a rX iv:mat h /41261v2[mat h.KT]2M ar25The Complex of Words and Nakaoka Stability Moritz C.Kerz mokerz@students.uni-mainz.de Abstract We give a new simple proof of the exactness of the complex of injective words and use it to prove Nakaoka’s homology stability for symmetric groups.The methods are generalized to show acyclicity in low degrees for the complex of words in ”general position”.1.An elementary proof of Nakaoka stability We will be concerned with the complex C ∗(m ),where C n (m )is the free abelian group generated by elements (x 1,...,x n ),where the x i are pairwise different natural numbers from 1to m .C 0(m )=Z .The differential is given by d (x 1,...,x n )=n j =1(−1)j +1(x 1,...,x j −1,x j +1,...,x n ).In discrete mathematics the following theorem is well established as the homol-ogy of C ∗(M )is given by the simplicial homology of the (shellable)poset of injective words.Shellability reduces the simplicial homology groups to those of a wedge of m-spheres.We refer to [1],[3],[7]for exact definitions and state-ments.Our proof of Theorem 1is new and rather straightforward compared to Farmer’s original elementary proof.Theorem 1(F.D.Farmer [3])The homology of C ∗(m )vanishes except in de-gree m .We have to introduce some notations which will be used throughout the paper.A chain c is called a term,if there exists an N ∈Z and x 1,...,x n ∈{1,...,m }with c =N (x 1,...,x n ).As C ∗(m )has a canonical basis all our sum decompostions correspond to parti-tions of the basis.We also speak about the appearance of numbers in a chain.For example 2appears in (2,3)+4(5,1)but 4does not.Although our complex C ∗(m )has no obvious cup product,we have a partially defined product.If c ∈C n (m ),c ′∈C l (m )are elementary c =N (x 1,...x n ),c =M (x ′1,...,x ′l ),we define c c ′:=N M (x 1,...,x n ,x ′1,...x ′l )if the numbers x 1,...,x n ,x ′1,...x ′l are paiwise different.This constuction extends bilinearly to arbitrary chains c ,c ′for which the numbers appearing in both of them are distinct.1There is a Leibniz rule for c∈C n(m)and c′∈C l(m)d(c c′)=d(c)c′+(−1)n c d(c′).Proof of Theorem1.The exactness in degree0is clear.For the rest we use induction on m.For the case m=2we have to check thatC2(2)−→C1(2)−→Zis exact.But ker(d:C1(2)→Z)is generated by elements of the form(x)−(x′)=d(x,x′)with x=x′.For the induction step we use a straightforward lemma:Lemma1If we have a number x∈{1,...,m}which does not appear in a cycle c∈C n(m),it is a boundary.Proof.According to the Leibniz rule c=c+(x)d(c)=d((x)c),since d(c)=0.Given an arbitrary cycle c of degree n<m we have to show that in order to apply the lemma we can eliminate a number from c by adding boundaries. Therefore we will push afixed number x∈{1,...,m}to the right until it vanishes from the cycle.If x appears somewhere in the cycle at thefirst index,writec= j(x)c j+c′Where the c j are elementary and c′does not have x at thefirst index.To each c j choose a number x j∈{1,...,m},x j=x which does not appear in c j.Thenc−d( j(x j)(x)c j)=c′+ j(x j)c j−(x j)(x)dc j.Now x does not appear at thefirst index anymore.The next steps until the vanishing of x are similar.Suppose x does not appear at thefirst i>0indices of c.Now we can writec= j s j(x)c j+c′.Where the c j are elementary and different,the s j have length i,x does not appear in s j and x does not appear at thefirst i+1entries of c′.One calculates: 0=d c= j (d s j)(x)c j+(−1)i s j c j+(−1)i+1s j(x)d c j +d c′.2If we forget those terms for which x does not appear at the i-th index,this equation implies d s j=0for all j.Since length(s j)=i<m−length((x)c j), there are by induction s′j with d s′j=s j and such that the following products make sense.z:= j s′j(x)c jIn the cyclec−d z=c′− j (−1)i+1s′j c j+(−1)i s′j(x)d c jx does not appear at thefirst i+1indices.Using the corresponding two hyperhomology spectral sequences for the natural action of the symmetric groupΣm on our complex C∗(m)(cf[2])one can now obtain a stability result due to Nakaoka.A similar proof was given by Maazen (thesis).Theorem2(Nakaoka[5])H m(Σn−1)=H m(Σn)for m<n/2.Proof.We use induction on n.It is well known for n=3H1(Σ2)=H1(Σ3)=Z/(2).For n≥4define C′l(n):=C l+1(n)for l≥0.ThenH m(Σn,Z)=H m(Σn,C′∗(n))when