# Differential Geometry

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Differential Geometry

ACKNOWLEDGEMENTS

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Order Paper NowI sincerely thank Almighty God, who has given me the good health, physical and mental

strength to undertake and accomplish this work.

I would like to thank and highly appreciate the professional support, guidance and

encouragement of my supervisor ————– who despite other personal and professional

duties was always available to offer guidance.

I would also pass my gratitude to my parents and relatives for their constant love, financial

support and encouragement throughout the study. I also pass my thanks to my friends who

helped me in various ways including (mention a few friends)—————————————-

———and all who contributed to the success of this research paper.

Thank you all.

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List of Authors

Helgason, Sigurdur -1962

Kobayashi, Shoshichi-, 2014.

Meyer, Mark, et al.- 2003.

Rutherford, D E. -2012.

Schoen, Richard, and Shing- 1994.

Wardle, K L. -2008.

Willmore, T J. -2013.

Willmore, Thomas James. -2013.

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Abstract

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Introduction

Differential geometry is discipline that uses integral calculus, differential calculus and

multilinear algebra to solve geometry tasks. The space curves, theory of plane and surface in a

three-dimensional space Euclidean formed the basis for the advancement of differential

geometry in 18 th and 19 th century. This research paper provides step by step solutions to

various differential geometry tasks

Body of the work

References

Differential Geometry

Differential geometry is discipline that uses integral calculus, differential calculus and

multilinear algebra to solve geometry tasks. The space curves, theory of plane and surface in a

three-dimensional space Euclidean formed the basis for the advancement of differential

geometry in 18 th and 19 th century. This research paper provides step by step solutions to

various differential geometry tasks.

1. A parameterized curve

Consider a parameterized curve:

x = f(v) z = g(v) v ∈ (a,b)

in the xz-plane, with f(v) > 0 everywhere. Let S be the surface of the revolution obtained by

the revolving curve around z-axis. If u ∈ [0,2π] is the angular coordinate, calculate the

induced metric on S ⊂ R 3 in local coordinates (u,v).

The parameterization of the surface is given by

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(x,y,z) = (f(v)cos(u),f(v)sin(u),g(v)).

Therefore

and

Hence

.

2. The rationally symmetric metric

Consider a rotationally symmetric metric:

dr 2 + (φ(r)) 2 dθ 2

where φ : [0,b) → [0,∞), with φ(0) = 0 and φ(r) > 0 for r > 0. To check for smoothness of the

metric at the origin we need to change from polar to Cartesian coordinates:

x = rcosθ y = rsinθ

Write the metric in (x,y) coordinates and show that the components g xx , g xy = g yx , and g yy are

smooth if and only if

φ 0 (0) = 1 and φ (even) (0) = 0.

Hint: Write φ(r) = rψ(r). Calculate g xx , etc., in terms of ψ(r). Show that smoothness at r = 0

implies ψ(0) = 1 and ψ (odd) (0) = 0.

Assume that φ(r) is analytic so that it has a power series expansion

φ(r) = a 1 r + a 2 r 2 + a 3 r 3 + …

since φ(0) = 0. The function ψ(r) has a power series expansion

ψ(r) = a 1 + a 2 r + a 3 r 2 + …

Now r = p x 2 + y 2 and θ = arctan(y/x). Therefore

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and

Therefore

.

The metric is smooth if and only if is smooth. We immediately see that a 1 = ψ(0) =

: Then

so a 2 = 0. In fact, a power series in r is a smooth function of x and y if and only if all the odd

degree terms vanish. By induction, this means that a 4 = 0, a 6 = 0, etc. This completes the proof.

3. Setting φ(r) = sinh(r) in the previous problem, defined for r ∈ [0,∞), gives a rotationally

symmetric metric on R 2 . Show that this is isometric to the Poincar´e disc, i.e., the ( open ) unit

disc in R 2 with the metric

.

To preserve the spherical symmetry the isometry should take the form

x = f(r)cosθ y = f(r)sinθ

for some bijective function f(r) : [0,∞) → [0,1). Then

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x2 + y2 = f2

dx = f 0 cosθdr − fsinθdθ dy =

f 0 sinθdr + fcosθdθ

and rewriting g in the coordinates (r,θ) we find

.

