LECTURE 21:CUT LOCUS1.THE CONJUGATE AND CUT LOCUSI The conjugate locus.Let (M,g)be a complete Riemannian manifold.Definition 1.l. For any p E M, we callCon(p) = [u E T,M I (d expp), is singular)the conjugate locus of p of p in T,M, and callCon(p) = (q E M I q is a conjugate point of palong some geodesic)the conjugate locus of p in M.Obviously Con(p)is a closed subset in TM.We first proveLemma 1.2. The conjugate locus Con(p) and Con(p) are measure zero sets in T,Mand M respectively.Proof. By Morse index theorem, for each u E S,M, the set [tu I t > 0] nCon(p) is口discrete. So the conclusion follows.As we have seen last time, thefirst conjugate point in each direction is important:a geodesic starting from p is locally length minimizing if the endpoint is before thefirst conjugate point, and is not locally length minimizing if the endpoint is afterthe first conjugate point. We define k : SpM -→ R U {+oo] to be the functionk(u) = inf(t > 0 | tu E Con(p)),and we set k(u) = +oo if there is no conjugate point in the direction v.Proposition1.3.The functionk is continuous.Proof. Suppose vi E S,M and vi -→ v. We want to prove k(ui) -→ k(u)We need an observation:Continuity Observation: Suppose i E SpM and vi → u. Fix land denotei=exp(tvi) and (t)=exp(tu) (0≤t≤l),By the smooth dependence of geodesics on initial values, con-verges uniformly to By using parallel transport [first from (t)
LECTURE 21: CUT LOCUS 1. The conjugate and cut Locus ¶ The conjugate locus. Let (M, g) be a complete Riemannian manifold. Definition 1.1. For any p ∈ M, we call Con( g p) = {v ∈ TpM | (d expp )v is singular} the conjugate locus of p of p in TpM, and call Con(p) = {q ∈ M | q is a conjugate point of palong some geodesic} the conjugate locus of p in M. Obviously Con( g p) is a closed subset in TxM. We first prove Lemma 1.2. The conjugate locus Con( g p) and Con(p) are measure zero sets in TpM and M respectively. Proof. By Morse index theorem, for each v ∈ SpM, the set {tv | t > 0} ∩ Con( g p) is discrete. So the conclusion follows. □ As we have seen last time, the first conjugate point in each direction is important: a geodesic γ starting from p is locally length minimizing if the endpoint is before the first conjugate point, and is not locally length minimizing if the endpoint is after the first conjugate point. We define k : SpM → R ∪ {+∞} to be the function k(v) = inf{t > 0 | tv ∈ Con( g p)}, and we set k(v) = +∞ if there is no conjugate point in the direction v. Proposition 1.3. The function k is continuous. Proof. Suppose vi ∈ SpM and vi → v. We want to prove k(vi) → k(v). We need an observation: Continuity Observation: Suppose vi ∈ SpM and vi → v. Fix l and denote γi = exp(tvi) and γ(t) = exp(tv) (0 ≤ t ≤ l), By the smooth dependence of geodesics on initial values, γi converges uniformly to γ. By using parallel transport [first from γi(t) 1
