Riemannian GEOMETRY

Contents

A smooth vector field Y = Y^{i}∂_{i} on R^{n} is the same thing as a smooth map Y = Y^{i}E_{i}: R^{n} → R^{n}. The derivative in the direction of (tangent) vector X of the map Y = Y^{i}E_{i} is X(Y) = X(Y^{i})E_{i}. For consistency sake we better declare the derivative of the vector field Y = Y^{i}∂_{i} to be \overline{∇}_{X}Y = X(Y^{i})∂_{i} to ensure that the diagram

[figure: C^{∞}(R^{n}, R^{n}); X; C^{∞}(R^{n}, R^{n}); T (R^{n}); _{\overline{∇}_{X}}; T (R^{n})]

commutes.

Date: 30th April 2005.

Smooth manifolds

1. Tangent vectors, cotangent vectors and tensors

Lemma 1.1. Let F: M^{m} → N^{n} be a smooth map. Suppose that (x^{1}, …, x^{m}) are local coordinates on M and (y^{1}, …, y^{n}) local coordinates on N. Then

(1.2)  F_{∗}(\frac{∂}{∂x^{j}}) = \frac{∂(y^{i}F)}{∂x^{j}} \frac{∂}{∂y^{i}} ,  1 ≤ j ≤ m,
(1.3)  F^{∗}(dy^{i}) = \frac{∂(y^{i}F)}{∂x^{j}}dx^{j} = d(y^{i}F),  1 ≤ i ≤ n

The (n × m)-matrix (\frac{∂(y^{i}F)}{∂x^{j}}) is the matrix for F_{∗} and the (m × n)-matrix (\frac{∂(y^{i}F)}{∂x^{j}})^{t} is the matrix for F^{∗}. In other words,

(F_{∗} \frac{∂}{∂x^{1}} … F_{∗} \frac{∂}{∂x^{m}} ) = (\frac{∂}{∂y^{1}} … \frac{∂}{∂y^{n}} ) (\frac{∂(y^{i}F)}{∂x^{j}})

(F^{∗}dy^{1} ; ⋮ ; F^{∗}dy^{n}) = (\frac{∂(y^{i}F)}{∂x^{j}}) ( dx^{1} ; ⋮ ; dx^{m})

Proof. F^{∗}(dy^{i})(\frac{∂}{∂x^{j}}) = dy^{i}(F_{∗}(\frac{∂}{∂x^{j}})) = F_{∗} (\frac{∂}{∂x^{j}} ) (y^{i}) = \frac{∂y^{i}F}{∂x^{j}}. □

2. The tangent bundle of a smooth manifold

3. Vector fields, covector fields, tensor fields, n-forms

Proposition 3.1. The differential d: C^{∞}(M) → T^{1}(M) is an R-linear map satisfying the derivational rules

d(uv) = (du)v + u(dv), d (\frac{u}{v}) = \frac{du}{v} − \frac{udv}{v^{2}} = \frac{vdu − udv}{v^{2}}

for all smooth functions u, v ∈ C^{∞}(M) (where v(p) ≠ 0 for all p ∈ M in the last formula). If F: M → N is a smooth map, then the diagram

\xymatrix{C^{∞}(M) & C^{∞}(N) \ar[l]_{F^{∗}} \\ d \ar[d] & d \ar[d] \\ T^{1}(M) & T^{1}(N) \ar[l]_{F^{∗}}}

commutes meaning that F^{∗}(du) = d(uF) for all u ∈ C^{∞}(N).

Example 3.2. Let s: R^{n+1} → R be the smooth map s(x) = |x|^{2} = ∑_{i=1}^{n+1}(x^{i})^{2}. Then s^{−1}(1) = S^{n} ⊂ R^{n} is the sphere of radius R. The differential ds = ∑ 2x^{i}dx^{i} is the linear map R^{n+1} = T_{p}R^{n+1} → R given by ds_{p}(v) = ∑ 2p^{i}v^{i} = 2⟨p, v⟩ with kernel ker ds_{p} = p^{⊥} at any point p ≠ 0. Let p be any point of S^{n} and ι_{∗}: T_{p}S^{n} → T_{p}R^{n+1} = R^{n+1} the linear map induced by the inclusion, ι. For any tangent vector X ∈ T_{p}S^{n}, ds(ι_{∗}X) = X(sι) = X(1) = 0. Hence the tangent space at p is the kernel of ds_{p},

T_{p}S^{n} = p^{⊥} ⊂ R^{n+1} = T_{p}R^{n+1}

and the tangent bundle of S^{n},

T S^{n} = {(p, v) ⊂ S^{n} × R^{n+1} | ⟨v, p⟩ = 0} ⊂ S^{n} × R^{n+1}

is the vector bundle whose fibre over any p ∈ S^{n} is p^{⊥}. A smooth vector field on S^{n} is a smooth map v: S^{n} → R^{n+1} such that v(p) ⊥ p for all p ∈ S^{n}. Show that any odd sphere has a vector field without zeros. Does S^{2} admit a smooth vector field with no zeros? Can you describe T RP^{n}? Can you describe T M if M = f^{−1}(0) consists of the manifold solutions to the equation f(x) = 0 for some smooth map f: R^{n+1} → R?

Definition 3.3. A smooth (\begin{matrix}k \\ ℓ \end{matrix})-tensor field on M is a smooth section of the tensor bundle

T_{ℓ}^{k}(M) → M.

Particular cases are • T_{0}^{0}(M) = C^{∞}(M)

• T_{1}^{0}(M) consists of vector fields on M

• T_{0}^{1}(M) consists of 1-forms on M Tensor fields admit

• T_{ℓ_{1}}^{k_{1}}(M) × T_{ℓ_{1}}^{k_{1}}(M) ⟶^{⊗} T_{ℓ_{1}+ℓ_{2}}^{k_{1}+k_{2}}(M) (tensor product of tensor fields)

• T_{ℓ+1}^{k+1}(M) ⟶^{tr} T_{ℓ}^{k}(M) (contraction of tensor fields)

Example 3.4. Let ω ∈ T_{0}^{1}(M) be a 1-form and X ∈ T_{1}^{0}(M) a vector field on M. Then X ⊗ ω ∈

T_{1}^{1}(M) is (\begin{matrix}1 \\ 1 \end{matrix})-tensor field with contraction tr(X ⊗ ω) = ω(X) (??).

In a coordinate patch any (\begin{matrix}k \\ ℓ \end{matrix})-tensor field A is (??) a C^{∞}(M)-linear combination

(3.5)  A = A_{i_{1}···i_{k}}^{j_{1}···j_{ℓ}}∂_{j_{1}} ⊗ ⋯ ∂_{j_{ℓ}} ⊗ dx^{i_{1}} ⊗ ⋯ dx^{i_{k}}

of tensor products of the basis tensor fields and basis 1-forms. The smooth functions A_{i_{1}···i_{k}}^{j_{1}···j_{ℓ}} are called the components of the tensor field A.

The tensor algebra of M is the graded algebra T^{∗}(M) = ∑_{k=0}^{∞} T^{k}(M) equipped with the tensor product T^{r}(M) × T^{s}(M) ⟶^{⊗} T^{r+s}(M). If F: M → N is a smooth map, F^{∗}: T^{k}(N) → T^{k}(M) is the linear map given by F^{∗}(A)(X_{1}, …, X_{k}) = A(F_{∗}X_{1}, …, F_{∗}X_{k}) for all A ∈ T^{k}(N) and all smooth vector fields X_{1}, …, X_{k} on M.

Lemma 3.6. T^{∗}(M) is a graded algebra. F^{∗}: T^{∗}(N) → T^{∗}(M) is a homomorphism of C^{∞}(N)- algebras: F^{∗}(aω ⊗ η) = F^{∗}(a)F^{∗}(ω) ⊗ F^{∗}(η).

Riemannian manifolds

Riemann’s idea was that in the infinitely small, on a scale much smaller than the the smallest particle, we do not know if Euclidean geometry is still in force. Therefore we better not assume that this is the case and instead open up for the possibility that in the infinitely small there may be other length functions, the may be other inner products on the tangent space! A Riemannian manifold is a smooth manifold equipped with inner product, which may or may not be the Euclidean inner product, on each tangent space.

4. Riemannian metric

Definition 4.1. A Riemannian metric on a smooth manifold M is a symmetric, positive definite

(\begin{matrix}2 \\ 0 \end{matrix})-tensor g ∈ T_{0}^{2}(M).

In a coordinate frame we may write

g = g_{ij}dx^{i} ⊗ dx^{j}, g_{ij} = g(∂_{i}, ∂_{j})

This means that g(U^{i}∂_{i}, V^{j}∂_{j}) = g_{ij}U^{i}V^{j} and in particular that

(4.1)  ⟨∂_{i}, ∂_{i}⟩ = g_{ii},  ⟨∂_{i}, ∂_{j}⟩ = g_{ij}

Note that there are only \frac{1}{2}n(n + 1) different functions here as g_{ij} = g_{ji} by symmetry.

Remark 4.2. Since the metric tensor is symmetric, it is traditional to write it in a basis of symmetric tensors. The symmetrization of ω ⊗ η is the tensor

ωη = \frac{1}{2}(ω ⊗ η + η ⊗ ω)

Note that ωη = ηω and that ω^{2} = ωω = ω ⊗ ω. Observe that

g = g_{ij}dx^{i}dx^{j} = 2 ∑_{i=1}^{n} g_{ii}(dx^{i})^{2} + 2 ∑_{1≤i<j≤n} g_{ij}dx^{i}dx^{j}

Lemma 4.3. Let F: M → N be an immersion and g a Riemannian metric on N. (1) F^{∗}g is a Riemannian metric on M.

(2) If g = g_{ij}dy^{i}dy^{j} in a coordinate frame on N, then

F^{∗}(g)|F^{−1}(U) = g_{ij}dy^{i}F dy^{j}F

Proof. It is a general fact that F^{∗}(g) is a smooth 2-form on M (??). F^{∗}(g) is symmetric because g is symmetric and it is positive definite because g is positive definite and F_{∗} is injective on each

fibre. F^{∗}(g_{ij}dy^{i}dy^{j}) =^{(??)} g_{ij}F^{∗}dy^{i}F^{∗}dy^{j} =^{(??)} g_{ij}dy^{i}F dy^{j}F. □

For instance if F: U → M is a parameterization (an inverse chart) of an open subset of M ⊂ R^{m}, then the pull-back of the induced metric on M is

(4.4)  F^{∗}(\overline{g}) = F^{∗}(δ_{ij}dx^{i}dx^{j}) = δ_{ij}dF^{i}dF^{j} = ∑_{i=1}^{m}(dF^{i})^{2}

These expressions are tensor fields living in the tensor algebra T^{∗}(M) of M.

Example 4.5. (Graphs) Let M ⊂ R^{n} × R be the graph of the smooth function f: M → R. Then s(x) = (x, f(x)) is a diffeomorphism so that the Riemannian manifold (M, ι^{∗}\overline{g}^{n+1}) is isometric to (R^{n}, s^{∗}\overline{g}^{n+1}) where the metric s^{∗}\overline{g}^{n+1} is

s^{∗}(∑_{i=1}^{n+1}(dx^{i})^{2}) = ∑_{i=1}^{n}(dx^{i})^{2} + (\frac{∂f}{∂x^{i}} dx^{i})^{2} = ∑_{i=1}^{n}(dx^{i})^{2} + \frac{∂f}{∂x^{i}} \frac{∂f}{∂x^{j}} dx^{i}dx^{j}

= (δ_{ij} + \frac{∂f}{∂x^{i}} \frac{∂f}{∂x^{j}} )dx^{i}dx^{j} = ∑_{i=1}^{n} (1 + (\frac{∂f}{∂x^{i}})^{2})(dx^{i})^{2} + 2 ∑_{1≤i<j≤n} \frac{∂f}{∂x^{i}} \frac{∂f}{∂x^{j}} dx^{i}dx^{j}

Example 4.6. let S_{+}^{2} be the upper hemisphere on S^{2} ⊂ R^{3} considered as the graph of the function f(x, y) = \sqrt{1 − x^{2} − y^{2}} defined on the unit ball B^{2} ⊂ R^{2}. Then (S_{+}^{2}, ι^{∗}\overline{g}^{3}) is isometric to (B^{2}, s^{∗}\overline{g}^{3}) where

s^{∗}\overline{g}^{3} = (dx)^{2} + (dy)^{2} + (\frac{−x}{\sqrt{1 − x^{2} − y^{2}}} dx + \frac{−y}{\sqrt{1 − x^{2} − y^{2}}} dy)^{2}

= \frac{1 − y^{2}}{1 − x^{2} − y^{2}} (dx)^{2} + \frac{1 − x^{2}}{1 − x^{2} − y^{2}} (dy)^{2} + \frac{2xy}{1 − x^{2} − y^{2}} dxdy

This means (??) that

⟨∂_{x}, ∂_{x}⟩ = \frac{1 − y^{2}}{1 − x^{2} − y^{2}} , ⟨∂_{y}, ∂_{y}⟩ = \frac{1 − x^{2}}{1 − x^{2} − y^{2}} , ⟨∂_{x}, ∂_{y}⟩ = \frac{xy}{1 − x^{2} − y^{2}}

at the point (x, y) ∈ B^{2}. In this metric, the basis tangent vectors, ∂_{x} and ∂_{y}, are not orthogonal at any point of the unit ball away from the axes. If we consider the curve γ(t) = (t, 0), −1 ≤ t ≤ 1, then the tangent vector γ_{∗}(\frac{d}{dt}) = \frac{d(xγ)}{dt}∂_{x} + \frac{d(yγ)}{dt}∂_{y} = ∂_{x} so that the length of this curve is

L_{s^{∗}(g)}(γ) = ∫_{−1}^{+1} |γ_{∗}(\frac{d}{dt})|dt = ∫_{−1}^{+1} \sqrt{\frac{1}{1 − t^{2}}} dt = π

What is the distance between (0, −1/2) and (1/2, 0)? What is the curve of shortest length between these two points?

Example 4.7. (Surface of revolution in R^{3})

Definition 4.8. A smooth map F: (M, g) → (N, h) between two Riemannian manifolds is an isometry if g = F^{∗}h; if g(X, Y) = h(F_{∗}X, F_{∗}Y) for all tangent vectors X, Y ∈ T_{p}M, p ∈ M.

Two Riemannian manifolds are isometric if we can deform one into the other by bending but not stretching. Is the upper unit hemisphere S_{+}^{2} ⊂ R^{3} isometric to the open unit ball B^{2} ⊂ R^{2}? Certainly, the diffeomorphism s: B^{2} → S_{+}^{2} from Example ?? is not an isometry as for instance ⟨∂_{x}, ∂_{x}⟩ ≠ 1 or because the tangent vectors ∂_{x} and ∂_{y} are not orthogonal throughout any open (nonempty) subspace of B^{2}. But there are many other diffeomorphisms and maybe we could find one that preserves the metrics? To decide if this is the case we need to find invariants of Riemannian metrics. Is S_{+}^{2} curved? And what does that mean? We need to develop some theory to answer these questions.

In order to decide if two given Riemannian manifolds are isometric we have to know have the metric tensor transforms under change of coordinate system? (Remember that the coordinate expression for a metric is an artefact of the coordinate system and not an intrinsic property of the metric.)

Lemma 4.9. Let ϕ: R^{n} → M and ψ: R^{n} → M be parameterizations of the same open subspace of M. If

ϕ^{∗}(g) = a_{ij}dx^{i}dx^{j}, ψ^{∗}(g) = b_{ij}dy^{i}dy^{j}

then

(a_{ij}) = (\frac{∂y^{i}}{∂x^{j}} )^{t} (b_{ij}) (\frac{∂y^{i}}{∂x^{j}} )

where y = ψϕ^{−1}.