m<n−1because of Theorem1.The other spectral sequence of the bi-complex gives for E1∗,∗:···H2(Σn,C′0)H2(Σn,C′1)···H2(Σn,C′n−2)H2(Σn,C′n−1)H1(Σn,C′0)H1(Σn,C′1)···H1(Σn,C′n−2)H1(Σn,C′n−1)H0(Σn,C′0)H0(Σn,C′1)···H0(Σn,C′n−2)H0(Σn,C′n−1)Using Shapiro’s Lemma we get:···H2(Σn−1)H2(Σn−2)···H2(Σ1)0H1(Σn−1)H1(Σn−2)···H1(Σ1)0H0(Σn−1)H0(Σn−2)···H0(Σ1)ZThe horizontal arrows can be computed as0,1,0,1,···,since they are the sums of the signs in d′1,d′2,···.We have E2i,0=H i(Σn−1)for i≥0and by induction E2i,j=0for i<n−j−1Finally,parts of the spectral sequence degenerate;E∞i,0=H i(Σn−1)for n−i>2, such that H i(Σn)=H i(Σn−1)for n/2>i.2.Words in general positionLet n:={i|1≤i≤n}.We associate to every(nonempty)set X a complex F(X)∗,the so-called complex of words with alphabet X,as follows:(F(X))n=Z<{f:n→X}>d n(f)=nk=1(−1)n+1f◦δk.Hereδk:[n]→[n+1]are the coface mapsδk(j)= j if1≤j<kj+1if k≤j.It is immediate that the homology of F(X)vanishes.For if c∈F(X)n is a cycle,c=d((x0)c)for arbitrary x0∈X(see Lemma1).For given X certain subcomplexes of F(X)can be used in hyperhomology spec-tral sequences as above,if their homology vanishes to some extent.These sub-groups are determined by conditions which one could call”general position conditions”.For applications cf[8],[9].It could be asked,how to explain the fact that the vanishing of homology is not affected by these conditions.In order to give a general result we translate the proof of Theorem1into our more complicated setting.Examples(i)Let X be afinite set.The complex of injective words is(F inj(X))n:={f∈F(X)n|f injective}.According to Theorem1the homology of this subcomplex is zero except in degree m=card(X)whererank(H m(F(X)∗))=(−1)m(1−m−1i=0(−1)i m(m−1)···(m−i)).This equals the number offixed point free permutations of X.(ii)Let F be afield and V an F-vector space.The complex of vectors in general position is(F gen(V))n:={f∈F(V)n|f in general position}.If F is infinite the complex is exact.A general vanishing result is contained in our main theorem.We introduce axioms for elements of a given(nonempty)set X to be in general position.4Definition1(General position)Let G l,m be relations in l+m variables from X;l>0,m≥0.The family(G l,m)l,m∈N is called a general position relation if the following properties are satisfied.Let x,y,z befinite sequences of elements of X of length l,m,n:(i)G n,m is symmetric in thefirst n and last m arguments.(ii)If G l+m,n(x,y;z)then G l,m+n(x;y,z).If G l,m+n(x;y,z)then G l,n(x;z).(iii)If G l,m+n(x;y,z)and G m,n(y;z)then G l+m,n(x,y;z).Forfixed G we say x is in G-general position to y iffG l,m(x;y).Given x∈F(X)l and y∈F(X)m we say x is in G-general position to y if every term of x is in G-general postition to every term in y(for l=0we demand nothing).Definition2Given afinite sequence b of elements of X the corresponding complex of chains in general position to b is(F G(X;b))n:={f∈F(X)n|f is in G-general position to b}.Before we can state the main theorem we have to introduce an invariant which determines an upper bound for the vanishing of the homology of F(X)∗.Definition3Given a general position relation G we define|G|to be the small-est natural number n≥0such that there is a sequence x of elements of X with length(x)=n such that there is no further element in X which is in general position to x.Examples(i)Example(i)is induced by saying x is in G inj-general position to y if theunderlying sets of x and y are disjoint and the entries of x are pairwise different.We have|G inj|=card(X)and F G inj(X)∗=(F inj(X))∗(ii)Example(ii)is induced by saying(x1,...,x i)is in G vec-general position to(y1,...,y j)if we have for all a k∈F,k=1,...,i+j and only dim(V) many of them nonvanishinga1x1+···+a i x i+a i+1y1+···+a i+j y j=0implies a k=0for all k∈{1,...,i}.If card(F)=∞or dim(V)=∞then|G vec|=∞.Unfortunately the exact value of|G vec|is not known for allfinite dimen-sional vector spaces overfinitefields.5Lemma2(a)For dim(V)≥card(F)we have|G vec|=dim(V)+1.