GEOMETRY OF MEASURES: PARTITIONS AND CONVEX BODIES … · 2017. 5. 4. · GEOMETRY OF MEASURES: PARTITIONS AND CONVEX BODIES 2 map. From standard facts of di erential geometry we

GEOMETRY OF MEASURES PARTITIONS AND CONVEX BODIES

ROMAN KARASEV

1 Introduction

In this course we start with several topological techniques that allow to partition sev-eral measures in a Euclidean space into equal parts or partition the space into parts ofprescribed measure These are classical results in discrete geometry and measure theoryand they find applications in a variety of problems

After that we give applications to point-line incidences and spanning trees with lowcrossing number following the excellent review of H Kaplan J Matousek and M Sharir [KMS12]The reader is also encouraged to read the review of L Guth on a similar topic [Guth13]

Then we discuss the monotone transportation and the BrunnndashMinkowski inequalityfollowing the nice course of K Ball [Ball04] We also consider the PrekopandashLeindlerinequality for log-concave measures the Minkowski theorem on facet areas the needledecomposition and the isoperimetric inequality for the Gaussian measure following inparticular the blog post of T Tao [Tao11] and the nice paper of F Nazarov M Sodinand A Volrsquoberg [NSV02]

We touch the topic of the isoperimetric inequality and concentration on the roundsphere as well as another result about the Gaussian measure known as the Sidak lemmaThen we give some simple facts about volumes of sections of a cube facet and vertexnumbers of centrally symmetric polytopes and sketch a proof of the Dvoretzky theoremfollowing another brief course of K Ball [Ball97] We also discuss the topological approachto the Dvoretzky theorem and recent positive and negative results in this direction

2 The BorsukndashUlam theorem

One common tool to prove results about partitions of measures is the classical BorsukndashUlam theorem [Bor33]

Theorem 21 For any continuous map f Sn rarr Rn there exists a pair of antipodalpoints xminusx isin Sn such that f(x) = f(minusx)

Proof By putting g(x) = f(x)minus f(minusx) we reduce this theorem to the following For anodd map g Sn rarr Rn there exists a point x isin Sn such that g(x) = 0 A map g is calledodd if g(minusx) = minusg(x) for any x

Then we consider a simple map g0 defined as follows If Sn is the unit sphere in Rn+1then g0 is the projection to a coordinate subspace Rn sub Rn+1 It is easy to observe thatfor g0 there is a unique antipodal pair x0minusx0 isin Sn that is mapped to zero Moreover atthis x0 (and minusx0) the Jacobian matrix Dg0 is nondegenerate

Assume that g does not map any point to zero Now we connect g0 and g by thehomotopy ht(x) = (1minust)g0(x)+tg(x) For any t the map ht(x) remains an odd continuous

2000 Mathematics Subject Classification 52C35 60D05Key words and phrases ham sandwich theorem monotone maps log-concavity isoperimetry

polytopesSupported by the Dynasty Foundation the Presidentrsquos of Russian Federation grant MD-35220121

the Federal Program ldquoScientific and scientific-pedagogical staff of innovative Russiardquo 2009ndash2013 and theRussian government project 11G34310053

GEOMETRY OF MEASURES PARTITIONS AND CONVEX BODIES 2

map From standard facts of differential geometry we may perturb the homotopy htslightly to obtain another homotopy ht(x) with the following properties

1) h0(x) is still equal to g0(x) 2) zero is a regular value for h Sn times I rarr Rn (I = [0 1]is the segment) and hminus1(0) is a one-dimensional submanifold Z sub Sntimes I with boundaryin Sn times partI 3) the map h1(x) may be not equal to g(x) but it still misses zero in Rn

Now starting from the unique pair (x0 0) (minusx0 0) isin partZ and trace this pair alongthe one-dimensional set Z This pair of point must finally arrive at some other pair(x1 t1) (minusx1 t1) sub Sntimes I but there is nowhere to arrive t1 = 1 is impossible becauseof the assumption (3) t1 = 0 would mean that the pair (x1minusx1) is the same as (x0minusx0)but with reversed order The latter is impossible because if (x0 0) and (minusx0 0) areconnected by a component of Z then the antipodal action (x t) 7rarr (minusx t) would have afixed point in this component which is wrong

Thus the assumption was wrong and we conclude that gminus1(0) is nonempty

Let us state another similar theorem

Theorem 22 Any odd map g Sn rarr Sn has odd degree

Proof The proof follows from taking quotient by the antipodal action RP n = SnZ2

and considering the induced map gprime RP n rarr RP n One may check that the mapgprimelowast H1(RP n) rarr H1(RP n) is an isomorphism Then from the explicit description of thecohomology Hlowast(RP nF2) = F2[w](wn+1) it follows that gprime induces an isomorphism inmodulo 2 cohomology and therefore its degree is odd

The above theorem has the following corollary due to H Hopf [Hopf44]

Theorem 23 Let M be a compact n-dimensional Riemannian manifold and δ gt 0 is apositive real number For any map f M rarr Rn there exist two points x y isinM connectedby a geodesic of length δ such that f(x) = f(y)

The proof is left to the reader Hint Consider the point x isin M such that f(x) isthe extremal point of the image f(M) Then for every direction ν isin TxM consider thegeodesic `(t ν) from x in the direction of ν and assuming the contrary construct twohomotopic maps from the set of directions (identified with Snminus1) to Snminus1 one of thembeing odd and the other being non-surjective (and therefore having zero degree)

For more information about the BorsukndashUlam theorem the reader is referred to thebook of Matousek [Mat03]

3 The ham sandwich theorem and its polynomial version

Now we are ready to prove the classical lsquoham sandwichrsquo theorem [ST42 Ste45]

Theorem 31 Let micro1 micron be probability measures in Rn that attain zero on everyhyperplane Then some hyperplane H partitions Rn into a pair of halfspaces H+ and Hminus

so that microi(H+) = microi(H

minus) = 12 for any i

Proof We put A = Rn to Rn+1 as the affine hyperplane defined by xn+1 = 1 Then forany unit vector ν isin Sn the inequality (ν y) ge 0 defines a halfspace H+

ν in A with thecomplement Hminusν For ν equal to (0 0plusmn1) those halfspaces become degenerate thatis coinciding with the empty set of with the whole A

Now we consider the map f Sn rarr Rn defined as follows

f(ν) = (micro1(H+ν ) micro1(H+

Be Theorem 21 there exist a pair νminusν isin Sn with microi(H+ν ) = microi(H

+minusν) = microi(H

minusν ) for any i

Since the total measure of A is 1 with respect to each microi we obtain microi(H+ν ) = microi(H

minusν ) =

12 for any i

Now we are going to consider more general partitions of the space We start from thesimplest case of the line R and consider the space of univariate polynomials of degree atmost d which we denote by Pd(R) For every f isin Pd(R) it is natural to consider the sets

H+f = x f(x) ge 0 and Hminusf = x f(x) le 0

We claim that for any d absolute continuous probability measures micro1 microd in R thereexists a polynomial f isin Pd(R) that splits (with R = H+

f cup Hminusf ) every measure into two

equal halves This fact is established by considering the moment map vd1 R rarr Rd thattakes t isin R to the vector (t t2 td) The images of the measures microi are defined and itis important that they attain zero in every halfspace this follows from the fact that theoriginal microi attain zero on every finite set Now we apply the ham sandwich theorem tothese measures in Rd and obtain an equipartitioning halfspace in Rd with equation

λ(x) ge 0

where λ is a linear function with possible constant term The function λ(vd1) then becomesa polynomial of degree at most d in one variable A nontrivial generalization of this one-dimensional fact for splitting into a given proportion α (1minusα) is given in [SW85] in thiscase the partitioning set has to be twice more complex than in the simple case α = 12

As an exercise the reader may try to prove another result in the line

Theorem 32 Assume f1 fn are integrable functions on the segment [0 1] Thenthere exists another function g orthogonal to every fi that only takes values plusmn1 and hasat most n discontinuity points

The general case of the polynomial ham sandwich theorem follows by considering the

Veronese map vdn Rn rarr R(d+nn ) minus 1 that takes an n-tuple (x1 xn) to the set of

all possible nonconstant monomials in xirsquos of degree at most d After counting suchmonomials we obtain

Theorem 33 Let n and d be positive integers and r =(d+nn

)minus 1 Then any r absolutely

continuous measures micro1 micror in Rn may be partitioned into equal halves simultaneouslyby a partition Rn = H+

f cupHminusf where f is a polynomial of degree at most d

This theorem has a version for partitioning finite point sets which is frequently neededin different problems

)minus 1 Then for any r finite

sets X1 Xr in Rn there exists a partition Rn = H+f cupH

minusf where f is a polynomial of

Proof Replace every point x isin Xi with a density distributed uniformly over a ball Bε(x)and sum those densities over all x isin Xi to obtain the density of the measure microi

Then apply Theorem 33 to microi and pass to the limit εrarr +0 It is easy to see that allpossible partitioning polynomials fε may be chosen to be contained in a bounded subsetof Pd(Rn) and therefore it is possible to select a limit polynomial f that will satisfy therequirements

4 Partitioning a single point set with successive polynomials cuts

In the review of Kaplan Matousek and Sharir [KMS12] the importance of the followingcorollary of the polynomial ham sandwich theorem is emphasized

Lemma 41 Let X be a finite set in Rn and r be a positive integer It is possible tofind a polynomial of degree at most Cnr

1n with the following property The set Z = x f(x) = 0 partitions Rn into connected components V1 VN so that |X cap Vi| le 1r|X|for every i

Proof We first use Theorem 34 to partition X into almost equal halves using the zero setZf1 of a linear function f1 Then we partition every part into equal halves with anotherzero set Zf2 of a function f2 which may be still chosen to be linear if n ge 2 Then we dothe same j times After that we have a collections of polynomials f1 fj and considertheir product f = f1f2 fj The zero set Zf partitions Rn into at least r = 2j connectedcomponents each containing at most 1r fraction of the set X

It remains to bound from above the degree of f On the i-th step we partitioned 2iminus1

sets and the required degree of the polynomials was at most (n2iminus1)1n The summationover i of this geometric progression gives the estimate

deg f le (nr)1n

1minus 2minus1n= Cnr

We proved the result for r powers of two for other r we could choose 2j to be the leastpower of two not less than r

Following [KMS12] we make several comments on this lemma Seemingly we parti-tioned the space into 2j parts but some parts could actually split into several connectedcomponent in that process So we actually do not control the number of parts The otherissue is that some points of X (and actually many of them) can lie on the set Zf and needa separate treatment in most applications

One may consider a simpler approach that gives partition into convex parts with largerintersection with lines We may partition a measure into equal halves with a line inarbitrary direction Then we can partition both parts simultaneously into equal quartersby the ham sandwich theorem on this step the partitioning line is unique Therefore weobtain a partition into 4 equal parts such that any line intersects (essentially intersectsin the interior) at most 3 of them Iterating this procedure hierarchically in k steps wepartition a measure into N = 4k parts and it is easy to see that any line intersects at most3k = N log 3 log 4 of them This estimate is asymptotically worse than the one obtainedwith polynomial cuts but is has an advantage that the parts are convex

When trying to generalize the above example to higher dimensions and intersectionswith hyperplanes we see that it is not trivial to find a convex equipartition of a singlemeasure so that every hyperplane does not intersect at least one of them in the interiorThe corresponding result is known as the YaondashYao theorem [YY85]

Theorem 42 It is possible to partition an absolutely continuous finite measure in Rn

into 2n equal convex parts so that any hyperplane does not intersect the interior of at leastone of the parts

Sketch of the proof We are not going to make the full proof because it is quite technicaland hard to visualize we only sketch the main ideas instead

The two-dimensional case is already proved Then we make induction on the dimensionand try to find a partition which is a twisted (in some sense) partition into coordinateorthants We select the basis e1 en and partition the measure micro into equal halves

with a hyperplane H perpendicular to e1 Then we consider all possible unit vectors vsuch that (v e1) gt 0 and project both halves of the measure onto H along v For everyone of the halves we obtain an (nminus 1)-dimensional YaondashYao partition and it is possibleto prove by induction that it is unique once the basis e2 en in H is selected

Then a version of the Brouwer fixed point theorem (a similar fact is Lemma 92 below)helps to prove that for some v the centers of the YaondashYao partitions for the two halvescoincide and give the new center c

This gives the required partition because every hyperplane H prime is either parallel to Hin this case everything is clear or intersects H in an affine (nminus 2)-subspace One of therays c+ tvtgt0 and cminus tvtgt0 is not touched by H prime and we select the corresponding halfH+ or Hminus let it be H+ without loss of generality Applying the inductive assumption tothe projection of micro|H+ along v and the intersection H capH prime we find a part in H+ whoseinterior is not intersected by H prime

To finish the proof one has to prove that the YaondashYao center c is defined uniquely bymicro and e1 en We omit these details

Then it is easy to iterate and obtain a partition into N equal convex parts so that any

hyperplane intersects at most Nlog(2nminus1)

log 2n of them in interiors The polynomial partitioncan give a better result it is possible to partition a measure into N equal parts so thatany hyperplane intersects at most O(N

nminus1n ) of them see [KMS12] for the details But for

the polynomial partition the parts may be non-convex and even not connected so convexpartitions are still useful in some cases The reader is referred to the paper of Bukh andHubard [BH11] for an interesting application of the YaondashYao theorem