Proof. Put P = (\frac{∂y^{i}}{∂x^{j}} ). Then (??)

y^{∗} (dy^{1} ; ⋮ ; dy^{n}) = P (dx^{1} ; ⋮ ; dx^{n})

Therefore

a_{ij}dx^{i}dx^{j} = ϕ^{∗}g = (ψy)^{∗}g = y^{∗}(ψ^{∗}g) = y^{∗}(b_{ij}dy^{i}dy^{j})

= y^{∗}((dy^{1} … dy^{n}) (b_{ij}) (dy^{1} ; ⋮ ; dy^{n}))

= (dx^{1} … dx^{n}) P^{t}(b_{ij})P (dx^{1} ; ⋮ ; dx^{n})

so that (a_{ij}) = P^{t}(b_{ij})P. □

Suppose that ϕ^{∗}g = a_{ij}dx^{i}dx^{j} for some parameterization ϕ. Is M locally flat? In other words, does there exist a re-parameterization ψ = ϕy of M such that ψ^{∗}g = δ_{ij}dy^{i}dy^{j}? Such a reparameterization exists if and only if the set of \frac{1}{2}n(n − 1) PDEs

(g_{ij}) = (\frac{∂y^{i}}{∂x^{j}} )^{t} (\frac{∂y^{i}}{∂x^{j}} ) ,

or equivalently,

(4.10)  g_{ij} = ∑_{k=1}^{n} \frac{∂y^{k}}{∂x^{i}} \frac{∂y^{k}}{∂x^{j}} ,  1 ≤ i ≤ j ≤ n,

has a solution y = (y^{1}, …, y^{n}). Riemann showed (in an essay that was never properly recognized) that (??) is equivalent to

\frac{∂^{2}y^{ℓ}}{∂x^{i}∂x^{j}} = Γ_{ij}^{k} \frac{∂y^{ℓ}}{∂x^{k}} , 1 ≤ i, j, ℓ ≤ n,

and thereby that (??) has a solution if and only if

R_{jkℓ}^{m} = 0, 1 ≤ j, k, ℓ, m ≤ n,

where the Γ_{ij}^{k} (the Christoffel symbols) and the R_{jkℓ}^{m} are certain functions defined in terms of the functions g_{ij}. This was the birth of the Riemann curvature tensor R_{jkℓ}^{m}! This direct approach, however, is not the one used today since it is conceptually simpler first to introduce a device called a connection that will enable us to work in a coordinate-free way on M.

Example 4.11. (Cylinders are flat.) Let γ(s) = (x(s), y(s)), a < s < b, be a smooth curve in R^{2} such that the tangent γ_{∗}(\frac{d}{ds}) = x′(s)∂_{x} + y′(s)∂_{y} ≠ 0 for all t. Let M ⊂ R^{3} be the cylinder over γ, the surface with parameterization ϕ(s, t) = (x(s), y(s), z(t)) where z(t) = t. Then (??),

ϕ^{∗}(\overline{g}^{3}) = (dx)^{2} + (dy)^{2} + (dz)^{2} = (x′(s)ds)^{2} + (y′(s)ds)^{2} + (dt)^{2} = |γ_{∗}(\frac{d}{ds})|^{2}(ds)^{2} + (dt)^{2}

What are Riemann’s equations in this case? Is there a solution?

5. The three model geometries

The model geometries are Euclidean geometry, spherical geometry, and hyperbolic geometry.

5.1. Euclidean geometry. Euclidean geometry is the geometry of the Riemannian manifold (R^{n}, \overline{g}^{n}) where

\overline{g}^{n} = δ_{ij}dx^{i}dx^{j} = ∑_{i=1}^{n}(dx^{i})^{2}

meaning that \overline{g}^{n}(U^{i}∂_{i}, V^{j}∂_{j}) = ∑_{i=1}^{n} U^{i}V^{i}. (The straight line over the g is to remind you of Euclidean geometry.)

The isometry group of Euclidean geometry

Isom(R^{n}, \overline{g}^{n}) = Aff(n) = R^{n} ⋊ O(n)

acts transitively on R^{n} by the rule (v, A)(x) = v + Ax. The isotropy subgroup at 0 ∈ R^{n} is O(n) and the projection

O(n) → Aff(n) = O(T R^{n}) → Aff(R^{n})/O(n) = R^{n}

is the unit n-frame bundle of Euclidean geometry R^{n}.

5.2. Spherical geometry. Spherical geometry is the geometry of the Riemannian manifold (S_{R}^{n}, g_{R}^{n}) where

S_{R}^{n} = {(ξ, τ) ∈ R^{n} × R | |ξ|^{2} + τ^{2} = R^{2}}

is the n-sphere of radius R and g_{R}^{n} = ι^{∗}( ∑_{i=1}^{n}(dξ_{i})^{2} + (dτ)^{2}) is the restriction of the Euclidean metric on R^{n} × R.

The isometry group of spherical geometry

Isom(S_{R}^{n}, g_{R}) = O(n + 1)

(The smooth action of O(n + 1) ⊂ Aff(n + 1) on R^{n+1} restricts to a smooth action on S^{n} – and in fact these are all isometries.) O(n + 1) acts transitively on S^{n}. The isotropy subgroup at N = (0, …, 0, 1) ∈ S^{n} is O(n) and the projection

O(n) → O(T S^{n}) = O(n + 1) → O(n + 1)/O(n) = S^{n}

is the unit n-frame bundle of spherical geometry S^{n}.

Proposition 5.1. Stereographic projection σ: S_{R}^{n} − N → R^{n} is given by

σ(ξ, τ) = \frac{R}{R − τ} ξ

and the inverse is given by

σ^{−1}(u) = (ξ(u), τ(u)),  ξ(u) = \frac{2R^{2}u}{|u|^{2} + R^{2}} ,  τ(u) = R\frac{|u|^{2} − R^{2}}{|u|^{2} + R^{2}}

Stereographic projection is a diffeomorphism.

Proof. Elementary. □

Proposition 5.2. The Riemannian manifold (S_{R}^{n}−N, g_{R}^{n}) is isometric to the Riemannian manifold (R^{n}, (σ^{−1})^{∗}g_{R}^{n}) where

(σ^{−1})^{∗}g_{R}^{n} = \frac{4R^{4}}{(R^{2} + |u|^{2})^{2}} \overline{g}^{n}

is conformally equivalent to the Euclidean metric.

Proof. By ??

(σ^{−1})^{∗}g_{R} = ∑_{j=1}^{n}(dξ^{j})^{2} + (dτ)^{2}

We will now use the derivation rules ??. The denominator of ξ and τ is |u|^{2} +R^{2} and d(|u|^{2} +R^{2}) = d(|u|^{2}) = ∑ 2u^{j}du^{j} which we shall write as 2⟨u, du⟩. Because

dξ^{j} = d (\frac{2R^{2}u}{|u|^{2} + R^{2}} ) = \frac{2R^{2}du^{j}}{|u|^{2} + R^{2}} − \frac{4R^{2}u^{j}⟨u, du⟩}{(|u|^{2} + R^{2})^{2}}
dτ = d (R\frac{|u|^{2} − R^{2}}{|u|^{2} + R^{2}} ) = R\frac{2⟨u, du⟩}{(|u|^{2} + R^{2})^{2}} − R\frac{(|u|^{2} − R^{2})2⟨u, du⟩}{(|u|^{2} + R^{2})^{2}}

= \frac{2R⟨u, du⟩(|u|^{2} + R^{2}) − 2R⟨u, du⟩(|u|^{2} − R^{2})}{(|u|^{2} + R^{2})^{2}}

= \frac{4R^{3}⟨u, du⟩}{(|u|^{2} + R^{2})^{2}}

we get

∑_{j=1}^{n}(dξ^{j})^{2} = ∑_{j=1}^{n} (\frac{4R^{4}(du^{j})^{2}}{|u|^{2} + R^{2}} + \frac{16R^{4}(u^{j})^{2}⟨u, du⟩}{(|u|^{2} + R^{2})^{4}} − \frac{16R^{4}u^{j}du^{j}⟨u, du⟩}{(|u|^{2} + R^{2})^{3}})

= \frac{4R^{4}(du^{j})^{2}}{(|u|^{2} + R^{2})^{2}} ∑_{j=1}^{n}(du^{j})^{2} + \frac{16R^{4}|u|^{2}⟨u, du⟩^{2}}{(|u|^{2} + R^{2})^{4}} − \frac{16R^{4}⟨u, du⟩^{2}}{(|u|^{2} + R^{2})^{3}}

= \frac{4R^{4}(du^{j})^{2}}{(|u|^{2} + R^{2})^{2}} \overline{g}^{n} + \frac{16R^{4}|u|^{2}⟨u, du⟩ − 16R^{4}|u|^{2}⟨u, du⟩^{2} − 16R^{6}⟨u, du⟩^{2}}{(|u|^{2} + R^{2})^{4}}

= \frac{4R^{4}(du^{j})^{2}}{(|u|^{2} + R^{2})^{2}} \overline{g}^{n} − \frac{16R^{6}⟨u, du⟩^{2}}{(|u|^{2} + R^{2})^{4}}

= \frac{4R^{4}(du^{j})^{2}}{(|u|^{2} + R^{2})^{2}} \overline{g}^{n} − (dτ)^{2}

□

5.3. Hyperbolic geometry. Hyperbolic geometry is the geometry of the Riemannian manifold (H_{R}^{n}, h_{R}^{n}) where

H_{R}^{n} = {(ξ, τ) ∈ R^{n} × R_{+} | |ξ|^{2} − τ^{2} = −R^{2}}

id the hyperbolic space n-space of radius R and h_{R}^{n} = ι^{∗}( ∑_{i=1}^{n}(dξ^{i})^{2} − (dτ)^{2}) is the restriction of the Minkowski metric on R^{n} × R. Let N = (0, …, 0, R) ∈ H_{R}^{n} be the north pole.

Remark 5.3. (Minkowski metric) Let m be the inner product on R^{n+1} with matrix D = diag(1, …, 1, −1). We can view m both as an inner product on the vector space R^{n} (⟨X, Y ⟩ = m(X, Y) = X^{t}DY) or as a Minkowski metric on the manifold R^{n+1} (⟨X^{i}∂_{i}, Y^{i}∂_{i}⟩ = m(X, Y)). For each p ∈ H_{R}^{n}, the tangent space

T_{p}H_{R}^{n} = p^{⊥} ⊂ R^{n+1} = T_{p}R^{n+1}

exactly as in ??.

Let

O(n, 1) = {A ∈ GL(n, R) | A^{t}DA = D}

be the group of linear automorphisms of R^{n+1} that preserve the inner product. The columns (or rows) of each A ∈ O(n, 1) form an orthogonal basis for R^{n+1} of vectors of length 1, …, 1, −1. The elements of the Lie group O(n, 1) preserve the subspace {(ξ, τ) ∈ R^{n+1} | |(ξ, τ)|^{2} = −R^{2}} of vectors of square length −R^{2}. This subspace has two connected components. Let O_{+}(n, 1) be the subgroup of O(n, 1) consisting of the elements that take the connected component H_{R}^{n} to itself. The Lie group O_{+}(n, 1) acts transitively on H_{R}^{n}; given any v ∈ H_{R}^{n}, we can find an A ∈ O_{+}(n, 1) whose last column is \frac{1}{R}v so that AN = v where N is the north pole. The isotropy subgroup at N consists of the A ∈ O_{+}(n, 1) whose last column is (0, …, 0, 1). The isotropy subgroup, isomorphic to O(n), act transitively on the tangent space T_{N}H_{R}^{n}. The projection

O(n) → O(T H_{R}^{n}) = O_{+}(n, 1) → O_{+}(n, 1)/O(n) = H_{R}^{n}

is the unit n-frame bundle of H_{R}^{n}.

Proposition 5.4. Hyperbolic stereographic projection π: H_{R}^{n} → B_{R}^{n} is given by

π(ξ, τ) = \frac{Rξ}{R + τ}

and its inverse is given by

π^{−1}(u) = (ξ(u), τ(u)), ξ(u) = \frac{2R^{2}u}{R^{2} − |u|^{2}} , τ(u) = R\frac{R^{2} + |u|^{2}}{R^{2} − |u|^{2}}

Hyperbolic stereographic projection is a diffeomorphism.

Proof. Elementary. □

Proposition 5.5. The Riemannian manifold (H_{R}^{n}, h_{R}^{n}) is isometric to the Riemannian manifold (B^{n}(R), (π^{−1})^{∗}h_{R}^{n}) where

(π^{−1})^{∗}(h_{R}^{n}) = \frac{4R^{4}}{(R^{2} − |u|^{2})^{2}} \overline{g}^{n}

is conformally equivalent to the Euclidean metric.

Proof. By ?? (applied to the Minkowski metric)

(σ^{−1})^{∗}h_{R} = ∑_{j=1}^{n}(dξ^{j})^{2} − (dτ)^{2}

Using that d(R^{2} − |u|^{2}) = −2⟨u, du⟩ and ?? we get

dξ^{j} = \frac{2R^{2}du^{j}}{R^{2} − |u|^{2}} + \frac{2R^{2}u^{j}2⟨u, du⟩}{(R^{2} − |u|^{2})^{2}}
dτ = R\frac{2⟨u, du⟩}{R^{2} − |u|^{2}} + R\frac{(R^{2} + |u|^{2})2⟨u, du⟩}{(R^{2} − |u|^{2})^{2}}

= \frac{2R^{3}⟨u, du⟩ − 2R|u|^{2}⟨u, du⟩ + 2R^{3}⟨u, du⟩ + 2R|u|^{2}⟨u, du⟩}{(R^{2} − |u|^{2})^{2}}

= \frac{4R^{3}⟨u, du⟩}{(R^{2} − |u|^{2})^{2}}

Therefore

∑_{j=1}^{n}(dξ^{j})^{2} = ∑_{j=1}^{n} (\frac{2R^{2}du^{j}}{R^{2} − |u|^{2}} + \frac{4R^{2}u^{j}⟨u, du⟩}{(R^{2} − |u|^{2})^{2}})^{2}

= \frac{4R^{4}}{(R^{2} − |u|^{2})^{2}} ∑_{j=1}^{n}(du^{j})^{2} + \frac{16R^{4}|u|^{2}⟨u, du⟩}{(R^{2} − |u|^{2})^{4}} + \frac{16R^{4}⟨u, du⟩^{2}}{(R^{2} − |u|^{2})^{3}}

= \frac{4R^{4}}{(R^{2} − |u|^{2})^{2}} \overline{g}^{n} + \frac{16R^{4}|u|^{2}⟨u, du⟩ + 16R^{6}⟨u, du⟩^{2} − 16R^{4}|u|^{2}⟨u, du⟩^{2}}{(R^{2} − |u|^{2})^{4}}

= \frac{4R^{4}}{(R^{2} − |u|^{2})^{2}} \overline{g}^{n} + \frac{16R^{6}⟨u, du⟩^{2}}{(R^{2} − |u|^{2})^{4}}

= \frac{4R^{4}}{(R^{2} − |u|^{2})^{2}} \overline{g}^{n} + (dτ)^{2}

□

Let U^{n} = {(x, y) ∈ R^{n−1} × R | y > 0} be the upper half plane in R^{n}.

Proposition 5.6. The Riemannian manifold (H_{R}^{n}, h_{R}^{n}) is isometric to the smooth manifold U^{n} equipped with the Riemannian metric R^{2} \frac{1}{y^{2}} \overline{g}^{n}.

Proof. A computation. □

Does the hyperbolic plane H^{2} embed isometrically in R^{3}? Any Riemannian ma embeds isometrically into some Euclidean space [?].

6. Connections

Let E → M be a smooth vector bundle over M and E(M) the C^{∞}(M)-module of smooth sections. A connection on E is a recipe for how to differentiate a section of E along a vector field.