For this to agree with the spherically symmetric metric we need

= 1 and i.e.,

and

(1)

(2) Solving the differential equation (1), we find i.e.,

.

To get f(0) = 0, the constant C must equal 0. One can then check that f(r) = tanh(r/2) also

satisfies the algebraic equation (2).

4. A proper affine function is a map f : R → R given by f(t) = bt+a for a ∈ R and b ∈ R> 0 . The set

of proper affine functions forms a Lie group: as a smooth manifold it is just the upper half

plane {(a,b) ∈ R 2 |b > 0}, and the group operation is given by composition.

a) Find a formula for the product (a,b)(c,d).

If f(t) = bt+a and g(t) = dt+c, then the composition (f ◦g)(t) = f(g(t)) = bdt+bc+a, so

(a,b)(c,d) = (bc + a,bd).

b) The identity element is e = (0,1). Find a formula for the left-invariant metric which takes

the value

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at e.

Let (a,b) be a point in the upper half plane U. The inverse of (a,b) is

(a,b) −1 = (−ab −1 ,b −1 ).

Let u = (u 1 ,u 2 ): v = (v 1 ,v 2 ) be two vectors in T (a,b) U. We first calculate the differ-

ential (dL(a,b)−1)(a,b). Let

γ(s) = (a,b) + s(u 1 ,u 2 ) = (a + su 1 ,b + su 2 )

be a path through (a,b) in the direction (u 1 ,u 2 ). Then

L ( a,b ) −1γ(s) = (−ab −1 ,b −1 )(a + su 1 ,b + su 2 )

= (b −1 (a + su 1 ) − ab −1 ,b −1 (b + su 2 ))

= (sb−1u1,1 + sb−1u2)

and thus

= (0,1) + sb −1 (u 1 ,u 2 )

−1

(dL(a,b)−1)(a,b)(u1,u2) = b (u1,u2). Therefore

h(u1,u2),(v1,v2)i(a,b) = h(dL ( a,b ) − 1 ) ( a,b)(u 1 ,u 2 ),(dL ( a,b ) − 1 ) ( a,b)(v 1 ,v 2 )i (0 ,1)

= hb−1(u1,u2),b−1(v1,v2)i(0,1)

= b−2(u1v1 + u2v2).

We see that the left-invariant metric on U is the hyperbolic metric

.

5. Let ∇ be the connection on R 2 that has vanishing Christoffel symbols Γ k ij with respect to

Cartesian coordinates (x,y). Calculate Christoffel symbols with respect to polar coordinates (r,θ).

In other words, calculate the coefficients of

, , , and ,

in the basis .

Since r = p x 2 + y 2 and θ = arctan(y/x), we find

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.

Dually, the bases of the tangent space are related by the inverse matrix

.

We can then calculate

and with a little more work

, and .

Note that the connection is torsion-free.

6. If M ⊂R N are submanifold of Euclidean space. We obtain Riemannian metric on M by

restricting the Euclidean metric, and this metric g has an associated Levi-Civita connection ∇.

Instead we could take the standard connection on R N and restrict it to a connection ∇ 0 on M.

Specifically, let Y and X be vector fields on M which extend to vector fields X and Y on a

neighbourhood of M in R N . Writing Y as (Y 1 ,…,Y N ), the standard connection on R N is given by:

∇ X Y = (D X Y 1 ,…,D X Y N ) ∈ TR N .

Then we defineto be the orthogonal projection of ∇ X Y to TR N | M ⊂ TR N .

Prove that ∇ 0 is the same as the Levi-Civita connection ∇ on M.

We first show that ∇ 0 is a connection on M. The equations follow automatically from the

corresponding equations for ∇:

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∇0fX+gY Z = f∇0XZ + g∇0Y Z

∇ 0 X (Y + Z) = ∇ 0 X Y + ∇ 0 X Z

=

=

= fproj(∇ X Y ) + X(f)proj(Y )

As required, proving that ∇ 0 is indeed a connection. Note that when we project, we are also

restricting to M. Moreover f| M = f, X(f)| M = X(f)| M (in general, value of Z(g) at a point p only

depends on the value Z p of the vector field at p), and then X(f)| M = X(f) (as X is a vector field on

M, we only need to differentiate f along M, not transversally; but f| M = f), etc.