5 The SzemeredindashTrotter theorem

We are going to apply Lemma 41 and deduce the SzemeredindashTrotter theorem aboutthe number of incidences between points and lines We start from the definition

Definition 51 Let P be a set of points and L be a set of lines in the plane Denote byI(PL) their incidence number that is the number of pairs (p `) isin P timesL such that p isin `

Theorem 52 In the plane I(PL) le C(|P |23|L|23 + |P | + |L|) for a suitable absoluteconstant C

Remark 53 This theorem is also valid for pseudolines that is subsets of the place thatbehave like lines in terms of their intersection Such a generalization so far seems to beout of reach of algebraic methods

Proof We start from a much weaker estimate which we are going to apply to differentsets of points and lines

Lemma 54 I(PL) le |L|+ |P |2

This lemma is proved by splitting L into two families one for the lines intersecting atmost one point of P and all other lines The details are left to the reader

Now we put m = |P | and n = |L| Take some parameter r whose value we will definelater We choose an algebraic set Z of degree O(

radicr) (from now on we use the notation

O(middot) to avoid different constants) that partitions R2 into connected sets V1 VN Let P0 = P cap Z and Pi = P cap Vi for any i Also denote by L0 the lines in L that

lie entirely on Z and denote by Li the lines in L that intersect Vi Note that Lirsquos arenot disjoint and |L0| le degZ = O(

radicr) The crucial fact is that every line from L L0

intersects Z in at most O(radicr) points and intersects at most O(

radicr) of the regions Vi

First we obviously estimate

I(P0 L0) le m|L0| = O(mradicr)

I(P0 Li) = nO(radicr) I(Pi L0) = 0

Summing up those obvious estimates we obtain O((m+n)radicr) in total It remains to use

Lemma 54 and boundsumi

I(Pi Li) lesumi

|Li|+ |Pi|2 le nO(radicr) +m2r

Now we make several observations The projective duality allows us to interchangepoints and lines and assume m le n Then Lemma 54 allows us to concentrate on the

caseradicn le m le n After that putting r = m43

n23 we make all the estimates made so far to

be of the form O(n23m23)

An interesting application of the SzemeredindashTrotter theorem is the sum-product es-timates We quote the simplest of them due to G Elekes For a subset A of a ringput

A+ A = a1 + a2 a1 a2 isin A A middot A = a1a2 a1 a2 isin A

Theorem 55 For any finite subset A of R we have

|A+ A| middot |A middot A| ge C|A|52

for an absolute constant C

Proof Consider the set of points in R2

P = (b+ c ac) a b c isin A

and the set of lines

L = y = a(xminus b) a b isin AObviously |L| = |A|2 and every ` isin L contains at least |A| points of P with given a andb and variable c Hence

|I(PL)| ge |A|3Now by the SzemeredindashTrotter theorem

|P | middot |L| ge C|I(PL)|32rArr |P | middot |A|2 ge C|A|92

and therefore |P | ge C|A|52 But P sube (A + A) times A middot A and we obtain the requiredinequality

Theorem 55 implies that at least one of the cardinalities |A + A| or |A middot A| is at leastradicC|A|54 The exponent 54 can be slightly improved through a more careful counting

see the details in [TaoVu10 Section 83] Actually P Erdos and E Szemeredi conjecturedthat it is possible to replace 54 with 2minus ε with arbitrarily small positive ε this remainsan open problem

6 Spanning trees with low crossing number

Sometimes it is important to estimate the number of parts for the partition in Lemma 41In order to do this we pass to the projective space and prove

Lemma 61 A hypersurface Z sub RP n of degree d partitions RP n into at most dn con-nected components

Proof Select a hyperplane H sub RP n that intersects Z generically Without loss ofgenerality assume H to be the hyperplane at infinity which is naturally RP nminus1

By the inductive assumption H intersects at most dnminus1 components of RP n Z Othercomponents C1 CN are bounded When considering the affine space Rn = RP n Hwe choose a degree d polynomial f with the set of zeros Z Every bounded componentCi has the property that on its boundary f vanishes and keeps sign in its interior Henceevery Ci has a maximum or a minimum of f in its interior Every critical point is a rootof the system of equations

partfpartx1

(x) = 0

partfpartxn

(x) = 0

These are algebraic equations of degree at most dminus1 each and therefore the system has atmost (dminus1)n solutions Of course to conclude this we need the solution set to be discreteThe general case is handled by bounding not the number of solutions but the number ofconnected components of solutions By an appropriate perturbation of the equations wemay split the components of its solution set into several isolated points each thus provingwhat we need

Now we have at most (d minus 1)n bounded components and at most dnminus1 unboundedcomponents The total number of components is therefore at most dn

For the affine case we note that every infinite component in projective setting may givetwo components in affine setting therefore in Rn Z we have at most (d minus 1)n + dnminus1

components (from the proof above) which is at most dn+1 always Hence in Lemma 41the number of components is actually of order r

For the number of connected components of the set Z itself there is a more preciseresult in the plane This number is bounded from above by the number

(degZminus1

)+1 by the

Harnack theorem [Har76] The reader may try to prove the Harnack theorem consideringthe real algebraic curve as a set of cycles on the corresponding complex algebraic curveof genus g =

(degZminus1

Now we give another application of Lemma 41 is the following theorem of Chazelleand Welzl [Wel88 Cha89 Wel92]

Theorem 62 Any finite set P sub R2 has a spanning tree T with the following propertyAny line ` (apart from a finite number of exceptions) intersects T in at most C

radic|P |

points

Proof The first observation is that it is sufficient to find an arcwise connected subset Xcontaining P and having small crossings with almost all lines Then it is easy to select atree T inside X that will still connect P and has crossings at most twice of the crossingsof X Then the edges of T may be replaced with straight line segments without increasingthe number of crossings The details of this reduction are left to the reader

Now we prove the following

Lemma 63 It is possible to find a set Y sup P with at most |P |2 connected components

and line crossing number at most Cradic|P |

The lemma is proved as follows Taking r = |P |C1 we obtain by Lemma 41 an

algebraic set Z of degree at most C2

radic|P |C1 that splits P into parts of size at most

C1 By Lemma 61 for sufficiently large C1 (but still an absolute constant) the number ofconnected components of R2 Z and therefore Z will be at most |P |2 Then in everycomponent Vi of R2 Z we have at most C1 points of P which we span by a tree Ti andattach this tree to the set Z Put Y = Z cup T1 cup middot middot middot cup TN Any line ` (apart from a finite

number of exceptions) will intersect Z at mostradic|P | times and will intersect at mostradic

|P | trees of Ti Hence this line will have at most (1 + C1)radic|P | points of intersection

with Y The lemma is provedNow we apply the lemma once then select a point in every component of Y thus

obtaining the set P2 with |P2| le 12|P | Then apply the lemma again to P2 pass toanother point set P3 with |P3| le 12|P2| and so on As it was in the proof of Lemma 41in log |P | number of steps we arrive at a connected set X = Y1cupY2cup spanning P Thenumber of crossings of X with a line ` is bounded from above by the sum of a geometricprogression with denominator 2minus12 and the leading term (1 +C1)

radic|P | so it is bounded

by Cradic|P | where C is another absolute constant

The reader is now referred to the review [KMS12] and a more advanced paper of Soly-mosi and Tao [ST12] for other interesting applications of Lemma 41

7 Counting point arrangements and polytopes in Rd

Lemma 61 of the previous section has interesting applications to estimating the numberof configurations of n points in Rd up to a certain equivalence of relation

Definition 71 Let x1 xn isin Rd be an ordered set of points We define its order typeto be the assignment of signs

sgn det(xi1 minus xi0 xid minus xi0)to all (d + 1)-tuples 1 le i0 lt middot middot middot lt id le n The configuration x1 xn is in generalposition if all those determinants are nonzero

Now we can prove

Theorem 72 The number of distinct order types for ordered sets of n points in generalposition in Rd is at most nd(d+1)n

Proof A configuration in general is characterized by nd coordinates of all its points Itsorder type depends on the signs of some

)polynomials of degree d each we denote

the product of these polynomials by P (x1 xn) this polynomial has degree d(nd+1

nd variablesIt is obvious that distinct order types of sets in general position must correspond to

distinct connected components of the zero set of P Hence by Lemma 61 and the remarkabout its affine case we have at most(

))nd+ 1 le nd(d+1)n

such connected components and order types

The above argument comes from the paper [GP86] of Jacob Eli Goodman and RichardPollack After that they easily observe that the number of combinatorially distinct sim-plicial polytopes on n vertices in dimension d is also at most nd(d+1)n The generalizationof this result to possibly not simplicial polytopes and some other improvements can befound in the paper [Alon86] of Noga Alon

In the case of d = 4 these results bound the number of polytopal triangulations of the3-sphere on n vertices Curiously (see [NW13] and the references therein) it is possible toconstruct much more triangulations of the 3-sphere of n vertices and conclude that mostof them are not coming from any 4-dimensional polytope

8 Chromatic number of graphs from hyperplane transversals

A wide range of applications of the appropriately generalized BorsukndashUlam theoremstarted when Laszlo Lovasz proved [Lov78] an estimate on the chromatic number of acertain graph known as the Kneser conjecture Let us give some definitions we denotethe segment of integers 1 n by [n]

Definition 81 Let G = (VE) be a graph with vertices V and edges E Its chromaticnumber χ(G) is the minimum number χ such that there exists a map V rarr [χ] sendingevery edge e isin E to two distinct numbers (colors)

Definition 82 The Kneser graph K(n k) has all k-element subsets of [n] as the sets ofvertices V and two vertices X1 X2 isin V form an edge if they are disjoint as subsets of [n]

A simple argument left as an exercise to the reader shows that it is possible to colorK(n k) in n minus 2k + 2 colors in a regular way The opposite bound χ(K(n k)) ge n minus2k + 2 was much harder to establish In order to prove it we follow the approach ofDolrsquonikov [Dol94] and first prove the hyperplane transversal theorem

Theorem 83 Let F1 Fn be families of convex compacta in Rn Assume that ev-ery two sets CC prime from the same family Fi have a common point Then there exists ahyperplane H intersecting all the sets of

Proof For every normal direction ν every set C isin⋃iF gives a segment of values of the

product ν middot x for x isin C For a given family Fi all these segments intersect pairwise andtherefore have a common point of intersection Let this point be di In other words allsets of Fi intersect the hyperplane

Hi(n) = x ν middot x = diActually the values di can be chosen to depend continuously on ν (if we choose the

middle of all candidates for example) and by definition they are also odd functions of nThe combinations d2 minus d1 dn minus d1 make an odd map Snminus1 rarr Rnminus1 which must takethe zero value according to the BorsukndashUlam theorem

Hence for some direction ν we can put d1 = d2 = middot middot middot = dn and the correspondinghyperplane will intersect all members of the family

Now we prove

Theorem 84 Let n ge 2k For the Kneser graph we have χ(K(n k)) = nminus 2k + 2

Proof The upper bounds is already mentioned so we prove the lower bound Assumethe contrary and consider a coloring of the graph in nminus 2k + 1 or less colors

Put the n vertices of the underlying set (from Definition 82) as a general position finitepoint set X in Rnminus2k+1 For any color i = 1 nminus2k+1 let Fi consist of all convex hullsof the k-element subsets of X that correspond to the ith color of the coloring of K(n k)Since the coloring is proper any two k-element subsets of X with the same color have acommon point and therefore these families Fi satisfy the assumptions of Theorem 83

Hence there is a hyperplane H touching every convex hull of every k-element subsetof X But this is impossible H can contain itself at most n minus 2k + 1 points of X fromthe general position assumption of 2k minus 1 remaining points some k must lie on one sideof H and therefore the convex hull of this k-tuple is not intersected by H This is acontradiction

It is in fact possible to find a much smaller induced subgraph of K(n k) having thesame chromatic number

Definition 85 Assume the ground set of n elements is arranged in a circle Its subset iscalled a Schrijver subset if it contains no two consecutive elements in the circular orderThe Schrijver graph S(n k) is a the subgraph of K(n k) induced on Schrijver sets

Theorem 86 Let n ge 2k For the Schrijver graph we have χ(S(n k)) = nminus 2k + 2

Proof We still need to use Dolnikovrsquos hyperplane transversal theorem but in a moresophisticated manner

Let the ground set X = [n] be identified with a subset of the circle S1 = R2πZ LetV be the space of trigonometric polynomials of order le k minus 1

kminus1sumj=1

(aj cos jx+ bj sin jx)

We will consider the restriction of functions on the circle S1 to the set X the set of allfunctions f X rarr R is then naturally identified with Rn A nonzero function in V cannothave more than 2kminus2 zeros in the circle and therefore its restriction to X is also nonzeroHence we may assume that V sub Rn and dimV = 2k minus 1 Let W be the orthogonalcomplement of V in Rn thus dimW = nminus 2k + 1

We have a natural map

I P rarr W lowast p 7rarr (f 7rarr f(p))

