The divisibility by β

111 Xⁿ⁻ⁱ. By the octahedral axiom the left column of the following diagram is distinguished:

X_i−1

//X^{n //}

=

Xⁿ⁻ⁱ⁺¹

X_i

//X^{n //}

Xⁿ⁻ⁱ

=

Σ^?TMF ^//X^{n−i+1 //}Xⁿ⁻ⁱ

Clearly, X_n = X and X₀ = 0, so X can be built up. The dual follows by the dual proof or Spanier-Whitehead-duality. The last thing to show is that for a building up sequence, the morphisms Σ^?TMF → ^Xi correspond to torsion elements x_i in π_∗X_i. By the triangle 8.1 this is equivalent to c(x_i) = 0. So suppose we had c(x): π_∗Σ^?TMF(2) → ^π∗X_i(2) non-zero. This is also non-zero if we tensor withQ, the quotient field ofTMF(2)_∗. There-fore, (π_∗X_i(2)⊗ Q)/c(x) has rank i−1. Hence, dim_Q(π_∗X_i+1(2)⊗ Q) ≤ i, which is a contradiction.

Thus, we proved Proposition 8.1.3.

Proof. Letx∈ ^E(M). By Lemma 8.3.4 and the contradiction assumption,σ(x)is a non-zero 3-torsion element in π_∗Σ⁴Mα. Thus, d(σ(x)) = 0 andσx = eαux for someux ∈ ^π∗M (for d see the end of Section 8.3). The element u_x is only well-defined up to the image ofr – therefore we can assume by the last proposition thatu_x is torsion. Hence u_x =^t_eαy_x since c(ux) = 0 for some yx ∈ ^π∗Σ⁴Mα by (8.1). By Lemma 5.2.1, we get thatσ(x) = βyx for some y_x ∈ ^π∗Mα. By the same argument, every torsion element in Mα is divisible by β and so we can repeat the process ify_x is not already inF₀.

Recall now that on the level of vector bundles, σ: M₀(2)→^Σ4Minduces the map σ_alg: Γ(f_∗f^∗O ⊗^π∗FM)→Γ(Eα⊗^ω−2⊗^π∗FM)

calledσin Section 8.2.

Corollary 8.5.4. The0-line of the DSS for M_α consists of permanent cycles.

Proof. We will use a rank argument: Let X ⊂ Γ(π_∗FMα) be the subgroup of perma-nent cycles. Then im(σ_alg) ⊂ ^X since the descent spectral sequence for M₀(2) collapses on E². Define a filtration on X by setting B_k = {^x ∈ ^{X : βk+1x} = 0 for somex ∈ F₀π_∗Mα reducing tox}^{. Sinceβ}operates trivially onM₀(2), we have im(σ_alg)⊂^B0. Hence X/B₀ is a subquotient of coker(σ_alg). The latter is anF3[∆^±3]-vector space of rank 3n for n the number of irreducible direct summands of π_∗FM – this is proven in the proof of Proposition 8.2.3 and at the end of Section 8.2. So, if X 6= Γ_∗(π_∗FMα), then X/B₀ is an F3[_∆^±3-vector space of rank smaller than 3n. We have 3ninvariant generators of the form

∆^j for j ∈ {^{0, 1, 2}} in the direct summands of π_∗M0(2) and we choose a basis g_i of the Z(3) span of these elements indexed by some index set I with |^I| = 3n. We know that σ(g_i) = βⁿⁱv_i for some v_i ∈ ^F0π_∗M_α withn_i maximal under all choices ofv_i; so there are 3n elements v_i. We assume that we have chosen inductively theg_i in the following way:

We order I in some way. The first of theg_i is chosen to be a primitive vector in the span of the ∆^j with maximal n_i. The (k+1)-st g_i is chosen to be one that is part of a basis of the span of the ∆^j together with the first k elements g_i and is among these one with the maximal n_i. This insures that σ(Σj∈^Ja_jg_j) = β^lv with a_j units and v 6= 0 always implies thatl≤ ⁿj for all jwitha_j 6=0.