Definition 6.1. A connection on E is a map

T (M) × E(M) ⟶^{∇} E(M)
(X, Y) → ∇_{X}Y

which is C^{∞}(M)-linear in X and R-linear in Y and satisfies the product rule

∇_{X}(fY) = (Xf)Y + f∇_{X}Y

for all f ∈ C^{∞}(M).

There always are connections, for instance the 0-connection given by ∇_{X}Y = 0.

Lemma 6.2. The value ∇_{X}Y (p) at the point p ∈ M only depends on Y in a neighborhood of p and X at p.

Proof. Let us first focus on Y -variable. By linearity, it is enough to show that if Y = 0 in a neighborhood U of p, then ∇_{X}Y (p) = 0. Choose a smooth bump function ϕ such that ϕ(p) = 1 and ϕ = 0 outside U. Then ϕY is the zero section so that

0 = ∇_{X}0 = ∇_{X}(ϕY) = (Xϕ)Y + ϕ∇_{X}Y

Evaluating at p, we get 0 = ∇_{X}Y (p) since Y (p) = 0 and ϕ(p) = 1.

Next, we focus on the X-variable. By an argument similar to the one just given, we first show that if X is 0 in a neighborhood of p, then 0 = ∇_{X}Y (p). Suppose now that we only know that X vanishes at the point p, X(p) = 0. Choose a moving frame E_{i} in a neighborhood of p. Extend the locally defined vector fields E_{i} to globally defined smooth vector fields. There are smooth functions X^{i} such that X = X^{i}E_{i} in a neighborhood of p. Then ∇_{X}Y (p) = ∇_{X^{i}E_{i}}Y (p) since X and X^{i}E_{i} are equal in a neighborhood of p. By C^{∞}(M)-linearity in X,

∇_{X^{i}E_{i}}Y = X^{i}∇_{E_{i}}Y

which evaluated at p is 0 since X^{i}(p) = 0 for all i. □

We are particularly interested in connections on the tangent bundle of M.

Definition 6.3. A linear connection is a connection T (M) × T (M) ⟶^{∇} T (M) on the tangent bundle of M.

Are there any linear connections, apart from the 0-connection, on M?

Example 6.4. The Euclidean connection \overline{∇}_{X}(Y^{j}∂_{j}) = X(Y^{j})∂_{j} is a nonzero linear connection on Euclidean space M = R^{n}. Note that \overline{∇}_{∂_{i}}∂_{j} = 0, 1 ≤ i, j ≤ n.

If we can construct connections on R^{n}, maybe we can define a connection in each coordinate patch on M and then put them together? What does a connection look like locally?

Lemma 6.5. Suppose that U is an open subspace of M that admit a moving frame E_{i}. The linear connection ∇ on M restricts to linear connection on U. If X = X^{i}E_{i} and Y = Y^{j}E_{j} are vector fields on U then

∇_{X}Y = (XY^{k} + X^{i}Y^{j}Γ_{ij}^{k})E_{k}

where the n^{3} smooth functions Γ_{ij}^{k} are the Christoffel symbols given by ∇_{E_{i}}E_{j} = Γ_{ij}^{k}E_{k}.

Proof. We compute

∇_{X}Y = ∇_{X}(Y^{j}E_{j}) = XY^{j}E_{j} + Y^{j}∇_{X}E_{j}XY^{j}E_{j} + Y^{j}∇_{X^{i}E_{i}}E_{j} = XY^{j}E_{j} + X^{i}Y^{j}∇_{E_{i}}E_{j}

= XY^{j}E_{j} + X^{i}Y^{j}Γ_{ij}^{k}E_{k} = XY^{k}E_{k} + X^{i}Y^{j}Γ_{ij}^{k}E_{k} = (XY^{k} + X^{i}Y^{j}Γ_{ij}^{k})E_{k}

where we use the Christoffel symbols. □

Thus we see that we can express ∇_{X}Y by means of the n^{3} smooth functions Γ_{ij}^{k}. Conversely, for any choice of n^{3} smooth functions Γ_{ij}^{k} we can define ∇_{X}Y, X, Y ∈ T (U) by the above formula and that will be a connection. So a linear connection on U is the same thing as a collection of n^{3} smooth functions on U.

In particular if M is a smooth manifold and {U_{α}} a smooth atlas on M then can find a linear connection ∇^{α} on each U_{α}. In order to construct a linear connection on M, let {ϕ_{α}} be a smooth partition of unity subordinate to {U_{α}}.

Theorem 6.6. (Existence of connections) Any smooth manifold M has many linear connections:

∇_{X}Y = ∑_{α} ϕ_{α}∇_{X}^{α}Y

is a linear connection on M.

Armed with a linear connection we know how to differentiate vector fields along vector fields. But as a bonus we can even differentiate arbitrary tensor fields along vector fields!

Lemma 6.7. (Existence and uniqueness of the covariant derivative of a tensor field) Let T (M) × T (M) ⟶^{∇} T (M) be a linear connection on M (??). Then there are unique connections

T (M) × T_{ℓ}^{k}(M) ⟶^{∇} T_{ℓ}^{k}(M)

on all tensor bundles T_{ℓ}^{k}(M) → M such that (1) ∇_{X}f = Xf for all f ∈ T_{0}^{0}(M) = C^{∞}(M)

(2) ∇_{X}Y is the given linear connection for all vector fields Y ∈ T_{0}^{1}(M)

(3) ∇_{X}(A ⊗ B) = ∇_{X}A ⊗ B + A ⊗ ∇_{X}B

(4) ∇_{X} tr A = tr ∇_{X}A for all A ∈ T_{ℓ+1}^{k+1}(M) Namely, for any 1-form ω, ∇_{X}ω is the 1-form given by

(6.8)  (∇_{X}ω)(Y) = X(ω(Y)) − ω(∇_{X}Y)

and, in general, for any (\begin{matrix}k \\ ℓ \end{matrix})-tensor A ∈ T_{ℓ}^{k}(M), ∇_{X}A is the (\begin{matrix}k \\ ℓ \end{matrix})-tensor given by

(6.9) (∇_{X}A)(ω^{1}, …, ω^{ℓ}, Y_{1}, …, Y_{k}) = X(A(ω^{1}, …, ω^{ℓ}, Y_{1}, …, Y_{k}))

− ∑_{i=1}^{k} A(ω^{1}, …, ω^{ℓ}, Y_{1}, …, ∇_{X}Y_{i}, …, Y_{k}) − ∑_{j=1}^{ℓ} A(ω^{1}, …, ∇_{X}ω_{j}, …, ω^{ℓ}, Y_{1}, …, Y_{k})

for any choice of k vector fields Y_{1}, …, Y_{k} and ℓ 1-forms ω^{1}, …, ω^{ℓ} on M.

Proof. Let’s assume that we have connections that satisfy items (1)–(4). What is the covariant derivative of a 1-form ω? For any two vector fields X and Y,

X(ω(Y)) =^{(1)} ∇_{X}(ω(Y)) =^{(??)} ∇_{X}(tr(ω ⊗ Y)) =^{(4)} tr ∇_{X}(ω ⊗ Y) =^{(3)} tr(∇_{X}ω ⊗ Y + ω ⊗ ∇_{X}Y)

=^{(??)} (∇_{X}ω)(Y) + ω(∇_{X}Y)

which is (??). So we are forced to define ∇_{X}ω as in (??). But then there is at most one possibility for ∇_{X}A since any tensor is a sum of a smooth function times tensor products of vector fields and 1-forms (combine the local expression (??) for a tensor with a smooth partition of unity). This shows uniqueness.

To show existence, use (??) to define the covariant derivative of a 1-form and then use (??) to define ∇_{X}A in general. Check that this definition satisfies (1)–(4). □

Definition 6.10. The total covariant derivative ∇: T_{ℓ}^{k}(M) → T_{ℓ}^{k+1}(M) is given by

∇A(ω^{1}, …, ω^{ℓ}, Y_{1}, …, Y_{k}, X) = (∇_{X}A)(ω^{1}, …, ω^{ℓ}, Y_{1}, …, Y_{k})

for all A ∈ T_{ℓ}^{k}(M).

Note that the total covariant derivative of the tensor field A ∈ T_{ℓ}^{k}(M) is zero if and only if the covariant derivative of A along all vector fields is zero: ∇A = 0 ⟺ ∀X ∈ T (M): ∇_{X}A = 0.

Example 6.11. There are total covariant derivatives

C^{∞}(M) = T_{0}^{0}(M) ⟶^{∇} T_{0}^{1}(M) ⟶^{∇} T_{0}^{2}(M) ⟶^{∇} T_{0}^{3}(M) ⟶^{∇} ⋯

If u ∈ C^{∞}(M) is a smooth function and X a vector field, then

(∇u)(X) = ∇_{X}(u) = X(u) = du(X)

so that ∇u = du.

The 2-form ∇^{2}u = ∇∇u is called the covariant Hessian of u. If X and Y a vector fields, then

(∇^{2}u)(Y, X) = (∇∇u)(Y, X) =^{(??)} (∇_{X}∇u)(Y) =^{(??)} X((∇u)(Y)) − ∇u(∇_{X}Y)

=^{∇u=du} X(Y (u)) − (∇_{X}Y)(u)

Consequently,

∇^{2}u = 0 ⟺ ∀X, Y ∈ T (M): Y (X(u)) = (∇_{Y} X)(u)

12 RIEMANNIAN GEOMETRY

Example 6.12. If g ∈ T_{0}^{2}(M) is the Riemannian metric, then ∇g is the 3-form given by

(∇g)(X, Y, Z) =^{(??)} (∇_{Z}g)(X, Y) =^{(??)} Zg(X, Y) − g(∇_{Z}X, Y) − g(X, ∇_{Z}Y)

for any three vector fields X, Y, Z. Consequently

∇g = 0 ⟺ ∀X, Y, Z ∈ T (M): Z⟨X, Y ⟩ = ⟨∇_{Z}X, Y ⟩ + ⟨X, ∇_{Z}Y ⟩

for all vector fields X, Y, Z. Example 6.13. There are total covariant derivatives

T (M) = T_{1}^{0}(M) ⟶^{∇} T_{1}^{1}(M) ⟶^{∇} T_{1}^{2}(M) ⟶^{∇} ⋯

If V ∈ T_{1}^{0}(M) is a vector field then ∇V is the (\begin{matrix}1 \\ 1 \end{matrix})-tensor given by

(∇V)(ω, X) = (∇_{X}V)(ω) = ω(∇_{X}V)

If V = V^{i}∂_{i} in local coordinates, then

(∇V)(dx^{i}, ∂_{j}) = dx^{i}(∇_{∂_{j}}V) = dx^{i}(∇_{∂_{j}}V^{k}∂_{k}) = dx^{i}(∂_{j}V^{k}∂_{k} + V^{k}Γ_{jk}^{ℓ}∂_{ℓ}) = ∂_{j}V^{i} + V^{k}Γ_{jk}^{i} and therefore

∇V = (∂_{j}V^{i} + V^{k}Γ_{jk}^{i})∂_{i} ⊗ dx^{j}

We say that the vector field V is parallel if ∇V = 0. Since

∇V = 0 ⟺ ∀X ∈ T (M): ∇_{X}V = 0,

V is parallel iff the covariant derivative of V along any vector field is zero. What tensor is ∇^{2}V? What does it mean if ∇^{2}V = 0?

7. Geodesics and parallel translation along curves

Let γ: I → M be a smooth curve on M. Definition 7.1. For any t ∈ I, the tangent vector

γ^{•} (t) = γ_{∗}(\frac{d}{dt}(t)) ∈ T_{γ(t)}M, γ^{•} (t)f = \frac{d(f ∘ γ)}{dt}(t), f ∈ C^{∞}(M)

is called the velocity vector of γ at the point γ(t).

If x is a coordinate system around γ(t_{0}) and xγ = (γ^{1}, …, γ^{n}) then

(7.2) γ^{•} (t) = \frac{dγ^{i}}{dt} (t)∂_{i}(γ(t))

as a special case of (??). Remark 7.3. We say that the curve γ, defined in an open neighborhood of 0, represents the tangent vector V ∈ T_{p}M if γ(0) = p and γ^{•} (0) = V. Then V f =γ^{•} (0)f = \frac{d}{dt}(fγ)(0) for any smooth function f ∈ C^{∞}(M) and F_{∗}V is represented by the image curve F γ for any smooth map F: M → N. Definition 7.4. A vector field along γ is a smooth map V : I → T M such that the diagram

\xymatrix{ & T M \ar[d] \\ I \ar[ur]^{V} \ar[r]_{γ} & M}

commutes. A vector field V along is extendible if V (t) = \overline{V} (γ(t)) for some vector field \overline{V} on a neighborhood of γ(I). The C^{∞}(I)-module of all vector fields along γ is denoted T (γ).

The velocity field γ^{•} (t) is an example of a vector field along γ.

The formulation of the lemma below makes use of ??.

Lemma 7.5 (Covariant differentiation along a curve). Let ∇ be a connection on M. There exists precisely one R-linear map D_{t}: T (γ) → T (γ) such that

D_{t}(fV) = \frac{df}{dt} V + fD_{t}V,  f ∈ C^{∞}(I), V ∈ T (γ)
D_{t}V = ∇_{γ^{•}(t)}\overline{V}  V extendible
• If E_{i} is a local frame around γ(t_{0}) and V = V^{j}E_{j}, then

D_{t}V = \frac{dV^{j}}{dt} E_{j} + V^{j}∇_{γ^{•}(t)}E_{j}

for t near t_{0}.

• If x is a coordinate system around γ(t_{0}) and V = V^{j}∂_{j}, then

D_{t}V = (\frac{dγ^{k}}{dt} + Γ_{ij}^{k} \frac{dγ^{i}}{dt} V^{j})∂_{k}

for t near t_{0}.

A curve on M is as close as possible to being a straight line if the curve at all times just continues in the direction of γ^{•} (t), if its velocity γ^{•} (t) does not change. No change means zero covariant derivative.

Definition 7.6. A smooth curve γ is a geodesic (with respect to the connection ∇) if the covariant derivative of its velocity field vanishes, D_{t} γ^{•}= 0.

A geodesic is a curve that follows its own nose. A geodesic is a curve with constant velocity. Light rays follow geodesics in space-time.

If x is a coordinate system around some point γ(t_{0}) of γ, then γ is a geodesic iff

(7.7)  \frac{d^{2}γ^{k}}{dt^{2}} + Γ_{ij}^{k} \frac{dγ^{i}}{dt} \frac{dγ^{j}}{dt} = 0

near t_{0}. In principle, we could determine geodesics by solving this equation. In practice, this is close to impossible. Instead we try to identify some properties that a geodesic must have (??, ??).

Proposition 7.8 (Existence and uniqueness of geodesics). Let p be a point on M and V ∈ T_{p}M a tangent vector at p There exists a unique maximal geodesic γ_{V} : I → M defined on an open interval containing 0 such that γ_{V} (0) = p and γ^{•}_{V} (0) = V.

Lemma 7.9 (Rescaling lemma). Let V ∈ T_{p}M be a tangent vector at p ∈ M and let c ∈ R be a real number. The geodesic γ_{cV} is defined at t iff the geodesic γ_{V} is defined at ct and then γ_{cV} (t) = γ_{V} (ct).

Proof. Consider the maximal geodesic γ_{V} : I → M. Put γ(t) = γ_{V} (ct) for all t ∈ c^{−1}I. Then γ(t) is a geodesic because it satisfies (??) in any coordinate system and it is defined at c^{−1}t. In fact,

γ(t) = γ_{cV} (t) as γ(0) = γ_{V} (0) = p and γ^{•} (0) = cV. □

A vector field along a curve is parallel if it doesn’t change; no change meaning zero covariant derivative. A curve is a geodesic if it has a parallel velocity field.

Definition 7.10. A vector field V along γ is parallel if it does not change along γ, D_{t}V = 0.

Proposition 7.11. Let γ be a curve on M. Suppose that p_{0} = γ(t_{0}) is a point on γ and V_{0} ∈ T_{p_{0}}M a tangent vector at that point. There exists precisely one parallel vector field V along γ such that V (t_{0}) = V_{0}.

