The document provides solutions to tutorial problems on differential geometry. It first shows that the differential of a function from a surface to 3D space is linear. It then calculates the Gauss map, Weingarten map, and principal curvatures for a sphere, surface of revolution, and other surfaces. The solutions involve parametrizing the surfaces and computing derivatives of the parametrizations.

1. Sample Assignment For Reference Only
Tutorial 1
1. Show that the differential pdG of a function G , from a surface S to 3
 is linear.
(proof) Let be , ( )G F C S
 . It is sufficient that for any ,   ,
( )d F G dF dG      .
Assume that , p pp S X T S  . Then
( ) ( )
( )
p p p p p
p p p p p
d F G X X F G X F X G
dF X dG X dF dG X
     
   
     
   
(end)
2. Calculate the Gauss map, the Wiengarten map and the principal curvatures for
(a) A sphere of radius, R ,
(solution) cos cos , cos sin , sinx R u v y R u v z R u   !
sin cos , sin sin , cos
cos sin , cos cos , 0
u u u
v v v
x R u v y R u v z R u
x R u v y R u v z
    
    
  
  
2 2 2
, 0, cosE R F G R u  
2 2
cosH EG F R u  
 
 
 
2
2
cos cos , cos sin , sin
sin sin , sin cos ,0
cos cos , cos sin ,0
u
uv
v
r R u v R u v R u
r R u v R u v
r R u v R u v
   
 
  



2 2
3 3 3
( , , ) cos , ( , , ) 0, ( , , ) cosu v u v uv u vu v
r r r R u r r r r r r R u   
        
2
2
3
2
3 3
2
2
1 cos
( , , )
cos
1
( , , ) 0
1 cos
( , , ) cos
cos
u v u
u v uv
u v v
R u
L r r r R
H R u
M r r r
H
R u
N r r r R u
H R u
     
  
     
  
  
  
-weingarten map:
3 2
2 4 2 3 2
1
0
cos 01 1
1cos 0 cos
0
GL FM GM FN R u R
FL EM FM ENEG F R u R u
R
 
     
      
           
 
A
-Gauss map:

2. Sample Assignment For Reference Only
2 2 2
2
2 2
2
(cos cos ,cos sin ,sin cos ),
(cos cos ,cos sin ,sin cos )
cos
u v
u v
u u
r r R u v u v u v
r r R
n u v u v u v
R ur r
  
 
 

 
 

 
-principle curvature:
1 2
1
k k
R
   .
(b) A surface of revolution given by the curve ( )x f z rotated about the z axis, and
(solution)
( )cos
( )sin
x f z
y f z
z z





 
( ( )cos , ( )sin , )r f z f z z 

2
2
2 2
2 2 2 2
( ( )cos , ( )sin ,1)
( ( )sin , ( )cos ,0)
( ) 1, 0, ( ) , ( ) ( ) 1
( ( )cos , ( )sin ,0)
( ( )sin , ( )cos ,0)
( (
z
z
z
z
r f z f z
r f z f z
E r f z F G r f z H EG F f z f z
r f z f z
r f z f z
r f z




 
 
 
 
  
  
                 
   
 
  
 


 



)cos , ( )sin ,0)f z 
2 2
2
( , , ) ( ) ( ), ( , , ) 0, ( , , ) ( )z z z zz
r r r f z f z r r r r r r f z    
   
        
2 2
2
2 2 2
( ) ( ) ( )
,
( ) ( ) 1 ( ) 1
0 ( ) ( )
0,
( ) ( ) 1 ( ) ( ) 1 ( ) 1
f z f z f z
L
f z f z f z
f z f z
M N
f z f z f z f z f z
 
    
  
      
    
-weingarten map:
2
2
2 2
2
2
2 3/ 2
2 1/ 2
( )
( ) 0
( ) 11
( )( )( ( ) 1)
0 ( ( ) 1)
( ) 1
( )
0
( ( ) 1)
0 ( ( ) 1) ( )
f z
f z
f z
f zf z f z
f z
f z
f z
f z
f z f z
 
  
     
   
   
 
  
 
    
A
-Gauss map:

3. Sample Assignment For Reference Only
 2
1
( )cos , ( )sin , ( ) 9 )
( ) ( ) 1
n f z f z f z f z
f z f z
    
 

-principle curvature:
1 22 3/ 2
( )
, ( ) ( ) 1
( ( ) 1)
f z
k k f z f z
f z

     
 