As in the proof of Theorem 84 we assume a coloring of the Schrijver k-tuples in nminus2k+1colors so that every two k-tuples of the same color intersect The images of the k-tuplesunder I have the property that every two k-tuples of the same color intersect and thenumber of colors equals the dimension of W lowast By Theorem 83 there exists a hyperplanein W lowast that intersects all the convex hulls of images of Schrijver k-tuples It is given bythe equation f = a where f isin (W lowast)lowast = W

Without loss of generality assume a ge 0 Look at the elements of X whose imageunder I lies in the halfspace f lt a we want to find a Schrijver k-tuple of such elementscontradicting the fact that f = a intersects all the convex hulls of Schrijver k-tuplesIn fact we will be done if we find a Schrijver k-tuple of the elements of X where f lt 0

Consider the subset N = p isin X f(p) lt 0 If this subset consists of at least ksegments of consecutive points (in the circular order) then we can take a point from eachsegment and they will make a Schrijver k-tuple Otherwise it is possible to have preciselykminus1 segments [u1 v1] [ukminus1 vkminus1] of the circle with endpoints not in X such that thepoints of X in these segments are precisely the points of N (some segments may containno point from X and are added just to have k minus 1 segments in total)

The system of 2k minus 2 equations

g(u1) = g(v1) = middot middot middot = g(ukminus1) = g(vkminus1) = 0

has a nonzero solution in V since dimV gt 2kminus 2 Since g cannot have more than 2kminus 2zeros it only changes sign at the points ui vi Hence we may choose g positive outsidethe union of the segments [u1 v1] [ukminus1 vkminus1] and negative inside the segments Bythe definition of N the product g(p)f(p) is non-negative on X and is positive on N Incase N = empty the product must also be positive on some point p isin X since f represents anonzero element of W In any casesum

pisinP

g(p)f(p) gt 0

that contradicts the orthogonality of g isin V and f isin W

Theorem 83 can also be generalized the following way

Theorem 87 Let F0 Fk be families of convex compacta in Rn Assume that everyn minus k + 1 or less number of sets from the same family Fi have a common point Thenthere exists a k-dimensional affine subspace L intersecting all the sets of

The proof of this result [Dol94] uses a more advanced topological fact Any nminusk sectionsof the canonical k-dimensional bundle over the real Grassmannian Gnk have a commonzero We sketch the proof as follows For a generic set of nminus k sections there is an oddnumber of such common zeros This is established by considering a certain ldquolinearrdquo set ofsections with precisely one nondegenerate common zero and then showing that the parityof the number of common zeros does not depend on the choice of the generic set of nminus ksections of the bundle This is essentially the same argument that shows that the degree(modulo 2) of a proper map between manifolds of the same dimension is well-defined

Arguing like in the proof of Theorem 84 it is possible to establish some results aboutchromatic numbers of certain hypergraphs see [ABMR11] for example Though the resultsfor hypergraphs are not so precise as in the case of graphs

More systematic topological approach of [Lov78] (see also the book [Mat03]) to thechromatic number of graphs works as follows For any two graphs GH consider thehomomorphisms G rarr H that is maps between the vertex sets V (G) and V (H) sendingevery edge of G to an edge of H There is a natural way to consider such homomorphismsas the vertex set of a simplicial (or cellular) complex Hom(GH)

Now we check that the definition of the chromatic number of G reads as the smallestχ such that there exists a homomorphism from G to Kχ where Kχ is the full graphof χ vertices Such a homomorphism induces a morphism of simplicial complexes c Hom(I2 G)rarr Hom(I2 Kχ) where I2 is the segment graph with two vertices and an edgebetween them Now the crucial facts are

bull We can interchange the vertices of I2 thus obtaining an involution on both thesimplicial complexes Hom(I2 G) and Hom(I2 Kχ)bull The complex Hom(I2 Kχ) has as faces pairs F1 F2 of disjoint subsets of [χ] and

therefore is combinatorially the (χminus1)-dimensional boundary of the crosspolytope(the higher dimensional octahedron) in Rχ This is the same as the (χ minus 1)-dimensional sphere with the standard involutionbull In some cases it is possible to map a sphere Sn of dimension larger than χ minus 1

to Hom(I2 G) so that this map respects the involution on the sphere Sn and onHom(I2 G) In particular it is possible when the simplicial complex Hom(I2 G)is (χminus 1)-connectedbull Then the existence of the map c Hom(I2 G) rarr Hom(I2 Kχ) commuting with

involution contradicts the BorsukndashUlam theorem

For more information the reader is referred to the book [Mat03]

9 Partition into prescribed parts

Here we stop using the ham sandwich theorem and introduce another technique ofmeasure partitions that has some useful consequences

Let us introduce the notion of a generalized Voronoi partition The standard Voronoipartition is defined as follows Start from a finite point set x1 xm sub Rn and put

Ri = x isin Rn forallj 6= i |xminus xi| le |xminus xj|

The sets Ri give a partition of Rn into convex parts A generalization of a Voronoipartition is obtained when we assign weights wi to every point xi and put

Ri = x isin Rn forallj 6= i |xminus xi|2 minus wi le |xminus xj|2 minus wj

In this case the inequalities in the definition are actually linear because the both partshave the same quadratic in x term |x|2

Having noticed the linear nature of the generalized Voronoi partition we may redefineit as follows Let us work with a finite dimensional linear space V and its dual V lowast For afinite set of linear forms λ1 λm isin V lowast and weights w1 wm isin R put

Ri = x isin Rn forallj 6= i λi(x) + wi ge λj(x) + wjDefinitely every generalized Voronoi partition has this kind of representation and thereader may check that the converse is also true With such a definition it is clear that theregions Ri are projections of facets of the convex polyhedron G+ sub V timesR defined by thesystem of linear inequalities

y ge λ1(x) + w1

y ge λm(x) + wm

We are going to establish the following fact

Theorem 91 Let micro be a probability measure on V that attains zero on every hyperplaneλ1 λm isin V lowast be a system of linear forms and α1 αm be positive integers with unitsum Then there exists weights w1 wm isin R such that the generalized Voronoi partitionRi corresponding to λi and wi has the following property

micro(Ri) = αi

Before proving it we exhibit an appropriate topological tool

Lemma 92 Assume that a continuous map f ∆rarr ∆ of the n-dimensional simplex toitself maps every face of ∆ to itself Then the map f is surjective

Proof We prove a stronger assertion the map f of the pair (∆ part∆) to itself has degree1 by induction

Since every facet parti∆ (for i = 0 n) is mapped to itself with degree 1 then therestriction of f to part∆ has degree 1 This means that the map flowast Hnminus1(part∆)rarr Hnminus1(part∆)is the identity From the long exact sequence of reduced homology groups

0 = Hn(∆) minusminusminusrarr Hn(∆ part∆)δminusminusminusrarr Hnminus1(part∆) minusminusminusrarr Hnminus1(∆) = 0

we obtain that flowast Hn(∆ part∆) rarr Hn(∆ part∆) also must be an isomorphism The lemmais proved

Proof of Theorem 91 Consider the barycentric coordinates t1 tm in an (m minus 1)-dimensional simplex ∆ Define the map f ∆rarr ∆ as follows For a point (t1 tm) con-sider the set of weights (minus1t1 minus1tm) and let f(t1 tm) be the set (micro(R1) micro(Rm))that corresponds to the partition Ri with given weights

It is easy to check that when some tirsquos vanish and we substitute wi = minusinfin the definitionremains valid and the map f is continuous up to the boundary of ∆ Moreover when tivanishes and wi turns to minusinfin the corresponding region Ri becomes empty and thereforef maps faces of ∆ to faces Now we apply Lemma 92 and conclude that the point(α1 αm) must be in the image of f

Lemma 92 also allows to prove several classical results Here is the Brouwer fixed pointtheorem

Theorem 93 Every continuous map f from a convex compactum to itself has a fixedpoint that is a point x such that f(x) = x

Proof Topologically every convex compactum is homeomorphic to a simplex ∆ of appro-priate dimension So we assume f ∆rarr ∆ We also assume that the center of ∆ is theorigin and its diameter is 1

If f(x) is never equal to x then from compactness |f(x) minus x| ge ε for some positiveconstant ε Now we replace f with another map

f(x) = (1minus ε2)f(x)

which still has no fixed points and maps ∆ to its interior Next we construct the con-tinuous map g ∆rarr ∆ as follows take the ray ρ(x) from f(x) towards x and mark the

first time it touches the boundary part∆ this is the point g(x) The assumptions that f hasno fixed points and maps the simplex to its interior guarantee that g(x) is continuousFrom the construction it is clear that g(x) = x for any x isin part∆ Therefore Lemma 92 isapplicable and g must be surjective But the construction also implies that its image ispart∆ and therefore we have a contradiction

Another application is the classical KnasterndashKuratowskindashMazurkievicz theorem

Theorem 94 Consider the n-dimensional simplex ∆ with facets part0∆ partn∆ AssumeXini=0 is a closed covering of ∆ such that any Xi does not intersect its respective parti∆Then the intersection

⋂ni=0Xi is not empty

Hint Replace the covering with the corresponding continuous partition of unity

10 Monotone maps

Theorem 91 has the following interpretation Let micro be a probability measure on V zero on hyperplanes and ν be a discrete measure on V lowast assigning to every λi from thefinite set λi sub V lowast the measure αi Then the convex piecewise linear function

u(x) = sup1leilem

(λi(x) + wi)

has the following property The (discontinuous) map f V rarr V lowast defined by f x 7rarr duxtransports the measure micro to the measure ν

Using the approximation of arbitrary measures by discrete measures we may replace νwith any ldquoreasonably goodrdquo measure on V lowast and obtain the following theorem

Theorem 101 For two absolutely continuous probability measures micro on V and ν on V lowast

there exits a convex function u V rarr R such that the map f x 7rarr dux sends micro to ν

The maps given by x 7rarr dux with a convex u are called monotone maps More pre-cise claims about the assumptions on micro and ν and continuity of the resulting map f inTheorem 101 can be found in the beautiful review [Ball04] If the map f is continuouslydifferentiable then its differential Df turns out to be a positive semidefinite quadraticform and the conclusion that micro is sent to ν just means that ρmicro = detDf middot ρν for thecorresponding densities of the measures

From the general properties of convex functions one easily deduces that

(101) 〈xminus y f(x)minus f(y)〉 ge 0

for a monotone map and the inequality becomes strict if x 6= y ρmicro is everywhere positiveand ρν is defined

Another description of a monotone map (see [Ball04] for details) for micro and ν is a mapthat sends micro to ν and maximizes the following ldquocost functionrdquo

C(f) =

〈x f(x)〉 dmicro

Also it is possible to describe the function u and its polar w as the solution to the linearoptimization problemint

u dmicro+

intV lowastw dν rarr min under constraints forallx isin V y isin V lowast u(x) + w(y) ge 〈x y〉

In the paper of M Gromov [Grom90] we find an example of an explicit constructionof a monotone map Let a measure micro be supported in a convex compactum K so thatconv suppmicro = K sub V and put for any k isin V lowast

u(k) = log

e〈kx〉 dmicro(x)

It can be checked by hand that the differential map f(k) = duk V lowast rarr V mapsV lowast precisely to the relative interior of K and its differential Df is a symmetric positivesemidefinite matrix which is positive definite if K has nonempty interior Therefore f isinjective and the volume of K is found as

volK =

intV lowast

detDf dk

This formula may be used as a starting point in studying mixed volumes see Section 13By straightforward differentiation we observe a curious fact The values f(k) and Df(k)are the first and the second moment of the measure e〈kx〉micro after its normalization

11 The BrunnndashMinkowski inequality and isoperimetry

An interesting application of monotone maps (following [Ball04]) is

Theorem 111 (The BrunnndashMinkowski inequality) Let A and B be open subsets of Rn

of finite volume each then

vol(A+B)1n ge volA1n + volB1n

where A+B denotes the Minkowski sum a+ b a isin A b isin B of A and B

Proof Consider the probability measures micro and ν distributed uniformly on A and Brespectively Let f be the monotone map between them this means that detDfx = VBVAat any point x isin A where we put VA = volA and VB = volB for brevity

Note that by (101) and the remark after it the map g(x) = x+ f(x) is injective on Aand therefore

vol(A+B) geintA

detDg dx =

det(id +Df(x)) dx

Now we are going to use the inequality det(X + Y )1n ge detX1n + detY 1n for positivesemidefinite n times n matrices (the BrunnndashMinkowski inequality for matrices) its proof issimply achieved by diagonalization and is left to the reader We conclude that

det(id +Df(x)) ge

and therefore

vol(A+B) ge

VA =(V

1nA + V

which is what we need

The most famous consequence of the BrunnndashMinkowski theorem is the isoperimetricinequality For an open subset A sub Rn we want to define its surface area volnminus1(partA)Several definitions are possible but for A with piecewise smooth boundary the followingformula (the Minkowski surface area) works

volnminus1(partA) = limhrarr+0

vol(A+Bh)minus volA

where Bh is the ball of radius h Let vn and sn be the volume and the surface area of theunit ball B1 in Rn

Theorem 112 For reasonable A we have

volnminus1(partA)

