Selçuk J. Appl. Math. Selçuk Journal of Vol. 14. No. 1. pp. 47-56, 2013 Applied Mathematics
Homology of a Prismatic Set B. Akyar
Department of Mathematics, Dokuz Eylül University, TR-35160 Izmir, Turkiye. e-mail:b edia.akyar@ deu.edu.tr
Received Date: December 24, 2011 Accepted Date: February 26, 2013
Abstract. We study a special type of prismatic sets namely the prismatic subdivision of a simplicial set. Prismatic sets have been used to construct a classifying map giving invariants in gauge theory. We give an algebraic inter-pretation of the prismatic subdivision and compute its homology group. Key words: Simplicial set; Prismatic subdivision; Spectral Sequences; Homology.
AMS Classi…cation: 18G30, 55U10, 55R20, 55U15. 1. Introduction
Prismatic sets have appeared in some di¤erent places (see Dupont-Ljungmann [4], Phillips-Stone [6]). In Akyar [1], an important special case of prismatic sets, namely the prismatic subdivision, has been introduced and studied in a topological sense. In the present paper, we give another approach to the prismatic subdivision in order to see the relation between the simplicial set and the corresponding prismatic subdivision in an algebraic sense.
We give an associated chain complex generated by the prismatic subdivision of a simplicial set with necessary di¤erentials. We show that two maps de-…ned between the simplicial set and the corresponding prismatic subdivision are homotopic. Moreover we show that the extensions of these homotopic maps correspond to the maps of bicomplexes involving the associated chain complex. Morever we calculate the homology group of the associated chain complex (Proposition 3.1, Proposition 3.2 and Proposition 3.3). They are the main results.
2. Prismatic Subdivision from an Algebraic Point of View We recall the de…nition of a simplicial set from Dupont [3].
De…nition 2.1. A simplicial set S = fSqg is a sequence of sets with face di : Sq ! Sq 1and degeneracy operators si: Sq ! Sq+1, i = 0; : : : ; q, satisfying the simplicial identities
didj = dj 1di : i < j djdi+1 : i j sisj = sj+1si : i j sjsi 1 : i > j and disj= 8 < : sj 1di : i < j id : i = j; i = j + 1 sjdi 1 : i > j + 1: Let p= f(t
0; : : : ; tp) 2 Rp+1jPiti= 1; ti 1g be a standard p-simplex given with barycentric coordinates. A prism is a product of simplices, that is, a set of the form q0:::qp= q0 qp.
De…nition 2.2. A prismatic set P is a sequence fPpg of (p+1)-multi-simplicial sets together with face operators
dk : Pp;q0;:::;qp! Pp 1;q0;:::;^qk;:::;qp
commuting with dij and sij (interpreting dkj = skj = id on the right) such that fPpg is a -set. If similarly there are given degeneracy operators
sk : Pp;q0;:::;qp! Pp+1;q0;:::;qk;qk;:::;qp
making fPpg a simplicial set, a prismatic set is called a strong prismatic set and we get an ordinary simplicial set (fPpg; dk; sk).
De…nition 2.3. We have for each p the thin realization jPpj =
G q0;:::;qp
q0:::qp P
p;q0;:::;qp=
with equivalence relation “ ” generated by the face and degeneracy maps "i
j : q0:::qi:::qp ! q0:::qi+1:::qp and ij : q0:::qi:::qp ! q0:::qi 1:::qp, respec-tively. Here fjPpjg is a -space hence it gives a fat realization
k jP j k = G p 0
p
by only using the face operators i di : q0:::qp Pp;q0;:::;qp !
