2 Definitions We begin by reviewing the construction and some basic properties of squeezed balls and spheres [5, 7]. For an integer n 1, let [n] denot

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1 Squeezed 2-Spheres and 3-Spheres are Hamiltonian Robert L. Hebble Carl W. Lee December 4, Introduction A (convex) d-polytope is said to be Hamiltonian if its edge-graph is. It is well-known that simple 3-polytopes are not, in general, Hamiltonian, though non-hamiltonian examples require at least 38 vertices [3, 4]. The question of whether all simple d-polytopes, d > 3, are Hamiltonian or not remains unsettled. Barnette (see [8]) conjectures that all simple 4-polytopes are Hamiltonian. In partial support of this conjecture, simple 4-prisms over 4-colorable (and hence all) 3-polytopes [8], and duals of cyclic d-polytopes [2, Section 17.2] are Hamiltonian. The duals of simple d-polytopes are simplicial d-polytopes, and their boundary complexes are therefore simplicial (d 1)-spheres. Thus it is natural to say that a pure simplicial (d 1)- complex has a Hamiltonian path if its facets (maximal faces) can be ordered F 1 ;:::;F m such that F i and F i+1 are adjacent (intersect in a set of cardinality d 1) for all 1» i» m 1. If, further, F 1 and F m are also adjacent, then we say that the simplicial complex has a Hamiltonian cycle or is Hamiltonian. In this paper we prove that all squeezed 2- and 3-spheres are Hamiltonian. In creating the collection of squeezed (d 1)-spheres, Kalai [5] establishes a large lower bound on the number of labeled simplicial (d 1)-spheres, and hence a substantial gap between the number of spheres and the number of simplicial convex d-polytopes. This collection extends the construction of [1] and thus contains at least one polytopal representative for every possible simplicial d-polytope f-vector. As a consequence, should there be a non-hamiltonian simple 4-polytope, it cannot be detected by its f-vector alone. 1

2 2 Definitions We begin by reviewing the construction and some basic properties of squeezed balls and spheres [5, 7]. For an integer n 1, let [n] denote the set f1;:::;ng. For a nonnegative integer d, [n] (d) is the set of all subsets of [n] of cardinality d. For F 2 [n] (d),we will write F i for F nfig and F + i for F [fig. For C 2 [n], define cone(c; 0) = ff +0:F 2 Cg. Let F; G 2 [n] (d) ;F = fa 1 ;:::;a d g;g= fb 1 ;:::;b d g, where a 1 < <a d and b 1 < < b d. We say F» p G if a i» b i for 1» i» d. Define F» L G if min(f 4 G) 2 F where F 4 G is the symmetric difference of the sets F and G. A collection ordered with respect to» L is said to be in lexicographic order. The reverse lexicographic order on [n] (d) is defined by F < RL G if max(f 4 G) 2 G. Define q(f; G) =minfj : a i = b i for all i jg taking q(f; G) =d +1ifa d 6= b d. Now suppose F is a proper subset of [n]. Let j =minfi 2 [n] :i 62 F g and k = maxfi 2 [n] : i 62 F g. We call f1;:::;j 1g the left set of F and fk +1;:::;ng the right set of F. If s; t 62 F, s; t 2 [n], s<t, but i 2 F for all s +1» i» t 1, then fs +1;:::;t 1g is a middle set of F. Note that left, right, or middle sets may beempty. Suppose j 2 F [n]. If j is not in the left set of F, define `(F; j) = maxfi 2 [n] : i < j; i 62 F g and L(F; j) = (F j) +`(F; j). If j is not in the right set of F, define r(f; j) = minfi 2 [n] : i > j; i 62 F g and R(F; j) = (F j) +r(f; j). These definitions correspond to pushing" a contiguous set of elements of F to the left" or right" respectively. For example if F is the set f1; 2; 4; 5; 6; 7g, thenl(f; 5) is the set f1; 2; 3; 4; 6; 7g and R(F; 4) is the set f1; 2; 5; 6; 7; 8g. Let d be a positive odd integer. Define F d (n) to be the collection of all members of [n] d+1 having even cardinality left, right, and middle sets. Note that for F 2 F d (N), if we write F = fa 1 ;:::;a d+1 g,we implicitly assume that the elements have been indexed so that a 1 < <a d+1. For even positive d, let F d (n) =cone(f d 1 ; 0). For positive d, consider a nonempty subcollection, I, off d [n] that is an initial set of F d [n] with respect to the partial order» p. Equivalently, iff 2 I, then L(F; j) 2 I for every j for which L(F; j) 2 F d [n]. Kalai [5] proves that the simplicial d-complex B(I) = fg : G F for some F 2 Ig is topologically a d-ball. He calls such a simplicial complex a squeezed ball. Note that the facets of B(I) are precisely the sets contained in I. Kalai observes that squeezed d-balls for even d are just those simplicial complexes of the form cone(b(j); 0), where J is some initial set with respect to» p of F d 1 (n). The boundary S(I) of a squeezed ball B(I) is topologically a (d 1)-sphere, and is called a squeezed sphere. The facets of S(I) are those subsets of B(I) of cardinality d that are contained in exactly one facet of B(I). 2

