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

# 2 Construction of $v_1$ self maps

We wish to construct a map $\Sigma ^{2p - 2} V(0) \to V(0)$ inducing multiplication by $v_1$ on $BP_*$ homology. The strategy is to construct a map $\mathbb {S}^{2p - 2} \to V(0)$ that induces multiplication by $v_1$ on $BP_*$ homology, and then extend it to a map $\Sigma ^{2p - 2}V(0) \to V(0)$ by obstruction theory.

First consider the $BP$ Adams–Novikov spectral sequence for $V(0)$. In degrees up to $2p - 2$, the spectral sequence looks like where $v_1 \in (2p - 2, 0)$ and $t_1 \in (2p - 3, 1)$. If $p = 2$, then we have an extra $t_1^2$ which will be right above $v_1$.

In either case, we see that there is no room for extra differentials. So we see that

Lemma

There is a map $\tilde{v}_1: \mathbb {S}^{2p - 2} \to V(0)$ that induces multiplication by $v_1$ on $BP_*$. If $p > 2$, then this map has order $p$ and is unique.

Since $V(0) = \mathbb {S}/p$, the map $\tilde{v}_1$ having order $p$ is the same as it extending to a map $\Sigma ^{2p - 2} V(0)$. Thus, we deduce that
Theorem

If $p > 2$, then there is a map $v_1: \Sigma ^{2p - 2} V(0) \to V(0)$ that induces multiplication by $v_1$ on $BP_*$.

In the case $p = 2$, we know $\pi _2 V(0) = \mathbb {Z}/4\mathbb {Z}$ or $\mathbb {Z}/2 \oplus \mathbb {Z}/2$. If it is $\mathbb {Z}/4\mathbb {Z}$, then this map has order $4$ and does not lift to a map $\Sigma ^2 V(0) \to V(0)$. This is indeed the case, as one can check using the $H\mathbb {F}_2$ Adams spectral sequence, so we do not have a $v_1$ self map at $p = 2$.