Equivariant Methods -...

Equivariant MethodsSinisa Vrecica

University of Belgrade

Computational Applications of Algebraic Topology

Topological methods in combinatorics,computational geometry, and the study of

algorithms

MSRI, October, 2006.

EQUIVARIANT METHODS – P. 1/45

G-spaces and equivariant maps

◮ X is aG-space ifG < HomeoX, i.e. if there is afamily of homeomorphisms of the spaceXdenoted by the elements of the groupG whichcompose as the elements of groupG multiply.We say thatG acts onX.

G-spaces and equivariant maps

◮ X is aG-space ifG < HomeoX, i.e. if there is afamily of homeomorphisms of the spaceXdenoted by the elements of the groupG whichcompose as the elements of groupG multiply.We say thatG acts onX.

◮ f : X → Y is an equivariant map (or aG-map) if(∀g ∈ G)(∀x ∈ X)f(g · x) = g · f(x),i.e. if f commutes with the action of the elementsof the groupG (we say thatf preservesG-symmetry).

Borsuk-Ulam theorem

◮ THEOREM: For every continuous mapf : Sn → R

n there is x ∈ Sn such thatf(−x) = f(x).

Borsuk-Ulam theorem

◮ THEOREM: For every Z/2-equivariant mapf : Sn → R

n there is x ∈ Sn such that f(x) = 0.

Borsuk-Ulam theorem

◮ THEOREM: For every Z/2-equivariant mapf : Sn → R

n there is x ∈ Sn such that f(x) = 0.

◮ THEOREM: There is no Z/2-equivariant mapf : Sn → Sn−1.

Borsuk-Ulam theorem, continued

◮ one of the most often applied topological resultsin geometry and combinatorics

◮ HAM -SANDWICH THEOREM: For every nmeasurable sets in R

n, there is a hyperplanedividing each of them in two parts of the samemeasure.

◮ HAM -SANDWICH THEOREM: For every nmeasurable sets in R

n, there is a hyperplanedividing each of them in two parts of the samemeasure.

◮ The excellent survey on the Borsuk-Ulamtheorem and its applications in Combinatoricsand Geometry is given inJ. MATOUŠEK, Using the Borsuk-Ulam Theorem,Springer, 2003.

Some other applications

◮ L. Lovász established the Kneser conjecture fromCombinatorics in 1978 showing:

THEOREM: However we divide the family of alln-subsets of the set of2n + k elements in thek + 1 subfamilies, there is a pair of disjointn-subsets in some of the subfamilies.

Some other applications

◮ L. Lovász established the Kneser conjecture fromCombinatorics in 1978 showing:

THEOREM: However we divide the family of alln-subsets of the set of2n + k elements in thek + 1 subfamilies, there is a pair of disjointn-subsets in some of the subfamilies.

◮ The proof is based on the extremely cleverapplication of the Borsuk-Ulam theorem.

Some other applications, continued

◮ I. Bárány, S. Shlosman and A. Szücs extended in1981 the Tverberg theorem from affine to thecontinuous case proving:

◮ THEOREM: Let p be a prime integer andN = (p − 1)(n + 1). For every continuous mapf : ∆N → R

n there arep disjoint faces of∆N

whosef -images intersect.

◮ THEOREM: Let p be a prime integer andN = (p − 1)(n + 1). For every continuous mapf : ∆N → R

n there arep disjoint faces of∆N

whosef -images intersect.

◮ M. Ozaydin (later K. Sarkaria, A. Volovikov)generalized the result for prime-powers, andshowed that the method does not work in othercases.

◮ N. Alon proved in 1987 Splitting necklacestheorem.

◮ D. Kozlov, in his lecture, said more about therecent resolution of the Lovász conjecture by E.Babson and himself.

Scheme

◮ X = configuration space= space of allcandidates

Scheme

◮ (Y, A) = test space with subspace

Scheme

◮ The symmetry of the problem induces an actionof some groupG both onX and(Y, A), makingthemG-spaces.

Scheme

◮ Test mapf : X → (Y, A) which is aG-map.

Scheme

◮ Test mapf : X → (Y, A) which is aG-map.

◮ x ∈ X is a solution to the problem ifff(x) ∈ A.

Scheme, continued

◮ If we show that for everyG-mapf ,im f ∩ A 6= ∅, the problem has a solution.

Scheme, continued

◮ Most oftenY = Rn andA = {0} (a subspace or

the subspace arrangement). In the former case theproblem reduces to the question whether for aG-mapf : X → R

n, the requirement0 ∈ im fhas to be satisfied.

