On the monotonicity of the expected volume of a random. Luis Rademacher Computer Science and Engineering The Ohio State University

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1 On the monotonicity of the expected volume of a random simplex Luis Rademacher Computer Science and Engineering The Ohio State University

2 Sylvester s problem 4 random points: what is the probability that they are in convex position? P : convex hull of n random points in convex n K body K. Which convex bodies K are extremal for EvolP K n volk

3 Questions by Meckes and Reitzner K M = EvolP K n cd EvolP M n M s weak conjecture: for n=d+1 (random simplex) there exists some c. M s strong conjecture: for n=d+1 and c=1 R s question: for arbitrary n and c=1.

4 Connectionwith slicing Slicing conjecture: every d-dimensional convex body of volume one has a hyperplane section of area >=c for some universal c. Equivalent: for K a d-dimensional convex body L K :=(deta(k)/(volk) 2 ) 1/2d hasauniversalupperbound. M s weak conjecture slicing M s weak conjecture: c>0suchthatforanypairk,m ofd-dimensionalconvexbodies K M = EvolP K d+1 c d EvolP M d+1

5 Main result About M s strong conjecture: True in dimension 1,2 False in dimension >=4 Strong numerical evidence for falsity in dim. 3. M s strong conjecture: ForanypairK,M ofd-dimensionalconvexbodies K M? = EvolP K d+1 EvolP M d+1

6 Busemann-Petty ( θ S d 1 )vol d 1 (K θ ) vol d 1 (L θ )? = volk cvoll Classical: when c=1, true iff dim<5 c slicing Similarity with our problem: Dimension-dependent answer, connection with slicing Difference with our problem: Ours has elementary solution, no Fourier analysis.

7 Questionby Vempala K M? = deta(k) c d deta(m) A(K) = covariance matrix of K Original question: for c=1.

8 c slicing: Questionby Vempala K M? = deta(k) c d deta(m) deta(k)/vol(k) 2 =L 2d K : easy from * and bounded isotropic constant : : 1. Let K,M be two convex bodies. Assumption with * imply L K c 1/2 d(k,m) L M : By affine invariance, w.l.o.g. K,M are in position such that K M d(k,m)k. Implies volm d(k,m) d volk. 2. Use Klartag s isomorphic slicing problem: given K,ε, there exists M s.t. d(k,m) 1+εand L M 1/ ǫ

9 Question by Vempala Connection with random simplexes: (µ(k)=centroid of K) ( deta(k)=d!e Xi K (volconvµ(k),x1,...,x d ) 2) = d! d+1 E( (volpd+1 K )2), I.e. second moment version of Meckes s question.

10 Second result We also solve Vempala s question for c=1 True in dimension 1,2 False in dimension >=3 Our solution to Vempala s question inspired our solution to Meckes s strong conjecture. Vempala s question(for c = 1): K L? = deta(k) deta(l)

11 Solutionto Vempala s question Intuition: extreme point near the centroid E (random simplexusing L\K) < E( using L)

12 Solution to Vempala s question Idea: Monotonicity of K adeta(k) holds for dimension difffor every K L there is a nonincreasing path of convex bodies from L to K. We will: define path, compute derivative along path, and study sign of derivative.

13 Solution to Vempala s question W.l.o.g. K is a polytope (by continuity, if there is a counterexample, then there is one where K is a polytope) Path : push hyperplanes parallel to facets of K in, one by one. L K

14 Solution to Vempala s question deta( ) continuous along path, piecewise continuously differentiable. Enough to compute derivative with respect intersection with moving halfspace. Enough to compute derivative in isotropic position (sign of derivative is invariant under affine transformations). K t =K {x:a T x t} d dt deta(k t)

15 Solution to Vempala s question Simple value of derivative in isotropic position: Proposition. Let K R d be an isotropic convex body. Let v R d be a unit vector. Let a=inf x K v x, b=sup x K v x. Let H t ={x R d :v x t}. LetK t =K H t,s t =K bdryh t. Then d dt deta(k t) = t=a ( d E X Sa X 2) vol d 1 S a volk.

16 Solution to Vempala s question Derivative implies dimension dependent condition: Lemma. Monotonicity under inclusion of K deta(k) holds for some dimensiondiffforanyisotropicconvex body K R d wehave dbd K. Proof: if part : condition implies negative derivative along path. d dt deta(k t) = t=a ( d E X Sa X 2) vol d 1 S a volk.

17 Proof (cont d) Only if part: There is an isotropic convex body L with a boundary point x at distance <d 1/2 from the origin. By an approximation argument can assume x is extreme point (keeping isotropy and distance condition) of new body L. Positive derivative as one pushes hyperplane in at x a little bit. If K is L truncated near x, deta(k) > deta(l).

18 When does the condition hold? Condition: for any isotropic convex body K dbd K. Milman-Pajor, Kannan-Lovász-Simonovits: For any isotropic convex body K: d+2 d B d K d(d+2)b d and this is best possible. I.e., condition fails iffd 3.

19 Solution to Meckes s strong conjecture Argument parallels case of det A(K): same path between K and L (push hyperplanes in), same derivative (i.e. with respect to moving hyperplane) Our derivative is a special case Crofton s differential equation: (from Kendall and Moran, Geometrical Probability, 1963)

20 Derivative Proposition (general derivative, Crofton). Let K R d be a convex body. Let v R d be a unit vector. Let a = inf x K v x, b = sup x K v x. Let H t = {x R d : v x t}. Let K t = K H t, S t = K bdryh t. Let f :(R d ) k Rbe asymmetriccontinuous function. Let X 1,...,X k berandom pointsink. Then d dt Ef(X ( 1,...,X k ) =k Ef(X 1,...,X k ) E ( ) ) vol d 1 S a f(x 1,...,X k ) X 1 S a t=a volk

21 Solution to Meckes s strong conjecture Dimension dependent condition (same proof): Lemma. Monotonicity under inclusion of K E Xi KvolconvX 0,...,X d holdsforsomedimensiondiffforanyconvexbodyk R d andanyx bdryk and X 0,...,X d randomink wehave EvolconvX 0,X 1,...,X d Evolconvx,X 1,...,X d.

22 When does the condition hold? Failsford 4: K=halfball,x=0,L=ballofsamevolumeasK. Ontheotherhand, EvolPd+1 EvolP K d+1 L (Blaschke-Groemer) = 1 ( ) d+1 κd+1 κ d(d+2) 1 (known) d! κ (d+1) 2ω d+1 κ d E K volconv(0,x 1,...,X d )=E Bd volconv(0,x 1,...,X d ) (symmety) = 1 ( ) d κd+1 2 (known) d! ω d+1 Combine to get κ d E K volconv(0,x 1,...,X d ) EvolP K d+1 2 d+2 d+1

23 When does the condition hold? true for d=1, easy (directly, without the condition)

24 When does the condition hold? Trueford=2: Theorem(Blaschke). Among all 2-dimensional convex bodies, E Xi K(volconv(X 0,X 1,X 2 )) vol(k) 1 12 withequalityiffk isatriangle. Lemma. LetK R 2 beaconvex body andletx bdryk. Then E X1,X 2 K(volconvx,X 1,X 2 ) volk 8 9π2. (1) Proof idea: Worst case, K is half ellipse, x its center. Then Steiner symmetrization, Blaschke s Schüttelung (shaking), symmetrization around x, Busemann s inequality.

25 When does the condition hold? d=3? The proof doesn t handle it, but numerical integration strongly suggests false (same counterexample).

26 Open questions Higher moments? (Reitzner) What about more than d+1 points? (technical) 3-D case Weak conjecture via Crofton s differential equation?