Notes on Riemannian Geometry

The first variational formula

  1. Lemma 6.3: symmetry lemma
  2. Lemma 6.4: every V is a variation field of some γ
  3. Proposition 6.5: the first variational formula
  4. Theorem 6.6: every minimizing curve is geodesic
  5. Corollary 6.7: γ critical point of L iff γ geodesic

How to get ε from the compactness property in Lemma 6.4: first consider a covering of γ ( [ a , b ] ) with uniformly normal neighborhood of each γ ( t ) and extract a finite covering. Let V max = max t [ a , b ] V ( t ) and let ε = inf δ n V max where δ n are the numbers associated to the selected uniformly normal neighborhoods. If we consider the geodesic σ starting at γ ( t ) and with inital speed V ( t ) V max then it is slower than the corresponding unit speed (which is well defined in a geodesic ball centered in γ ( t ) and of radius ε . V max ) so is defined at least between times ± ε . V max . Hence Γ ( s , t ) = exp γ ( t ) ( sV ( t ) ) = σ ( s . V max ) is well defined.

The Gauß' Lemma and its consequences

  1. Theorem 6.8: Gauß' Lemma
  2. Corollary 6.9: grad r = r
  3. Corollary 6.10: in geodesic ball, the radial geodesic is the unique minimizing curve.
  4. Corollary 6.11: r ( x ) = d ( p , x )
  5. Theorem 6.12: every riemannian geodesic is locally minimizing
  6. Theorem 6: every minimizing curve is geodesic (alternative proof)

Hopf & Rinow

We consider the following statements:

  1. M is compact
  2. M complete as metric space
  3. M geodesically complete i.e. for all p M , exp p is defined on the whole tangent space T p M .
  4. Any two points p , q in M can be joined by a minimizing segment.

We also introduce the weaker forms C' and D' where we replace in C and D "for all p" by "there exists a p". We have the following graph of implications ( C C ' and D D ' are trivial):

1 A 2 B 1->2 3 C 1->3 6.16 2->3 4 D 2->4 6.15 3->4 5 C' 3->5 6 D' 4->6 5->2 6.14 5->6

The open disc is an example of a non-complete manifold with property D, the converse direction in corollary 6.15 is false. Similarly, the converse of corollary 6.16 is false, because n is an example of a noncompact complete riemannian manifold.

Theorem 6.13 shows that B and C are equivalent in three steps:

  1. B C
  2. C ' D ' . Note that the same p is used, so actually we can generalize to C D
  3. C ' B . We use D' to build the sequence γ i (so that d i is a Cauchy sequence). However we don't have D A as stated in corollary 6.15, because we also need C to ensure that exp p V is well-defined.

A B holds because any compact metric space is complete. This should allow to get corollaries 6.15 and 6.16 from the graph.

The Curvature Tensor - Flat manifolds

  1. Proposition 7.1: curvature endomorphism is tensorial
  2. Lemma 7.2: curvature endomorphism and tensor are intrisic properties
  3. Theorem 7.3: manifold flat iff curvature tensor vanishes

In proof of theorem 7.3, "So it suffices to show that k + 1 ( i E j ) = 0 " is because we already know i E j = 0 on M k , so if i E j is parallel along the k + 1 direction then i E j = 0 on the whole M k + 1 .

Symmetries of Curvature Tensor

  1. Proposition 7.4: symmetries of curvature tensor

The rotation angle theorem

  1. Theorem 9.1: the rotation angle theorem

The Gauß-Bonnet Formula

  1. Lemma 9.2: γ polygon in M , Rot ( γ ) = 2 π
  2. Theorem 9.3: the Gauß-Bonnet formula.
  3. Corollary 9.4: using K = 0 , κ n = 0 , ε i = π α i we get i α i = π
  4. Corollary 9.5: using K = 0 , κ N = 1 R and ε i = 0 , we get L = 2 π R .
  5. Corollary 9.6: K = 0 and ε i = 0
  6. Another example given during the lecture for the sphere : let Area ( α ) be the area of a digon of interior angle α . It is proportional to α and by evaluating (for example) at α = π 2 we get Area ( α ) = 2 α . If D is the area of the triangle given by the intersection of three such digons, and A, B, C denotes the area of the remaining pieces, we have: A + B + C + D = 4 π 2 and A + D = 2 α , B + D = 2 β and C + D = 2 γ . Finally we get A + B + C + D = 2 π = 2 ( α + β + γ + D ) so D = α + β + γ π . This is also what we get using the Gauß-Bonnet formula, with K = 0 , κ N = 1 1 and ε 1 = π α , ε 2 = π β and ε 3 = π γ .

End of the proof of theorem theorem 9.3, as given in the lecture. First we define:

  1. Ω f A = φ ( Ω ) ( f φ -1 ) det g x 1 x 2
  2. K = Rm ( 1 , 2 , 2 , 1 ) det g
  3. ( X , Y ) = X ( ω ( Y ) ) Y ( ω ( X ) ) ω ( [ X , Y ] )
  4. γ P x + Q y = a b P γ 1 t + Q γ 2 t t

We construct an orthonormal basis E 1 E 2 = A 1 2 using Gram-Schmidt. From definition 2 and 3, we check det g ( det A ) 2 = det g E = 1 and K = ω ( E 1 , E 2 ) = ( det A ) ω ( 1 , 2 ) and consequently, by definition 1:

Ω K A = φ ( Ω ) ω ( 1 , 2 ) x 1 x 2

For coordinate frames, [ 1 , 2 ] = 0 so ω ( 1 , 2 ) = ω 2 x 1 ω 1 x 2 where ω i = ω ( i ) . Using Green's Theorem and definition 4 we get the expected result:

Ω K A = a b ω 1 γ 1 t + ω 2 γ 2 t t = a b ω ( γ 1 t 1 + γ 2 t 2 ) t = a b ω ( γ ˙ ( t ) ) t

The Gauß-Bonnet Theorem

  1. Theorem 9.7: The Gauß-Bonnet theorem
  2. Corollary: χ ( M ) does not depend on triangulation. χ ( M ) = 0 for the torus, χ ( M ) = 2 for the sphere. The only possible values are 2 2 g (stated but not proved during the lecture).
  3. If there is a nonzero vector field, then χ ( M ) = 0 . (hence there is no nonzero vector on the sphere) proof: If X is a nonzero vector field on the whole manifold, we can define E 1 = X X and E 2 such that ( E 1 , E 2 ) is orthonormal. We have a 1-form ω such that Y E 1 = ω ( Y ) E 2 and Y E 2 = ω ( Y ) E 1 for any vector field Y , as in the proof of the Gauß-Bonnet Formula. M K A = M ω = i Δ i ω = i Δ i ω = 0 ( ω is defined on the whole manifold and as in the proof of the Gauß-Bonnet Theorem, the integrals on the edges cancel out).
  4. If K 0 , there is no simple digon. proof: suppose the contrary, let α 1 , α 2 > 0 be the interior angles. Then ε i = π α < π , κ n = 0 and M K A = 2 π ε 1 ε 2 > 0 .
  5. If K > 0 , any two geodesics intersect. proof: otherwise we can cut the manifold into three disjoint connected pieces Ω 1 , Ω 2 , Ω 3 using this geodesics. We have χ ( Ω i ) = 1 2 π ω i K A > 0 and since it is an integer, χ ( Ω i ) 1 . Hence χ ( M ) = i = 1 3 χ ( Ω i ) 3 , which is a contradiction (according to the possible values of 2).

Only theorem 9.7 is from Lee's book, the others have only been given during the lecture. The other propositions of chapter 9 have not been seen during the lecture.

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