Scheme, continued

◮ Most oftenY = Rn andA = {0} (a subspace or

the subspace arrangement). In the former case theproblem reduces to the question whether for aG-mapf : X → R

n, the requirement0 ∈ im fhas to be satisfied.

◮ SinceRn \ {0} ≃ Sn−1, this is equivalent with

the claim that there is noG-mapf : X → Sn−1.

Methods I - Dold’s theorem

◮ DOLD’ S THEOREM: If the spaceX isn-connected, the spaceY is n-dimensional andthe groupG acts freely on these spaces, thenthere is noG-map fromX to Y .

◮ Here, the spaceX is n-connected if its homologygroupsH0(X), H1(X), ..., Hn(X) and itsfundamental groupπ1(X) are trivial.

◮ The action of the groupG on the spaceX is freeif for any x ∈ X and anyg ∈ G the equalityg · x = x impliesg = e, the neutral element ofG.

Dold’s theorem, continued

◮ The Dold’s theorem reduces, in many cases, theconsidered problem to the determination ofconnectivity of the configuration space.

◮ For example, for the continuous Tverbergtheorem, the configuration space was a cellcomplex whose cells were products ofp disjointsimplices of∆N (deleted product), and the keytechnical step in the proof was the establishing ofhigh connectivity of this complex.

◮ Using deleted joins instead of deleted products,K. Sarkaria showed that the proof could besubstantially simplified, but this could not bedone in some other examples.

◮ In more complicated cases high connectivity ofthe configuration space could be establishedusing the nerve lemma or some spectral sequence.

◮ For the colored Tverberg theorem, theconfiguration space was the join of chessboardcomplexes and it was important to determinetheir connectivity.

◮ COLORED TVERBERG THEOREM: Givenn + 1collections of2p − 1 pointsC0, C1, ..., Cn in R

n,there arep simplices with vertices of differentcolors, which intersect.

◮ Configuration space is([2p − 1]∗(d+1))∗pδ = ([2p − 1]∗pδ )∗(d+1).

◮ [2p − 1]∗pδ is an(p − 2)-connected chessboardcomplex. This object is important in Discrete andComputational Geometry, Representation Theoryetc.

◮ Actually, we obtained more, the continuousversion of the theorem.

◮ Using the above, it is shown that the theorem istrue for non-primep with 4p − 3 points of eachcolor.

Methods II - characteristic classes

◮ Sometimes a vector spaceY is naturally assignedto everyx ∈ X in such way that we obtain thevector bundle overX. It is appropriate in suchcases (instead of aG-map missing the origin) toconsider nowhere-zero section of that bundle.

Methods II - characteristic classes

◮ Sometimes a vector spaceY is naturally assignedto everyx ∈ X in such way that we obtain thevector bundle overX. It is appropriate in suchcases (instead of aG-map missing the origin) toconsider nowhere-zero section of that bundle.

◮ It is well known that such section does not exist ifcertain characteristic class (Euler,Stiefel-Whitney) of that bundle does not vanish.

Characteristic classes, continued

◮ The non-vanishing of some characteristic classesof the bundles over Grassmann manifold, Stiefelmanifold,... could be established.

◮ A link between equivariant maps and sections ofvector bundles is provided by:

THEOREM: Let X be a principalG-bundle overB andY aG-vector space. Then the sections ofthe induced vector bundleX ×G Y are in 1-1correspondence with theG-mapsf : X → Y .

◮ THEOREM: Let µ0, µ1, ..., µk be a collection ofσ-additive probability measures inRn. Thenthere exist ak-dimensional planeF such thatevery closed halfspaceH containingF satisfiesµi(H) ≥ 1

n−k+1 for everyi = 0, 1, ..., k.

◮ k = n − 1 is the ham-sandwich theorem, andk = 0 is the Rado’s center-point theorem.

n−k+1 for everyi = 0, 1, ..., k.

◮ k = n − 1 is the ham-sandwich theorem, andk = 0 is the Rado’s center-point theorem.

◮ The configuration space is modelled by theintersection pointsx of k-planesF with theirorthogonal complementsL ∈ Grn−k(R

n).EQUIVARIANT METHODS – P. 17/45

.....................................

◮ So, the configuration space is the total space ofthe canonical bundle over the Grassmannmanifold.

◮ By projecting the measures to(n − k)-subspaceL and using the Rado’s theorem, we see that foreveryL and every measureµi there is a pointxL

such that the planeFi containingxLi orthogonal

to L satisfies the claim of the theorem forµi.

