with LPC, then it must satisfy the following. Letx∗∈M be the midpoint of a minimal geodesic connectingx0andy0and u, v ∈ Tx∗M be unit vectors with u ⊥v. Letγ
represent the geodesic issuing fromx∗in direction u and let Vtrepresent the parallel
transport ofvalongγ. Then the following holds.
b(γ (t)),γt˙ =λt+ b(x∗),u (81)
whereλis as in Lemma47, and
b(γ (t)),Vt = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ b(x∗), vcos√K t+ T u, vsin √ K t √ K if K >0, b(x∗), v + T u, vt if K =0, b(x∗), vcosh√−K t+ T u, vsinh √ −K t √ −K if K <0. (82)
where the matrix T given by Ti j = ∇eib(x∗),ej −λei,ejis a skew-symmetric
matrix.
Proof To see (81), note that d
dtb(γ (t)),γt˙ = ∇γ˙tb(γ (t)),γt˙ =S(γt˙,γt˙)=λ.
Take anyx ∈ B(x0,r)andy ∈ B(y0,r)such thatx∗ ∈ H(x,y). Since H(x,y)is the fixed point set of the isometry fx,y, it is therefore a totally geodesic submanifold of M. Letκ denote the Killing vectorfield for which (71) holds at timet =0 with
F0= fx,y. Take any unit speed geodesicγ passing throughx∗and lying inH(x,y). (Note that, if a geodesic lies inH(x,y)for a short time, it should lie inH(x,y)for all time. See, for example, the proof of Proposition 24 of [32], p. 145.)
Let(nt :t ≥0)be the parallel transport of the vector normal to the hypersurface
H(x,y)at x∗ along the geodesicγ. Note that, as H(x,y)is totally geodesic, the second fundamental form vanishes identically on H(x,y)[26, Exercise 8.4]. This fact implies that parallel transportation of a vectorv ∈ Tx∗H(x,y)with respect to the induced metric onH(x,y)agrees with parallel transportation ofvin the ambient manifold M [26, Lemma 8.5]. Thus,nt is precisely the direction that is reversed at γ (t)by fx,y.
Equation (71) gives us
b(γ (t)),nt = 1
2κ(γ (t)),nt. (83) Differentiating the above twice with respect totalong the geodesicγ, and using the fact that∇γ (˙t)nt = 0 becausent was defined using parallel transport alongγ, we obtain D2tb(γ (t)),nt = 1 2D 2 tκ(γ (t)),nt
(usingDt as shorthand for covariant differentiation∇γ˙ along the geodesicγ) which, along with (73) and (74), gives
d2
dt2b(γ (t)),nt + K
2κ(γ (t)),nt =0. (84) Consequently Eq. (83) shows that the functiont→ b(γ (t)),ntsatisfies the follow- ing differential equation
d2
dt2b(γ (t)),nt +Kb(γ (t)),nt =0. (85)
For any geodesicγpassing throughx∗, not necessarily lying inH(x,y), and for any parallel vectorfield Vt alongγ orthogonal toγt˙, a similar technique uses (71), (73) and (74) to give us
d2
dt2b(γ (t)),Vt +Kb(γ (t)),Vt
= d2
dt2b(fx,y◦γ (t)),d fx,y(Vt) +Kb(fx,y◦γ (t)),d fx,y(Vt). (86)
Now, following the lines of the proof of Lemma37, we can iteratively compose the isometries in
S= fx,y∈G : x∈B(x0,r),y∈B(y0,r),dist(x,x∗)=dist(y,x∗) = 1 2dist(x,y)
to deduce that the closed subgroup of isometries G∗ generated byS is the whole isotropy group ofx∗inG. Further, from Step 1 and Step 2 in the proof of Lemma37, it can be seen that for any pair of linearly independent unit vectorsu, v∈Tx∗M, there is a sequence of isometries{Fk}k≥1such that for eachk,Fkis a composition of isometries inS, dFk fixes vectors inTx∗M that are orthogonal to{u, v}, and dFk(u) → v as
k→ ∞.