3 Kalai proves that the reverse lexicographic ordering of the facets of a squeezed ball B(I) is a shelling order. By examining this shelling, one can readily characterize the facets of S(I) [7]. Proposition 2.1 Suppose d is odd and B(I) is a squeezed d-ball. Let F = fa 1 ;:::;a d+1 g2i be afacet of B(I) and a i 2 F. Then F a i is a facet of S(I) if and only if: 1. i is even and a i is in the left set of F, or 2. i is odd and R(F; a i ) 62 I. Suppose d is even and B(I) =cone(b(j); 0) is a squeezed d-ball. cardinality d is a facet of S(I) if and only if: Then F 2 B(I) of 1. F is a facet of B(J), or 2. F = G +0 where G is a facet of S(J). 3 Squeezed 2-Spheres are Hamiltonian Let S be any simplicial (d 1)-complex with vertex set V, F be a facet of S, and v =2 V. The simplicial complex resulting from the stellar subdivision of F is (S nffg) [fg + v : G ρ F; G 6= F g. A simplicial (d 1)-complex is a stacked sphere if it can be obtained from the boundary of a d-simplex, fg ρ H : G 6= Hg (where H has cardinality d + 1), by a sequence of stellar subdivisions of facets. Stacked spheres are all polytopal, and are precisely the boundary complexes of stacked polytopes. Stacked polytopes are dual to truncation polytopes, those obtained from a d-simplex by a sequence of vertex truncations. An easy proof by induction shows that stacked spheres are Hamiltonian (and in fact stacked 2-spheres admit precisely three Hamiltonian cycles [6]). We will show that squeezed 2-spheres are Hamiltonian by showing that they are stacked. In this section we use the notation v i instead of just i when referring to a vertex of a squeezed ball or sphere. We will also abbreviate a set fv i ;v j ;v k ;v`g by v i v j v k v`, with the convention that i<j<k<`. Theorem 3.1 Squeezed 2-spheres are stacked. 3

