1 Preliminary Technical Remark 2 Geodesics, Exponential Map, Hopt

Math 6397 Riemannian Geometry II The Jacobi Field

1 Preliminary Technical Remark

We start with a preliminary technical remark. Let (M, g) be a m-dimensional manifold with Levi-Civita connection 5. Let H be either I or I × I, where I is an interval. Let f : H → M be a smooth map. We consider the bundle f ∗(TM) over H, and we introduce a connection f ∗5 on f ∗(TM) by putting, ∗ for every X ∈ TxH, Y a section f (TM), ∗ (f 5)X Y := 5f∗(X)Y. In the sequel, instead of (f ∗5), we shall simply write 5, since the map will be clear from the context. A section of f ∗(TM) is called a vector ﬁelds along f. An important role will be played by vector ﬁelds along γ : I → M, i.e. section of γ∗(TM).

2 Geodesics, Exponential map, Hopt-Rinow theorem

Let (M, g) be an m-dimensional Riemannian manifold and 5 the Riemannian connection. A diﬀerentiable curve γ :[a, b] → M on M is geodesic in M 0 if the tangent vector ﬁeld γ (t) = γ∗(d/dt) is parallel along the curve γ, i.e. 0 i 5γ0(t)γ (t) = 0. Suppose that γ is given in a coordinate chart (U, x ) by xi = xi ◦ γ(t), 1 ≤ i ≤ m. Then γ is geodesic if and only if 2 i m j k d x X i dx dx 2 + Γkj = 0 1 ≤ i ≤ m. dt j,k=1 dt dt Since 0 0 2 0 0 0 2 0 0 (d/dt) < γ (t), γ (t) > := γ (t) < γ (t), γ (t) > = 2 < 5γ0(t)γ (t), γ (t) >= 0, the length kγ0(t)k is independent of t. Thus Z t s(t) = kγ0(u)kdu = kγ0(t)k(t − a), a i.e., the arc-length s(t) of γ is a linear function of the parameter t. The parameter t is actually arc-length if and only if kγ0(t)k = 1, in which we say that the geodesic is normalized. 1 According the theorem about existence of the solution for (system) ODEs, given a point P ∈ M and v ∈ TP (M) with kvk < δ (some positive constant), there exists a unique geodesics C(t, P, v):(−, ) → M (or we just 0 write Cv(t)) such that Cv(0) = P , Cv(0) = v. We make the following very important remark: we always have Cλv(t) = Cv(λt), because if we set α(u) = Cv(λu), then

dα dC d2α d2C = λ v , = λ2 v . du dt du2 dt2

Thus, since Cv is a geodesic, α is also a geodesic and satisfies that α(0) = 0 P, α (0) = λv, so by the uniqueness, Cv(λt) = Cλv(t). This proves the remark. Note that Cv(λt) is defined on (−/λ, /λ). The remark permits us to make the interval of definition of a geodesic uniformly large in a neighborhood of P ∈ M, say on (−2, 2). So we can define the exponential map expp : TpM → M (note that expp may not be defined on the whole space of TpM) by v 7→ γ(1; p, v), v ∈ TpM, where γv(t) := γ(t; q, v) (or we just write γ(t)) defined on (−2, 2) is the 0 unique geodesic that satisfies γ(0) = p, γ (0) = v. Geometrically, expp(v) is a point of M obtained by going out the length equal to kvk, starting from p, along the geodesic γ which passes through p with velocity equal to v/kvk(note that γv/kvk(kvk) = γv(1)).

Let us calculate (exp)∗0. Let I0 : TpM → T0(TpM) be the canonical d isomorphisim, i.e. I0(v) = dt (tv)|t=0. For, v ∈ B(0, ) ⊂ TpM, recall that γv(t) = γtv(1), then d (exp) (v) := (exp) (I (v)) = exp (tv)| ∗0 ∗0 0 dt p t=0 d = γ (1)| dt tv t=0 d = γ (t)| dt v t=0 = v.

So (exp)∗0 is the identity map from Tp(M) to Tp(M). In particular, it is non-singular at 0. By the inverse function theorem, it implies that expp is a

2 local diﬀeomorphism in a neighborhood of 0 on TpM, i.e., there exists > 0 such that expP : B(0p, ) ⊂ TP (M) → M is a diﬀeomorphism onto its image.

Recall that TM is the set of pairs (p, v) where p ∈ M and v ∈ TpM. If {xi} are the local coordinates around p, then v = yi∂/∂xi. Thus, we can take (x1, . . . , xm, y1, . . . , ym) as coordinates of (p, v) in TM. In what follows we identify the tangent space T (TMp)v to TMp at v ∈ TpM with TpM itself by identifying ∂/∂yi with ∂/∂xi.

Lemma (Gauss). Let p ∈ M and v ∈ TpM such that exppv is deﬁned. Let ∼ w ∈ TpM = T (TpM)v. Then

< (expp)∗v(v), (expp)∗v(w) >=< v, w > .

The proof uses (at least implicitly) the theory of Jacobi ﬁeld, so it will be proved in the next section.

Hopf-Rinow Theorem. Let (M, g) be a complete Riemannian manifold. Then for every p ∈ M, the exponential map expp is deﬁned everywhere in TpM.

Proof. If not, i.e. there is some point p ∈ M and unit-vector v ∈ TpM such that the geodesic γ(t) = expp(tv) is not deﬁned at some t0 > 0. WLOG, we assume that expp(tv) is deﬁned on [0, t0). Take a increasing ti ∈ [0, t0) such that ti → t0. Then, for every > 0, there exists N such that for i ≥ j > N, 0 ≤ ti − tj < . Since the geodesic γ has arc-length as its parameter,

d(γ(ti), γ(tj)) ≤ L(γ|[ti,tj ]) = ti − tj < .

