Solutions of Homework 9
The solution of the second part of the exercise 2 and the solution of the exercise 4 (which are not
compulsory) see at the end of this file.
1. Calculate the shape operator for the sphere x2 + y 2 + z 2 = R2 :
(
r(ϕ, θ)
x = R sin θ cos ϕ
y = R sin θ sin ϕ .
z = R cos θ
Calculate principal curvatures, Gaussian and mean curvature for this sphere.
Solution We use results of calculations of vectors rθ , rϕ and of unit normal vector n(θ, ϕ) from the
exercise 1 of the Homework 8. By the definition (see lecture notes) the action of shape operator on any
tangent vector v is given by the formula Sv = −∂v S. Hence for basis vectors rθ = ∂θ , rϕ = ∂ϕ (see the
calculations in the Solutions of Homework 8) we have
µ
Srθ = −∂θ n(θ, ϕ) = −∂θ
and
µ
Srϕ = −∂ϕ n(θ, ϕ) = −∂ϕ
r(θ, ϕ)
R
r(θ, ϕ)
R
¶
µ
=−
¶
µ
=−
∂θ r(θ, ϕ)
R
¶
∂ϕ r(θ, ϕ)
R
=−
rθ
R
¶
=−
rϕ
R
I
We see that shape operator is equal to S = − R
, where I is an identity operator. Its matrix in the basis
∂θ , ∂ϕ is equal to
µ 1
¶
0
R
−
.
0 R1
In the case if we choose the opposite direction for unit normal vector then we will come to the same answer
just with changing the signs: if n → −n, S → −S.
We see that principal curvatures, i.e. eigenvalues of shape operator are the same:
λ1 = λ2 = −
1
1
, i.e. κ1 = κ2 = −
R
R
(if we choose the opposite sign for n then κ1 = κ2 = R1 ). Thus we can calculate Gaussian and mean
curvature: Gaussian curvature
1
K = κ1 · κ2 = det S = 2 .
R
Mean curvature
2
H = κ1 + κ2 = Tr S = − .
R
If we choose the opposite sign for n then S → −S, principal curvatures change the sign, Gaussian curvature
K = κ1 · κ2 does not change but mean curvature H = κ1 + κ2 will change the sign: if n → −n then H = R2 .
2. Calculate the shape operator for the cylinder x2 + y 2 = R2 :
(
r(h, ϕ)
x = R cos ϕ
y = R sin ϕ .
z=h
Calculate principal curvatures, Gaussian and mean curvature for this cylinder.
†
What values the normal curvature of an arbitrary curve on the cylinder of radius R can take? (You
may consider first horizontal circle, vertical line and helix.)
1
Solution
To calculate the shape operator for the cylinder we use results of calculations of vectors rh , rϕ and of unit
normal vector n(ϕ, h) from the exercise 1b of the Homework 8. By the definition the action of shape operator
on any tangent vector v is given by the formula Sv = −∂v S. Hence for basis vectors rϕ = ∂ϕ , rh = ∂h (see
the calculations in the Solutions of Homework 8) we have


cos ϕ
Srh = −∂h n(ϕ, h) = −∂h  sin ϕ  = 0
0
and

 
sin ϕ
cos ϕ
rϕ
Srϕ = −∂ϕ n(ϕ, h) = −∂ϕ  sin ϕ  =  − cos ϕ  = −
R
0
0




−R sin ϕ
cos ϕ
(Recall that n(h, ϕ) =  sin ϕ  and rϕ =  R cos ϕ  (See the solutions of Homework 8).)
0
0
br
For an arbitrary tangent vector X = arh + brϕ , SX = − Rϕ . Shape operator transforms tangent vectors
to tangent vectors. Its matrix in the basis rh , rϕ equals to
µ
¶
0 0
−
0 R1