4 Proof. Let B be a squeezed 3-ball with vertex set fv 1 ;:::;v n g. Its facets are faces of cardinality 4, and its subfacets are faces of cardinality 3. denote the boundary complex of B. The facets are those subfacets of B that are contained in exactly one facet of B. We will prove the theorem by induction on the number of vertices Note that is the boundary of a simplex, we are done. We say a facet F of B is of type j if j =minfi : v i 2 F g. See, for example, the 3-ball in Figure 1, in which we abbreviate v i as i. Let k be the maximum type of a facet of B. F F F F F ffl F ffl F ffl F ffl F F F Figure 1: The facets of a squeezed 3-ball, with its type 2 facets indicated by ffl A useful consequence of the fact that B is squeezed is that every set of the form v i v i+1 v j v j+1 where j» k + 2 is a facet of B. Case I: Assume k = 1. By the definition of a squeezed ball, the facets of B must be of the form v 1 v 2 v 3 v 4 ;v 1 v 2 v 4 v 5,...,v 1 v 2 v n 1 v n. Thus they each contain vertices 1 and 2, and each facet of the form v 1 v 2 v j 1 v j meets the facet v 1 v 2 v j 2 ;v j 1 in the common face v 1 v 2 v j 1, 5» j» n. Starting with v 1 v 2 v 3 v 4, with the addition of each facet v 1 v 2 v j 1 v j, 5» j» changes by a stellar subdivision of the sphere facet v 1 v 2 v j 1. is a stacked sphere. Case II: Suppose k 2. Let F be the lexicographically-least facet of type k. So F = v k v k+1 v k+2 v k+3. Note that no other facets of type k contain v k+2. For example, in Figure 1, k =3,F = F 10, and v k+2 = v 5. We first show that v k+2 is contained in exactly three facets For 1» j» k 1 let A j = v j v j+1 v k+2 v k+3 and B j = v j v j+1 v k+1 v k+2. Both A j and B j are facets of B since B 4

5 is squeezed. The set of facets of B containing v k+2 is ff; A 1 ;B 1 ;:::;A k 1 ;B k 1 g. Now, for 1» j» k 1, we have v j v j+1 v k+2 = A j B j. So v j v j+1 v k+2 For 2» j» k 1, v j v k+1 v k+2 = B j B j 1. Also, for the same j, we have v j v k+2 v k+3 = A j A j 1. Thus v j v k+1 v k+2 ;v j v k+2 v k+3 for 2» j» k 1. Finally, v k v k+2 v k+3 = A k 1 F, and v k v k+1 v k+2 = B k 1 F. So v k v k+1 v k+2 ;v k v k+2 v k+3 This leaves v k+2 only in the three facets v 1 v k+1 v k+2, v 1 v k+2 v k+3 and v k+1 v k+2 v k+3 Next, we construct a new simplicial complex from the ball B. Its facets are obtained from the facets of B by removing F and, for each 1» j» k 1, replacing A j and B j with C j = v j v j+1 v k+1 v k+3. This is a squeezed ball B 0 with vertices v 0 1;:::;v 0 n 1 where v 0 j = v j for 1» j» k + 1 and v 0 j = v j+1 for k +2» j» n 1. See the example in Figure 2. B B A B A F B C C Figure 2: The facets of B and B 0 The collection F(@B) of facets not containing v k+2 is identical to the collection F(@B 0 ) of facets 0 apart from v 1 v k+1 v k+3. This correspondence is detailed in Figure 3. Case 1: Suppose G = v i v i+1 v j v j+1 2 B and G v i is a facet not containing v k+2. Then R(G; v i ) 62 B and necessarily j > k +2 (using the fact that B is squeezed). So G 62 ff; A 1 ;B 1 ;:::;A k 1 ;B k 1 g and R(G; v i ) 62 fc 1 ;:::;C k 1 g. Therefore G 2 B 0, R(G; v i ) 62 B 0, and G v i 0. Case 2: Suppose G = v 1 v 2 v j v j+1 2 B and G v 2 is a facet not containing v k+2. Then G 62 ff; A 1 ;B 1 ;:::;A k 1 ;B k 1 g. Therefore G 2 B 0 and G v

