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We know well this short exact sequence $$ 1 \to \mathbb{Z}_2 \to Spin(N) \to SO(N) \to 1. $$ The $j$-th Stiefel-Whitney class of the associated vector bundle of $SO(N)$, as $w_j(V_{SO(N)})$, can be nontrivial in general,

$$ w_j(V_{SO(N)}) \in H^j(BG,\mathbb{Z}_2)\overset{?}{=}H^j(BSO(N),\mathbb{Z}_2). $$ Here I suppose the $G=SO(N)$. We can either use the topological cohomology $H^j(BG,\mathbb{Z}_2)$ of classifying space $BG$, or consider the group cohomology H$^j(G,\mathbb{Z}_2)$ of the group $G$.

If the corresponding vector bundle of is the tangent bundle of the base manifold $M$, we have $$ w_j(M) \in H^j(M,\mathbb{Z}_2). $$

Questions: In either case,

when we lift from

  • $V_{SO(N)}$ to $V_{Spin(N)}$.

or

  • The base manifold $M$ with the SO-structure ($SO(N)$) to the $M'$ with the Spin-structure ($Spin(N)$),

how do we see explicitly that

  • the nontrivial $w_j(V_{SO(N)}) \in H^j(BG,\mathbb{Z}_2)\overset{?}{=}H^j(BSO(N),\mathbb{Z}_2)$ and
  • the nontrivial $w_j(M) \in H^j(M,\mathbb{Z}_2)$,

when the $w_j$ is inflated from $SO$ to $Spin$, and $w_j$ becomes its trivialization (a coboundary or just vanish)? Namely, how to show:

  • $w_j(V_{Spin(N)})=0$.

  • $w_j(M')=0$.

I know the statements are almost obviously true by definition, but can we precisely write down a (de Rham?) manifold-cocycle or group-cocycle on the simplicial complex (well-triangulated), such that we can show both - $w_j$ becomes a coboundary (instead of a cocycle) when lifting to $Spin(N)$? Namely,

  • $w_j(V_{Spin(N)})=\delta \beta_{j-1}(V_{Spin(N)})$.

  • $w_j(M')=\delta \beta_{j-1}(M')$.

The best answer to me is also to introduce the proper way to write down the $j$-cocycle $w_j(V_{SO(N)})$ and $w_j(M)$ on the SO or Spin manifold; and see how they can be split to the $(j-1)$-cochains when pulling backward from $Spin(N) \to SO(N)$.

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