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?


Let me explain why the condition is not the triviality of the difference cocycle. Maybe an important point to note is that the condition with the difference cocycle is not about the vanishing of an obstruction, it is about the enumeration of possible lifts of maps from one Postnikov truncation to the next.

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.

  • $\begingroup$ 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. --- $\endgroup$ – Warlock of Firetop Mountain Jul 6 at 19:59
  • 1
    $\begingroup$ 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? $\endgroup$ – Warlock of Firetop Mountain Jul 6 at 19:59
  • 1
    $\begingroup$ 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. $\endgroup$ – Gustavo Granja Jul 8 at 6:35
  • 1
    $\begingroup$ Thanks, I can now accept the answer. $\endgroup$ – Warlock of Firetop Mountain Jul 8 at 15:11

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.