(c) The surface of revolution about the z  axis of a circle in the xz  plane with center
( ,0,0)d with radius r d .
(solution)
 
 
2 22 2 2
( cos )cos
( cos )sin
sin
sin cos , sin sin , cos
( cos )sin ,( cos )cos ,0
, 0, ( cos ) , ( cos )
( cos )cos cos , ( cos
u
v
u v
u v
x d r u v
y d r u v
z r u
r r u v r u v r u
r d r u v d r u v
E r r F G r d r u H EG F r d r u
r r r d r u u v r d r
 

 
 
  
   
         
     


 
 
 
2
2
2
)cos sin , ( cos )sin
( cos cos , cos sin , sin )
( sin sin , sin cos ,0)
( ( cos )cos , ( cos )sin ,0)
( cos ) ( cos )cos
, 0, cos
( cos ) ( cos )
u
uv
v
u u v r d r u u
r r u v r u v r u
r r u v r u v
r d r u v d r u v
r d r u r d r u u
L r M N u
r d r u r d r u
 
   
 
    
 
        
 



-weingarten map:
2
2 2 2
2
1
0
( cos ) ( ) 01
cos( cos ) 0 cos 0
( cos )
d r u r r
ur d r u r u
d r u
 
   
   
       
A
-Gauss map:
(cos sin ,cos sin ,sin )n u v u v u 

-principle curvature:
1 2 2
1 cos
,
( cos )
u
k k
r d r u
 

(d) The surface parametrized by
 3 2 3 2 2 2
( , ) /3 , /3 ,r u v u u uv v v vu u v      .
(solution)

4. Sample Assignment For Reference Only
 
2 2
2 2
2 2
2 2 2 2 2 2 2 2 2 2
2 2 2 2 2 2 2
2
(1 ,2 ,2 )
(2 ,1 , 2 )
(1 ) , 0, (1 ) , (1 )
( 2 ,2 ,2), (2 ,2 ,0), (2 , 2 , 2)
2 ( 1),2 ( 1),1 ( )
4(1
u
v
u
uvu v
u v
r u v uv u
r uv v u v
E r u v F G u v H u v
r u v r v u r u v
r r u u v v u v u v
u v
L
  
   
          
     
        
 
 



  
 
2
2 2 2 2 2
2 2
2 2 2 2 2
) 4
, 0,
(1 ) 1
4(1 ) 4
(1 ) 1
M
u v u v
u v
N
u v u v
  
   
  
  
   
-weingarten map:
2 2 32 2
2 2 4 2 2
2 2 3
4
0
(1 )4(1 ) 01
4(1 ) 0 4(1 )
0
(1 )
u vu v
u v u v
u v
 
          
     
   
A
-Gauss map:
2 2 2 2 2 2
2 2 2
2 2
2 2 2 2 2 2
1
( 2 (1 ),2 (1 ),1 ( ))
(1 )
2 2 1
, ,
1 1 1
n u u v v u v u v
u v
u v u v
u v u v u v
       
 
  
  
      

-principle curvature:
1 22 2 3 2 2 3
4 4
,
(1 ) (1 )
k k
u v u v
  
   
Tutorial 2
1. Show that the second fundamental form II p is symmetric.
(proof) we take two arbitrary tangent vectors , pT S   and two arbitrary real number
,   . Then we have, first of all:
1
2 1 1 1 2
self-adjoint symmetrical
( , ) ( ( ), ) ( , ( )) ( ( ), ) ( , )
A
A A A

                     ,
which means that 2 is symmetrical. (end)

5. Sample Assignment For Reference Only
2. Show that the elementary symmetric functions 1 1( , , )i nS k k  are the coefficient of
i
x in
the expansion of 1 1(1 ) (1 )nk x k x  .
(proof) When 2n  ,
2 2
1 2 1 2 1 2 1 2(1 )(1 ) 1 ( ) 1k x k x k k x k k x S x S x         .
Therefor 0 1 1 2 1 2 2 1 2 1 21, ( , ) , ( , )S S k k k k S k k k k    .
When 3n  ,
1 2 3 1 2 3
2 3
1 2 2 3 1 3 1 2 3
(1 )(1 )(1 ) 1 ( )
( )
k x k x k x k k k x
k k k k k k x k k k x
       
   
So 0 1 1 2 3 2 1 2 2 3 1 3 3 1 2 31, , ,S S k k k S k k k k k k S k k k        .
…
1 2
1 2
2
1 2
1
1 2
(1 )(1 ) (1 ) 1
n
n i i i
i i i
n
n
k x k x k x x k x k k
x k k k
 
