A subexponential size RP n
aa r X i v : . [ m a t h . C O ] S e p A SUBEXPONENTIAL SIZE RP n KARIM ADIPRASITO ♠ , SERGEY AVVAKUMOV ♣ , AND ROMAN KARASEV ♦ Dedicated to Wolfgang K¨uhnel on the occasion of his 70th birthday.
Abstract.
We address a long-standing and long-investigated problem in combinatorialtopology, and break the exponential barrier for triangulations of real projective space,constructing a trianglation of RP n of size e ( + o (1)) √ n log n . Introduction
While in general, every smooth manifold allows for a triangulation, it is a notoriouslyhard problem to construct small triangulations of manifolds, and usually poses a difficultchallenge. And so, outside of special cases, there are few cases known where the upperbounds and lower bounds come even close to each other.Let us focus on minimality in terms of the number of vertices that a triangulation ofa given manifold would have, a study Banchoff and K¨uhnel initiated [K¨uh95]. The bestlower bounds in this area are usually either homological, or homotopic in nature. Indeed, ◦ it is clear that the number of vertices cannot be lower than the ball-category, or themore studied Lusternik-Schnirelmann category [CLOT03]. In particular ◦ it is bounded from below in terms of the cup length of the space in question. In fact,it is easy to show and observed by Arnoux and Marin that for a space of cup length n ,one needs (cid:0) n (cid:1) vertices [AM91].Finally ◦ Murai gave a lower bound in terms of the Betti numbers of (closed and orientable)manifolds [Mur15], which was simplified and generalized to general manifolds by Adipr-asito and Yashfe [Adi18, AY20]. This bound in general is not so good for interestingmanifolds, as it seems insensitive to any interesting multiplicative structure in the co-homology ring, let alone homotopy.On the example of RP n , the bound by Arnoux and Marin is best. Still, the best con-struction so far is essentially still K¨uehnel’s observation that the barycentric subdivisionof the n + 1-simplex yields a triangulation of the n -sphere on 2 n +1 − Z / Z -action, such that antipodal vertices are at distance at least 3 from each other. Thisyields a triangulation of RP n of size 2 n − ♠ Has received funding from the European Research Council under the European Unions SeventhFramework Programme ERC Grant agreement ERC StG 716424 - CASe and the Israel Science Founda-tion under ISF Grant 1050/16. ♣ Has received funding from the European Research Council under the European Unions SeventhFramework Programme ERC Grant agreement ERC StG 716424 - CASe. ♦ Supported by the Federal professorship program grant 1.456.2016/1.4 and the Russian Foundationfor Basic Research grants 18-01-00036 and 19-01-00169. RP n Acknowledgment.
The second author wishes to thank Arseniy Akopyan for helpfuldiscussions. 2.
Main result
Theorem 2.1.
For all positive integers n , there exists a convex centrally symmetric n -dimensional polytope P such that: • All the vertices of P lie on the unit sphere. • For any vertex A ∈ P , if F, F ′ ⊂ P are faces with A ∈ F and − A ∈ F ′ , then F ∩ F ′ = ∅ . • The number of vertices of P is less than e ( + o (1)) √ n log n . Corollary 2.2.
For all positive integers n , there exists a triangulation of R P n − with atmost e ( + o (1)) √ n log n vertices.Proof. Let P be the polytope whose existence is guaranteed by Theorem 2.1.The group Z acts on ∂P by central symmetries. Let S be a Z -equivariant, i.e.,symmetric, triangulation of ∂P . Such triangulation may, for example, be obtained byconsequently pulling pairs of opposite vertices of ∂P . The triangulated sphere S has thesame number of vertices as P .From the properties of P , we get that the closed stars of any two opposite verticesof S are disjoint. Hence, the quotient S/ Z is a simplicial complex. This complex ishomeomorphic to R P n − . (cid:3) Proof of Theorem 2.1
Our proof is constructive, and proceeds by constructing a Delaunay triangulation ofthe sphere [Del28].Let V be a subset of the set of non-empty subsets of { , . . . , n } . We identify any A ∈ V with a unit vector in R n whose endpoint has its i th coordinate equal to 1 / p | A | if i ∈ A and 0 otherwise.A centrally symmetric polytope P ( V ) is the convex hull of the endpoints of the vectors V ⊔ − V .Our proof of Theorem 2.1 consists of the following three claims. By h· , ·i we denote theinner product. Claim 3.1.
