Rectangles, integer vectors and hyperplanes of the hypercube
RRectangles, integer vectors and hyperplanes of the hypercube
E. Gioan , I. P. Silva
Abstract.
We introduce a family of nonnegative integer vectors - primitive vectors- defining hyperplanes of the real affine cube over C n := {− , } n and study theirproperties with respect to the rectangles of the cube. As a consequence we give ashort proof that, for small dimensions ( n ≤ We consider C n := {− , } n . A f f ( C n ) denotes the oriented matroid of theaffine dependencies of C n over R , the real affine cube .The oriented matroid A f f ( C n ) can be defined in several equivalentways. We recall its definition in terms of the family of signed cocircuits.A signed cocircuit of A f f ( C n ) is an ordered pair X = ( X + , X − )satisfying the following two conditions:1) H := C n \ ( X + ∪ X − ) is the set of all elements of C n spanning the sameaffine hyperplane H of R n .2) If the hyperplane H := C n \ ( X + ∪ X − ) is defined by a linear equation H : x . h = b ( h , b ) ∈ Z n +1 then either X + := { v ∈ C n : v . h > b } and X − := { v ∈ C n : v . h < b } or X + := { v ∈ C n : v . h < b } and X − := { v ∈ C n : v . h > b } .To every hyperplane H : x . h = b of the matroid A f f ( C n ) is associatedthe pair of opposite signed cocircuits, X = ( X + , X − ) and − X = ( X − , X + ).This pair encodes the 2-partition of C n \ H = X + (cid:93) X − into the points lyingon each one of the open half-spaces of R n defined by the affine hyperplanespanned by H.The oriented matroid A f f ( C n ) is defined by the collection of all itssigned cocircuits.We suggest that the reader consults [2] as general reference on orientedmatroids and [5] as general reference on matroids.The (signed) rectangles of C n are the shortest (signed) circuits of1 a r X i v : . [ m a t h . C O ] S e p f f ( C n ). They are the signed sets of the form ± R with R = ( { u , v } , { u (cid:48) , v (cid:48) } ) = u + v + u (cid:48) − v (cid:48) − , where ( u , u (cid:48) , v , v (cid:48) ) are the vertices of a rectangle of R n withdiagonals uv , u (cid:48) v (cid:48) . We denote by R the family of the signed rectangles thereal affine cube.The Facets of C n are the subsets: H i + := { v ∈ C n : v i = 1 } and H i − := { v ∈ C n : v i = − } , i = 1 , . . . , n. The skew-facets of C n are thesubsets: H ij + := { v ∈ C n : v i = v j } and H ij − := { v ∈ C n : v i = − v j } , ≤ i < j ≤ n. . They are hyperplanes of the real affine cube as well as itsshortest cocircuits.The signed cocircuits of A f f ( C n ) complementary of the facets and ofthe skew facets are denoted respectively: X i + = ( H i − , ∅ ), X i − = ( H i + , ∅ )and X ij + = ( H i + ∩ H j − , H i − ∩ H j + ), X ij − = ( H i + ∩ H j + , H i − ∩ H j − ).We denote F the family of the signed cocircuits complementary of thefacets of the skew facets of the real cube A f f ( C n ). Definition 1.1
An oriented cube ( canonically oriented cube of ([7])) is anoriented matroid over C n containing as signed circuits the signed rectanglesof R and, as signed cocircuits, the signed cocircuits of F . In this note we give a short proof of the following theorem (Theorem3.1) whose proof is mentioned in ([7]).
Theorem
For n ≤ the real affine cube A f f ( C n ) is the unique orientedcube. This is a small step towards an answer to the next Question 1. A posi-tive answer to this question with the results of [7] would imply the existenceof a purely combinatorial characterization of affine/linear dependencies of ± Question. ([7])
Is the real cube the unique oriented cube?
Our proof of the theorem uses the encoding of A f f ( C n ) in terms ofthe family H n of non-negative integer vectors of N n +10 that defines its hyper-planes and signed cocircuits up to symmetries of the non-oriented matroidof the real affine cube (see [6]). This is briefly recalled in section 3, wherethe complete proof is presented.In the next section 2 we define primitive vectors of C n and from therea recursive family of hyperplanes - primitive hyperplanes of A f f ( C n ) whosebehaviour with respect to the net of signed rectangles implies that the cor-responding signed cocircuits of the real cube must be signed cocircuits ofevery oriented cube. 2e point out that as a direct consequence of the main theorem weobtain (for n ≤
7) a new proof of the following conjecture of M. Las Vergnasalso open for n > Las Vergnas cube conjecture ([4]): A f f ( C n ) is the unique orientationof the (non-oriented) real affine cube. This conjecture was verified computationally for n ≤
7, by J. Bokowskietal [3] .
