Cha, J.C. (Jae)

Friedl, S. (Stefan)

Funke, F. (Florian)

Münstersches Informations und Archivsystem für Multimediale Inhalte

The Grothendieck group of polytopes and norms

CHA, JAE CHOON

FRIEDL, STEFAN

FUNKE, FLORIAN

포항공과대학교

Cha, J.

Friedl, S.

Funke, F.

MPG.PuRe

Mu¨nster J. of Math. 10 (2017), 75–81 Mu¨nster Journal of MathematicsDOI 10.17879/33249451813urn:nbn:de:hbz:6-33249452071c© Mu¨nster J. of Math. 2017The Grothendieck groupof polytopes and normsJae Choon Cha, Stefan Friedl, and Florian Funke(Communicated by Tadeusz Januszkiewicz)Abstract. Polytopes in Rn with integral vertices form a monoid under the Minkowski sum,and the Grothendieck construction gives rise to an abelian group. Symmetric polytopesgenerate a subgroup. Similarly, difference bodies (which we refer to as norms) also generatea subgroup. We show that for every n the two subgroups agree.1. IntroductionIn this paper we define a polytope in Rn to be the convex hull of a finitesubset of Rn. If the finite subset lies in the lattice Zn in Rn, then we say thatthe polytope is integral.We denote by P(n) the set of all integral polytopes in Rn. Given twopolytopes P and Q in Rn, the Minkowski sum of P and Q is defined to be thepolytopeP +Q :={p+ q | p ∈ P and q ∈ Q}.Under the Minkowski sum P(n) becomes an abelian monoid, where the iden-tity element is the polytope consisting of the origin. We denote by G(n) theGrothendieck group of the monoid P(n). (See Section 2.1 for details.)We introduce a few more definitions:• The reflection in the origin of a polytope P is P := {−x | x ∈ P}.• A polytope P is symmetric if P = P . Symmetric polytopes form a sub-monoid Psym(n) ⊂ P(n) and a subgroup Gsym(n) ⊂ G(n).• An (integral) polytope P is an (integral) norm if there exists an (integral)polytope Q such that P = Q+Q. (What we call norms are often referredThe first author was supported by NRF grants 2013067043 and 2013053914. The sec-ond author was supported by the SFB 1085 ‘Higher invariants’, funded by the DeutscheForschungsgemeinschaft (DFG). The third author was supported by the GRK 1150 ‘Homo-topy and Cohomology’ funded by the DFG, the Max Planck Institute for Mathematics, andthe Deutsche Telekom Stiftung.76 Jae Choon Cha, Stefan Friedl, and Florian Funketo as difference bodies, but in light of Section 2.2 we prefer the non-standard name of a norm.) Integral norms form a submonoid Pnorm(n) ⊂P(n) and they generate a subgroup Gnorm(n) ⊂ G(n).Clearly a polytope that is a norm is also symmetric. In the real setting theconverse holds. More precisely, any symmetric polytope P can be written asP = 12P + 12P = 12P + 12P .This shows that symmetric polytopes are also norms.In the remainder of the paper we study only integral polytopes and inte-gral norms. Since any integral norm is symmetric, it follows that Pnorm(n) ⊂Psym(n) and Gnorm(n) ⊂ Gsym(n) for any n. We address the question whetherall symmetric integral polytopes are integral norms. The question arises natu-rally on its own, and in addition, there is a motivation from the study of grouprings. See Section 2.2 for a related discussion.Every one-dimensional symmetric integral polytope P is of the form P =[−x, x] for some x ∈ Z≥0. It can be written as P = Q +Q where Q = [0, x].This shows that every one-dimensional symmetric integral polytope is in factan integral norm. Thus Pnorm(1) = Psym(1) and Gnorm(1) = Gsym(1).The situation is more subtle in dimension two and higher. First of all wehave the following elementary lemma.Lemma 1.1. For any n ≥ 2 we have Psym(n) 6= Pnorm(n).Our main result is that, to our surprise, the situation is very different ifone considers the Grothendieck group. More precisely, we have the followingtheorem.Theorem 1.2. For any n we have Gsym(n) = Gnorm(n).The polytope group G(n) has garnered a lot of interest over the last fewyears. For example the second author, Wolfgang Lu¨ck and Stephan Tillmann[2, 3, 4] associate to an L2-acyclic group π that satisfies the Atiyah conjecturean element in G(n), where