◮ By projecting the measures to(n − k)-subspaceL and using the Rado’s theorem, we see that foreveryL and every measureµi there is a pointxL

such that the planeFi containingxLi orthogonal

to L satisfies the claim of the theorem forµi.

◮ The question is, whether for someL all thesepoints coincide.

◮ This is equivalent with the question whether forsomeL the differencesxL

1 − xL0 , ..., xL

k − xL0 all

vanish.

1 − xL0 , ..., xL

k − xL0 all

vanish.

◮ For this, it is enough to prove that the Whitneysumγ ⊕ · · · ⊕ γ of k copies of the canonicalbundle over the Grassmann manifold Grn−k(R

n)does not admit a nowhere zero section.

1 − xL0 , ..., xL

k − xL0 all

vanish.

◮ For this, it is enough to prove that the Whitneysumγ ⊕ · · · ⊕ γ of k copies of the canonicalbundle over the Grassmann manifold Grn−k(R

n)does not admit a nowhere zero section.

◮ Now, it suffices to prove thatwn(γ)k is a non-zeroelement inH∗(Grn−k(R

n); Z/2).

◮ This turns out to be equivalent with the followingalgebraic claim:

CLAIM : The monomial(t1t2 · · · tn−k)k does not

belong to the ideal generated by the symmetricmonomials in the variabless1, ..., sk, t1, ..., tn−k.

◮ This turns out to be equivalent with the followingalgebraic claim:

CLAIM : The monomial(t1t2 · · · tn−k)k does not

belong to the ideal generated by the symmetricmonomials in the variabless1, ..., sk, t1, ..., tn−k.

◮ There is a number of ways to prove this claim.We provided a topological proof using theideal-valued cohomological index theory ofFadell and Husseini.

Methods III - index theory

◮ The universalG-bundleEG → BG consists of acontractibleG-spaceEG on which the groupGacts freely and the orbit spaceBG = EG/G.

Methods III - index theory

◮ The universalG-bundleEG → BG consists of acontractibleG-spaceEG on which the groupGacts freely and the orbit spaceBG = EG/G.

◮ The equivariant mapf : X → Y induces (aftermultiplying byEG and taking the orbits ofGaction) the commutative diagram:

X ×G EG Y ×G EG

Index theory, continued

◮ In cohomology we have:

H∗G(X) H∗

H∗(BG)

p∗1 p∗2

Figure 4:

◮ The kernels ofp∗1 andp∗2 are the ideals in thecohomology ring of the classifying spaceBG ofthe groupG and are called indices and denotedby IndG(X) and IndG(Y ).

◮ From the above commutative diagram it followsIndG(Y ) ⊆ IndG(X).

◮ If we could determine these ideals and show thatthis relation does not hold, the obtainedcontradiction would show that there is noequivariant map fromX to Y .

Methods IV - obstruction theory

◮ If X is a CW-complex and the spaceY \ A is(n − 1)-connected, there is aG-mapf fromn-skeleton ofX to Y \ A.

◮ To some(n + 1)-cell en+1 from every orbit, weassign the homotopy class of the composition ofthe characteristic mapϕen+1 : Sn → Xn of thiscell with f and extend this equivariantly.

◮ In such way we obtain cochaino(f) ∈ Cn+1

G (X; πn(Y )).

Obstruction theory, continued

◮ This cochain is a cocycle and the mapf can beextended over the(n + 1)-skeleton ofX (aftermodifying onn-cells, but not on(n − 1)-skeletonof X) iff its cohomology class (called theobstruction class) equals zero.

◮ If X andY are manifolds and we have a"generic" mapg : X → Y , the Poincaré dual tothis class is represented by the fundamental classof the submanifoldg−1(A) of X. So, it suffices toprove that this class does not vanish.

◮ If the codimension of theG-invariantsubmanifoldA in the manifoldY equals thedimension of the manifoldX, the submanifoldg−1(A) is a finite set, and it is enough to checkthat it contains an odd number of points.

◮ The cases wheng−1(A) is at least1-dimensionalare much more complicated.

Equipartition problem

◮ In 1960, B. Grünbaum posed the following

QUESTION: When do anyj measurable sets inR

d admit an equipartition byk hyperplanes?

◮ If this is the case we say that the triple(d, j, k) isadmissible. This question could be reformulatedin the following way.

