Is there a quick way to distinguish between a wedge of spheres and a suspended projective space?

Question

I remember reading about the following example a while back in one of Steenrod's papers on cohomology operations. If we look at $S^3\vee S^5$ and $\Sigma \mathbb{C} P^2$, these spaces have isomorphic cohomology rings over $\mathbb{Z}$, because cup products in the cohomology of a suspension are always zero. However, because Steenrod squares are stable, we have a non-trivial $Sq^2$ acting on $H^3(\Sigma \mathbb{C} P^2)$, and thus these spaces cannot be homotopy-equivalent.

Is there another quick, or elementary, way to distinguish these spaces?

Answer 1

We can also distinguish them using homotopy groups. Note that $\mathbb{C}P^2$ is the $4$-skeleton of $\mathbb{C}P^\infty$ which is a $K(\mathbb{Z},2)$, and it follows that $\pi_3(\mathbb{C}P^2)\cong \pi_3(\mathbb{C}P^\infty)=0$. Since $\mathbb{C}P^2$ is simply connected, the suspension map $\pi_3(\mathbb{C}P^2)\to\pi_4(\Sigma\mathbb{C}P^2)$ is surjective by the Freudenthal suspension theorem, so $\pi_4(\Sigma\mathbb{C}P^2)$ is trivial.

On the other hand, $\pi_4(S^3)\cong \mathbb{Z}/2$ and $S^3$ is a retract of $S^3\vee S^5$, so $\pi_4(S^3\vee S^5)$ is nontrivial. Thus $S^3\vee S^5$ cannot be homotopy equivalent to $\Sigma\mathbb{C}P^2$.

(Of course, this raises the question of how we know that $\pi_4(S^3)$ is nontrivial, which probably cannot be shown by any means that are easier than the Steenrod square argument you mentioned, and indeed some commonly used ways to prove it are very closely related to the fact that $Sq^2$ is nontrivial on $H^2(\mathbb{C}P^2)$.)

Answer 2

Since we were having fun I thought I'd get this down. It seemed suitably ridiculous.

There is a $7$-connected map $\iota:\Sigma\mathbb{C}P^2\rightarrow SU_3$ given by sending a complex line to the generalised reflection through that line (this map is actually discussed at some point in Steenrod and Epstein when they talk about the cell-structures of the classical groups). Assume that there is a homotopy equivalence between $S^3\vee S^5$ and $\Sigma\mathbb{C}P^2$. Then there is a map $w:S^7\rightarrow \Sigma\mathbb{C}P^2$ representing the Whitehead product that attaches the top cell of $S^3\times S^5$, so in particular there is a cofiber sequence

$$S^7\xrightarrow{w} \Sigma\mathbb{C}P^2\rightarrow S^3\times S^5.$$

Then since $\pi_7SU_3=0$ there is a null-homotopy $\iota\circ w\simeq\ast$ and this gives rise to an extension $\hat\iota:S^3\times S^5\rightarrow SU_3$. Since $H^*SU_3\cong\Lambda(x_3,x_5)$ is an exterior algebra on classes satisfying $\iota^*x_3=\sigma x$, and $\iota^*x_5=\sigma x^2\in H^*\Sigma\mathbb{C}P^2$ we find that $\hat\iota $ induces an isomorphism on cohomology. Since both spaces are simply connected this map must in fact be a homotopy equivalence

$$\hat\iota:S^3\times S^5\simeq SU_3.$$

But this is absurd, since in this case $S^5$ retracts off of the H-space $SU_3$ and so must itself have an H-structure. But this cannot be, for in resolving the Hopf invariant one problem Adams proved that the only spheres which support H-structures are $S^0$, $S^1$, $S^3$ and $S^7$. Hence we have a contradiction.