8. The Riemannian connection

On a Riemannian manifold there is a preferred connection.

Theorem 8.1 (The fundamental theorem of Riemannian geometry). A Riemannian manifold admits precisely one symmetric connection compatible with the metric.

This particular connection is called the Riemannian connection or the Levi–Civitta connection.

Definition 8.2. The connection ∇ is symmetric if

∇_{X}Y − ∇_{Y} X = [X, Y]

for all vector fields X, Y ∈ T (M).

Lemma 8.3. Let E_{i} be a local moving frame such that [E_{i}, E_{j}] = 0, 1 ≤ i, j ≤ n (for instance E_{i} = ∂_{i} could be a coordinate frame). Then ∇ is symmetric if and only if

Γ_{ij}^{k} = Γ_{ji}^{k}, 1 ≤ i, j, k ≤ n

Proof. ∇_{E_{i}}E_{j} − ∇_{E_{j}}E_{i} = (Γ_{ij}^{k} − Γ_{ji}^{k})E_{k}. □

Definition 8.4. The connection ∇ is compatible with the metric g if

X ⟨Y, Z⟩ = ⟨∇_{X}Y, Z⟩ + ⟨X, ∇_{X}Z⟩

for any three vector fields X, Y, Z ∈ T (M).

Lemma 8.5. Let E_{i} be a local moving frame. Then ∇ is compatible with the metric g if and only if

E_{k}g_{ij} = Γ_{ki}^{ℓ}g_{jℓ} + Γ_{kj}^{ℓ}g_{iℓ}, 1 ≤ i, j, k, ℓ ≤ n

Proof. E_{k} ⟨E_{i}, E_{j}⟩ − ⟨∇_{E_{k}}E_{i}, E_{j}⟩ − ⟨E_{i}, ∇_{E_{k}}E_{j}⟩ = E_{k}g_{ij} − Γ_{ki}^{ℓ}g_{jℓ} − Γ_{kj}^{ℓ}g_{iℓ}. □

Proof of Theorem ??. We first show uniqueness. Assume that ∇ is a symmetric connection that is compatible with g. Then

X ⟨Y, Z⟩ =^{??} ⟨∇_{X}Y, Z⟩ + ⟨Y, ∇_{X}Z⟩ =^{??} ⟨∇_{X}Y, Z⟩ + ⟨Y, ∇_{Z}X⟩ + ⟨Y, [X, Z]⟩

for any three vector fields X, Y, Z ∈ T (M). Permute X, Y, Z cyclically, obtain

X ⟨Y, Z⟩ − Z ⟨X, Y ⟩ + Y ⟨X, Z⟩ = 2 ⟨∇_{X}Y, Z⟩ + ⟨Y, [X, Z]⟩ − ⟨X, [Z, Y]⟩ + ⟨Z, [Y, X]⟩

and conclude that

(8.6) ⟨∇_{X}Y, Z⟩ = \frac{1}{2} (X ⟨Y, Z⟩ − Z ⟨X, Y ⟩ + Y ⟨X, Z⟩ − ⟨Y, [X, Z]⟩ + ⟨X, [Z, Y]⟩ − ⟨Z, [Y, X]⟩)

This equation shows that ∇, if it exists, is determined by the metric.

Next, we show existence. We will define ∇ in any open submanifold where we have a moving frame E_{i} with [E_{i}, E_{j}] = 0 (for instance in a chart domain). The only possibility is to put

(8.7)  Γ_{ij}^{k} = \frac{1}{2}g^{kℓ} (E_{i}g_{jℓ} − E_{ℓ}g_{ij} + E_{j}g_{iℓ})

because

(8.8)  Γ_{ij}^{k}g_{kℓ} = ⟨∇_{E_{i}}E_{j}, E_{ℓ}⟩ = \frac{1}{2} (E_{i}g_{jℓ} − E_{ℓ}g_{ij} + E_{j}g_{iℓ})

by equation (??). Since Γ_{ij}^{k} = Γ_{ji}^{k} and

Γ_{ki}^{ℓ}g_{jℓ} + Γ_{iℓ}^{ℓ}g_{iℓ} = Γ_{ki}^{ℓ}g_{ℓj} + Γ_{iℓ}^{ℓ}g_{ℓi} =^{??} \frac{1}{2} (E_{k}g_{ij} − E_{j}g_{ik} + E_{i}g_{jk}) + \frac{1}{2} (E_{k}g_{ij} − E_{i}g_{jk} + E_{j}g_{ik}) = E_{k}g_{ij}

this connection ∇ is symmetric and compatible with g (??, ??). □

Example 8.9. The Euclidean connection \overline{∇} (??) is the Riemannian connection on Euclidean space R^{n} for it is symmetric and compatible with the Euclidean metric \overline{g} (??) since Γ_{ij}^{k} = 0.

What is the Riemannian connection in spherical and hyperbolic space?

Lemma 8.10. Let ∇ be a connection and g a metric on M. The following conditions are equivalent:

(1) ∇ and g are compatible (??)

(2) ∇g = 0

(3) \frac{d}{dt} ⟨V, W ⟩ = ⟨D_{t}V, W ⟩ + ⟨V, D_{t}W ⟩ whenever V, W are vector fields along a smooth curve (4) ⟨V, W ⟩ is constant whenever V, W are parallel vector fields along a smooth curve

(5) Parallel transport is an isometry Proof. (1) ⟺ (2): ??

(3) ⟹ (4) ⟹ (5) : Obvious.

(5) ⟹ (3) : Let P_{1}, …, P_{n} be parallel vector fields that are orthonormal at one point of the curve and hence orthonormal at any point. Write V = V^{i}P_{i} and W = W^{i}P_{i}. Then ⟨V, W ⟩ = ∑ V^{i}W^{i} and (??) D_{t}V = \frac{dV^{i}}{dt} P_{i} and D_{t}W = \frac{dW^{i}}{dt} P_{i}. Hence

⟨D_{t}V, W ⟩ + ⟨V, D_{t}W ⟩ = ∑ \frac{dV^{i}}{dt} W^{i} + ∑ V^{i} \frac{dW^{i}}{dt} = \frac{d}{dt} ∑ V^{i}W^{i} = \frac{d}{dt} ⟨V, W ⟩

(3) ⟹ (1) : Let Y and Z be smooth vector fields on M and let γ be a smooth curve, γ(0) = p,

γ^{•} (0) = X(p). Then

X_{p} ⟨Y, Z⟩ =γ^{•} (0) ⟨Y, Z⟩ = \frac{d}{dt} ⟨Y, Z⟩ = ⟨D_{t}Y, Z⟩ + ⟨Y, D_{t}Z⟩ = ⟨∇_{X_{p}}Y, Z⟩ + ⟨Y, ∇_{X_{p}}Z⟩

(1) ⟹ (3) : Let p = γ(0), choose a vector field X with X(p) =γ^{•} (0), and choose an orthonormal moving frame, E_{i}, around p. Then

0 = X ⟨E_{i}, E_{j}⟩ = ⟨∇_{X}E_{i}, E_{j}⟩ + ⟨E_{i}, ∇_{X}E_{j}⟩

since ∇ and g are compatible Write V = V^{i}E_{i} and W = W^{j}E_{j}. Then D_{t}V =V^{i}^{•} E_{i} + V^{i}∇_{X}E_{i}, and similarly for W, at the point p (??). Therefore,

⟨D_{t}V, W ⟩ + ⟨V, D_{t}W ⟩ = (V^{i}^{•} W^{j} + V^{i} W^{j}^{•}) ⟨E_{i}, E_{j}⟩ + V^{i}W^{j}(⟨∇_{X}E_{i}, E_{j}⟩ + ⟨E_{i}, ∇_{X}E_{j}⟩)

= ∑(V^{i}^{•} W^{i} + V^{i} W^{i}^{•}) = \frac{d}{dt} ∑ V^{i}W^{i} = \frac{d}{dt} ⟨V, W ⟩

□

Corollary 8.11. Riemannian geodesics have constant speed.

Proof. \frac{d}{dt}| γ^{•} (t)|^{2} = \frac{d}{dt}⟨γ^{•} (t), γ^{•} (t)⟩ = 2⟨D_{t} γ^{•} (t), γ^{•} (t)⟩ = 0. □

9. Connections on submanifolds and pull-back connections

Let (\overline{M}, \overline{g}) be a Riemannian manifold and M ⊂ \overline{M} be an embedded submanifold. Suppose that we have a connection ∇ on \overline{M}. How can we obtain a connection on the submanifold M?

For any tangent vector X_{p} ∈ T_{p}\overline{M}, let X_{p}^{T} ∈ T_{p}M denote the orthogonal projection of X_{p}.

Proposition 9.1 (Existence of uniqueness of tangential connections). There exists precisely one connection ∇^{T} on M such that

∇_{X}^{T}Y = (∇\overline{X} \overline{Y} )^{T}

whenever \overline{X}, \overline{Y} are vector fields on \overline{M} and X, Y their restrictions to M. If ∇ is the Riemannian connection on \overline{M}, then ∇^{T} is the Riemannian connection on M.

Lemma 9.2. Let ∇ be a connection and let X, Y ∈ T (M) be vector fields on M. Then ∇_{X}Y (p) only depends on X(p) and Y along a curve tangent to X(p).

Proof. ∇_{X}Y (p) = ∇_{γ^{•}(0)}Y (p) = D_{t}Y (0) for any curve γ: (−ε, ε) → M with γ(0) = p and γ^{•} (0) =

X(p). □

Proposition 9.3 (Pull-back connections). Let ϕ: M → \overline{M} be a diffeomorphism and ∇ a connection on \overline{M}.

Let (also) ∇ be the map that makes the diagram

\xymatrix{T (M) × T (M) \ar[d]_{ϕ_{∗}×ϕ_{∗}}^{≅} \ar@{-->}[r]^{∇} & T (M) \ar[d]^{ϕ_{∗}}_{≅} \\ T (\overline{M}) × T (\overline{M}) \ar[r]^{∇} & T (\overline{M})}

commutative. Thus ϕ_{∗}∇_{X}Y = ∇_{ϕ_{∗}X}ϕ_{∗}Y for all X, Y ∈ T (M).

(1) ∇ is a connection on M.

(2) Covariant differentiation wrt ∇ makes the diagram

\xymatrix{T (γ) \ar[r]^{D_{t}} \ar[d]_{ϕ_{∗}}^{≅} & T (γ) \ar[d]_{ϕ_{∗}}^{≅} \\ T (ϕγ) \ar[r]_{D_{t}} & T (ϕγ)}

commutative. Thus ϕ_{∗}D_{t}V = D_{t}ϕ_{∗}V for any vector field V along the curve γ in M.

(3) Assume that M and \overline{M} are Riemannian manifolds and that ϕ is an isometry. If \overline{∇} the Riemannian connection on \overline{M}, then ∇ is the Riemannian connection on M. Thus ϕ_{∗}∇_{X}Y = \overline{∇}_{ϕ_{∗}X}ϕ_{∗} for all X, Y ∈ T (M).

Corollary 9.4. Isometries of Riemannian manifolds take Riemannian geodesics to Riemannian geodesics.

Proof. Let ϕ: (M, g) → (\overline{M}, \overline{g}) be an isometry of Riemannian manifolds. Then the connection on M is the pull-back of the connection on \overline{M} (??.(??)). Let γ_{V}, V ∈ T_{p}M, be a geodesic on M. Then

D_{t}((ϕγ_{V})^{•}) = D_{t}(ϕ_{∗} γ^{•}) =^{??.(??)} ϕ_{∗}D_{t}(γ^{•}) = ϕ_{∗}0 = 0

so that ϕγ_{V} = γ_{ϕ_{∗}V}. □

10. Geodesics in the three geometries

We determine the geodesics in Euclidean, spherical, and hyperbolic geometry.

10.1. Euclidean geometry. The Riemannian connection on R^{n} is the Euclidean connection (??).

Let p = (0, …, 0) ∈ R^{n} and V = (1, 0, …, 0) ∈ T_{p}R^{n}. We know that there is a unique maximal geodesic γ_{V} running through p with velocity V. We also know that ϕγ_{V} = γ_{V} for any isometry ϕ ∈ O(n) of R^{n} preserving (p, V). The map ϕ(ξ_{1}, ξ_{2}, ξ_{3}, …, ξ_{n}) = (ξ_{1}, −ξ_{2}, ξ_{3}, …, ξ_{n}) is such an isometry (it is a diffeomorphism and it preserves |ξ|). Thus γ_{V} must have ξ^{2}γ_{V} = 0. Similarly, ξ^{3}γ_{V} = 0, …, ξ^{n}γ_{V} = 0. Thus γ_{V} must run along the ξ^{1}-axis. Since it has constant speed and γ_{V}^{•} (0) = V, we must have

γ_{V} (t) = (t, 0, …, 0)

This was just one geodesic! But since R^{n} is homogeneous and isotropic, we have in fact determined all geodesics: The geodesics in Euclidean geometry are the straight lines. For any point not on a geodesic there is a unique geodesic passing through that point parallel to the given geodesic.

10.2. Spherical geometry. The Riemannian connection on

S_{R}^{n} = {(ξ, τ) ∈ R^{n} × R | |ξ|^{2} + τ^{2} = R^{2}} ⊂ R^{n+1}

is the tangential connection (??) arising from the Euclidean connection on ambient R^{n+1}.

Let N = (0, …, R) ∈ S_{R}^{n} be the North Pole and V = (1, 0, …, 0) ∈ T_{N}§_{R}^{n}. What is γ_{V}, the geodesic running through N with velocity V? Using the isometries that change sign on ξ_{i} for 2 ≤ i ≤ n we see, as above, that γ_{V} must run in the intersection of S_{R}^{n} and the ξ^{1}τ-plane. Thus

γ_{V} (t) = (R sin(t/R), 0, …, 0, R cos(t/R))

We conclude that the geodesics in spherical geometry are great circles, the intersection of S_{R}^{n} with planes through the origin. For any point not on a geodesic there is a no geodesic passing through that point parallel to the given geodesic.

10.3. Hyperbolic geometry. The Riemannian connection on

H_{R}^{n} = {(ξ, τ) ∈ R^{n} × R_{+} | |ξ|^{2} − τ^{2} = −R^{2}} ⊂ R^{n+1}

is the tangential connection (??) arising from the Euclidean connection, considered as the Riemannian connection of Minkowski metric |(ξ, τ)|^{2} = |ξ|^{2} − τ^{2}, on ambient R^{n+1}.

Let N = (0, …, R) ∈ S_{R}^{n} be the North Pole and V = (1, 0, …, 0) ∈ T_{N}§_{R}^{n}. What is γ_{V}, the geodesic running through N with velocity V? Using the isometries that change sign on ξ_{i} for 2 ≤ i ≤ n we see, as above, that γ_{V} must run in the intersection of H_{R}^{n} and the ξ^{1}τ-plane. Thus

γ_{V} (t) = (R sinh(t/R), 0, …, 0, R cosh(t/R))

We conclude that the geodesics in hyperbolic geometry are great hyperbolas, the intersection of H_{R}^{n} with planes through the origin. (The isometry group O(n, 1)_{+} takes planes through the origin, u^{⊥}, to planes through the origin.) For any point not on a geodesic there are uncountably many geodesic passing through that point parallel to the given geodesic.

11. The exponential map and normal coordinates

Let M be a Riemannian manifold with Riemannian connection ∇ (??). Put

E = {V ∈ T M | γ_{V} is defined at 1}

where γ_{V}, V ∈ T_{p}M, is the geodesic through γ(0) = p with velocity γ^{•} (0) = V (??). We define

(11.1)  exp: E → M by exp(V) = γ_{V} (1)

meaning that exp(V) is obtained by following the geodesic with initial velocity vector V for one time unit. We let exp_{p} denote the restriction of exp to E_{p} = E ∩ T_{p}M.