2LECTURE21:CUTLOCUSto %(0) = p along %, then from p = (0) to (t) along one can iden-tify each T%(t)M with T(t)M. As a result, we get an identification,:vo→voFurthermore, for any X Evo, again by smoothdependence, 1((X), (X)) →I(X,X).Case 1: k(u) = c< +oo We want to prove k(vi) → k(u) = c.. Take I = c+ e. Then ind() ≥ 1, i.e. I(X, X) < 0 for some X E vo. Bycontinuity observation above, I(-'(X), -'(X) < 0 for i large enough.So ind(%) ≥ 1 and thus k(v) ≤ l + e for i large enough..Take l = c-e.Sincevo is infinite dimensional, we can't conclude that I% ispositive definite directly from the fact I is positive definite.However, fromthe proof of Morse index theorem, the maximal negative definite space of 1?can be taken to be a subspace inT = [X e vo [ X is Jacobi on each [ti, ti+i]]or any other direct sum complement ofT2 = [V Vo I X(t)) =...= X(th) = 0)where 0 = to < ti <... < tk < tk+i = l are chosen so that ([ti,ti+il) C Uj.and Uo,Ui,...,Uk are strongly convex open subsets that cover . Since converges to uniformly,we have ([ti,ti+il)CU,forilarge enough.Wemay use the same partition for all . Now we have ;(T2")=T, and thusVO. =亚-1(V0) = 亚-1(T 由 T) = ()-1(T)田Tso ind(%) also equals the maximal dimension of subspace in (,)-1(T) onwhich1%is negativedefinite.Now suppose1%(X,X)<0for,(X)Twhichcanbetaken sothatsupX=1,thenwemaytake a convergentsubsequence of ;(X,) T and conclude the existence of X 0 withI(X,X)≤ 0, which contradicts with the fact I is positive definite. So weget k(vi) ≥l -e for i large enough.Case 2: k(u) = +oo Suppose to the contrary that k(vi) has a bounded subsequence.Without less of generality, suppose k(vs)≤ c for all i.Take l = c.By the same口argument above we get a contradiction.As a consequence we getCorollary 1.4. The set of first conjugate points of p in T,M is closed.Proof.If tivi are first conjugate points of p and tiui-u,thenk(u/lol) = lim k(os) = lim t, = [0l-口So is the first conjugate point of p in the direction u/lul
2 LECTURE 21: CUT LOCUS to γi(0) = p along γi , then from p = γ(0) to γ(t) along γ] one can identify each Tγi(t)M with Tγ(t)M. As a result, we get an identification Ψi : V 0 γi → V0 γ . Furthermore, for any X ∈ V0 γ , again by smooth dependence, I γi (Ψ−1 i (X), Ψ −1 i (X)) → I γ (X, X). Case 1: k(v) = c < +∞ We want to prove k(vi) → k(v) = c. • Take l = c + ε. Then ind(γ) ≥ 1, i.e. I γ (X, X) < 0 for some X ∈ V0 γ . By continuity observation above, I γi (Ψ−1 i (X), Ψ −1 i (X)) < 0 for i large enough. So ind(γi) ≥ 1 and thus k(vi) ≤ l + ε for i large enough. • Take l = c−ε. Since V 0 γ is infinite dimensional, we can’t conclude that I γi is positive definite directly from the fact I γ is positive definite. However, from the proof of Morse index theorem, the maximal negative definite space of I γ can be taken to be a subspace in T γ 1 = {X ∈ V0 γ | X is Jacobi on each [ti , ti+1]}, or any other direct sum complement of T γ 2 = {V ∈ V0 γ | X(t1) = · · · = X(tk) = 0}, where 0 = t0 < t1 < · · · < tk < tk+1 = l are chosen so that γ([tj , tj+1]) ⊂ Uj , and U0, U1, · · · , Uk are strongly convex open subsets that cover γ. Since γi converges to γ uniformly, we have γi([tj , tj+1]) ⊂ Uj for i large enough. We may use the same partition for all γi . Now we have Ψi(T γi 2 ) = T γ 2 , and thus V 0 γi = Ψ−1 i (V 0 γ ) = Ψ−1 (T γ 1 ⊕ T γ 2 ) = (Ψi) −1 (T γ 1 ) ⊕ T γi 2 , so ind(γi) also equals the maximal dimension of subspace in (Ψi) −1 (T γ 1 ) on which I γi is negative definite. Now suppose I γi (Xi , Xi) < 0 for Ψi(Xi) ∈ T γ 1 which can be taken so that sup |Xi | = 1, then we may take a convergent subsequence of Ψi(Xi) ∈ T γ 1 and conclude the existence of X ̸= 0 with I γ (X, X) ≤ 0, which contradicts with the fact I γ is positive definite. So we get k(vi) ≥ l − ε for i large enough. Case 2: k(v) = +∞ Suppose to the contrary that k(vi) has a bounded subsequence. Without less of generality, suppose k(vi) ≤ c for all i. Take l = c. By the same argument above we get a contradiction. □ As a consequence we get Corollary 1.4. The set of first conjugate points of p in TpM is closed. Proof. If tivi are first conjugate points of p and tivi → v, then k(v/|v|) = limi→∞ k(vi) = limi→∞ ti = |v|. So v is the first conjugate point of p in the direction v/|v|. □