6 0 ) Case 1: v i v i+1 v j v j+1 v i v i v i+1 v j v j+1 v i Case 2: v 1 v 2 v j v j+1 v 2 v 1 v 2 v j v j+1 v 2 Case 3a: v k v k+1 v k+2 v k+3 v k+2 v k 1 v k v k+1 v k+3 v k 1 Case 3b: v i v i+1 v k+2 v k+3 v k+2 (i<k) v i v k v i+1 v k+1 v k+3 v k+1 Case 3c: v i v i+1 v j v j+1 v j (j 6= k +2) v i v i+1 v j v j+1 v j Case 4: v 1 v 2 v 3 v 4 v 4 v 1 v 2 v 3 v 4 v 4 Figure 3: Correspondence between F(@B) and F(@B 0 ) Case 3a: Suppose G = v k v k+1 v k+2 v k+3 2 B and G v k+2 is a facet not containing v k+2. Then G = F and H = v k v k+1 v k+3 v k+4 = R(G; v k+2 ) 62 B. Now C k 1 = v k 1 v k v k+1 v k+3 2 B 0 and R(C k 1 ;v k 1 ) (with respect to B 0 ) is H 62 B 0. Therefore C k 1 v k 1 0. Case 3b: Suppose G = v i v i+1 v k+2 v k+3 2 B and G v k+2 is a facet not containing v k+2, where i < k. Then G = A i and H = v i v i+1 v k+3 v k+4 = R(G; v k+2 ) 62 B. Now C i = v i v i+1 v k+1 v k+3 2 B 0 and R(C i ;v k+1 ) (with respect to B 0 ) is H 62 B 0. Therefore C i v k+1 0. Case 3c: Suppose G = v i v i+1 v j v j+1 2 B and G v j is a facet not containing v k+2, where j 6= k + 2. Obviously v k+2 62 G v j implies j 6= k +1 either. So G 62 ff; A 1 ;B 1 ;:::;A k 1 ;B k 1 g. Also R(G; v j ) 62 B, so necessarily j > k +2 (using the fact that B is squeezed). So R(G; v j ) 62 fc 1 ;:::;C k 1 g. Therefore G 2 B 0, R(G; v j ) 62 B 0, and G v j 0. Case 4: Suppose G = v 1 v 2 v 3 v 4. Then G 2 B, G 2 B 0, G v 4 and G v 4 0. It is straightforward to check that all possibilities of facets in F(@B 0 )have been accounted for, establishing the correspondence between F(@B) andf(@b 0 ). This analysis implies is obtained 0 by the stellar subdivision of the facet v 1 v k+1 v k+3. 0 is a squeezed 2-sphere with one fewer vertex and by the induction 0 is stacked. is stacked and hence, by induction, squeezed 2-spheres are stacked. 2 6

7 Corollary 3.2 Squeezed 2-spheres are Hamiltonian. 4 Squeezed 3-Spheres are Hamiltonian In this section we show every squeezed 3-sphere is Hamiltonian by exhibiting an explicit ordering of its facets. We first show that the shelling order of the facets of a squeezed 2- sphere described in [7] is actually a Hamiltonian path. We then use this and the fact that a squeezed 4-ball B is a squeezed 3-ball joined to the point 0 to get a path involving the containing 0. Then we insert the remaining facets to form a Hamiltonian cycle The shelling order for squeezed (d 1)-spheres in [7] specializes to boundaries of squeezed 3-balls B in the following way. Let F = a 1 a 2 a 3 a 4 be a facet of B. (Implicit in this notation is that F = fa 1 ;a 2 ;a 3 ;a 3 g where a 1 <a 2 <a 3 <a 4.) Suppose F a k The only such facets when k is even are 123 = and 12a 3 a 4 2. The shelling order begins by ordering these even" facets lexicographically. In Figure 4 we list the facets of a 3-ball B and the ordering of the facets of the form F a k where k is even. B H H H H H H H Figure 4: A squeezed 3-ball B and the ordering of the even" facets 7

