GAUSS MAP
ANDRÉ NEVES
1. Problems
(1) Let α : I −→ S be a curve in S parametrized by arc length so that
λ1 λ2 > 0 at the point α(0). Show that
|α00 (0)| ≥ min{|λ1 |, |λ2 |},
where λ1 , λ2 are the principal curvatures of the second fundamental
form.
(2) (Exercise 4 of 3.2 of Do Carmo) Assume that S is a surface with
principal curvatures satisfying |k1 | ≤ 1, |k2 | ≤ 1 everywhere. Is is
true that if C is a curve in S then it has |~k| ≤ 1?
(3) (Exercise 8 of 3.2 of Do Carmo) Describe the image of the Gauss
map for the following surfaces.
(a) S = {z = x2 + y 2 };
(b) S = {x2 + y 2 − z 2 = 1};
(c) S = {x2 + y 2 = cosh z 2 }.
(4) (Exercise 5 a), 5b), 5c) and 5d) of 3.3 of Do Carmo) Consider the
surface S with chart
x(u, v) = (u − u3 /3 + uv 2 , v − v 3 /3 + vu2 , u2 − v 2 ).
(a) Show that
g11 = g22 (1 + u2 + v 2 )2 ,
g12 = 0.
(b) Show that
A11 = 2,
A12 = A21 = 0, A22 = −2.
(c) The principal curvatures are
λ1 = 2(1 + u2 + v 2 )−2 and λ2 = −2(1 + u2 + v 2 )−2 .
(d) The coordinate curves are lines of curvature where we say α(t)
a curve in S is a line of curvature if α0 (t) is an eigenvector of A
at α(t) for all t.
1
2
ANDRÉ NEVES
(5) Consider a surface of revolution parametrized by
φ(θ, s) = (r(s) cos θ, r(s) sin θ, z(s))
(r0 (s))2
where
Show that
+ (z 0 (s))2 = 1 for all s.
A22
∂φ
= −r(s)z 0 (s),
∂θ
∂φ ∂φ
A12 = A21 = A
.
= 0,
∂s
∂θ
∂φ ∂φ
.
=A
= z 0 (s)r00 (s) − z 00 (s)r0 (s)
∂s
∂s
A11 = A
∂φ
∂θ
.
(6) (Do Carmo, 2.5, Ex 1) Compute the metric (gij )i,j=1,2 for the following surfaces
(a) φ(u, v) = (a sin u cos v, b sin u sin v, c cos u);
(b) φ(u, v) = (au cos v, bu sin v, u2 );
(c) φ(u, v) = (au cosh v, bu sinh v, u2 );
(d) φ(u, v) = (a sinh u cos v, b sinh u sin v, c cosh u).
(7) (Do Carmo, 2.5, Ex 5) Show that the area A of a bounded region R
of the surface {z = f (x, y)} is given by
Z q
1 + (∂x f )2 + (∂y f )2 dxdy,
A=
Ω
where Ω is the projection of R on the xy-plane.
(8) (Do Carmo, 2.5, Ex 11) Let S be a surface of revolution and C its
generation curve (see Exercise 5 above). Let S be the arc length of
C and denote by ρ = ρ(s) the distance of the rotation axis of the
point C corresponding to s.
(a) Show that the area of S is
Z l
A = 2π
ρ(s)ds,
0
where l is the length of C.
(b) Apply part (a) to compute the area of a torus of revolution.
(9) (Do Carmo, 2.5, Ex 12) Consider a surface S which is a tube of
radius r around a curve α which has non-vanishing curvature. Parametrize the surface S and then show that its area is 2πrlength(α).
(10) Let S be a connected surface such that λ1 (p) = λ2 (p) = λ(p) for all
p ∈ S.
GAUSS MAP
3
i) Show that there is some constant λ0 such that λ(p) = λ0 for all
p.
ii) Show that if λ above is identically zero, then S is contained
inside a plane.
2. Solutions
(1) Let p = α(0) and denote by e1 , e2 in Tp S the principal directions of
A, the second fundamental form, meaning that A(e1 ) = λ1 e1 and
A(e2 ) = λ1 e2 and e1 , e2 form an orthonormal basis for Tp S. Because
α0 (0) ∈ Tp S and α0 (0).α0 (0) = 1, one has the existence of some θ for
which
α0 (0) = cos θe1 + sin θe2 .