In the case if we choose the opposite direction for unit normal vector then we will come to the same answer
just with changing the signs: if n → −n, S → −S.
We see that principal curvatures, i.e. eigenvalues of shape operator are:
κ1 = 0, κ2 = −
1
R
(if we choose the opposite sign for n then κ1 = κ2 = R1 ). Thus we can calculate Gaussian and mean
curvature: Gaussian curvature
K = κ1 · κ2 = det S = 0 .
Mean curvature
1
.
R
If we choose the opposite sign for n then S → −S, principal curvatures change the sign, Gaussian curvature
K = κ1 · κ2 remains the same but mean curvature H = κ1 + κ2 will change the sign: if n → −n then H = R1 .
H = κ1 + κ2 = Tr S = −
Solution of the last part † of this problem see in the end of this file
3. Calculate the shape operator for the cone x2 + y 2 − k 2 z 2 = 0:
(
x = kh cos ϕ
r(h, ϕ)
y = kh sin ϕ .
z=h
Calculate principal curvatures, Gaussian and mean curvature for this cone.
Solution
To calculate the shape operator for the cone we use the results of calculations of vectors rh , rϕ and
of unit normal vector n(ϕ, h) from the exercise 1c of the Homework 8. Recall that for the cone cone
x2 + y 2 − k 2 z 2 = 0,
(
x = kh cos ϕ
r(ϕ, h)
y = kh sin ϕ
z=h
2
and




k cos ϕ
−kh sin ϕ
∂r
∂r
rh =
=  k sin ϕ  , rϕ =
=  kh cos ϕ 
∂h
∂ϕ
1
0


cos ϕ
and normal unit vector n = √k12 +1  sin ϕ  (See for the detail the solution of Homework 8).
−k
By the definition the action of shape operator on any tangent vector v is given by the formula Sv =
−∂v S. Hence for basis vectors rh = ∂h , rϕ = ∂ϕ


cos ϕ
1
Srh = −∂h n(ϕ, h) = − √
∂h  sin ϕ  = 0
k2 + 1
−k
and



sin ϕ
cos ϕ
1
 − cos ϕ  =
Srϕ = −∂ϕ n(ϕ, h) = −∂ϕ = − √
∂ϕ  sin ϕ  = √
2+1
k2 + 1
k
−k
0


