Math 120A Discussion Session Week 6 Notes November 6, 2018

Today we’ll quickly discuss some solutions to last week’s extended example before working on some similar problems in groups. The goal is to prepare ourselves for Friday’s midterm.

Stereographic projection 2 3 Throughout, we’ll use the coordinate patch x: R → R defined by  2u 2v u2 + v2 − 1 x(u, v) = , , . u2 + v2 + 1 u2 + v2 + 1 u2 + v2 + 1

We won’t repeat all of these solutions in section, but will focus on the last two problems.

1. Verify that the image of x is contained in S2. What is the image precisely?

2 2 (Solution) We need to show that kx(u, v)k = 1 for all (u, v) ∈ R . Indeed, 4u2 + 4v2 + u4 + 2u2v2 + v4 − 2u2 − 2v2 + 1 kx(u, v)k2 = (u2 + v2 + 1)2 u4 + 2u2v2 + v4 + 2u2 + 2v2 + 1 = = 1. (u2 + v2 + 1)2

The image of x is S2 with the north pole (0, 0, 1) removed. This is most easily seen by considering the geometric construction that led us to x, but we can verify that the north pole is not in the image by noticing that this would require u2 + v2 − 1 = u2 + v2 + 1, an obvious absurdity. ♦

2. Give a formula for the inverse x−1(x, y, z). (Hint: Produce this map geometrically.)

2 −1 2 (Solution) Given (x, y, z) ∈ S , x (x, y, z) should be the point in R with the property that the line connecting x−1(x, y, z) to (0, 0, 1) intersects S2 at (0, 0, 1). Now consider the line passing through (0, 0, 1) and (x, y, z), parametrized by

(1 − t)(0, 0, 1) + t(x, y, z) = (tx, ty, 1 + t(z − 1)).

2 −1 This will intersect R when 1 + t(z − 1) = 0, which is to say when t = (1 − z) . (We won’t have z = 1, since (x, y, z) can’t be the north pole.) So

 x y  x−1(x, y, z) = , . 1 − z 1 − z

♦

2 3 3. Define y: R → R to be the south pole stereographic projection. This is defined similarly to x, but with the line passing through (0, 0, −1) instead of (0, 0, 1). Give a formula for y(u, v).

1 (Solution) For this we can just reflect everything through the xy-plane — the line that connected (u, v) to (0, 0, 1) will now pass through (0, 0, −1). So

 2u 2v 1 − u2 − v2  y(u, v) = , , . u2 + v2 + 1 u2 + v2 + 1 u2 + v2 + 1

♦

4. Compute the coordinate transformation x−1 ◦ y(u, v) and give a geometric description of this map.

(Solution) We have

 2u 2v 1 − u2 − v2  x−1 ◦ y(u, v) = x−1 , , . u2 + v2 + 1 u2 + v2 + 1 u2 + v2 + 1

Since z = (1 − u2 − v2)(u2 + v2 + 1) we have 1 − z = (2u2 + 2v2)/(u2 + v2 + 1). So

x 2u u2 + v2 + 1 u = · = . 1 − z u2 + v2 + 1 2u2 + 2v2 u2 + v2 Similarly, y/(1 − z) = v/(u2 + v2). So

 u v  x−1 ◦ y(u, v) = , . u2 + v2 u2 + v2

2 This map takes a nonzero point in R and scales it by the reciprocal of the square of its magnitude. Geometrically, this map inverts the plane over the unit circle, swapping the regions inside and outside the unit circle. ♦

5. Compute the metric coefficients g11, g12, g21, g22 for x. (Recall that gij = hxi, xji.) Where on S2 are these functions maximized? (Just argue heuristically.) Does this make sense?

