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Let $M$ be (say smooth) manifold. From the short exact sequence of groups $0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}_2 \to 0$ (where the first map is multiplication by $2$) one obtain long exact sequence in cohomology. In particular one obtains the connecting map $\beta:H^2(M,\mathbb{Z}_2) \to H^3(M,\mathbb{Z})$. This map is called the Bockstein homomorphism. Define $W_3(M):=\beta(w_2(M))$ where $w_2(M)$ is the second Stiefel-Whitney class. The class $W_3(M)$ is called the third integral Stiefel-Whitney class.
On the other hand there is another class in $H^3(M,\mathbb{Z})$ which at the first sight has nothing to do with $W_3(M)$: it is called Dixmier-Douady class and is defined in terms of bundles of simple $C^*$-algebras.

It turns out that these two classes coincide: this is proved in this paper by Plymen-see Theorem 2.8. However the proof relies on another result. The author gives precise reference:

Marry P. ,,Varietes spinorielles. Geometrie riemannienne en dimension 4'', Seminaire Arthur Besse, CEDIC, Paris 1981

however I was unable to find it (even if I could, unfortunately I don't speak French). So

I would like to understand why $W_3(M)=\delta(M)$, in particular understand the last two lines of the case ''i)'' in the proof of Theorem 2.8 in the above paper.

EDIT: The relevant bundle for defining $\delta(M)$ is the (even part of) the complex Clifford bundle of the tangent bundle. Recall that the complex Clifford algebras are isomorphic to either $M_{2^n}(\mathbb{C})$ or to $M_{2^n}(\mathbb{C}) \oplus M_{2^n}(\mathbb{C})$ thus the even part of Clifford algebra is always a simple algebra.

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    $\begingroup$ Which bundle (or gerbe) do you take the DD class of? $\endgroup$ Dec 30, 2017 at 16:13
  • $\begingroup$ Any such class is a dixmier douady class of a bundle of simple $C^*$-algebra. So you can find a bundle of algebra such that this holds. Moreover the construction of such a bundle is quitte explicite when your class is described in $H^2(M,U)$ where U is the sheaf of continuous complex function of module 1, which is your case here. $\endgroup$ Dec 30, 2017 at 17:24
  • $\begingroup$ The relevant bundle is the clifford bundle, see the edit. $\endgroup$
    – truebaran
    Dec 30, 2017 at 23:02

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