Therefore
A(α0 (0)).α0 (0) = cos θA(e1 ).α0 (0) + sin θA(e2 ).α0 (0)
= λ1 cos θe1 .α0 (0) + λ2 sin θe2 .α0 (0)
= λ1 cos2 θ + λ2 sin2 θ + (λ1 + λ2 )e1 .e2 cos θ sin θ
= λ1 cos2 θ + λ2 sin2 θ.
Either λ1 , λ2 are both positive or both negative and so in either case
we have
|A(α0 (0)).α0 (0)| = |λ1 | cos2 θ + |λ2 | sin2 θ.
If N denotes the unit normal to S we saw in class that α00 (0).N =
A(α0 (0)).α0 (0) and thus, denoting by σ = min{|λ1 |, |λ2 |}, we have
|α00 (0)| ≥ |α00 (0).N | = |A(α0 (0)).α0 (0)| = |λ1 | cos2 θ + |λ2 | sin2 θ
≥ σ cos2 θ + σ sin2 θ = σ = min{|λ1 |, |λ2 |}.
(2) No. The plane P = {z = 0} has principal curvatures zero but the
curve α(t) = (cos t, sin t, 0) doe not have curvature zero.
(3) Before I solve it let me make some remarks. If Rθ : R3 −→ R3
denotes a rotation by angle θ which fixes the z-axis, then all surfaces
we consider have that Rθ (S) = S, i.e., the surfaces are invariant
under rotations which fix the z-axis. The consequence of this remark
is that if N (p) denotes the unit normal vector at the point p ∈ S,
then N (Rθ (p)) = Rθ (N (p)). In other words, the image of the Gauss
map is invariant under rotations which fix the z-axis.
Each of the surfaces is of the form S = {f (x, y, z) = 0} and thus
N = ∇f |∇f |−1 . Our previous remark means it suffices to look at
N (x, 0, z).
4
ANDRÉ NEVES
a) f (x, y, z) = z − x2 − y 2 and thus
N (x, 0, z) = (−2x, 0, 1)(4x2 + 1)−1/2 ,
where z = x2 and x ∈ R. It is simple to see that
x −→ (−2x, 0, 1)(4x2 + 1)−1/2
maps the real line into S 2 ∩ {y = 0, z > 0}, where S 2 denotes
the unit sphere. Because the image of the Gauss map must be
invariant under rotations which fix the z-axis we obtain that
the image must be S 2 ∩ {z > 0}, i..e, the northern hemisphere.
b) f (x, y, z) = z 2 + 1 − x2 − y 2 and thus
N (x, 0, z) = (−x, 0, z)(x2 + z 2 )−1/2 ,
where 1 + z 2 = x2 . The curve 1 + z 2 = x2 , y = 0 has two
connected components and, without loss
√ of generality, we can
consider just the one given by x = 1 + z 2 , y = 0. Why is
that? Well, if C1 , C2 are the two connected components then
C2 = Rπ (C1 ), i.e, one connected component can be obtained
by the other if we rotate it 180 degrees around the z-axis and
so, for the purpose of computing the image of the Gauss map,
it is enough to just consider C1 because we already know that
the image will be invariant by rotations which keep the z-axis
fixed.
In
√ sum, it suffices to see what is the image of (recall x =
1 + z2)
p
N (x, 0, z) = (− 1 + z 2 , 0, z)(1 + 2z 2 )−1/2 , z ∈ R.
We have
lim N (x, 0, z) = (−1, 0, 1)2−1/2 , lim N (x, 0, z) = (−1, 0, −1)2−1/2
z→+∞
z→−∞
and
z
1
1
−√ < √
< √ for all z ∈ R.
2
2
2
1 + 2z
Therefore the image of N (x, 0, z) is the part of the great circle
S 2 ∩ {y = 0} which lies in the sector
{(cos θ, 0, sin θ) | 3π/4 < θ < 5π/4}.
The image of the Gauss map is then
{x2 + y 2 + z 2 = 1} ∩ {|z| < 2−1/2 }.
GAUSS MAP
5
c) f (x, y, z) = cosh2 z − x2 − y 2 and thus
N (x, 0, z) = (−x, 0, sinh z cosh z)(x2 + sinh2 z cosh2 z)−1/2 ,
where cosh2 z = x2 . The curve cosh2 z = x2 , y = 0 has two
connected components and so it is enough to consider the one
given by x = cosh z, y = 0. In this case the expression simplifies
to
N (x, 0, z) = (−1, 0, sinh z)(1 + sinh2 z)−1/2 = (−1, 0, sinh z) cosh−1 z.
