Construction of $v_1$ and $v_2$ self-mapsConstruction of $v_2$ self maps

# 3 Construction of $v_2$ self maps

We next attempt to construct $v_2$ self maps. It turns out there is no $v_2$ self map when $p = 3$, so in this section we will exclusively concentrate on the case $p > 5$.

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

If we further find that this map has order $p$, then it factors through $\Sigma ^{2p^2 - 2} V(0)$. This is the same as saying there are no elements in the $t - s = 2p^2 - 2$ column apart from (multiples of) $v_2$.

Finally, to show that this descends to a map from $\Sigma ^{2p^2 - 2} V(1)$, we precompose with the (suspension of the) $v_1$ self map of $V(0)$ to get a map

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

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

$[\mathbb {S}^{2p^2 + 2p - 3}, V(1)] \to [\Sigma ^{2p^2 + 2p - 4} V(0), V(1)] \to [\mathbb {S}^{2p^2 + 2p - 4}, V(1)]$

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

So to show that $v_2$ exists, we have to prove the following:

Theorem

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

$t - s = 2p^2 - 3, \quad 2p^2 + 2p - 3,\quad 2p^2 + 2p - 4.$

Moreover, the column $t - s = 2p^2 - 2$ is generated by $v_2$ of order $p$.

So we will have to compute the Adams–Novikov spectral sequence up to $t - s \leq 2p^2 + 2p - 3$. In Ravenel's green book, the computation was done up to $\sim p^3$, but for our range, we can get away with doing some simple counting.

In this range, the generators in $BP_*BP$ that show up in the cobar complex are $t_1, t_2$ and $v_2$. Note that there are two ways we can multiply $t_1$ — either in $BP_* BP$ itself, or as $t_1 \otimes t_1$ in the cobar complex. In either case, any appearance of $t_1$ will contribute at least $2p - 3$ to $t - s$. Similarly, $t_2$ and $v_2$ contribute $2p^2 - 3$ and $2p^2 - 2$ respectively. The assumption that $p \geq 5$ allows us to perform the following difficult computation:

Theorem

If $p \geq 5$, then $2p - 3 \geq 7$.

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

 Element $t - s$ Terms involving only $t_1$ ?? $v_2$ $2p^2 - 2$ $t_2$ $2p^2 - 3$ $t_1 v_2$ $2p^2 + 2p - 5$ $t_1 t_2$ $2p^2 + 2p - 5$ $t_1 \otimes t_2$ $2p^2 + 2p - 6$ $t_2 \otimes t_1$ $2p^2 + 2p - 6$

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

The element $t_1$ is a primitive element, and the following lemma is convenient:

Lemma

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

$\operatorname{Ext}_\Gamma (\mathbb {F}_p, \mathbb {F}_p) = E(h_i: i = 0, 1, \ldots ) \otimes P(b_i: i = 0, 1, \ldots )$

where

$h_i = x^{p^i} \in \operatorname{Ext}^1,\quad b_i = \sum _{0 < j < p} \frac{1}{p} \binom {p}{j} x^{jp^i} \otimes x^{(p - j)p^i} \in \operatorname{Ext}^2.$

The presence of elements not involving $t_1$ will not increase the number of purely $t_1$ cohomology classes, but may kill off some if the coboundary of some term is purely $t_1$ (e.g. $\mathrm{d}(t_2)$). So it is enough to see that there are no product of the $h_i$ and $b_i$ fall into the relevant columns.

In degrees $t - s \leq 2p^2 + 2p - 4$, we have generators

$1 \in \operatorname{Ext}^{0, 0}, \quad h_0 \in \operatorname{Ext}^{1, 2p - 2},\quad h_1 \in \operatorname{Ext}^{1, 2p^2 - 2p},\quad b_0 \in \operatorname{Ext}^{2, 2p^2 - 2p},$

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 $p = 3$, we have $2p^2 + 2p - 3 = 21$, and $h_1 b_0 \in \operatorname{Ext}^{3, 24}$ is a non-trivial element with $t - s = 2p^2 + 2p - 3$.