Proposition 11.2 (Properties of the exponential map). Let exp: E → M be the exponential map on the manifold M.

(1) E is an open subspace of T M and exp: E → M is a smooth map.

(2) E_{p} is star-shaped around 0 for each p ∈ M.

(3) exp_{p}: E_{p} → M takes straight lines through 0 ∈ T_{p}M to geodesics through p: exp_{p}(tV) = γ_{V} (t) (where both functions are defined for the same set of ts).

(4) (exp_{p})_{∗}: T_{0}T_{p}M = T_{p}M → T_{p}M is the identity map.

(5) The exponential map commutes with isometries: The diagram

\xymatrix{T_{p}M \ar[d]_{_{exp_{p}}} \ar[r]^{ϕ_{∗}} & T_{p}\overline{M} \ar[d]^{_{exp_{ϕ(p)}}} \\ M \ar[r]_{ϕ} & \overline{M}}

commutes for any isometry ϕ: M → \overline{M}.

Proof. We defer the proof of (1). Let V ∈ T_{p}M and t ∈ R. Then

tV ∈ E_{p} ⟺ exp_{p} is defined at tV ⟺ γ_{tV} is defined at 1 ⟺ γ_{V} is defined at t

and for such a t, exp_{p}(tV) = γ_{tV} (1) = γ_{V} (t) by the Rescaling lemma (??). In particular,

V ∈ E_{p} ⟺ γ_{V} is defined at 1 ⟹ γ_{V} is defined at s ⟺ sV ∈ E_{p}

when 0 < s ≤ 1. Thus E_{p} is star-shaped around 0. Assuming that E_{p} is open in T_{p}M and that exp_{p} is smooth we now compute the differential of exp_{p}. Since the tangent vector V ∈ T_{0}T_{p}M = T_{p}M is (??) represented by the image curve t → tV, (exp_{p})_{∗} is represented by curve exp(tV) = γ_{V} (t)

with γ^{•}_{V} (0) = V. Thus (exp_{p})_{∗}V = V. If ϕ: M → \overline{M} is an isometry, then ϕγ_{V} = γ_{ϕ_{∗}V} (??) so that ϕ exp_{p}(V) = ϕγ_{V} (1) = γ_{ϕ_{∗}V} (1) = exp_{ϕ(p)}(ϕ_{∗}V).

The geodesic vector field G on T M is the defined by

G(V)f = \frac{d}{dt}|_{t=0}(f γ^{•}_{V} (t)), f ∈ C^{∞}(T M)

where γ^{•}_{V}(t) is curve on T M obtained by taking the velocity of the geodesic γ_{V}. Note that

γ_{V} (t_{0} + t) = γ_{γ^{•}_{V} (t_{0})}(t) so that

G(γ^{•}_{V} (t_{0}))f = \frac{d}{dt}|_{t=0}f(γ^{•}_{V} (t + t_{0})) = \frac{d}{dt}|_{t=t_{0}}f(γ^{•}_{V} (t))

where t_{0} is an arbitrary point in the open interval of definition for γ_{V}. This means that integral curves for the vector field G are velocity fields along geodesics. By a general theorem, there exists a smooth map θ: O → T M, defined on an open subspace O ⊂ R × T M containing 0 × T M, such that θ(t, V) is the maximal integral curve for G through V ∈ T M at time t = 0. Hence

E = {V ∈ T M | (1, V) ∈ O} = i_{1}^{−1}O, i_{1}(V) = (1, V),

is open and exp(V) = πθ(1, V) is smooth as a composition of smooth maps (π: T M → M is the projection). □

Example 11.3. Let N = (0, 0, 1) be the North Pole of S^{2} ⊂ R^{3} and let V be a unit vector in the tangent space T_{N}S^{2} = N^{⊥}. The exponential map exp_{N}: T_{N}S^{2} = N^{⊥} → S^{2} takes the unit speed radial line tV to the unit speed geodesic whose trace is the intersection of S^{2} with the plane through 0 ∈ R^{3} containing N and V.

Let U_{p} ⊂ T_{p}M be an open subset, star-shaped around 0, of the tangent space such that exp_{p}: U_{p} → U is a diffeomorphism between U_{p} and an open subset U of M. Choose an orthonormal basis E_{i} for T_{p}M (with inner product g_{p}) and an orthonormal basis e_{i} for R^{n} (with standard inner product \overline{g}). Let E: (R^{n}, \overline{g}) → (T_{p}M, g_{p}) be the isometry given by Ee_{i} = E_{i}. Then

(11.4)  x = E^{−1} ∘ exp_{p}^{−1} : M ⊃ U → x(U) ⊂ R^{n}  \xymatrix{M \ar[rr]_{x} & T_{p}M \ar[l]_{^{exp^{p}}} & R^{n} \ar[l]_{E}}

are normal coordinates around p. (We will often forget to mention E so that V can stand for a tangent vector V ∈ T_{p}M as well as a vector V ∈ R^{n}.) The smooth function

(11.5)  r: U − p → R,  r(q) = |x(q)| = \sqrt{∑ x^{i}(q)^{2}} = |V |_{g} (exp_{p}(V) = q)

is the radial distance function and the unit radial vector field on U − p, denoted

(11.6)  \frac{∂}{∂r} (q),  q ∈ U − p,

is the vector field formed by the velocity vectors of the unit speed radial geodesic; to any q ∈ U − p it associates the velocity of the unit speed radial geodesic through q.

If B_{R}(0) ⊂ U_{p}, then

\overline{B}_{R}(p) = exp_{p}(\overline{B}_{R}(0)) = \{q ∈ x(U) | r(q) ≤ R\}
B_{R}(p) = exp_{p}(B_{R}(0)) = {q ∈ x(U) | r(q) < R}
S_{R}(p) = exp_{p}(∂B_{R}(0)) = {q ∈ x(U) | r(q) = R}

is a (closed) geodesic ball, respectively, a geodesic sphere around p. (Make a drawing of the situation!)

Proposition 11.7 (Properties of normal coordinates). Let x: U → x(U) be normal coordinates (??) around p.

(1) x(γ_{EV} (t)) = tV for all V ∈ T_{p}M and for all small t. (In normal coordinates, the geodesics through p are straight lines through 0.)

(2) x(p) = 0, \frac{∂}{∂x^{i}} (p) = E_{i}, g_{ij}(p) = δ_{ij}, Γ_{ij}^{k}(p) = 0, ∂_{k}g_{ij}(p) = 0.

(3) \frac{∂}{∂r} = \frac{x^{i}(q)}{r(q)} \frac{∂}{∂x^{i}} for all q ∈ U − p.

Proof. (1) x^{−1}(tV) = exp_{p}(tEV) = γ_{EV} (t).

(2) x^{−1}(0) = exp_{p}(0) = p. Let γ(t) = x^{−1}(te_{i}) be the ith coordinate axis. Then γ^{•} (t) = \frac{∂}{∂x^{i}} (γ(t)) in general (??). In this case, γ(t) = x^{−1}(te_{i}) = exp_{p}(tE_{i}) = γ_{E_{i}}(t) is the geodesic through γ(0) = p

with velocity γ^{•} (0) = E_{i}. Thus E_{i} = \frac{∂}{∂x^{i}} (p). The components of the Riemannian metric g at p are g_{ij}(p) = g(\frac{∂}{∂x^{i}} (p), \frac{∂}{∂x^{j}} (p)) = g(E_{i}, E_{j}) = δ_{ij}. The radial curve γ(t) = x^{−1}(t(e_{i} + e_{j})) is a geodesic so its coordinates γ^{k} satisfy the ODEs (??) which in this particular case means that Γ_{ij}^{k}(γ(t)) = 0 for all t. For t = 0, we get Γ_{ij}^{k}(p) = 0. In other words, ∇_{∂_{i}}∂_{j}(p) = 0 for all i, j. Then also

∂_{k}g_{ij}(p) = ∂_{k} ⟨∂_{i}, ∂_{j}⟩ (p) = ⟨∇_{∂_{k}}∂_{i}(p), ∂_{j}(p)⟩ + ⟨∂_{i}(p), ∇_{∂_{k}}∂_{j}(p)⟩ = 0 + 0 = 0

since the Riemannian connection is compatible with the Riemannian metric (??).

(3) Let q ∈ U − p. Suppose that x(q) = V ∈ R^{n}. Then x^{i}(q) = V^{i} and r(q) = |x(q)| = |V | (??). The unit speed radial geodesic from p to q is γ(t) = exp_{p}(tV/|V |), 0 ≤ t ≤ |V |. Its velocity vector

is γ^{•} (t) = \frac{V^{i}}{|V |} \frac{∂}{∂x^{i}} (γ(t)) (??); at q, in particular, its velocity is \frac{V^{i}}{|V |} \frac{∂}{∂x^{i}} (q) = \frac{x^{i}(q)}{r(q)} \frac{∂}{∂x^{i}} (q). □

Let Γ: (−ε, ε) × [a, b] → M be a variation of the curve γ(t) = Γ(0, t). We write (s, t) for points in (−ε, ε)×[a, b] (so that the interval [a, b] is placed on the vertical t-axis!). Let Γ_{s}(t) = Γ(s, t) = Γ^{t}(s) so that Γ_{s} is a a curve in the t-direction (a main curve) and Γ^{t} is a curve in the s-direction (a transverse curve). Let

∂_{t}Γ(s, t) = \frac{d}{dt}Γ_{s}(t) =Γ^{•}_{s} (t), ∂_{s}Γ(s, t) = \frac{d}{ds}Γ^{t}(s) =Γ^{t}^{•} (s)

be the velocities of the main, respectively, the transverse curves. We may view the main velocity field ∂_{t}Γ as a vector field along a transverse curve Γ^{t} and consider its covariant derivative D_{s}∂_{t}Γ along Γ^{t}. Similarly, we may view the transverse velocity field ∂_{s}Γ as a vector field along a main curve Γ_{s} and consider its covariant derivative D_{t}∂_{s}Γ along Γ_{s}.

Lemma 11.8 (Symmetry Lemma). D_{s}∂_{t}Γ = D_{t}∂_{s}Γ or D_{s} Γ^{•}_{s}= D_{t} Γ^{t}^{•}.

Proof. This is a local question. Choose a coordinate system x around Γ(s_{0}, t_{0}). In local coordinates,

xΓ(s, t) = (Γ^{1}(s, t), …, Γ^{n}(s, t)) and (??) ∂_{t} =Γ_{s}^{•}= \frac{∂Γ^{i}}{∂t} ∂_{i}, ∂_{s} =Γ^{t}^{•}= \frac{∂Γ^{i}}{∂s} ∂_{i}. In local coordinates there is a formula for the covariant derivative along a curve (??). Using that formula, we get that

D_{s}∂_{t}Γ = (\frac{∂^{2}Γ^{k}}{∂s∂t} + Γ_{ij}^{k} \frac{∂Γ^{i}}{∂s} \frac{∂Γ^{j}}{∂t} ) ∂_{k}

and there is a similar formula for D_{t}∂Γ_{t} except that the s and t swap places. The point is now that the Riemannian connection is symmetric so that the Christoffel symbols are symmetric in i and j (??). □

The variational field

(11.9)  V (t) =Γ^{t}^{•} (t, 0)

is the restriction to γ(t) of the transverse vector field Γ^{t}^{•}.

Lemma 11.10 (First Variation of smooth curves). \frac{d}{ds}L(Γ_{s})(0) = ⟨V, γ^{•}⟩ |_{a}^{b} −∫_{a}^{b} ⟨V, D_{t} γ^{•}⟩ dt when γ(t) is a unit speed curve.

Proof. We differentiate the function s → L(Γ_{s}) and then evaluate the result at s = 0. Using that the connection is compatible with the metric (??) and The Symmetry lemma we get

\frac{d}{ds}L(Γ_{s}) = \frac{d}{ds} ∫_{a}^{b} | Γ_{s}^{•} (t)|dt = ∫_{a}^{b} \frac{∂}{∂s} ⟨Γ^{•}_{s}, Γ^{•}_{s}⟩^{1/2} dt =^{??} ∫_{a}^{b} \frac{1}{| Γ^{•}_{s} |} ⟨D_{s} Γ^{•}_{s}, Γ^{•}_{s}⟩ dt

=^{??} ∫_{a}^{b} \frac{1}{| Γ^{•}_{s} |} ⟨D_{t} Γ^{t}^{•}, Γ^{•}_{s}⟩ dt

When s = 0, Γ^{•}_{s}=Γ^{•}_{0}=γ^{•}, | γ^{•} | = 1, and Γ^{•}_{s} (0, t) = V (t) so that

\frac{d}{ds}L(Γ_{s})(0) = ∫_{a}^{b} ⟨D_{t}V, γ^{•}⟩ dt =^{??} ∫_{a}^{b} (\frac{d}{dt} ⟨V, γ^{•}⟩ − ⟨V, D_{t} γ^{•}⟩) dt

= ⟨V (b), γ^{•} (b)⟩ − ⟨V (a), γ^{•} (a)⟩ − ∫_{a}^{b} ⟨V, D_{t} γ^{•}⟩ dt

□

Lemma 11.11 (Gauss lemma). In a normal neighborhood of a point p we have:

(1) The geodesic spheres are orthogonal to the geodesic rays.

(2) ⟨\frac{∂}{∂r} , Y ⟩ = Y (r) = dr(Y) for any tangent vector Y ∈ T_{q}M, q ≠ p (In other words, grad r = \frac{∂}{∂r}).

Proof. Let q = exp_{p}(V) where V ≠ 0. Then x(q) = V and r(q) = |V | = R. The claim is that

W ⊥ V ⟹ (exp_{p})_{∗}W ⊥γ^{•}_{V} (1)

for any W ∈ T_{V} T_{p}M = T_{p}M (where V^{⊥} = T_{V} S_{R}(0)). Let σ(s) be a curve in S_{R}(0) that represents W, σ(0) = V, σ^{•} (0) = W. Consider the variation Γ(s, t) = exp_{p}(tσ(s)) of γ_{V} (t) = exp_{p}(tV). Put S = ∂_{s}Γ and T = ∂_{t}Γ. Note that the curves in the t-direction, t → exp_{p}(tσ(s)), are geodesics of velocity T and speed |T | = |σ(s)| = R. Hence D_{t}T = 0 and |T | = R is constant. It follows that

\frac{∂}{∂t}⟨S, T ⟩ =^{(??)} ⟨D_{t}S, T ⟩ + ⟨S, D_{t}T ⟩ = ⟨D_{t}S, T ⟩ =^{(??)} ⟨D_{s}T, T ⟩ =^{(??)} \frac{1}{2} \frac{∂}{∂s}|T |^{2} = 0

where we use that the connection is compatible with the metric and symmetric. Thus

⟨S, T ⟩(0, 0) = ⟨S, T ⟩(0, 1)

since ⟨S, T ⟩(s, t) is independent of t. We know compute

S(0, 0) =Γ^{0}^{•} (0) = 0, S(0, 1) =Γ^{1}^{•} (0) = (exp_{p})_{∗}W, T (0, 1) =γ^{•}_{V} (1)

as Γ^{0}(s) = p, Γ^{1}(s) = exp_{p}(σ(s)), and Γ_{0}(t) = exp_{p}(tV) = γ_{V} (t). It follows that

0 = ⟨S, T ⟩(0, 0) = ⟨S, T ⟩(0, 1) = ⟨(exp_{p})_{∗}W, γ^{•}_{V} (1)⟩

which is the first item of the lemma.