)nminus1n

Proof From the BrunnndashMinkowski inequality we obtain

vol(A+Bh) ge (volA1n + v1nn h)n

and thereforevolnminus1(partA) ge n(volA)

nminus1n v1n

Differentiating the equality volBh = vnhn we obtain sn = nvn and therefore

volnminus1(partA) ge snvn

(volA)nminus1n v1n

which is equivalent to the required inequality

Remark 113 A more careful argument can show that the equality is attained only if Ais a ball

For another approach to the BrunnndashMinkowski inequality the reader is referred to thepaper of M Gromov [Grom90] and the post of T Tao [Tao11] That approach uses uppertriangular Jacobi matrices in place of positive semidefinite matrices and turns out to beuseful in several generalizations of the BrunnndashMinkowski inequality

Let us mention other consequences of the BrunnndashMinkowski inequality

Corollary 114 Let A and B be centrally symmetric convex bodies in Rn Then thevolume vol(A+ x) capB is maximal if x = 0

Proof Observe that

A capB supe 1

2((A+ x) capB + (Aminus x) capB)

then apply the BrunnndashMinkowsky inequality

Theorem 115 (The RogersndashShepard inequality) For any convex A

vol(Aminus A) le(

Proof Definitely AminusA = a1 minus a2 a1 a2 isin A is a projection of AtimesA sube R2n onto theRn under the map π (a b) 7rarr a minus b The fibers of this projection over x isin Rn are thesets (a1 a2) isin Atimes A such that a1 minus a2 = x that is a1 = a2 + x Hence up to translationthe fiber over x is A cap (A+ x)

When A cap (A+ x1) and A cap (A+ x2) are both nonempty then

A cap(A+

2(x1 + x2)

)supe 1

2(A cap (A+ x1) + A cap (A+ x2))

Hence vol(A cap (A+ x))1n is a concave function on AminusA If we introduce the norm xwith the unit ball Aminus A then from concavity

vol(A cap (A+ x)) ge (1minus x)n volA

Now integrate this over x to obtain

vol2nAtimes A = (volA)2 =

intAminusA

vol(A cap (A+ x)) dx ge

ge volA middotintAminusA

(1minus x)n dx = volA middotintAminusA

int (1minusx)n

1 dy dx =

= volA middotint 1

intxle1minusy1n

1 dx dy = volA middotint 1

(1minus y1n)n vol(Aminus A) dy =

= volA middot vol(Aminus A) middotint 1

(1minus y1n)n dy

Let us calculate the last integral by substitution y = tn and using the beta-functionint 1

(1minus y1n)n dy =

(1minus t)nntnminus1 dt = nB(n+ 1 n) =nΓ(n+ 1)Γ(n)

Γ(2n+ 1)=

=nn(nminus 1)

)minus1

Substituting this into the previous inequality we complete the proof

12 Log-concavity

Let us discuss logarithmically concave measures (abbreviated ldquolog-concaverdquo) and theirproperties see also [Ball04 Tao11] A log-concave measure on Rn is a measure havingdensity ρ such that log ρ(x) is a concave function though there are more general definitionsfor measures having no density Following the usual convention we allow values minusinfin forconcave functions corresponding to 0 for log-concave functions

The log-concavity is expressed by the inequality

(121) ρ((1minus t)x1 + tx2) ge ρ(x1)1minustρ(x2)t

Moreover for continuous density ρ it is sufficient to check the case t = 12 that is

(x1 + x2

)geradicρ(x1)ρ(x2)

The main result about log-concave measures is the PrekopandashLeindler inequality

Theorem 121 If a measure micro is log-concave and π Rm rarr Rn is a linear surjectionthen πlowastmicro is log-concave

After the trivial observation that a Cartesian product of log-concave measures is log-concave we immediately obtain

Corollary 122 The convolution of two log-concave measures is log-concave

Proof of Theorem 121 It sufficient to consider the case when π drops the dimension by1 and use induction Moreover since the concavity property is essentially one-dimensionalit suffices to consider the case of n = 1 and π (x y) 7rarr x where y isin R

Now we have to prove thatintRρ((1minus t)x1 + tx2 y) dy ge

(intRρ(x1 y) dy

)1minust

middot(int

Rρ(x2 y) dy

)tunder the assumption

ρ((1minus t)x1 + tx2 (1minus t)y1 + ty2) ge ρ(x1 y1)1minustρ(x2 y2)t

Put for brevity

f(y) = ρ(x1 y) g(y) = ρ(x2 y) h(y) = ρ((1minus t)x1 + tx2 y)

and also

intRf(y) dy G =

intRg(y) dy H =

intRh(y) dy

Consider the monotone map ϕ R rarr R such that is transports f(y)Fdy to g(y)Gdythat is for all y

minusinfinf(y) dy =

int ϕ(y)

minusinfing(y) dy

It follows that ϕprime(y) = f(y)Gg(ϕ(y))F

As y runs from minusinfin to +infin the value ϕ(y) does the same

therefore ψ(y) = (1minus t)y + tϕ(y) also runs monotonically from minusinfin to +infin So we write

intRh(ψ(y)) dψ(y) =

intRh(ψ(y))

(1minus t+

tf(y)G

g(ϕ(y))F

Using the assumption h(ψ(y)) ge f(y)1minustg(ϕ(y))t we obtain

H ge F 1minustGt

(f(y)G

Fg(ϕ(y))

)1minust((1minus t)g(ϕ(y))

and using the mean inequality(

(1minus t)g(ϕ(y))G

+ tf(y)F

)ge(f(y)F

)tmiddot(g(ϕ(y))G

)1minustwe conclude

H ge F 1minustGt

Fdy = F 1minustGt

Now we make several observations Every measure with constant density inside a convexbody and zero density outside is log-concave by definition So any its projection is alsolog-concave

Then we start from two convex bodies A sub Rn B sub Rn and put the into Rn+1 asAprime = Atimes 0 and Bprime = B times 1 Note that the convex hull convAprime cupBprime contains the set((1 minus t)A + tB) times t Therefore by projection to the last coordinate and log-concavitywe obtain

vol((1minus t)A+ tB) ge volA1minust volBt

this inequality is the dimension-independent version of the BrunnndashMinkowski inequalityIndeed it actually implies the standard BrunnndashMinkowski inequality as follows ReplacingA and B with their homothetic copies 1

1minustA and 1tB and using the homogeneity of the

volume we rewrite

vol(A+B) ge 1

(1minus t)(1minust)nttnvolA1minust volBt

The reader is invited to check that after finding the maximum of the right hand side in twe arrive again at the BrunnndashMinkowski inequality

If fact the inequality

(122) micro((1minus t)A+ tB) ge microA1minustmicroBt

holds for arbitrary log-concave measure micro and convex bodies A and B with almost thesame proof as for vol = micro This inequality (122) is sometimes considered as the definitionof log-concavity which is also applicable to measures with no density The reader maycheck that for a measure with density the two definitions are equivalent

Actually the case of possibly non-convexA andB in the dimension-independent BrunnndashMinkowski inequality is also valid and is contained in the following version of the PrekopandashLeindler inequality (see also [Tao11] for generalizations to nilpotent Lie groups) alsoknown as the functional BrunnndashMinkowski inequality

Theorem 123 Assume f g h are nonnegative densities in Rn such that for some t isin[0 1] and x y isin Rn

h((1minus t)x+ ty) ge f(x)1minustg(y)t

Then intRn

h(x) ge(int

f(x) dx

)1minust

middot(int

g(y) dy

Sketch of the proof One way of proving this theorem is to show that the property h((1minust)x+ ty) ge f(x)1minustg(y)t is preserved under projections dropping dimension by 1 like wedid in the proof of Theorem 121

Another approach is to consider a monotone map ϕ sending the measure f(x)dx tog(x)dx and then consider the monotone map ψ x 7rarr (1minust)x+tϕ(x) and write inequalitiesfor h((1minus t)x+ tϕ(x)) similarly to the proof of Theorem 121

Another result that benefits from log-concavity is the Minkowski theorem on facet areasWe split it into the simple direct theorem and the harder converse theorem

Theorem 124 If P sub Rn is a polytope ν1 νm are its facet normals and A1 Amare its respective facet surface areas then

(123) A1ν1 + A2ν2 + middot middot middot+ Amνm = 0

Proof Consider a constant vector field v it has zero divergence and by the Gauss theoremits flux through partP is zero That is

A1(ν1 v) + A2(ν2 v) + middot middot middot+ Am(νm v) = 0

Since v is arbitrary the result follows

Theorem 125 For any prescribed set of normals ν1 νm spanning Rn and positivefacet surface areas A1 Am satisfying (123) there exists a unique (up to translations)polytope P corresponding to these data

Sketch of the proof Introduce real variables t1 tm and consider the polytope

(124) P (t) = x isin Rn foralli = 1 m (x νi) le tiSince a positive combination of νirsquos equals zero then P (t) is always bounded The equa-tion (124) may treat xjrsquos and tirsquos as variables and define an (n + m)-dimensional poly-

hedron P this way P (t) is a parameterized n-dimensional section of P By the PrekopandashLeindler inequality the function f(t) = volP (t) is log-concave in t The

standard geometric differentiation reasoning shows that its logarithmic derivative equals

d log f(t) =1

f(t)(A1(t)dt1 + middot middot middot+ Am(t)dtm)

where Ai(t) is the surface area of the corresponding facet The map ϕ(t) = d log f(t) istherefore ldquominusrdquo monotone and its image must be convex

Definitely the image of ϕ(t) satisfies (123) (divided by f(t) = volP ) We have to checkthat for every positive vector y satisfying

(125) y1ν1 + y2ν2 + middot middot middot+ ynνn = 0

there exists a proportional vector in the image of ϕ Denote the set of nonnegative ysatisfying (125) by Q The polyhedron Q is a cone centered in 0 and from convexityof the image of ϕ and the linear programming considerations to prove what we need itis sufficient to check that every extremal ray of Q intersects the image of ϕ But suchan extremal ray corresponds to a vector y with at most n + 1 positive coordinates putI = i yi gt 0

It may happen that I contains less than n indexes We avoid such degenerate cases byperturbing slightly the set of normals ν1 νm the general case then can be deduced bygoing to the limit So it is sufficient to consider the case |I| = n + 1 It is easy to checkthat such vectors y correspond to simplices x foralli isin I (x νi) le ti that are obviouslycontained in the image of ϕ Hence all nonnegative vectors y satisfying (125) are in theimage of ϕ up to scaling

The uniqueness of P follows from the fact that the image of ϕ is essentially (m minus n)-dimensional and ϕ is monotone hence the preimage ϕminus1(y) is n dimensional and consistsof a set of polytopes taken to each other by translations

In this section we presented the analytical approach to log-concavity and the readeris invited to read the review of RP Stanley [Stan89] about algebraic point of view onlog-concavity One simple fact left as an exercise for the reader is If a univariate monicpolynomial P (x) has all roots real and negative then its coefficients from a log-concavesequence

In a wonderful way the algebraic approach returns with a proof of the AlexandrovndashFenchel inequality for mixed volumes of convex bodies which in turn gives another proofof the BrunnndashMinkowski inequality A typical example of an algebraic log-concavity resultis

Theorem 126 Let X be an n-dimensional normal projective algebraic variety and L andM be ample divisor classes on X Then the sequence v0 vn of intersection numbers

vk = (L L︸︷︷︸k

M M︸︷︷︸nminusk

is log-concave

Sketch of the proof For corresponding facts from algebraic geometry please consult thetextbook [GH78]

We have to prove the inequality

(126) (L L︸︷︷︸k

)2 ge (L L︸︷︷︸kminus1

M M︸︷︷︸nminusk+1

) middot(L L︸︷︷︸k+1

M M︸︷︷︸nminuskminus1

After multiplying L and M by some positive integers we assume that L and M correspondto projective embeddings of X By an appropriate version of the Bertini theorem wechoose the hyperplane sections (in L) for the first k minus 1 of L in the intersection formulaand the hyperplane sections (in M) for the last n minus k minus 1 of M so that X reduces to anormal surface S and over this surface we have to prove

(LM)2 ge (LL) middot(MM)

By the Hodge theorem the intersection form of an algebraic surface has positive index 1then this inequality is just the inverse CauchyndashSchwarz inequality

The general form of (126) is

(127) (L1 L2 L3 Ln)2 ge (L1 L1 L3 Ln) middot(L2 L2 L3 Ln)

which is the algebraic form of the AlexandrovndashFenchel inequality

13 Mixed volumes

Let us make a brief discussion of mixed volumes following [Grom90] The crucial factis

Theorem 131 Let K1 Kn be convex bodies in Rn the expression

vol(t1K1 + middot middot middot+ tnKn)

for nonnegative ti is a polynomial in tirsquos of degree n and the coefficient at t1 tn dividedby n is called the mixed volume of K1 Kn and is denoted by MV(K1 Kn)

Proof Consider the monotone maps fi from Rn to the respective Ki with potentials

ui(k) = log

e〈kx〉 dmicroi(x)

where microi are some measures with convex hulls of support equal to the respective Ki Themap

f(k) = t1f1(k) + tnfn(k)

is also monotone with potential u(k) = t1u1(k) + middot middot middot+ tnun(k) It is a simple fact aboutconvex functions that for continuously differentiable u(k) that image of the differential

map du(k) = f(k) is convex Therefore the image K of f(k) is convex obviously K subeK = t1K1 + middot middot middot+ tnKn