      
 
 
 
so 1 2
1 2
0 1 2 1 1
1
1, , , ,
n
i i i n n
i i i
S S k S k k S k k
 
       .
3. Calculate the frames for the sphere based on
(a) the standard parameterization
(solution)
(cos cos , cos sin , sin )
( sin cos , sin sin , cos )
( cos sin , cos cos , 0)
u
v
r u v u v u
r u v u v u
r u v u v

  
 



(b) stereographic projection
4. Calculate frames for
(a) The torus
(solution) the equation
( cos )cos
( cos )sin
sin
x a b u v
y a b u v
z b u
  

  
 
.
( , , ) ( sin cos , sin sin , cos )
( , , ) ( ( cos )sin ,( cos )cos ,0)
u u u u
v v v v
r x y z b u v b u v b u
r x y z a b u v a b u v


   
    

  

  
(b) The catenoid

6. Sample Assignment For Reference Only
cosh cos , cosh sin ,
sinh cos ,sinh sin ,1
cosh sin , cosh cos ,0
u
v
u u
x a v y a v z u
a a
u u
r v v
a a
u u
r a v a v
a a
  
 
  
 
 
  
 


Tutorial 3
1. Calculate the first fundamental form for
(a) The sphere of radius, R ,
(solution) ( cos cos , cos sin , sin )r R u v R u v R u

,

 
 
sin cos , sin sin , cos
cos sin , cos cos ,0
u
v
r
r R u v R u v R u
u
r
r R u v R u v
v

   


   





2 22 2 2
11 12 22, 0, cosu u v vg r R g r r g r R u     
  
.
Therefor the first fundamental form is the following:
2 2 2 2 2
cosR du R udv .
(b) The torus with inner radius, r and outer radius, R
(solution) the equation
( cos )cos
( cos )sin
sin
x a b u v
y a b u v
z b u
  

  
 
Where ,
2 2
R r R r
a b
 
  . Therefor
cos cos
2 2
cos sin
2 2
sin
2
R r R r
x u v
R r R r
y u v
R r
z u
  
   
  
   
   
  

 

.
i.e.

7. Sample Assignment For Reference Only
sin cos , sin sin , cos
2 2 2
cos sin , cos cos , 0
2 2 2 2
u u u
v v v
r R r R R r
x u v y u v z u
R r R r R r R r
x u v y u v z
  
  
      
        
   
  
  
2
2 2 2
11 22 12
( ) 1
, [( ) ( )cos ], 0
4 4
u u u
R r
g x y z g R r R r u g

           
Therefor the first fundamental form is  2 2 2 21
( ) [( ) ( )cos ]
4
R r du R r R r u dv     .
2. Use your answers to the previous question to find the length of
(a) A curve from the north pole of the sphere that winds twice around the sphere before
ending up at the south pole
(b) A curve that winds three times around the small randius for each time around the
major radius
3. In lectures we calculated 1E and shows that for the inertial frame
1 1 2
1 2 2 1 2
1 2
( , )
( , ) ( , )
( , )
x r x x
X x x x s x x
q x x
 
 
  
 
 
then 11(0,0) 0r  . By calculating 2 1 2 1, , ,E F F G , and 2G , show that all the second
derivatives of r and s are zero at (0,0).
4. Show that 2
12 22 11 11 22 12
1 1
2 2
F F G q q q    .
Tutorial 4
1. Construct an atlas for
(a) The torus
(b) the cylinder
from the charts for the circle from the lectures.
(solution) (a) Let
1
S be the circumference and
1 1
M S S  .
1 1
,U S V S 
: [ 1,1] ( , )U U  
: [ 1,1] ( , )V V  

8. Sample Assignment For Reference Only
 the atlas is {( , )}U V    .
(b) Let
1
S be the circumference and I be the open interval,
1 1
M S S  .
1
,
: [ 1,1]
: ( , )
U S V I
U
V a b


 
 

{( , )}U V    
2. Show that the function on the sphere that outputs the z  coordinate of the point is
differentiable.
(proof) the spherical co-ordinates
cos cos
cos sin
sin
x a u v
y a u v
z a u
 

 
 
cos
dz
a u
du
 . Therefor the function is differentiable.
3. Show that function on the real projective plane given by the angle the line makes with the
xy  plane is differentiable.
(proof)
2 2 2
:( , , ) arcsin ( 0)
a
f x y z xyz
x y z
 