n = b1(π). This invariant has been examined bythe third author and Dawid Kielak for free-by-cyclic groups in [6]. Further-more the third author showed in [5, Thm. 4.4] that G(n) is a free abeliangroup by constructing an explicit basis and proved a statement dual to that ofTheorem 1.2, see [5, Thm. 6.3].2. Preliminaries2.1. The polytope group. Let n ∈ N. It is straight-forward to show thatthe monoid P(n) of integral polytopes has the cancellation property, i.e. forpolytopes P ,Q,R ∈ P(n) with P +Q = P+R we have Q = R. (For instancesee [9, Lem. 3.1.8].)On P(n) ×P(n), define (P ,Q) ∼ (P ′,Q′) if P + Q′ = P ′ +Q. This is anequivalence relation since P(n) has the cancellation property. Let G(n) be theMu¨nster Journal of Mathematics Vol. 10 (2017), 75–81The Grothendieck group of polytopes and norms 77set of equivalence classes. It is straight-forward to see that G(n) is an abeliangroup under(P ,Q) + (P ′,Q′) := (P + P ′,Q+Q′).It is referred to as the Grothendieck group of P(n). It is also straight-forwardto see that the mapP(n)→ G(n), P 7→ (P , 0)is a monomorphism. We will use this monomorphism to identify P(n) with itsimage in G(n). As usual, given P and Q ∈ P(n), we write P −Q = (P ,Q).2.2. Motivation: The marked polytope for a group ring element.Here we discuss a motivation which leads us to consider integral polytopes.Let G be a finitely generated group. An integral polytope in H1(G;R) is theconvex hull of a finite number of points in Im{H1(G;Z)→ H1(G;R)}. All theconcepts and definitions for polytopes in Rn generalize in an obvious way topolytopes in H1(G;R). In particular, we can consider the monoid P(H) ofintegral polytopes in H := H1(G;R) and we can consider the correspondinggroup G(H).We denote by ǫ : G → H = H1(G;R) the canonical map. Given a nonzeroelementf =∑g∈Gfgg ∈ Z[G] (fg ∈ Z)we refer toP(f) := convex hull of{ǫ(g) | g ∈ G with fg 6= 0}⊂ H = H1(G;R)as the polytope of f . Now suppose that the ring Z[G] is a domain, i.e. it hasno nonzero element which is a left or right zero-divisor. Conjecturally this isprecisely the case when G is torsion-free. It is straight-forward to see that inthis case the mapP : Z[G] \ {0} → P(H), f 7→ P(f)is a monoid homomorphism. We refer to [4, Lem. 3.2] for details.Now let G be a group that is torsion-free elementary amenable. It followsfrom [7, Thm. 1.4] that the group ring Z[G] is a domain. Furthermore by[1, Cor. 6.3] the ring Z[G] satisfies the Ore condition, that is, for any twononzero elements x, y ∈ Z[G] there exist nonzero elements p, q ∈ Z[G] suchthat xp = yq. This implies that Z[G] admits a ‘naive’ ring of fractions K(G),which usually is referred to as the Ore localization of Z[G]. We refer to [8, §4.4]for details. In the following we denote by K(G)×ab the abelianization of themultiplicative group K(G)× = K(G) \ {0}. It is straight-forward to see thatthe above map P : Z[G] \ {0} → P(H) extends to a group homomorphismP : K(G)×ab → G(H).This group homomorphism, and a generalization thereof to groups that satisfythe Atiyah conjecture, are used in [2, 3, 4] to study elements in K(G)×ab.Mu¨nster Journal of Mathematics Vol. 10 (2017), 75–8178 Jae Choon Cha, Stefan Friedl, and Florian FunkeThe inversion g 7→ g−1 on G extends linearly to the standard involutionf 7→ f on Z[G]. It extends naturally to an involution on K(G) and on K(G)×ab.A norm in K(G)×ab is an element that can be written as f ·f for some f ∈ K(G).The above group homomorphism P : K(G)×ab → G(H) sends norms to norms.Our paper grew out of an attempt to detect elements in K(K)×ab that are notnorms.3. Proofs3.1. Proof of Lemma 1.1. Let P be a polytope in Rn. A face of P is anysubset of P of the formF ={p ∈ P | φ(p) = max{φ(p′) | p′ ∈ P}},where φ : Rn → R is a homomorphism. Recall that a face of P is a polytope inits own right. In fact, a face of P is the convex hull of a proper subset of thevertex set of P . We call a polytope contained in a face a subface of P . Thenwe have the following elementary lemma.Lemma 3.2. Let P and Q be