TITLE: Third differential in Atiyah Hirzebruch spectral sequence
QUESTION [26 upvotes]: 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.

REPLY [21 votes]: 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.