We have v_i ∈ ^Bni sinceβσ(g_i) =σ(βg_i) =0. Suppose, there exists anv⁰_iπ_∗Mα with the same reduction v⁰_i = v_i in the zero-line, but βⁿⁱv⁰_i = 0. Then exists an x ∈ ^π∗M_α of higher filtration such thatv⁰_i =v_i−^{x. Sincex}is torsion, it is by the (proof of the) last corollary of the formβ^lv forv ∈ ^F0π_∗Mα. Thus, β^l+niv = βⁿⁱx = βⁿⁱv_i = σ(g_i) in contradiction to the maximality ofn_i. Thus, v_i ∈/ ^Bn_i−¹.

Since⊕i≥¹B_i/B_i−1∼= X/B₀, there is ak∈ NandJ ⊂ ^{I such thatv}j ∈^Bk−^Bk−¹and the (v_j)_j∈J are linear dependent overF3 inB_k/B_k−1. That is, there exista_j ∈ {^1,−¹}^{such that} Σja_jv_j ∈ ^Bk−¹. As above, this impliesβ^l+kv=β^kΣja_jv_j =σ(Σja_jg_j) =0 for somev∈ ^π∗Mα

andl> 0 and thusv= 0. HenceΣja_jg_j ∈ ^im(c). But since 1 and −1 are units inZ(3), we haveΣja_jg_j∈ ^E(M), which is a contradiction to our main contradiction assumption.

Notation 8.5.5. We recollect the notation from the last proof for the rest of the chapter: We have an index setI of cardinality 3n, indexing elementsg_i ∈^π∗M₀(2)⊂^π∗M(2)spanning E(M)in the sense that every element inE(M)is of the form∑a_ig_i for a_i ∈ Z(3). We have numbers n_i and elementsv_i ∈ ^F0π_∗M_α such that σ(g_i) = βⁿⁱv_i. Thev_i reduce by the last proof to a basis {^vi}^{of cokerσ}alg. Note that the v_i are (thus, since im(r) = im(σ)by the

113 proof of Proposition 8.2.3) not in im(r_∗)and can be modified by elements in im(r_∗)so that thev_i are in the span of the elements of the form∆^j in H⁰(M^;π∗Mα)by Proposition 8.2.3 and the fact that β·^im(r) =0.

Corollary 8.5.6. The1-line of the DSS of M consists of permanent cycles.

Proof. The map ^teαin the triangle (8.1) in the introduction induces as in Theorem 6.4.3 a morphism of descent spectral sequences, which is exactly ^teαon E². This implies that the whole first line of the descent spectral sequence in Mconsists of permanent cycles (which, of course, cannot be boundaries) since^teα: Γ(π_∗FMα)∼=Γ(π_∗FM⊗^Eα)→ ^H1(M^;π∗+4FM) is surjective (as H¹(M^;π∗FM⊗ ^f∗f^∗O^ω−2) =0).

In the rest of this section, we want first to investigate how many times an element might be divided by β and then investigate in detail how the torsion exactly looks like.

Before we begin with this, we have to compute a Toda bracket:

Lemma 8.5.7. The Toda bracketheα,β⁴, 3i(where we vieweαagain as a map Σ⁷TMF → ^TMFα) contains±{^3∆2}^.

Proof. We first want to check that the Toda bracket is actually defined. Since β²α = 0 in π_∗TMF, we see thatβ²eα∈^π27TMFα is mapped to zero in the exact sequence

π_∗TMF→^π∗TMFα →^π∗−⁴TMF

and is thus the image of an element a ∈ ^π27TMF. The only non-zero elements in this degree are±{α∆}^.5 These are annihilated byβ² and thusβ⁴eα=0 and the Toda bracket is defined.