q0:::^qi:::qp
Pp 1;q0;:::;^qi;:::;qp which act on
q0:::qp as the projection inducing a structure
of a simplicial space on fjPpjg. The further equivalence relation on k jP j k is generated by
("it; s; ) (t; is; di ); t 2 p 1; s 2 q0:::qp; 2 Pp;q0;:::;qp:
Now, let us give the prismatic triangulation (the prismatic subdivision) PpSq0;:::;qp
of a simplicial set S:
It is explicitly de…ned by PpSq0;:::;qp := Sq0+ +qp+p, where q0+ + qp =
n. For its thin realization jP S j, the face operators are de…ned by di j := dq0+ +qi 1+i+j, j = 0; : : : ; qi, i = 0; : : : ; p and the degeneracy operators are
de…ned by si
j := sq0+ +qi 1+i+j, j = 0; : : : ; qi, i = 0; : : : ; p. The face maps
are the operators corresponding to the inclusions q0+ +^qi+ +qp+p 1! n+p
deleting the qi+ 1 basis vectors with indices q0+ + qi 1+ i; : : : ; q0+ + qi + i. For the sequences of spaces fjP S jg, we obtain the fat realization jj jP S j jj =Fp 0 p jPpS j= where jPpS j =
F
p 0 q0:::qp Sn+p= and the face operators jdij : jPpS j ! jPp 1S j are given by jdij = i di, i = 0; : : : ; p. Let us give the construction of a bicomplex of P as follows:
A bicomplex C ; (P ) of P is a family f(Cp;n(P ); @H; @V)g of modules with the horizontal boundary map @H : Cp;n(P ) ! Cp 1;n(P ) and the vertical boundary map @V : Cp;n(P ) ! Cp;n 1(P ) such that @H@H = 0; @V@V = 0 and @H@V + @V@H = 0, where Cp;n(P ) =Lq0+ +qp=nCp;q0;:::;qp(P ) is the associated chain
complex C (P ) generated by PpSq0;:::;qp. Here the vertical boundary map is
de…ned using boundary maps in the …bre direction
@Fi: Cp;q0;:::;qp(P ) ! Cp;q0;:::;qi 1;:::;qp(P )
given by @Fi = P( 1)jdij, where if qi = 0 then @Fi = 0. The total vertical boundary map is then
@V = @F0+ ( 1)q0+1@F1+ + ( 1)q0+ +qp 1+p@Fp = p+1 X i=1 ( 1)i 1+ Pi 2 j=0qj @Fi:
The horizontal boundary map is de…ned by
@H = @0+ ( 1)q0+1@1+ + ( 1)q0+ +qp 1+p@p = p X k=0 ( 1)q0+ +qk 1+k@ k where @k = 0 : if qk> 0 dk : if qk= 0
so that @ = @V+@His the total boundary map in the total complex C (P ) which is the cellular chain complex for the geometric realization. Thus C (P ) is a
bigraded Z-module with the two di¤erentials @V and @H. Since Cp;n(P ) = L
q0+ +qp=nZ[PpSq0;:::;qp] =
L
q0+ +qp=nZSp+n(q0;:::;qp), there is a natural
mapLq0+ +qp=nZSp+n(q0;:::;qp)! Cp+n(S) and also a map Cp;n(P ) ! Cp+n(S).
Note.The sign convention for the boundary map @ is determined by the follow-ing diagram
P S v! ES & #
S
where EpS = S S, (p + 1-times), is another (p + 1)-multi-simplicial set given with the face operators di : EpS ! Ep 1S projecting on the i-th factor and the degeneracy operators si: EpS ! Ep+1S repeating the i-th factor. The sign conventions for the boundary map in E S and in S give a motivation to get the correct sign for the one in PpSq0;:::;qpsince the di¤erentials are preserved
under the maps in the diagram above. So the di¤erential @ can be thought of asPn+pj=0( 1)jdj.
Remark 2.1. i)Let S be the simplicial set mand x 2 Smthere is a simplicial map fx: m! S de…ned by fx(0; : : : ; m) = x for a generator m= (0; : : : ; m) 2
m, that is, f
x(0; : : : ; ^{; : : : ; m) = dix and fx(0; : : : ; i; i; : : : ; m) = six.
ii)By using i) it is enough to construct chain maps for S = m. We can de…ne the chain maps u0; u1 as u0 : Sp ! P0Sq0 = Sm = Sq0 and u1 : Sp !