8 The shelling order then continues with the boundary facets of the form F a k where k is odd in the following manner. Suppose the facets of B in reverse lexicographic order are F 1 ;F 2 ;:::;F m. Define G p = ff k ;F k+1 ;:::;F`g to be those facets for which a 4 = p. In each of these groups G p, find the minimum i so that F i a 3 is a boundary facet and order the odd" facets associated with this group as F i a 3 ;F i+1 a 3 ;:::;F` 1 a 3 ;F` a 3 ;F` a 1. Let G Λ p be this ordered list of these odd" facets Note that it is possible that G Λ p is empty (but not if p = n, the number of vertices of B), and that if G Λ p has only one facet, it is F` a 1. Now order all of the odd" facets by listing them in the order G Λ n; G Λ n 1; G Λ n 2;::: and so on until G Λ n t = ;. It is a fact that if G Λ p = ; then G Λ i = ; for all i<p. In Figure 5, we list the facets of a 3-ball B and the ordering of the facets the form F a k where k is odd. B H H H H H H H Figure 5: A squeezed 3-ball B and the ordering of the seven remaining odd" facets Proposition 4.1 The above shelling order of squeezed 2-spheres is a Hamiltonian path. Proof. Let F a k be a facet where F 2 B. We will show, for each possibility ofk and F,which adjacent facet G b immediately precedes F a k in the shelling order. 8

9 Suppose k is even. If F = 1234 and k =4,then F a 4 = F 4 is the first facet in the shelling order. If F =1234and k =2,then F 2 is the second facet in the shelling order, and is adjacent to F 4. Otherwise, since k is even, F = 12a 3 a 4, a 3 > 3; and k = 2. Let G = L(F; a 4 ). Now G 2 is adjacent tof 2 and it is evident from the shelling order that G 2istheimmediate predecessor of F 2. Suppose k is odd. Case I: k =1. That is, we are considering a facet of the form F a 1. Now either F a 3 is a facet or it is not a facet. If F a 3 is a facet thenf a 3 and F a 1 are adjacent, and it is evident from the shelling order that F a 3 is the immediate predecessor of F a 1. Otherwise, F a 3 is not afacetof@b. Thus G = R(F; a 3 ) 2 B. Now F a 1 implies that R(F; a 1 ) 62 B, which implies that R(G; a 1 )=R(R(F; a 3 );a 1 ) 62 B since B is squeezed. Thus G a 1 = R(F; a 3 ) a 1 is a facet Now G a 1 is adjacent tof a 1, and it is evident from the shelling order that G a 1 is the immediate predecessor of F a 1. Case II: k = 3. That is, we are considering a facet of the form F a 3 where F 2 B. Recall that F a 3 afacetof@b implies that R(F; a 3 ) 62 B. Now eitherf =12a 3 a 4 or L(F; a 2 ) 2 B (since B is squeezed). If F = 12a 3 a 4, then R(F; a 3 ) 62 B forces a 3 = n 1 and a 4 = n. So F a 3 is the first odd" facet in the shelling order. Thus F 2 immediately precedes F a 3 in the shelling order, and it is also adjacent. Now suppose L(F; a 2 ) 2 B. Subcase 1: G = L(F; a 2 ) and G a 3 is a facet Easily G a 3 is adjacent to F a 3, and in the shelling order G a 3 is the immediate predecessor of F a 3. Subcase 2: L(F; a 2 ) a 3 is not a facet Thus R(L(F; a 2 );a 3 ) 2 B. Let b be the smallest element of G, where G = R(L(F; a 2 );a 3 ). Now G b is a facet since R(R(L(F; a 2 );a 3 );b) = R(F; a 3 ) 62 B. Also, G b is adjacent to F a 3, and G b is the immediate predecessor of F a 3. Thus the shelling order is Hamiltonian path. 2 We are now ready to construct Hamiltonian cycles for squeezed 3-spheres. Theorem 4.2 Squeezed 3-spheres are Hamiltonian. Proof. Let B be a squeezed 4-ball be the associated squeezed 3-sphere. Now B = cone(b 0 ; 0) for some squeezed 3-ball B 0. The above shelling order 0 induces an ordering of the facets containing 0 that is a Hamiltonian path. The last facet of this path is a facet of the form F a 1 where ffl F 2 B, 9