The element β⁴eα in the E²-term of the DSS of TMF_α is a permanent cycle (since eα is in DSS(TMFα)and β⁴ is one in DSS(TMF)) and can only be hit by ad9-differential from

±^∆2: Column 48 in lines below 9 consists only of line zero elements and by Scholium 8.2.4 and the fact that im(r) consists of permanent cycles, the existence of a non-trivial differential implies a non-trivial differential from∆². Using Theorem 6.4.1, we could use that Massey products converge to Toda brackets and get the result.

Alternatively, one can use the definition of the Toda bracket and sees that it suffices to prove that the lift ofβ⁴ ∈^π40TMFin the exact sequence

π₄₈TMFα →^π48TMF₀(2)→^π40TMF

is±∆² ∈ ^π48TMF₀(2). Indeed, these span the non-trivial elements in π₄₈TMF₀(2)which are mapped trivially into the zero line of the DSS ofTMF modulo the image of π₄₈TMFα

(as can be seen, for example, by an im(r)-argument).

Lemma 8.5.8. All n_iare smaller than4.

5One can check that βeα is non-zero and therefore ais non-zero as well. But this is not needed for our argument.

Proof. Assume we have an x ∈ ^E(M) such that σ(x) = β⁴z, which is automatically 6= 0 since else xwould be in the image ofc. Look at the following diagram:

TMF ^{3 //}

TMF

=

=

%%K

KK KK KK KK K

TMF ^β

4 //

yytttttt=tttt

Σ⁻⁴⁰TMF

}}{{{{{{z{{

=

TMF ^{σ(x) //}

M_α

=

Σ⁻⁴⁸TMFα //

tz

||xxxxxxxxx

Σ⁻⁴⁸TMF₀(2)

zz //Σ⁻⁴⁰TMF ^{eα //}

~~}}}}}}z}} Σ⁻⁴⁷TMFα

tz

||xxxxxxxxx

Σ⁻⁴M ^//M₀(2) ^//M_α ^{teα //}Σ⁻³M

Here we use the isomorphism DTMF_α = Hom_TMF(TMF_α,TMF) ∼= Σ⁻⁴TMF_α, under which_tzcorresponds tozas in Lemma 4.2.6 (withk=7 andZ= TMF_α). The Toda bracket heα,β⁴, 3i^contains {3∆²}. Therefore, we get that h^teα,σ(x), 3i^{contains b} = tz({3∆²}). We havec(b) =3x⁰ by the definition of the Toda bracket with σ(x) = σ(x⁰). The elementx⁰ is invariant (since 3x⁰is), but is not in the image ofc(sinceσ(x⁰)6=0). Hence, the correspond-ing element x⁰ in H⁰(M^;π∗FM) ∼= π_∗(M(2))^S3 cannot be a permanent cycle in DSS(M) and hence is not in the image of r. By Proposition 8.2.3, we can find an y ∈ ^π∗M(2) with c_alg(r_alg(y) +x⁰) a generator of a direct summand. Set x⁰⁰ = x⁰+cr(y) ∈ ^π∗M(2). This is clearly an invariant generator. We have that c(b+3r(y)) = 3x⁰⁰. Furthermore, for w:=_tz(1) +r(∆⁻²y)∈^π∗M, the following holds:

3∆²c(w) =c({3∆²}^w) = 3x⁰+cr({3∆²}∆⁻²y)

= 3x⁰+cr(3y)

= 3x⁰⁰.

Hence,c(w) =_∆⁻²x⁰⁰, which is an invariant generator. This is a contradiction to our global contradiction hypothesis.

We assume now thatπ_∗FM has only summands of the form Oand its shifts. Our aim for the rest of the section is to understand the torsion in π_∗M andπ_∗M_α. The arguments will resemble these of our argumentations in the low rank cases. We have that^teα(v_i) =αw_i for some elements w_i in the 0-line of the E²-term of the descent spectral sequence of M.