Take any geodesicγ issuing fromx∗and lying inH(x,y)for somex ∈B(x0,r),
y∈B(y0,r)and letnt denote the parallel vectorfield alongγthat is inverted by fx,y. LetG∈Gbe a composition of isometries inSwhich fixγand letv=dG(n0). LetVtv
denote the parallel transport ofvalongγ. AsGis an isometry, [26, Proposition 5.6 (b)] impliesG∗nt =Vtv. Applying (86) at each composition corresponding toG, we get
d2
dt2b(γ (t)),nt +Kb(γ (t)),nt =
d2
dt2b(γ (t)),Vtv +Kb(γ (t)),Vtv. (87)
By (85), the left hand side of the above is zero. Thus, the right hand side should vanish too. Solving this gives (82) withVvin place ofV and the given matrixT.
Now, consider any parallel vectorfieldVtalongγwhich is orthogonal toγt˙. By the discussion following the definition ofS, there exists a sequence of isometries{Fk}k≥1
such that eachFkis a composition of isometries inS,Fkfixesγ, and dFk(n0)→V0 ask→ ∞. AsFkfixesx∗for eachk, by [30, p. 7], we can choose a subsequencekl such thatFkl →FinGasl→ ∞. WriteV
(k)
t =Fk∗nt. By [30, Lemma 4], for each
t ≥0,V(kl)
t →dF(nt)inTγ (t)Masl → ∞. In particular, dF(n0)=V0, and asFis an isometry fixingγ, dF(nt)=Vtfor allt≥0. Thus, we haveVt(kl)→VtinTγ (t)M for eacht ≥0. From the discussion in the previous paragraph, (82) holds withV(kl)
in place ofV for eachl ≥1. Takingl→ ∞, we obtain (82) for the vectorfieldV. Finally, take any pair of unit vectorsu, v∈ Tx∗M satisfyingu ⊥v. Letσ be the geodesic issuing fromx∗such thatσ(˙ 0)=u. We can obtain a sequence of isometries
{Gk}k≥1such that eachGkis a composition of isometries inSand dGk(γ (˙ 0))→u ask → ∞. Writeuk =dGk(γ (˙ 0))and letσk be the geodesic issuing fromx∗in the directionuk. Denote byVtv,k andVtvthe parallel transport ofvalongσk andσ respectively. By the previous discussion, we know that (82) holds withVv,kin place of V andσk in place ofγ for eachk ≥ 1. Observe that for each fixedt ≥ 0, both sides of (82) depend continuously onuandv(this observation for the left hand side follows from the fact that the solution to the geodesic and parallel transport equations depends continuously on the initial data). Thus, we can takek→ ∞to get (82) with
The fact thatT is skew-symmetric follows from the observation thatSx∗(ei,ej)= λei,ej(by Lemma47) and therefore
∇eib(x∗),ej −λei,ej = 1 2 ∇eib(x∗),ej − ∇ejb(x∗),ei .
Since M is a maximally symmetric space (by Theorem38), the dimension of its set of Killing vectorfields is d(d2+1). Thus, for any vectorw ∈ Tx∗M and any skew- symmetric matrixT, there exists a unique Killing vectorfield K withK(x∗) = w
and∇eiK(x∗),ej = Ti j. Moreover, as every Killing vectorfield is a Jacobi field
(i.e. satisfies (73)), it follows thatKsatisfies the following equation analogous to (82), for unit vectorsu, v∈Tx∗M withu⊥v.
K(γ (t)),Vt = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ w, vcos√K t+ T u, vsin √ K t √ K ifK >0, w, v + T u, vt ifK =0, w, vcosh√−K t+ T u, vsinh √ −K t √ −K ifK <0. (88)
Thus, if we setKx∗as the Killing vectorfield uniquely determined byw=b(x∗)and
Ti j = ∇eib(x∗),ej −λei,ej, we see from Lemmas47and48that the vectorfield
bcan be written as
b=Dλx∗+Kx∗ (89)
whereDλx∗is thedilation vectorfield aboutx∗with dilation coefficientλdefined as
Dxλ∗(γ (t))=λtγ (˙ t) (90) for any geodesicγ issuing fromx∗. Now, we claim that dilation vectorfields do not arise in the case of non-zero-curvature.