We are interested in nonnegative integer vectors h ∈ N n and howthey stratify the vertices of the cube C n ∈ R n into parallel levels (strata)orthogonal to h . Definition 2.1 ( h -levels of C n ) Given h ∈ N n we consider | h | := (cid:80) ni =1 h i .For every a ∈ { , . . . , | h |} the a -level of h , denoted S a ( h ) or simply S a , isthe set of vertices of the cube C n defined as: S a = S a ( h ) := { v ∈ C n : h . v = | h | − a } Clearly S | h |− a = − S a , in particular S = { } and S | h | = {− } .Eventually S a = ∅ . Definition 2.2 ( Realizable h - rectangles) An h - rectangle is a sequence of four natural numbers r = ( a ≤ b ≤ c ≤ d ) such that d = b + c − a and ≤ a and d ≤ | h | . We distinguish between , , and - rectangles according to the number of different h -levels theycross.A -rectangle is of the form r = ( a = a = a = a ) , a -rectangle is of theform r = ( a = a < b = b ) , -rectangle is of the form r = ( a < b = b < c ) and a -rectangle is of the form r = ( a < b < c < d ) ,An h -rectangle r = ( a ≤ b ≤ c ≤ d ) is realizable if there is a signed geometricrectangle, R = v + a v − b v − c v + d , of A f f ( C n ) such that v a ∈ S a ( h ) , v b ∈ S b ( h ) , v c ∈ S c ( h ) and v d ∈ S d ( h ) . Definition 2.3 (Embedding of levels)
Given two h − levels , S a , S b we saythat S a is embedded in S b , written S a (cid:44) → S b , if there is a spanning tree Γ a of the complete graph K ( S a ) such that for every edge { u , u (cid:48) } of Γ a ⊆ S a there is a pair of vertices v , v (cid:48) ∈ S b such that R = u + u (cid:48) − v (cid:48) − v + is a signedrectangle of the cube. Remark.
Notice that if S a is embedded in S b , S a (cid:44) → S b , then the affine spanof S a must be parallel to the affine span of S b . The notion of embedding oflevels is from this point view a combinatorial version of parallelism in R n . Notation.
We usually identify an element v ∈ C n with the subset α ⊆ [ n ]of its negative entries, more precisely with the sequence of the elements of α written by increasing order. Example: (1 , , , ∈ C ≡ ∅ , (1 , − , − , ∈ C ≡ h ∈ N n − and g = ( h , g ) ∈ N n . The g -levels and h -levels are related inthe following way, for every a ∈ N , 0 ≤ a ≤ | g | : S a ( g ) = S a ( h ) n + ∪ S a − g ( h ) n − where S a ( h ) n + = { ( v , ∈ C n : v ∈ S a ( h ) } and S a − g ( h ) n − = { ( v , − ∈ C n : v ∈ S a − g ( h ) } . Examples
1) The vector h = ( , , , ) has 6 levels: S = ∅ , S = { , , } , S = { , , , } , S = { , , , } , S = { , , } , and S = { } . Observe that for every a (cid:54) = 2 one has S a (cid:44) → S , as represented inFigure 2. Note also that S (cid:44) → S but S (cid:54) (cid:44) → S .2) The vector h = ( , , , ) has 9 levels. No level S b = S b ( h ) satisfies theproperty ∀ a (cid:54) = 2 , S a (cid:44) → S b . 4igure 2: The (1,1,1,2)-levels and embeddings of the levels S a , a (cid:54) = 2 in S . Definition 2.4 (primitive vectors)
A vector h ∈ N n is called a primitive vector if all its 3 and 4- numericalrectangles are realizable. Examples.