We claim that (the open convex set) K actually equals K up to boundary To provethis it is sufficient to compare their support functions

h(pK) = supxisinK〈p x〉 h(p K) = sup

xisinK〈p x〉

Take k = αp it is easy to see that for α rarr +infin the measure e〈kx〉microi(x) on Ki getsconcentrated near the points of Ki where the linear form 〈p x〉 attains its maximumHence 〈p fi(αp)〉 rarr h(pKi) for α rarr +infin Since the support functions are obviouslyadditive with respect to the Minkowski sum we obtain

〈p f(αp)〉 rarr h(pK)

when α rarr +infin Since h(p K) ge 〈p f(αp)〉 by definition there must be an equality

h(p K) = h(pK)

Now we can calculate vol K = volK

(131) vol(t1K1 + middot middot middot+ tnKn) =

det(t1Df1(k) + middot middot middot+ tnDfn(k)) dk

Obviously the determinant under the integral is a polynomial of degree n and the resultfollows

Actually the mixed volume MV(K1 Kn) is positive the reader may deduce it from(131) noting that Dfirsquos are positive definite In fact a stronger fact the AlexandrovndashFenchel inequality holds

Theorem 132 For any convex bodies K1 Kn sub Rn we have

MV(K1 K2 Kn)2 ge MV(K1 K1 K3 Kn) middotMV(K2 K2 K3 Kn)

It seems that there is no easy way to prove this inequality One way is to relate themixed volumes to intersection numbers of ample divisors over a toric variety and thenprove (127) arguing similarly to the proof of Theorem 126 see [Ful93] The other wayis to prove an analogue of this inequality for positive definite matrices and their mixeddiscriminants and then use (131) along with other nontrivial observations see [Grom90]

It makes sense to mention the Bernstein theorem [Bern76] connecting the mixed vol-umes and intersection numbers of divisors First for every Laurent polynomial in nvariables

P (x) =sumk

where we use the notation xk = xk11 xknn we define the Newton polytope

N(P ) = convk isin Zn ck 6= 0Now the theorem reads

Theorem 133 The system of equations

P1(x) = 0

P2(x) = 0

Pn(x) = 0

for x isin (Clowast)n has either an infinite number of solutions or a finite number of solutionsnot exceeding n MV(N(P1) N(Pn)) For a generic choice of the coefficients ck of thepolynomials keeping the Newton polytopes the same the number of solutions is preciselyn MV(N(P1) N(Pn))

In [Bern76] besides this fact a certain sufficient condition in terms of faces of theNewton polytopes was given that guarantees that the set of solution consists of preciselyn MV(N(P1) N(Pn)) points without using the term generic

We do not give a proof of this theorem here An elementary but technical reasoningcan be found in the original paper [Bern76] A more conceptual approach using symplec-ticKahler geometry can be found in [Ati83]

14 The BlaschkendashSantalo inequality

We give an application of the PrekopandashLeindler inequality (in form of Theorem 123)to the BlaschkendashSantalo inequality Following the paper of Lehec [Leh09A] we start fromproving the functional version of this inequality

Theorem 141 Assume f and g are nonnegative measure densities such that

f(x)g(y) le eminus(xy)

for any x y isin Rn Also assume thatintRn yg(y) dy = 0 and

intRn xf(x) dx converges Thenint

f(x) dx middotintRn

g(y) dy le (2π)n

We are going to make the proof in several steps First we observe that it is sufficientto prove the following

Lemma 142 For any nonnegative measure density f(x) with convergingintRn xf(x) dx

there exists z isin Rn such that for any other nonegative density g(x) with convergingintRn xg(x) dx the inequality

f(x+ z)g(y) le eminus(xy)

for any x y isin Rn implies intRn

f(x) dx middotintRn

g(y) dy le (2π)n

Informally the lemma replaces the mass centering condition by an arbitrary centeringcondition for one of the functions independent of the other function

Proof of Theorem 141 assuming Lemma 142 Suppose we have an appropriate z fromLemma 142 From the assumption of Theorem 141 it follows that

f(z + x)g(y)e(zy) le eminus(zy)minus(xy)+(yz) = eminus(xy)

Hence intRn

f(x) dx middotintRn

g(y)e(zy) dy le (2π)n

Let us estimate e(yz) ge 1 + (y z) and integrate this inequality after multiplying by g(y)intRn

g(y) + (y z)g(y) dy leintRn

g(y)e(zy) dy

Taking into account thatintRn yg(y) dy = 0 we obtainint

g(y) dy leintRn

g(y)e(zy) dy

which implies the required inequality

In order to prove Lemma 142 we start with proving the corresponding one-dimensionalfact

Lemma 143 Let f(x) and g(y) be nonnegative densities on the half-line R+ If f(x)g(y) leeminusxy for any x y isin R+ thenint +infin

f(x) dx middotint +infin

g(y) dy le π

Proof Put u(s) = f(es)es v(t) = g(et)et and w(r) = eminuse2r2er From the assumptions it

follows that for any s t isin R

)= eminuse

s+t2e(s+t)2 geradicf(es)g(et)e(s+t)2 =

radicu(s)v(t)

Now the one-dimensional case of Theorem 123 impliesint +infin

g(y) dy =

int +infin

minusinfinu(s) ds middot

int +infin

minusinfinv(t) dt le

le(int +infin

minusinfinw(r) dr

(int +infin

eminusz22 dz

Proof of Lemma 142 The proof is by induction Assume the normalizationintRn f(x) dx =

1 which can be achieved by multiplying f(x) by a constant and dividing g(y) by the sameconstant

The one-dimensional case follows from Lemma 143 as follows choose z to be themedian of f After a shift of z to 0 we have to prove that the inequality

f(x)g(y) le eminusxy

implies int +infin

minusinfing(y) dy le 2π

But Lemma 143 is applicable to f(x) and g(y) in the range x y gt 0 and to f(minusx) andg(minusy) in the same range Summing up the results and noting thatint 0

minusinfinf(x) dx =

int +infin

f(x) dx = 12

we obtain the required inequalityNow we use induction as follows Let micro be the measure with density f(x) We can

partition micro into equal halves H+ and Hminus with a hyperplane H orthogonal to the lastbasis vector en Since we may translate f without loss of generality we assume that H+

and Hminus are defined by xn ge 0 and xn le 0 respectivelyLet c+ and cminus be mass centers of micro|H+ and micro|Hminus After another translation of f we

also assume that the segment [cminus c+] intersects H at the origin Now let v be the vectorparallel to [cminus c+] and normalized so that (v en) = 1 Then we consider the linearoperators A and B defined by

Ae1 = e1 Aenminus1 = enminus1 Aen = v B = (Aminus1)T

so that (AxBy) = (x y) for any x y isin Rn Note that the determinants of these mapsequal 1 and A maps H to itself while B may not map H to itself Consider the functionsof xprime yprime isin H

F (xprime) =

int +infin

f(xprime + sv) ds G(yprime) =

int +infin

g(Bxprime + ten) dt

From the normalizationintHF (xprime) dx = 12 Also the mass center of F (xprime) is zero by the

selection of vector v The assumption can be rewritten

f(xprime + sv)g(Byprime + ten) le eminus(xprime+svByprime+ten) = eminus(Axprime+sAenByprime+ten) =

= eminus(xprimeyprime)minus(xprimeten)minus(sAenByprime)minusst(ven) = eminus(xprimeyprime)minusst

We now fix xprime and yprime and consider f(xprime+sv) and g(Byprime+ten)e(xprimeyprime) as functions of positivevariables s and t By Lemma 143 we obtain

F (xprime)G(yprime) le π

2eminus(xprimeyprime)

Now we invoke the inductive assumptions interchanging F (xprime) and G(yprime) taking intoaccount Lemma 142 the fact

intHxprimeF (xprime) dxprime = 0 implies thatint

F (xprime) dxprime middotintH

G(yprime) dyprime le π

2(2π)nminus1

SinceintHF (xprime) dx = 12 we obtainint

g(y) dy =

G(yprime) dyprime le π(2π)nminus1

Similarly inverting en and v we obtainintBHminus

g(y) dy le π(2π)nminus1

and it remains to sum these inequalities to obtainintRn

g(y) dy le (2π)n

Remark 144 A finer argument shows that it is possible to satisfy the assumption [cminus c+] en after a careful choice of the coordinate system In this case v = en the operators A Bequal the identity operator and the formulas get considerably simpler This assumptionis satisfied if we choose the hyperplane H = xn = 0 so that to minimize the integralint

dist(xH)f(x) dx

By varying the normal of H and its constant term we obtain thatintH+

xf(x) dxminusintHminus

xf(x) dx perp H and

f(x) dxminusintHminus

f(x) dx = 0

which is exactly what we need

Another approach (see [Leh09B]) to Lemma 142 invokes the YaondashYao theorem (The-orem 42) to partition Rn into 2n convex cones A1 A2n with center at z so thatintAif(x) dx = 2minusn for every i After translating z to the origin one observes that the

characterizing property of the YaondashYao partition means that the space Rn is covered bythe family Ai 2n

i=1 of polar conesThen one invokes the logarithmic form of the PrekopandashLeindler inequality

Lemma 145 Let f g h be nonnegative absolute integrable function on the positive coneRn

+ Assume that

h(radicx1y1

radicxnyn) ge

radicf(x)g(y)

for any x y isin Rn+ Thenint

f(x) dx middotintRn+

g(y) dy le

(intRn+

h(z) dz

Proof Substitute

f(s1 sn) = f(es1 esn)es1+middotmiddotmiddot+sn

g(t1 tn) = g(et1 etn)et1+middotmiddotmiddot+tn

h(r1 rn) = h(er1 ern)er1+middotmiddotmiddot+rn

It is easy to check that for any s t isin Rn(h

ge f(s)g(t)

Then Theorem 123 implies thatintRn

f(s) ds middotintRn

g(t) dt le(int

h(r) dr

that is equivalent to what we need

Now Lemma 145 applied to h(z) = eminus|z|22 gives for any YaondashYao cone Ai the inequalityint

minusAig(y) dy le πn

It remains to sum over i = 1 2n to prove Lemma 142Now we deduce the classical form of the BlaschkendashSantalo inequality for convex bodies

Corollary 146 Let K be a convex body in Rn with mass center at the origin at let

K = y isin Rn forallx isin K (x y) le 1be its polar body Then

volK middot volK le v2n

where vn is the volume of the unit ball in Rn

Proof Consider the corresponding ldquonormsrdquo (note that x may not coincide with minusx)x = minr ge 0 x isin rK x = minr ge 0 x isin rK

The definition of the polar body means that for any x y isin Rn

(x y) le x middot yNow we introduce two functions

f(x) = eminusx22 g(y) = eminusy

and check thatf(x)g(y) = eminusx

22minusy22 le eminus(xy)

Also f(x) has mass center at the origin Hence by Theorem 141 the product of theirintegrals is at most (2π)n Now we calculate by changing the integration orderint

f(x) dx =

int f(x)

1 dydx =

volK(minus2 log y)n2 dy = cn volK

for the constant cn =int 1

0(minus2 log y)n2 dy The same holds for g(y)int

g(y) dy = cn volK

It remains to calculate cn this is simple if we take the unit ball in place of K and theEuclidean norm |x| in place of x

(2π)n2 =

eminus|x|22 dx = cnvn

volK middot volK le (2π)n

It is a famous problem (the Mahler conjecture) to establish the lower bound for centrallysymmetric convex bodies

volK middot volK ge 4n

which is an equality for the cube and the crosspolytope Some known partial results onthis conjecture are summarized in the blog post by Tao [Tao07] The best known resultis established by Greg Kuperberg [Ku08]

(141) volK middot volK ge πn

15 Needle decomposition

Now we are going to consider an interesting tool in studying inequalities for measuresthe needle decomposition The reader is referred to [NSV02] for a deeper review of thissubject

The main result is the following theorem

Theorem 151 For any two absolutely continuous finite measures micro and ν on Rn sup-ported in a bounded convex set S and a prescribed ε gt 0 it is possible to partition S intosome number of convex pieces P1 PN so that for every Pi

micro(Pi)

micro(Rn)=

ν(Pi)

ν(Rn)

and one of the alternatives hold Either Pi can be included into a ε-neighborhood of a lineor not But the total measure micro + ν of the parts satisfying the latter alternative is lessthan ε

A piece Pi in this theorem has small deviation from a line so it looks like an almost1-dimensional ldquoneedlerdquo This justifies the name ldquoneedle decompositionrdquo

Sketch of the proof We start from the case n = 2 We can partition the two measureswith a line into equal halves Then we can partition every part with another line so thatboth measures are partitioned into equal fourths Then we proceed this way many timesIs we impose an additional technical assumption say that the sum of densities of micro andν is separated from zero on S then it is easy to check that after an appropriate numberof steps all the parts Pi become ε-needles (ε-close to lines)

The technical assumption on positive density can be avoided as follows Take somepositive δ and break the parts Pi into two categories Those for which micro(Pi) + ν(Pi) gtδ volPi the ldquoessentialrdquo parts and all the other ldquoinessentialrdquo parts For the essential partsafter a definite number of steps depending on ε and δ we conclude that they are ε-needles For the inessential parts we observe that their total measure micro + ν is at mostδ volS which can be chosen to be less than ε

