Solutions to homework 5

Ma 109b Geometry and Topology Winter 2010
HOMEWORK 5 SOLUTIONS
1. Note that in each of these cases we need to find a, b, c, u, v so that the differential
is injective. We do this by showing xu ∧ xv = 0. Note that it is not necessary
to check that the map is injective since the inverse function theorem tells us
that we’ll get a parametrization around each point at which the differential is
injective. So we do some calculation:
a)
xu = (a cos u cos v, b cos u sin v, −c sin u)
xv = (−a sin u sin v, b sin u cos v, 0)
xu ∧ xv = sin u(−bc sin u cos v, ac sin u sin v, ab cos u).
We can take a, b, c = 0 and u = 0.
b)
We can take a, b = 0 and u = 0
c)
We can take a, b = 0 and u = 0
E = a 2 cos 2 u cos 2 v + b 2 cos 2 u sin 2 v + c sin 2 v
F = (b 2 − a 2 ) sin u cos u sin v cos v
G = a 2 sin 2 u sin 2 v + b 2 sin 2 u cos 2 v
xu = (a cos v, b sin v, 2u)
xv = (−au sin v, bu cos v, 0)
xu ∧ xv = u(−2bu cos v, −2au sin v, ab)
E = a 2 cos 2 v + b 2 sin 2 v + 4u 2
F = (b 2 − a 2 )u cos v cos v
G = a 2 u 2 sin 2 v + b 2 u 2 cos 2 v
xu = (a cosh v, b sinh u, 2u)
xv = (au sinh v, bu cosh v, 0)
xu ∧ xv = u(−2bu cos v, −2au sin v, ab)
E = a 2 cosh 2 v + b 2 sinh 2 v + 4u 2
F = (a 2 + b 2 )u sinh v cosh v
G = a 2 u 2 sinh 2 v + b 2 u 2 cosh 2 v
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Ma 109b Geometry and Topology Winter 2010
d)
xu = (a cosh u cos v, b cosh u sin v, c sinh u)
xv = (−a sinh u sin v, b sinh u, 0)
xu ∧ xv = sinh u(−bc sinh u cos v, −ac sinh u sin v, ab cosh u)
We can take a, b, c = 0 and u = 0.
E = a 2 cosh 2 u cos 2 v + b cosh 2 sin 2 v + c 2 sinh 2 u
F = (b 2 − a 2 ) sinh u cosh u sin v cos v
G = a 2 sinh 2 u sin 2 v + b 2 sinh 2 u cos 2 v
2. Assume that S1 is orientable and pick an atlas {(Ui, xi)} compatible with this
orientation. So the change of coordinates between any neighborhoods must have
positive determinant Then {(Ui, φ ◦ xi)} is an atlas for S2. We check that this
induces an orientation. Indeed, for Ui, Uj such that φ ◦ xi(Ui) ∩ φ ◦ xj(Uj) = ⊘,
the change of coordinates is
hence has positive determinant.
(φ ◦ xj) −1 ◦ (φ ◦ xi) = x −1
j
Since φ is a diffeomorphism the same argument works in the other direction.
So we’ve shown that a diffeomorphism induces a choice of orientation, which
implies that a surface is orientable iff any diffeomorphic surface is orientable.
To show that a diffeomorphism can change orientation consider the plane parametrized
by x(u, v) = (u, v, 0). Then the Jacobian of the diffeomorphism (u, v, 0) →
(u, −v, 0) has negative determinant so it reverses orientation.
3. We just need to check that (N ◦ α) ′ (t) is nonzero everywhere. But (N ◦ α) ′ (t) =
dN(α ′ (t)), α ′ (t) is nonzero everywhere and dN is an invertible linear map away
from planar and parabolic points.
4. a) Write
w1 = ae1 + be2
w2 = ce1 + de2
where e1, e2 are principal directions with corresponding to the principal curvatures
k1, k2. Then with k1 = −k2 we have
< dN(w1), dN(w2) >=< ak1e1+bk2e2, ck1e1+dk2e2 >= ack 2 1+bdk 2 2 = −k2k2 < w1, w2 > .
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◦ xi

Ma 109b Geometry and Topology Winter 2010
b) If v and w are the tangent vectors at p of two curves intersecting at p we
have
< dN(v), dN(w) >
||dN(v)|| · ||dN(w)|| =
=
< dN(v), dN(w) >
< dN(v), dN(v) > 1/2 · < dN(w), dN(w) > 1/2
−K < v, w >
√ −K < v, v > 1/2 · √ −K < dN(w), dN(w) > 1/2
= < v, w >
||v|| · ||w|| .
5. a)Suppose that the tractrix is parametrized by (x(t), y(t)). The tangent line at
(x(t), y(t)) is given in terms of (u, v) by
u = x′ (t)
y ′ (v − y(t)),
(t)
so we find that the (square of the) distance between (x(t), y(t)) and the xintercept
of this line is
1 = y(t) 2 ([ x′ (t)
y ′ (t) ]2 + 1).
We’ll assume that the curve is parametrized by arc length, that is y ′ (t) 2 +
x ′ (t) 2 = 1 and simplify this to
So we can take
for non-negative t.
x(t) =
y(t) 2 = y ′ (t) 2 .
y(t) = e −t
t
0
√ 1 − e −2t dt
b) As in p.76 of do Carmo, the whole surface for t > 0 is regular.
We can use as parametrizations
for t > 0, 0 < θ < 2π and −π < θ < π.
c) See do Carmo, p.161.
(x(t), y(t) cos θ, y(t) sin θ)
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