◮ PROBLEM: Determine∆(j, k), the smallestdimensiond such that the triple(d, j, k) isadmissible.

Equipartition problem, continued

◮ k = 1 is the ham-sandwich theorem. It says that(d, d, 1) is admissible or∆(d, 1) = d.

◮ What about the casej = 1 (one measurable set)?

◮ ∆(1, 2) = 2 is simple and∆(1, 3) = 3 is obtainedby H. Hadwiger in 1966. He also showed∆(2, 2) = 3 and∆(2, 3) = 5.

◮ E. Ramos, building on some previous results,introduced new ideas (a generalization of theBorsuk-Ulam theorem to the product of sphereswith a combinatorial proof) and obtained somenew and general results about the admissibletriples or the functiond = ∆(j, k).

◮ By considering the system ofj interval measuresconcentrated on the moment curve, he showedthat for admissible triples the inequalitydk ≥ j(2k − 1) has to be satisfied, or in other

words∆(j, k) ≥ j(2k−1)k

.EQUIVARIANT METHODS – P. 30/45

◮ Besides that, he also improved the upper boundsfor the functiond = ∆(j, k) whenj is a power of2. He also proved e.g. that∆(1, 4) ≤ 5,∆(5, 2) ≤ 9 and∆(3, 3) ≤ 9.

◮ The only obtained general upper bound∆(j, k) ≤ j2k−1 is obtained by the successiveapplication of the ham-sandwich theorem.

◮ We havej(2k−1)k

≤ ∆(j, k) ≤ j2k−1, and one isinclined to believe that the lower bound is tight.

◮ The most interesting cases left open are to decidewhether the triples(4, 1, 4), (8, 5, 2) and(7, 3, 3)are admissible.

◮ Let me present, as the illustration, some resultsfrom

P. Mani, S. V. and R. Živaljevic, Topology andcombinatorics of partition of masses byhyperplanes, Adv. in Math., to appear

◮ Let me present, as the illustration, some resultsfrom

P. Mani, S. V. and R. Živaljevic, Topology andcombinatorics of partition of masses byhyperplanes, Adv. in Math., to appear

◮ The geometrical question will be reduced to thetopological one and the latter will be resolvedusing the combinatorial argument. EQUIVARIANT METHODS – P. 32/45

Configuration space

◮ We embedRd in Rd+1 at the level1, i.e.

Rd ≈ R

d × {1} → Rd+1.

Configuration space

Rd ≈ R

d × {1} → Rd+1.

◮ an oriented hyperplaneH in Rd × {1} ↔

an orientedd-subspaceLH in Rd+1 ↔

a vectorxH in Sd

Configuration space

Rd ≈ R

d × {1} → Rd+1.

◮ an oriented hyperplaneH in Rd × {1} ↔

an orientedd-subspaceLH in Rd+1 ↔

a vectorxH in Sd

◮ So, the configuration space isSd × · · · × Sd = (Sd)k.

Configuration space

(0, 1)

Rd × {1}

Figure 5:EQUIVARIANT METHODS – P. 34/45

Test space and test map

◮ k orientedd-subspaces inRd+1 determine2k

hyperorthants.

◮ For every measurable set, the measures of itsintersections with these hyperorthants sum up to1.

hyperorthants.

◮ If we denoteV = {v ∈ R2k

i vi = 1}, the testspace isY = V j, and the invariant subspace is aone-point setA = {( 1

2k , ...,12k )}

hyperorthants.

◮ If we denoteV = {v ∈ R2k

i vi = 1}, the testspace isY = V j, and the invariant subspace is aone-point setA = {( 1

2k , ...,12k )}

◮ Trivially Y \ A ≃ Sj(2k−1)−1.EQUIVARIANT METHODS – P. 35/45

Group of symmetries

◮ The group of symmetries is the Weyl groupWk = (Z/2)⊕k

⋊ Sk, the symmetry group of ak-dimensional cube.

Group of symmetries

◮ The involutions fromZ/2 send the correspondingcoordinates to their antipodal points in theconfiguration space, and the elements of thesymmetric group permute the coordinates.All elements permute the above hyperorthantsand so act on the test space by the permutations.

Group of symmetries

◮ The involutions fromZ/2 send the correspondingcoordinates to their antipodal points in theconfiguration space, and the elements of thesymmetric group permute the coordinates.All elements permute the above hyperorthantsand so act on the test space by the permutations.

◮ In the case of interest in this lecturek = 2, we geta dihedral groupD8.