10 ffl F =0a 1 a 2 a 3 a 4 where a j = a 1 +(j 1) for j =2; 3; 4, and ffl there does not exist a facet G of the 4-ball satisfying the above two conditions for which F < RL G (otherwise, F a 1 would not be a facet). See Figure F Figure 6: The facets of a squeezed 4-ball Let S 0 = fg a i : G 2 B, G a i and 0 2 G a i g, S 1 = fg 0 : G 2 B and G» RL F g, and S 2 = fg 0:G2 B and F < RL Gg. Order the facets in S 0 using the reverse of ordering induced by the shelling 0. Note that this ordering begins with the facet F a 1 and ends with the facet Let S 1 = S 14 [ S 15 [ :::[ S 1a4 where S 1k = fg : G 2 S 1 and k = max Gg. Order the facets within each S 1k from lexicographically greatest to lexicographically least if a 4 k is an odd number and from lexicographically least to lexicographically greatest if a 4 k is an even number. Concatenate these orderings in the sequence S 14 ;S 15 ;:::;S 1a4. This results in a Hamiltonian path of the facets in S 1 that begins with the facet 1234 and ends with the facet F 0. Note that in the example of Figure 7 we have S 0 consisting of facets F 1 through F 14, S 1 consisting of facets F 15 through F 20, S 14 consisting of the facet F 15, S 15 consisting of facets F 16 through F 17, S 16 consisting of facets F 18 through F 20,anda 4 =6. 10

11 F F F F F F F F F F F F F F F F F F F F Figure 7: Some of the facets of the boundary of the squeezed 4-ball Now concatenating these paths forms a cycle, H 1, of the facets in S 0 [ S 1. The remaining facets, S 2,of@B will be inserted" into H 1 carefully maintaining the cycle of facets at each insertion. Now S 2 consists of facets of the form Q = b 1 b 2 b 3 b 4 where 0 < b 1» a 1 and a 3 < b 3. (Otherwise, Q +0» RL F or R(F; a 1 ) 2 B contradicting the definition of F.) Let W i = fq = b 1 b 2 b 3 b 4 : Q 2 S 2 and b 3 = a 3 + ig. For each oddi, consider W i and W i+1. If W i+1 = ; then set Wi 0 = ;. Suppose W i+1 6= ;. Let P i+1 = max <RL fq : Q 2 W i+1 g and P i = L(P i+1 ;b 4 ) 2 W i where P i+1 = b 1 b 2 b 3 b 4. Let Wi 0 = W i+1 [fp : P 2 W i and P» RL P i g. See Figure 8, in which P 2 is F 5 0 and P 1 is F 2 0. Listing the elements of W i Wi 0 from least to greatest in reverse lexicographic order, and the remainder of the elements of Wi 0 from greatest to least in reverse lexicographic order, 11

12 F F F F F F F F Figure 8: The 4-ball: V 1 = ff 1 ;F 2 ;F 3 g and V 2 = V 1+1 = ff 4 ;F 5 g. we have Wi 0 = f12bb 3 ; 23bb 3 ;:::;b 1 b 2 bb 3 = P i ;b 1 b 2 b 3 b 4 = P i+1 ;L(P i+1 ;b 2 );:::;12b 3 b 4 g, where b = b 3 1=`(P i+1 ;b 4 ). Note that the above ordering is a Hamiltonian path of the facets in Wi 0. Now 012bb 3 and 012b 3 b 4 2 B and hence 01bb 3 and 01b 3 b 4 2 S 0. Also 01bb 3 is adjacent to 01b 3 b 4 in H 1. Insert Wi 0 in the above order between 01bb 3 and 01b 3 b 4 in H 1. Do this for each i for which Wi 0 6= ; to form a Hamiltonian cycle, H 2, on the facets in S 0 [ S 1 [fp : P 2 Wi 0 for some ig. Now we must explain what to do with the facets in S 2 [ i odd Wi 0. Let P belong to S 2 [ i odd Wi 0. Thus P 2 W i for some odd i, and P i < RL P if W i+1 6= ;. Note that S 2 [ i odd Wi 0 contains all of W i if W i+1 = ; and i is odd. Suppose i (odd) is such that W i+1 6= ;. Let Q 1 < RL Q 2 < RL < RL Q m be the facets in W i such that P i < RL Q j for all j. For each j, let G j = 0b j 1b j 2b 3 b 4, where P i = b 1 b 2 b 3 b 4, and Q j = b j 1b j 2b 3 b 4. Now for each j, G j b 3 is a facet since P i+1 =max <RL fq : Q 2 W i+1 g < RL R(G j ;b 3 ) 0 and hence R(G j ;b 3 ) 62 B. Note that for 1» j» m 1, G j b 3 is adjacent tog j+1 b 3. Also G m b m 1 is adjacent to G m b 3 in H 2. For each odd j < m insert G j 0 and G j+1 0 between G j b 3 and G j+1 b 3 in H 2. Observe that G j 0 is adjacent tog j+1 0sinceG j = L(G j+1 ;b j+1 2 ). If m is odd, place G m 0between G m b m 1 and G m b 3. Do this for each i for which W i+1 6= ;. Finally, for the possibly one i for which W i 6= ; and W i+1 = ;, let W i = fq 1 ;:::;Q m g 12