3LECTURE21:CUTLOCUSTThe cut locus.Let be the normal geodesic in (M, g) with (O) = p and (O) = w. Suppose(M, g) is complete so that can be defined on R. Let's concentrate for t > O, whichcorresponds to the part of the geodesic in the direction .For t small lio,t is lengthminimizing between (O) and (t).For general t, it may happen that either lo.ty islength minimizing between (O) and (t) for all t > 0, or there exists to such thatlo,t is no longer length minimizing between (O) and (t) for all t > to.Definition 1.5.Let (M,g)be a completeRiemannian manifold, p M a point,and : [0,oo) → M a normal geodesic with (0) = p. Ifto := sup(t / ([0,t)) is a minimizing geodesic] < +oo,then we will call (to) the cut point of p along .. The cut locus of p in M is defined to be the set Cut(p) of all cut points of palong all geodesics that start from p. The cut locus of p in T,M is defined to be the set Cut(p) of all vectors E T,M so that expp(u) is a cut point.Remark. If M is compact, then Cut(p)+ for all p.Erample.On Rm and Hm (endowed with the canonical metrics), there exists onlyone normal minimizing geodesic joining any two given points.So Cut(p)= forall p.Erample.For Sm with the roundmetric, Cut(p)=(p) for any p E M, wherep =-pis the antipodal point of p. Note that p is also the first conjugate point of p.Erample. For the cylinder S' × IR endowed with the canonical metric, if p = (eioo, zo),then Cut(p) = [(ei(8o+n), z) I z e R) is the vertical line “opposite to p". Note thatp has no conjugate points at all.By definition we haveLemma 1.6. For any g Cut(p), there erists a unique minimizing geodesic joiningp to q.Proof. If there exist two minimizing geodesics , of length I joining p to q, then is minimizing between p and q, and is no longer minimizing after q:The curve defined by connecting with lz,+e) is a piecewisesmooth but not smooth curve connecting p to (l +e) whose lengthis l +e.But according to the first variation formula, any piecewisesmooth but not smooth curve is not a minimizing curve[c.f. Corollary2.7 in Lecture 16]. We conclude that l(o,+) is also not a minimizingcurve, since it has the same length as .口So q E Cut(p)
LECTURE 21: CUT LOCUS 3 ¶ The cut locus. Let γ be the normal geodesic in (M, g) with γ(0) = p and ˙γ(0) = v. Suppose (M, g) is complete so that γ can be defined on R. Let’s concentrate for t > 0, which corresponds to the part of the geodesic in the direction v. For t small γ|[0,t] is length minimizing between γ(0) and γ(t). For general t, it may happen that either γ|[0,t] is length minimizing between γ(0) and γ(t) for all t > 0, or there exists t0 such that γ|[0,t] is no longer length minimizing between γ(0) and γ(t) for all t > t0. Definition 1.5. Let (M, g) be a complete Riemannian manifold, p ∈ M a point, and γ : [0,∞) → M a normal geodesic with γ(0) = p. If t0 := sup{t | γ([0, t]) is a minimizing geodesic} < +∞, then we will call γ(t0) the cut point of p along γ. • The cut locus of p in M is defined to be the set