Thew_i can be chosen to span theZ(3)-span of elements of the form∆ⁱsince these generate H⁰_∗(M^;π∗FM)/ ker(α) and ^teα is surjective onto H¹(M^;π∗FM). All elements β^kαw_i are permanent cycles since the β^k are permanent cycles in DSS(TMF). The elements β^kαw_i for k ≥ ⁿi must be boundaries since^teα(β^kv_i)is zero. We know that the w_i support non-trivial differentials d_piw_i. Hence, alsod_pi(β^kw_i)6=0 in E^pi by Lemma 8.3.5 since the 1-line consists of permanent cycles. All in all, this implies that all torsion in π_∗M is detected by the αβ^kw_i for k < n_i. Note that this also implies that all elements of the form eαβ^kv_i cannot be permanent cycles in DSS(M_α) since β^k+1w_i = ^t_eα(_eαβ^kv_i) is not a permanent cycle. Therefore, theβ^kv_i span all the torsion inπ_∗M_α.

115 Suppose that some linear combination Σi∈^I0a_iβ^kv_i is a boundary fork < n_i with I⁰ ⊂ ^I non-empty and a_i ∈ {±¹}^{. Since} Σia_iβ^kv_i 6= 0 (by the linear independence statements in the proof of Corollary 8.5.4), β^kΣia_iv_i must be detected by a permanent cycle of the form Σjb_jβ^mv_j. Assume β^kΣia_iv_i = β^mΣjb_jv_j and setn to be the maximum of the n_i fori ∈ ^I0. Then σ(Σiwithni=na_igI) = βⁿΣia_iv_i = β^n+m−kΣjb_jv_j. As in the proof of Corollary 8.5.4, this implies m = k, which is not true. Thus, β^kΣia_iv_i 6= β^mΣjb_jv_j and their difference x is detected a permanent cycle of the formΣνc_nuv_nu. As before,x 6=Σνc_nuv_nuand is detected by a permanent cycle of even higher filtration and so on. Since the filtration is bounded by the last lemma, at some point we get an equality, which implies a contradiction as before.

Thus,Σi∈^I0a_iβ^kv_i is no boundary fork <n_i with I⁰ ⊂ ^I non-empty and a_i ∈ {±¹}^.

Suppose now that some linear combinationΣi∈^I0a_iβ^kiv_i is in the image ofσ fork_i < n_i with I⁰ ⊂ ^I non-empty and a_i ∈ {±¹}^{. Then} Σia_iβ^kv_i has to be a boundary for k the minimum of thek_i witha_i 6=0∈F3. Thus,Σia_iβ^kv_i =0 inE², which impliesΣia_iβ^kv_i =0.

Thus, arguing as in Corollary 8.5.4, Σia_iv_i ∈ ^Bk−¹, which is a contradiction to that the v_i are linear independent inH⁰(M^;π∗FMα)/B_k−1. ThusΣia_iβ^kiv_i cannot be in im(σ).

Since we know thus that the F3-span of the β^kv_i for k < n_i gets mapped injectively into the torsion ofπ_∗M by^teα, we know by rank comparison that noαβ^kw_i is a boundary.

We set{^αβkwi}^:=^t_eα(β^kv_i), which is detected by αβ^kw_i and is therefore in strict filtration 2k+1.

All in all, we have thus proven the following proposition:

Proposition 8.5.9. Letπ_∗FM have only summands of the formOand its shifts. Then the torsion ofπ_∗M is anF3-vector space with basis given by{^αβkwi}^{with k} < n_i and i ∈ I. The torsion of π_∗Mα is anF3-vector space with basis given byβ^kv_i with k≤ ⁿi and i∈ ^I.

Warning8.5.10. Similar to {^α∆} ∈^π27TMF, the notation {^αβkwi}does not entail that this element is divisible byα. But it is true thatβ^k{^αwi}={^αβkwi}^.

Im Dokument United Elliptic Homology (Seite 111-115)