We have
lim N (x, 0, z) = (0, 0, 1), lim N (x, 0, z) = (0, 0, −1)
z→+∞
z→−∞
and
sinh z
< 1 for all z ∈ R.
cosh z
Thus the image of N (x, 0, z) is S 2 ∩ {y = 0} ∩ x < 0 and the
image of the Gauss map is the sphere minus the north pole and
minus the south pole.
−1 <
(a)
a)
∂x
∂x
= (1 − u2 + v 2 , 2uv, 2u),
= (2uv, 1 − v 2 + u2 , −2v)
∂u
∂v
Thus
∂x ∂x
g11 =
.
= (1 − u2 + v 2 )2 + 4u2 v 2 + 4u2 = (1 + u2 + v 2 )2
∂u ∂u
∂x ∂x
.
= 2uv(1 − u2 + v 2 ) + 2uv(1 − v 2 + u2 ) − 4uv = 0
∂u ∂v
∂x ∂x
.
= (1 + u2 − v 2 )2 + 4u2 v 2 + 4v 2 = (1 + u2 + v 2 )2
=
∂v ∂v
b)
g12 =
g22
∂2x
∂2x
∂2x
=
(−2u,
2v,
2),
=
(2u,
−2v,
−2),
= (2v, 2u, 0)
(∂u)2
(∂v)2
∂u∂v
To compute the exterior normal we do
∂x ∂x
×
∂u
∂v
3
= (−2u − 2uv 2 − 2u, 2v 3 + 2vu2 + 2v, 1 − 2u2 v 2 − u4 − v 4 )
X=
= (1 + u2 + v 2 )(−2u, 2v, 1 − u2 − v 2 )
and so the normal is given by
N=
X
= (1 + u2 + v 2 )−1 (−2u, 2v, 1 − u2 − v 2 ).
|X|
6
ANDRÉ NEVES
Finally, we obtain that
e = N.
∂2x
= 2(1 + u2 + v 2 )−1 .(1 + u2 + v 2 ) = 2
(∂u)2
∂2x
= (1 + u2 + v 2 )−1 (−2uv + 2uv) = 0
∂u∂v
∂2x
g = N.
= (1 + u2 + v 2 )−1 (−2(1 + u2 + v 2 )) = −2.
(∂v)2
f = N.
a11
a21
c) The principal curvatures are the eigenvalues of
−1
a12
e f
g11 g12
=
a22
g12 g22
f g
2 0
(1 + u2 + v 2 )−2
0
=
0 −2
0
(1 + u2 + v 2 )−2
2(1 + u2 + v 2 )−2
0
=
.
0
−2(1 + u2 + v 2 )−2
Because this matrix is already in diagonal form we obtain
that the eigenvalues are
λ1 = 2(1 + u2 + v 2 )−2 and λ2 = −2(1 + u2 + v 2 )−2
d) The coordinate curves are the curves given by
u(t) = x(x0 + t, y0 ) and v(t) = x(x0 , y0 + t).
We know that
∂x
∂x
(x0 + t, y0 ), v 0 (t) =
(x0 , y0 + t).
u0 (t) =
∂u
∂v
Hence we have
∂x
∂x
2
∂x
2
A(u0 (t)) = a11
+ a12
=
=
u0 (t)
2
2
2
2
∂u
∂v
(1 + u + v ) ∂u
(1 + u + v 2 )2
and
∂x
∂x
2
∂x
2
A(v 0 (t)) = a21
+ a22
=−
=−
v 0 (t).
2
2
2
2
∂u
∂v
(1 + u + v ) ∂v
(1 + u + v 2 )2
Thus the coordinate curves are lines of curvature.
• We have
∂φ
∂φ
= (−r(s) sin θ, r(s) cos θ, 0),
= (r0 (s) cos θ, r0 (s) sin θ, z 0 (s))
∂θ
∂s
and so
∂2φ
= (−r(s) cos θ, −r(s) sin θ, 0)
(∂θ)2
GAUSS MAP
7
∂2φ
= (−r0 (s) sin θ, r0 (s) cos θ, 0)
∂θ∂s
∂2φ
= (r00 (s) cos θ, r00 (s) sin θ, z 00 (s))
(∂s)2
To compute the normal we do
∂φ ∂φ
×
= (z 0 r cos θ, zr0 sin θ, −rr0 )
∂θ
∂s
and thus
X
N=
= (z 0 cos θ, z 0 sin θ, −r0 ).
|X|
X=
Therefore
A11 =
∂2φ
.N = −rz 0 ,
(∂θ)2
A12 =
∂2φ
.N = 0,
∂θ∂s
A22 =
∂2φ
.N = r00 z −z 00 r
(∂s)2
(b) We have
g11 =
∂φ ∂x
. (u, v),
∂u ∂u
(a)
g12 =
∂φ ∂x
. (u, v),
∂u ∂v
g22 =
∂φ ∂x
. (u, v).