(b)For dim(V)=2we have|G vec|=card(F)+1.Proof of(a).Let n=dim(V)+1and(e i)1≤i≤n−1be a basis of V.We show|G vec|≥nfirst.Otherwise we had a sequence(x i)1≤i≤n−1,x i∈V, such that there does not exist a vector x∈V in G vec-general position to(x i). This can only be true if(x i)is a basis of V.But if it is a basis the element x1+x2+···+x n−1would be in general position to it.Contradiction.Now the sequence f=(e1,...,e n−1,e1+e2+···+e n−1)is maximal in the sense of Definition3.To see this we have to show there is no vector x= a1e1+···+a n−1e n−1,a i∈F,in G vec-general position to f.If there is an index j such that a j=0,x is obviously not in general postion to f.But if no coefficient in x vanishes two of them have to be equal since dim(V)≥card(F). In case a1=a2we can writex=a1(e1+···+e n−1)+(a3−a1)e3+···(a n−1−a1)e n−1.This shows f is in fact maximal.Proof of(b).For dim(V)=2we have x∈V in G vec-general position to y∈V iffx=0=y and[x]=[y]∈P(V).So|Gen vec|is simply the number of rational points in P(V)/F.Theorem3For a set X with general position condition G and afinite se-quence a=(a1,...,a l)of elements of X the corresponding homology groups H m(F G(X;a))vanish for m≤(|G|−l−1)/2.In contrast to Theorem1one should notice that it is in general not possible to erase the factor1/2from the inequality of the theorem.Proof.Denote the degree by m.The exactness at m=0is trivial.We proceed by induction on m≥1.Let c be a cycle in F m(X;a1,...,a l).Case m=1:We can suppose c=(x)−(x′).Because1≤(|G|−l−1)/2we have l+2<|G| so that there exists y∈X in G-general position to(x,x′,a1,...,a l). According to Definition1(iii)(y,x,x′)is in G-general position to(a1,...,a l) and it is allowed to write d((y)c)=c.Induction step:Choose x∈X in G-general position to(a1,...,a l).The simplest case is c in G-general position to(x,a1,...,a l).Here we can apply a construction similar to Lemma1;we have c=d((x)c),since(x)c is in G-general position to(a1,...,a l).We reduce to this case by changing c by boundaries.To be more precise we introduce a number I(c)∈{0,...,m}which is m iffthe above applies,that is6c in G-general position to(x,a1,...,a l).By adding boundaries be will see that we can increase I(c).For d∈F G(X;a)n we define I(d)∈{0,...,n}as the the greatest natural number i≤n such thatπi(v)is in G-general position to(x,π′i(v),a1,...,a l) for any term v of d(πi denotes the projection to thefirst i entries andπ′i the projection to the last n−i entries).Reduction to I(c)>0:Suppose I(c)=0.Let c= j x j with x j elementary.Choose for every j y j∈X with y j in G-general position to(x,x j,a1,...,a l).This is possible since length(x,x j,a1,...,a l)=1+m+l<|G|.Clearly I(c−d( j(y j)x j))>0. Now suppose m>I(c)>0:Writec= j s j x j+x′such that exactly those terms v of c for which I(v)>I(c)are in x′,length(s j)=I(c)and all x j are elementary and pairwise different.Lemma3d(s j)=0for all j.Proof.The terms v of d(c)such that I(v)<I(c)are exactly the terms ofd(s j)x j.In order to see this,notice that for a term v of d(c)we have I(v)=(v),a1,...,a l). I(c)−1iffthe I(c)-th entry of v is not in G-general position to(x,π′I(c)Now projecting the identity0=d(c)to the terms v with I(v)<I(c)we get 0= j s j x j.Lemma3is proven. According to our assumtion s j is in G-general position to(x,x j,a1,...,a l).By induction on m we know there exist s′j with d(s′j)=s j and s′j in G-general position to(x,x j,a1,...,a l),because2i+[(m−i)+l+1]≤2m+l≤|G|−1. NowI(c−d( j s′j x j))=I(x′+(−1)I(c)s′j d(x j))>I(c).Thisfinishes the reduction to I(c)=m and therefore the induction step for length(c)=m is accomplished.Theorem3could be useful in the generalization of Suslin’s GL-stability[8]tofinitefields.A thorough treatment seems to indicate the following result:(i)Given m≥0and n≥m we have H m(GL n(F))=H m(GL n+1(F))for almostallfinitefields F.(ii)For given m≥0H m(GL m−1(F))→H m(GL m(F))is surjective for almostallfinitefields F.7In fact(i)has been proven–using other methods–by Quillen[6].He did even show,that the statement is true for allfields with more then2elements.Similar results with weaker bounds are due to Maazen and van der Kallen[4].AcknowledgementThe above results were obtained in the Seminar M¨u ller-Stach2004at Mainz.I am indebted to Stefan M¨u ller-Stach,Volkmar Welker,W.van der Kallen, Philippe Elbaz-Vincent and Oliver Petras for their support.References[1]Bj¨o rner,Anders;Wachs,Michelle On lexicographically shellable posets. 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