Now consider the case n gt 2 For an (n minus 2)-dimensional linear subspace L sub Rn wemake the 2-dimensional partition of the 2-plane RnL into parts so that the essential partsare εn-needles and the total measure of the inessential parts is small The correspondingpartition in Rn is a partition with hyperplanes parallel to L

Then the trick is to repeat this construction for different Lrsquos passing sufficiently closeto any given (n minus 2)-dimensional direction also requiring that the total measure of theinessential parts on the jrsquoth step is at most ε2j The measures micro and ν get partitionedinto many equal pieces and it is possible to check that most of the parts cannot beessentially two-dimensional (contain a two-dimensional disk of a prescribed size) whilethe inessential parts together have total measure micro+ν less than ε Indeed if some essentialpart Pi has a 2-dimensional disk D of radius εn inside then we take the orthogonal tothe disk linear subspace LD sub Rn and observe the following Since we made hyperplanecuts almost parallel to LD on some stage this disk could not survive for the essentialparts

The useful observation is as follows For every ε-needle part Pi we choose the line`i to which it is ε-close If the measures micro were log-concave then their restrictions microiand νi to the convex body Pi remain log-concave Since Pi is almost one-dimensionalthen it makes sense to consider the projections of microi and νi to `i which become log-concave measures in the line by Theorem 121 And if the densities of micro and ν werecontinuous then this measure on the line approximates the original measures on Pi with an

appropriate precision Of course by approximating measures the assumption of continuityfor the density can be dropped in most practical applications To put it short the needledecomposition can sometimes reduce questions about pairs of log-concave measures onRn to pairs of log-concave measures in the line

There is a more sophisticated version of the needle decompositions called pancakedecomposition used for example in [Grom03] Informally when the number of measuresto partition increases then the ldquoessential dimensionrdquo of the parts increases accordingly

16 Isoperimetry for the Gaussian measure

Let us apply the needle decomposition to establish the isoperimetric inequality for aGaussian measure on Rn After some rescaling any such measure obtains the densityeminusπ|x|

2 which we like for its simplicity and normalization

intRn e

minusπ|x|2 dx = 1 The cru-cial property of the Gaussian measure is that the density of its orthogonal projection isproportional to restriction of its density to any line

Since the notion of the surface area does not make much sense for Gaussian measureswe formulate the ldquointegratedrdquo form of the isoperimetric inequality

Theorem 161 For a Gaussian measure micro on Rn consider an open subset U and ahalfspace H of the same measure micro(H) = micro(U) Then for their ε-neighborhoods we have

micro(Uε) ge micro(Hε)

Remark 162 Note that the value micro(Hε) depends on micro(H) and ε and does not depend onthe dimension n

So we see that for the standard measure in Rn the ldquomost roundrdquo body is the ball andfor the Gaussian measure the ldquomost roundrdquo body is the halfspace

Sketch of the proof Let U = Rn U be the complement of U After making the needledecomposition for two restrictions micro|U and micro|U we reduce the problem to a one-dimensionalproblem preserving the equality

micro(U cap Pi)micro(Pi)

= micro(U) = micro(H)

The measure micro on Pi after projection to `i becomes a log-concave measure with densityρ Moreover it is strongly log-concave in the following sense

ρ((1minus t)x1 + tx2) ge ρ(x1)1minustρ(x2)teπt(1minust)|x1minusx2|2

It is easy to check that the proof of Theorem 121 can be modified to prove that the stronglog-concavity is preserved under projections

Now everything reduces to the following

Lemma 163 Let micro be the probability measure on the line with density eminusπx2

and ν bea strongly log-concave probability measure on the line Consider an open subset U and ahalfline H such that micro(H) = ν(U) Then for their ε-neighborhoods we have

ν(Uε) ge micro(Hε)

The proof of this lemma is left as a technical exercise for the reader It makes sense toapproximate U with a union of intervals and then show that these intervals can be movedto the left or to the right decreasing (ν(Uε))

primeε Finally all the intervals can be merged

into a single one either (minusinfin t) or (t+infin) In this case it is easy to verify that the value

(ν(Uε))primeε = ρ(t) is at least eminusπx

2 where

ν(U) =

minusinfineminusπξ

17 Isoperimetry and concentration on the sphere

It is interesting and important that on the round sphere Sn sube Rn+1 the correspondingversion of Theorem 161 also holds

Theorem 171 For a the uniform probability measure σ on Sn consider an open subsetU and a halfspace H sub Rn+1 such that σ(HcapSn) = σ(U) Then for their ε-neighborhoods(in the geodesic path metric of Sn) we have

σ(Uε) ge σ((H cap Sn)ε)

This means in particular that the minimal (nminus1)-dimensional volume of partU for givenσ(U) is attained at spherical caps like H cap Sn We do not give a proof here because itdepends on some deep results in Riemannian geometry see the appendix to [MS86] byM Gromov for example The crucial notion here is the Ricci curvature of the Riemannianmetric which equals nminus 1 for the unit n-dimensional round sphere

Another heuristic evidence for Theorem 171 is that for large dimensions n the Gauss-ian measure with density eminusπ|x|

2on Rn gets concentrated near the round sphere of radiusradic

nπ After rescaling byradicπn we see that it ldquoapproachesrdquo the measure σ and therefore

the isoperimetric inequality for the sphere is ldquoasymptoticallyrdquo correctThe isoperimetric inequality has the following famous consequence known as the con-

centration of measure phenomenon on the sphere

Theorem 172 Let U sub Sn have measure σ(U) ge 12 Then the measure of the neigh-borhood σ(Uε) becomes almost 1 for ε of order

radicn in particular the following estimate

σ(Uε) ge 1minus eminus(nminus1)ε2

In [Led01] it is shown that this theorem can be deduced from the isoperimetry of theGaussian measure using some analytical technicalities Following [GM2001] we give asimple proof of a weaker fact

Proof of a weaker assertion Consider the complement V = Sn Uε note that the spher-ical distance between U and V is at least ε Now we pass to subsets of the unit ball Bn+1

defined as follows

U0 = x isin Bn+1 x|x| isin U V0 = x isin Bn+1 x|x| isin V for their volumes we have (v = vn+1 is the volume of the unit ball here)

volU0 = vσ(U) volV0 = vσ(V )

Consider the set X = 12(U0 + V0) simple trigonometry shows that X consists of vectors

with lengths at most cos ε2 and therefore

volX le v cosn+1 ε

(1minus sin2 ε

)n+12 le veminus

(n+1) sin2 ε2

The BrunnndashMinkowski inequality in particular gives volV0 le volX and therefore

vσ(V ) le veminus(n+1) sin2 ε

which implies

σ(Uε) ge 1minus eminus(n+1) sin2 ε

A usual way to use the concentration on the sphere is the following (Levyrsquos lemma)

Corollary 173 Let f be a 1-Lipschitz (|f(x)minus f(y)| le dist(x y)) function on Sn thenthe measure of the set where f(x) differs from its median value Mf is estimated by

σx |f(x)minusMf | ge ε le 2eminus(nminus1)ε2

Informally on most of the sphere the function f differs from Mf by at most O( 1radicn)

Much more information about the concentration phenomenon on the sphere and manyother metric-measure spaces can be found in the book [Led01]

18 More remarks on isoperimetry

There are many other instances of the isoperimetric inequality Here we just give abrief sketch of them for more details see the book [Lub94]

First the isoperimetric inequality on Riemannian manifolds is connected through theCheegerndashBuser inequalities to the smallest positive eigenvalue of the Laplace operator(the Laplacian) Let us give some explanations without proofs The Laplace operator fordifferential forms on a smooth closed manifold M is defined as

∆ = ddlowast + dlowastd

where dlowast is the formal adjoint to the d operator on differential forms It is easy to checkthe characteristic property of the Laplace operatorint

(ω∆ω)ν =

|dω|2ν +

|dlowastω|2ν

where ν is the volume form associated with the Riemannian structure For functions thisreduces to ∆f = dlowastdf and int

f∆fν =

|df |2ν

From this formula (more precisely for its version with f∆g) it is clear that ∆ is self-adjoint and non-negative Moreover since we assume M to be compact and connectedthe only harmonic functions (ie satisfying ∆f = 0) are constant functions Thereforeon the subspace

L20(M) =

f isin L2(M)

fν = 0

the Laplace operator is strictly positive By some standard tools of functional analysis itcan be shown that ∆minus1 is a compact operator on L2

0(M) and the smallest eigenvalue of∆|L2

0(M) makes sense and is denoted by λ1(M) Therefore λ1(M) is the largest positivenumber satisfying the followingint

|df |2ν ge λ1(M) middotintM

|f |2ν for all f such that

fν = 0

The connection to the isoperimetric inequalities now becomes more clear since for apartition M = AcupB into two subsets we can consider a functions almost constant on Aalmost constant on B and changing in a reasonable way near the boundary partA = partBAfter normalizing it to have a zero mean we may apply the definition of λ1(M) anddeduce some kind of isoperimetric inequality showing that partA = partB is sufficiently large

It is also possible to show the converse By considering the sublevel sets Mc = x isinM f(x) le c and applying a certain kind of isoperimetric inequality to them a lowerbound for λ1(M) follows by careful integration over c

One particular case when the above construction simplifies greatly is the case of theisoperimetry on graphs which is the main topic of the book [Lub94] First the isoperi-metric constant of a graph G = G(E V ) is the number c such that

|E(AB)| ge cmin|A| |B|for every partition of the set of vertices V = A cup B where E(AB) denotes the set ofedges between A and B It can be easily proved that c(G) can be related to the smallestpositive eigenvalue of the graph Laplacian this is left as an exercise for the reader Thegraph Laplacian is defined as follows For a function f on vertices we put

∆f(y) = deg yf(y)minussum

(xy)isinE

The only functions corresponding to the zero eigenvalue are constants and it is easy toshow that ∆ is self-adjoint and positive on the functions with zero mean The correspond-ing minimal positive eigenvalue is denoted by λ1(G)

Now we make the following definition

Definition 181 A family of graphs Gn is called a family of expanders is |V (Gn)| rarr infin asnrarrinfin the degrees of their vertices are uniformly bounded deg v le k for any v isin V (Gn)and their isoperimetric constants c(G) (or the numbers λ1(G)) are uniformly bounded bya positive number c

Informally the expander graphs are quantitatively more than connected while havingthe bounded degree of any vertex The theory of expander graphs is large and interestingso here we only mention the main facts without proofs

First if we try to construct a bipartite graph on two sets of size n and connect anyvertex on the left hand side to some k vertices on the right hand side randomly thenas n tends to infinity the probability to have of the inequality c(G) ge c gt 0 tends to 1if c lt k2 This means that appropriately constructed ldquorandom graphsrdquo are expandersand this fact has serious practical applications

Second an explicit construction of expanders is rather difficult The most usual wayis to consider an infinite discrete group Γ with a set of generators S and build its Cayleygraph G(Γ S) Then take the set of quotients ΓNn by a family of normal subgroupssuch that |ΓNn| rarr infin and consider the induced graph on ΓNn It turns out that someproperties of the group Γ and its representations on the Hilbert space (the ldquoKazhdanproperty Trdquo or other similar properties) allow to prove that the constructed family ofgraphs is a family of expanders For the details (and many other interesting stuff like theBanachndashTarski paradox) the reader is referred to the already mentioned book [Lub94] orother sources

19 Sidakrsquos lemma

Here we are going to prove a useful fact about Gaussian measures known as Sidakrsquoslemma We define a centrally symmetric strip in Rn as the set S = x |λ(x)| le w forsome λ isin Rnlowast and w isin R+

Theorem 191 Let A be a centrally symmetric convex body in Rn and S be a centrallysymmetric strip in Rn Then for a Gaussian probability measure micro we have

micro(A) middot micro(S) le micro(A cap S)

The key idea of the proof is to introduce a new probability measure ν(X) = micro(XcapS)micro(S)

The inequality that we want to obtain is formalized in the following

Definition 192 Suppose micro and ν are two probability measures on Rn We say that νis more peaked than micro if for any centrally symmetric convex body A

ν(A) ge micro(A)

Now the proof of Theorem 191 consist of two lemmas

Lemma 193 If ν is more peaked than micro then for any centrally symmetric log-concavedensity f we have int

f dν geintRn

f dmicro

Proof Rewrite the integralintRn

f dν =

int f(x)

1 dνdy =

int f(x)

ν(Cy)dy

where Cy = x f(x) le y For log-concave and centrally symmetric f the sets Cy areconvex and centrally symmetric so the inequality follows by integration in y

Lemma 194 If the measure ν is more peaked than micro as measures on Rn and τ is afinite centrally symmetric log-concave measure on Rm then ν times τ is more peaked thanmicrotimes τ on Rn+m

Proof Denote by x and y the points in Rn and Rm respectively Consider a centrallysymmetric convex body A sub Rn+m The measure 1times τ is log-concave and its restrictionτ prime to A is also log-concave and centrally symmetric Hence by Theorem 121 the density

f(x) =

inty(xy)isinA

is log-concave and centrally symmetric since it is the density of the projection of τ prime toRn