The casek = 2

◮ We obtained the explicit combinatorial algorithmwhich in terms of some combinatorial functionsdecides if aD8-map exists, when treating thetriples of the form(6m − 1, 4m − 1, 2) and(6m + 2, 4m + 1, 2).

The casek = 2

◮ We restrict our attention here to the case of thetriple (8, 5, 2).

The casek = 2

◮ We restrict our attention here to the case of thetriple (8, 5, 2).

◮ We use the obstruction theory approach first, andfor a generic mapg take the test map induced bythe5 intervals placed along the moment curve inR

8.EQUIVARIANT METHODS – P. 37/45

(8, 5, 2) is admissible

◮ It turns out that the Poincaré dual of theobstruction class is the fundamental class of theunion of three circles inHD8

1 (S8 × S8;Z).

◮ SinceS8 × S8 is 7-connected,HD8

1 (S8 × S8;Z) ∼= H1(D8;Z).

1 (S8 × S8;Z).

◮ SinceS8 × S8 is 7-connected,HD8

1 (S8 × S8;Z) ∼= H1(D8;Z).

◮ By representing the obtained three circles by thesequences of the signed words in two letters, andby identifying the latter as the elements of thehomology of the dihedral group, we get that theobstruction class is non-trivial.

Identification of the singular set

B(+ +)aab (+ -)bAab ( baAb+ -) (+ -)baaB (+ -)Bbaa

Figure 6: Metamorphoses of curves

◮ So, there is no equivariant map, and the triple(8, 5, 2) is admissible, or∆(5, 2) = 8.

◮ For the triple(4, 1, 4), R. Živaljevic (TAMS, toappear) showed that an equivariant map doesexist, and so this approach does not provide ananswer to the original question.

◮ The same is very likely to be true for the triple(7, 3, 3).

Index theory approach

◮ In this approach we work with the subgroupH = (Z/2)⊕k of the symmetry groupWk.

◮ We knowH∗(BH; F2) = F2[x1, ..., xk] and theindices are IndH((Sd)k) = (xd+1

1 , ..., xd+1k ) and

IndH(S(Y )) = ((Pk(x1, ..., xk))j), where

Pk(x1, ..., xk) =x1 · · ·xk(x1+x2) · · · (xk−1+xk) · · · (x1+· · ·+xk)is a Dickson polynomial.

◮ We knowH∗(BH; F2) = F2[x1, ..., xk] and theindices are IndH((Sd)k) = (xd+1

1 , ..., xd+1k ) and

IndH(S(Y )) = ((Pk(x1, ..., xk))j), where

Pk(x1, ..., xk) =x1 · · ·xk(x1+x2) · · · (xk−1+xk) · · · (x1+· · ·+xk)is a Dickson polynomial.

◮ OverF2, a Dickson polynomial has also anotherdescription and as a consequence we get:

THEOREM: Let

Pk = Det

x1 x21 x4

1 . . . x2k−1

x2 x22 x4

2 . . . x2k−1

2... ... ... ... ...

xk x2k x4

k . . . x2k−1

∈ F2[x1, . . . , xk]

be a Dickson polynomial. Thenj measures inRd

admit an equipartition byk hyperplanes if

(Pk)j /∈ Ideal{xd+1

1 , . . . , xd+1k }.

◮ Let j = 2q + r, 0 ≤ r ≤ 2q − 1. Then the aboveconsiderations give∆(2q + r, k) ≤ 2k+q−1 + r.

◮ This upper estimate is strictly better than theprevious one ifr 6= 0 and equals the previous oneif r = 0.

◮ As a consequence we get∆(2q+1 − 1, 2) = 3 · 2q − 1,∆(2q+1 − 2, 2) = 3 · 2q − 3 + ε and∆(2q+1 − 3, 2) = 3 · 2q − 4 + η, whereε, η ∈ {0, 1}.

◮ Here are some particular examples whichillustrate the power of the previous theorem. Thebest previously known upper bounds are given inparentheses.

17 ≤ ∆(7, 3) ≤ 19 (28),14 ≤ ∆(6, 3) ≤ 18 (24),35 ≤ ∆(15, 3) ≤ 39 (60),33 ≤ ∆(14, 3) ≤ 38 (56),27 ≤ ∆(7, 4) ≤ 35 (56),23 ≤ ∆(6, 4) ≤ 34 (48),

57 ≤ ∆(15, 4) ≤ 71 (120),53 ≤ ∆(14, 4) ≤ 70 (112)

THANK YOUFOR YOUR

ATTENTION!

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