AALBORG UNIVERSITY. The trace space of the k-skeleton of the n-cube. Iver Ottosen. Department of Mathematical Sciences. Aalborg University

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1 AALBORG UNIVERSITY The trace space of the k-skeleton of the n-cube by Iver Ottosen R January 2014 Department of Mathematical Sciences Aalborg University Fredrik Bajers Vej 7 G DK Aalborg Øst Denmark Phone: Telefax: URL: e ISSN On-line version ISSN

2 The trace space of the k-skeleton of the n-cube Iver Ottosen January 10, 2014 Department of Mathematical Sciences, Aalborg University, AC Meyers Vænge 15, 2450 København SV, Denmark Abstract We show that the space of directed paths on the k-skeleton of the n-cube is homotopy equivalent to the nerve of a certain category of flags of finite sets 1 Introduction In [D] Dijkstra introduced the so-called PV-model for linear concurrent programs with semaphores The space of execution paths of such a program is a directed space One can study it using the methods of directed algebraic topology introduced by Fajstrup, Goubault, Grandis and Raussen (see [FRG], [G]) We will consider the following problem: Assume that s is a semaphore (that is a shared resource) of capacity k Run n copies of the program P s V s in parallel P s V s P s V s P s V s What is the homotopy type of the associated space of execution paths? Let us describe the space of execution paths precisely Let I = [0, 1] denote the unit interval In our setting, a directed path is a continuous curve γ : I I n with weakly increasing coordinate functions The semaphore s is modeled by open subintervals I i =]a i, b i [ I for 1 i n A point x in I n is forbidden if the set {i x i I i } has more than k elements Thus the state space of the program is X n (k) = {x I n #{i x i I i } k} and we want to study the space of directed paths form 0 to 1 in this state space, or its homotopy equivalent trace space consisting of directed paths up to reparametrizations P (X n (k))(0, 1) T (X n (k))(0, 1) When k = 0, the trace space is empty When k n, we have a contractible trace space So we assume that 1 k n 1, since these are the nontrivial cases Note that for k = n 1 the trace space is homotopy equivalent to the sphere S n 2 since this is a special case of [R] 1

3 2 A covering of the state space The k-skeleton of the n-cube is the special state space where a i = 0 and b i = 1 for all i: I n (k) = {x I n #{i 0 < x i < 1} k} By a continuous deformation of each factor of I n we obtain: Proposition 21 The inclusion I(k) n Xn (k) induces a homotopy equivalence of trace spaces T (I(k))(0, n 1) T (X(k))(0, n 1) Proof For 0 < a < b < 1 we define a continuous map H (a,b) : I I I by 0, 0 x sa, x sa H (a,b) (s, x) =, sa x sb + 1 s, s(b a)+1 s 1, sb + 1 s x 1 If we fix s I to any value we have a weakly increasing map of the variable x So we can write H (a,b) : I I I Define r (a,b) : I I by r (a,b) (x) = H (a,b) (1, x) and observe that H (a,b) (0, x) = x We now form Cartesian products of these types of maps H : I I n I n ; H(s, x) = (H (a1,b 1 )(s, x 1 ),, H (an,b n)(s, x n )), r : I n I n ; r(x) = H(1, x) = (r (a1,b 1 )(x 1 ),, r (an,b n)(x n )), so that H is a homotopy from id I n to r Note that r restricts to a map r : X n (k) In (k) and let ι : In (k) Xn (k) denote the inclusion We also have restrictions H 1 and H 2 of the homotopy H with the following properties: H 1 : I I n (k) I n (k), H 1 (0, x) = x, H 1 (1, x) = r ι(x), H 2 : I X n (k) X n (k), H 2 (0, x) = x, H 2 (1, x) = ι r(x) Thus we have a d-homotopy equivalence I(k) n Xn (k) Via precomposition we get a homotopy equivalence of associated directed path spaces and hence a homotopy equivalence of trace spaces We will now cover I(k) n by appropriate subsets, such that the nerve lemma can be applied in a similar fashion as in [R] Let [1 : m] denote the set of integers from 1 to m Definition 22 For non-negative integers j and i 1, i 2,, i r [1 : n] we define the following (possibly empty) subspace of the n-cube: X #j i 1,i 2,,ı r = {x I n x i1 = x i2 = = x is = 0 and #{i x i = 1} j} Furthermore, X #j denotes the subset of the n-cube consisting of points with at least j coordinates equal to 1 and X i1,i 2,,i r = X #0 i 1,i 2,,i r 2