∂v ∂v
g11 = a2 cos2 u cos2 v + b2 cos2 u sin2 v + c2 sin2 u
g12 = (b2 − a2 ) cos u cos v sin u sin v
g22 = sin2 u(a2 sin2 v + b2 cos2 v)
(b)
g11 = a2 cos2 v + b2 sin2 v + 4u2
g12 = (b2 − a2 )u cos v sin v
g22 = u2 (a2 sin2 v + b2 cos2 v)
(c)
g11 = a2 cosh2 v + b2 sinh2 v + 4u2
g12 = (b2 + a2 )u cosh v sinh v
g22 = u2 (a2 sinh2 v + b2 cosh2 v)
(d)
g11 = a2 cosh2 u cos2 v + b2 cosh2 u sin2 v + c2 sinh2 u
g12 = (b2 − a2 ) cosh u cos v sinh u sin v
g22 = sinh2 u(a2 sin2 v + b2 cos2 v)
8
ANDRÉ NEVES
(c) We are using the chart
φ : Q −→ R3
φ(x, y) = (x, y, f (x, y)).
Thus
∂φ
= (1, 0, fx ),
∂x
g11
∂φ
= (0, 1, fy )
∂y
and so
∂φ ∂φ
∂φ ∂φ
∂φ ∂φ
=
.
= 1 + fx2 , g12 =
.
= fx fy , g22 =
.
= 1 + fy2 .
∂x ∂x
∂x ∂y
∂y ∂y
Hence, from class,
Z q
Z q
2
A=
g11 g12 − g12 dxdy =
1 + fx2 + fy2 dxdy.
R
R
(4) The surface S in question is parametrized by
φ : (0, 2π) × (0, l) −→ R3
φ(θ, s) = (ρ(s) cos θ, ρ(s) sin θ, g(s)),
where we are assuming that the curve C on the plane given by C(s) =
(ρ(s), g(s)) is parametrized by arc-length, meaning C 0 .C 0 = 1. We
also assume that ρ > 0. We have
∂φ
∂φ
= (−ρ(s) sin θ, ρ(s) cos θ, 0),
= (ρ0 (s) cos θ, ρ0 (s) sin θ, g 0 (s))
∂θ
∂s
and so
∂φ ∂φ
∂φ ∂φ
.
= ρ2 (s), g12 =
.
= 0,
g11 =
∂θ ∂θ
∂θ ∂s
∂φ ∂φ
.
= (ρ0 (s))2 + (g 0 (s))2 = C 0 .C 0 = 1.
∂s ∂s
Using the formula for the area we have
Z
Z 2π Z l
Z l
q
2
A=
g11 g22 − g12 dθds =
ρ(s)dθds = 2π
ρ(s)ds.
g22 =
[0,2π]×[0,l]
0
0
0
Now to part b). You should make a picture to convince yourself
of this, but for a torus of revolution the curve C should be a circe of
radius r centered at a point (a, b) and, without loss of generality, we
can assume that b = 0, i.e, C(s) = (a + r cos(s/r), r sin(s/r)) where
0 < s < 2πr. Applying the above formula we get
Z 2πr
Z 2πr
A = 2π
ρ(s)ds = 2π
(a + r cos(s/r))ds = 4πra.
0
0
GAUSS MAP
9
(5) Exercise 11: We start by parametrizing the surface S. This is not
totally obvious but it makes sense. You should run what I do below
when α is a curve of radius 1 in the xy-plane (in which case S is a
nice looking torus if inner radius r.)
Parametrize α by arc length. We known that α00 (s) = ~k is orthogonal to α0 (s) and set n = α00 /kα00 |. Consider the vector b(s) so that
α0 , n, b form a positive orthonormal basis of R3 .
