Math 120B, Spring 2010 - Homework #4
Isometries, ruled surfaces, more on the Lie bracket, and the covariant derivative
Due on Friday, April 30 (in class)
Please note that this problem set is on two pages. Solve the following six problems:
From Chapter 5 of M&P: 10.4, 10.5, 10.6 (10 points each).
Problem 4A (ruled surfaces). (30 points) Ruled surface can be “swept out” by straight lines. More
precisely, they are surfaces that can be parametrized as x(t, v) = α(t) + vβ(t) for given curves α : (a, b) → R3
and β : (a, b) → R3 (we will ignore the possibility of self-intersections, which make x not one-to-one; also,
there may be “singular points” where kx1 × x2 k = 0). Curve α is called directrix of x. Note that coordinate
curves x(t0 , v), for fixed t0 ∈ (a, b), are straight lines (called the rulings and often indicated with Lt0 ).
(a) Let α(s) , (cos s, sin s, 0) be the arclength parametrization of the circle S1 ; let β(s) , α0 (s) + e3 ,
where e3 = (0, 0, 1). Verify that x(s, v) = α(s)+vβ(s) is a parametrization of the one-sheet hyperboloid
of revolution x2 + y 2 − z 2 = 1 (see Figure 1; do not worry about injectivity).
Figure 1. The hyperboloid x2 + y 2 − z 2 = 1 with the rulings; the directrix α is the “belt”.
(b) Prove that the hyperboloid x2 + y 2 − z 2 = 1 is actually doubly ruled ; that is, through each point there
are two straight lines that lie on the surface. In other words: what is another possible choice of β that
results in a different set of rulings?
(c) Let x(t, v) = α(t) + vβ(t) and assume w.l.o.g. that kβ(t)k ≡ 1. Also assume that1 β̇(t) 6= 0 for
all t. Prove that there exists a curve on the ruled surface of the form γ(t) = α(t) + r(t)β(t) such
that hγ̇(t), β̇(t)i ≡ 0; this line is called line of striction. Hint: find function r(t) in function of α(t),
β(t) and their derivatives. What is the line of striction in the example of part (a)?
(d) If γ(t) is the line of striction of x(t, v) = α(t) + vβ(t), prove that y(τ, u)
= γ(τ ) + uβ(τ ) is a
reparametrization of x. That is, find a function G(t, v) = τ (t, v), u(t, v) such that y ◦ G = x;
verify that det JG 6= 0. Note that in y we are using the line of striction γ as directrix.
(e) Consider the ruled surface y(τ, u) = γ(τ ) + uβ(τ ), where γ is the line of striction, kβ(τ )k ≡ 1 and
[γ̇, β, β̇]
β̇(τ ) 6= 0 for all τ . Prove that y1 (τ, u) × y2 (τ, u) = η(τ )β̇(τ ) + u β̇(τ ) × β(τ ), where η(τ ) ,
kβ̇k2
is called distribution parameter . Show that any point on the line of striction where γ̇ = 0 is a singular
point (i.e. a point where ky1 × y2 k = 0).
2
η (τ )
(f ) Finally, show that Gaussian curvature for y is K(τ, u) = − [η2 (τ
and it is therefore the case that
)+u2 ]2
K ≤ 0 on all the surface (Hint: first verify that L22 = 0; if this is the case, in order to calculate det Λ
you do not need to compute L11 ). Where is K equal to zero (that is, along which rulings)?
1
Ruled surfaces with β̇(t) ≡ 0 are called cylindrical . Ruled surfaces with β̇(t) 6= 0 for all t are called non-cylindrical.
1
Problem 4B (more on the Lie bracket). (20 points) For a scalar function f : U → R and a vector
field X = X i xi ∈ X(M ) we may define the new vector field f X as follows (we are using Einstein’s summation
convention):
(f X)(u1 , u2 ) , f (u1 , u2 )X i (u1 , u2 )xi (u1 , u2 ),
(1)
(so on each tangent space f X parallel to X, but its length and orientation may be modified). Also, a vector
field X = X i xi ∈ X(M ) may be interpreted as “directional derivative” for a scalar function h : U → R, i.e. we
define the new scalar function2 :
∂h
∂h
X(h) , X i i ,
or more precisely:
X(h) (u1 , u2 ) , X i (u1 , u2 ) i (u1 , u2 ).
(2)
∂u
∂u
Prove that, for any scalar functions f, h : U → R and any smooth vector fields X, Y ∈ X(M ):
(a) X(f h) = f (X(h)) + h(X(f )), and X(f + h) = X(f ) + X(h) (these are a scalar functions);
(b) [X, Y](f ) = X(Y(f )) − Y(X(f )) (this is also a scalar function);
(c) [f X, Y] = f [X, Y] − (Y(f ))X (this is a vector field!).
Problem 4C (the covariant derivative). (20 points) Let x : U → R3 be a simple surface and Y = Y i xi be
a vector field on M = x(U). Fix a point (a, b) ∈ U and let P = x(a, b). Let X = X i xi (a, b) be a tangent vector
in space TP M (X may or may not be part of a vector field). Consider an arbitrary curve α(t) = x(u1 (t), u2 (t))
on M with the properties that α(0) = P and α̇(0) = X, i.e. (u̇1 (0), u̇2 (0)) = (X 1 , X 2 ) (see Figure 2). Finally,
define y(t) = Y(u1 (t), u2 (t)): note that y is the vector field Y along curve α, and we have y(t) ∈ Tα(t) M for
all t. Prove (using index notation and Einstein’s summation convention) that
dy ∂Y k
i
k
T
(0) = X j
(a,
b)
+
Y
(a,
b)
Γ
(a,
b)
xk (a, b)
(3)
ij
dt
∂uj
where T (·) is the tangential component of a vector (introduced in Problem 3C of Homework #3). The righthand side of (3) is called the covariant derivative at P = x(a, b) of vector field Y ∈ X(M ) with respect to
vector X ∈ TP M and is indicated by ∇X Y(a, b) (it is the tangential component of the derivative of Y in
the direction of X and it is independent of the choice of α, as long as α̇(0) = X). Prove that when both X
and Y are vector fields it is the case that ∇X Y − ∇Y X ≡ [X, Y] at all points on the surface M .
Figure 2. Illustration of the covariant derivative ∇X Y as the projection of
dy
(0) on tangent space TP M .
dt
2
The multiplication by a scalar function f and the action of X on h are defined more rigorously, and perhaps more clearly,
as follows. Functions f and h are actually scalar functions defined on the surface M , i.e. we have that f : M → R and
h : M → R (e.g. h(P ), P ∈ M , may represent the surface temperature at point P ). Given a coordinate patch for the surface
x : U → M (which we assume surjective for simplicity) let X be a vector field on M . The multiplication by a scalar is defined as
(f X)(u1 , u2 ) , (f ◦ x)(u1 , u2 ) X i (u1 , u2 ) xi (u1 , u2 ), while the directional derivative of the scalar function h is X(h) , X i ∂(h◦x)
,
∂ui
`
´
1
2
1
2
2
i.e. X(h) (u1 , u2 ) , X i (u1 , u2 ) ∂(h◦x)
(u
,
u
),
with
(u
,
u
)
∈
U.
When
M
=
R
the
latter
definition
coincides
with
the
classical
i
∂u
∂h
notion of directional derivative of a scalar function h in the direction of vector X, given by ∂X
= X · ∇h. In our simplified
1
2
definitions (1) and (2) we are essentially identifying points P ∈ M with their coordinates (u , u ) ∈ U.
2