Next we show that ∇ 0 is compatible with the Riemannian metric on M.

proj(∇ X Z)i

(You should think carefully about why each equality holds, perhaps writing things in local

coordinates to see more clearly. For example, the first equality holds because Y and Z give Y

and Z when we restrict to M, so hY ,Zi gives hY,Zi when restricted to M. For the fifth equality,

Z lies in TM ⊂ TR N , so when we take the inner product with ∇ X Y we can ignore the component

of ∇ X Y which is orthogonal to TM; equivalently, we can project ∇ X Y to TM.)

Finally, we show that ∇ 0 is torsion-free.

= proj(∇ X Y ) + proj(∇ Y X)

= proj([X,Y ])

This time we will use local coordinates. If {x 1 ,…,x n } is local coordinates on M and complete to

local coordinates {x 1 ,…,x n ,x n+1 ,…,x N } on R N . Then

proj( ∇ X f Y )

proj( f ∇ X Y + X ( f ) Y )

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where X n+1 ,…,X N all vanish on M since X| M = X. There is a similar expression for

Y . Now

!

proj([X,Y ]) = proj

where a second equality follows by the fact that projection to TM means that we keep only the

terms involving, and the third equality follows from the fact that X i and Y i

vanish on M for i = n + 1,…,N. It follows that ∇ 0 is torsion-free.

In conclusion, we have shown that ∇ 0 is a torsion-free connection on M which is compatible

with the Riemannian metric. By uniqueness of such connections, ∇ 0 must be the LeviCivita

connection on M.

8. Consider the upper half-plane

with the hyperbolic metric

.

a) Calculate the Christoffel symbols of the hyperbolic metric. Letting (x 1 ,x 2 ) = (x,y), we see

that g ij = δ ij /y 2 . Substituting this into

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we find and all other Christoffel symbols:

b) Let v 0 be the vector (0 and let be the curve t →7(t,1).

Find the parallel transport of v 0 along c, i.e., find a formula for the parallel vector field

that satisfies V (0) = v 0 .

Let. Then

where the last equality is due to the fact that y always equals 1 on the curve c. The

unique solution with initial condition is

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.

9. Geodesics on a surface of revolution: Consider the surface of a revolution S given by the

parameterization

(u,v) 7→ (f(v)cosu,f(v)sinu,g(v))

where f ,g are smooth functions with f(v) > 0 and (f 0 (v)) 2 + (g 0 (v)) 2 > 0 for all v. In Homework

1 we calculated the metric on S induced by the Euclidean metric on R 3 :

a. Derive the local form of the geodesic equations:

Let (x 1 ,x 2 ) = (u,v). As in problem 3, the only non-vanishing Christoffel symbols are

, and Γ 2 22 , because the metric is diagonal and independent of the first

coordinate u. We find

where 0 always denotes derivatives with respect to v. The result follows by substituting these

Christoffel symbols into the geodesic equations.

b. Fix a geodesic γ. A parallel is a circle given by v = constant (and hence also r 2 = x 2 + y 2 =

constant). Suppose that at γ(t), the geodesic γ meets a parallel of radius r(t) at an angle β(t).

Prove that

r(t)cosβ(t)

is independent of t. This is called Clairaut’s relation.

Let us first find a formula for cosβ(t). Suppose the parallel and the geodesic meet at the point

(x,y,z) = (fcosu,fsinu,g). The unit tangent to the parallel is given by

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.

The velocity vector field of the geodesic is given by

.

Rescaling t if necessary, we can assume that geodesic is normalized (parametrized by length

arc), so that the velocity vector has length one. The inner product of these two unit tangent

vectors then gives

and multiplying by r(t) = p x 2 + y 2 = f gives

.

Now

which vanishes by the first geodesic equation. Therefore r(t)cosβ(t) is constant (independent

of t) as required.

* c) The paraboloid is the surface of revolution built from a parabola (f(v),g(v)) = (v,v 2 ). Use

Clairaut’s relation to show that a geodesic on the paraboloid intersects itself an infinite

number of times (unless it is given by u = constant).

Firstly, substituting f = v and g = v 2 into the geodesic equations gives

.

