Construction of v1v_1 and v2v_2 self-maps — Construction of v2v_2 self maps

3 Construction of v2v_2 self maps

We next attempt to construct v2v_2 self maps. It turns out there is no v2v_2 self map when p=3p = 3, so in this section we will exclusively concentrate on the case p>5p > 5.

We first sketch the argument to see how far in the Adams–Novikov spectral sequence we have to go. There is an element v2Ext0,2p22(BP,BP/(p,v1))v_2 \in \operatorname{Ext}^{0, 2p^2 - 2}(BP_*, BP_*/(p, v_1)). We want to show this survives to give a map S2p22V(1)\mathbb {S}^{2p^2 - 2} \to V(1), and so we will have to understand the ts=2p23t - s = 2p^2 - 3 column, which we will find to be empty.

If we further find that this map has order pp, then it factors through Σ2p22V(0)\Sigma ^{2p^2 - 2} V(0). This is the same as saying there are no elements in the ts=2p22t - s = 2p^2 - 2 column apart from (multiples of) v2v_2.

Finally, to show that this descends to a map from Σ2p22V(1)\Sigma ^{2p^2 - 2} V(1), we precompose with the (suspension of the) v1v_1 self map of V(0)V(0) to get a map

Σ2p2+2p4V(0)Σ2p22v1Σ2p22V(0)V(1), \Sigma ^{2p^2 + 2p - 4} V(0) \overset {\Sigma ^{2p^2 - 2} v_1}{\longrightarrow } \Sigma ^{2p^2 - 2} V(0) \longrightarrow V(1),

and we have to show that this vanishes. We shall show that [Σ2p2+2p4V(0),V(1)]=0[\Sigma ^{2p^2 + 2p - 4} V(0), V(1)] = 0 using the long exact sequence

[S2p2+2p3,V(1)][Σ2p2+2p4V(0),V(1)][S2p2+2p4,V(1)] [\mathbb {S}^{2p^2 + 2p - 3}, V(1)] \to [\Sigma ^{2p^2 + 2p - 4} V(0), V(1)] \to [\mathbb {S}^{2p^2 + 2p - 4}, V(1)]

given by the cofiber sequence SpSV(0)\mathbb {S}\overset {p}{\rightarrow } \mathbb {S}\rightarrow V(0).

So to show that v2v_2 exists, we have to prove the following:

Theorem

Suppose p>3p > 3. Then Exts,t(BP/(p,v1))=0\operatorname{Ext}^{s, t}(BP_*/(p, v_1)) = 0 for

ts=2p23,2p2+2p3,2p2+2p4. t - s = 2p^2 - 3, \quad 2p^2 + 2p - 3,\quad 2p^2 + 2p - 4.

Moreover, the column ts=2p22t - s = 2p^2 - 2 is generated by v2v_2 of order pp.

So we will have to compute the Adams–Novikov spectral sequence up to ts2p2+2p3t - s \leq 2p^2 + 2p - 3. In Ravenel's green book, the computation was done up to p3\sim p^3, but for our range, we can get away with doing some simple counting.

In this range, the generators in BPBPBP_*BP that show up in the cobar complex are t1,t2t_1, t_2 and v2v_2. Note that there are two ways we can multiply t1t_1 — either in BPBPBP_* BP itself, or as t1t1t_1 \otimes t_1 in the cobar complex. In either case, any appearance of t1t_1 will contribute at least 2p32p - 3 to tst - s. Similarly, t2t_2 and v2v_2 contribute 2p232p^2 - 3 and 2p222p^2 - 2 respectively. The assumption that p5p \geq 5 allows us to perform the following difficult computation:

Theorem

If p5p \geq 5, then 2p372p - 3 \geq 7.

Thus, we can enumerate all the terms that appear in the cobar complex in the range ts2p2+2p3t - s \leq 2p^2 + 2p - 3:

Element

tst - s

Terms involving only t1t_1

??

v2v_2

2p222p^2 - 2

t2t_2

2p232p^2 - 3

t1v2t_1 v_2

2p2+2p52p^2 + 2p - 5

t1t2t_1 t_2

2p2+2p52p^2 + 2p - 5

t1t2t_1 \otimes t_2

2p2+2p62p^2 + 2p - 6

t2t1t_2 \otimes t_1

2p2+2p62p^2 + 2p - 6

The term t2t_2 is in a problematic column, but it shall not concern us for two reasons. Firstly, it has s=1s = 1, so it wouldn't get hit by our differentials. Secondly, we can calculate that d(t2)=t1t1p\mathrm{d}(t_2) = t_1 \otimes t_1^ p (e.g. see 4.3.15 of Ravenel's Green book), and so t2t_2 does not actually appear in the Adams spectral sequence. So we would be done if we can show that there are no purely t1t_1 terms appearing in the four columns of the theorem.

The element t1t_1 is a primitive element, and the following lemma is convenient:

Lemma

Let Γ=P(x)\Gamma = P(x) be a Hopf algebra over Fp\mathbb {F}_ p (p>2p > 2) on one primitive generator in even degree. Then

ExtΓ(Fp,Fp)=E(hi:i=0,1,)P(bi:i=0,1,) \operatorname{Ext}_\Gamma (\mathbb {F}_ p, \mathbb {F}_ p) = E(h_ i: i = 0, 1, \ldots ) \otimes P(b_ i: i = 0, 1, \ldots )

where

hi=xpiExt1,bi=0<j<p1p(pj)xjpix(pj)piExt2. h_ i = x^{p^ i} \in \operatorname{Ext}^1,\quad b_ i = \sum _{0 < j < p} \frac{1}{p} \binom {p}{j} x^{jp^ i} \otimes x^{(p - j)p^ i} \in \operatorname{Ext}^2.\fakeqed

The presence of elements not involving t1t_1 will not increase the number of purely t1t_1 cohomology classes, but may kill off some if the coboundary of some term is purely t1t_1 (e.g. d(t2)\mathrm{d}(t_2)). So it is enough to see that there are no product of the hih_ i and bib_ i fall into the relevant columns.

In degrees ts2p2+2p4t - s \leq 2p^2 + 2p - 4, we have generators

1Ext0,0,h0Ext1,2p2,h1Ext1,2p22p,b0Ext2,2p22p, 1 \in \operatorname{Ext}^{0, 0}, \quad h_0 \in \operatorname{Ext}^{1, 2p - 2},\quad h_1 \in \operatorname{Ext}^{1, 2p^2 - 2p},\quad b_0 \in \operatorname{Ext}^{2, 2p^2 - 2p},

We see that no product of these can enter the column we care about. So we are done.

As a side note, in the case p=3p = 3, we have 2p2+2p3=212p^2 + 2p - 3 = 21, and h1b0Ext3,24h_1 b_0 \in \operatorname{Ext}^{3, 24} is a non-trivial element with ts=2p2+2p3t - s = 2p^2 + 2p - 3.