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Unformatted text preview: e to squeeze through this bottleneck because {if the gallon ows nonrecurrently{ it
has an intrinsic positive crosssectional ux that can never diminish.
Proof of the Pinchedcusp Lemma. Supposing not conservative, there is some cueball set S
of positive volume and a positive time so that
;1) ( S) is disjoint from S. (15) Moreover, S can be taken to lie to the left of some vertical line, say, left of x = 1. After deleting a
nullset we can assume that for all v 2 S,
t limsup xcoord
t!1 v = +1 : (16) If not, then (by dropping to a positivemass subset) all cueballs in S forever stay left of some line, x = 100
say, and ( 1;1)(S) would be a invariant set of nite volumeto which Poincare Recurrence
would apply, contradicting (15).
Consequently the situation is as gure 17 illustrates. For each positive number x, let Lx denote
the set of cueballs with footpoint on the vertical linesegment from x; 0 up to x; f (x) and which
point to the right, ie. their directions are between =2 and =2. This \line" is a 2dimensional
subset of cueball space. Because of (16), eventually tS will have positive volume lying to the
right of L1 . Thus there exists a time t0 such that ux( ) is positive, where
:= L1 \ t0 ( S) : (This follows from breaking L1 into countably many pieces U and applying (13b) to each.) In addition, since
assertion (15) is ow invariant, we may conclude that \ ;1) ( ) is empty.
Strategy. We have progressed from a nonrecurrent set S of positive volume to a nonrecurrent
crosssection of positive ux. We will obtain a contradiction by examining the bad set B := v 2
Printed: February 14, 1995 R (v ) = 1
Filename is Article/18Billiards/billiards.ams.tex 13 Billiards inside a cusp of cueballs which never come back to .
y=f(x) S v Σ Bottleneck B T U L(v )
B
x L1 Lx Figure 17 A pinchedcusp. To make them visible, sets and B are shown thickenedthey are
actually subsets of \line" L1 . The forward trajectory of a cueball v 2 B will never again touch
(although it might conceivably hit L1 elsewhere) and will sooner or later cross any given Lx . But if Lx is chosen to be a su ciently small bottleneck, not all of B will be able to squeeze
through. Every such v, as (16) reminds, eventually hits any particular line Lx to the right of L1 . Thus the
induced map TB L is de ned on all of B and maps it into Lx. The owinvariance of ux now
gives the key inequality that
x ux(Lx ) ux(B) ; for all x > 1. But ux(Lx ) is proportional to f (x), by (11b). Thus the above inequality will atly contradict
that f () is pinched against the xaxis, if we can rule out ux(B) being zero. In order to do this,
we now look at those cueballs whose behavior is antithetical to the bad set.
The in nitelyoften set. Consider the set I of cueballs which, under T , return to in nitely
often,
I :=
B T 1(B) T 2 (B) T 3 (B) : : : :
Of necessity, T maps I into I and so RI is nite on all of I. But I
and, since \ ;1)( )
is empty, we see that RI must in fact be everywhere less than the constant . Consequently, no
cueball v 2 I could satisfy (16). Thus I is empty and, in consequence,
= B T 1(B) T 2 (B) T 3(B) : : : :
But ux( ) > 0, and so some T n (B) has positive ux. The third part of the Flux Proposition,
(13c), implies that ux B
ux T n (B) . Thus ux(B) is positive, completing the proof of the
Pinchedcusp Theorem.
1 Appendix
The observation that a Poincaresection provides a fast proof of conservativity of the billiard ow in a convex
measure cusp, arose in a discussion with my colleague Albert Fathi. That argumen...
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