Now we observe that

ν times τ(A) =

f(x) dν geintRn

f(x) dmicro = microtimes τ(A)

by Lemma 193

And the proof of Theorem 191 is complete by the following obvious

Lemma 195 If micro is the Gaussian measure on the line and S is a centrally symmetricsegment then the measure defined by

ν(X) =micro(X cap S)

micro(S)

is more peaked than micro

Then it suffices to take the Cartesian product of the result of this last lemma with theGaussian measure on Rnminus1 to obtain Theorem 191

Remark 196 Note that it is conjectured (the Gaussian correlation conjecture) that The-orem 191 can be generalized to the case of two arbitrary centrally symmetric convexbodies A and B with the same inequality micro(A) middot micro(B) le micro(A capB)

Another result related to the notion of ldquomore peakedrdquo is the lower bound for the sectionvolume of the unit cube Qn = [minus12 12]n

Theorem 197 Let L be a k-dimensional linear subspace of Rn Then the k-dimensionalvolume of the section L capQn is at least 1

Remark 198 The upper bounds on the section volume are harder to prove It is knownthat any hyperplane section of Qn has area at most

radic2 This result was extended to lower

values of k but for some pairs (k n) the precise upper bound is still not known See thereview of K Ball [Ball01] for the details of this and many other interesting facts

Proof Consider the uniform measure ν on Qn and compare it with the Gaussian measuremicro with density eminusπ|x|

2 Actually ν is more peaked than micro the proof is reduced to the

one-dimensional case by applying Lemma 194 and noting that any product of two suchmeasures (in possibly different dimensions) is log-concave

Now take an ε-neighborhood Lε of L By the definition of ldquomore peakedrdquo we obtain

ν(Lε) ge micro(Lε)

This can be decoded as

vol(Lε capQn) geintBnminusk

eminusπ|x|2

where Bnminuskε is the (nminus k)-dimensional ball of radius ε The right hand side for εrarr +0 is

asymptotically vnminuskεnminusk And the trick is that the k-dimensional volume of L capQn may

be defined (following Minkowski) to be

volk L capQn = limεrarr+0

vol(L capQn)εvnminuskεnminusk

It remains to note that the difference between vol(L cap Qn)ε and volLε cap Qn is o(εnminusk)(the proof is left to the reader) and the result then follows

20 Centrally symmetric polytopes

A usual way to apply the Sidak lemma (Theorem 191) is to analyze the behavior ofa centrally symmetric convex polytope in terms of its facets Any such polytope K isdefined by a system of inequalities

|(ni x)| le wi i = 1 N

where ni are unit normal vectors to facets of K and wi are positive reals Each suchinequality defines a strip and therefore Theorem 191 is applicable For example thefollowing lemma holds

Lemma 201 Assume a centrally symmetric convex polytope K sub Rn contains the unitball B and has 2N facets Then it intersects more than a half of the sphere S of radius

cradic

(for sufficiently large N) where c gt 0 is some absolute constant

Sketch of the proof Choose a Gaussian measure with density(απ

)n2eminusα|x|

2 An easy es-

timate using integration by parts shows that any strip Pi defined by |(ni x)| le wi hasmeasure (here we use wi ge 1)

micro(Pi) geint 1

minus1

radicα

πeminusαx

dx ge 1minus 1radicπα

eminusα

It is known (for example from the direct calculation of moments with gamma functionand the Chebyshev inequality) that this Gaussian measure is mostly concentrated aroundthe radius |x| =

radicnα

By Theorem 191

micro(K) = micro(P1 cap middot middot middot cap PN) ge micro(P1) middot middot middot middot middot micro(PN) ge(

1minus 1radicπα

eminusα)N

so in order to prove the lemma (that K intersects more than a half of a sphere of radius

cradic

) we have to take α of order logN and check that micro(K) is greater than some

absolute positive constant that is(1minus 1radic

παeminusα)Nge c2

(201) N log

(1minus 1radic

παeminusα)ge c3

for a negative constant c3 Then we observe the value α = c logN with some c lt 1satisfies this inequality for sufficiently large N

Using Lemma 215 from the next section we conclude (the FigielndashLindenstraussndashMilmantheorem)

Theorem 202 Let K be a centrally symmetric convex polytope with 2N facets and 2Mvertices then

logN middot logM ge γn

for some absolute constant γ gt 0

Proof By Lemma 215 we assume that K contains the unit ball B and is contained inthe ball

radicnB The dual body Klowast defined by

Klowast = x isin Rn forally isin K (x y) le 1is therefore contained in B and contains 1radic

nB By Lemma 201 K intersects more than a

half of the sphere of radius r = cradic

and Klowast intersects more than a half of the sphere

of radius rlowast = cradic

Note that the inequality rlowastr le 1 is what we need to prove Assume the contraryrlowastr gt 1 Then for any x with |x| = r consider its renormalized xlowast = rlowast x|x| Note that it

cannot happen that x isin K and xlowast isin K simultaneously because (x xlowast) gt 1 Hence eitherK intersects at most half of the sphere Sr or Klowast intersects at most half of the sphere Srlowast which is a contradiction

Small values of N and M not suitable for Lemma 201 may be considered separatelyin a similar fashion

In fact the above lemma and theorem can be proved without the Sidak lemma byestimating the surface areas of spherical caps The idea is that a spherical cap may bevery close to a halfsphere in terms of distance and still have very small surface area Thenwe observe that the complement to a strip on a sphere is a pair of caps of small areaand if the sum of all areas is at most half of the area of the sphere then Lemma 201is established This cap area estimate corresponds to the value α = c logN while thestronger estimate (201) was not fully used in the above proof

However the approach with caps requires writing down and estimating some integralsof trigonometric functions while the Sidak lemma allows us to work with simple expo-nential expressions Which is more important a careful use of the Sidak lemma allowedA Barvinok [Barv11] to prove the following strengthening of Theorem 202 Under theassumption of the above theorem the inequality N le αn implies M ge eβn for somepositive β = β(α)

Another result estimates from below the volume of a centrally symmetric polytopecontaining the unit ball

Lemma 203 Assume a centrally symmetric convex polytope K sub Rn contains the unitball B and has 2N facets Then its volume is at least(

Proof The intersection of K with the ball bounded by the sphere from 201 is at leasthalf of this ball by volume

Corollary 204 Let K sub Rn be a polytope with N vertices on the unit ball B Then

volKge(

Proof We can replace K with convK cup minusK to make it centrally symmetric At thisstep the volume of K increases and the volume of K decreases So assume K centrallysymmetric with 2N vertices Note that the polar K has 2N facets and then

(volK)2

volK middot volKge (volK)2

where we used Corollary 146 and Lemma 203

The same approach with more care allows to prove that logN can be replaced withlog(Nn)+1 as was shown by Gluskin in [Glu1988] Barany and Furedi in [BF87] proveda similar result using different technique and deduced the following Suppose we have ablack box that for any point x isin Rn either asserts that x is inside the unknown convexbody K or provides a separating hyperlane for K and x From the above results weconclude that an algorithm estimating the volume of K with such a black box needs ahuge number of queries to the black box But it turns out that practically the volumecan be efficiently estimated with a randomized algorithm

Another famous problem about centrally symmetric polytopes is to prove that the totalnumber of faces of all dimensions (including the polytope itself) is at least 3n The readermay easily check that this bound is attained for the cube or its dual the crosspolytopeIn the case of simple or simplicial polytopes Stanley proved [Stan87] this conjecture usingthe correspondence between the linear space spanned by the faces of a polytope up to acertain equivalence relation and the cohomology ring of the corresponding algebraic (toric)variety Let us give a sketch for those knowing or willing to learn some toric geometryThe symmetry of a polytope makes an action of the involution ι on the polytope Pand its corresponding toric variety XP Let T = (Clowast)n be the torus acting on XP Then it remains to compare the two descriptions of the T -equivariant cohomology of XP HlowastT (XP ) = Hlowast(XP ) otimes Hlowast(BT ) from the collapsing spectral sequence on the one handand HlowastT (XP ) is the StanleyndashReisner ring of P that is something defined combinatoriallyby P From this it follows that the Hilbert series of the trace of ι on those spaces satisfies

h(ιHlowastT (XP )) = h(ιHlowastT (XP )) middot h(ιHlowast(BT )) =h(ιHlowastT (XP ))

(1 + t)n= 1

Hence h(ιHlowast(XP )) = (1 + t)n and from this it is possible to deduce the lower bounds onthe numbers of faces calculated from hP (t) = h(ιHlowast(XP )) see the details in [Stan87]and [Cox11]

21 Dvoretzkyrsquos theorem

We are going to discuss one of the most famous applications of the concentration phe-nomenon on the sphere the Dvoretzky theorem

Theorem 211 For a positive real ε and a positive integer k there exists another positiveinteger n(k ε) with the following property If middot is any norm on Rn then there exists aEuclidean norm | middot | on Rn and a k-dimensional linear subspace L sube Rn such that

|x| le x le (1 + ε)|x|

The proof needs several lemmas of which we prove all but one First we will needδ-nets on the sphere Skminus1

Definition 212 A finite set X sub Skminus1 is a δ-net if for every x isin Skminus1

dist(xX) le δ

In what follows we assume for simplicity that δ le π4 and assume k to be sufficientlylarge

Lemma 213 There exists a δ-net in Skminus1 of size at most k4kminus1

δkminus1

Proof Find an inclusion maximal set of disjoint spherical caps of radius δ2 (balls ingeodesic metric) in Skminus1 Let X be the set of their centers Since we cannot add anyother spherical cap of radius δ2 to the set every point of Skminus1 is at distance at most δfrom some x isin X and therefore X is a δ-net

Now we estimate |X| comparing the total surface area of the caps which is at least|X|vkminus1(δ4)kminus1 (here we use that δ is not too big) to the surface area of the spheresk = kvk Hence

|X| le kvk4kminus1

vkminus1δkminus1le k4kminus1

δkminus1

here we use that vk le vkminus1 for k ge 6 which can be seen from the explicit formula

vk = πk2

Γ(k2+1)

Lemma 214 If X is a δ-net in Skminus1 and δ lt π4 then every xprime isin Skminus1 can be expressedas a positive linear combination of vectors from X with sum of coefficients at most

radic2

Proof Let us prove that the ball Bprime of radius 1radic

2 is contained in convX Assuming thecontrary by the HahnndashBanach theorem we obtain a vector y with |y| = 1

radic2 such that

the setHy = x (x y) ge (y y)

has no common interior point with convX Then it is easy to see that the normalizedy|y| is at geodesic distance at least π4 gt δ from X

So we conclude that Skminus1 is insideradic

2 convX which is equivalent to what we need

Lemma 215 Let K be a centrally symmetric convex body in Rn Then there exists acentrally symmetric ellipsoid E such that

E sube K suberadicnE

Proof Take as E the ellipsoid of maximal volume contained in K the John ellipsoidAfter a linear transformation assume that E is a unit ball If there exist a point x isin Kwith |x| gt

radicn than it can be shown by a straightforward calculation that after stretching

E in the direction of x and shrinking it in the orthogonal to x directions E can increaseits volume while remaining inside the set conv(E cup x cup minusx) sube B

Now in Theorem 211 we consider some norm on Rn By Lemma 215 we choose theellipsoid E and define | middot | to be the Euclidean norm with E as the unit ball Then weconsider f(x) = x as a function of the sphere Snminus1 the conclusion of Lemma 215 readsas

1radicnle f(x) le 1

on the sphere The condition f(x) le 1 (using the triangle inequality for the norm) meansthat f is 1-Lipschitz on Snminus1 Let M be the median of f For the following lemma weonly give a sketch of a proof

Lemma 216 Under the above assumptions on middot and | middot | generated by the Johnellipsoid the median M of middot has the lower bound

radiclog n

with some absolute constant c

See the discussion of this lemma in [Ball97] A different approach to this difficultyis to select an orthonormal (relative to | middot |) base ei such that ei ge nminusi

n(this is the

DvoretzkyndashRogers lemma) then choose the linear span of the first n2 of these vectorsas the new space to work in and establish an analogue of Lemma 216 for this subspaceThis is established using averaging of the expression x1e1 + middot middot middot+ xnen over all possiblechanges of signs of the coordinates xi along with the triangle inequality for the normThe details are given for example in [Led01 Section 35]

Proof of Theorem 211 assuming Lemma 216 By Corollary 173 (here σ is the probabil-ity measure on Snminus1) for the set

C = x isin Snminus1 |x minusM | geMε8we have

σ(C) le 2eminus(nminus2)M2ε2

128 le 2eminusc2ε2 logn

Now we take δ = ε4 and choose a δ-net X on Skminus1 the latter is considered as the unitsphere of the coordinate subspace Rk sub Rn Applying a random rotation ρ to X we see

the following For any xi isin X the probability of the event xi isin C is at most 2eminusc2ε2 logn

128 If in total

2eminusc2ε2 logn

128 |X| lt 1

then for some random rotation the whole X gets inside Snminus1 C Then we choose L tobe the image ρ(Rk) and denote by S(L) the unit sphere of L From the bound on |X| ofLemma 213 the inequality condition is satisfied when

k16kminus1

εkminus1eminus

c2ε2 logn128 lt 12

which is indeed true for sufficiently large nBy Lemma 214 we conclude that for any x isin S(L) after a rescaling making the median