The vectors (1 , , , , , , C .The next Proposition gives, in particular, a recursive construction of primi-tive vectors of R n . Proposition 2.1 (Properties of primitive vectors)
Let h = ( h , . . . , h n ) be a primitive vector of C n . Then:(i) For every ≤ a ≤ | h | , the a -level S a of h is nonempty , moreover for a (cid:54) = 0 , | h | , every level S a contains at least two elements.(ii) For every g ∈ N , ≤ g ≤ | h | + 1 , g = ( h , g ) is a primitive vector of C n +1 . Proof. (i) Imediate from the definition, since every 3-rectangle must be re-alizable.(ii) We have S a ( g ) = S a ( h )( n + 1) + ∪ S a − g ( h )( n + 1) − .Let r = a < b ≤ c < d be a 3 or 4- rectangle of g = ( h , g ). If d < | h | or a > g the rectangle r is certainly h - realizable. In the first case with elements of S a ( h )( n + 1) + , in the second case with elements of S a − g ( h )( n + 1) − .5f a < g and d > | h | , then, since a + d = b + c > | h | we can guarantee that c ≥ g . Consider the rectangle r (cid:48) = ( a < b ; c − g < d − g ). If r (cid:48) is a 2-rectanglethen a = c − g and b = d − g . Take v a ∈ S a ( h )( n + 1) + and v b ∈ S b ( h )( n +1) + . Then v c = ( v a , − ∈ S c ( g ) and v d = ( v b , − ∈ S d ( g ) and R (cid:48) =(( v a , , ( v b , v c , , ( v d , r (cid:48) is a 3 or 4 h -rectangle then itis realizable in S a ( h )( n + 1) + . Let R (cid:48) = (( v a , , ( v b , v c − g , , ( v d − g , r (cid:48) then R = (( v a , , ( v b , v c − g , − , ( v d − g , − r in C n . Theorem 2.1 (primitive vectors and oriented cubes)
Let M = M ( C n ) be an oriented cube and cl : 2 C n −→ C n its closureoperator.Assume that there is a primitive vector h of C n such that some h -level S b has the following property: ( E ) ∀ a (cid:54) = b S a (cid:44) → S b Then either(i) cl ( S b ) = C n , or(ii) cl ( S b ) = S b is a hyperplane of M whose complement is the support ofthe pair ± X b of signed cocircuits of M defined by X b = ( ∪ ab S a ) . Proof.
First notice that, by definition of embedding, the hypothesis that S b satisfies condition ( E ) implies that if one element v ∈ S a , a (cid:54) = b belongs to cl ( S b ) then S a ⊂ cl ( S b ). On the other hand if some element v ∈ S a , a (cid:54) = b is not cl ( S b ) then, by orthogonality with the signed rectangles of the em-bedding, R = ( u + a , u (cid:48) − a , v (cid:48) − b , v + b ), all the elements of S a must have the samesign in any signed covector V = ( V + , V − ), complementary of cl ( S b ), i.e. onemust have either S a ⊆ V + or S a ⊆ V − . With these remarks in mind wenow prove the theorem. Claim 1. if there is v ∈ cl ( S b ) such that v ∈ S b − ∪ S b +1 then cl ( S b ) = C n . We may assume n > b (cid:54) = 0 , | h | .Assume that there is v ∈ S b − ∩ cl ( S b ). In this case, by the above remarks S b − ⊂ cl ( S b ). And because the 3-rectangle r = ( b − < b = b < b + 1)is realizable, orthogonality of the signed covectors complementary of cl ( S b )with any geometric realization R = u + b − u − b v − b v + b +1 of r , implies that v b +1 must be in cl ( S b ) and therefore S b +1 ⊆ cl ( S b ).Similarly if v ∈ S b +1 ∩ cl ( S b ) orthogonality with the same geometric re-alization R of the 3-rectangle r = ( b − < b = b < b + 1) implies that S b +1 ∪ S b − ⊆ cl ( S b ). So, if some v ∈ S b − ∪ S b +1 in the closure of S b thenboth levels S b − and S b +1 are contained in cl ( S b ).6ow, once S b +1 ∪ S b − ⊆ cl ( S b ), considering the sequence of geometric real-izations of the 3-rectangles r b = ( b < b + 1 = b + 1 < b + 2), r b +1 = ( b + 1 b S a ). This provessimultaneously that cl ( S b ) = S b and that X b must be a cocircuit of M .If ( S b − ∪ S b +1 ) ∩ cl ( S b ) = ∅ then cl ( S b ) is a flat of M whose comple-ment is the support of (signed) covetors of M . Using a realization R =( u + b − , u − b , v − b , v + b +1 ) of the 3-rectangle r = ( b − < b = b < b +1) we concludethat S b − and S b +1 must have different signs in any covector V = ( V + , V − )complementary of cl ( S b ). We may assume w.l.o.g. that S b − ⊆ V + and that S b +1 ⊆ V − .Consider the sequence of 3-rectangles with the ”upper” vertex in S b or S b − defined by: r b +2 = ( b < b + 1 = b + 1 < b + 2), s b +3 = ( b − < b + 1 = b + 1
Let h be a primitive vector of C n − for which there exists b ∈ N satisfyingthe following two conditions:(h-i) ∀ a ∈ N , < a < | h | , a (cid:54) = b S a ( h ) (cid:44) → S b ( h ) (h-ii) The signed set X b := ( ∪ ab S a ( h )) is a signed cocir-cuit of every oriented cube M ( C n − ) Then, every vector g = ( h , g ) ∈ N n , with ≤ g ≤ b , satisfies the followingtwo properties:(g-i) ∀ a ∈ N , < a < | g | , a (cid:54) = b S a ( g ) (cid:44) → S b ( g ) .(g-ii) The signed set ˜ X b = ( ∪ ab S a ( g )) is a signed cocir-cuit of every oriented cube M ( C n ) . Proof.