So {γ(tj)} is a Cauchy sequence. By the complete assumption, it conver- gences to q ∈ M. So limt→t0 γ(t) = q. Deﬁne γ(t0) = q, then the deﬁnition of γ can be extended across t0, which contradicts the choice of t0.

Corollary. At every two points p, q in a complete Riemannian manifold, there exists a geodesic which connects p and q.

3 3 Jacobi ﬁeld

Let γ :[a, b] → M be a geodesic, and J = J(t) be a tangent vector ﬁeld along γ. Denote by

0 5J J (t) = 5 ∂ J = 5γ0 J = , ∂t dt

2 00 5 J J (t) = 5 ∂ 5 ∂ J = 5γ0 5γ0 J = . ∂t ∂t dt2

Definition. Let γ :[a, b] → M be a geodesic on an m-dimensional Riemma- nian manifold (M, g). A vector field J = J(t) along γ is called Jacobi field if it satisfies the following Jacobi equation:

J 00(t) + R(J(t), γ0(t))γ0(t) = 0.

It is obvious that if J is a Jacobi ﬁeld along γ, then J ⊥ := J− < J, T > T is also a Jacobi ﬁeld along γ where T = γ0.

0 Using the compatibility with the metric and the fact that 5 ∂ γ (t) = 0, ∂t we compute d2 < J, γ0 > = < J 00(t), γ0 > dt2 = − < R(J, γ0)γ0, γ0 > = −R(J, γ0, γ0, γ0) = 0, by the symmetries of the curvature tensor. Thus, the Jacobi ﬁeld J has the following important property: d2 < J, γ0 >≡ 0 dt2 so < J(t), γ0(t) > is a linear function of t, i.e. < J(t), γ0(t) >= λt + µ for some constants λ and µ.

Next, we want to ﬁnd a local expression the the Jacobi equation. Let γ be a geodesic on M. We express the Jacobi ﬁeld equation in terms of the

4 local coordinate. Choose an orthonormal basis {ei|p} for TpM, and extend it to a parallel orthonormal frame along all γ. Then

0 ei(t) = 5 ∂ ei(t) = 0. ∂t Let J(t) be a Jacobi ﬁeld along γ. Write

m X i J(t) = J (t)ei(t). i=1 Then m m 0 X i0 00 X i00 J (t) = J (t)ei(t),J (t) = J (t)ei(t). i=1 i=1 0 Pm 0k Write γ (t) = k=1 γ ek, then Thus

00 0 0 X i00 X i j 0k 0l J (t) + R(J(t), γ (t))γ (t) = {J (t) + Rljk(γ(t))J γ γ }ei(t). i j Hence the Jacobi equation is reduced to

i00 X i j 0k 0l J (t) = Rljk(γ(t))J γ γ , 1 ≤ im. j

Theorem 3.1. Let γ :[a, b] → M be a geodesic on an m-dimensional Riemmanian manifold (M, g). Then for every v, w ∈ Tγ(a)M, there exists a unique Jacobi ﬁeld J(t) along γ satisfying

J(a) = v, J 0(a) = w.

There are always two trivial Jacobi fields along any geodesic γ, which can 0 0 0 0 be written down immediately: Because 5 ∂ γ (t) = 0 and R(γ , γ )γ = 0 by ∂t 0 antisymmetry of R, the vector field J0(t) = γ (t) satisfies the Jacobi equation 0 0 0 with J0(0) = γ (0),J0(0) = 0. Similarly, J1(t) = tγ (t) is a Jacobi field 0 0 satisfying J1(0) = 0,J1(0) = γ (0). It is easy to see that J0 is the variation field of the variation F (t, u) = γ(t + u), and J1 is the variation field of the variation F (t, u) = γ(teu). Therefore, these two Jacobi fields just reflect the possible reparametrizations of γ, and don’t tell us anything about the behavior of geodesics other than γ itself.

5 To distinguish these two trivial cases from more informative ones, we make the following deﬁnitions.

Deﬁnition If J is orthogonal to γ everywhere, then J is called a normal Jacobi ﬁeld.

Theorem 3.2. Let γ : I → M be a geodesic on an m-dimensional Riemma- nian manifold (M, g), and let a ∈ I.

(a) A Jacobi ﬁeld J along γ is normal if and only if

J(a) ⊥ γ0(a),J 0(a) ⊥ γ0(a).

(b) Any Jacobi ﬁeld orthogonal to γ0 at two points is normal.

Proof. As we showed earlier, f(t) :=< J(t), γ0(t) > is a linear function of t. Note that f(a) =< J(a), γ0(a) > and f 0(a) =< J 0(a), γ0(a) >. Thus J(a) and J 0(a) are orthogonal to γ0(a) if and only if f and its ﬁrst order derivative vanish at a, which happens if and only if f ≡ 0. Similarly, if J is orthogonal to γ0 at two points, then f vanishes at two points, so f ≡ 0. This proves the theorem.

Next we show that the Jacobi ﬁelds arise naturally from the variation of the geodesic γ through neighborhouring geodesics. Let now γ :[a, b] → M be a geodesic. A variation of γ is a diﬀerential map F (t, u):[a, b] × (−, ) → M with F (t, 0) = γ(t) for all t ∈ [a, b]. F is said to be a geodesic variation if for every u ∈ (−, ), the curve γu := F (·, u) is geodesic.

A typical geodesic variation can be constructed as follows: we assume that a = 0 and let p = γ(0). For every v, w ∈ TpM, take a smooth curve σ = σ(u), u ∈ (−, ) such that

σ(0) = p, σ0(0) = v.

Also, take a parallel translation of w and γ0(0) along the curve σ to get two vector ﬁelds W (u) and T (u), u ∈ (−, ) which are parallel along σ. Let X(u) = T (u) + uW (u) (note: X(0) = T (0) = γ0(0),X0(u) = W (u) since

6 X(u),W (u) are parallel). We then “recover” the curve γu from X(u) by taking the exponential map γu := expσ(u)(tX(u)), i.e. for (t, u) ∈ [0, b] × (−, ), let F (t, u) = expσ(u) t(T (u) + uW (u)), then, for every u ∈ (−, ), the curve γu := F (·, u) is geodesic, and γ0 = γ, so F is a geodesic variation of the geodesic γ.

