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.

An interesting thing to note, for people finding this question in the future, is that the same sequence for $KU/2$, that is $H^*(X, KU/2^0(pt. ))\to H^{*+3}(X, KU/2^0(pt.)$ is infact $Sq^2Sq^1+Sq^3$. The nontriviality of the Bockstein on $Sq^2Sq^1$ obstructs the lifting of its effect to the integral sequence. – Pax Kivimae Mar 6 '17 at 19:58
This follows from the following considerations:
This differential in the AtiyahHirzebruch 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).
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 mod2 cohomology to integer cohomology. This comes from a calculation of the cohomology of EilenbergMac Lane spaces, which describe all possible cohomology operations; for n sufficiently large we have $H^{n+3}(K(\mathbb{Z},n)) = \mathbb{Z}/2$.
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 Kgroups). I believe that you can find this for $\mathbb{RP}^2 \times \mathbb{RP}^4$; perhaps someone more industrious can flesh this out?

But computing the K group of $\mathbb{R}P^k$ needs to apply AHSS. Actually at the appendix of the paper by Atiyah, titled Analytic cycle on complex manifold, He computed $d_{2p−1}$ in terms of Steenrod ppower by attaching a cell to some$\Sigma^k\matbb{C}P^n$ and using some relation of differentials and chern character which I don't quite understand. But I expect for d3 there should be some straight forward example – Sam Nariman Apr 24 '11 at 5:11

Why do the differentials for the AHSS have to be a stable cohomology operation? Do you have a reference? – 54321user Aug 15 '17 at 3:51

@54321user I believe that I learned that from Adams. To be a stable cohomology operation it merely needs to (a) be a natural transformation (which is true because the AtiyahHirzebruch spectral sequence is natural), and (b) be compatible with the suspension isomorphism (which one can show straightforwardly when the (based) space is a point and then progressively bootstrapped up to the result for a general space). – Tyler Lawson Aug 18 '17 at 23:07
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 AtiyahHirzebruch 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.
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 nonzero $d_3$ differential in the AHSS for $\mathbb{R}P^2 \times \mathbb{R}P^4$.
The Ktheory 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 $yx = 0$ 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 $yx = 0$ in the AHSS start (or end) at trivial groups. So there must be a nonzero differential on the 3rd page. It's not clear to me which one it is, but I haven't thought about the multiplicative structure.

But computing the K group of $\mathbb{R}P^k$ needs to apply AHSS. Actually at the appendix of the paper by Atiyah, titled Analytic cycle on complex manifold, He computed $d_{2p1}$ in terms of Steenrod ppower by attaching a cell to some $\Sigma^k \matbb{C}P^n$ and using some relation of differentials and chern character which I don't quite understand. But I expect for $d_3$ there should be some straight forward example. – Sam Nariman Apr 24 '11 at 5:06

Sam, are you saying that my argument might be circular, because one would need to understand the $d_3$ differential already in order to compute the Ktheory of $RP^2$ and $RP^4$? In both cases, there's no room for a nonzero differential (on any page). One does have to solve the extension problem in order to see that $K^0 (RP^4)$ is $Z\oplus Z/4$ and not $Z\oplus Z/2 \oplus Z/2$. It seems unlikely to me that this could require knowing something about $d_3$, since it's zero in this case. – Dan Ramras Apr 24 '11 at 6:07

According to this question math.stackexchange.com/questions/26438/… math.stackexchange, there's a computation of $K^*(RP^n)$ on pp. 100110 of Atiyah's Ktheory book. I'm not sure what's involved; I don't think I've ever gone through the computation. – Dan Ramras Apr 24 '11 at 6:09

For what I wrote above, one really does need to know that the torsion group is $Z/4$ and not $Z/2\oplus Z/2$. I was actually specifically interested in figuring out how to solve this extension problem the day this question appeared; I'll look into it further. – Dan Ramras Apr 24 '11 at 6:11

In this case, you can solve the extension problem using Chern classes; namely, you show twice a line bundle is not stably trivial by showing that it has nontrivial
$w_2$
. – Tyler Lawson Apr 24 '11 at 8:39