# Obstruction to homotopy, cohomology operations and Dold-Whitney theorem

I am reading the famous paper by Dold and Whitney "Classification of Oriented Sphere Bundles Over A 4-Complex".

I'll state their theorem for the case of SO(3) bundles

Classification Theorem:Let $$B_1, B_2$$ be principal $$SO(3)$$ bundles over a complex $$K$$ of dimension at most 4. let $$h_i:K\to G_n\simeq BSO(3)$$ be the classifying map for $$B_i$$. Assume they have the same second Stiefel-Whitney class $$w_2(B_1)=w_2(B_2)=: w_2$$ (then we can assume that $$h_1$$ and $$h_2$$ agree on the 3-skeleton $$K^3$$). Let $$d(h_1,h_2) \in H^4(K;\pi_4(G_n))$$ be the difference cocycle cohomology class.
The bundles $$B_1,B_2$$ are equivalent iif there exists a cohomology class $$x\in H^1(K,\mathbb{Z}/2)$$ such that

$$(1) \quad \quad d(h_1,h_2) = \beta x \cup \beta x + \beta (x\cup w_2)$$

where $$\beta$$ is the Bockstein homomorphism associated with the coefficient sequence $$0\to \mathbb{Z} \overset{2\cdot}{\to} \mathbb{Z}\to \mathbb{Z}/2\to 0$$ and $$\pi_3(G_n)$$ has been identified with $$\mathbb{Z}$$.

Usually we require the cohomology class of the cocycle to $$d(h_1,h_2)$$ to be $$0$$, why in (1) they require that condition instead?

Explanation of the question I have seen stated in others book on obstruction theory that $$h_1,h_2:K^n\to Y$$ are homotopic rel to $$K^{n-2}$$ iif $$[d(h_1,h_2)]=0$$ (under the hypothesis of $$Y$$ being simple, I don't know if this is the point) this seems to be very different from $$(1)$$.

Cohomology operations. Dold & Whitney appeal to a theorem from Eilenberg MacLane's paper "On the groups $$H(\pi,n),\quad III:$$ Operations and Obstructions", theorem 14.2. I looked into it and indeed they don't require the difference cocycle to be null-cohomologous. I only explained myself that the RHS of $$(1)$$ comes from the cohomology operation induced by the $$k$$-invariant (I guess Postnikov invariant) of the target space.

Can someone explain better the setup and theorem 14.2 in Eilenberg MacLane's paper?

If we denote by $$\tau_{\leq n} {\rm BSO}(3)$$ the Postnikov truncation which kills all homotopy above degree $$n$$, then we have the usual fiber sequence of obstruction theory $$K(\pi_4{\rm BSO}(3),4)\to \tau_{\leq 4}{\rm BSO}(3) \to \tau_{\leq 3}{\rm BSO}(3).$$ This gives rise to an exact sequence of pointed sets of homotopy classes $$[X_+,\Omega\tau_{\leq 3}{\rm BSO}(3)]\to [X_+,K(\pi_4{\rm BSO}(3),4)]\to [X_+,\tau_{\leq 4}{\rm BSO}(3)]\to [X_+,\tau_{\leq 3}{\rm BSO}(3)].$$ Note that $${\rm BSO}(3)$$ is simply connected (hence also simple), so there is no difference between pointed and unpointed maps here. Since $$\tau_{\leq 3}{\rm BSO}(3)=K(\mathbb{Z}/2,2)$$, we can rewrite that sequence $${\rm H}^1(X,\mathbb{Z}/2)\to {\rm H}^4(X,\mathbb{Z})\to [X_+,{\rm BSO}(3)]\to {\rm H}^2(X,\mathbb{Z}/2).$$ The set $$[X,{\rm BSO}(3)]$$ gives the isomorphism classes of oriented rank 3 bundles, the last map takes a bundle to its second Stiefel-Whitney class. The additional invariant lives in a quotient group of $$H^4(X,\mathbb{Z})$$ and the first map $${\rm H}^1(X,\mathbb{Z}/2)\to {\rm H}^4(X,\mathbb{Z})$$ takes a class $$x$$ to $$\beta x\cup \beta x+\beta(x\cup w_2)$$. So the difference cocycle is a class in $${\rm H}^4$$, but the bundle is still trivial if the cocycle lies in the image of $${\rm H}^1(X,\mathbb{Z}/2)$$ which explains the condition.
• If I understand well, for a fixed lift $h_1\in[X,\tau_{\leq 4}BSO(3)]$, the map $H^4(X,\mathbb{Z})\to [X,\tau_{\leq 4} BSO(3)]$ sends a cocycle $\omega\mapsto h_2$ such that the difference cocycle $d(h_1,h_2)=\omega \in H^4$, is this correct? And by the exactness of the sequence we get that the lifts with the same second Stiefel-Whitney class are exactly the image of this map. Moreover the map has a kernel given by $\beta x \cup \beta x + \beta (x\cup w_2)$ hence all these difference cocycles give rise to homotopic lifts. --- – Warlock of Firetop Mountain Jul 6 at 19:59
• This is a very nice picture, but I am still missing something. Given the two maps $h_1,h_2:X\to BSO(3)$, suppose they coincide over the three skeleton, so are the same in $[X,\tau_{\leq 3} BSO(3)]$. Now (from say Mosher-Tangora's book on cohomology op. page 7) I know that $h_1$ and $h_2$ are homotopic rel to the 2-skeleton iif $[d(h_1,h_2)]=0$ this is the same as having being two homotopic lifts, what am I missing here? – Warlock of Firetop Mountain Jul 6 at 19:59
• If $[d(h_1,h_2)]\neq 0$ then the maps will not be homotopic via a homotopy which is constant on the $2$-skeleton. Eilenberg and MacLane are computing the effect on the difference cocycle of changing the homotopy on the $2$-skeleton. – Gustavo Granja Jul 8 at 6:35