 
2 2 2 2 2
2 2 2 2 2
2 2
2 2 2
( )
( )
f xz
x x y x y z
f yz
y x y x y z
x yf
z x y z

 
   

 
   

 
  
Because 0xyz  , the function is differentiable.
Tutorial 5
1. Write the coordinate vector-fields for cartesian coordinates on 2
 ,
x


and
y


in terms of the polar coordinate vector fields

9. Sample Assignment For Reference Only
r


and



(solution) 2 2( , ) cos
, , , tan , arctan
( , ) sin
x x r x r y y
r x y
y y r y r x x
 
 
 
  
    
  
2 2
cos
cos , sin
sin cos
,
r x r r
x r yx y
x r y r

 
   
 
   
 
 
  
 
Therefor
sin
cos
cos
sin
r
x x r x r r
r
y y r y r r
 

 
 

 
      
      
      
      
      
      
2. Calculate the vector-field transformation between stereographic coordinates and the
angular coordinates on the sphere,
2
S
Tutorial 6
Let , ,A y z B x z C x y
z y z x y x
     
     
     
.
1. Calculate the Lie derivative of B with respect to A.
(solution)

10. Sample Assignment For Reference Only
2 2 2 2
2
2
[ , ]AL B A B AB BA
y z x z x z y z
z y z x z x z y
y x z z x z
z z x y z x
x y z z y z
z z y x z y
z
yx y yz zx z
z z x z x y z
  
             
           
             
        
       
        
        
      
        
     
     
       
2 2 2 2
2
2
y x
z
xy x xz zy z
z z y z y x z x y
y x
x y


     
     
        
 
 
 
2. Show Af Bf Cf  for
2 2 2
( , , )f x y z x y z   .
(proof)
2 2 2
( , , )f x y z x y z  
2 2 2
2 2 2
2 2 2
( ) ( 2 ) 2 0
( ) ( 2 ) 2 0
( ) 2 2 0
Af y z x y z y z z y
z y
Bf x z x y z x z z x
z x
Cf x y x y z x y y x
y x
  
         
  
  
         
  
  
         
  
3. Use that fact to sketch the curves of the one parameter groups associated with ,A B , and
C .
(solution) curve C :
( )
( )
x x z
y y z
 

 
(where z is auxiliary variable)
The one parameter groups are 1 2: , :f z y f y x  .
1 2 1 1 2( ), ( ( )) ( )y f z x f f z f f z   
2 2 2
( , , ) ( ) ( )f x y z x z y z z  
( 2 ) 2 0
f f dy
Af y z y z z y
z y dz
 
       
 
…………………………………..(1)
( 2 ) 2 0
dx
Bf x z z x
dz
      …………………………………..(2)
2 2 0
dy dx
Cf x y y x
dz dz
      …………………………………..(3)

11. Sample Assignment For Reference Only
From (3),
dy dx
dx dz
 .
(1), (2) 1
dy dx
dx dy dz
dz dz
     
The tangent vector of the curve
( )
( )
x t
y y
z t





 
, (1,1,1) 

1
2
1
1
dy
y z c
dz
dx
x z c
dz
   
   
1
2
x t c
y t c
z t
 

  
 
When 1 2 0c c  , the curve C is the line parallel to 

and passing (0,0,0).
In general, the curve C is the line parallel to 

and passing 1 2( , ,0)c c .
Tutorial 7
Let M be a two-dimensional manifold with coordinates 1x and 2x . The Christoffel symbols for a
connection  are identically zero except for
1 1 2
12 21 2 11 2 2tan , cos sinx x x      
1. Calculate XY for
1 2 1
1
,
cos
X Y
x x x
 
 
 
.
(solution) j i
X i j
Y Y x
x

  

1 2 1 2
2
1
1, 0, , 0
cos
X X Y Y
x
   
j
j j k
i iki
Y
Y Y
x

   


12. Sample Assignment For Reference Only
1
1 1 1 1 2
1 11 12 21
2
2 2 1 2 2
1 11 12 2 2 21
2
1
1 1 1 1 2
2 21 22 22 2
2 2
2
2 2 1 2 2
2 21 222
0 0 tan 0 0
1
0 cos sin 0 sin
cos
1 1
( tan ) 0 0
cos cos
0 0 0 0
Y
Y Y Y x
x
Y
Y Y Y x x x
x x
Y
Y Y Y x
x x x x
Y
Y Y Y
x

          


          

  
            
   

         

1 2
1 2 22 2 2
1 sin sini j
X i j
Y X Y X Y x x
x x x x
   
       
   