Cut(p) of all cut points of p along all geodesics that start from p • The cut locus of p in TpM is defined to be the set Cut( ^p) of all vectors v ∈ TpM so that expp (v) is a cut point. Remark. If M is compact, then Cut(p) ̸= ∅ for all p. Example. On R m and Hm (endowed with the canonical metrics), there exists only one normal minimizing geodesic joining any two given points. So Cut(p) = ∅ for all p. Example. For S m with the round metric, Cut(p) = {p¯} for any p ∈ M, where ¯p = −p is the antipodal point of p. Note that ¯p is also the first conjugate point of p. Example. For the cylinder S 1×R endowed with the canonical metric, if p = (e iθ0 , z0), then Cut(p) = {(e i(θ0+π) , z) | z ∈ R} is the vertical line “opposite to p”. Note that p has no conjugate points at all. By definition we have Lemma 1.6. For any q ̸∈ Cut(p), there exists a unique minimizing geodesic joining p to q. Proof. If there exist two minimizing geodesics γ, σ of length l joining p to q, then γ is minimizing between p and q, and is no longer minimizing after q: The curve ¯γ defined by connecting σ with γ|[l,l+ε] is a piecewise smooth but not smooth curve connecting p to γ(l + ε) whose length is l + ε. But according to the first variation formula, any piecewise smooth but not smooth curve is not a minimizing curve[c.f. Corollary 2.7 in Lecture 16]. We conclude that γ|[0,l+ε] is also not a minimizing curve, since it has the same length as ¯γ. So q ∈ Cut(p). □
4LECTURE21:CUTLOCUSI Cut points v.s. first conjugate points.The following theorem relates cut points with first conjugate points:Theorem 1.7. Suppose (to) is the cut point of p = (0) along a normal geodesic, then at least one of the following assertion holds:(1)(to)is the first conjugatepoint of p along.(2) (to) is the first point along so that there erists another normal geodesic + from p to (to) with length L(o) = to = L(l(o,tol).Proof. Take a decreasing sequence ti -→ tt. Let oi be a normal minimizing geodesicconnecting p to (t:). Then by definition of cut point, si := L(oi) < t,. Note that[o;(O)) is a sequence in the unit sphere SpM. By passing to a subsequence, wemayassumeo;(O)→X,E SpM.Let be the normal geodesic with o(O) =p,o(O) =Xp. Then by continuity, is a minimizing geodesic connecting p to (to),thus L(o) = to.Case1:X,=(0).SinceS<t,we haveso;(0)+t(0).Butexpp(s;o;(0)) =;(si) =(t)=expp(t,(0)),so exp, is not a local diffeomorphism near to(0). So (to) is a conjugate point ofp. Obviously it is the first conjugate point, otherwise ([o,to]) is not minimizingCase 2: X + (O). Then g is a geodesic that is different from .We haveto = L(llo,tol) ≤ L(o) = lim L(oi) ≤ lim t; = to.So L(o)=to.To show that(to)is thefirst point alongwith this property, weargue by contradiction.If there exists at<to and a normal geodesic connectingp to () so that L(a) = t, then by the argument in the proof of Lemma 1.6, lo.talis not a minimizing curve. This contradicts with the definition of cut point.Corollary 1.8. If q E Cut(p), then p E Cut(q).Proof.If g is the cut point of p along , then is minimizing between p and q.It follows that the “opposite geodesic"- is also minimizing between q and p.Moreover, by the theorem above, either q is the first conjugate point of p along ,or there exists a different normal geodesic joint p to q which has length L(o) =口dist(p,g). In both cases -is no longer minimizing after p. So p E Cut(g).Remark. One can show that the function f : SM → R U [oo] defined byJ to,if p,x,(to) is the cut point of p along ,f(p,Xp) =+oo, if p has no cut point along p,Xp.is a continuous function. It follows that Cut(p) is a closed subset in M. It hasmeasure zero since there is at most one cut point in each directionin M. Note thatby definition f ≤k
