8
$\begingroup$

Let $E^4$ be the two stage Postnikov space appearing in the homotopy type of the classifying space $G/PL$. One of its properties is that it only has two nontrivial homotopy groups $\pi_2(E)=Z/2Z$ and $\pi_4(E)=Z$. Here $Z$ denotes the integers.

In Lopez de Medrano's book on involutions, it is stated without proof that the homotopy classes of maps $[CP^2, G/PL]$ is isomorphic to $Z$.

Anyone know a proof? He gives a passing hint that it has to do with the k-invariant used to glue the Postnikov pieces together.

Thanks.

$\endgroup$

2 Answers 2

8
$\begingroup$

In general, suppose we want to compute homotopy classes $\pi_0 [X, Y]$ of maps $X \to Y$. Assume for simplicity that $Y$ is simply connected, which is the case here. Then we can use the Postnikov tower of $Y$; write the $n$-truncated part of it as $Y_n$. If we know $\pi_0 [X, Y_n]$, we can compute $\pi_0 [X, Y_{n+1}]$ by taking each element of $\pi_0 [X, Y_n]$ and computing how it lifts to $Y_{n+1}$. There is a fiber sequence

$$B^{n+1} \pi_{n+1} Y \to Y_{n+1} \to Y_n$$

which, given that $Y$ is simply connected, is principal; this means that the obstruction to lifting a map $X \to Y_n$ to a map $X \to Y_{n+1}$ is a cohomology class in $H^{n+2}(X, \pi_{n+1} Y)$, the pullback of the Postnikov invariant in $H^{n+2}(Y_n, \pi_{n+1} Y)$ classifying the above fiber sequence. When this obstruction vanishes, the set of lifts is classified by $H^{n+1}(X, \pi_{n+1} Y)$. In particular, if $X$ is a $d$-dimensional CW complex and $n \ge d$, then the obstruction classes always vanish and lifts are always unique, which gives

$$\pi_0 [X, Y] \cong \pi_0 [X, Y_n].$$

In our case, $X = \mathbb{CP}^2$ is a $4$-dimensional CW complex and so we only need to go up to $n = 4$. We have $Y_2 \cong B^2 \mathbb{Z}_2$, so

$$\pi_0 [X, Y_2] \cong H^2(\mathbb{CP}^2, \mathbb{Z}_2) \cong \mathbb{Z}_2$$

by universal coefficients. The same is true of $Y_3 \cong Y_2$. The obstruction to lifting to $Y_4$ is a class in $H^5(\mathbb{CP}^2, \mathbb{Z})$, which again vanishes. The set of lifts is $H^4(\mathbb{CP}^2, \mathbb{Z}) \cong \mathbb{Z}$, and we don't need to go beyond this. This gives

$$\pi_0 [X, Y_4] \cong \pi_0 [X, Y] \cong \mathbb{Z} \times \mathbb{Z}_2$$

as sets; I assume the claim is that we should in fact get $\mathbb{Z}$ the abelian group using the infinite loop space structure on $G/PL$, but these computations aren't enough to get at that. With some refinement I believe the above computations show that, as a group, $\pi_0 [X, Y]$ fits into a short exact sequence

$$0 \to \mathbb{Z} \to \pi_0 [X, Y] \to \mathbb{Z}_2 \to 0$$

but I don't know how to push them to distinguish $\mathbb{Z}$ from $\mathbb{Z} \times \mathbb{Z}_2$.

$\endgroup$
2
  • $\begingroup$ I think you meant to say that $H^2({\mathbb CP}^2,{\mathbb Z}_2)$ is ${\mathbb Z}_2$, not $0$. $\endgroup$ Jul 24, 2016 at 2:48
  • $\begingroup$ @Allen: yes, sorry, I got confused. I am also not sure how to access the abelian group structure coming from the infinite loop space structure on $G/PL$. $\endgroup$ Jul 24, 2016 at 3:47
7
$\begingroup$

This is the second half of an answer completing the first half given by @Qiaochu Yuan.

The information we need is that the $k$-invariant is $\beta Sq^2$, and the space $E$ has the $H$-space structure of the fiber of $\beta Sq^2: K(\mathbb Z/2,2)\to K(\mathbb Z,5)$. (Madsen-Milgram Theorem 4.34, but they do not prove this, only refer to Sullivan.)

Since all maps between the EM-spaces are $H$-maps, the short exact sequence one obtains is a short exact sequence of groups.

Let $F$ be the fiber of $\beta: K(\mathbb Z/2,4)\to K(\mathbb Z,5)$, in fact $F=K(\mathbb Z/2,4)$.

The fiber sequence $K(\mathbb Z,4)\to F \to K(\mathbb Z/2,4)$ gives rise to another short exact sequence when applying $[\mathbb C P^2,-]$ to it, namely $0\to \mathbb Z \to \mathbb Z \to \mathbb Z/2\to 0$. Now $Sq^2$ induces a map of the short exact sequences which is the identity on the $\mathbb Z$-terms on the left and also the identity on the $\mathbb Z/2$ terms on the right, since $Sq^2:H^2(\mathbb C P^2,\mathbb Z/2)\to H^4(\mathbb C P^2,\mathbb Z/2)$ is an isomorphism. It follows that the group we are interested in is isomorphic to $\mathbb Z$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.