The purpose of this post is to show the existence of a phantom map ${\phi: \Sigma^\infty \mathbb{CP}^\infty \rightarrow \Sigma^\infty S^3}$. To say that is a phantom map means that ${\phi}$ is nonzero, but ${\phi|_{\Sigma^\infty \mathbb{CP}^n}}$ is nontrivial for each ${n}$: in particular, ${\phi}$ induces the zero map in any homology theory. (This is essentially following a paper of Brayton Gray, “Spaces of the same $n$-type for all $n$.”)

In order to do this, we note that ${\Sigma^\infty \mathbb{CP}^\infty }$ is the homotopy colimit of the spectra ${\Sigma^\infty \mathbb{CP}^n}$, and consequently we have a Milnor exact sequence

$\displaystyle 0 \rightarrow \varprojlim^1 [ \Sigma \Sigma^\infty \mathbb{CP}^n, S^3] \rightarrow [\Sigma^\infty \mathbb{CP}^\infty, S^3] \rightarrow \varprojlim [\Sigma^\infty \mathbb{CP}^n, S^3] \rightarrow 0.$

For convenience, we have abbreviated ${\Sigma^\infty S^3}$ to simply ${S^3}$.

Our goal will be to show that the ${\varprojlim^1}$ term is nonzero. It will then follow that we have a “phantom” map ${\Sigma^\infty \mathbb{CP}^\infty \rightarrow S^3}$, as desired.

1. Failure of the Mittag-Leffler condition

To do so, we consider the inverse system of groups ${[\Sigma \Sigma^\infty \mathbb{CP}^n, S^3] = \left\{\mathbb{CP}^n, S^2\right\}}$ where ${\left\{\right\}}$ denotes homotopy classes of stable maps. The claim is that this does not satisfy the Mittag-Leffler condition. In fact, let’s look at the images of the maps

$\displaystyle \left\{\mathbb{CP}^n, S^2 \right\} \rightarrow \left\{\mathbb{CP}^1 , S^2\right\} \simeq \mathbb{Z}.$

Rational stable cohomotopy is the same as rational cohomology, so the map is a ${\mathbb{Q}}$-isomorphism: in particular, the image has finite index. However, the claim is that given any class in ${\left\{\mathbb{CP}^1, S^2\right\}}$, it is not in the image of ${\left\{\mathbb{CP}^n, S^2\right\}}$ for ${n}$ sufficiently large.

In fact, choose a class ${x \in \left\{\mathbb{CP}^1, S^2\right\}}$; suppose ${x}$ induces multiplication by ${m}$ on ${H^2}$. Suppose ${n \gg m}$, and we had a stable map in ${\left\{\mathbb{CP}^n , S^2 \right\}}$ restricting to ${x}$ on ${\mathbb{CP}^1}$. Then we would have a map of suspension spectra

$\displaystyle \Sigma^\infty \mathbb{CP}^n \rightarrow \Sigma^\infty S^2$

inducing multiplication by ${m}$ on ${H^2}$. But, for any odd prime ${p}$, the Steenrod ${p}$th power ${\mathcal{P}^1}$ acts on ${H^2(\mathbb{CP}^n)}$ by raising the generating class to the power ${p}$. If ${n > p > m}$, then ${\mathcal{P}^1}$ acts nontrivially on the ${x^*(\iota_n) }$ for ${\iota_n }$ the generator of ${H^2(S^2; \mathbb{Z}/p)}$. On the contrary, it acts trivially on ${H^2(S^2; \mathbb{Z}/p)}$. This is a contradiction, and we find that there is no stable map ${\mathbb{CP}^n \rightarrow S^2}$ restricting to a degree ${m}$ map ${\mathbb{CP}^1 \rightarrow S^2}$, if ${n}$ is sufficiently large (bigger than the first prime after ${m}$).

So we’ve seen that the inverse system ${\left\{\mathbb{CP}^n, S^2\right\}}$ does not satisfy the ML condition. In the next section, we will show that this implies the ${\varprojlim^1}$ term is actually nonzero, which will prove the existence of a phantom map.

2. A lemmas on inverse limits

An inverse system of abelian groups satisfying the Mittag-Leffler condition has vanishing $\varprojlim^1$; the purpose of the next lemma is to prove a weak converse.

Lemma 1 Let ${\dots \rightarrow G_n \rightarrow G_{n-1} \rightarrow \dots}$ be an inverse system of countable abelian groups, and suppose that it does not satisfy the Mittag-Leffler condition. Then ${\varprojlim^1 G_n \neq 0}$.

Let’s assume for simplicity that the ML condition is not satisfied at the zeroth stage: that is, that the decreasing sequence of subgroups

$\displaystyle \mathrm{Im}(G_k \rightarrow G_0 ) \subset G_0$

does not stabilize. Then we have a map of inverse systems

$\displaystyle \left\{G_k\right\} \rightarrow \left\{\mathrm{Im}(G_k \rightarrow G_0)\right\}$

which is a surjection, and consequently induces a surjection on ${\varprojlim^1}$ (because there is no ${\varprojlim^2}$ for a ${\mathbb{Z}_{\geq 0}}$-indexed tower of abelian groups). Consequently, it suffices to prove the lemma for an inverse system of injections.

So let’s do that. Suppose we have an inverse system of countable abelian groups

$\displaystyle \dots \hookrightarrow G_n \hookrightarrow G_{n-1} \hookrightarrow \dots \hookrightarrow G_0.$

We have a short exact sequence of inverse systems

$\displaystyle 0 \rightarrow \left\{G_n\right\} \rightarrow \left\{G_0\right\} \rightarrow \left\{G_0/G_n\right\} \rightarrow 0,$

and if ${\varprojlim^1 \left\{G_n\right\} =0}$, then we have a short exact sequence

$\displaystyle 0 \rightarrow \varprojlim G_n \rightarrow G_0 \rightarrow \varprojlim G_0/G_n \rightarrow 0.$

But ${\varprojlim G_0/G_n}$ is a complete topological group under the inverse limit topology. Also, there is a neighborhood basis at ${0}$ consisting of the images of ${G_m}$ for each ${m}$ (these are nonzero because the sequence does not stabilize), and in particular the inverse limit is not discrete, and has the property that the complement of any point is dense. By the Baire category theorem, it follows that ${\varprojlim G_0/G_n}$ is actually uncountable. This contradicts the existence of a surjection ${G_0 \rightarrow \varprojlim G_0/G_n}$.