(Solution) First we need the derivatives xu and xv. We have

 2v2 − 2u2 + 2 −4uv 4u  x (u, v) = , , . u (u2 + v2 + 1)2 (u2 + v2 + 1)2 (u2 + v2 + 1)2

Symmetrically we also have

 −4uv 2u2 − 2v2 + 2 4v  x (u, v) = , , . v (u2 + v2 + 1)2 (u2 + v2 + 1)2 (u2 + v2 + 1)2

Then (2v2 − 2u2 + 2)2 + (−4uv)2 + (4u)2 g (u, v) = hx (u, v), x (u, v)i = 11 u u (u2 + v2 + 1)4 4v4 + 8u2v2 + 4u4 + 8v2 + 8u2 + 4 4 = = . (u2 + v2 + 1)4 (u2 + v2 + 1)2

2 Similarly, 4 g (u, v) = hx (u, v), x (u, v)i = . 22 u u (u2 + v2 + 1)2

For g12 and g21 we have

g21(u, v) = g12(u, v) = hxu(u, v), xv(u, v)i (2v2 − 2u2 + 2)(−4uv) + (−4uv)(2u2 − 2v2 + 2) + (4u)(4v) = (u2 + v2 + 1)2 −16uv + 16uv = = 0. (u2 + v2 + 1)2

The vanishing of these coefficients means that xu and xv give us an orthogonal (but not typically orthonormal) basis for the tangent space at x(u, v). Notice that g11 and g22 are maximized when u = v = 0 — at the south pole. This is when x is doing the greatest amount of stretching of the plane. As u and v grow without bound these coefficients head towards zero and x tries to fit a neighborhood of infinity into a small region around the north pole. ♦

6. Use the metric coefficients to compute the arc length of the equator of S2. Compute the arclength of the circle determined by z = c for some constant −1 < c < 1.

(Solution) We’ll do the second part. The equator is the special case c = 0. In the coordinates induced by x the circle z = c is given by

u2 + v2 − 1 1 + c = c ⇒ u2 + v2 = . u2 + v2 + 1 1 − c Let’s write k = p(1 + c)/(1 − c) for brevity. Then we can parametrize the circle u2 +v2 = k2 by α(t) = (k cos(t), k sin(t)) for t ∈ [0, 2π]. Then the arclength of our curve is given by

Z Z 2π Z 2π 0 p 0 2 0 0 0 2 ds = k(x ◦ α) (t)kdt = g11(u (t)) + 2g12u (t)v (t) + g22(v (t)) dt. x(α) 0 0 We have u0(t) = −k sin(t) and v0(t) = k cos(t), and we computed the metric coefficients above. So the arclength is Z 2π Z 2π p 2 2 √ g11(−k sin(t)) + g22(k cos(t)) dt = k g11dt, 0 0 2 −2 since g22 = g11. Now g11(α(t)) = 4(k + 1) , so the arclength is Z 2π 2k 4πk 2 dt = 2 . 0 k + 1 k + 1 We can simplify this a bit by noticing that 1 + c 2 k2 + 1 = + 1 = . 1 − c 1 − c

3 So (k2 + 1)−1 = (1 − c)/2, meaning that the arclength of our circle is √ 1 + c 1 − c √ √ p 4π √ = 2π 1 + c 1 − c = 2π 1 − c2. 1 − c 2 We have some reality checks here. For the equator we have c = 0, so the arclength is 2π, as expected. As we head towards the north or south pole, where c = 1 or c = −1, respectively, the arclength will tend towards 0. ♦

Group work 2 2 3 This is another extended example. Throughout we’re considering the cylinder x + y = 1 in R . Consider the parametrizations

1 2 1 1 2 1 2 x(u , u ) = (cos u , sin u , u ), −π < u < π, u ∈ R and 1 2 1 p 2 1 2 y(v , v ) = (v , 1 − (v1)2, v ), −1 < v < 1, v ∈ R.

1. Find the matrix (gij) of metric coefficients for x. 2. Define f : V → U by f(v1, v2) = (arccos(v1), v2), where U and V are the domains of x and y, respectively. Verify that y = x ◦ f. Then we have the formula t (gij) = Jf (gij)Jf ,

where Jf is the Jacobian of f,(gij) is the matrix of metric coefficients for x, and (gij) is the matrix of metric coefficients for y. Use this formula to compute (gij). 3. Consider the curve on the cylinder parametrized by

α(t) = (cos t, sin t, t), 0 < t < π.

This curve lies in the overlap of our two coordinate patches. Compute its arclength twice — once in each set of coordinates.

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