Let ˜ M = M ( C n ) be an oriented cube. Consider vectors h ∈ N n − and g = ( h , g ) ∈ N n in the conditions of the Theorem. Consider ˜ H := cl ( S b ( g )).We prove the theorem in three steps. (I) ˜ H is a hyperplane of ˜ M , whose pair of signed cocircuits restricts in thefacet H n + of ˜ M to the pair of signed cocircuit ± X b of S b ( h ). (II) g satisfies ( g − i ). (III) The unique extension of the cocircuit X b of the facet H n + to a signedcovector of ˜ M complementary of S b ( g ) and orthogonal to the rectangles of C n is the signed set ˜ X b defined ( g − ii ). (I) Since h satisfies ( h − i ) we know that S b − g ( h ) (cid:44) → S b ( h ) implying alsothat S b − g ( h ) n − (cid:44) → S b ( h ) n + . Consequently the ranks of S b ( g ) and S b ( h ) n + are related by:(1) rk ( S b ( g )) = rk ( S b ( h ) n + ) + 1 . The restriction of the canonically oriented cube ˜ M to a facet is an orientedcube of rank rk ( ˜ M ) −
1. The set H := S b ( h ) n + is, by condition ( h − i ) ahyperplane of the facet H n + of ˜ M and therefore a hyperline of ˜ M implyingthat rk ( H ) = rk ( ˜ M ) −
2. Replacing in (1) we have:(2) rk ( S b ( g )) = rk ( H ) + 1 = rk ( ˜ M ) − H := cl ˜ M ( S b ( g )) must be a hyperplane of ˜ M and its complementthe support of a pair of signed cocircuits of ˜ M . Moreover, one of thesesigned cocircuits of ˜ M must restrict in the facet H n + to the signed cocircuit X b := ( ∪ ab S a ( h ) n + ) completing the proof of Step (I).8 II) g satisfies ( g − i ) . Notice that for every a, ≤ a < g , S a ( g ) = S a ( h ) n + and since h satisfies( h − i ) it is clear that in this case we have: S a ( g ) n + (cid:44) → S b ( h ) n + and therefore S a ( g ) n + (cid:44) → S b ( g ). For a > | h | , since S a ( g ) = S a − g ( h ) n − (cid:44) → S b ( h ) n + andthe result in this case is also a direct consequence of the fact that h satisfies( h − i )For a, g ≤ a ≤ | h | , S a ( g ) = S a ( h ) n + ∪ S a − g ( h ) n − intersects both facets H n + and H n − . The hypothesis that h satisfies ( h − i ) guarantees that S a ( g ) n + (cid:44) → S b ( h ) n + and also that S a − g ( h ) n − (cid:44) → S b ( h ) n + so, in orderto conclude that S a ( g ) (cid:44) → S b ( g ) it is enough to prove that there is a ge-ometric realization of the 2-rectangle r = ( a = a ; b = b ) of the form R = (( u a , + , ( u a − g , − − , ( v b − g , − − , ( v b , + ) with ( u a , ∈ S a ( h ) n + ,( u a − g , − ∈ S a − g ( h ) n − , ( v b − g , − ∈ S b ( h ) n − and ( v b , ∈ S b ( h ) n + .Consider the 3 or 4 rectangle of r (cid:48) = ( a − g < a ; b − g < b ) of h . The vector h is a primitive vector of C n − so there is a geometric realization R (cid:48) =( u + a − g , u − a ; v − b − g , v + b ) of this rectangle in C n − leading the desired realizationof the 2-rectangle r in C n and concluding the proof that g satisfies ( g − i ). (III) g satisfies ( g − ii )We know from ( I ) that there is a signed cocircuit ˜ X of ˜ M complementaryof