The parametrization is
x(s, v) = α(s) + r(n(s) cos v + b(s) sin v),
0 < v < 2π, 0 < s < l
To compute the area we need to compute the metric. Let’s recalling all we know about α0 , n and b.
(a) α0 .α0 = 1 and n.α0 = 0α. Also n.n = 1 and so n0 .n = 0. For the
same reason b0 .b = 0.
(b) The fact that b(s).n(s) = 0 implies, by differentiation, that
n0 .b + b0 .n = 0.
(c) Less obvious is that b0 .α0 = 0. This follows because α00 is parallel
to n and so b.α00 = 0. Hence, by differentiating, we get
b.α0 = 0 =⇒ b0 α0 + b.α00 = 0 =⇒ b0 .α0 = −b.α00 = 0.
(d) Because α0 , n, b form an orthonormal basis we have
n0 = (n0 .α0 )α0 + (n0 .n)n + (n0 .b)b = (n0 .α0 )α0 + (n0 .b)b
and
b0 = (b0 .α0 )α0 + (b0 .n)n + (b0 .b)b = (b0 .α0 )α0 − (b.n0 )n = −(b.n0 )n.
Using this we obtain
∂x
= α0 (s) + r(n0 (s) cos v + b0 (s) sin v)
∂s
∂x
= r(−n(s) sin v + b(s) cos v),
∂v
Recalling that
n0 = (n0 .α0 )α0 + (n0 .b)b
b0 = −(b.n0 )n
we get
∂x
= (1 + r cos vn0 .α0 )α0 (s) + r cos v(n0 .b)b − r sin v(n0 .b)n.
∂s
Thus, because α0 , n, b form an orthonormal basis, we have
g11 = (1 + r cos v(n0 .α0 ))2 + r2 (n0 .b)2 ,
g12 = r2 (n0 .b),
g22 = r2
and so
q
2 = r|1 + r cos v(n0 .α0 )| = r + r 2 cos v(n0 .α0 ),
g11 g22 − g12
10
ANDRÉ NEVES
where we assume that n0 .α0 > 0 (we can always choose n so that this
will be true because we are assuming that α00 never vanishes and
n0 .α0 = −n.α00 ). Thus
Z l Z 2π q
Z l Z 2π
2
A=
g11 g22 − g12 dvds = r
1 + r cos v(n0 .α0 )dvds
0
0
Now
Z lZ
0
0
2π
cos v(n0 .α0 )dvds =
Z
0
2π
Z
0
0
l
cos vdv
(n0 .α0 )ds = 0
0
and so
Z lZ
2π
1dvds = 2πrl,
A=r
0
0
where l is the length of α because we are parametrizing by arc length.
(6) Given X, Y ∈ Tp S, we have defined A(X, Y ) = −DX N.Y , where
N is a choice of unit normal to S. Hence, if λ1 (p) = λ2 (p) = λ(p),
then this means A(X, Y ) = λ(p)X.Y , which is equivalent to say that
DX N = −λ(p)X for all X ∈ Tp S (why?).
If φ is a chart with coordinates (x1 , x2 ) then we have for i = 1, 2.
∂(N ◦ φ)
∂φ
= D∂xi φ N = −λ(p)
∂xi
∂xi
Thus, differentiating again
∂ 2 (N ◦ φ)
∂λ ∂φ
∂2φ
− λ(p)
=−
.
∂xi ∂xj
∂xj ∂xi
∂xi ∂xj
Thus
∂ 2 (N ◦ φ)
∂2φ
∂λ ∂φ
+ λ(p)
=−
.
∂xi ∂xj
∂xi ∂xj
∂xj ∂xi
The left hand side is invariant under switching i with j and so the
right hand side must be as well. Therefore
∂λ ∂φ
∂λ ∂φ
−
=0
∂xj ∂xi ∂xi ∂xj
∂λ
but the vectors involved are linearly independent which means ∂x
=
j
0 for j = 1, 2. So λ ◦ φ is constant, which means λ is constant.
To prove (ii) we note that from (i) we have X = N ◦ φ constant
in U because λ is identically zero. Hence, if we set f (x) = X.φ(x),,
then for i = 1, 2
∂f
∂φ
∂φ
= X.
= N ◦ φ.
= 0.
∂xi
∂xi
∂xi
Thus f is constant in U which means φ(U ) is contained in the plane
{~x ∈ R3 : X.~x = c}
GAUSS MAP
11
for some c. The arbitrariness of φ and the fact S is connected imply
that all S is contained in the same plane.