4 LECTURE 21: CUT LOCUS ¶ Cut points v.s. first conjugate points. The following theorem relates cut points with first conjugate points: Theorem 1.7. Suppose γ(t0) is the cut point of p = γ(0) along a normal geodesic γ, then at least one of the following assertion holds: (1) γ(t0) is the first conjugate point of p along γ. (2) γ(t0) is the first point along γ so that there exists another normal geodesic σ ̸= γ from p to γ(t0) with length L(σ) = t0 = L(γ|[0,t0]). Proof. Take a decreasing sequence ti → t + 0 . Let σi be a normal minimizing geodesic connecting p to γ(ti). Then by definition of cut point, si := L(σi) < ti . Note that {σ˙ i(0)} is a sequence in the unit sphere SpM. By passing to a subsequence, we may assume ˙σi(0) → Xp ∈ SpM. Let σ be the normal geodesic with σ(0) = p, σ˙(0) = Xp. Then by continuity, σ is a minimizing geodesic connecting p to γ(t0), thus L(σ) = t0. Case 1: Xp = ˙γ(0). Since si < ti , we have siσ˙ i(0) ̸= tiγ˙(0). But expp (siσ˙ i(0)) = σi(si) = γ(ti) = expp (tiγ˙(0)), so expp is not a local diffeomorphism near t0γ˙(0). So γ(t0) is a conjugate point of p. Obviously it is the first conjugate point, otherwise γ([0, t0]) is not minimizing. Case 2: X ̸= ˙γ(0). Then σ is a geodesic that is different from γ. We have t0 = L(γ|[0,t0]) ≤ L(σ) = limi L(σi) ≤ lim i ti = t0. So L(σ) = t0. To show that γ(t0) is the first point along γ with this property, we argue by contradiction. If there exists a t < t ¯ 0 and a normal geodesic ¯σ connecting p to γ(t¯) so that L(¯σ) = t¯, then by the argument in the proof of Lemma 1.6, γ|[0,t0] is not a minimizing curve. This contradicts with the definition of cut point. □ Corollary 1.8. If q ∈ Cut(p), then p ∈ Cut(q). Proof. If q is the cut point of p along γ, then γ is minimizing between p and q. It follows that the “opposite geodesic” −γ is also minimizing between q and p. Moreover, by the theorem above, either q is the first conjugate point of p along γ, or there exists a different normal geodesic σ joint p to q which has length L(σ) = dist(p, q). In both cases −γ is no longer minimizing after p. So p ∈ Cut(q). □ Remark. One can show that the function f : SM → R ∪ {∞} defined by f(p, Xp) = t0, if γp,Xp (t0) is the cut point of p along γ, +∞, if p has no cut point along γp,Xp . is a continuous function. It follows that Cut(p) is a closed subset in M. It has measure zero since there is at most one cut point in each directionin M. Note that by definition f ≤ k
LECTURE 21: CUT LOCUS52.THEDISTANCEFUNCTIONI Smoothness of Distance Function.Nowlet'sfixpEM and considerthedistancefunctiondp : M → R, dp(q) = dist(p,q)As we have already seen, d, is a continuous function. However, it is not hard to seethat d & C(M). In fact, d, is never smooth at p.Erample. Consider (S?, gs). Let p = -p be the antipodal point of p. Then for qnear p, d,(qg) = π - dp(q). It follows that d, is also not smooth at p.Theorem2.1.Thefunctiond,issmooth onM/Cut(p)U(p).Moreover,for eachqEM/Cut(p)U(p),if we let betheunique normal