the hyperplane ˜ H whose restriction to the facet H n + is X b . The fact that g satisfies ( g − i ), proved in (II), implies directly that the unique extensionof X b to H n − \ S b − g ( h ) n − orthogonal to the rectangles of ˜ M must satisfythe conditions: ∪ a | h | the fact that they are embeddedin S b ( g ) implies that all the elements in each level will have the same signhowever some more arguing is needed before concluding that they must allbe negative in ˜ X .We consider separately two cases: Case 1) a ≤ b and case 2) a > b . Case 1)
In this case consider the 3-rectangle of g : r = (2 b − a < b = b < a ). If the rectangle r (cid:48) = ( b − g < a − g ; 2 b − a < b ) of h is a 3or 4 rectangle of h then, since h is primitive, r (cid:48) is realizable. Considera realization R (cid:48) = ( u b − g , v a − g ; v (cid:48) b − a , u (cid:48) b ) of r (cid:48) in C n − . Then clearly R = ( v (cid:48) b − a , + ( u (cid:48) b , − ( u b − g , − − ( v a − g , − + is a geometric real-ization of r in C n and by orthogonality with this circuit in any exten-sion of X b to H n − \ S b − g ( h ) n − the sign of ( v a − g , − S a ( g ) ⊆ ˜X − . In the case r (cid:48) is a 2-rectangle of h , whichoccurs when g = a − b , take u b − g ∈ S b − g ( h ) and v b ∈ S b ( h ). Clearly R = ( u b − g , + ( v b , − ( u b − g , − − ( v b , − + is a geometric realization of r in C n and we also conclude that S a ( g ) ⊆ X − .9 ase 2) In this case r = ( | h | − ( a − b ) < b < | h | < a ) is a 4-rectangleof g (notice that b < | h | < a ) and r (cid:48) = ( b − g < a − g ; | h | − ( a − b ) < | h | ) is a rectangle of h . If r (cid:48) is a 3 or 4-rectangle of h then, since h isprimitive, there is a realization ( u b − g , v a − g ; v (cid:48)| h |− ( a − b ) , u (cid:48)| h | ) of r (cid:48) in C n − and R = v (cid:48)| h |− ( a − b ) , + ( u b − g , − − ( u (cid:48)| h | , − ( v a − g − + ) is geometricrealization of r . By orthogonality with this circuit we conclude as beforethat S a ( g ) ⊆ ˜ X − . The case r (cid:48) is a 2-rectangle the argument is similar as inthe previous case, leading to the conclusion that the unique extension of X b to H n − \ S b − g ( h ) n − orthogonal to the rectangles of ˜ M is the signed vector˜ X b of the theorem, thus proving that g satisfies ( g − ii ). (cid:3) The next Corollary whose proof is left to the reader restates Theorem3.2 as a recursive procedure for constructing from signed cocircuits of everycanonically oriented cube over C n signed cocircuits of every canonicallyoriented cube over C n +1 . Corollary 2.1
Let h ∈ N n0 be a primitive vector such that the signed cocir-cuit of A f f ( C n ) defined by S b ( h ) : x . h = | h | − b is a signed cocircuit ofevery oriented cube M ( C n ) . For every c ≤ b let g := ( h , b − c ) , then thesigned cocircuit of A f f ( C n +1 ) defined by S b ( g ) : g . x = | g | − b is a signedcocircuit of every oriented cube M ( C n +1 ) . We are now ready to prove the main theorem.