4 Notation 23 Inj n m denotes the set of injective maps from [1 : m] to [1 : n] Definition 24 For α Inj n n k we define the subspace U α I n (k) by U α = X #(n k) X #(n k 1) α(1) X #(n k 2) α(1),α(2) X α(1),α(2),,α(n k) Proposition 25 Let denote the coordinate-vise maximum operation on R n and let denote the minimum operation on the integers For all α Inj n n k and r, s there is an inclusion X #(n k r) α(1),,α(r) X#(n k s) α(1),,α(s) X#(n k r s) α(1),,α(r s) Thus U α is -closed and so is any intersection of U α s Proof Assume that x X #(n k r) α(1),,α(r) and y X#(n k s) α(1),,α(s) Then the point x y has at least (n k r) (n k s) = n k r s coordinates which are equal to 1 and (x y) α(1) = = (x y) α(r s) = 0 By Proposition 28 of [R] we have the following: Corollary 26 The trace space T ( α A U α )(0, 1) is either empty or contractible for every nonempty family of injections A Inj n n k Theorem 27 T (I n (k))(0, 1) = T (U α )(0, 1) α Inj n n k Proof For K, L [1 : n] we introduce the notation U L K = {x I n i K : x i = 0, j L : x j = 1} Let γ P (I n (k) )(0, 1) Choose t 1 > 0 minimal such that j : γ j (t 1 ) = 1 Define sets We have J 1 = {j [1 : n] γ j (t 1 ) = 1}, K 1 = {l [1 : n] γ l (t) = 0 for t < t 1 } n k K 1, 1 J 1 k, K 1 J 1 = Inductively, choose t i > t i 1 minimal such that γ j (t i ) = 1 for some j [1 : n] with j / J 1 J i 1 Define sets We have J i = {j [1 : n] γ j (t i ) = 1, j / J 1 J i 1 }, K i = {l [1 : n] γ l (t) = 0 for t < t i } n k K i + J 1 J i 1, 1 J i k, K i (J 1 J i ) = 3

5 Stop at t m > t m 1 when γ i (t) = 1 for all t t m and all j [1 : n] Note hat m n We now have K 1 K 2 K m = and γ([0, t 1 [) U K 1, γ([t 1, t 2 [) U J 1 K 2,, γ([t i 1, t i ]) U J 1 J i 1 K i,, γ([t m 1, t m ]) U J 1 J m 1 K m, γ([t m, 1]) U [1:n] Next, we form an appropriate subsequence Choose i 1 such that n k K i1 but n k > K i1 +1 Put A 1 = K i1 and B 1 = We have A 1 B 1 =, n k A 1 + B 1 Inductively, choose i j > i j 1 such that K ij = K ij 1 +1 but K ij +1 K ij 1 +1 Put A j = K ij and B j = J 1 J ij 1 For some s we have K is 1 +1 = Here we finish by A s =, B s = J 1 J is 1 We now have a strictly decreasing sequence A 1 A 2 A s = and a sequence = B 1 B 2 B s such that for all i Furthermore, we have such that A i B i =, n k A i + B i γ([0, t i1 [) X B 1 A 1,, γ([t ij 1, t ij [) X B j A j,, γ([t s, 1]) X Bs A s γ([0, 1]) s Put n i = A i and choose α Inj n n k such that X B j A j {α(1),, α(n i )} = A i, for 2 i s 1 and {α(1),, α(n k)} A 1 Then γ([0, 1]) U α As in [R], we need a covering of I n (k) by open subsets of In Choose 0 < ɛ < 1 3 Replace the conditions x i = 0 and x j = 1 by x i < ɛ and x j > 1 ɛ respectively in the above definition of the covering Then one gets an open covering and an analogue of Proposition 212 in [R] holds 3 A simplicial complex model of the trace space Definition 31 Let Ck n denote the poset Ck n = {A Inj n n k A, T ( U α )(0, 1) } The partial order is inclusion For an object A Ck n we have a simplex in RInjn n k = R n!/k! as follows: α A A 1 = {(x α ) x α 0 for all α, 4 xα = 1, x α = 0 if α / A}