Let ∂F (t, u) ∂ ! T˜(t, u) = : F | , ∂t ∗ ∂t (t,u) ∂F (t, u) ∂ ! U˜(t, u) = : F | . ∂u ∗ ∂u (t,u) ˜ ˜ Then T is the tangent vector to the curve γu := F (·, u) and U is the tangent vector to the section γt := F (t, ·), and its restriction U(t) = U˜(t, 0) is called a variation vector ﬁeld of the variation F .

Theorem 3.3 Let γ :[a, b] → M be a geodesic on an m-dimensional Riem- manian manifold (M, g). Then the variation vector ﬁeld U of its geodesic variation is a Jacobi ﬁeld along γ.

Proof. Let F be the geodesic variation of γ. We make the following claim: for every vector ﬁeld V along γ,

5 ∂ 5 ∂ V − 5 ∂ 5 ∂ V = R(U,˜ T˜)V. ∂u ∂t ∂t ∂u This is a local issue, so we can compute it in any local coordinates. Writing ∂ V (t, u) = V i(t, u) , ∂xi then i ∂V (t, u) ∂ i ∂ 5 ∂ V = + V 5 ∂ . ∂t ∂t ∂xi ∂t ∂xi Therefore,

2 i i i ∂ V (t, u) ∂ ∂V (t, u) ∂ ∂V (t, u) ∂ i ∂ 5 ∂ 5 ∂ V = + 5 ∂ + 5 ∂ +V 5 ∂ 5 ∂ . ∂u ∂t ∂u∂t ∂xi ∂t ∂u ∂xi ∂u ∂t ∂xi ∂u ∂t ∂xi

7 Interchanging 5 ∂ and 5 ∂ and substracting, we see that all terms, except ∂t ∂u the last canceled:

i ∂ ∂ 5 ∂ 5 ∂ V − 5 ∂ 5 ∂ V = V (5 ∂ 5 ∂ − 5 ∂ 5 ∂ ). ∂u ∂t ∂t ∂u ∂u ∂t ∂xi ∂t ∂u ∂xi Now we need to compute the commutator in parentheses. If we write the coordinate functions of F as xj(t, u), then

∂xk ∂ ∂xk ∂ T˜ = ; U˜ = . ∂t ∂xk ∂u ∂xk Note that ∂ ∂ ∂xk ∂ ∂ ∂ 5 i = 5T˜ i = 5 i . ∂t ∂x ∂x ∂t ∂xk ∂x Hence,

∂ ∂xk ∂ ! ∂ ∂ ∂ ∂ 5 5 i = 5 5 i ∂u ∂t ∂x ∂u ∂t ∂xk ∂x ∂2xk ∂ ∂xk ∂ ! ∂ ∂ = 5 i + 5U˜ 5 i ∂u∂t ∂xk ∂x ∂t ∂xk ∂x ∂2xk ∂ ∂xk ∂xj ∂ ∂ ∂ ∂ = 5 i + 5 5 i . ∂u∂t ∂xk ∂x ∂t ∂u ∂xj ∂xk ∂x Interchanging t with u and j with k, and substituting, we ﬁnd that the ﬁrst terms canceled out, and we get

∂ ∂ ∂xk ∂xj ∂ ∂ ! ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ 5 5 i − 5 5 i = 5 5 i − 5 5 i ∂u ∂t ∂x ∂t ∂u ∂x ∂t ∂u ∂xj ∂xk ∂x ∂xk ∂xj ∂x ∂xk ∂xj ∂ ∂ ! ∂ = R , ∂t ∂u ∂xj ∂xk ∂xi ∂ = R(U,˜ T˜) . ∂xi Hence

5 ∂ 5 ∂ V − 5 ∂ 5 ∂ V = R(U,˜ T˜)V. ∂u ∂t ∂t ∂u

8 This proves the claim. Since F is a geodesic variation, 5 ∂ T˜ = 0, so 5 ∂ 5 ∂ ∂t ∂s ∂t T˜ = 0. By the claim,

0 = = 5 ∂ 5 ∂ T˜ ∂u ∂t

= 5 ∂ 5 ∂ T˜ + R(U,˜ T˜)T˜ ∂t ∂u

= 5 ∂ 5 ∂ U˜ + R(U,˜ T˜)T,˜ ∂t ∂t where the last step follows from the fact that 5 is torsion free, so 5 ∂ T˜ = ∂u 5 ∂ U˜. Letting u = 0, then ∂t

0 5 ∂ 5 ∂ U = R(U, γ )γ. ∂t ∂t So U is a Jacobi ﬁeld along γ. This ﬁnishes the proof.

One of the important result about Jacobi ﬁeld is the following, which is the inverse statement of Theorem 3.3:

Theorem 3.4 Let γ :[a, b] → M be a geodesic on a complete Riemmanian manifold (M, g). Let J = J(t) be a Jacobi ﬁeld along γ. Then J must be a variation vector ﬁeld of some geodesic variation of γ.

Proof. We repeat the construction of F (t, u) again, as we already discussed above. WLOG, we assume that a = 0 and let p = γ(0). For every v = 0 J(0), w = J (0) ∈ TpM, take a smooth curve σ = σ(u), u ∈ (−, ) such that

σ(0) = p, σ0(0) = v.