Theorem 3.1
For n ≤ the oriented matroid A f f ( C n ) is the unique ori-ented cube. We recall from [7] that every oriented cube over C n must have rank n +1. The next Proposition about matroids then guarantees that in order toprove theorem 3.1 we ”only” have to prove that every signed cocircuit of A f f ( C n ) must be a signed cocircuit of every oriented cube M ( C n ). Proposition 3.1
Consider two matroids M , M (cid:48) , witout loops, over thesame set E and with the same rank r . Let H , H (cid:48) denote the families ofhyperplanes respectively of M and M (cid:48) and assume that H ⊆ H (cid:48) . Then H = H (cid:48) and M = M (cid:48) . roof. Suppose that there is a hyperplane H (cid:48) ∈ H (cid:48) \ H . Let B = { h , . . . , h r − } be a basis of H (cid:48) in M (cid:48) . Since | B | < r , B is contained insome hyperplane H ∈ H of M . The assumption that H ⊂ H (cid:48) implies that H ∈ H (cid:48) and in M (cid:48) we have B ⊂ H ∩ H (cid:48) . Now H ∩ H (cid:48) is a flat of M (cid:48) whoserank is less then the rank of H (cid:48) contradicting the assumption that B is abasis of H (cid:48) .The explicit definition of A f f ( C n ) for n ≤ H n of non-negative integer vectors, up to automorphisms of the class of orientations(i.e. of the unsigned arrangement of hyperplanes representing the orientedmatroid) H n := { ( h ≤ h ≤ . . . ≤ h n ≤ h n +1 ) ∈ N n +10 : gcd ( h , . . . , h n +1 ) = 1 and x . ( h , . . . , h n ) = h n +1 defines a hyperplane of Af f ( C n ) } . Each vector ( h ≤ h ≤ . . . ≤ h n ≤ h n +1 ) ∈ H n of the form (0 α , h α , . . . h α k k )with α ≥ α , . . . , α k ≥ α + α + . . . + α k = n + 1 determines ex-actly ( n +1)! α ! ...α k ! n − α distinct hyperplanes (and pairs of signed cocircuits ) of A f f ( C n ).We recall that there is a natural procedure to generate vectors of H n fromvectors of H n − and a recursive family G n ⊆ H n that we briefly recall from[6]. Definition 3.1 (the family G n ) [6] Given h = ( h ≤ h ≤ . . . ≤ h n ) ∈ H n − . For i = 1 , . . . n let h i :=( h , . . . , h i − , h i +1 , . . . , h n ) , then the level S b ( h i ) with b = | h i |− h i is a hyper-plane of A f f ( C n − ) and for every c ≤ b , such that S c ( h i ) (cid:54) = ∅ the vector g := ( h i , b − c, | h i | − b − c ) is a vector of H n .We denote by G n the family of vectors of H n which is obtained (up to re-ordering of entries) in this way. The next Theorem gives the explicit definition of H n as well as therelation between H n and G n , for n ≤ Theorem A ([6])Consider H n := { ( h ≤ . . . ≤ h n ≤ h n +1 ) ∈ N n +10 : gcd ( h , . . . , h n +1 ) =1 and ( h , . . . , h n ) . x = h n +1 def ines a hyperplane of A f f ( C n ) } . The list H n as well as its sublist G n of definition 3.1 is the following for n ≤ H = G = { (1 , } ; H = G = { (0 , , } ; H = G = { (0 , , , , (1 , , , } ; 11 = G = { (0 , , , , , (0 , , , , , (1 , , , , } ; H = G = { (0 , , , , , , (0 , , , , , , (0 , , , , , , (1 , , , , , , (1 , , , , , , (1 , , , , , , (1 , , , , , } ; H = G = { (0 , , , , , , , (0 , , , , , , , (0 , , , , , , (0 , , , , , , , (0 , , , , , , , (0 , , , , , , , (0 , , , , , , , (1 , , , , , , , (1 , , , , , , , (1 , , , , , , , (1 , , , , , , , (1 , , , , , , , (1 , , , , , , , (1 , , , , , , , (1 , , , , , , , (1 , , , , , , , (1 , , , , , , , (1 , , , , , , , (1 , , , , , , , (1 , , , , , , , , , , , , } ; H = G ∪{ (1 , , , , , , , , (1 , , , , , , , , (1 , , , , , , , , (1 , , , , , , , , (1 , , , , , , , , (1 , , , , , , , , (1 , , , , , , , , (2 , , , , , , , , (2 , , , , , , , , (2 , , , , , , , , (2 , , , , , , , , (2 , , , , , , , } . We are now ready to prove the main Theorem.
Proof of Theorem 3.1.