6 We define functors D : (C n k ) op Top; A T ( α A U α )(0, 1), E : C n k Top; A A 1 An argument similar to the one given [R], proof of Theorem 35, shows that the trace space of I n (k) is homotopy equivalent to the colimit of E and to the nerve of Cn k Thus, we have the following result: Theorem 32 There is a homotopy equivalence T (I n (k))(0, 1) N (C n k ) We will now give a better description of the category Ck n subsets can be described as follows: An intersection of Lemma 33 Let α 1, α 2,, α r Inj n n k Then one has U αj = n k i=0 X #(n k i) α 1 (1),,α 1 (i),α 2 (1),,α 2 (i),,α r(1),,α r(i) Proof We insert the expression defining U αj and obtain U αj = (X #(n k) X #(n k 1) = X #(n k) α j (1) (X #(n k 1) α j (1) X #(n k 2) α j (1),α j (2) X α j (1),α j (2),,α j (n k)) X #(n k 2) α j (1),α j (2) X α j (1),α j (2),,α j (n k)) since the term X #(n k) appears in every factor and when we intersect it by another summand, we get a subset of X #(n k) By a similar argument, we can rewrite the above as X #(n k) X #(n k 1) α 1 (1),α 2 (1),,α r(1) (X #(n k 2) We continue this way, and get the desired result α j (1),α j (2) X α j (1),α j (2),,α j (n k)) We can now formulate a useful criterion for whether the trace space of an intersection of subsets is empty or not Corollary 34 Let α 1, α 2,, α r Inj n n k Then one has T ( U αj )(0, 1) i : #{α 1 (1),, α 1 (i), α 2 (1),, α 2 (i),, α r (1),, α r (i)} i + k 1 5

7 Proof The end point 1 lies in X #(n k) and not in any other summand The starting point 0 lies only in X α1 (1),,α 1 (n k),,α r(1),,α r(n k) There exists a directed path from 0 which moves further ahead precisely when #{α 1 (1),, α 1 (n k),, α r (1),, α r (n k)} n 1 Assuming that, we can arrive at a point in X #1 α 1 (1),,α 1 (n k 1),,α r(1),,α r(n k 1) There exists a directed path further ahead from this point precisely when 1 + #{α 1 (1),, α 1 (n k 1),, α r (1),, α r (n k 1)} n 1 Assuming that, we can get to a point in X #2 α 1 (1),,α 1 (n k 2),,α r(1),,α r(n k 2) and so on If we have arrived at a point in X #(n k), we can always move further to the end point 1 Definition 35 For A Inj n n k we let Im i (A) = α A{α(1), α(2),, α(i)} Theorem 36 The poset C n k has the following purely combinatorial description: C n k = {A Inj n n k i : i Im i (A) i + k 1} 4 The flag-category model Definition 41 The flag category is the poset Fk n = {(F 1 F 2 F n k ) F n k [1 : n], i : i F i i + k 1} The partial order is given by F G i : F i G i Note that there is a forgetful functor U : C n k F n k ; A (Im 1 (A), Im 2 (A),, Im n k (A)) We now define a functor, which turns out to be the left adjoint of U Definition 42 The functor V : F n k Cn k is defined by V (F ) = {α Inj n n k U({α}) F } Note that V (F ) is a final among the objects A with property U(A) = F Theorem 43 V : F n k Cn k is the left adjoint of the forgetful functor U : Cn k F n k Thus there is a homotopy equivalence N (C n k ) N (U) N (V ) N (F n k ), N (Cn k ) N (F n k ) 6

8 Proof The composite UV is the identity on Fk n, so the counit of the adjunction is simply the identity Id UV Id F n k By definition of V we have an inclusion A V U(A) for all objects A C n k This defines the unit of the adjunction Id C n k V U A functor induces a map of nerves and a natural transformation of functors induces a homotopy between such maps Thus an adjoint pair of functors induces a homotopy equivalence By combining Theorems 32 and 43 we get our main result: Theorem 44 There is a homotopy equivalence T (I n (k))(0, 1) N (F n k ) References [D] [FRG] [G] [R] E W Dijkstra, Co-operating sequential processes, from: Programming Languages, (F Genuys, editor), Academic Press, New York (1968), L Fajstrup, M Raussen, E Goubault, Algebraic topology and concurrency, Theor Comput Sci 357 (2006), no 1 3, M Grandis, Directed algebraic topology, models of non-reversible worlds, Cambridge University Press (2009) M Raussen, Simplicial models of trace spaces, Algebraic & Geometric Topology 10 (2010),