1547671870-The_Ricci_Flow__Chow

3. BUSEMANN FUNCTIONS 309

and notice that the triangle inequality proves for all s 2 so that

Cs,Ts =d(a,1(s)) ~d(a(Ts 0 ),1(s))

(B. 7) ~ d ( C\'. ( Ts 0 ) , I (so)) + d (! (So) , I ( S)) = S - E.

Let !3s,o : [O, Cs,o] -+ Mn be a minimal geodesic such that !3s,O = a (0) =

x and !3s,o (Cs,o) = I (s), where Cs,O ~ L (!3s,o) = d (x, I (s)). Because

a (0), a (£) E IHI., =Mn\ UsE(O,oo) B (! (s), s), we have Ts E (0, £) and

(B.8) Cs,Ts < S ~ Cs,O·

In particular, we can apply the first variation formula at Ts to conclude
that a (Ts) J_ /Js,Ts (0). Thus we have constructed a right triangle with
vertices I ( s) , a (0) , a (Ts) and sides /3s,o, al [0,Ts], /3s,Ts. Since sect (g) 2 0, the
Toponogov comparison theorem implies that the hypotenuse length satisfies
the inequality

(B.9) Cs,O = L (/3s,o) ~ V L (aJ[O,Ts])

2

+ L (f3s,TJ

2

=VT}+ £~,Ts"

Recalling that Ts E [O, £] and combining inequalities (B.7), (B.8), and (B.9),
we see that IHI 1 can fail to be totally convex only if

S2 ~ T; + c;,Ts ~ g2 + c;,Ts ~ g2 + (s - c)2

for all s 2 so. Since£ and E are independent of s, this is impossible. D

3.3. Constructing totally convex sublevel sets. We are now ready
to construct the sublevel sets of the Busemann function associated to a point
OE Mn.

DEFINITION B.55. If 1 : [0, oo) -+ Mn is a ray and s E [0, oo), the

shifted ray Is : [O, oo) -+Mn is defined for all s' E [O, oo) by

Is ( s') ~ I ( s' + s).

Note that 10 = ')'. It is worth noting some elementary properties of the

half spaces associated to a shifted ray.

LEMMA B.56. Ifs ~ t, then lffi 1 t ~ lffi 1 s.

PROOF. For all r E [O, oo ), we have the inclusion

Hence

B (! ( t + r) , r) ~ B (! ( s + ( t - s + r)) , t - s + r) ~ IIB,,s.

IIB,,t ~ LJ B (! ( t + r) , r) ~ IIB,,s.

rE(O,oo)

LEMMA B.57. Whenever 0 < s < t,

IIB,,s = {x E Mn: d(x,IIB,,t) < t - s}.

D