Consider an oriented cube M ( C n ).In order to prove that every signed cocircuit defined by a vector h ∈H n is a signed cocircuit of M ( C n ) it is enough to prove that one of thesigned cocircuits defined by an equation h i . x = h i is a signed cocircuit of M ( C n ). This is a consequence of the behaviour of the families of signedsets, R and F , under the symmetries of the real cube ([7]). Cases n ≤ . For n ≤ h ∈ H n , h i is a primitive vector for every i = 1 , . . . n +1.Theorem 3.2. then guarantees that all the hyperplanes and correspondingsigned cocircuits of A f f ( C n ) defined by vectors of G n +1 = H n +1 must besigned cocircuits of M ( C n ) and therefore, by Proposition 3.2 M ( C n ) = A f f ( C n ). Case n = 7 . Claim 2.A)
The signed cocircuits of A f f ( C ) defined by the vectos of G must be signed cocircuits of M ( C ) . The arguments applied in the previous cases to vectors of H leadto the conclusion that the hyperplanes and signed cocircuits of A f f ( C )defined by 109 of the 131 vectors of G must be hyperplanes and signedcocircuits of M ( C ). The remaining 22 vectors of G arise from the fol-lowing 5 vectors of H : (1 , , , , , , , (1 , , , , , , , (1 , , , , , , , (1 , , , , , ,
5) and (1 , , , , , , . h i that is not primitive.In the case of the first 4 vectors, the restriction h , in the case of thefifth the restriction h so we can not apply Theorem 3.2 to conclude thatthe hyperplanes of G n obtained extending the hyperplanes h i . x = h i = S b ( h i ) , with b = | h i |− h i in the facet H + with an h i -level in the facet H − must be signed cocircuits of M ( C ) needs some further verification.Actually in all the cases the proof goes in the same way, generalizingthe proof of theorem 3.2:First we verify that in all the 5 cases the h i -levels embed in the level S b ( h i ) : h i . x = b ( b = | h i |− h i ).Secondly, for all c, ≤ c ≤ b for which g := ( h i , b − c, | h i |− b − c ) is oneof the 109 vectors of G n arising from primitive restrictions of vectors of H there is nothing to prove. We retain those c (cid:48) s (e.g. c ≥ h ) for which g is oneof the remaining 22 vectors of G n . For those we verify that S a ( g ) (cid:44) → S b ( g ).Note that for that it is enough to prove that there is a 2-rectangle whoseedges in each level connect the facets H + and H − .The embeddings S a ( g ) (cid:44) → S b ( g ) imply: first, that for every v ∈ S c ( h i )7 − the hyperplane cl ( S b ( h i )7 + ∪ v ) of M ( C ) must contain all theelements of S c ( h i )7 − . Next, that the only way of extending the signedcocircuit of S b ( h i ) in the facet H + to the complementary of S c ( h i )7 − inthe facet H − , orthogonally to the signed rectangles, must be the signedcocircuit of A f f ( C ) complementary of S b ( g ).Once Claim 2A. is proved we start proving: Claim 2.B)
All the signed cocircuits of A f f ( C ) determined by the vec-tors of H \ G must be signed cocircuits of M ( C )Of these twelve vectors eight differ in exactly one entry from a vectorof G , meaning that they span hyperplanes of R n parallel to the affine spanof some hyperplane of A f f ( C ) that we already know is a hyperplane of M ( C ).These eight vectors, together with the vector of G differing in oneentry are:(1 , , , , , , , , (1 , , , , , , , and (1 , , , , , , ,
7) all dif-fering in one entry from (1 , , , , , , , ∈ G ,(1 , , , , , , ,
8) and (1 , , , , , , , ∈ G ,(2 , , , , , , ,
5) and (1 , , , , , , , ∈ G ,(2 , , , , , , ,
7) and (1 , , , , , , , ∈ G .In all the cases, let h be a restriction of the vector of G defining ahyperplane S c ( h ) that we already know must be in M ( C ) and let S b ( h ) bethe level corresponding to the hyperplane of A f f ( C ) whose pair of signed13ocircuits we want to conclude must be in M ( C ). In order to do so weproceed in all the cases in the same way:(i) verify that in M , rk ( S b ( h )) ≤