13 with Q 1 < RL < RL Q m and let G j =0b j 1b j 2b 3 b 4 where Q j = b j 1b j 2b 3 b 4. For each j, G j b 3 is a facet since W i+1 = ;. We can see that for 1» j» m 1, we have G j b 3 is adjacent to G j+1 b 3 and G m b m 1 is adjacent to G m b 3 in H 2. Now for each odd j < m insert G j 0 and G j+1 0between G j b 3 and G j+1 b 3 in H 2. Note that G j 0 is adjacent to G j+1 0 since G j = L(G j+1 ;b j+1 2 ). If m is odd, place G m 0between G m b m 1 and G m b 3. Doing all this creates a Hamiltonian cycle H of the facets of the squeezed 2 Figure 9 depicts the Hamiltonian cycle of Theorem 4.2 of the boundary facets of the 4-ball in Figure 6. 5 Remarks As mentioned at the beginning, the fact that the set of squeezed 3-spheres contains at least one polytopal representative for each simplicial f-vector implies that the f-vector cannot be the sole obstacle for a simple 4-polytope to be Hamiltonian. It would be nice to know whether the higher dimensional squeezed spheres are Hamiltonian as well. Despite the perhaps annoyingly specialized arguments developed to tackle this particular class of objects, squeezed spheres provide a fertile ground for testing or extending properties of simplicial polytopes and exploring the boundary between polytopal and non-polytopal spheres. 13

14 F F F F F F F Figure 9: A Hamiltonian cycle of the facets of the boundary of the squeezed 4-ball 14

15 References [1] L. J. Billera and C. W. Lee, A proof of the sufficiency of McMullen's conditions for f-vectors of simplicial convex polytopes, J. Combin. Theory Ser. A 31 (1981) [2] B. Grünbaum, Convex Polytopes, vol. 16 of Pure Appl. Math., Interscience Publishers [John Wiley & Sons, Inc.], New York, [3] D. A. Holton and B. D. McKay, The smallest non-hamiltonian 3-connected cubic planar graphs have 38 vertices, J. Combin. Theory Ser. B 45 (1988) [4], Erratum: The smallest non-hamiltonian 3-connected cubic planar graphs have 38 vertices" [J. Combin. Theory Ser. B 45 (1988) ; MR 90f:05090], J. Combin. Theory Ser. B 47 (1989) 248. [5] G. Kalai, Many triangulated spheres, Discrete Comput. Geom. 3 (1988) [6] J. Lederberg, Hamilton circuits of convex trivalent polyhedra (up to 18 vertices) Amer. Math. Monthly 74 (1967) [7] C. W. Lee, Kalai's squeezed spheres are shellable, Discrete Comput. Geom. 24 (2000) [8] M. Rosenfeld and D. Barnette, Hamiltonian circuits in certain prisms, Discrete Math. 5 (1973) Information on Authors Robert L. Hebble Kentucky State University Frankfort, Kentucky rhebble@gwmail.kysu.edu Carl W. Lee University of Kentucky Lexington, Kentucky lee@ms.uky.edu 15