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New examples of oriented matroids with 2 disconnected realization spaces 1 0 2 Yasuyuki Tsukamoto ∗ n a January 13, 2012 J 2 1 Abstract ] O We construct oriented matroids of rank 3 on 13 points whose realiza- C tion spaces are disconnected. They are defined on smaller points than . the known examples with this property. Moreover, we construct the one h on 13 points whose realization space is a connected and non-irreducible t a semialgebraic variety. m [ 1 Oriented Matroids and Matrices 1 v Throughout this section, we fix positive integers r and n. 0 Let X = (x ,...,x ) Rrn be a real (r,n) matrix of rank r, and E = 6 1 n ∈ 5 1,...,n be a set of labels of the columns of X. For such matrix X, a map { } 2 χ can be defined as X . 1 0 χX :Er →{−1,0,+1}, χX(i1,...,ir):=sgndet(xi1,...,xir). 2 1 The map χ is called the chirotope of X. The chirotope χ encodes the infor- X X : mation onthe combinatorialtype which is called the oriented matroid of X. In v i this case, the oriented matroid determined by χ is of rank r on E. X X We noteforsomepropertieswhichthe chirotopeχ ofamatrixX satisfies. X r a 1. χ is not identically zero. X 2. χ is alternating, i.e. χ (i ,...,i )=sgn(σ)χ (i ,...,i ) X X σ(1) σ(r) X 1 r for all i ,...,i E and all permutation σ. 1 r ∈ 3. For all i ,...,i ,j ,...,j E such that 1 r 1 r ∈ χX(jk,i2,...,ir)·χX(j1,...,jk−1,i1,jk+1,...,jr)≥0 for k =1,...,r, we have χ (i ,...,i ) χ (j ,...,j ) 0. X 1 r · X 1 r ≥ ∗Department of Mathematics, Kyoto University, Sakyo-ku, Kyoto 606-8502, Japan [email protected] 1 The third property follows from the identity det(x ,...,x ) det(y ,...,y ) 1 r 1 r · r = det(yk,x2,...,xr) det(y1,...,yk−1,x1,yk+1,...,yr), · Xk=1 for all x ,...,x ,y ,...,y Rr. 1 r 1 r ∈ Generally, an oriented matroid of rank r on E (n points) is defined by a mapχ:Er 1,0,+1 ,which satisfies the above three properties([1]). The →{− } map χ is also called the chirotope of an oriented matroid. We use the notation (E,χ) for an oriented matroid which is on the set E and is defined by the M chirotope χ. An oriented matroid (E,χ) is called realizable or constructible, if there M exists a matrix X such that χ = χ . Not all oriented matroids are realizable, X but we don’t consider non-realizable case in this paper. Definition1.1. Arealization ofanorientedmatroid = (E,χ)isamatrix M M X such that χ =χ or χ = χ. X X − ′ Two realizations X,X of is called linearly equivalent, if there exists a linear transformation A GLM(r,R) such that X′ = AX. Here we have the ∈ equation χ =sgn(detA) χ . X′ · X Definition 1.2. The realization space ( ) of an oriented matroid is the R M M setofalllinearly equivalentclasses of realizations of , inthequotienttopology induced from Rrn. M Our motivation is as follows: In 1956, Ringel asked whether the realization spaces ( )arenecessarilyconnected[4]. Itisknownthateveryorientedma- R M troid on less than 9 points has a contractible realization space. In 1988, Mnëv showed that ( ) can be homotopy equivalent to