minimizinggeodesic frompto q,thenthegradient of dpat qis(Vdp)(q) =9(dp(q))Proof. For each q E M \Cut(p) U (p), let be the unique normal minimizinggeodesic from p to q and denote X = 9(0) e S,M. LetA= (L()X9 /qE M \Cut(p) U (p))Then A c T,M \ (0) is an open set andexp, : A-→ M / Cut(p) U (p)is smooth. Moreover, at each vector in A, expp is non-singular and thus a localdiffeomorphism. Since exp, is globally one-to-one on A, it is a diffeomorphism fromA to M /Cut(p)U (p).It follows thatexpp' : M \Cut(p) U (p) → A C T,M\[0]]is smooth. Thus d,(q) = [exp,'(q)| is smooth on M / Cut(p) U (p)To calculate its gradient at q, we choose any X, e T,M and let o(s) be a smoothcurve in M /Cut(p)U (p) tangent to Xg at q = o(O). Now we consider the variationof so that be the unique minimizing geodesic from p to o(s). Observe that thevariation field vector [which is a Jacobi fieldl of this variation at the point g is exactlyX.. So according to the first variation formula,( -(0 (0 -((),ds口It follows that (Vdp)(q) =9(d,(q).Remarks. One can show that if there exists two minimizing geodesic from p to q,then d, is not differentiable at q
LECTURE 21: CUT LOCUS 5 2. The distance function ¶ Smoothness of Distance Function. Now let’s fix p ∈ M and consider the distance function dp : M → R, dp(q) = dist(p, q). As we have already seen, dp is a continuous function. However, it is not hard to see that dp ̸∈ C ∞(M). In fact, dp is never smooth at p. Example. Consider (S 2 , gS2 ). Let ¯p = −p be the antipodal point of p. Then for q near ¯p, dp(q) = π − dp¯(q). It follows that dp is also not smooth at ¯p. Theorem 2.1. The function dp is smooth on M \ Cut(p)∪ {p}. Moreover, for each q ∈ M \ Cut(p) ∪ {p}, if we let γ q be the unique normal minimizing geodesic from p to q, then the gradient of dp at q is (∇dp)(q) = ˙γ q (dp(q)). Proof. For each q ∈ M \ Cut(p) ∪ {p}, let γ q be the unique normal minimizing geodesic from p to q and denote Xq = ˙γ q (0) ∈ SpM. Let A = {L(γ q )X q | q ∈ M \ Cut(p) ∪ {p}}. Then A ⊂ TpM \ {0} is an open set and expp : A → M \ Cut(p) ∪ {p} is smooth. Moreover, at each vector in A, expp is non-singular and thus a local diffeomorphism. Since expp is globally one-to-one on A, it is a diffeomorphism from A to M \ Cut(p) ∪ {p}. It follows that exp−1 p : M \ Cut(p) ∪ {p} → A ⊂ TpM \ {0}] is smooth. Thus dp(q) = | exp−1 p (q)| is smooth on M \ Cut(p) ∪ {p}. To calculate its gradient at q, we choose any Xq ∈ TqM and let σ(s) be a smooth curve in M \ Cut(p)∪ {p} tangent to Xq at q = σ(0). Now we consider the variation of γ q so that γ q s be the unique minimizing geodesic from p to σ(s). Observe that the variation field vector [which is a Jacobi field] of this variation at the point q is exactly Xq. So according to the first variation formula, Xq(dp) = d ds s=0 dp(σ(s)) = d ds s=0 L(γ q s ) = ⟨Xq, γ˙ q (dp(q))⟩. It follows that (∇dp)(q) = ˙γ q (dp(q)). □ Remarks. One can show that if there exists two minimizing geodesic from p to q, then dp is not differentiable at q