X(u),W (u) are parallel). We then “recover” the curve γu from X(u) by taking the exponential map γu := expσ(u)(tX(u)), i.e. for (t, u) ∈ [0, b] × (−, ), let F (t, u) = expσ(u) t(T (u) + uW (u)), then, for every u ∈ (−, ), the curve γu := F (·, u) is geodesic, and γ0 = γ, so F is a geodesic variation of the geodesic γ. The tangent vectors to variation

9 curve and the section curve are ∂ ! T˜(t, u) = F | , ∗ ∂t (t,u)

∂ ! U˜(t, u) = F | . ∗ ∂u (t,u) From the discussion in the beginning of the section, we have ˜ U = U|u=0 is a Jacobi ﬁelds along γ, and

∂ ! ∂ U(0) = F | = | (F (0, u) = σ0(0) = v, ∗ ∂u (0,0) ∂u u=0 and 0 ˜ ˜ ˜ U (0) = 5 ∂ U|(0,0) = 5 ∂ T |(0,0) = 5σ0(0)T |(0,u). ∂t ∂u But ∂ ! T˜(0, u) = F | = (exp ) (T (u) + uW (u)) = T (u) + uW (u). ∗ ∂t (0,u) σ(u) ∗0

Hence 0 U (0) = 5σ0(0)(T (u) + uW (u)) = W (0) = w, where, we used the property that T (u) and W (u) are parallel along σ. From above, we see that U = U(t) is a Jacobi ﬁeld along γ satisfying U(0) = v = J(0),U 0(0) = w = J 0(0). By the uniqueness result, we have

J(t) = U(t).

Hence J(t) is a variation vector ﬁeld of the geodesic variation F of γ. This ﬁnishes the proof.

The proof above actually gives an explicit expression of the geodesic variation F , as well as the Jacobi ﬁeld J. If J(0) = 0, then we can take σ(u) as the constant curve (point) σ(u) = p = γ(0). So the geodesic variation of 0 0 γ becomes F (t, u) = expγ(0) t(γ (0) + uJ (0)). Hence we get the following

10 very important result which shows how Jacobi ﬁelds can be used to study the derivative of the exponential map!.

Corollary 3.2. Let γ : [0, b] → M be a geodesic on a complete Riemmanian manifold (M, g). Let J = J(t) be a Jacobi ﬁeld along γ with J(0) = 0. Then Hence 0 J(t) = t(expγ(0))∗,tγ0(0)(J (0)), (1) i.e. the derivative of the exponential map expp : TpM → M evaluated at the 0 0 point tγ (0) ∈ TpM and applied to tJ (0). Proof. As in the proof of above theorem,

0 0 F (t, u) = expγ(0) t(γ (0) + uJ (0)) is the geodesic variation of γ, and the variation vector ﬁeld ∂ ! U(t) = F | ∗ ∂u (t,0) is a Jacobian ﬁeld with U(0) = 0,U 0(0) = J 0(0), so by the uniqueness theorem, U = J. We know calculate U as follows: consider the map 0 0 φ : [0, b]×(−, ) → Tγ(0)M by φ(t, u) = t(γ (0)+uJ (0)), so F = expγ(0) ◦φ, then by the chan rule ∂ ! J(t) = U(t) = F | ∗ ∂u (t,0) ∂ ! = (exp ) ◦ φ (t, u) | γ(0) ∗,φ(t,0) ∗ ∂u (t,0) 0 0 = (expγ(0))∗,tγ0(0)(tJ (0)) = t(expγ(0))∗,tγ0(0)(J (0)). This proves the corollary.

As application of the derivative, we prove the Gauss’s lemma proved above.

Proof of the Gauss’ lemma. We now prove the Gauss’ lemma stated earlier. ∼ By the result above, for v ∈ Tp(M), w ∈ TpM = Tv(TpM), consider the 0 geodesic γ(t) := expp tv, then γ(0) = p and γ (0) = v. Let F (t, u) = exp(t(v + uw)).

11 Then J(t) = (expγ(0))∗tv(tw) (which is the variation vector ﬁeld of F ) is a 0 Jacobi ﬁeld along the curve γ(t) := expp tv, and J (0) = w. Hence

< v, w >=< γ0(0),J 0(0) > .

Using the property for the Jacobi ﬁelds we proved earlier,

d2 < J(t), γ0(t) >= 0, dt2 so < J(t), γ0(t) > is a linear function, say < J(t), γ0(t) >= at + b, so, < J 0(t), γ0(t) >= a which is a constant (using γ is a geodesic) and b =< J(0), γ0(0) >= 0. Therefore,

< v, w > = < γ0(0),J 0(0) >=< J 0(1), γ0(1) >= a =< J(1), γ0(1) >

= (expγ(0))∗v(w), (expp)∗,vv > . This proves Gauss’ lemma.

As another application, we prove

Corollary 3.3 Let p ∈ M and v ∈ TpM be contained in the domain of deﬁnition of expp, and c(t) = expptv. Let the piecewise sommth curve γ : [0, 1] → TpM be likewise contained in the domain of deﬁnition of expp, and assume that γ(0) = 0, γ(1) = v. Then

kvk = L(expptv|t∈[0,1]) ≤ L(expp ◦γ), and the equality holds if and only if γ is diﬀerent from the curve tv, t ∈ [0, 1] only by a reparameterization.

Proof. We shall show that any piecewise sommth curve γ : [0, 1] → TpM with γ(0) = 0 satisﬁes

L(expp ◦γ) ≥ kγ(1)k.

Write γ(t) = r(t)φ(t), r(t) ∈ R and φ(t) ∈ TpM, with kφ(t)k ≡ 1 (polar coordinates in TpM). We have

γ0(t) = r0(t)φ(t) + r(t)φ0(t)

12 and < φ, φ0 >≡ 0. Thus, by Corollary 3.2, also

0 < exp∗,γ(t)φ(t), exp∗,γ(t)φ (t) >= 0, kexp∗,γ(t)φ(t)k = kφ(t)k = 1, and it follows that 0 0 kexp∗,γ(t)γ (t)k ≥ |r (t)|. Hence Z 1 Z 1 0 0 L(expp ◦γ) = kexp∗,γ(t)γ (t)kdt ≥ |r (t)|dt ≥ r(1) − r(0) = kγ(1)k. 0 0 This proves the statement.