We will only need the first equation, which as we saw in part b) is equivalent to Clairaut’s

relation. Separating u and v

and integrating gives

+ const. Therefore

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for some constant C.

Without loss of generality, let’s assume that r(t) = v achieves its minimum value v 0 at t = 0. At

this point the geodesic is tangent to the parallel and β(0) = 0. Since r(t)cosβ(t) must be

constant, we see that for t > 0, β(t) > 0 and r(t) = v is increasing (the geodesic is “climbing”).

On the other hand, for t < 0, β(t) < 0, and geodesic is “climbing” by reverse direction. These

two directions will intersect each other infinitely many times provided the geodesic keeps

circling the surface, i.e., provided u →∞ as t →∞ (and by symmetry, u →−∞ as t →−∞).

Since the z-coordinate on the surface is g = v 2 , we must have

as the increase in the z-coordinate cannot be greater than the distance covered by the geodesic

in time t (recall that the geodesic is parametrized by arc length). Therefore

and

+ const.

We conclude that u(t) has the desired asymptotic behaviour as t →∞.

10. Killing fields: A Killing vector field X ∈X(M) is an infinitesimal isometry, meaning that the

flow ψ t that it generates (locally and for small t) is a local isometry.

a) For M = R n , we can think of a vector field as a map X : R n →R n (by identifying the tangent

space at every point of R n with R n itself). We call X a linear field if this map is given by an n ×

n matrix. Prove that a linear field is a Killing field if and only if this matrix is anti-symmetric.

Suppose X is given by an n × n matrix A, so that the vector field at a point (x 1 ,…,x n ) is given by

.

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The equation for a flow line (x 1 (t),…,x n (t)) is then with solution

.

Thus the flow ψ t is given by exp(tA) (which is defined using the power series for exp). Now

exp(tA) preserves the metric on R n if and only if exp(tA) is in the orthogonal group O(n), if

and only if A is anti-symmetric.

b) If a Killing field X vanishes at point p ∈ M, show that X is tangent to the geodesic spheres

around p (equivalently, the geodesic spheres are preserved by the flow ψ t ).

If X vanishes at p, then p will be a fixed point of the flow ψ t generated by X. Now for small r,

the geodesic sphere of radius r centred at p is precisely the locus of points q which are

distance r from p. If X is a Killing field then ψ t must be an isometry, and thus

d(p,ψ t (q)) = d(ψ t (p),ψ t (q)) = d(p,q) = r.

Therefore ψ t (q) also lies on the geodesic sphere of radius r centred at p. In particular, the

curve t 7→ ψ t (q) lies on the geodesic sphere and differentiating shows that X is tangent to the

sphere.

* c) The Killing equation: Show that X is the Killing vector field if only

h∇ Y X,Zi + h∇ Z X,Y i = 0

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for all the vector fields Y and Z ∈X(M).

A hint for part c) can be found in the textbook, page 82.

If X vanishes identically then there is nothing to prove, so choose a point p at which X(p) 6=

0. We will prove the equation holds at p, and therefore by continuity it will hold everywhere.

Since X(p) 6= 0 we can find a hypersurface (codimension one submanifold) S ⊂ M which is

transversal to X at p, and also in a small neighbourhood of p. Let then (x 1 ,…,x n−1 ) be the local

coordinates on S. There is a map

S × (−δ,δ) → M

which takes a point (q,t) to ψ t (q) where ψ t is the flow generated by X. This map is a local

diffeomorphism in a neighbourhood of p, and this yields local coordinates (x 1 ,…,x n−1 ,t) on M in

a neighbourhood of p such that X = ∂ ∂t .

Now for 1 ≤ i,j ≤ n − 1 we calculate

where the first equality follows from the fact that ∇ is torsion-free. Now in local coordinates

the flow ψ t of X given by

(x 1 ,..,x n−1 ,0) 7→ (x 1 ,..,x n−1 ,t)

so

.

Since ψ t is an isometry, we must have

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Therefore is independent of t and

.

We have shown that

h∇ Y X,Zi + h∇ Z X,Y i = 0

for basis vectors Y and Z (the case when Y or Z is ∂ ∂t can be proved similarly). Since this

equation is C ∞ (M)-linear in both Y and Z, it must hold for all vector fields Y and Z ∈X(M).

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