We now know that there is an orthogonal decomposition T_{q}M = R\frac{∂}{∂r} (q) ⊕ T_{q}S_{R}(p) for we have

already seen that \frac{∂}{∂r} (q) is the unit vector proportional to γ^{•}_{V} (1). Any Y ∈ T_{q}M therefore admits an orthogonal decomposition of the form Y = α\frac{∂}{∂r} (q) + X where α ∈ R and X ∈ T_{q}S_{R}(p) is tangent to the geodesic sphere. Hence

⟨\frac{∂}{∂r} (q), Y ⟩ = α|\frac{∂}{∂r} (q)|^{2} = α

because X(r) = 0 as r is constant on S_{R}(p). On the other hand,

Y_{∗}(r) = (α\frac{∂}{∂r} (q) + X)(r) = α\frac{∂}{∂r} (q)(r)

where

\frac{∂}{∂r} (q)(r) = (\frac{x^{i}}{r} ∂_{i}) (∑(x^{i})^{2})^{1/2} = ∑ \frac{x^{i}}{r} \frac{2x^{i}}{2r} = \frac{r^{2}}{r^{2}} = 1

and we have proved also the second item of the lemma. □

31.03.05

12. The Riemann distance function

Let M be a connected Riemannian manifold.

Definition 12.1. A regular curve on M is a smooth map γ: [a, b] → M such that γ^{•} (t) ≠ 0 for all t ∈ [a, b]. The real number

Lγ = ∫_{a}^{b} | γ^{•} (t)|dt

is the length of the regular curve γ.

A piecewise regular curve on M is a continuous map γ: [a, b] → M such that γ|[a_{i−1}, a_{i}] is regular for some subdivision a = a_{0} < a_{1} < ⋯ < a_{n} = b of [a, b]. The real number

Lγ = ∑ L(γ|[a_{i−1}, a_{i}])

is the length of the piecewise regular curve γ.

A piecewise regular curve γ on M has a velocity vector γ^{•} (t) at all points t which is not one of the break points a_{i}. At a break point a_{i}, we let

Δ_{i} γ^{•}=γ^{•} (a_{i}^{+})− γ^{•} (a_{i}^{−})

denote the jump between the velocity from the left, γ^{•} (a_{i}^{−}) ∈ T_{γ(a_{i})}, and from the right, γ^{•} (a_{i}^{+}).

Proposition 12.2. (1) The length of a (piecewise) regular curve is invariant under repara-

metrization.

(2) Any regular curve has a unit speed parameterization.

Proof. Let γ: [a, b] → M be a regular curve.

(1) Let t: [c, d] → [a, b] be a bijective smooth map with t′(s) ≠ 0 for all s ∈ [c, d]. Then

(γ ∘ t)^{•}(s) = t′(s) γ^{•} (t(s))

so that

∫_{c}^{d} |(γ ∘ t)^{•}(s)|ds = ± ∫_{c}^{d} | γ^{•} (t(s))|t′(s)ds = ∫_{a}^{b} | γ^{•} (t)|dt

where the + applies if t′ > 0 and the − applies if t′ < 0.

(2) Let s: [a, b] → [0, L(γ)] be the smooth map s(t) = ∫_{a}^{t} | γ^{•} (t)|dt. Then s′(t) = | γ^{•} (t)| by the

Fundamental theorem of Calculus. Let t be the inverse function. Then (γ ∘t)^{•}(s) = t′(s) γ^{•} (t(s)) = \frac{1}{s′(t)} γ^{•} (t) is a unit speed curve. □

Lemma 12.3. Any two points in M can be connected by a piecewise regular curve.

Proof. Since connected and locally path-connected spaces are path-connected, M is path-connected [?, §17]. Given any two points, p and q in M, there exists a continuous curve γ: [0, 1] → M connecting them. By the Lesbesgue number lemma [?, §19], there is a subdivision 0 = a_{0} < a_{1} < ⋯ < a_{n} = 1 of [0, 1] such that γ([a_{i−1}, a_{i}]) is contained in a coordinate neighborhood x: U → R^{n} such that x(U) is a ball. Replace γ([a_{i−1}, a_{i}]) by a smooth curve within this coordinate neighborhood between the two end-points. □

Definition 12.4. The function d: M × M → [0, ∞) given by

d(p, q) = inf{L(γ) | γ is a piecewise regular curve from p to q}

is the Riemann distance function.

We all know from general topology that all topological manifolds are metrizable. On a Riemannian manifold we can construct an explicit metric.

Lemma 12.5. d is a metric on the topological space M.

Let Γ: [−ϵ, ϵ] × [a, b] → M be a fixed endpoint variation of the unit speed piecewise regular curve γ(t) = Γ(0, t).

For each s, the length of the piecewise regular curve Γ_{s} is L(Γ_{s}). What is the rate of change of the length of the main curves near γ (hoping that this length function is smooth)?

Theorem 12.6 (First variation formula for piecewise smooth curves). Let γ be a unit speed piecewise regular curve and Γ any piecewise smooth variation of γ. Then

\frac{d}{ds}|_{s=0}L(Γ_{s}) = − ∫_{a}^{b} ⟨V, D_{t} γ^{•}⟩ dt − ∑_{i=1}^{n−1} ⟨V (a_{i}), Δ_{i} γ^{•}⟩

where V is the variational field along γ.

Proof. Since L(Γ_{s}) = ∑ L(Γ_{s}|[a_{i−1}, a_{i}]) is a sum, just add the contributions from each subinterval [a_{i−1}, a_{i}] where we are in the smooth situation (??). Remember that the endpoints are fixed under the variation so that the variational field is 0 at the endpoints. □

The formula shows that the length decreases when we vary γ in the direction of the jumps at the points a_{i} or vary γ is the direction of the acceleration D_{t} γ^{•} between the points a_{i}.

Lemma 12.7. Any vector field V (which vanishes at the endpoints) along a piecewise smooth curve is the variational field of some (fixed endpoint) variation of the curve.

Proof. Thanks to compactness, we can find ϵ > 0 so that ±ϵV (t) ∈ E ⊂ T M for all t ∈ [a, b]. Let Γ(s, t) exp(sV (t)) for (s, t) ∈ (−ϵ, ϵ) × [a, b]. Then Γ is a piecewise smooth variation of γ whose transverse curves are geodesics with velocity \frac{d}{ds} |_{s=0}Γ(s, t) = \frac{d}{ds} |_{s=0} exp(sV (t)) =^{??} V (t). □

Corollary 12.8. (1) Piecewise regular minimizing curves of constant speed are geodesics. (2) Geodesics are locally minimizing curves.

Proof of first part of ??. Let γ be a minimizing curve. For any vector field V along γ, the expression on the right hand side of the equation in ?? is 0 as L(γ_{s}) has a minimum at s = 0. Use this to show first that D_{t} γ^{•} (t_{0}) = 0 at any point which is not a break point. Thus γ is a piecewise geodesic. Next show that there are no break points so that γ is in fact a geodesic. □

What we showed was in fact that geodesics are critical points of the functional L.

Is there a minimizing curve between any two points of M? Are minimizing curves unique? No, there are uncountably many minimizing curves between the North Pole and the South Pole. Or look at the situation where you want to go to the other shore of a lake, there are usually two possibilities. Only if two points are sufficiently close, then there is in fact a unique minimizing curve between them.

Theorem 12.9. Let p be a point of M and \overline{B}_{R}(p) a closed geodesic ball around p.

(1) For any point q ∈ \overline{B}_{R}(p) in the geodesic ball there is a unique minimizing curve from p to q, namely the radial geodesic. Then d(p, q) = r(q) where r is the radial distance function (??).

(2) For any point q ∉ \overline{B}_{R}(p) outside the geodesic ball there is a point x ∈ S_{R}(p) such that d(p, q) = R + d(x, q). Then d(p, q) > R.

Proof. (1) Suppose that q ∈ \overline{B}_{R}(p) lies in the geodesic ball around p and that r(q) = r where r is the radial distance function (??) for that ball. The radial unit speed geodesic from p to q is γ(t) = exp_{p}(tV), t ∈ [0, r], where V ∈ T_{p}M is the unit vector with exp_{p}(rV) = q. This curve has length r so that d(p, q) ≤ r.

Now let σ: [a, b] → M be any piecewise regular unit speed curve from p to q. Let a_{0} be the last

Make a drawing! point where r(σ(t)) = 0 and b_{0} the first point after a_{0} such that r(σ(t)) = r. Then σ|[a_{0}, b_{0}] runs inside the closed geodesic ball of radius r. On the interval (a_{0}, b_{0}] we decompose the velocity

(12.10)  σ^{•} (t) = α(t)\frac{∂}{∂r} (α(t)) + X(t)

into its radial component along the unit radial vector field (??) and a component X(t) ⊥ \frac{∂}{∂r}. Then

α(t) = ⟨σ^{•} (t), \frac{∂}{∂r} ⟩ =^{??}σ^{•} (t)(r) = \frac{d}{dt}(rσ), 1 = | σ^{•} (t)|^{2} = |α(t)|^{2} + |X(t)|^{2} ≥ |α(t)|^{2}

for t ∈ (a_{0}, b_{0}] and therefore

b − a = L(σ) ≥ L(σ|[a_{0}, b_{0}]) = b_{0} − a_{0} ≥ lim_{δ→0} ∫_{a_{0}+δ}^{b_{0}} α(t)dt = lim_{δ→0} ∫_{a_{0}+δ}^{b_{0}} \frac{d}{dt}(rσ)dt

= lim_{δ→0}(r(q) − r(σ(a_{0} + δ))) = r(q) = r

Since σ was an arbitrary curve, we have shown that the radial geodesic γ is minimizing and that d(p, q) = R.

We will now show that γ is the only minimizing curve (up to reparametrization). Suppose that σ: [0, r] → M is some minimizing unit speed curve from p to q. Then the inequalities in the above computation are in fact equalities so that α(t) = 1 for all t. Then X(t) = 0 and (??) says that σ(t) is an integral curve through p for the radial unit vector field. So is γ. Uniqueness of integral curves implies that σ(t) = γ(t) for all t ∈ [0, r].

(2) Suppose that q ∉ \overline{B}_{R}(p). By compactness, there is a point x ∈ S_{R}(p) such that d(x, q) is minimal. Then d(p, q) ≤ d(p, x) + d(x, q) = R + d(x, q). Suppose that d(p, q) < d(p, x) + d(x, q) = R + d(x, q). Then there exists a piecewise smooth curve σ connecting p and q of length L(σ) < R + d(x, q). Let σ_{1} be the first part of σ that runs entirely inside the closed geodesic ball from p to a point on S_{R}(p), and let σ_{2} be the last part of σ that runs entirely outside the closed geodesic ball from a point of S_{R}(p) to q. Then

R + d(x, q) > L(σ) ≥ L(σ_{1}) + L(σ_{2}) ≥ R + L(σ_{2})

because L(σ_{1}) ≥ R by the first part of this theorem. Now L(σ_{2}) < d(x, q) so that the start-point of σ_{2} is a point on S_{R}(p) that is closer to q than x. Contradiction! □

05.04.05

A smooth curve γ: I → M is locally minimizing if any t_{0} ∈ I has a neighborhood such that γ|[t_{1}, t_{2}] is minimizing for all t_{1} < t_{2} in this neighborhood.

Proof of second part of ??. Let γ be a geodesic and γ(t_{0}) a point on γ. Choose a uniformly geodesic neighborhood W around γ(t_{0}). The preimage γ^{−1}(W) is a union of open intervals. Let I_{0} ⊂ I be the interval containing t_{0}. If t_{1}, t_{2} ∈ I_{0} then γ|[t_{1}, t_{2}] is geodesic in W ⊂ B_{δ}(γ(t_{1})) through γ(t_{1}) so it is a radial geodesic in B_{δ}(γ(t_{1})), hence (??) minimizing. □

Also when we go beyond where exp_{p} is injective we can sometimes find minimizing curves, they may no longer be unique, though. (Look at curves on S^{2} from N to S.)

Lemma 12.11. Suppose that there is a point p ∈ M such that the exponential map at p is defined on the whole tangent space T_{p}M. Then there is a minimizing curve, of the form t → exp_{p}(tV), 0 ≤ t ≤ d(p, q) for some unit vector V ∈ T_{p}M, from p to any other point in M.

Proof. Let q be some point different from p and let T = d(p, q) > 0 be the distanec between p and q.

Choose a closed geodesic ball, \overline{B}_{R}(p), around p. We may assume that q is outside this ball, ie T > R, for otherwise we already know that there exists a minimizing curve from p to q (??). Let x be a point on S_{R}(p) that realizes the distance between S_{R}(p) and q and let γ be the unit speed radial geodesic from p through x. By assumption, γ is defined for all t ≥ 0. The miracle is that γ goes through q: γ(T) = q.

To see this, consider the set

S = {b ∈ [0, T] | d(p, q) = d(p, γ(b)) + d(γ(b), q)}

By using the continuity of the distance function d one can show that S is closed (take a sequence of points in S). From ?? we know that [0, R] ⊂ S. Let A = sup S and put y = γ(A). Then T = d(p, q) = d(p, y) + d(y, q) = A + d(y, q) as A ∈ S. Suppose that A < T. Choose a closed geodesic ball \overline{B}_{δ}(y) around y where 0 < δ < T − A. Let z be the point on the geodesic sphere S_{δ}(y) such that d(y, q) = δ + d(z, q) (??) and let τ be the unique radial unit speed geodesic from y to z. The piecewise smooth curve γ|[0, A] ∪ τ from p to z has length A + δ and as

d(p, z) ≥ d(p, q) − d(z, q) = T − (d(y, q) − δ) = T − (T − A − δ) = A + δ

it is minimizing, hence a geodesic (??), in particular a smooth curve with no breaks. Thus γ|[0, A] and τ must fit together to form the geodesic γ|[0, A+δ] by uniqueness of geodesics. Now z = γ(A+δ) and d(p, z) = A + δ so that

d(p, z) + d(z, q) = (A + δ) + (T − A − δ) = T = d(p, q)

which contradicts that A is the supremum of S.

Now we know that T ∈ S so that

T = d(p, q) = d(p, γ(T)) + d(γ(T), q) = L(γ|[0, T]) + d(γ(T), q) = T + d(γ(T), q)

so that d(γ(T), q) = 0 and γ(T) = q. □

Now comes a definition that will only be used for a very short time!

Definition 12.12. A Riemannian manifold is geodesically complete if all maximal geodesics are defined for all of R.

07.04.05

R^{2} is geodesically complete, B_{1}^{2}(0) is not.

Recall that a metric space is complete if all Cauchy sequences converge. Compact metric spaces are complete.

Theorem 12.13 (Hopf-Rinow). M is complete as a metric space ⟺ M is geodesically complete.

Proof. Suppose that M is complete as a metric space. The claim is that all geodesics are defined for all time. Suppose that there is some unit speed maximal geodesic γ: (a, b) → M that cannot be extended beyond b. Let t_{i} ∈ (a, b) be an increasing sequence of points converging to b. Then γ(t_{i}) is Cauchy for d(γ(t_{i}), γ(t_{j})) ≤ L(γ|[t_{i}, t_{j}]) = |t_{i} − t_{j}|. Let q = lim q(t_{i}). Choose a uniformly geodesic neighborhood W around q and a δ > 0 such that W ⊂ B_{δ}(p) for all p ∈ W. This means that any unit speed geodesic through a point of W exists at least in a time span of δ. Choose t_{j} so that t_{j} > b − δ and q(t_{j}) ∈ W. Then we can extend the geodesic γ near t_{j} for at least time δ beyond t_{j}. Contradiction!

Next sssume that there is a point p ∈ M such that exp_{p} is defined for all T_{p}M. Let (q_{i}) be a Cauchy sequence. We can assume that p ≠ q_{i} for all i (if not, throw away some of the q_{i}). Choose V_{i} ∈ T_{p}M such that t → exp_{p}(tV_{i}), 0 ≤ t ≤ 1, is a minimizing radial geodesic from p to q_{i} (??). Then d(p, q_{i}) = ∫_{0}^{1} |V_{i}|dt = |V_{i}|. Since Cauchy sequences are bounded, the sequence (V_{i}) is bounded in T_{p}M. Any bounded sequence in the inner product space T_{p}M contains a convergent subsequence by compactness. Suppose that V_{i_{k}} → V ∈ T_{p}M. Then q_{i_{k}} = exp_{p}(V_{i_{k}}) → exp_{p}(V) ∈ M by continuity of exp_{p}. Any Cauchy sequence containing a convergent subsequence is itself convergent. Thus lim q_{i} = exp_{p}(V). □

In the future we will not bother to say ‘geodesically complete’ but just say ‘complete’.