Remark: Corollary 4.3.3 by no means implies that the geodesic c(t) = exp +ptv is the shortest connection between its end points. It only is shorter than any other curve that is the exponential image of a curve with the same initial and end points as the ray tv, 0 ≤ t ≤ 1.

4 Jacobi Fields of the Manifolds with Con- stant Sectional Curvature

Let (M, g) be a Riemannian manifold with constant sectional curvature C. From HW7, Problem 3 in last sememster, we know that R(Z, W, X, Y ) =< R(X,Y )Z, W >= −c(< X, Z >< Y, W > − < X, W >< Y, Z >), or equivalently, R(X,Y )Z = −c(< X, Z > Y − < Y, Z > X). Let γ be a geodesic with unit-speed in M. Substitute this into the Jacobi equation, we find a normal Jacobi field J satisfies

0 = J 00 − C(< J, γ0 > γ0− < γ0, γ0 > J) = J 00 + CJ.

Choosing a parallel normal vector ﬁeld E along γ and setting J(t) = u(t)E(t), and assuming that J(0) = 0, then the Jacobi equation becomes that

u00(t) + cu(t) = 0 has fundamental√ solutions with u(0) = 0 as follows: if c > 0, then√ u(t) = √1 √1 c sin ct, if c = 0, then u(t) = t, if c < 0, then u(t) = −c sinh −ct. So we proved the following statement: Let (M, g) be a Riemannian manifold

13 with constant sectional curvature C. Let γ be a geodesic with unit-speed in M. Then the normal Jacobi field along along γ with J(0) = 0 are precisely the vector fields J(t) = u(t)E(t), where E is any parallel normal√ vector field along γ, and u(t) is as follows: if c > 0, then u(t) = √1 sin ct, if c = 0, √ c √1 then u(t) = t, if c < 0, then u(t) = −c sinh −ct.

5 Conjugate points

Let (M, g) be a complete m-dimensional Riemannian manifold. According to the Hopf-Rinow theorem, for every p ∈ M, the exponential map is deﬁned on TpM everywhere. Because, (expp)∗0 the derivative of expp at the zero vector 0, is the identity map on Tp(M) = T0(TpM), so it is non-singular in a neighborhood of 0. But, in general, sometimes, it may be (expp)∗ may be singular at some points other than zero. Let us look at the case of unit-sphere in Rn+1.

Let n+1 M = Sn = {x ∈ Rn+1 | X(xα)2 = 1}. α=1 Then (M, g) is a space with constant curvature (the sectional curvature is 1). We know that for every p ∈ Sn, all geodesic curves starting from p converge to q = −p, and the length of these geodesics are all equal to π. Hence, on n−1 S (π) ⊂ Tp(M), the image under the map expp is the single point set {q}. In particular, for every smooth curve σ(t) on Sn−1(π),

expp(σ(t)) = q, hence 0 (expp)∗σ(t)(σ (t)) = 0. n−1 This means that the map expp is degenerate on S (π) everywhere. Motivated from above example, we introduce the following deﬁnition.

Deﬁnition. Let (M, g) be a complete m-dimensional Riemannian manifold.

Let p ∈ M, v ∈ TpM. If the exponential map expp is degenerate at v, i.e. there exists w ∈ TpM = Tv(TpM), such that

(expp)∗v(w) = 0,

14 then we call q = expp(v) ∈ M is the conjugate point of p along the curve γ(t) = expp(tv).

Hence, by the inverse function theorem,

Theorem. Suppose p ∈ M, v ∈ TpM, and q = exppv. Then expp is local diﬀeomorphism in a neighborhood of v if and only if q is not conjugate to p along the geodesic γ(t) = expp(tv), t ∈ [0, 1].

The following theorem uses Jacobi ﬁeld to describe the conjugate points.

Theorem. Let γ : [0, b] → M be a geodesic on a complete Riemannian manifold (M, g). p = γ(0), q = γ(b). Then q is a conjugate point of p along the curve γ if and only if there exists a nonzero Jacobi ﬁeld J = J(t), satisfying J(0) = J(b) = 0.

Proof. First of all, for every t ∈ [0, b], we have

0 γ(t) = expp(tγ (0)).

Let J = J(t) be a nonzero Jacobi ﬁeld along γ with J(0) = J(b) = 0. By Theorem 3.2, J 0(0) 6= 0. From the corollary 3.2 above,

0 J(t) = t(expγ(0))∗tγ0(0)(J (0)). Because J(b) = 0, we have

0 (expp)∗bγ0(0)(J (0)) = 0.

0 From the deﬁnition, q = γ(b) = expp(bγ (0) is a conjugate point of p along the curve γ.

Conversely, let q be a conjugate point of p along the curve γ. Then there exists a non-zero vector w ∈ TpM such that

(expp)∗bγ0(0)(w) = 0.

15 For every t ∈ [0, b], let

J(t) = t(expp)∗tγ0(0)(w), then J = J(t) is a Jacobi ﬁeld along the curve γ with J(0) = J(b) = 0. This ﬁnishes the proof.

Corollary. Let p, q ∈ M. If q = γ(b) is the conjugate point of p = γ(0) along the geodesic γ(t), t ∈ [0, b], then p is the conjugate point of q along the geodesic γ˜(t) = γ(b − t).

Theorem. Let γ : [0, b] → M be a geodesic on a complete Riemannian manifold (M, g). p = γ(0), q = γ(b). If q is not a conjugate point of p along the curve γ, then for every v ∈ TpM, w ∈ TqM, there exists a unique Jacobi ﬁeld with J(0) = v, J(b) = w.