polytopes in Rn. Then any face of P is, up totranslation, a subface of P +Q.Proof. Given a face F of P , there exists φ ∈ Hom(Rn,R) such thatF ={p ∈ P | φ(p) = max{φ(p′) | p′ ∈ P}}.If we letG ={q ∈ Q | φ(q) = max{φ(q′) | q′ ∈ Q}},then F +G is a face of P +Q which contains a translate of F . Now we are in a position to prove Lemma 1.1.Proof of Lemma 1.1. We first show that Psym(2) 6= Pnorm(2). Let k ∈ N.We denote by P the integral two-dimensional symmetric polytope spanned by(k, 0), (k, 1), (−k, 0) and (−k,−1). We want to show that P is not an integralnorm. We denote by X the integral polytope spanned by (0, 0) and (2k, 1),and by Y the integral polytope spanned by (0, 0) and (0, 1).Suppose P is an integral norm. Thus we can write P = Q + Q, where Qis an integral polytope. By Lemma 3.2, each face of Q is, up to translation, asubface of P . This implies that, up to translation, each face of Q is a subface ofX or of Y. Since neither X or Y admits one-dimensional integral subpolytopes,we see that, up to translation, each face of Q is either X , Y or a point. Inparticular, up to translation, Q equals either {0}, X , Y or X + Y. But it isstraight-forward to verify that in each case Q+Q 6= P .This shows that Psym(2) 6= Pnorm(2). Now let n ≥ 3. We consider the twomapsΦ : R2 → Rn, (x, y) 7→ (x, y, 0),Ψ : Rn → R2, (x1, . . . , xn) 7→ (x1, x2).Mu¨nster Journal of Mathematics Vol. 10 (2017), 75–81The Grothendieck group of polytopes and norms 79Both maps induce homomorphisms on the polytope monoids that map sym-metric polytopes to symmetric polytopes and integral norms to integral norms.Clearly Ψ is a splitting of Φ. Now it follows that if P is an integral symmet-ric polytope in R2 that is not a norm, then Φ(P) ⊂ Rn is also a symmetricpolytope that is not an integral norm. 3.3. Proof of Theorem 1.2. In this section we will prove Theorem 1.2. Sogiven any n we want to show that Gsym(n) = Gnorm(n). This is equivalent toshowing that given any integral polytope P in Rn there exist integral polytopesQ and R such thatP +Q+Q = R+R.The key idea is to prove this statement by induction on n where we performthe induction step by cutting along a hyperplane.In the proof of Theorem 1.2 we will use the following definitions and nota-tions. A hyperplane H in Rn can be written as H = {x ∈ Rn | x · v = 0} forsome v ∈ Rn. We define the halves of a real polytope P ⊂ Rn with respect Hto beP+ :={x ∈ P | x · v ≥ 0}, P− :={x ∈ P | x · v ≤ 0}.Informally speaking, when H meets P in a proper subset, P+ and P− areobtained by cutting P alongH . We remark that P+ and P− may be exchangeddepending on the choice of v, but it will not cause any issue for our purpose.It is known that each of P+, P− and P ∩H is a real polytope whenever it isnonempty.Lemma 3.4 (Normalization by a hyperplane). Suppose P ∈ Rn is a symmetricpolytope and H ⊂ Rn is a hyperplane. Let P+ and P− be the halves of P withrespect to H. ThenP+ + P+ = P− + P− = P+ + P− = P + (P ∩H).Proof. Since P is symmetric, P± = P∓. Therefore it suffices to show thatP+ + P− = P + (P ∩H).Each x ∈ P lies in either P+ or P−. If x ∈ P+, then sinceP ∩H = P+ ∩ P− ⊂ P−,we have {x}+ (P ∩H) ⊂ P+ +P−. By symmetry, {x}+ (P ∩H) ⊂ P+ + P−when x ∈ P−. It follows that P + (P ∩H) ⊂ P+ + P−.For the reverse inclusion, suppose x ∈ P+ and y ∈ P−. Since P is convex,there is t ∈ [0, 1] such that the point z := tx+(1− t)y lies on P ∩H . Considerp := (1 − t)x + ty. Since P is convex, p ∈ P . Therefore x + y = p + z lies inP + (P ∩H). This shows P+ + P− ⊂ P + (P ∩H). We can not directly apply Lemma 3.4 to an integral polytope, since ingeneral, given an integral polytope P there does not exist a hyperplane, suchthat P ∩H is again integral. To overcome this, the following is useful:Mu¨nster Journal of Mathematics Vol. 10 (2017), 75–8180 Jae Choon Cha, Stefan Friedl, and Florian FunkeLemma 3.5 (Vertical stretching). Let P be an integral polytope in Rn. Denoteby dZ the line segment in Rn joining the origin and the point (0, . . . , 0, d). Asusual, identify