.
2. Write down the equations for parallel transport for this connection.
(solution) for parallel transport, 2 2
0 sin 0X Y x
x

   

Assume that the vector field ( )Y t parallel transport according to the curve r .
:r 1 1
2 2
( )
( )
x x t
x x t
 

 
!
2 2
2
2 22
1 11
2
sin
0
( )cos
( )
sin 0
x dx
x t cx dt
x t cdx
x
dt

    
 
 

3. Combine them into a single equation and write down the solution.
(solution)
4. Pick a starting point and vector and solve for the coefficients in the solution.
5. Calculate the torsion of this connection.
(solution) k k k
ij ij jiT     : torsion tensor
0k
ij  . torsion=0
Tutorial 8
1. Write the standard metric for the sphere in terms of the coordinates  and  .
(solution)
cos cos
cos sin
sin
x
y
z
 
 

 

 
 
the standard metric:
( sin cos , sin sin ,cos )
( cos sin ,cos cos ,0)
r
r


    
   
  
 


2 2 2 2 2
11 12 221, 0, cos , cosg g g dS d d       
2. Write the standard metric for the torus in terms of the toroidal and poloidal angles.

13. Sample Assignment For Reference Only
(solution)
2 2 2
11 12 22
2 2 2 2 2 2
( cos )cos
( cos )sin
sin
(sinh cos ,sinh sin ,1)
( cosh sin , cosh cos ,0)
sinh 1, 0, cosh
(sinh 1) ( cosh )
u
v
x a b u v
y a b u v
z b u
u u
r u v
a a
u u
r a u a v
a a
u u
g g g a
a a
u u
dS du a dv
a a
  

  
 

 
   
  


3. Consider the metric
3 2 2
g dw dt dz   and the coordinate transformations
( , ) cosh( )cos( )
( , ) cosh( )sin( )
( , ) sinh( )
z x y x y
t x y x y
w x y x
 
 




(a) Calculate
2 2 2
z t w 
(b) Express g in the new coordinates
(solution)
cosh( )sin( )sinh( )cos( )
sinh( )sin( ) , cosh( )cos( )
cosh( ) 0
yx
x y
x y
z x yz x y
t x y t x y
w x w
    
     
 
    

   
     
(a)
2 2 2
cosh(2 )z t w x   
(b)
x y
x y
x y
dz z dx z dy
dw w dx w dy
dt t dx t dy
   

   
   
2 2 2
2 2 2 22 2 2 2
2 22 2
2 2 2 2 2 22 2
2 2 2 2
2 ( 2 )
( 2 )
( ) 2( ) ( )
(cosh ( ) sinh ( ))
x x y y x x y y
x x y y
x x x x y x y x y y y y
g dw dt dz
w dx w w dxdy w dy t dx t t dxdy t dy
z dx z z dxdy z dy
w t z dx w w t t z z dxdy w t z dy
x x dx  
   
            
      
                    
  
 
2 2 2
2 2 2 2 2 2
cosh ( )
[cosh ( ) sinh ( )] cosh ( )
x dy
x x dx x dy
 
   

  
4. Express the metric for Minkowski space
2 2 2 2
0 0 0 0g cdt dx dy dz    in terms of new
coordinates

14. Sample Assignment For Reference Only
0
0
0
cos( )
sin( )
t t
x r t
y r t
z z
 
 

 
 

(solution)
2 2 2 2
0 0 0 0g cdt dx dy dz   
0 00
00 0
0 0 0
0 00
0 10
sin( )0 sin( )
, ,
0 cos( ) cos( )
1 00
tz
z t
z t
z t
t tt
x r tx x r t
y y r t y r t
z zz




    
    
    
           
  
        
      
0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
sin( ) sin( ) sin( )( )
cos( ) cos( ) cos( )( )
t z
t z
t z
t z
dt t d t dt t dz dt
dx x d x dt x dz r t d r t dt r t d dt
dy y d y dt y dz r t d r t dt r t d dt
dz z d z dt z dz dz





          
          

     
             
           
     
2 2 2 2
0 0 0 0
2 2 2 2 2 2 2 2 2 2 2 2
2 2 2 2 2 2
2 2 2 2 2 2 2
sin ( )( 2 ) cos ( )( 2 )
( 2 )
( ) 2
g cdt dx dy dz
cdt r t d d dt dt r t d d dt dt dz
cdt r d d dt dt dz
c r dt r d r d dt dz
           
   
   
    
         
    
    