TITLE: Why is the constant relating the chern class and curvature form always $2\pi i$?
QUESTION [12 upvotes]: I'm reading Milnor's book on Characteristic Classes. In Appendix C, Milnor shows the invariant polynomial of the curvature form and the Chern class differ by powers of $2\pi i$. He first shows that the first Chern class and the trace of the curvature form are multiples of each other by a constant $a$. He calculates this constant for a bundle on a sphere, and finds it is $2\pi i$. Milnor says it's enough to calculate $a$ for one specific case, and that this constant is the same for all vector bundles.
My question is: why is it enough to find $a$ for one specific case? Is there any reason why the constant can't be different for different bundles?
The proof is done on page 306.

REPLY [2 votes]: This is meant as a supplement to anomaly’s elegant answer above. In what follows, all manifolds are assumed to be closed and oriented.

For each smooth manifold $ M $, let

$ \mathscr{L}(M;\mathbb{C}) $ denote the set of all isomorphism classes of smooth (complex) line bundles over $ M $;
$ {H^{*}}(M;\mathbb{C}) $ denote the cohomology ring of $ M $.

Hence, we may view $ \mathscr{L}(\bullet;\mathbb{C}) $ and $ {H^{*}}(\bullet;\mathbb{C}) $ as contravariant functors from the category of smooth manifolds to the category of sets.
Milnor essentially establishes that $ \Omega_{12} $ is a natural transformation from $ \mathcal{L}(\bullet;\mathbb{C}) $ to $ {H^{*}}(\bullet;\mathbb{C}) $, i.e., for each morphism $ f: M \to N $ of smooth manifolds, we have the following commutative diagram:
$$
\require{AMScd}
\begin{CD}
\mathscr{L}(N;\mathbb{C}) @>{f^{*}}>> \mathscr{L}(M;\mathbb{C}) \\
@V{{\Omega_{12}}(N)}VV                @VV{{\Omega_{12}}(M)}V \\
{H^{*}}(N;\mathbb{C})     @>>{f^{*}}> {H^{*}}(M;\mathbb{C})
\end{CD}
$$
It is a well-known result (see Theorem 14.5 of Milnor’s and Stasheff’s book) that any element of the cohomology ring $ {H^{*}}(\mathbb{C P}^{\infty};\mathbb{C}) $ is a polynomial in the first Chern class of the tautological line bundle $ \gamma^{1} $ over $ \mathbb{C P}^{\infty} $. Having said so, let $ P \in \mathbb{C}[X] $ be a polynomial such that
$$
[{\Omega_{12}}(\mathbb{C P}^{\infty})](\gamma^{1}) = P({c_{1}}(\gamma^{1})).
$$
Let $ M $ be any smooth $ 2 $-manifold and $ E $ any smooth line bundle over $ M $. As $ \gamma^{1} $ is a universal line bundle, there exists a smooth mapping $ f: M \to \mathbb{C P}^{\infty} $ such that $ E = {f^{*}}(\gamma^{1}) $. The naturality of the first Chern class $ c_{1} $ then yields
\begin{align}
    [{\Omega_{12}}(M)](E)
& = [{\Omega_{12}}(M)]({f^{*}}(\gamma^{1})) \\
& = {f^{*}}([{\Omega_{12}}(\mathbb{C P}^{\infty})](\gamma^{1})) \\
& = {f^{*}}(P({c_{1}}(\gamma^{1}))) \\
& = P({f^{*}}({c_{1}}(\gamma^{1}))) \\
& = P({c_{1}}({f^{*}}(\gamma^{1}))) \\
& = P({c_{1}}(E)).
\end{align}
Now, $ [{\Omega_{12}}(M)](E) \in {H^{2}}(M;\mathbb{C}) $, so $ P $ is a linear polynomial with no constant term, i.e., $ P = \alpha X $ for some $ \alpha \in \mathbb{C} $. Therefore, for all smooth $ 2 $-manifolds $ M $ and all smooth line bundles $ E $ over $ M $,
$$
[{\Omega_{12}}(M)](E) = \alpha \cdot {c_{1}}(E).
$$
Conclusion: We can determine the value of $ \alpha $ by considering the cotangent bundle $ T^{*} \mathbb{S}^{2} $ over $ \mathbb{S}^{2} $.