C is a set S = {z ∈ C : |z − w| = r} for r > 0 and w = a + bi ∈ C. Thus π−^1 (S) is the set of points (u 1 , u 2 , u 3 ) ∈ S^2 satisfying |π(u 1 , u 2 , u 3 ) − w| = r. Squaring this equation and using the formula for π, I obtain
r^2 =
u 1 + iu 2 1 − u 3
− (a + bi)
2
u^21 + u^22 (1 − u 3 )^2
−u 1 a + u 2 b 1 − u 3
1 + u 3 1 − u 3
−u 1 a + u 2 b 1 − u 3
Clearing the denominators and rearranging then gives −u 1 a + u 2 b + u 3 (1 + r^2 − a^2 − b^2 ) = r^2 − a^2 − b^2 , which on dividing through by the right side yields v · (u 1 , u 2 , u 3 ) = 1
where v = (−a,b,1+r
(^2) −a (^2) −b (^2) ) r^2 −a^2 −b^2.^ That is,^ u^ ∈^ S
(^2) satisfies π(u) ∈ S if and only if u lies on the plane v · u = 1, which meets S^2 in a circle.
- the preimage of a line in C is a circle in S^2 passing through (0, 0 , 1).
Solution. Let L = {x + iy ∈ C^2 : ax + by = c} be the line in question. Then u ∈ S^2 satisfies π(u) ∈ L if and only if au 1 + bu 2 = c(1 − u 3 ), or rather au 1 + bu 2 + cu 3 = c which defines a plane and therefore meets S^2 in a circle. One checks directly, moreover, that (u 1 , u 2 , u 3 ) = (0, 0 , 1) lies in this plane.
It might help to remember that every circle on S^2 is obtained by intersecting S^2 with a plane.
Problem 3. Recall from class that (one dimensional) complex projective space P^1 is defined to to be the quotient of C^2 \ {(0, 0)} by the equivalence relation (z, w) ∼ (z′, w′) if and only if each pair is a non-zero complex multiple of the other, i.e. iff each pair lies on the same complex line through (0, 0). I denote the equivalence class of (z, w) by [z, w] and refer to (z, w) as homogeneous coordinates of the point [z, w] ∈ P^1. As I discussed in class, the map π : P^1 → Cˆ given by π[z, w] = w/z (= ∞ if z = 0) is a homeomorphism. Recall also that A homogeneous map of C^2 is a map F : C^2 → C^2 such that F (z, w) = (F 1 (z, w), F 2 (z, w))
akzk^ + · · · + a 0 , Q(z) = bℓzℓ^ + · · · + b 0 I find
f (1/z) =
zℓ zk
ak + · · · + a 0 zk bℓ + · · · + b 0 zℓ^
= zℓ−kg(z)
where g is holomorphic near 0. That is, 0 is a root of f (1/z) with multiplicity ℓ − k = deg Q − deg P. Hence ∞ is a zero of f (z) with the same multiplicity. I conclude that f has exactly deg P + (deg Q − deg P ) = ℓ = deg Q = d zeroes.