We actually proved that the conditions

(1) M is metric complete

(2) All maximal geodesics in M are defined on R

(3) All maximal geodesics through one point of M are defined on R are equivalent and that any of these conditions imply

(4) There is a minimizing curve between any two points of M.

13. The Riemann curvature tensor

The Riemann cuvature tensor is the obstruction to flatness.

Definition 13.1. The Riemann curvature endomorphism is the C^{∞}(M)-multilinear map

\begin{aligned}T(M) × T(M) × T(M) & → T(M) \\ (X, Y, Z) & → R(X, Y)Z := ∇_{X}∇_{Y} Z − ∇_{Y} ∇_{X}Z − ∇_{[X,Y]}Z \end{aligned}

and the Riemann curvature tensor is the corresponding (??) (3, 1)-tensor R given by R(X, Y, Z, ω) = ω(R(X, Y)Z).

Proposition 13.2. The map from ?? is C^{∞}(M)-multilinear.

Proof. The funtion is clearly R-multilinear.

Note that [fX, Y] = f[X, Y] − Y (f)X by computing [fX, Y](g). The computation

∇_{fX}∇_{Y} Z − ∇_{Y} ∇_{fX}Z − ∇_{[fX,Y]}Z = f∇_{X}∇_{Y} Z − ∇_{Y} (f∇_{X}Z) − f∇_{[X,Y]}Z + Y (f)∇_{X}Z

= f∇_{X}∇_{Y} Z−Y (f)∇_{X}Z−f∇_{Y} ∇_{X}Z−f∇_{[X,Y]}Z+Y (f)∇_{X}Z = f(∇_{X}∇_{Y} Z−∇_{Y} ∇_{X}Z−∇_{[X,Y]}Z)

shows that the function is C^{∞}(M)-linear in the X-variable.

The function is anti-symmetric in the X and Y -variables, so is is also C^{∞}(M)-linear in the Y -variable.

A direct computation as above shows that is also C^{∞}(M)-linear in the Z-variable. □

Equivalently, the Riemann curvature tensor is the (4, 0)-tensor field Rm given by

Rm(X, Y, Z, W) = ⟨R(X, Y)Z, W ⟩

In local coordinates (x^{i}), the cmponents of the curvature tensor

R = R_{ijk}^{ℓ}dx^{i} ⊗ dx^{j} ⊗ dx^{k} ⊗ ∂_{ℓ}, Rm = R_{ijkℓ}dx^{i} ⊗ dx^{j} ⊗ dx^{k} ⊗ dx^{k}

are given by

R(∂_{i}, ∂_{j})∂_{k} = R_{ijk}^{ℓ}∂_{ℓ}

so that

R_{ijkℓ} = ⟨R(∂_{i}, ∂_{j})∂_{k}, ∂_{ℓ}⟩ = ⟨R_{ijk}^{m}∂_{m}, ∂_{ℓ}⟩ = g_{ℓm}R_{ijk}^{m}, R_{ijk}^{m} = g^{ℓm}R_{ijkℓ}

Example 13.3. In euclidean geometry (R^{n}, \overline{g}), ∇_{X}(Z^{k}∂_{k}) = X(Z^{k})∂_{k}, and an easy computation shows that R = 0. On S^{2} with spherical coordinates the curvature tensor is nonzero in that R(∂_{θ}, ∂_{ϕ})∂_{ϕ} = −∂_{θ}.

Lemma 13.4 (Symmetries in the Riemann curvature tensor). Let R and Rm be the curvature tensor of a Riemannian manifold.

(1) R(X, Y)Z = −R(Y, X)Z or Rm(X, Y, Z, W) = Rm(Y, X, Z, W)

(2) Rm(X, Y, Z, W) = −Rm(X, Y, W, Z)
(3) Rm(X, Y, Z, W) = −Rm(Z, W, X, Y)

In a local coordinate frame these identities are equivalent to (1) R_{ijkℓ} = −R_{jikℓ}

(2) R_{ijkℓ} = −R_{ijℓk}
(3) R_{ijkℓ} = −R_{kℓij}

Proof. Each item is a more or less clever calculation.

(1) Clear.

(2) It is enough to show that Rm(X, Y, Z, Z) = 0 as we see by expanding Rm(X, Y, Z + W, Z + W) using multilinearity. Now

XY |Z|^{2} − Y X|Z|^{2} = [X, Y]|Z|^{2}

⟺^{??} 2(⟨∇_{X}∇_{Y} Z, Z⟩ + ⟨∇_{Y} Z, ∇_{X}Z⟩) − 2(⟨∇_{Y} ∇_{X}Z, Z⟩ + ⟨∇_{X}Z, ∇_{Y} Z⟩) = 2 ⟨∇_{[X,Y]}Z, Z⟩

⟺ ⟨∇_{X}∇_{Y} Z, Z⟩ − ⟨∇_{Y} ∇_{X}Z, Z⟩ = ⟨∇_{[X,Y]}Z, Z⟩ ⟺ Rm(X, Y, Z, Z) = 0

□

Proposition 13.5. The Riemann curvature tensor Rm is invariant under local isometries.

Proof. If ϕ: M → M̃ is a (local) isometry then R̃m(ϕ_{∗}X, ϕ_{∗}Y, ϕ_{∗}Z, ϕ_{∗}W) = Rm(X, Y, Z, W) by (??.(3)). □

Example 13.6. The curvature tensor in euclidean geometry is R = 0. The curvature tensor of S^{2} is not zero as, for instance, R(\frac{∂}{∂ϕ}, \frac{∂}{∂θ})\frac{∂}{∂ϕ} = −\frac{∂}{∂θ} so that Rm(\frac{∂}{∂ϕ}, \frac{∂}{∂θ} , \frac{∂}{∂θ}) = g_{11} = −R^{2} sin^{2} ϕ using spherical coordinates as in [?, Exercise 5.7]. Thus S^{2} is not locally isometric to R^{2}.

A Riemannian manifold is flat if any point has a neighborhood isometric to an open subspace of R^{n}.

Theorem 13.7. M is flat ⟺ R = 0.

Proof. We have already establishe one direction. What remains is to show that if R = 0 is a neighborhood of a point p then there are coordinates (y^{i}) near p so that g_{ij} = δ_{ij} in these coordinates.

Choose an orthonormal basis (E_{1}, …, E_{n}) for T_{p}M and corresponding normal coordinates (x^{i}) around p such that ∂_{i}(p) = E_{i} (??). Since R = 0 we have

∇_{∂_{i}}∇_{∂_{j}}Z = ∇_{∂_{j}}∇_{∂_{i}}Z, 1 ≤ i, j ≤ n,

for [∂_{i}, ∂_{j}] = 0.

Consider one of the tangent vectors, E_{j}, at p. Extend E_{j} to a vector field in the coordinate neighborhood by letting E_{j}(x^{1}, …, x^{n}) be the parallel translation of E_{j} along the coordinate curves (0, …, 0) → (x^{1}, 0, …, 0) → (x^{1}, x^{2}, 0, …, 0) → ⋯ → (x^{1}, x^{2}, …, x^{n}). The resulting vector field is actually smooth.

Since parallel translation preserves the inner product we have for instance that

⟨E_{i}(x^{1}, x^{2}, 0, …, 0), E_{j}(x^{1}, x^{2}, 0, …, 0)⟩ = ⟨E_{i}(x^{1}, , 0, …, 0), E_{j}(x^{1}, 0, …, 0)⟩

= ⟨E_{i}(0, …, 0), E_{j}(0, …, 0)⟩ = ⟨E_{i}, E_{j}⟩ = δ_{ij}

In general, we see in this way that (E_{1}, …, E_{n}) is an orthonormal moving frame.

We claim now that the vector field E_{j} is parallel, ie that ∇_{∂_{i}}E_{j} = 0 for all i. At any point (x^{1}, 0, …, 0) on the x^{1}-axis we have that

∇_{∂_{1}}E_{j}(x^{1}, 0, …, 0) = 0

because the vectorfield E_{j} by construction is parallel along the x^{1}-axis. Similarly, at any point (x^{1}, x^{2}, 0, …, 0) in the x^{1}x^{2}-plane we have that

∇_{∂_{2}}E_{j}(x^{1}, x^{2}, …, 0) = 0

because the vector field E_{j} by construction is parallel along any line in the x^{1}x^{2}-plane parallel to the x^{2}-axis. We also have that ∇_{∂_{1}}E_{j} is parallel along any such line because

∇_{∂_{2}}∇_{∂_{1}}E_{j} = ∇_{∂_{1}}∇_{∂_{2}}E_{j} = ∇_{∂_{1}}(0) = 0

at any point in the x^{1}x^{2}-plane. (This is where we use the assumption that R = 0.) Now, parallel translation is a linear isomorphism between tangent spaces, so since the vector field ∇_{∂_{1}}E_{j} vanishes at the point (x^{1}, 0, …, 0) at the start of the curve, ∇_{∂_{1}}E_{j} vanishes too at the point (x^{1}, x^{2}, 0, …, 0) which is the end of the curve. We have now proved that

\begin{matrix}∇_{∂_{1}}E_{j}(x^{1}, 0, …, 0) = 0 \\ ∇_{∂_{1}}E_{j}(x^{1}, x^{2}, 0, …, 0) = 0, ∇_{∂_{2}}E_{j}(x^{1}, x^{2}, 0, …, 0) = 0 \end{matrix}

Continuing this process we conclude after n steps that

∇_{∂_{1}}E_{j}(x^{1}, …, x^{n}) = 0, ∇_{∂_{1}}E_{j}(x^{1}, …, x^{n}) = 0, …, ∇_{∂_{n}}E_{j}(x^{1}, …, x^{n}) = 0,

meaning that E_{j} is parallel.

Since the vector fields E_{j} are all parallel vector fields, symmetry (??) implies that [E_{i}, E_{j}] = 0, that they commute. A theorem of differential geometry says that the vector fields E_{i} are coordinate vector fields for some coordinate system (y^{i}) near p. □

14.04.05 Space-times Hall’s book on relativity [?] is in the library. Hawking and Ellis [?] or O’Neill [?] are also a

possibilities.

Multilinear Algebra

We review some vector space constructions from linear algebra.

14. Tensors

Let V be a finite dimensonal real vector space and V^{∗} = Hom(V, R) the dual space of linear forms on V.

Definition 14.1. The (\begin{matrix}k \\ ℓ \end{matrix})-tensor space of V is the vector space T_{ℓ}^{k}(V) of all R-multilinear

homomorphisms

\underbrace{V^{∗} × ⋯ × V^{∗}}_{ℓ} × \underbrace{V × ⋯ × V}_{k} → R

An element of T_{ℓ}^{k}(V) is called a (\begin{matrix}k \\ ℓ \end{matrix})-tensor.

The tensor product of A ∈ T_{ℓ_{1}}^{k_{1}}(V) and B ∈ T_{ℓ_{2}}^{k_{2}}(V), is the multilinear map A ⊗ B ∈ T_{ℓ_{1}+ℓ_{2}}^{k_{1}+k_{2}}(V) given as the composite

\underbrace{V^{∗} × ⋯ × V^{∗}}_{ℓ_{1}+ℓ_{2}} × \underbrace{V × ⋯ × V}_{k_{1}+k_{2}} = \underbrace{V^{∗} × ⋯ × V^{∗}}_{ℓ_{1}} × \underbrace{V × ⋯ × V}_{k_{1}} × \underbrace{V^{∗} × ⋯ × V^{∗}}_{ℓ_{2}} × \underbrace{V × ⋯ × V}_{k_{2}}

⟶^{A×B} R × R ⟶^{·} R

where · is multiplication in the ring R.

A (\begin{matrix}k \\ ℓ \end{matrix})-tensor A ∈ T_{ℓ}^{k}(V) eats ℓ covectors ω^{1}, …, ω^{ℓ} ∈ V^{∗} and k vectors v_{1}, …, v_{k} ∈ V and

spits out a real number A(ω^{1}, …, ω^{ℓ}, v_{1}, …, v_{k}) ∈ R.

Some special cases of tensor spaces are • T_{0}^{0}(V) = R

• T_{0}^{1}(V) = Hom(V, R) = V^{∗}

• T_{1}^{0}(V) = Hom(V^{∗}, R) = Hom(Hom(V, R), R) ≅ V

• T_{ℓ}^{k}(V) ≅ T_{k}^{ℓ}(V^{∗})

• T_{0}^{k} is a contravariant and T_{ℓ}^{0} a covariant endofunctor of the category of finite dimensional

real vector spaces. The tensor product is a bilinear map

T_{ℓ_{1}}^{k_{1}}(V) × T_{ℓ_{1}}^{k_{1}}(V) ⟶^{⊗} T_{ℓ_{1}+ℓ_{2}}^{k_{1}+k_{2}}(V)
(A, B) → A ⊗ B

and

\begin{matrix}(A ⊗ B)(ω^{1}, …, ω^{ℓ_{1}}, ω^{ℓ_{1}+1}, …, ω^{ℓ_{1}+ℓ_{2}}, v_{1}, …, v_{k_{1}}, v_{k_{1}+1}, …, v_{k_{1}+k_{2}}) \\ = A(ω^{1}, …, ω^{ℓ_{1}}, v_{1}, …, v_{k_{1}})B(ω^{ℓ_{1}+1}, …, ω^{ℓ_{1}+ℓ_{2}}, v_{k_{1}+1}, …, v_{k_{1}+k_{2}}) \end{matrix}

Lemma 14.2. dim T_{ℓ}^{k}(V) = (dim V)^{k+ℓ}

Proof. Let E_{1}, …, E_{n} be a basis for V and ϕ^{1}, …, ϕ^{n} the dual basis for V^{∗} given by ϕ^{i}(E_{j}) = δ_{i}^{j} where n = dim V. Then the set

(14.3)  {E_{j_{1}} ⊗ ⋯ ⊗ E_{j_{ℓ}} ⊗ ϕ^{i_{1}} ⊗ ⋯ ⊗ ϕ^{i_{k}} | 1 ≤ i_{1}, …, i_{k}, j_{1}, …, j_{ℓ} ≤ n}

is a basis for T_{ℓ}^{k}(V): These multilinear maps are clearly linearly independent and since

(14.4) A = ∑ A_{i_{1}···i_{k}}^{j_{1}···j_{ℓ}}E_{j_{1}} ⊗ ⋯ ⊗ E_{j_{ℓ}} ⊗ ϕ^{i_{1}} ⊗ ⋯ ⊗ ϕ^{i_{k}}, A_{i_{1}···i_{k}}^{j_{1}···j_{ℓ}} = A(ϕ^{j_{1}}, …, ϕ^{j_{ℓ}}, E_{i_{1}}, …, E_{i_{k}})

for any multilinear map A ∈ T_{ℓ}^{k}(V), they generate T_{ℓ}^{k}(A) □

Lemma 14.5. (Contraction of tensors) For each k and ℓ there exits a unique R-linear map

tr: T_{ℓ+1}^{k+1}(V) → T_{ℓ}^{k}(V)

such that tr(v ⊗ A ⊗ ϕ) = ϕ(v)A for all v ∈ V = T_{1}^{0}(V), A ∈ T_{ℓ}^{k}(V) and ϕ ∈ V^{∗} = T_{0}^{1}(V).

