# HOMEWORK 1 Let γ : [−L, L] → R 2 represent a smooth closed curve

```HOMEWORK 1
ANDRÉ NEVES
Let γ : [−L, L] → R2 represent a smooth closed curve parametrized counterclockwise (i.e., N points towards the interior of the curve) and parametrized
by arc-length.
(1) If θ(t) is a smooth function that stands for the angle between γ 0 (t)
and (1, 0). Show that k(t) = θ0 (t).
(2) Assume that k(t) &gt; 0 for all −L ≤ t ≤ L and that γ has no self
intersections, i.e., γ is injective when restricted to (0, L).
Show that if γ(0) = 0, and N (0) = (0, 1), then
γ ⊆ {(x, y) : y ≥ 0}.
HINT: You may want to consider the function f (t) = γ(t).N (0)
and look at its critical points.
(3) If you drop the requirement that γ has no self intersections in Exercise (2), does the conclusion of Exercise (2) still hold?
Let S be a smooth connected surface in space.
(4) If F : S → R3 is a smooth map with DFp = 0 for all p, show that F
is constant.
(5) Suppose that 0 ∈ S and S ⊂ {z ≥ 0}. Show that T0 S = {z = 0}.
Solutions
1 (1): In this case we have γ 0 (t) = (cos θ(t), sin θ(t)) and so N (t) =
(− sin θ(t), cos θ(t)). Using this, we have
k(t) = γ 00 (t).N (t) = θ0 (t)(− sin θ(t), cos θ(t)).N (t) = θ0 (t).
1 (2): Because N (0) = (0, 1) we have that θ(0) = 0.
Set θ0 = θ(−L). Because θ(t) is increasing we must have θ0 &lt; 0.
The hypothesis that γ has no self intersections is used to show that θ(L) =
θ0 + 2π (note that in general what we know is that θ(L) = 2πk + θ0 as the
curve is closed). Indeed because γ has no self intersections its winding
number has to be 2π and so
Z
Z L
2π = k(t)ds =
θ0 (t)dt = θ(L) − θ(−L).
γ
−L
It is only at this stage that the no self intersection hypothesis is used and
this is an important step (as the next question shows). The conclusion is
1
2
ANDRÉ NEVES
that the range of the function theta is [θ0 , θ0 + 2π] and that 0 is contained
in this interval.
Let f (t) = γ(t).(0, 1), −L ≤ t ≤ L. If f 0 (t0 ) = 0 then
γ 0 (t0 ).(0, 1) = 0 =⇒ sin θ(t0 ) = 0 =⇒ θ(t0 ) = kπ, k ∈ N.
. The interval [θ0 , θ0 + 2π] contains at most two multiples of π (either −π
and 0 or 0 and π.). In other words, f has exactly two critical points and
t0 = 0 is one of them.
Furthermore,
f 00 (t0 ) = k(t0 )N (t0 ).(0, 1) = k(t0 ) cos θ(t0 ) = &plusmn;k(t0 ).
Hence, t0 = 0 is a local minimum (θ(0) = 0) and the other critical point is
necessarily a local maximum.
Putting all this together, f : [−L, L] → R is such that f (0) = 0, f (L) =
f (−L), f has only two critical points: a local minimum at the origin and
a local maximum somewhere else. A moment of thought shows that this
implies that f (t) ≥ 0 for all t, which is what we wanted to show.
1 (3): No.
2 (1): Choose p, q points in S. Because S is connected there is a curve
α connecting p = α(0) to q = α(1). Set f (t) = F (α(t)). Then
f 0 (t) = dFα(t) (α0 (t)) = 0
for all 0 ≤ t ≤ 1
and thus f is constant, which means F (p) = F (q). The arbitrariness of p, q
implies that F is constant.
A Cautionary Tale: Some of you said that because dFp = 0 it meant that
the 3 &times; 3 matrix that represents dF was 0 and so F is constant. That is not
true. Think of S = {|~x| = 1} and F : S → R3 given by F (~x) = |~x|(1, 0, 0).
Then F is constant along S but is not constant as a function in R3 .
2 (2): Let’s start with an important comment: Surely S is graphical
near the origin over some plane P . Of course you can assume that P is the
xy-plane. But if you do that you can no longer assume that S ⊂ {z ≥ 0}
because now you have changed your coordinate systems. What was before
the half-space {z ≥ 0} becomes now some other half-space. Hence if you
assume that P is the xy-plane, then you have that S ⊂ {ax + by + cz ≥ 0}
for some a, b, c, and so you are not doing much.
The correct way of solving this is as follows: Pick X ∈ T0 S, and so X =
0
α (0) for some curve α ⊂ S. Set f (t) = α(t).(0, 0, 1). Because S ⊂ {z ≥ 0}
we have that f (t) ≥ 0 for all t and f (0) = 0. Thus
0 = f 0 (0) = α0 (0).(0, 0, 1) = X.(0, 0, 1).
Therefore (0, 0, 1) is perpendicular to every X ∈ T0 S which means that
T0 S = {z = 0}.
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