PpS0;:::;0 = Sp = Sm, where m = n + p, n = q0+ + qp. So the elements in PpSq0;:::;qp are denoted by (i0; : : : ; iq0; iq0+1; : : : ; iq0+q1+1; : : : ) (increasing
se-quence). Consequently, u0 and u1 are de…ned as u0(0; : : : ; m) = (0; 1; : : : ; m) and u1(0; : : : ; m) = (0; 1; : : : ; m).
iii)The simplicial set k jS j k = k 1k jS j, where 1 is given with one element p = (0; : : : ; p) in each degree and the boundary operator is de…ned as @i p = p 1, i = 0; : : : ; p. k 1k = Fp o p= , where is an equivalence relation which is de…ned by "it t, t 2 p 1. Hence C ( 1) is the chain group with only one generator in each degree, namely p= (0; : : : ; p) 2 Cp( 1) and
@ p= p X i=0 ( 1)i(0; : : : ; ^{; : : : ; p) = p X i=0 ( 1)i p 1 Thus @ p= 0 : p is odd or p = 0 p 1 : p = 2k; k = 1; : : :
Now let us consider the maps C (S) u0 u1
C ; (P ) and …nd the extensions aw0 and aw1of u0and u1, respectively as awi: C ( 1) C (S) ! C ; (P ), i = 0; 1, on the chain level.
Lemma 2.1. Let S be the simplicial set n. There exists a map between bicomplexes aw0: C ( 1) C (S) ! C ; (P ) de…ned by aw0(p (0; : : : ; n)) = X 0 i0 ::: ip 1 n ( 1)pq0+(p 1)q1+ +qp 1 (0; : : : ; i0; i0; : : : ; i1; : : : ; ip 1; : : : ; ip): In particular, aw0 is a chain map of the total complexes.
Proof. We need to show that @Haw0= aw0@H and @Vaw0= aw0@V. It follows directly from the de…nition of aw0. Moreover one can easily show that aw0is a chain map of the total complexes for all p. For further details see Akyar [2]. Lemma 2.2. Let S be a simplicial set. There exists a chain map aw1 : C ( 1) C (S) ! C ; (P ), for S = n, de…ned by aw1(p (0; : : : ; n)) = X 0 i0 ::: ip 1 n ( 1)pq0+ +qp 1 (0; : : : ; i0; i0; : : : ; ip 1; : : : ; ip):
Proof. It follows from the de…nition of aw1. See Akyar [2].
Remark 2.2. One can easily see that aw0(0 x) = u0by taking S = n and p = 0 as follows:
aw0(0 (0; : : : ; n)) = X 0 i0 n
(0; : : : ; i0) = (0; : : : ; n) = u0(0; : : : ; n):
Similarly, the extension aw1 of u1becomes when p = 0 aw1(0 (0; : : : ; n)) =
X 0 i0 n
(0; : : : ; i0) = (0; : : : ; n) = u1(0; : : : ; n):
3. Homology of a Prismatic Set
Now we examine the homology H(C (P )). For the preparation of this calcu-lation we recall some de…nitions and facts about spectral sequences from Mac Lane [5].
A Z-bigraded module is a family E = fEp;ng of modules, one for each pair of indices p; n 2 Z. A di¤erential d : E ! E of bidegree ( r; r 1) is a family of
homomorphisms fd : Ep;n! Ep r;n+r 1g, one for each p; n, with d2= 0. Then H(E) = H(E; d) is the bigraded module fHp;n(E)g de…ned as
Hp;n(E) = Ker(d : Ep;n! Ep r;n+r 1)=d(Ep+r;n r+1):
E is made into a Z-graded module E = fEmg with total degree m by Em =Pp+n=mEp;n, d induces a di¤erential d : Em ! Em 1 with the usual degree -1 and H(fEmg; d) is the graded module obtained from Hp;n(E) as Hm=Pp+n=mHp;n.
De…nition 3.1. A spectral sequence E = fEr; drg is a sequence E2; E3; : : : of Z-bigraded modules, each with di¤erentials dr : Erp;n ! Erp r;n+r 1, r = 2; 3; : : : of bidegree ( r; r 1) and with isomorphism
H(Er; dr) = Er+1; r = 2; 3; : : :
Each Er+1is the bigraded homology module of the preceding (Er; dr).
De…nition 3.2. A …ltration Fp = FpC of chain complexes (C (P ); @) of R-modules, one for each p 2 Z, is given by 0 F0C F1C : : : FpC : : : C , where R is a commutative ring with unit. This …ltration is always assumed to be …nite in each degree.
