# Why is the constant relating the chern class and curvature form always $2\pi i$?

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.

• I believe that the trick here is the Leray-Hirsch Theorem. – Berrick Caleb Fillmore Jun 24 '14 at 20:08

The point of the proof is to show that what Milnor calls $\Omega_{12}(M)$ is a characteristic class for complex line bundles $\zeta\to M^2$. It's not an arbitrary element of $H^*(M, \mathbb{C})$; it's a map from bundles to cohomology that's natural and defined in terms of a certain classifying space. On p. 298, Milnor proves that certain forms induce characteristic classes; he then proves on p 306 that $\Omega_{12}(M)$ is a form of that type and thus defines a characteristic class. The only characteristic classes for complex line bundles are of the form $\zeta \to \alpha c(\zeta)\in H^*(X, \mathbb{C}) = H^*(X)\otimes \mathbb{C}$ for some fixed $\alpha\in\mathbb{C}$ (see, for example, the axiomatic definition of the Chern class--- it's determined by its value on the tautological bundle of $\mathbb{CP}^\infty$), and going through the computation for the case of $M = S^2, \zeta = T^*S^2$ determines $\alpha$.

• Let me see if I understand. The characteristic class maps bundles to cohomology. The class in question must be a constant times the chern class. Thus, the constant a must be independent of the bundle chosen to evaluate at, since the characteristic class is a map from bundles to cohomology and not the cohomology itself. – Aurey Jun 25 '14 at 7:36
• Right: A characteristic class is a specific kind of map from bundles over $M$ to $H^*(M, \pi)$ for some group of coefficients $\pi$. Every complex line bundle (assuming paracompactness, etc.) is a pullback $f^*\gamma^1$, where $\gamma^1\to \mathbb{CP}^\infty$ is the tautological line bundle. By naturality, any characteristic class $\eta$ satisfies $\eta = f^* \eta(\gamma^1)$. But $\eta(\gamma^1)\in H^*(\mathbb{CP}^\infty) = \mathbb{Z}$, so any two classes $\eta, \eta'$ differ by a constant multiple. – anomaly Jun 25 '14 at 7:43

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 $\mathbb{S}^{2}$.