Measuring Ellipsoids 1

Measuring Ellipsoids 1

Igor Rivin Temple University 2

What is an ellipsoid? E = {x E n Ax = 1}, where A is a non-singular linear transformation of E n. Remark that Ax = Ax, Ax = x, A t Ax. The matrix Q = A t A is a positive definite matrix, with eigenvalues λ 1,..., λ n, whose (positive) square roots σ 1,..., σ n are the so-called singular values of A. Their geometric significance is that the semi-axes of E are the quantities a i = 1/σ i. After an orthogonal change of coordinates, we can write E = {x E n n x 2 i λ i = i=1 n i=1 x 2 i σ2 i = n i=1 x 2 i a 2 i = 1}. 3

Volume of an Ellipsoid vol E = κ n det A = κ n n a i, i=1 where κ n is the volume of the unit ball in E n. 4

k-th Mean Curvature m k ( K, x) ( ) n 1 m k k (x) = s k (k 1 (x),..., k n 1 (x)). k-th integral mean curvature M k M k = K m k (x)da, 5

Steiner s Formula for Area of Parallel Surface (at distance ρ) vol n 1 K ρ = n 1 k=0 ( ) n 1 M k k ( K)ρ k, The volume of a neighborhood of thickness ρ vol Kρ = vol K + ρ 0 vol n 1 K τ dτ. 6

Another formula for M k Let G n,r be the Grassmanian of affine r-planes in E n, with a suitably normalized invariant measure µ. Then µ ( {L r G n,r L r K } ) = ω n 2... ω n r 1 M (n r)ω r 1... ω r 1 ( K), (1) 0 where ω k is the surface area of the unit sphere in E k+1. 7

Theorem 1 (Archimedes axiom). M( K) is monotonic under inclusion. That is, if K 1 K 2, then M i ( K 1 ) M i ( K 2 ). The inequality is strict if K 2 \K 1 has non-empty interior, and i < n 1, where n is the dimension of the ambient Euclidean space. 8

Integral Mean Curvature for Polytopes Let P be a convex polytope in E n. ( ) n 1 M i i ( P) = α ( f )vol n i 1 f, codim. i faces f of P (2) where α ( f ) is the exterior angle at f, described as follows: Consider the Gauss map, which maps each point p of P to the set of outer normals to the support planes to P passing through p. The image of all of P will be the unit sphere S n 1 E n, and the combinatorial structure of P will induce a dual cell decomposition C of S n 1, in particular, the image of a codimension-i face f of P will be an i-dimensional totally geodesic face f of C. The i-dimensional area of that face is the exterior angle at f. 9

Polyhedra which tile space Suppose that translates of P tile E n, so that P is a fundamental domain for a free action of a group G of translations on E. In particular, G acts on P, preserving the combinatorial structure. Let the quotient by P G. Then ( ) n 1 M i i = ω i vol n 1 i f. codim. i faces f of P G In particular, if P is a rectangular parallelopiped: P = [0, l 1 ] [0, l 2 ] [0, l n ], then we obtain M i (P) = ω i s n 1 i (l 1,..., l n ). 10

Inscribed and Circumscribed Cubes The cube C i with vertices ( ) ±1, ±1,, ±1 n n n inscribed in the unit sphere, while the cube C o with vertices (±1, ±1,, ±1) is circumscribed around the unit sphere. is If E is an ellipsoid with semi-axes a 1,..., a n, the preimages of P i of C i and P o of C o are, respectively, inscribed and circumscribed (rectangular) parallelopipeds. From the Archimedean theorem, it follows that the mean curvatures of E are contained between those of P i and P o. 11

Which means that: M i (P o ) ( n ) n 1 i = M i ( P n ) M (E) M i (P o ), or ( ) n 1 i 2 ωi s n 1 i (l 1,..., l n ) ( n n 1 ) M i (E) i and M i (E) 2 n 1 iω is n 1 i (l 1,..., l n ) ). ( n 1 i In three dimensions, the upper bound of is sharp, as shown by examining ellipsoids with l 1 = 1, l 2 = 1, l 3 1. The lower bound is not sharp; the sharp bound is given by Polya and Szegö in a 1948 paper. 12

Area of the Equidistant Surface Theorem 2. Let f (ρ) = 1 n n (ρ + 2l ρ i ) ρ n 2 n l i. i=1 Then the area of the equidistant surface E ρ is bounded as follows: i=1 c n f (ρ) vol n 1 E ρ C n f (ρ), where C n /c n ( n ) n 1 Γ ((n + 1)/2) 2π (n+1)/2. (estimate for volume follows immediately). 13

Lattice Points Consider an ellipsoid E, and consider the number N(e) of points of the integer lattice in E. How do we estimate (E) = N(E) vol E? By a fairly obvious argument argument, (E) is dominated by the volume of a tubular neighborhood of radius (n) around E. We have: (E) C n n 0 f (ρ)dρ. (3) 14

John ellipsoid It is a well-known theorem of Fritz John that for any convex body K, there exists an ellipsoid E K, such that E K /n K E K for centrally-symmetric K, E K /n can be improved to E K / n. We can estimate the symmetric functions of the semi-axes (and hence the semi-axes themselves) of the John ellipsoid in terms of the mean curvature integrals of K, and vice versa. 15