Let Fp= FpC be a …ltration as in the de…nition above. There is an associated spectral sequence fEr
p;ng, r = 0; 1; : : : ; 1, which is a sequence of bigraded chain complexes with di¤erentials
dr: Erp;n! Erp r;n+r 1such that Er+1p;n= Kerdr=dr(Erp+r;n r+1) where p; n 0 and if for each k there exists a t such that FtCi= Cifor i k then dp;nr = 0 for r > t and p + n k, so that Et+1p;n = Et+2p;n = = E1p;n. In particular
E1p;n= Hp+n(Fp=Fp 1) and d1: E1
p;n! E1p 1;n is the boundary map in the exact sequence 0 ! Fp 1=Fp 2! Fp=Fp 2! Fp=Fp 1! 0:
Moreover Er converges to H(C (P )), that is, there is an induced …ltration Fp(H(C (P ))) = Im(H(Fp; ) ! H(C (P )))
such that
0 F0(H(C (P ))) F1(H(C (P ))) : : : Fp(H(C (P ))) : : : H(C (P )) and there is a canonical isomorphism
The total chain complex C (P ) has two …ltrations 0F pCm= M k p Ck;m k and00FnCm= M k n Cm k;k;
associated to the spectral sequences f0Erp;n;0drg and f00Ern;p;00drg both con-verging to H(C (P )). Here
0E1
p;n= Hn(Cp; (P ); @V); 00E1n;p= Hp(C ;n(P ); @H) and
(3.1) 0E2p;n= Hp(Hn(C ; (P ); @V); @H); 00E2n;p= Hn(Hp(C ; (P ); @H); @V):
Proposition 3.1. Let P be the prismatic set associated to a simplicial set S. Suppose
Hp(C ;n(P )) = 0 for p > 0 and n = 0; : : : :
Let C 1;n(P ) = coker(@H: C1;n(P ) ! C0;n(P )). Then the edge map e induced by the projection C1;m(P ) ! C0;m(P ) ! C 1;m(P ) is a homology isomorphism e : H(C (P )) ! H(C 1; (P ); @V).
Proof. By using the second formula in (3.1) we get 00E2
n;p=
0 : p > 0 Hn(C 1; (P )) : p = 0 and for r 2,00E2
n;p=00Ern;p. On the other hand for p = 0 E1m;0= FmHm=Fm 1Hm= Hm(C (P )): Thus Hm(C (P )) = E1m;0= Hm(C 1; (P )), where m = n + p. Proposition 3.2. We have
H (C (P ); @) = H (C (S)):
Proof. There is a map between two bicomplexes (3.2) aw0: C ( 1) C (S) ! C ; (P ):
We have given that aw0 is a chain map. On the left hand side of aw0, we have a double complex and the di¤erential @H is the identity on C (S) and for this
double complex we have 0E1p;n = Hn(Cp; (P ); @V) which is independent of p since in (3.2),0E1p; is the same for all p. By the …rst formula in (3.1), we have
0E2
p;n= 0 : p > 0
Hn(C (S)) : p = 0:
Then H (C (P ); @) = H (jP Sj; @V). Since jP Sj and jS j are homeomorphic, H (C (P )) = H (C (S)).
Proposition 3.3. We have00E1n;p= Hp(C ;n(P ); @H) where 00E1
n;p= 0 : n > 0
Hp(C (S)) : n = 0:
Proof. Let us …rst explain some notations. We have Cp;n(P ) = M q0+ +qp=nZ[Pp Sq0;:::;qp] = M q0+ +qp=nZSn+p(q0;:::;qp) : Let us take S as a standard m-simplex, that is, Sm= mand the elements in the double complex are denoted by the following partition (i0; : : : ; iq0;
iq0+1; : : : ; iq0+q1+1; : : : ) with p + 1 groupings, where n + p = m. One can
easily show that H1(C ;0(P )) = H(C (S)) since Cp;0(P ) = Lq0+ +qp=0Sp.
On the other hand, one can also easily show that H1(C ;1(P )) = 0. We have Cp;1(P ) =Lq0+ +qp=1Sp+1. An element in this chain complex is denoted by
(0; : : : ; 1; : : : ; 0)
| {z }
p+1 tim es
, where 1 is in the i-th place.
