Third differential in Atiyah Hirzebruch spectral sequence

Question

Does any one know why $d_3: H^* (X, K^0(point))\rightarrow H^{*+3}(X,K^0(point))$ is actually extended $Sq^3$ to $\mathbb{Z} $ coefficient.

Answer 1

This follows from the following considerations:

1. This differential in the Atiyah-Hirzebruch spectral sequence must be a stable cohomology operation for general nonsense reasons (the first nonvanishing differential always is, no matter what the generalized cohomology theory is).

2. There are exactly two stable cohomology operations $H^*(X) \to H^{*+3}(X)$ with integer coefficients. One of them is zero, and the other is $\beta \circ Sq^2 \circ r$, where $r$ is reduction mod 2 and $\beta$ is the Bockstein from mod-2 cohomology to integer cohomology. This comes from a calculation of the cohomology of Eilenberg-Mac Lane spaces, which describe all possible cohomology operations; for n sufficiently large we have $H^{n+3}(K(\mathbb{Z},n)) = \mathbb{Z}/2$.

3. The $d_3$ differential is not the zero cohomology operation. For this, it suffices to find one space for which this differential is nontrivial (and you can find this by actually calculating the complex K-groups). I believe that you can find this for $\mathbb{RP}^2 \times \mathbb{RP}^4$; perhaps someone more industrious can flesh this out?

Answer 2

A pretty direct argument was given by Frank Adams in the proof of 16.6 (page 336) in part III of his 1974 Chicago lectures (MR0402720). Thinking of the Atiyah--Hirzebruch spectral sequence for $K^*(X)$ as arising from the Postnikov tower $\{P^n ku\}$ of ku (Adams calls this spectrum $bu$), row 0 and row 2 come from the layers $HZ$ and $\Sigma^2 HZ$, sitting in a cofiber sequence $\Sigma^2 HZ \to P^2 ku \to HZ \to \Sigma^3 HZ$. The $d_3$-differential is induced by the third map, i.e., the first $k$-invariant of $ku$. To see that the $k$-invariant is the integral lift $\beta Sq^2$ of $Sq^3$, i.e., not zero, Adams looks at the third space in the $ku$-spectrum, namely $SU$, and notices that $\beta Sq^2 \ne 0$ in $H^6(K(Z, 3); Z)$ but $H^6(SU; Z) = 0$. This implies that the $k$-invariant is nonzero.

Answer 3

Okay, I can't pass up the chance to try and be more industrious than Tyler (this is really a comment on Tyler's answer).

I'll try to explain why there's a non-zero $d_3$ differential in the AHSS for $\mathbb{R}P^2 \times \mathbb{R}P^4$.

The K-theory of $\mathbb{R}P^{2k}$ is $\mathbb{Z} \oplus \mathbb{Z}/2^k$ in dimension 0 and is trivial in dimension 1. Applying the Kunneth Theorem tells you that $K^0 (\mathbb{R}P^2 \times \mathbb{R}P^4) = \mathbb{Z}\oplus \mathbb{Z}/2\oplus \mathbb{Z}/4\oplus \mathbb{Z}/2$. Comparing with the cohomology of $\mathbb{R}P^2 \times \mathbb{R}P^4$, one sees that one factor of $\mathbb{Z}/2$ appearing on the line $y-x = 0$ in the E_2 page of the AHSS has to be killed by a differential.

After the 3rd page, all differentials coming in (or out) of the line $y-x = 0$ in the AHSS start (or end) at trivial groups. So there must be a non-zero differential on the 3rd page. It's not clear to me which one it is, but I haven't thought about the multiplicative structure.