7. For that, if necessary using the rectan-gles, we reduce to 7 the number of elements needed to span all the elementsof that h -level.(ii) verify that for all a (cid:54) = b , S a ( h ) (cid:44) → S b ( h ). This implies, in particular,that the hyperplane S c ( h ) is embedded in S b ( h ) and consequently, for anelement v a ∈ S a we have rk ( cl ( S b ∪ v a )) ≤ rk ( S b ) + 1. On the otherhand the embedding S a (cid:44) → S b implies that rk ( cl ( S b ∪ v a )) = rk ( cl ( S b ∪ S a )) = rk ( M ) = 8 and we conclude that rk ( S b ) = 7 and cl ( S b ) must be ahyperplane of M .(iii) the fact that for all a (cid:54) = b , S a (cid:44) → S b implies that in the signed cocircuitcomplementary of cl ( S b ) all the elements of S a must all have the same sign,or be all contained in the hyperplane cl ( S b ). In the eight cases a ”ladder”of signed rectangles leads the conclusion that the pair of signed cocircuitsof A f f ( C ) complementary of S b is the unique signature of the complementof S b orthogonal to the rectangles of C and therefore, these cocircuits of A f f ( C ) must be cocircuits of M .Of the four remaining vectors of H \ G the following two:(1 , , , , , , ,
9) and (1 , , , , , , ,
6) differ in exactly one entryfrom the vectors, respectively, (1 , , , , , , ,
7) and (1 , , , , , , ,
6) of H whose signed cocircuits we already know must be in M and we pro-ceed exactly as before in order to conclude that the corresponding signedcocircuits of A f f ( C ) must be signed cocircuits of M too.Finally we are left with the two vectors (1 , , , , , , ,
5) and(1 , , , , , , , S b ( h ) := h . x = h ( b = | h |− h ) mustbe cocircuits of M . In both cases the hyperplane contains exactly sevenelements of C , therefore we know that rk ( cl ( S b )) ≤ cl ( S b ) = S b is a hyperplane whose complement,signed as in A f f ( C ) must be signed cocircuits of M we proceed as before:first, we verify that ∀ a (cid:54) = b, S a (cid:44) → S b . Next use the net of signed rectanglesto conclude that if one element of a different level S a belongs to cl ( S b ) then cl ( S b ) = C , implying that cl ( S b ) = S b . Then, essentially the same ladderof rectangles proves the unicity of the signature, concluding the Proof ofTheorem 3.1. 14 Final Remarks
In this note we defined recursively a family of nonnegative integer vectorsorthogonal to hyperplanes of the real cube - primitive vectors. We provedthat the corresponding primitive hyperplanes and signed cocircuits of thereal cube A f f ( C n ) must be hyperplanes and signed cocircuits of every ori-ented cube. The proof relied upon a direct characterization of the behaviourof these nonnegative integer vectors with respect to the net of rectangles of A f f ( C n ).Although explicit descriptions of the affine cube for n = 8 have beencomputed [1], and our methods apply to this case, an approach of this nextcase without computer aid would still be too lengthy. A better understand-ing of the real affine cube and its symmetries is needed.For instance, note that the hyperplanes defined by primitive vectorsare strictly contained in the family G n (definition 3.1) which for n ≥ H n . Although we think it must be true, wewere not able to prove that the hyperplanes and signed cocircuits determinedby the vectors of G n must be hyperplanes and signed cocircuits of everyoriented cube. This would, in particular, simplify further our proof of thecase n = 7. References [1] O. Aschholzer, F. Aurenhammer, Classifying hyperplanes in hyper-cubes, SIAM J. Discrete Math. (1) (1996), 225-232.[2] Bj¨orner, M. Las Vergnas, B. Sturmfels, N. White, G. Ziegler, OrientedMatroids , Encyclopedia of Math. Appl., vol. 46, Cambridge UniversityPress, 1993.[3] J.Bokowski, A. Guedes de Oliveira, U. Thiemann, A. Veloso da Costa,On the cube problem of Las Vergnas,
Geom. Dedicata (1) (1996),25-43.[4] M. Las Vergnas, J.P. Roudneff, I. Sal¨aun, Regular polytopes, preprint1989, 12 pp.(unpublished).[5] J. Oxley, Matroid Theory , 2nd ed, Oxford University Press 2011.[6] I.P.F. da Silva, Recursivity and geometry of the hypercube,
Lin. Alg.and its Appl. , (2005), 223-233.[7] I.P.F. da Silva, Cubes and orientability, Discrete Mathematics (2008), 3574-3585. 15 . Gioan e-mail adress: [email protected] (UMR 5506-CC477), 161,r. Ada 34095 Montpellier Cedex 5 -FRANCE