Now we introduce a …ltration Fp = FpC (P ) of the total complex C (P ) as 0 F0C (P ) F1C (P ) : : : FpC (P ) : : : C (P ). Let us use Fi(S) for Fi(C (P )) for short, i = 0; : : : . Hence F0(S) is given by the partitions of the form (0; : : : ; 0; 1), S1(1) d0 S2(0;1) d0 d1 S3(0;0;1) d0 d1+d2 S4(0;0;0;1)
By the chain contraction s0we get H(F0(S)) = 0. On the other hand F1(S)=F0(S) is given by the partitions of the form (0; : : : ; 1; 0), i.e., the complex F1(S) is given by
S1(1) S2(0;1) S3(0;0;1) S4(0;0;0;1): : :
S2(1;0) S3(0;1;0) S4(0;0;1;0): : :
F0(S) is a sub-complex of F1(S) and by the quotient group F1(S)=F0(S), we have a short exact sequence
The quotient gives us S2(1;0) S3(0;1;0) S4(0;0;1;0) : : : and by the contrac-tion s0 we get H(F1(S)=F0(S)) = 0. Thus H(F1(S)) = 0.
In order to make the characterization of the …ltration clear, we need some no-tations: Let us consider the elements in PpSq0;:::;qp, where q0+ + qp = 1.
Fi(S) will be characterized by putting at most i zeros after 1 in the sequence, i.e., at most one is 1 and the rest are 0. One can see H(C (P )) = 0 by using an induction on the …ltration, since Cn(P ) = F1i=0FiCn(P ) and H(C (P )) = L
iH(Fi(C (P )). Hence H(C ;1(P )) = 0.
We can also show that H(C ;2(P )) = 0. Although H(C ;n(P )) = 0 for n > 2, it is quite complicated to write the …ltration. Therefore, we will only show that this result is true when n = 2 and one can follow the same idea for n > 2 . We know that Cp;2(P ) =L1p=0Sp+2 = Cp;2(P ) = C(2)(P ) C(1;1)(P ), here we have two di¤erent types of elements (0; : : : ; 2; : : : ; 0) and (0; : : : ; 0; 1; 1) coming from C(2)(P ) and C(1;1)(P ), respectively. For the …rst type of elements, we will exactly follow the same idea as before when n = 1. The …ltration Fi(S) will be characterized by putting at most i zeros after 2 in the sequence. The rest follows from the case when n = 1. Thus H(C(2)(P )) = 0. For the second type of elements C(1;1)(P ), let us start with F
0(S) given by the partition of the forms (0; : : : ; 1; 1) as S3(1;1) d0 S4(0;1;1) d0 d1 S5(0;0;1;1) d0 d1+d2 S6(0;0;0;1;1)
Here the chain contraction is s0 so we get H(F0(S)) = 0. On the other hand F1(S)=F0(S) is given by the partitions (0; : : : ; 1; 0; 1) and (0; : : : ; 0; 1; 1; 0) then F1(S) is given by
S3(1;1) S4(0;1;1) S5(0;0;1;1) S6(0;0;0;1;1): : :
S4(1;0;1) S5(0;1;0;1) S6(0;0;1;0;1): : : S4(1;1;0) S5(0;1;1;0) S6(0;0;1;1;0): : :
By the same idea as before we have H(F1(S)=F0(S)) = 0 and H(F0(S)) = 0 then using the short exact sequence we get H(F1(S)) = 0. By using an induction on i in FiS, one can see that H(FiC(1;1)(P )) = 0. In general we can determine the …ltration FiS by putting at most k zeros after 1 in the q0 th; i k zeros after 1 in the qk+1 st places in FiS=Fi 1S, where k = 0; : : : ; i. Thus we get H(C(1;1)(P )) = 0. We already know Cp;2(P ) = C(2)(P ) C(1;1)(P ) and H(C ;2(P )) = H(C(2)(P )) H(C(1;1)(P )) = 0. By using the same idea for the …ltration, one can see H(C ;n(P )) = 0 for n > 2.
References
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