an arbitrary semialgebraic R M variety [2]. His result implies that they can have arbitrary complicated topo- logical types. In particular, there exist oriented matroids with disconnected realization spaces. Suvorov and Righter-Gebert constructed such examples of orientedmatroidsofrank3on14points,in1988andin1996respectively[5,3]. Howeveritisunknownwhichisthesmallestnumberofpointsonwhichoriented matroids can have disconnected realization spaces. See [1] for more historical comments. One of the main results of this paper is the following. Theorem 1.3. There exist oriented matroids of rank 3 on 13 points whose realization spaces are disconnected. Realization spaces of oriented matroids are semialgebraic varieties. So it makes sense whether a realization space is irreducible or not. As far as the author knows, no example of an oriented matroid with a connected and non- irreducible realization space has been explicitly given. 2 Theorem 1.4. There exists oriented matroid of rank 3 on 13 points whose realization space is connected and non-irreducible. Acknowledgment. I would like to thank Masahiko Yoshinaga for valuable discussions and comments. I also thank Yukiko Konishi for comments on the manuscript. 2 Construction of the examples Throughout this section, we set E = 1,...,13 . We will define three chirotopes χ−{,χ0 and χ}+. We will see that (E,χ−) M and (E,χ+) have disconnected realization spaces, and (E,χ0) has a con- M M nected and non-irreducible realization space. LetX(s,t,u)bea real(3,13)matrixwiththreeparameterss,t,u R given ∈ by X(s,t,u):=(x ,...,x ) 1 13 1 0 0 1 s s 0 1 1 st s+t u st+su − − = 0 1 0 1 0 1 t t u t t u+su − 0 0 1 1 1 1 1 1 0 1 su 1 u+su  − − s+t st s2u s(t u+su) − − − t t u+su . − 1 su 1 u+su − −  This is a consequence of the computation of the following construction sequence. Bothoperations“ ”and“ ”canbecomputedintermsofthestandard crossproduct“ ”inR3. T∨hewhol∧econstructiondepends onlyonthe choiceof the three param×eters s,t,u R. ∈ x =t(1,0,0), x =t(0,1,0), x =t(0,0,1), x =t(1,1,1), 1 2 3 4 x =s x +x , 5 1 3 · x =(x x ) (x x ), 6 1 4 2 5 ∨ ∧ ∨ x =t x +x , 7 2 3 · x =(x x ) (x x ), 8 1 7 2 4 ∨ ∧ ∨ x =u x +x , 9 2 1 · x =(x x ) (x x ), 10 7 9 3 6 ∨ ∧ ∨ x =(x x ) (x x ), 11 4 5 8 9 ∨ ∧ ∨ x =(x x ) (x x ), 12 1 10 4 5 ∨ ∧ ∨ x =(x x ) (x x ). 13 3 6 1 11 ∨ ∧ ∨ 3 We setX =X 1,1,1 . The chirotopeχǫ is the alternatingmapsuchthat 0 2 2 3 (cid:0) (cid:1) ǫ if (i,j,k)=(9,12,13), χǫ(i,j,k)= (cid:26) χ (i,j,k) otherwise, X0 for all (i,j,k) E3(i<j <k), ∈ where ǫ ,0,+ . ∈{− } The oriented matroid which we will study is ǫ := (E,χǫ). M M Remark 2.1. We can replace X with X 1,1,u′ where u′ is chosen from 0 2 2 R 1,0,1,1,3,2,3 . We will study the c(cid:0)ase 0 <(cid:1)u′ < 1. If we choose u′ \{− 2 2 } 2 otherwise, we can get other oriented matroids with disconnected realization spaces. In the construction sequence, we need no assumption on the collinearity of x ,x ,x . Hence every realization of ǫ is linearly equivalent to a matrix 9 12 13 M X(s,t,u) for certain s,t,u, up to