Proof. Recall that from Corollary 3.2, we know that

0 J(t) = t(expγ(0))∗tγ0(0)(J (0)) is a Jacobi ﬁeld along γ with J(0) = 0. So to construct Jacobi ﬁeld with J(0) = v, J(b) = w, we construct J1 with J1(0) = 0,J1(b) = w and J2(0) = v, J2(b) = 0 respectively. To do so, we need pass the condition “J1(b) = w to 0 J1(b) =w ˜”.

0 0 From the assumption, since γ(t) = expp(tγ (0)) so q = γ(b) = expp(bγ (0)), the map (expp)∗bγ0(0) is non-singular, so it is a linear isomorphism. So there existsw ˜ ∈ Tbγ0(0)(TpM), such that

w = (expp)∗bγ0(0)(w ˜).

For every t ∈ [0, b], let t J1(t) = (expp)∗tγ0(0)(w ˜), b

Then J1 is Jacobi ﬁeld along the curve γ with

J1(0) = 0,J1(b) = (expp)∗bγ0(0)(w ˜) = w.

16 On the other hand, since p is not a conjugate point of q, there exists a Jacobi ﬁeld J2 along the curve γ with J2(0) = v, J2(b) = 0. Let

J(t) = J1(t) + J2(t).

Then J is Jacobi ﬁeld along the curve γ with J(0) = v, J(b) = w.

We now prove the uniqueness. If J˜ 6= J. Then J¯ = J˜ − J is a Jacobi ﬁeld along the curve γ with J¯(0) = J¯(b) = 0. This contradicts with the assumption that q is a conjugate point of p. This ﬁnishes the proof.

6 The ﬁrst and second variation formulas

Let γ :[a, b] → M be a diﬀerentiable curve with kγ0(t)k =const. The lenght of γ is Z b l = L(γ) = kγ0(t)kdt = |γ0(0)|(b − a). a A variation of γ is a diﬀerential map F (t, u):[a, b] × (−, ) → M with

F (t, 0) = γ(t) for all t ∈ [a, b]. Write γu := F (·, u), and its arc-length is

Z b 0 L(u) = L(γu) = kγu(t)kdt. a Let ∂ ! T˜(t, u) = F | , ∗ ∂t (t,u) ∂ ! U˜(t, u) = F | ,U = U˜| . ∗ ∂u (t,u) u=0 Then U = U(t) is a variation vector ﬁeld of the variation of F . Then the ﬁrst variation formula is

Z b < 5 ∂ U,˜ T˜ > 0 d ∂t L (u) = L(γu) = dt. du a < T,˜ T˜ > Hence Z b 0 d −1 L (0) = L(γu)|u=0 = −l < 5 ∂ U, T > dt. du a ∂t

17 In particular, it shows that a geodesic is the critical point of the arc-length functional. The geodesic is normal (namely, the parameter t is its arc-length) if |γ0(t)| = 1, i.e. l = b − a.

Using the the ﬁrst variation formula, we can show that every minimizing curve is geodeisc when it it given by a unit-speed parametrization.

The the second variation formula is

d2 L00(0) = L(γ )| du2 u u=0 Z b ( ) b − a ∂ ˜ 0 0 0 0 0 = < 5 ∂ U|u=0, γ > + < R(U, γ )U, γ > + dt l a ∂t ∂u (b − a)3 Z b − 2 dt, l3 a

0 ⊥ where U = 5 ∂ U = 5γ0 U. In order to simplify the above. Let U be the ∂t normal component of U, which is orthogonal to γ0, i.e.

(b − a)2 U ⊥ = U − < U, γ0 > γ0. l2 Then 2 0 (b − a) U 0 = U ⊥ + γ0. l2 So

2 0 0 2(b − a) 0 = + l2 (b − a)4 + 2< γ0, γ0 > l4 2 0 0 (b − a) 0 = + 2, l2 where we used

⊥0 0 ∂ ⊥ 0 ⊥ 0 = − = 0. ∂t

18 Hence, we have that the second variation formula is b − a L00(0) = < 5 U, γ0 > |b l U a b − a Z b n 0 o + |U ⊥ |2+ < R(γ0,U ⊥)γ0,U ⊥ > dt. l a ˜ where 5U U = 5 ∂ U|u=0. In particular, if F (a, u) = γ(a),F (b, u) = γ(b) ∂u (i.e. the variation has the ﬁxed end points), then we have

b − a Z b n 0 o L00(0) = |U ⊥ |2+ < R(γ0,U ⊥)γ0,U ⊥ > dt. l a Assume that F is a normal variation of γ with ﬁxed end points, i.e. U is normal to γ everywhere. Then

b − a Z b n o L00(0) = |U 0|2+ < R(γ0,U)γ0, U > dt. l a Using d = − , dt and U(0) = U(1) = 0, then

b − a Z b L00(0) = − dt. l a Note that this actually related to the Jacobi ﬁeld.

It should come as no surprise that the second variation depends only on the normal component of U; intuitively, the tangential components of U contributes only to a reparametrization of γ, and the length is independent of parametrization. For this reason, we general apply the second variation formula only to variations whose variation ﬁelds are proper (ﬁx-end points variation) and normal.

We deﬁne a symmetric bilinear form I, called the index form, on the space of proper normal vector ﬁelds along γ by

Z b I(V,W ) = {< V 0,W 0 > + < R(γ0,V )γ0, W >}dt. a

19 You should think of I(V,W ) as sort of “Hessian” or second order derivative of the length functional. Then, for every proper variation of γ,

L00(0) = I(U ⊥,U ⊥).

Because every proper normal vector ﬁeld along γ is the variation ﬁeld of some proper variation, it can be rephrased on terms of the index form in the following way

Corollary. If F is a proper variation of a geodesic γ whose variation field is a proper normal variation field V , the second variation of L(γ) is I(V,V ), i.e.L00(0) = I(V,V ). In particular, if γ is minimizing, then I(V,V ) ≥ 0 for any proper normal vector field along γ.