M equal 1

x le sumxiisinX

cixi leradic

2 maxxiisinXxi le

radic2(1 + ε8)

Hence again using the triangle inequality for the norm middot on S(L) has Lipschitz constantat most

radic2(1 + ε8) It follows that the values of middot on S(L) are between

(1minus ε8)minus ε4 middotradic

2(1 + ε8)

and(1 + ε8) + ε4 middot

radic2(1 + ε8)

For sufficiently small ε after a slight rescaling of middot we obtain the inequality

|x| le x le (1 + ε)|x|for any x isin L

22 Topological and algebraic Dvoretzky type results

It was noted in [Mil88] that a simple proof for the Dvoretzky theorem would followfrom the following topological conjecture of Knaster [Kna47]

Conjecture 221 For any finite set X sub Snminus1 with |X| = n and any continuous functionf Snminus1 rarr R it is possible to find a rotation ρ such that

f(ρx1) = middot middot middot = f(ρxn)

where x1 xn are the points of X

Some cases of this conjecture were confirmed when X is an orthonormal basis whenn = 3 when |X| is prime and the points of X form a two-dimensional regular polygonsee the discussion in [DK11] for more details and references But unexpectedly in thepaper of Kashin and Szarek [KS03] a counterexample was constructed using some knowl-edge about sections of a cube and similar things A greatly simplified exposition of thiscounterexample is given in [Mat10 Miniature 32]

Of course if the conjecture were true we could take a δ-net on Skminus1 as X of size at most(4δ

)kby Lemma 213 Then assuming Conjecture 221 for n ge

)kwe could rotate X

and rescale the Euclidean norm to have the equality x = |x| for any x isin X The restwould follow from Lemma 214

This approach could still pass if we establish the weak Knaster conjecture with a rea-sonable estimate for the function n(m)

Conjecture 222 There exists a function n = n(m) with the following property Forany finite set x1 xm sub Snminus1 of size k and any continuous function f Snminus1 rarr R itis possible to find a rotation ρ such that

f(ρx1) = middot middot middot = f(ρxm)

About this generalized conjecture almost as little is known as about the original oneIn fact already for m = 4 the existence of n(m) is only known for some very particulartypes of sets x1 xm

Another direction of Dvoretzky-type results is the ldquoalgebraic Dvoretzky theoremrdquo from [DK11]

Theorem 223 For an even positive integer d and a positive integer k there exists n(d k)such that for any homogeneous polynomial f of degree d on Rn where n ge n(d k) thereexists a linear k-subspace L sube Rn such that f |L is proportional to the d2-th power of thestandard quadratic form

Q = x21 + x2

2 + middot middot middot+ x2n

This theorem is very similar to the Dvoretzky theorem but unlike the latter it givesa precisely ldquoroundrdquo section for a polynomial Unfortunately the topological tools usedin its proof do not give any explicit bound on n(d k) Moreover it is not clear how todeduce the Dvoretzky theorem from this result even if the bound were reasonable

On the other hand for odd degree d it is proved with elementary topology [DK11] thatany homogeneous polynomial of degree d in n variables vanishes on some linear k-subspace

with n = k +(d+kminus1d

) This fact is originally due to BJ Birch [Bir57] who established it

by algebraic tools with a worse estimate for n(d k)Let us sketch the proof of the Birch theorem for a homogeneous polynomial P of degree

d in n variables The space of all orthonormal k-frames in Rn is parameterized by theStiefel manifold Vnk For any frame (e1 ek) the polynomial

Pe(t1 tk) = P (t1e1 + t2e2 + middot middot middot+ tkek)

is a degree d homogeneous polynomial in k variables The space W of all such polynomialshas dimension

(d+kminus1d

) The correspondence e 7rarr Pe makes an odd map from Vnk to W

Here odd is understood for Vnk so that along with a frame (e1 ek) we can considerthe frame

minus(e1 ek) = (minuse1 minusek)Then one form of the generalized BorsukndashUlam theorem (see [Mat03]) asserts that someframe is mapped to zero (that is exactly what we need) provided the space Vnk is

(dimW minus 1)-connected This is indeed the case when n ge k +(d+kminus1d

A similar proof works for the following result about ldquomaking a convex function sym-metricrdquo

Theorem 224 Let X sub Skminus1 be a centrally symmetric subset with |X| = 2m Ifn ge m + k and f Snminus1 rarr R is a continuous function then it is possible to find anisometric copy ρ(X) sub Snminus1 such that for any x isin ρ(X)

f(x) = f(minusx)

References

[Alon86] N Alon The number of polytopes configurations and real matroids Mathematika 3362ndash711986

[ABMR11] JL Arocha J Bracho L Montejano JL Ramırez Alfonsın Transversals to the convex hullsof all k-sets of discrete subsets of Rn Journal of Combinatorial Theory Series A 118(1)197ndash2072011

[Ati83] MF Atiyah Angular momentum convex polyhedra and algebraic geometry Proceedings of theEdinburgh Mathematical Society (2) 26(2)121ndash133 1983

[Ball97] K Ball An elementary introduction to modern convex geometry Flavors of Geometry MSRIPublications 31 1997

[Ball01] K Ball Convex geometry and functional analysis Handbook of the Geometry of Banach Spaces161ndash194 2001

[Ball04] K Ball An elementary introduction to monotone transportation Geometric Aspects of Func-tional Analysis Lecture Notes in Mathematics 1850 99ndash106 2004

[BF87] I Barany Z Furedi Computing the volume is difficult Discrete and Computational Geometry2(1)319ndash326 1987

[Barv11] A Barvinok A bound for the number of vertices of a polytope with applications Arxiv preprintarXiv11082871 2011

[Bern76] DN Bernstein The number of roots of a system of equations Functional Anal Appl 9(3)183ndash185 1976

[Bir57] BJ Birch Homogeneous forms of odd degree in a large number of variables Mathematika 4102ndash105 1957

[Bor33] K Borsuk Drei Satze uber die n-dimensionale euklidische Sphare Fundam Math 20(1)177ndash1901933

[BH11] B Bukh A Hubard Space crossing numbers Arxiv preprint arXiv11021275 2011[Cha89] B Chazelle E Welzl Quasi-optimal range searching in spaces of finite VC-dimension Discrete

Comput Geom 4467ndash489 1989[Cox11] D Cox J Little H Schenck Toric Varieties Graduate Studies in Mathematics 124 American

Mathematical Society 2011[Dol94] VL Dolrsquonikov Transversals of families of sets in Rn and a connection between the Helly and

Borsuk theorems Sb Math 79(1)93ndash107 1994

[DK11] VL Dolrsquonikov RN Karasev Dvoretzky type theorems for multivariate polynomials and sectionsof convex bodies Geometric And Functional Analysis 21(2)301ndash318 2011

[Ful93] W Fulton Introduction to toric varieties Annals of Mathematics Studies 131 Princeton Uni-versity Press Princeton NJ 1993

[GM2001] A A Giannopoulos V D Milman Euclidean structure in finite dimensional normed spacesHandbook of the Geometry of Banach Spaces 1707ndash779 2001

[Glu1988] ED Gluskin Extremal properties of orthogonal parallelepipeds and their applications to thegeometry of Banach spaces (in Russian) Mat Sbornik 136(178)85ndash96 1988

[GH78] P Griffiths J Harris Principles of algebraic geometry Wiley-Interscience New YorkndashChichesterndashBrisbanendashToronto 1978

[GP86] J E Goodman R Pollack Upper bounds for configurations and polytopes in Rd DiscreteComput Geom 1219ndash227 1986

[Grom90] M Gromov Convex sets and Kahler manifolds IHES wwwihesfr gromovPDF[68]pdf1990

[Grom03] M Gromov Isoperimetry of waists and concentration of maps Geometric and FunctionalAnalysis 13178ndash215 2003

[Guth13] L Guth Unexpected applications of polynomials in combinatoricsmathmitedu lguthExpositionerdossurveypdf 2013

[Har76] C G A Harnack Uber Vieltheiligkeit der ebenen algebraischen Curven Math Ann 10189ndash1991876

[Hopf44] H Hopf Eine Verallgemeinerung bekannter Abbildungs und Uberdeckungssatze PortugaliaeMath 4129ndash139 1944

[KMS12] H Kaplan J Matousek M Sharir Simple proofs of classical theorems in discrete geometry viathe GuthndashKatz polynomial partitioning technique Discrete Comput Geom 48(3)499ndash517 2012

[KS03] B S Kashin S J Szarek The Knaster problem and the geometry of high-dimensional cubesComptes Rendus Mathematique 336(11)931ndash936 2003

[Kna47] B Knaster Problem 4 Colloq Math 3030ndash31 1947[Ku08] G Kuperberg From the Mahler conjecture to Gauss linking integrals Geom Funct Anal

18(3)870ndash892 2008[Led01] M Ledoux The concentration of measure phenomenon Amer Math Soc 2001[Leh09A] J Lehec A direct proof of the functional Santalo inequality Comptes Rendus Mathematique

347(1)55ndash58 2009[Leh09B] J Lehec Partitions and functional Santalo inequalities Archiv der Mathematik 92(1)89ndash94

2009[Lov78] L Lovasz Kneserrsquos conjecture chromatic number and homotopy Journal of Combinatorial The-

ory Series A 25319ndash324 1978[Lub94] A Lubotzky Discrete groups expanding graphs and invariant measures Birkhauser 1994[Mat03] J Matousek Using the Borsuk-Ulam theorem Lectures on topological methods in combinatorics

and geometry Springer Verlag 2003[Mat10] J Matousek Thirty-three miniatures Mathematical and algorithmic applications of linear alge-

bra Student Mathematical Library 53 American Mathematical Society Providence RI 2010[MS86] V D Milman G Schechtman Asymptotic theory of finite dimensional normed spaces Lecture

Notes in Mathematics 1200 Springer Verlag 1986[Mil88] V D Milman A few observations on the connections between local theory and some other fields

Geometric Aspects of Functional Analysis Lecture Notes in Mathematics 1317 283ndash289 1988[NSV02] F Nazarov M Sodin A Volrsquoberg The geometric Kannan-Lovasz-Simonovits lemma

dimension-free estimates for the distribution of the values of polynomials and the distribution ofthe zeros of random analytic functions St Petersbg Math J 14(2)351ndash366 2002

[NW13] E Nevo S Wilson How many n-vertex triangulations does the 3-sphere have Arxiv preprintarXiv13111641 (2013)

[Sid67] Z Sidak Rectangular confidence regions for the means of multivariate normal distributionsJournal of the American Statistical Association 62(318)626ndash633 1967

[ST12] J Solymosi and T Tao An incidence theorem in higher dimensions Discrete Comput Geom48(2)255ndash280 2012

[Stan87] RP Stanley On the number of faces of centrally-symmetric simplicial polytopes Graphs andCombinatorics 3(1)55-66 1987

[Stan89] RP Stanley Log-concave and unimodal sequences in algebra combinatorics and geometryAnnals of the New York Academy of Sciences 576500ndash535 1989

[Ste45] H Steinhaus Sur la division des ensembles de espace par les plans et des ensembles plans par lescercles Fundam Math 33245ndash263 1945

[ST42] A H Stone and J W Tukey Generalized ldquosandwichrdquo theorems Duke Mathematical Journal9(2)356ndash359 1942

[SW85] W Stromquist and D Woodall Sets on which several measures agree J Math Anal Appl108241ndash248 1985

[ST83] E Szemeredi and W T Trotter Extremal problems in discrete geometry Combinatorica 3381ndash392 1983

[Tao07] T Tao Open question the Mahler conjecture on convex bodiesterrytaowordpresscom20070308open-problem-the-mahler-conjecture-on-convex-bodies2007

[TaoVu10] T Tao V H Vu Additive combinatorics Cambridge studies in advanced mathematics 1052010

[Tao11] T Tao The BrunnndashMinkowski inequality for nilpotent groupsterrytaowordpresscom20110916the-brunn-minkowski-inequality-for-nilpotent-groups 2011

[Wel88] E Welzl Partition trees for triangle counting and other range searching problems Proc 4thAnnu ACM Sympos Comput Geom 23ndash33 1988

[Wel92] E Welzl On spanning trees with low crossing numbers Data structures and efficient algorithmsLecture Notes in Computer Science vol 594 Springer Verlag 1992 233ndash249

[YY85] A C Yao F F Yao A general approach to d-dimensional geometric queries Proceedings ofthe seventeenth annual ACM symposium on Theory of computing STOC 85 New York NY USAACM 1985 163ndash168

Roman Karasev Dept of Mathematics Moscow Institute of Physics and TechnologyInstitutskiy per 9 Dolgoprudny Russia 141700

Roman Karasev Institute for Information Transmission Problems RAS Bolshoy Karetnyper 19 Moscow Russia 127994

Roman Karasev Laboratory of Discrete and Computational Geometry YaroslavlrsquoState University Sovetskaya st 14 Yaroslavlrsquo Russia 150000

E-mail address r n karasevmailru

URL httpwwwrkarasevruen

1 Introduction

5 The SzemereacutedindashTrotter theorem

10 Monotone maps

12 Log-concavity

13 Mixed volumes

14 The BlaschkendashSantaloacute inequality

19 Šidaacuteks lemma

21 Dvoretzkys theorem