6LECTURE21:CUTLOCUSTHessian ofthe distance function.By using the second variation formula one can calculate the Hessian of dp onM Cut(p) U (p). Recall that the Hessian of a smooth function f is(V2f)q(Xq, Ya) = (XqYf - Vx,Y)f = Vx((Vf,Y)) - (Vf, Vx,Y)= (Vx, Vf,Y).Now Let s : [0, i] -→ M be geodesic variation of by minimizing geodesics withs(O) = p [so its variation field X is a normal Jacobi field along with X(0) = 0]. Then(V2dp)g(Xg, Ya) = 《(VxVdp)g, Ya) = (Vx(g), Ya) = (Vg()X, Ya), VYq E T,MSoweprovedProposition 2.2. Suppose q Cut(p)U(p]. Let : [0, 1] → M be the unique lengthminimizing normal geodesic connecting p to g, and let X be a normal Jacobi fieldalong with X(O) =0. Denote X,=X(t). Then for any Yg ET,M,(V2dp)g(Xg, Yg) = (Vs()X, Yg)Here are two special cases that will be quite useful later:Corollary 2.3. Suppose q Cut(p) U (p], and : [0, 1] -→ M the unique lengthminimizing normal geodesic connecting p to q.(1) For any Yq E T,M, (V2dp)q((U), Ya) = (V()%, Ya) = 0.(2) For any normal Jacobi field X along with X(O) = 0,(V2dp)g(Xq, Xq) = (V(g)X,Xa) = I(X, X).Note that the “singularity" of dp at the point p is not too bad: one can alwaysremove the singularity at pby considering the function d, instead.So it is reasonableto study (V?dp)p. In general, for any smooth function f e C(M) and Xp TpM,if we let be the geodesic (t) = exp,(tX,), then "the second order derivative of falong " isd2ddd(Vf, ) = (V,Vf, ) = (V2 f)(0)(%, ),dizf o(t) =dtdi(fo)=dtSo we getLemma 2.4. (f 0)"(t) = (V2f)(t)(%, ).On the other hand, if X, is a normal vector, i.e. is a normal geodesic, thenfor t small enough we have dp((t) = t2. So we get, for X, E SpM,(V2d)p(Xp, Xp) = 2g(Xp, Xp)and thus by polarization, we getProposition2.5.TheHessianofdatp is(V2dg)p = 2g
6 LECTURE 21: CUT LOCUS ¶ Hessian of the distance function. By using the second variation formula one can calculate the Hessian of dp on M \ Cut(p) ∪ {p}. Recall that the Hessian of a smooth function f is (∇2 f)q(Xq, Yq) = (XqY f − ∇Xq Y )f = ∇Xq (⟨∇f, Y ⟩) − ⟨∇f, ∇Xq Y ⟩ = ⟨∇Xq∇f, Y ⟩. Now Let γs : [0, l] → M be geodesic variation of γ by minimizing geodesics with γs(0) = p [so its variation field X is a ✿✿✿✿✿✿ normal✿✿✿✿✿✿ Jacobi✿✿✿✿✿ field along γ with X(0) = 0]. Then (∇2 dp)q(Xq, Yq) = ⟨(∇X∇dp)q, Yq⟩ = ⟨∇X(q)γ˙ q , Yq⟩ = ⟨∇γ˙ q(l)X, Yq⟩, ∀Yq ∈ TqM. So we proved Proposition 2.2. Suppose q ̸∈ Cut(p)∪ {p}. Let γ : [0, l] → M be the unique length minimizing normal geodesic connecting p to q, and let X be a normal Jacobi field along γ with X(0) = 0. Denote Xq = X(l). Then for any Yq ∈ TqM, (∇2 dp)q(Xq, Yq) = ⟨∇γ˙ (l)X, Yq⟩. Here are two special cases that will be quite useful later: Corollary 2.3. Suppose q ̸∈ Cut(p) ∪ {p}, and γ : [0, l] → M the unique length minimizing normal geodesic connecting p to q. (1) For any Yq ∈ TqM, (∇2dp)q( ˙γ(l), Yq) = ⟨∇γ˙ (l)γ, Y ˙ q⟩ = 0. (2) For any normal Jacobi field X along γ with X(0) = 0, (∇2 dp)q(Xq, Xq) = ⟨∇γ˙ (q)X, Xq⟩ = I(X, X). Note that the “singularity” of dp at the point p is not too bad: one can always remove the singularity at p by considering the function d 2 p instead. So it is reasonable to study (∇2d 2 p )p. In general, for any smooth function f ∈ C ∞(M) and Xp ∈ TpM, if we let γ be the geodesic γ(t) = expp (tXp), then “the second order derivative of f along γ” is d 2 dt2 f ◦ γ(t) = d dt d dt(f ◦ γ) = d dt⟨∇f, γ˙⟩ = ⟨∇γ˙ ∇f, γ˙⟩ = (∇2 f)γ(t)( ˙γ, γ˙). So we get Lemma 2.4. (f ◦ γ) ′′(t) = (∇2 f)γ(t)( ˙γ, γ˙). On the other hand, if Xp is a normal vector, i.e. γ is a normal geodesic, then for t small enough we have d 2 p (γ(t)) = t 2 . So we get, for Xp ∈ SpM, (∇2 d 2 p )p(Xp, Xp) = 2g(Xp, Xp) and thus by polarization, we get Proposition 2.5. The Hessian of d 2 p at p is (∇2 d 2 p )p = 2g