−kh sin ϕ
1
rϕ
− √
since rϕ =  kh cos ϕ  .
k k2 + 1 h
0

1
We see that for an arbitrary tangent vector X = arh + brϕ ,
SX = S(arh + brϕ ) = −
b
√
rϕ .
kh k 2 + 1
. Shape operator transforms tangent vectors to tangent vectors. Its matrix in the basis rh , rϕ equals to
µ
¶
0
0
−
0 hk√11+k2
In the case if we choose the opposite direction for unit normal vector then we will come to the same answer
just with changing the signs: if n → −n, S → −S.
We see that principal curvatures, i.e. eigenvalues of shape operator are:
κ1 = 0, κ2 = −
1
√
hk 1 + k 2
(if we choose the opposite sign for n then κ1 = κ2 = hk√11+k2 ). Thus we can calculate Gaussian and mean
curvature: Gaussian curvature
K = κ1 · κ2 = det S = 0 .
Mean curvature
H = κ1 + κ2 = Tr S = −
1
.
hk 1 + k 2
√
If we choose the opposite sign for n then S → −S, principal curvatures change the sign, Gaussian curvature
K = κ+ · κ− remains the same but mean curvature H = κ+ + κ− will change the sign: if n → −n then
H = hk√11+k2 .
4
†
Calculate the shape operator, Gaussian and mean curvature for the surface
(x = u
r(u, v)
y=v
.
z = F (u, v)
3
(1)
at the origin, i.e. at the point u = v = 0 in the case if F = Au2 + 2Buv + Cv 2 .
Consider the case if F (u, v) = uv (the surface is ”saddle”).
See the solution in the end of this file.
5 Assume that the action of the shape operator at the tangent coordinate vectors ru = ∂u , rv = ∂v at the
given point p of the surface r = r(u, v) is defined by the relations: S(∂u ) = 2∂u +2∂v and S(∂v ) = −∂u +5∂v .
Calculate principal curvatures, Gaussian and mean curvatures of the surface at this point.
Solution
We see that the matrix of the shape operator in the basis ∂u , ∂v is equal to
¶
µ
2 −1
S=
2 5
Hence Gaussian curvature K = det S = 12 and mean curvature H = Tr S = 7. To calculate principal
curvatures k1 , k2 note that
½
k1 + k2 = H = 7
k1 · k2 = K = 12
Hence k1 = 3, k2 = 4; κ1 , k2 are eigenvalues of the shape operator.
6 † Let A and B be two given points on the sphere with radius R with spherical coordinates {θA , ϕA }
and {θB , ϕB }.
a† ) Show that the shortest curve joining these points is the arc of the great circle.
b† ) Find the length of this curve.
This problem is the same as the last question of the last problem in the Homework 8 (see the solutions
of the Homework 8).
Solution of the last part † of the problem 2 (not compulsory)
For the points of the circle C1 normal curvature is equal (up to a sign) to 1/R. Indeed consider
point moving around C1 with constant speed (x = R cos ωt, y = R sin ωt, z = h0 ). Speed is equal to
v = ωR and the acceleration vector a = −Rω 2 cos ωt∂x − Rω 2 sin ωt∂y is orthogonal to the surface: a =
ω 2 Rn, where we choose unit normal vector to be n = −(x/R, y/R, 0) at the points (x, y, z) of the cylinder
(n = (− cos ωt, − sin ωt, 0)). Normal curvature is equal to κn = (a, n)/(v, v) = ω 2 R/ω 2 R2 = R1 . If we
choose n = +(x/R, y/R, 0) = (cos ωt, sin ωt, 0) then normal curvature would change a sign: n → −n,
(a, n) → −(a, n) and κnormal → −κnormal = − R1 .
†Consider now a point moving around the curve C2 (helix) with constant speed: x = R cos ωt, y =
R sin ωt, z = vt. Then velocity vector is equal to v = −Rω sin ωt∂x + Rω cos ωt∂y + v∂z and a =
√
−Rω 2 cos ωt∂x − Rω 2 sin ωt∂y (it is normal (centripetal) acceleration ). We see that v = ω 2 R2 + v 2 ,
an = (a, n) = ω 2 R (we choose n = −(x/R, y/R, 0) at the point (x, y, z) of the cylinder) and normal curvature is equal to
ω2 R
κn = 2 2
ω R + v2
†
(In the case if we change n → −n then normal curvature will change a sign too) One can see that
0 < κnormal =
ω2 R
1
ω2 R
≤
=
2
2
2
2
2
ω R +v
ω R
R
On the cylinder normal curvature varies from zero (for straight line) to 1/R: 0 ≤ κnormal ≤
4
1
R
For the straight line C3 normal curvature obviously is equal to 0: e.g. if particle moves on C3 with
constant speed, then its acceleration is equal to zero.
† Now consider an arbitrary curve on the cylinder. We know that normal curvature takes values in
the interval between principal curvatures. According to calculation of shape operator we see that principal
curvatures for cylinder are equal to
1
κ− = 0, κ+ = .
R
£
¤
Hence for an arbitrary curve on the cylinder (2) normal curvature takes values in the interval 0, R1 . Of
course this can be shown straightforwardly∗∗ .
Remark Note that one can consider on the cylinder x2 + y 2 = R2 a circle of very small radius r. The
curvature (usual curvature which we studied before) of this circle will be equal to 1/r. We see that usual
curvature of curve can be very big, but normal curvature cannot be bigger than 1/R.
Solution of the exercise 4 (not compulsory)
Solution
By the definition the action of shape operator on any tangent vector v is given by the formula Sv =
−∂v S. Hence for basis vectors ru = ∂u and rv = ∂v we have
Sru = −
∂n(u, v)
∂n(u, v)
, Srv = −
∂u
∂u
To calculate these vectors in the origin (at the point u = v = 0) we need to know the value of normal unit
vector field n(u, v) in the vicinity of the origin. Recall that the normal unit vector field n(u, v) equals to