The next Proposition gives another expression for I, which makes the role of the Jacobi equation more evident.

Proposition 5.1. For any pair of proper normal vector ﬁelds V,W along a geodesic segment γ,

Z b k 00 0 0 X 0 I(V,W ) = − < V + R(V, γ )γ , W > dt − < 4iV ,W (ai) >, a i=1

0 where {ai} are the points where V is not smooth, and 4iV is the jump on 0 V at t = ai.

Proof. On any interval [ai−1, ai] where V and W are smooth, d < V 0, W >=< V 00, W > + < V 0,W 0 > . dt Thus, by the fundamental theorem of calculus,

Z ai Z ai < V 0,W 0 > dt = − < V 00, W > + < V 0, W > |ai . ai−1 ai−1 ai−1

Summing over i, and noting that W is continuous at t = ai, and W (a) = W (b) = 0, we get the identity. This ﬁnishes the proof.

20 7 Geodesics do not minimize past conjugate points

In this section, we use the second variation to prove another extremely important fact about conjugate points: No geodesic is minimizing past its ﬁrst conjugate point.

Theorem. If γ is a geodesic segment from p to q that has an interior conjugate point to p, then there exists a proper normal vector ﬁeld X along γ such that I(X,X) < 0. In particular, γ is not minimizing.

Proof. Suppose γ : [0, b] → M is a unit-speed parametrization of γ and γ(a) is the first conjugate to γ(0) for some 0 < a < b. This means that there is a non-trivial normal Jacobi field J along γ|[0,a] that vanishes at t = 0 and t = a. Define a vector field V along γ by V (t) = J(t) for t ∈ [0, a], V (t) = 0 for t ∈ [a, b]. and X(t) = 0 for t ∈ [a + , b]. This is a proper, normal, piecewise smooth vector field along γ.

Let W (t) be a smooth proper normal vector field along γ such that W (b) is equal to the jump 4V 0 at t = a. Such a vector field is easily constructed in a local coordinates and extended to all γ by a bump function. Note that 4V 0 = −J 0(a) is not zero, because otherwise J would be a Jacobi field satisfying J(a) = J 0(a) = 0, and thus would be identically zero.

For small positive , let X = V + W . Then

2 I(X,X) = I(V,V ) + 2I(V,W ) + I(W, W ). Since V satisﬁes the Jacobi equation on each subinterval [0, a] and [a, b], and V (a) = 0, by the Proposition above,

Z a Z b I(V,V ) = − < V 00 + R(V, γ0)γ0, V > dt − < V 00 + R(V, γ0)γ0, V > dt 0 a − < 4V 0,V (a) >= − < 4V 0,V (a) > . Similarly, I(V,W ) = − < 4V 0,W (b) >= −|W (b)|2. Thus 2 2 I(X,X) = −2|W (b)| + I(W, W ).

21 If we choose small enough, this is strictly negative. This proves the theorem.

In contrast to the above theorem, we prove

Theorem. Let γ :[a, b] → M be geodesic. It there does not exist a point conjugate to γ(a) along γ, then there exists > 0 with the property that for any piecewise smooth curve g :[a, b] → M with g(a) = γ(a), g(b) = γ(b), d(g(t), γ(t)) < for all t ∈ [a, b], we have that L(g) ≥ L(c) with equality if and only if g is a reparametrization of c.

Proof. We want to apply Corollary 3.2.3. WLOG, we assume that a = 0, b = 1 and v := γ(0). Since there are no points conjugate to γ(a), the exponential map expp is of maximal rank along any radical curve tv, 0 ≤ t ≤ 1. Thus, by inverse function theorem, for each t, expp is diffeomorphism in a suitable neighborhood of tv. We cover {tv, 0 ≤ t ≤ 1} by finitely many such neighborhoods {Ωi, 1 ≤ i ≤ k} and let Ui := exppΩi. Let us assume that tv ∈ Ωi for ti−1 ≤ t ≤ ti. If > 0 is sufficiently small, we have for any curve g : [0, 1] → M satisfying the assumption d(g(t), γ(t)) < , g([ti−1, ti]) ⊂ Ui. −1 Let c(t) = (expp|Ωi ) (g(t)) for ti−1 ≤ t ≤ ti, then c : [0, 1] → TpM with expp ◦ c = g, c(0) = 0, c(1) = v. Thus Corollary 3.2.3 implies our statement.

8 Curvature and topology

By means of the second variation of geodesics we can prove the following

Theorem (Myers and Bonnet). Let (M, g) be a complete Riemannian manifold. If

(i) (Myers) r(M) ≥ (n − 1)k for some constant k > 0, or

√(ii) (Bonnet) KM ≥ k > 0, then the diameter of M satisfying d(M) ≤ π/ k, so that M is actually compact.

Proof. Suppose the contrary. There there are points p, q ∈ M and (Hopf- Rinow theorem) a minimal unit-speed curve γ connecting p and q with length

22 √ l > π/ k. Fix a normalized minimal geodesic γ : [0, l] → M (with unit- speed) and let {ei} be an orthonormal basis of parallel ﬁelds along γ such 0 that e1 = γ = T . Let Wi(t) = sin(πt/l)ei(t) be vector ﬁelds along γ. Then 0 π 00 π2 Wi (t) = l cos(πt/l)ei(t),Wi (t) = − l2 sin(πt/l)ei(t). Hence Z l 2 I(Wi,Wi) = − < 5T Wi − R(T,Wi)T,Wi > dt 0 Z l 2 2 π = (sin(πt/l)) { + R(T, ei)T, ei >}dt 0 l2 from which it follows that n Z l 2 X 2 π I(Wi,Wi) = (sin(πt/l)) {(n − 1) 2 − r(T,T )}dt. i=2 0 l √ So if r(M) ≥ (n − 1)k > 0 and l > π/ k, then the sum therefore at least one summand must be negative. But this is impossible since some I(Wi,Wi) < 0 implies that the second variation of arclength is in some direction negative, so that γ cannot be minimal. This proves the theorem.