Can we do better? 16

Cauchy s formula Let K be a convex body in E n. Let u S n 1 be a unit vector, and let us define V u (K) to be the (unsigned) n 1-dimensional volume of the orthogonal projection of K in the direction U. Cauchy s formula then states that V n 1 ( K) = n 1 ω n 2 V u(k) dσ S n 1 = (n 1) ω n 1 ω V u(k) dσ, n 2 S n 1 where dσ denotes the standard area element on the unit sphere. 17

Projected areas of an Ellipsoid In the case where K = E is an ellipsoid, given by n E = {x E n i=1 q 2 i x2 i = 1} there are several ways of computing V u. The result is: ( ni=1 u 2 ) V u (E) = κ n 1 i q2 i ni=1 q i. (4) 18

Since V n (E) = κ n ni=1 q i, we can rewrite Cauchy s formula (4) for E in the form: R(E) def = V n 1( E) V n (E) = n S n 1 n i=1 u 2 i q2 i dσ, (5) where R(E) is the isoperimetric ratio of E. 19

Amazing, but true! Theorem 3. The ratio R(E) is a norm on the vectors q of inverse lengths of semiaxes (q = (q 1,..., q n ).) Proof. The integrand in the formula (5) is a norm. Corollary 4. There exist constants c n,p, C n,p, such that c n,p q p R(q) C n,p q p, where q p is the L p norm of q. 20

How can we compute integrals ove S n? Theorem 5. Let f (x 1,..., x n ) be a homogeneous function on E n of degree d (in other words, f (λx 1,..., λx n ) = λ d f (x 1,..., x n ).) Then ( ) ( ) n + d n Γ 2 f dσ = Γ E ( f (X S n 1 2 1,..., X n ) ), where X 1,..., X n are independent random variables with probability density e x2. 21

Application to Ellipsoids R(E) = n S n 1 = n Γ ( n 2 ) Γ ( n+1 2 n u 2 i q2 i dσ )E i=1 ( ) q 2 1 X 1 + + q 2 n X n where X i is a Gaussian with variance 1/2., 22

A class of Hypergeometric Functions First, we need a definition: Definition 6. Let a, b 1,..., b n, c, x 1,..., x n be complex numbers, with x i < 1, i = 1,..., n, Ra > 0, R(c a) > 0. We then define the Lauricella Hypergeometric Function F D (a; b 1,..., b n ; c; x 1,..., x n ) as follows: F D (a; b 1,..., b n ; c; x 1,..., x n ) = Γ(c) Γ(a)Γ(c a) 1 0 u a 1 (1 u) c a 1 n (1 ux i ) b i du. (6) i=1 We also have the series expansion: F D (a; b 1,..., b n ; c; x 1,..., x n ) = (a) ni=1 m1 + +m n (b i ) mi m 1 =0 m n =0 valid whenever x i < 1, i. (c) m1 + +m n n i=1 x m i i m i!, (7) 23

Now, we can write ( ) R(E) = n Γ2 n 2 Γ ( ) α 2 n+1 2 n q 2 ( j 2 F D 1/2; η 1j,..., η nj ; n + 1 ) 2 ; 1 αq2 1,..., 1 αq2 n j=1 where η ij = 1/2 + δ ij, and α is a positive parameter satisfying 1 αq 2 j < 1., (8) 24

Low Dimensions In two dimensions, the circumference of an ellipse with semiaxes a b is: ( ) C(a, b) = 4aE π/2, 1 b 2 /a 2. In three dimensions, the surface area of an ellipsoid with semiaxes a b c is given by ( A(a, b, c) = 2π c 2 bc 2 ) + a a 2 c 2F(φ, m) + b 2 c 2 E(φ, m), where m = a2 (b 2 c 2 ) b 2 (a 2 c 2 ), and φ = arcsin 1 c2. a 2 25

Law of Large Numbers Theorem 7. Let q 1,..., q n,... be a sequence of positive numbers such that ni=1 q 4 i lim n ( ni=1 q 2 ) = 0. 2 i Let E n be the ellipsoid in E n with semiaxes a 1 = 1/q 1,..., a n = 1/q n. Then lim n Γ ( ) n+1 2 Γ ( ) n 2 R(E n ) n 12 ni=1 q 2 i = 1. 26

Corollary 8. Let a 1,..., a n,... be such that 0 < c 1 a i /a j c 2 <, for any i, j. Let E n be the ellipsoid with major semi-axes a 1,..., a n. Then lim n Γ ( n+1 2 Γ ( ) n 2 ) n R(E n ) 1 2 ni=1 1 a 2 i = 1. 27

Finite Dimensions We know that R(E) is a norm on the vector q = (q 1,..., q n ) let us agree to write q R def = R(E) n = S n 1 n q 2 i x2 i dσ. i=1 where q is the vector of inverses of the major semi-axes of E. We know that for any p > 0, c n,p q p q R C n,p q p, for some dimensional constants c n,p, C n,p. we will give good estimates on the constants c n,2 and C n,2. How good? 28

3Γ ( n 2 ) 2Γ ( n+1 2 ) q R q ( ) Γ n2 ( ). (9) πγ n+1 2 29

First Basic Inequality Lemma 9. Let F(x) be a probability distribution, and let M a (F) def = E F ( x a ) denote the absolute moments of F (we will abuse notation in the sequel by referring to the absolute moments of a random variable as well as those of its distribution function). Further, let 0 β α. Then M β (F) 1 + M α (F). 30

Second Basic Inequality n n S n 1 i=1 q 2 i u2 i dσ i=1 q 2 i S n 1 u 1 dσ. (holds by concavity of square root). 31