−Fu
1
 −Fv 
n(u, v) = p
1 + Fu2 + Fv2
1
(see the Solution of Homework 8). Hence



−Fu
∂n(u, v)
∂ 
1
 −Fv  ,
p
Sru = −
=−
∂u
∂u
1 + Fu2 + Fv2
1
∗∗
Show straightforwardly that for an arbitrary curve on the cylinder (2) normal curvature takes values
£
¤
in the interval 0, R1 . For convenience we choose normal unit vector attached at the point (x, y, z) of the
cylinder n = −(x/R, y/R, 0). (In this case normal curvature of circle and helix were positive). Consider
an arbitrary smooth curve r = r(ϕ(t), h(t)) on the cylinder (2) and take any point P = r = (x, y, z) on it
(x2 + y 2 + z 2 = R2 ). Let v be velocity vector at this point, v = (vx , vy , vz ). Consider expansion of the vector
v on horizontal and vertical components. v = vhorisont +vvertical , vhorisontal = vx ∂x +vy ∂y , vvertical = vz ∂z .
Normal (centripetal) acceleration an is equal to an = −an n, where an =
normal curvature:
κnormal =
2
vhorisontlal
R
(v, v)
=
|vhorisontal |2
R
=
2
vx
+vy2
R .
We have for
2
vx2 + vy2
vhorisontlal
1
≤ .
=
2
2
R(vhorisontlal
R(vx2 + vy2 + vz2 )
R
+ vvertical
)
We see that normal curvature could be less or equal than 1/R (normal curvature of the circle) and bigger
or equal than 0 (normal curvature for straight line) (See also the example in Lecture notes.)
5
µ
On the other hand if F = Au2 + 2Buv + Cv 2 then it is easy to see that
u = v = 0. Hence at the origin

∂
∂u
¶
√
1
1+Fu2 +Fv2
= 0 at the point

−Fu
∂n(u, v)
∂ 
1
 −Fv  =
p
Sru = −
=−
∂u
∂u
1 + Fu2 + Fv2
1

 

−Fu
Fuu
∂ 
1
−p
−Fv  =  Fvu 
1 + Fu2 + Fv2 ∂u
1
0
since √
1
1+Fu2 +Fv2

= 0 at the origin.
Analogously we come to


 

−Fu
Fuv
∂ 
∂n(u, v)
1
 −Fv  =  −Fvv 
p
=−
Srv = −
∂v
∂v
1 + Fu2 + Fv2
1
0
Now recall that tangent vectors ru , rv are the following
 
 
1
0
∂r(u, v)  
∂r(u, v)  
ru =
=
0
, rv =
=
1
∂u
∂v
Fu
Fv
(See the exercise 1d) in the Homework 8 and the solution of this exercise in the Solutions of Homework 8)
hence at the origin (at the point u = v = 0) for F = Au2 + 2Buv + Cv 2
 
 
1
0
¯
¯
ru ¯u=v=0 ==  0  , rv ¯u=v=0 ==  1 
0
0
Hence we have that at the origin


Fuu
∂n(u, v)
Sru = −
= −  Fvu  = Fuu ru + Fuv rv
∂u
0


Fuv
∂n(u, v)
Srv = −
= −  Fvv  = Fuv ru + Fvv rv
∂v
0
i.e. matrix of the shape operator S is
µ
S=
Fuu
Fvu
Fuv
Fvv
¶
µ
=
2A
2B
2B
2C
¶
.
Gaussian curvature at the origin equals to
2
= 4AC − 4B 2 . .
K = det S = Fuu Fvv − Fuv
Mean curvature
H = Tr S = Fuu + Fvv = 2A + 2C .
For the special case of the saddle shape operator
µ
¶ µ ¶
Fuu Fuv
01
S=
=
Fvu Fvv
10
Gaussian curvature at the origin equals to
2
K = det S = Fuu Fvv − Fuv
= −1 .
Mean curvature
H = Tr S = Fuu + Fvv = 0 .
6