Proof. The only possibility is the linear map whose values on the basis (??) is

tr(E_{j} ⊗ E_{j_{1}} ⊗ ⋯ ⊗ E_{j_{ℓ}} ⊗ ϕ^{i_{1}} ⊗ ⋯ ⊗ ϕ^{i_{k}} ⊗ ϕ^{i}) = δ_{i}^{j}E_{j_{1}} ⊗ ⋯ ⊗ E_{j_{ℓ}} ⊗ ϕ^{i_{1}} ⊗ ⋯ ⊗ ϕ^{i_{k}}

and this map actually has the property because

tr(v ⊗ A ⊗ ω) = tr(v^{i}E_{i} ⊗ A ⊗ ω_{j}ϕ^{j}) = ω_{j}v^{i}δ_{i}^{j}A = ω(v)A

for all v ∈ V and ω ∈ V^{∗}. □

Lemma 14.6. The relation

ωB(ω^{1}, …, ω^{ℓ}, v_{1}, …, v_{k}) = A(ω, ω^{1}, …, ω^{ℓ}, v_{1}, …, v_{k})

is an isomorphism between the vector space T_{k}^{ℓ+1}(V) of all multilinear homomorphisms

\underbrace{V^{∗} × ⋯ × V^{∗}}_{ℓ+1} × \underbrace{V × ⋯ × V}_{k} ⟶^{A} R

and the vector space of all multilinear homomorphisms

\underbrace{V^{∗} × ⋯ × V^{∗}}_{ℓ} × \underbrace{V × ⋯ × V}_{k} ⟶^{B} V

In particular, T_{1}^{1}(V) ≅ Hom(V, V).

The tensor algebra of V is the graded algebra T^{∗}(V) = ⨁_{k=0}^{∞} T^{k}(V) with tensor product as product. T^{∗} is a contravariant functor.

15. Tensors of inner product spaces

Suppose that V has an inner product g = ⟨ , ⟩ ∈ T^{2}(V). The inner product induces isomorphisms

⟨v, ω^{♯}⟩ = ω(v)  \xymatrix{V \ar[r]^{♭} & V^{∗} \ar[l]^{♯}}  v^{♭}(u) = ⟨u, v⟩

or, more generally,

\xymatrix{\underbrace{V^{∗} × ⋯ × V^{∗}}_{ℓ} ×V × \underbrace{V × V ⋯ × V}_{k} \ar[r]^{♭} & \underbrace{V^{∗} × ⋯ × V^{∗}}_{ℓ} ×V^{∗} × \underbrace{V × V ⋯ × V}_{k} \ar[l]^{♯}}

inverse to each other, which in turn induces isomorphisms

\xymatrix{T_{ℓ}^{k+1}(V) \ar[r]^{♯} & T_{ℓ+1}^{k}(V) \ar[l]^{♭}}

given by

A^{♭}(ϕ^{1}, …, ϕ^{ℓ}, v_{1}, v_{2}, …, v_{k+1}) = A(ϕ^{1}, …, ϕ^{ℓ}, v_{1}^{♭}, v_{2}, …, v_{k+1}),  A ∈ T_{ℓ+1}^{k}(V)
B^{♯}(ϕ^{1}, …, ϕ^{ℓ}, ϕ^{ℓ+1}, v_{1}, …, v_{k}) = B(ϕ^{1}, …, ϕ^{ℓ}, (ϕ^{ℓ+1})^{♯}, v_{1}, …, v_{k}),  B ∈ T_{ℓ}^{k+1}(V)

and again inverse to each other.

Definition 15.1. The trace with respect to g is the linear map

tr_{g}: T_{ℓ}^{k+1}(V) ⟶^{♯} T_{ℓ+1}^{k}(V) ⟶^{tr} T_{ℓ}^{k−1}(V)

where we dualize the first V -variable and take the trace with respect to the new V^{∗} and the next V -variable.

16. Coordinate expressions

Let V be a real vector space with basis E_{1}, …, E_{n} and dual basis ϕ^{1}, …, ϕ^{n} so that the set (??) is a basis for the tensor space T_{ℓ}^{k}(V). For any tensor A ∈ T_{ℓ}^{k}(V), the coordinates of A (??) with respect to this basis,

A_{i_{1}···i_{k}}^{j_{1}···j_{ℓ}} = A(ϕ^{j_{1}}, …, ϕ^{j_{ℓ}}, E_{i_{1}}, …, E_{i_{k}}),

are called the components of A.

16.1. Kronecker’s δ. The components of the (\begin{matrix}1 \\ 1 \end{matrix})-tensor δ ∈ T_{1}^{1}(V) given by δ(ω, v) = ω(v) are

(16.2)  δ_{i}^{j} = δ(ϕ^{j}, E_{i}) = ϕ^{j}(E_{i}) = \begin{cases}1 & i = j \\ 0 & i ≠ j \end{cases}

16.3. Endomorphisms. Let A: V → V be an endomorphism of V. The components of A, viewed (??) as the (\begin{matrix}1 \\ 1 \end{matrix})-tensor A ∈ T_{1}^{1}(V) given by A(ω, v) = ω(Av), are A_{i}^{j} = A(ϕ^{j}, E_{i}) = ϕ^{j}AE_{i}.

16.4. Tensor product. If A ∈ T_{ℓ_{1}}^{k_{1}}(V) and B ∈ T_{ℓ_{2}}^{k_{2}}(V) then

(16.5)  (A ⊗ B)_{i_{1}···i_{k_{1}} i_{k_{1}+1}···i_{k_{1}+k_{2}}}^{j_{1}···j_{ℓ_{1}} j_{ℓ_{1}+1}···j_{ℓ_{1}+ℓ_{2}}} = A_{i_{1}···i_{k_{1}}}^{j_{1}···j_{ℓ_{1}}} B_{i_{k_{1}+1}···i_{k_{1}+k_{2}}}^{j_{ℓ_{1}+1}···j_{ℓ_{1}+ℓ_{2}}}

16.6. Trace. If the tensor A ∈ T_{k}^{ℓ+1}(V) has components A_{i_{1}···i_{k+1}}^{j_{1}···j_{ℓ+1}} meaning that

A = ∑ A_{i_{1}···i_{k+1}}^{j_{1}···j_{ℓ+1}}E_{j_{1}} ⊗ ⋯ ⊗ E_{j_{ℓ+1}} ⊗ ϕ^{i_{1}} ⊗ ⋯ ⊗ ϕ^{i_{k+1}}

then

tr(A) = ∑ A_{i_{1}···i_{k+1}}^{j_{1}···j_{ℓ+1}} tr(E_{j_{1}} ⊗ E_{j_{2}} ⊗ ⋯ ⊗ E_{j_{ℓ+1}} ⊗ ϕ^{i_{1}} ⊗ ⋯ ⊗ ⊗ϕ^{i_{k}} ⊗ ϕ^{i_{k+1}})
= ∑ A_{i_{1}···i_{k+1}}^{j_{1}···j_{ℓ+1}}δ_{i_{k+1}}^{j_{1}}E_{j_{2}} ⊗ ⋯ ⊗ E_{j_{ℓ+1}} ⊗ ϕ^{i_{1}} ⊗ ⋯ ⊗ ϕ^{i_{k}}
= ∑ A_{i_{1}···i_{k}m}^{mj_{2}···j_{ℓ+1}}E_{j_{2}} ⊗ ⋯ ⊗ E_{j_{ℓ+1}} ⊗ ϕ^{i_{1}} ⊗ ⋯ ⊗ ϕ^{i_{k}}

which means that the components of tr(A) ∈ T_{ℓ}^{k}(V) are

(16.7)  tr(A)_{i_{1}···i_{k}}^{j_{2}···j_{ℓ+1}} = A_{i_{1}···i_{k}m}^{mj_{2}···j_{ℓ+1}}

a sum of the components of A. 16.8. Inner product. The inner product on V is a symmetric (2, 0)-tensor g ∈ T_{0}^{2}(V) whose components

(16.9)  g_{ij} = g(E_{i}, E_{j}) = ⟨E_{i}, E_{j}⟩

form a symmetric matrix (g_{ij}). 16.10. Raising and lowering of the index. Since

(v^{j}E_{j})^{♭} = v^{j}E_{j}^{♭} = v^{j}E_{j}^{♭}(E_{i})ϕ^{i} = v^{j} ⟨E_{i}, E_{j}⟩ ϕ^{i} = g_{ij}v^{j}ϕ^{i}

we may write

(v^{j}E_{j})^{♭} = v_{i}ϕ^{i} where v_{i} = g_{ij}v^{j}, (ω_{i}ϕ^{i})^{♯} = ω^{j}E_{j} where ω^{j} = g^{ij}ω_{i}

because the linear map V ⟶^{♭} V^{∗} has matrix (g_{ij}) so that the inverse map V ⟵^{♯} V^{∗} has (g_{ij})^{−1} = (g^{ij}) as its matrix. We get v_{i} from v^{j} by lowering the index and ω^{j} from ω_{i} by raising the index.

More generally, if A ∈ T_{ℓ+1}^{k}(V), then the components of A^{♭} ∈ T_{ℓ}^{k+1}(V) are

(A^{♭})_{ii_{1}···i_{k}}^{j_{1}···j_{ℓ}} = g_{ij}A_{i_{1}···i_{k}}^{j_{1}···j_{ℓ}j}

because

A^{♭}(ϕ^{j_{1}}, …, ϕ^{j_{ℓ}}, E_{i}, E_{i_{1}}, …, E_{i_{k}}) = A(ϕ^{j_{1}}, …, ϕ^{j_{ℓ}}, E_{i}^{♭}, E_{i_{1}}, …, E_{i_{k}})

= g_{ij}A(ϕ^{j_{1}}, …, ϕ^{j_{ℓ}}, ϕ^{j}, E_{i_{1}}, …, E_{i_{k}}) = g_{ij}A_{i_{1}···i_{k}}^{j_{1}···j_{ℓ}j}

and if B ∈ T_{ℓ}^{k+1}(V), then the components of B^{♯} ∈ T_{ℓ+1}^{k}(V) are

(16.12)  (B^{♯})_{i_{1}···i_{k}}^{j_{1}···j_{ℓ}j} = g^{ij}B_{ii_{1}···i_{k}}^{j_{1}···j_{ℓ}}

because

B^{♯}(ϕ^{j_{1}}, …, ϕ^{j_{ℓ}}, ϕ^{j}, E_{i_{1}}, …, E_{i_{k}}) = B(ϕ^{j_{1}}, …, ϕ^{j_{ℓ}}, (ϕ^{j})^{♯}, E_{i_{1}}, …, E_{i_{k}}) = g^{ij}B(ϕ^{j_{1}}, …, ϕ^{j_{ℓ}}, E_{i}, E_{i_{1}}, …, E_{i_{k}}) = g^{ij}B_{ii_{1}···i_{k}}^{j_{1}···j_{ℓ}}

Riemannian GEOMETRY 31

16.13. Trace with respect to g. If B ∈ T_{ℓ}^{k+1}(V) has components B_{i_{1}···i_{k+1}}^{j_{1}···j_{ℓ}} then the components of tr_{g}(B) ∈ T_{ℓ}^{k−1}(V) are

tr_{g}(B)_{i_{3}···i_{k+1}}^{j_{1}···j_{ℓ}} =^{(??)} (B^{♯})_{ji_{3}···i_{k+1}}^{j_{1}···j_{ℓ}j} =^{(??)} g^{ij}B_{iji_{3}···i_{k+1}}^{j_{1}···j_{ℓ}}

if we remember to take the trace at the right places. In particular, tr_{g}(B) = g^{ij}B_{ij} ∈ R for any (2, 0)-tensor B ∈ T_{0}^{2}(V).

Non-euclidean geometry

Historical account of non-euclidean geometry.

Axiom 16.14 (Parallel axiom of euclidean geometry). For any straight line L and any point P outside L there is a unique line through P that does not meet L.

Axiom 16.15 (Parallel axiom of spherical geometry). For any straight line L and any point P outside L there is no line through P that does not meet L.

Axiom 16.16 (Parallel axiom of hyperbolic geometry). For any straight line L and any point P outside L there are at least two lines through P that do not meet L.

17. The hyperbolic plane

The hyperbolic plane of radius R is the Riemannian manifold

H_{R}^{2} = {(ξ, τ) ∈ R^{2} × R_{+} | |ξ|^{2} − τ^{2} = −R^{2}} ⊂ R^{2} × R = R^{3}

with metric induced from the Minkowski metric on R^{3}. Equivalently, the hyperbolic plane of radius R is the disc of radius R

B_{R}^{2}(0) = {u ∈ R^{2} | |u| < R}

equipped with the metric induced from H_{R}^{2} under hyperbolic stereographic projection

π: H_{R}^{2} → B_{R}^{2}(0), π^{−1}(u) = (\frac{2R^{2}u}{R^{2} − |u|^{2}} , R\frac{R^{2} + |u|^{2}}{R^{2} − |u|^{2}} )

which is a diffeomorphism.

Stereograhic projection induces a correspondence between geodesics in the two models since isometries preserve geodesics. Since the geodesics in the hyperboloid model H_{R}^{2} are the curves

{(ξ, τ) ∈ H_{R}^{2} | (ξ, τ) · (α, β) = 0}, R^{3} ∋ (α, β) ≠ 0,

the geodesics in the disc model are the curves

{u ∈ B_{R}^{2}(0) | π^{−1}(u) · (α, β) = 0} = {u ∈ B_{R}^{2}(0) | 2Ru · α + (R^{2} + |u|^{2})β = 0}

parametrized as constant speed curves. If β = 0, this is a straight line through 0. If β ≠ 0, we replace (α, β) by a parallel vector of the form (−α, 1) and get

{u ∈ B_{R}^{2}(0) | |u − Rα|^{2} + |Rα|^{2} = R^{2}} = {u ∈ B_{R}^{2}(0) | |u − Rα|^{2} = R^{2}(1 − |Rα|^{2})}

If |α| ≥ 1, this is the empty set. When |α| < 1, this is part of a circle with center Rα and radius R\sqrt{1 − |α|^{2}}. If we let u_{0} denote either of the two points where the circle meets the boundary of the disc, then we see that

|u_{0} − Rα|^{2} + |Rα|^{2} = |u_{0}|^{2}

which means that u_{0} − Rα ⊥ u_{0} so that the center, Rα, of the circle lies on the tangent to the boundary circle at u_{0}. We conclude that the geodesics in the ball model are straight lines through 0 and circular arcs meeting the boudary circle in right angles.

The geodescis crossing the boundary of the hyperbolic plane of radius 1 at the point (1, 0) are • circular arcs with radius V > 0 and center (1, ±V) on T

• the straight line through (1, 0) perpendicular to T where T is the tangent at (1, 0) to the boundary circle.

Definition 17.1. The Voronoï distance between the point P and the circle c with centre C and radius R is V (c, P) = \sqrt{|CP |^{2} − R^{2}}.

If P is outside the circle, V (c, P) is the length of the tangent segments through P; V (c, P) = 0 when P is on the circle; and V (c, P) is purely imaginary when P is inside the circle.

The centers of the geodesic circular arcs through a point P_{0} in the hyperbolic plane B_{R}^{2}(0) of radius R are the points P for which

|P_{0}P | = V (∂B_{R}^{2}(0), P)

meaning that the euclidean distance to P_{0} equals the Voronoï distance to the boundary circle ∂B_{R}^{2}(0). This locus is the perpendicular bisector of P_{0}P′_{0} where P′_{0} is the point where the radius through P_{0} crosses the boundary circle.

Figure 1. Geodesics in the hyperbolic plane

Figure 2. Two lines through P that do not meet the green line!

If P_{1} and P_{2} are two given points in B_{R}^{2}(0), not on the same diameter, then the center of the geodesic through P_{1} and P_{2} is the point Q for which

|P_{1}Q| = V (∂B_{R}^{2}(0), Q) = |P_{2}Q|

meaning that Q is the intersection of the two bisecting lines associated to P_{1} and P_{2} as just described.

Figure 3. The unique geodesic through two given points