# A question about the Chern-Weil construction of Euler class

I'm reading Appendix C of "Characteristic Classes" by Milnor & Stasheff and something confuses me. When proving that the Pfaffian of the curvature form (of an oriented $2n$-plane bundle $\xi$ over a smooth manifold $M$, with a metric and a connection compatible with that metric) represents a multiple of the Euler class of $\xi$, the authors say: (here $K$ stands for curvature tensor)

The notation $Pf(K(\tilde{\gamma}))$ suggests that $\tilde{\gamma}$ has a connection on it, thus the base space must be a smooth manifold. On the other hand, $\tilde{\gamma}$ is required to be "universal", so it should be the tautological bundle over the oriented Grassmannian $\tilde{G}_{2n}(\mathbb{R}^{\infty})$, but then the base space is not a manifold, seemingly a contradiction. So what should this bundle $\tilde{\gamma}$ be?

In the second line of the quoted passage, it says that $\tilde\gamma$ should be `universal in dimensions $\leq 4n$'. This means that it's universal for bundles over complexes of dimension at most $4n$, and so we can take a finite dimensional oriented Grassmannian $\tilde{G}_{2n}(R^N)$ for sufficiently large $N$. This is a manifold, and the restriction of $\tilde\gamma$ has a connection.
You can read about $n$-universal bundles in Steenrod's book, Fiber Bundles.
• Thank you for the reference. Some more questions: How does the structure of $H^*(\tilde{G}_{2n}(\mathbb{R}^N);\mathbb{R})$ guarantees that $a^2=b^2$ implies $a=\pm b$? Steenrod's book only considers finite complexes, so how is $\tilde{\gamma}$ universal for bundles over manifolds of dimension $\leq 4n$?(I guess it's possible to extend bundle maps on finite subcomplexes to a countable complex using a weak topology argument?) – AaronS May 10 '18 at 13:45