Suppose that V satisfies the following properties:(1) { i } ∈ V for all i ∈ { , . . . , n } .(2) If A ∈ V and | A | > , then A \ i ∈ V for any i ∈ A .(3) Let A, B ∈ V be vertices and X ∈ S n − be a unit vector with non-negative coordi-nates such that h A, B i = 0 and h A, X i = h B, X i . Then there is C ∈ V such that h C, X i < h A, X i = h B, X i .Then P ( V ) satisfies the conditions of Theorem 2.1, except maybe the condition on thenumber of vertices.Proof. By the property (1), P ( V ) is an n -dimensional polytope.Let A ∈ V be a vertex and F, F ′ ⊂ P ( V ) be faces with A ∈ F and − A ∈ F ′ . We needto prove that F ∩ F ′ = ∅ . Assume to the contrary, that there is a vertex B ∈ F ∩ F ′ .Without the loss of generality, we may assume that B ∈ V (as opposed to B ∈ − V ).By the property (2), we have that the projection of P ( V ) to any coordinate hyperplanelies in P ( V ). Which means, that(*) every facet of P ( V ) belongs to a coordinate orthant. SUBEXPONENTIAL SIZE RP n From (*), we get that every two vertices of P ( V ) sharing a facet can be connected byan arc not longer than π . So, the shortest arcs connecting A to B and B to − A are bothnot longer than π . Since A and − A are opposite, it means that the arcs are exactly π long, so h A, B i = 0.Both A and B lie in the positive orthant, so by (*) there exists a facet G ∋ A, B also lying in the positive orthant. Let X be the outer normal of G , its coordinates arenon-negative. We already know that h A, B i = 0. So, we can use the property (3) to find C ∈ V such that h C, X i < h A, X i = h B, X i . The existence of such vertex C contradictsto G being a facet of P ( V ). (cid:3) Claim 3.2.
Suppose that V satisfies the following properties:(1) { i } ∈ V for all i ∈ { , . . . , n } .(2) If A ∈ V and | A | > , then A \ i ∈ V for any i ∈ A .(3) Foe every A, B ∈ V with A ∩ B = ∅ , there are i ∈ A and j ∈ B such that either(3a) B ⊔ i ∈ V and A ⊔ j \ i ∈ V ,or(3b) A ⊔ j ∈ V and B ⊔ i \ j ∈ V .Then V satisfies the requirements of Claim 3.1.Proof. We only need to check the requirement (3) of Claim 3.1.Denote the coordinates of X by x , . . . , x n .Without the loss of generality, we may assume that A = { , . . . , a } and that x ≤ x ≤· · · ≤ x a .The part h A, B i = 0 of the condition (2) of Claim 3.1 implies A ∩ B = ∅ . So, withoutthe loss of generality, we may assume that (3b) is satisfied, i.e., there are i ∈ A and j ∈ B such that A ⊔ j ∈ V and B ⊔ i \ j ∈ V .If x i > x j , then C := B ⊔ i \ j ∈ V is as required. So, we can assume that x j ≥ x i ≥ x .Denote A := A \ ∈ V and A := A ⊔ j ∈ V .We have h X, A i = x + · · · + x a √ a , h X, A i = x + · · · + x a √ a − , h X, A i = ( x + x j ) + x + · · · + x a √ a + 1 ≥ x + x + · · · + x a √ a + 1 . Denote f ( α ) := αx + x + · · · + x a √ a − α = x √ a − α + ( x − x ) + · · · + ( x a − x ) √ a − α . We have that h X, A i = f (1) , h X, A i = f (0) , h X, A i ≥ f (2) . Consider the function g ( β ) := x β + ( x − x ) + · · · + ( x a − x ) β . Using 0 ≤ x ≤ x ≤ · · · ≤ x a we get that g is a convex function when β >