Rn−1 with the hyperplane of points with last coordinate zeroin Rn. Then for all sufficiently large d > 0,(P + dZ + dZ) ∩ Rn−1is integral.Proof. Denote by π : Rn → Rn−1 the projection map which forgets the last co-ordinate. Denote by v1, . . . , vk ∈ Zn the vertices of P . Let wi = π(vi) ∈ Rn−1and write vi = (wi, ai) with ai ∈ Z. Let Y be the convex hull of {w1, . . . , wk}.Suppose d satisfies d > |ai| for all i. Now it suffices to prove the following:Claim. (P + dZ + dZ) ∩ Rn−1 = Y.Obviously we have(P + dZ + dZ) ∩ Rn−1 = π((P + dZ + dZ) ∩ Rn−1)⊂ π(P + dZ + dZ).Since π(dZ) = {0}, we deduce π(P + dZ + dZ) = π(P). Since the projectionof the convex hull of a set is the convex hull of the projection of the set, wehave π(P) = Y. It follows that (P + dZ + dZ) ∩ Rn−1 ⊂ Y.For the reverse inclusion, observe that (wi, ai±d) ∈ P+dZ+dZ for each i.Note that one of ai ± d is negative and the other is positive. By convexity wehave (wi, 0) ∈ P + dZ + dZ. Once again by convexity, it follows thatY ⊂ (P + dZ + dZ) ∩ Rn−1. Now we are ready for the proof of the main result of this paper.Proof of Theorem 1.2. We prove the theorem by induction on n. For n = 0,the statement is trivial. Suppose the conclusion holds for n− 1, and supposeP is a symmetric integral polytope in Rn. As above we identify Rn−1 with thehyperplane of points with last coordinate zero in Rn. By Lemma 3.5, there isd ∈ N such that (P + dZ + dZ)∩Rn−1 is an integral polytope in Rn−1. WriteY = P + dZ + dZ for brevity. Since Y is symmetric, Y ∩ Rn−1 is symmetrictoo. Therefore, by the induction hypothesis, there are integral polytopes Qand R in Rn−1 such that(Y ∩Rn−1) +Q+Q = R+R.By Lemma 3.4, we haveY + (Y ∩Rn−1) = Y+ + Y+,where Y+ denotes a half of Y with respect to the hyperplane Rn−1. SinceY ∩Rn−1 is integral, we also deduce that Y+ is an integral polytope. From theabove equations, it follows thatP + (dZ +Q) + (dZ +Q) = (Y+ +R) + (Y+ +R). Acknowledgments. We wish to thank Matthias Nagel for helpful conversa-tions. We are also grateful to the referee for helpful feedback.Mu¨nster Journal of Mathematics Vol. 10 (2017), 75–81The Grothendieck group of polytopes and norms 81References[1] J. Dodziuk et al., Approximating L2-invariants and the Atiyah conjecture, Comm. PureAppl. Math. 56 (2003), no. 7, 839–873. MR1990479[2] S. Friedl and W. Lu¨ck, L2-Euler characteristics and the Thurston norm, preprint (2016).[3] S. Friedl and W. Lu¨ck, Universal L2-torsion, polytopes and applications to 3-manifolds,preprint (2016).[4] S. Friedl and S. Tillmann, Two-generator one-relator groups and marked polytopes,preprint (2015), arXiv:1501.03489.[5] F. Funke, The integral polytope group, preprint (2016), arXiv:1605.01217.[6] F. Funke and D. Kielak, Alexander and Thurston norms, and the Bieri–Neumann–Strebelinvariants for free-by-cyclic groups, preprint (2016), arXiv:1605.09067.[7] P. H. Kropholler, P. A. Linnell and J. A. Moody, Applications of a new K-theoretictheorem to soluble group rings, Proc. Amer. Math. Soc. 104 (1988), no. 3, 675–684.MR0964842[8] D. S. Passman, The algebraic structure of group rings, Pure Appl. Math., Wiley-Interscience, New York, 1977. MR0470211[9] R. Schneider, Convex bodies: The Brunn–Minkowski theory, Encyclopedia Math. Appl.,44, Cambridge Univ. Press, Cambridge, 1993. MR1216521Received February 14, 2016; accepted June 16, 2016Jae Choon ChaDepartment of Mathematics, POSTECH,Pohang 37673, Republic of KoreaE-mail: jccha@postech.ac.krSchool of Mathematics, Korea Institute for Advanced Study,Seoul 02455, Republic of KoreaStefan FriedlFakulta¨t fu¨r Mathematik, Universita¨t Regensburg,Universita¨tsstr. 31, 93040 Regensburg, GermanyE-mail: sfriedl@gmail.comFlorian FunkeMathematisches Institut, Universita¨t Bonn,Endenicher Allee 60, 53115 Bonn, GermanyE-mail: ffunke@math.uni-bonn.deMu¨nster Journal of Mathematics Vol. 10 (2017), 75–81

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The Grothendieck group of polytopes and norms

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