Prove
Lemma 5 ([Endpoint correction]). For a smooth field V=(w,z) on [a,c] , I_ [a,c]…Lemma 4 ([Tangential invariance]). The integrand defining I_J^⊥(V) is the index…Proposition
Use inward coordinates at a finite endpoint. If τ=0 there, then and give τ=-κp_1^2/2 . If p_1≠0 , the boundary point is a strict local maximum of τ , and nearby times have τ<0 . If p_1=0 , then the scalar equation τ=-F'(τ)/2 has the constant solution τ=0 with these initial data. Smooth ODE uniqueness forces this solution locally. Neither possibility can be an endpoint of a nonempty open-neck component. This proves transversality and also handles a boundary grazing time. In the second case the corresponding geodesic is the fiber geodesic in the boundary torus, so uniqueness keeps it in that torus for all time. If v vanishes at an interior time and F' also vanishes there, uniqueness again makes t constant. Then the solution stays at this height for all time, so the passage is ℝ . Otherwise such a zero is a strict extremum of t . Reflecting the interval coordinate if necessary makes it a maximum t_* with c:=F'(t_*)>0 . Concavity gives F'(t)≥c for 0≤t≤t_* . On each side of the turn, the trajectory moves toward t=0 and cannot turn again, because v=-F'(t)/2≤-c/2 . It reaches 0 in finite time: after the turn, for example, t(s)≤t_* -c(s-s_*)^2/4 as long as it stays in the neck. Time reversal and uniqueness make the two halves equal. Translating the parameter gives. Concavity also gives F'(t)≤F'(0)=κp_1^2 , proving . Since F'(t_*)>0 , p_1 cannot be zero. Finally, v_0^2=F(t_*)-F(0)=∫_0^ t_* F'(u)\,du ≤κp_1^2t_*≤κp_1^2, which proves. It remains to classify passages with v nowhere zero. Here t is strictly monotone. If both endpoints are finite, transversality and monotonicity force the two different boundary tori. At an infinite endpoint, bounded monotonicity gives a limit t_∞ . The identity v^2=1-F(t) shows that |v| has a limit, which must be zero because t is bounded. Thus F(t_∞)=1 . Since v tends to -F'(t_∞)/2 , this limit must also be zero, or v could not tend to zero. If t_∞ were a boundary height, and F'(t_∞)=0 would force p_1=0 in that end frame. Then F would be constant and equal to 1 throughout its end portion, giving v=0 there, a contradiction. Hence the limiting height is interior. A nonconstant monotone passage cannot have two infinite endpoints. Indeed, its two limiting heights would be distinct, with F=1 at both. Concavity would give F≥1 between them, whereas gives F<1 at every traversed height. This contradiction finishes the classification.