0. If g isnot strictly convex, then it’s linear.Assume that g is constant. Then x = · · · = x a = 0, meaning that h X, A i = 0.Because X is unit, we get that x k > k . From (1), we have { k } ∈ V and h X, { k }i > h X, A i , meaning that C := { k } is as required.So, g is either strictly convex or is linear and non-constant. SUBEXPONENTIAL SIZE RP n By definition, f ( α ) = g ( √ a − α ) , so f (0) = g ( √ a − , f (1) = g ( √ a ) , and f (2) = g ( √ a + 1) . From 0 ≤ √ a − < √ a < √ a + 1 and because g is either strictly convex or non-constantlinear, we get that either f (0) > f (1) or f (2) > f (1). Meaning that either h X, A i > h X, A i , or h X, A i > h X, A i , so either C := A or C := A is as required. (cid:3) Claim 3.3.
There is a set V of size at most e ( + o (1)) √ n log n satisfying the requirements ofClaim 3.2.Proof. Partition the set { , . . . , n } into several disjoint groups. Let V be the set of subsetsof { , . . . , n } , whose intersection with every group, except maybe one, contains not morethan 1 element.The maximal group M ( A ) of A is the group whose intersection with A is maximalin size (pick one, if there are several such groups). Note, that the union of A with anyelement of M ( A ) is still in V .Suppose that either A ∩ M ( B ) or B ∩ M ( A ) is non-empty. Without the loss of generality,let us assume that A ∩ M ( B ) is non-empty. Pick any elements i ∈ A ∩ M ( B ) and j ∈ B ∩ M ( B ). Then i and j satisfy (3a).The remaining case is when both A ∩ M ( B ) and B ∩ M ( A ) are empty. Pick any elements i ∈ A ∩ M ( A ) and j ∈ B ∩ M ( B ). Then i and j satisfy both (3a) and (3b).It remains to pick the groups so that the size of V is as required. We pick k groups ofalmost equal size, each group not larger than s := ⌈ nk ⌉ .Let us analyze the size of V . For an element A ∈ V we have • k choices of the maximal group M ( A ), • at most 2 s choices of which elements of M ( A ) to add to A , • at most s + 1 choices for each of the k − V multiple times. In total we get | V | < s ( s + 1) k − k. Putting k = s = √ n we get the required asymptotic | V | < √ n √ n √ n √ n = e ( + o (1)) √ n log n . (cid:3) Remark . Our choice of k = s = √ n in the proof of Claim 3.3 is slightly subopti-mal. Consequently, the bound on the number of vertices in Theorem 2.1 can be slightlyimproved, but only by a subpolynomial factor, which to state precisely we deemed notrelevant. Remark . The set of combinatorial conditions on V in the statement of Claim 3.2 isnot the only one which makes the claim work. Although, other conditions give a worsebound on the number of vertices. In general, we don’t know what conditions on V arenecessary for our proof of Theorem 2.1 to go through. References [Adi18] Karim Adiprasito,
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Department of Mathematical Sciences, University of Copenhagen,Copenhagen, Denmark
E-mail address : [email protected], [email protected] Sergey Avvakumov, Department of Mathematical Sciences, University of Copenhagen,Copenhagen, Denmark
E-mail address : [email protected] Roman Karasev, Moscow Institute of Physics and Technology, Institutskiy per. 9,Dolgoprudny, Russia 141700Institute for Information Transmission Problems RAS, Bolshoy Karetny per. 19,Moscow, Russia 127994
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