Remark: (a) In case (ii) (i.e. the Bonnet’s condition), we can just take W (t) = sin(πt/l)E(t), where E(t) is any parallel normal unit vector ﬁeld along γ.

(b) The sphere Sn(r) := {x ∈ Rn+1 | |x| = r} of radius r has (sectional) 1 n−1 curvature r2 , hence Ricci curvature r2 and diameter πr. The theorem above means that if If M has Ricci curvature not less than the one of Sn(r), then the diameter of M is at most one of Sn(r).

Theorem (Cartan-Hadama). Let (M, g) be a complete Riemnannian manifold with nonpositive sectional curvature. Then M has no conjugate points.

Proof. Let γ : [0, l] → M be a normalized geodesic, and X be a vector ﬁeld, normal to γ, then Z l 2 2 I(X,X) = {| 5T X| − |X| KM (X ∧ T )}ds ≥ 0. 0 Then by the corollary above, M has no conjugate points. This proves the theorem.

23 9 Some Comparison Theorems

We’ll prove, in this section, that an upper bound on sectional curvature produces a lower bound on Jacobi fields. In particular, we recall that we three model space Sn, Rn and Hn of curvature 1, 0, -1. Let γ(t) be a geodesic 0 n n n with |γ (t)| = 1, w ∈ Tγ(0)M where M ∈ {S , R ,H }. Then the Jacobi field J(t) along γ with J(0) = 0,J 0(0) = w is given by (sin t)W, tW, (sinh t)W respectively, where W is the parallel vector field along γ with W (0) = w. So it is natural to compare with these spaces.

Our starting point is the following very classical comparison theorem for ordinary diﬀerential equations.

Theorem (Sturm Comparison Theorem). Suppose u and v are diﬀeren- tiable real-valued functions on [0, t], twice diﬀerentiable on (0,T ), and u > 0 on (0,T ). Suppose further that u and v satisfy u”(t)+a(a)u(t) = 0, v”(t)+a(t)v(t) ≥ 0, u(0) = v(0) = 0, u0(0) = v0(0) > 0 for some function a : [0, t] → R. Then v(t) ≥ u(t) on [0,T ].

Proof. Consider the function f(t) = v(t)/u(t) defined on (0,T ). It follows 0 0 from l’Hopital’s rule that limt→0 f(t) = v (0)/u (0) = 1. Since f is differ- entiable on (0,T ), if we could show that f 0 ≥ 0 there it would follow from elementary calculus that f ≥ 1 and therefore v ≥ u on (0,T ), and by conti- nuity also on [0,T ]. Differentiating d v v0u − vu0 = . dt u u2 Thus to show that f 0 ≥ 0 it would suffice to show that v0u − vu0 ≥ 0. Since v0(0)u(0) − v(0)u0(0) = 0, we need only show this expression has no nonnegative derivative. Differentiating again and substituting the ODE for u, d (v0u − vu0) = v00u + v0u − v0u0 − vu00 = v00u + avu ≥ 0. dt This proves the theorem.

Theorem (Jacobi Field Comparison Theorem). Suppose that (M, g) is a Riemannian manifold with all sectional curvature bounded above by a

24 constant C. If γ is a unit speed geodesic in M, and J is any normal Jacobi ﬁeld along γ such that J(0) = 0, then

(1) If C = 0, |J(t)| ≥ t|J 0(0)| for t ≥ 0;

1 t 0 (2) If C = R2 , |J(t)| ≥ R(sin R )|J (0)| for 0 ≤ t ≤ πR;

1 t 0 If C = − R2 , |J(t)| ≥ R(sinh R )|J (0)| for t ≥ 0; Proof. The function |J(t)| is smooth wherever J(t) 6= 0. Using the Jacobi equation, we compute

d2 d < J 0, J > |J| = dt2 dt < J, J >1/2 < J 00, J > < J 0,J 0 > < J 0, J > = + − < J, J >1/2 < J, J >1/2 < J, J >3/2 < R(J, γ0)γ0, J > |J 0|2 | < J 0, J >2 = + − . |J| |J| |J|3

Using the Schwarz inequality, < J 0, J >2≤ |J 0|2|J|2, so the sum of the last two terms above is nonnegative. Thus

d2 R(J, γ0, γ0,J) |J| ≥ − . dt2 |J|

Since < J, γ0 >= 0 and |γ0| = 1, R(J, γ0, γ0,J)/|J|2 is the sectional curvature of the plane spanned by J and γ0. Therefore our assumption on the sectional curvature of M guarantees that R(J, γ0, γ0,J) ≤ C|J|2, so

d2 |J| ≥ −C|J| dt2 wherever |J| > 0. We wish to use the Sturm comparison theorem to compare |J| with the 00 t solution u to u + Cu = 0, where u(t) = t if C = 0, u(t) = R sin R if 2 t 2 C = 1/R , and u(t) = R sinh R if C = −1/R . To do so, we need to arrange d|J|/dt = 1 at t = 0, because u0(0) = 1. Multiplying J by a positive constant, we may assume, WLOG, that |J 0(0)| = 1. Also J can be written near t = 0

25 as J(t) = tW (t) where W is a smooth vector ﬁeld (the proof is omitted). Therefore d |J(t)| − |J(0)| |J||t=0 = lim dt t→0 t t|W (t) = lim = |W (0)| = |J 0(0)| = 1. t→0 t Now the Sturm comparison theorem implies that |J| ≥ u, provided |J| is nonzero (to ensure that it is smooth). The fact that d|J|/dt = 1 at t = 0 means that |J| > 0 on some interval (0, ), and |J| cannot attain its ﬁrst zero before u does without contradicting the estimate |J| ≥ u. Thus |J| ≥ u as long as u ≥ 0, which proves the theorem.