multiplication on each column with positive scalar. Moreover,we have the rational isomorphism ∗(χǫ) (0, )12 = ( ǫ), R × ∞ ∼R M where ∗(χǫ) := X(s,t,u) R3·13 χ = χǫ . Thus we have only to R { ∈ | X(s,t,u) } prove that the set ∗(χǫ) is disconnected (resp. non-irreducible) to show that R the realization space ( ǫ) is disconnected (resp. non-irreducible). R M The equation χ =χǫ means that X(s,t,u) sgndet(x ,x ,x )=χǫ(i,j,k), for all (i,j,k) E3. (1) i j k ∈ Wewrite someofthemwhichgivethe equationsontheparameterss,t,u. Note that for all (i,j,k) E3( i,j,k = 9,12,13 ),the sign is given by ∈ { }6 { } χǫ(i,j,k)=sgndet(x ,x ,x ) . i j k s=t=1/2,u=1/3 | From the equation sgndet(x ,x ,x ) = sgn(s) = sgn(1/2)= +1, we get s >0. 2 3 5 Similarly, we get det(x ,x ,x )=1 s>0, therefore 2 5 4 − 0<s<1. (2) From the equations det(x ,x ,x )=t>0,det(x ,x ,x )=1 t>0, we get 1 7 3 1 4 7 − 0<t<1. (3) Moreover,we have the inequalities det(x ,x ,x )=u>0, (4) 1 9 3 det(x ,x ,x )=1 t u>0, (5) 4 7 9 − − det(x ,x ,x )=t u>0, (6) 3 9 8 − det(x ,x ,x )=s t2 (1 s)u >0, (7) 5 13 7 − − det(x ,x ,x )=(1(cid:0) s) (1 t)2(cid:1) su >0. (8) 6 12 8 − − − (cid:0) (cid:1) 4 2 6 4 9 10 12 7 8 13 11 3 1 5 Figure 1: Column vectors of X . 0 From the equation det(x ,x ,x )=u(1 2s)(1 2t+tu su), we get 9 12 13 − − − sgn u(1 2s)(1 2t+tu su) =ǫ. (9) − − − (cid:0) (cid:1) Conversely,if we have Eqs. (2) - (9), then we get (1). We can interpret a (3,13) matrix as the set of vectors x ,...,x R3. 1 13 { } ⊂ After we normalize the last coordinate for x (i E 1,2,9 ), we can visualize i the matrix on the affine plane (x,y,1) R3 =∈R2\.{Figure}1 shows the affine { ∈ } ∼ image of X . See Figures 2, 3 for realizations of ǫ. 0 M Proof of Theorem 1.3. We prove that ∗(χ−) and ∗(χ+) are disconnected. R R From Eqs. (2) - (9), we obtain 0<s<1, 0<u<t<1 u, ∗(χ−)=(s,t,u) R3 (1 t)2 su>0, t2 (1 s−)u>0, , R ∼ ∈ −(1 2−s)(1 2t+t−u s−u)<0  − − −   0<s<1, 0<u<t<1 u, ∗(χ+)=(s,t,u) R3 (1 t)2 su>0, t2 (1 s−)u>0, . R ∼ ∈ −(1 2−s)(1 2t+t−u s−u)>0  − − − First, we show that ∗(χ−) is disconnected, more precisely, consisting of R two connected components, by proving the next proposition. Proposition 2.2. ∗(χ−)= (s,t,u) R3 0<s<1/2 ,0<u<min 1 t,(1−t)2, 2t−1 R ∼(cid:26) ∈ 1/2<t<1 (cid:26) − s t s (cid:27)(cid:27) − 1/2<s<1 t2 1 2t (s,t,u) R3 , 0<u<min t, , − . ∪(cid:26) ∈ 0<t<1/2 (cid:26) 1 s s t (cid:27)(cid:27) − − 5 2 2 4 4 6 6 9 9 10 12 7 8 13 11 10 12 7 8 11 13 5 1 5 1 3 3 Figure 2: Realization of −(on the left) and that of +(on the right). M M 2 2 4 4 6 6 9 9 10 12 7 8 10 12 13 11 7 8 13 11 5 1 5 1 3 3 Figure 3: Realizations of 0. M 6 Proof. There are two cases 1 2s>0, 1 2t+tu su<0, − − − (1 2s)(1 2t+tu su)<0  or − − − ⇔ 1 2s<0, 1 2t+tu su>0. − − − Note that  (2 u)(2t 1)= 2(1 2t+tu su)+u(1 2s), (10) − − − − − − t2 (1 s)u= (1 2t+tu su)+(1 t)(1 t u), (11) − − − − − − − − (1 t)2 su=(1 2t+tu su)+t(t u). (12) − − − − − ( ) For the case 1 2s>0 and 1 2t+tu su<0, the inequality 2t 1>0 ⊂ − − − − follows from Eq. (10). Since we have 0<s<1/2<t<1, we get 1 2t+tu su<0, − − (1 t)2 2t 1  (1 t)2 su>0, u<min 1 t, − , − . (13)  1 −t u−>0 ⇔ (cid:26) − s t−s (cid:27) − −  For the other case 1 2s < 0, similarly, we get 1 2t > 0 from Eq. (10). − − Since we have 0<t<1/2<s<1, we get 1 2t+tu su>0, − − t2 1 2t  t2 (1 s)u>0, u<min t, , − . (14)  t −u>−0 ⇔ (cid:26) 1−s s−t (cid:27) −  ( ) For the component 0<s<1/2<t<1, the inequalities 1 2t+tu su< ⊃ − − 0, (1 t)2 su>0, 1 t u>0followfrom(13). Thuswegett2 (1 s)u>0 − − − − − − from Eq. (11). The inequality u<t holds because t>1/2 and u<1 t. − For the other component 0 < t < 1/2 < s < 1, similarly, we get the inequalities 1 2t+tu su > 0, t2 (1 s)u > 0, t u > 0 from (14), and − − − − − (1 t)2 su > 0 from Eq. (12). Last, we get u < 1 t from t < 1/2 and − − − u<t. For the set ∗(χ+), we have the following proposition. R Proposition 2.3. ∗(χ+)= (s,t,u) R3 0<s<1/2,0<u<1/2, (1 s)u<t< 1−su R ∼(cid:26) ∈ (cid:12) (1 u)2 (1 s)u>0, − 2 u (cid:27) (cid:12) − − − p − (s,t,u) R3(cid:12)(cid:12) 1/2<s<1,0<u<1/2, 1−su <t<1 √su . ∪(cid:26) ∈ (cid:12) (1 u)2 su>0, 2 u − (cid:27) (cid:12) − − − (cid:12) The proof is similar to(cid:12) that of Proposition 2.2 and omitted. Proof of Theorem 1.4. We show that ∗(χ0) consistsof twoirreducible compo- R nents whose intersection is not empty. From Eqs. (2) - (9), we get 0<s<1, 0<u<t<1 u, ∗(χ0)=(s,t,u) R3 (1 t)2 su>0, t2 (1 s−)u>0, . R ∼ ∈ −(1 2−s)(1 2t+t−u s−u)=0  − − −   7 Here we have the decomposition ∗(χ0)= (s,t,u) R3 0<t<1, 0<u<2t2, u<2(1 t)2, 1 2s=0 R ∼n ∈ (cid:12) − − o (s,t,u) R(cid:12)(cid:12)3 0<s<1, 0<u<1/2, (1−u)2−su>0, . ∪(cid:26) ∈ (1 u)2 (1 s)u>0, 1 2t+tu su=0 (cid:27) − − − − − The intersection of the two irreducible components is the set X 1,1,u 0<u< 1 = (s,t,u) R3 s=t= 1, 0<u< 1 . n (cid:0)2 2 (cid:1)(cid:12)(cid:12) 2o∼n ∈ (cid:12)(cid:12) 2 2o The proof is also s(cid:12)imilar to that of Proposition 2.(cid:12)2 and omitted. Figure3showstworealizationsof 0. Ontheleft,itshowstheaffineimage M ofX 1,3,1 ,ontheirreduciblecomponent1 2s=0. Ontheright,theimage 2 8 4 − of X(cid:0)3,11,2(cid:1) , so it is on the other component 1 2t+tu su=0. They can 4 24 7 − − be de(cid:0)formed (cid:1)continuously to each other via X 1,1,u (0<u< 1). 2 2 2 (cid:0) (cid:1) References [1] A. Björner, M. Las Vergnas, B. Sturmfels, N. White, G. Ziegler, Oriented Matroids, Encyclopedia of Mathematics 46, Cambridge University Press 1993. [2] N.E. Mnëv, The universality theorems on the classification problem of con- figuration varieties and convex polytopes varieties, in: Viro, O.Ya. (ed.): Topology and Geometry - Rohlin Seminar, Lecture Notes in Mathematics 1346, Springer, Heidelberg, 1988,527-544. [3] J. Richter-Gebert, Two interesting oriented matroids, Documenta Mathe- matica 1, 1996, 137-148. [4] G. Ringel, Teilungen der Ebene durch Geraden oder topologische Geraden, Math. Zeitschrift 64, 1956, 79-102. [5] P.Y.Suvorov,Isotopic but not rigidly isotopic plane system of straight lines, in: Viro, O.Ya. (ed.): Topology and Geometry - Rohlin Seminar, Lecture Notes in Mathematics 1346, Springer, Heidelberg, 1988, 545-556. 8