Stable homotopy groups of complex projective plane

Question

We know that there is a cofiber sequence $S^3\xrightarrow{\eta}S^2\to\mathbb{C}\mathbb{P}^2$. It's easy to know that $\pi_3^s(\mathbb{C}\mathbb{P}^2)=0$ so there is a surjection $$\partial:\pi_7^s(S^2\wedge\mathbb{C}\mathbb{P}^2)\to\pi_7^s(\mathbb{C}\mathbb{P}^2\wedge\mathbb{C}\mathbb{P}^2)$$ by the long exact sequence of $\eta$. The former group $\pi_7^s(S^2\wedge\mathbb{C}\mathbb{P}^2)$ is $\mathbb{Z}/12$, generated by the second hopf element. I wonder whether $\partial$ is an isomorphism?

Answer 1

The question is equivalent to asking what the multiplication-by-$\eta$-map $\pi_4\mathbb{CP}^2 \to \pi_5 \mathbb{CP}^2$ is (which can be rewritten as $\pi_2\mathbb{S}/\eta \to \pi_3\mathbb{S}/\eta$). The source is generated by the lift of $2 \in \pi_2S^2$ to $\pi_2 \mathbb{S}/\eta$. Thus, the question translates to the question what the Toda bracket $\langle \eta, 2, \eta\rangle$ is (the translation is classical, see e.g. Lemma 4.1 in my thesis for an account). As $\eta$ is $2$-torsion, only the $2$-local computation is relevant. As follows e.g. by a Massey product computation in the Adams spectral sequence, this Toda bracket is here the coset $2\nu + 4\pi_3\mathbb{S}$ (see e.g. Proposition 6 in Chapter 17 of Mosher--Tangora). Thus, $2$-locally, $\pi_4\mathbb{CP}^2 \to \pi_5 \mathbb{CP}^2$ has image generated by $2\nu$, while $3$-locally it is zero. Thus $\partial$ is an isomorphism $3$-locally and has image $\mathbb{Z/2}$ when working $2$-locally. This implies $\pi_3(C\eta \wedge C\eta) \cong \mathbb{Z}/6$.

In general, primary information (i.e. multiplication by elements) on a cone like $\mathbb{S}/\eta$ translates into secondary information (i.e. $3$-fold Toda brackets) on $\mathbb{S}$.

Edit: I have corrected above my earlier erroneous Toda bracket computation as caught by Nanjun Yang and corrected also the result.

Answer 2

$\newcommand{\Z}{\mathbb Z}\newcommand{\cA}{\mathcal A}\newcommand{\Sq}{\mathrm{Sq}}\newcommand{\CP}{\mathbb{CP}}\newcommand{\Ext}{\mathrm{Ext}}$It's possible to run the Adams spectral sequence directly to show that the $2$-torsion subgroup of $\pi_7^s(\CP^2\wedge\CP^2)$ is isomorphic to $\Z/2$, so that $\partial$ can't be an isomorphism. The calculation is a little longer than Lennart Meier's answer above, but it is more elementary. In what follows, all cohomology is with $\Z/2$ coefficients.

As an $\cA$-module, $\tilde H^\ast(\CP^2\wedge\CP^2)$ looks like $\tilde H^\ast(\CP^2)\oplus \Sigma^2\tilde H^\ast(\CP^2)$, with the lowest- and highest-degree elements joined by a $\Sq^4$. There is a short exact sequence of $\cA$-modules \begin{equation} 0\longrightarrow \Sigma^4\tilde H^\ast(\CP^2)\longrightarrow \tilde H^\ast(\CP^2\wedge\CP^2)\longrightarrow \Sigma^2 \tilde H^\ast(\CP^2)\longrightarrow 0. \end{equation} Here's a picture of that short exact sequence.

A short exact sequence of $\cA$-modules induces a long exact sequence of Ext groups. Often one computes this long exact sequence by drawing an Adams chart for the Ext of both the sub and the quotient; the boundary map has bidegree $(-1, 1)$. Beaudry and Campbell, section 4.6, give some examples of this technique. To use this, we need to know $\Ext_\cA(\tilde H^\ast(\CP^2), \Z/2)$; you can look it up in Hood Chatham and Dexter Chua's Adams spectral sequence calculator (this is for $\Sigma^{-2}\tilde H^\ast(\CP^2)$, so shift everything to the right by $2$), but it is also possible to compute it by hand in the range needed using a similar long exact sequence associated to the short exact sequence $0\to\Sigma^4\Z/2\to\tilde H^\ast(\CP^2)\to\Sigma^2\Z/2\to 0$.

Thus the long exact sequence in Ext looks like this:

We're done once we figure out whether the boundary map in red is nonzero: either way, there are no differentials to or from the $7$-line, and there are no possible hidden extensions on the $7$-line. We will show the red boundary map is nonzero by showing the map $$i^\ast\colon \Ext_\cA^{1,9}(\tilde H^\ast(\CP^2\wedge\CP^2), \Z/2) \longrightarrow \Ext_\cA^{1,9}(\Sigma^4\tilde H^\ast(\CP^2), \Z/2),$$ which is induced by $i\colon \Sigma^4\tilde H^\ast(\CP^2)\to\tilde H^\ast(\CP^2\wedge\CP^2)$, vanishes; exactness then forces the red boundary map to be injective, so the $7$-line of the $E_2$-page for $\tilde H^\ast(\CP^2\wedge\CP^2)$ has only a single $\Z/2$.

Now let's show $i^\ast$ vanishes. $\Ext_\cA^{1,9}(\Sigma^4\tilde H^\ast(\CP^2), \Z/2)\cong\Z/2$ and is generated by the upside-down question mark extension:

Therefore if $i^\ast\ne 0$, there is some extension $0\to\Sigma^9\Z/2\to M\to \tilde H^*(\CP^2\wedge\CP^2)\to 0$ whose pullback along $i$ is the upside-down question mark extension. The only way for this to work would be the following diagram:

However, this choice of $M$ is not a valid $\cA$-module: in $M$, $\Sq^5x = \Sq^1\Sq^4x \ne 0$, but $\Sq^5 = \Sq^4\Sq^1 + \Sq^2\Sq^1\Sq^2$, and $\Sq^4\Sq^1x = \Sq^2\Sq^1\Sq^2x = 0$. Therefore $i^\ast = 0$.