Choose any connection form ω in P, and let Ω be the associated curvature form; i.e., Ω = D ω {\displaystyle \Omega =D\omega } , the exterior covariant derivative of ω. If f ∈ C [ g ] G {\displaystyle f\in \mathbb {C} [{\mathfrak {g}}]^{G}} is a homogeneous polynomial function of degree k; i.e., f ( a x ) = a k f ( x ) {\displaystyle f(ax)=a^{k}f(x)} for any complex number a and x in g {\displaystyle {\mathfrak {g}}} , then, viewing f as a symmetric multilinear functional on ∏ 1 k g {\textstyle \prod _{1}^{k}{\mathfrak {g}}} (see the ring of polynomial functions), let
be the (scalar-valued) 2k-form on P given by
where vi are tangent vectors to P, ϵ σ {\displaystyle \epsilon _{\sigma }} is the sign of the permutation σ {\displaystyle \sigma } in the symmetric group on 2k numbers S 2 k {\displaystyle {\mathfrak {S}}_{2k}} (see Lie algebra-valued forms#Operations as well as Pfaffian).
If, moreover, f is invariant; i.e., f ( Ad g x ) = f ( x ) {\displaystyle f(\operatorname {Ad} _{g}x)=f(x)} , then one can show that f ( Ω ) {\displaystyle f(\Omega )} is a closed form, it descends to a unique form on M and that the de Rham cohomology class of the form is independent of ω {\displaystyle \omega } . First, that f ( Ω ) {\displaystyle f(\Omega )} is a closed form follows from the next two lemmas:1
Indeed, Bianchi's second identity says D Ω = 0 {\displaystyle D\Omega =0} and, since D is a graded derivation, D f ( Ω ) = 0. {\displaystyle Df(\Omega )=0.} Finally, Lemma 1 says f ( Ω ) {\displaystyle f(\Omega )} satisfies the hypothesis of Lemma 2.
To see Lemma 2, let π : P → M {\displaystyle \pi \colon P\to M} be the projection and h be the projection of T u P {\displaystyle T_{u}P} onto the horizontal subspace. Then Lemma 2 is a consequence of the fact that d π ( h v ) = d π ( v ) {\displaystyle d\pi (hv)=d\pi (v)} (the kernel of d π {\displaystyle d\pi } is precisely the vertical subspace.) As for Lemma 1, first note
which is because R g ∗ Ω = Ad g − 1 Ω {\displaystyle R_{g}^{*}\Omega =\operatorname {Ad} _{g^{-1}}\Omega } and f is invariant. Thus, one can define f ¯ ( Ω ) {\displaystyle {\overline {f}}(\Omega )} by the formula:
where v i {\displaystyle v_{i}} are any lifts of v i ¯ {\displaystyle {\overline {v_{i}}}} : d π ( v i ) = v ¯ i {\displaystyle d\pi (v_{i})={\overline {v}}_{i}} .
Next, we show that the de Rham cohomology class of f ¯ ( Ω ) {\displaystyle {\overline {f}}(\Omega )} on M is independent of a choice of connection.2 Let ω 0 , ω 1 {\displaystyle \omega _{0},\omega _{1}} be arbitrary connection forms on P and let p : P × R → P {\displaystyle p\colon P\times \mathbb {R} \to P} be the projection. Put
where t is a smooth function on P × R {\displaystyle P\times \mathbb {R} } given by ( x , s ) ↦ s {\displaystyle (x,s)\mapsto s} . Let Ω ′ , Ω 0 , Ω 1 {\displaystyle \Omega ',\Omega _{0},\Omega _{1}} be the curvature forms of ω ′ , ω 0 , ω 1 {\displaystyle \omega ',\omega _{0},\omega _{1}} . Let i s : M → M × R , x ↦ ( x , s ) {\displaystyle i_{s}:M\to M\times \mathbb {R} ,\,x\mapsto (x,s)} be the inclusions. Then i 0 {\displaystyle i_{0}} is homotopic to i 1 {\displaystyle i_{1}} . Thus, i 0 ∗ f ¯ ( Ω ′ ) {\displaystyle i_{0}^{*}{\overline {f}}(\Omega ')} and i 1 ∗ f ¯ ( Ω ′ ) {\displaystyle i_{1}^{*}{\overline {f}}(\Omega ')} belong to the same de Rham cohomology class by the homotopy invariance of de Rham cohomology. Finally, by naturality and by uniqueness of descending,
and the same for Ω 1 {\displaystyle \Omega _{1}} . Hence, f ¯ ( Ω 0 ) , f ¯ ( Ω 1 ) {\displaystyle {\overline {f}}(\Omega _{0}),{\overline {f}}(\Omega _{1})} belong to the same cohomology class.
The construction thus gives the linear map: (cf. Lemma 1)
In fact, one can check that the map thus obtained:
is an algebra homomorphism.
Let G = GL n ( C ) {\displaystyle G=\operatorname {GL} _{n}(\mathbb {C} )} and g = g l n ( C ) {\displaystyle {\mathfrak {g}}={\mathfrak {gl}}_{n}(\mathbb {C} )} its Lie algebra. For each x in g {\displaystyle {\mathfrak {g}}} , we can consider its characteristic polynomial in t:3
where i is the square root of -1. Then f k {\displaystyle f_{k}} are invariant polynomials on g {\displaystyle {\mathfrak {g}}} , since the left-hand side of the equation is. The k-th Chern class of a smooth complex-vector bundle E of rank n on a manifold M:
is given as the image of f k {\displaystyle f_{k}} under the Chern–Weil homomorphism defined by E (or more precisely the frame bundle of E). If t = 1, then det ( I − x 2 π i ) = 1 + f 1 ( x ) + ⋯ + f n ( x ) {\displaystyle \det \left(I-{x \over 2\pi i}\right)=1+f_{1}(x)+\cdots +f_{n}(x)} is an invariant polynomial. The total Chern class of E is the image of this polynomial; that is,
Directly from the definition, one can show that c j {\displaystyle c_{j}} and c given above satisfy the axioms of Chern classes. For example, for the Whitney sum formula, we consider
where we wrote Ω {\displaystyle \Omega } for the curvature 2-form on M of the vector bundle E (so it is the descendent of the curvature form on the frame bundle of E). The Chern–Weil homomorphism is the same if one uses this Ω {\displaystyle \Omega } . Now, suppose E is a direct sum of vector bundles E i {\displaystyle E_{i}} 's and Ω i {\displaystyle \Omega _{i}} the curvature form of E i {\displaystyle E_{i}} so that, in the matrix term, Ω {\displaystyle \Omega } is the block diagonal matrix with ΩI's on the diagonal. Then, since det ( I − t Ω 2 π i ) = det ( I − t Ω 1 2 π i ) ∧ ⋯ ∧ det ( I − t Ω m 2 π i ) {\textstyle \det(I-t{\frac {\Omega }{2\pi i}})=\det(I-t{\frac {\Omega _{1}}{2\pi i}})\wedge \dots \wedge \det(I-t{\frac {\Omega _{m}}{2\pi i}})} , we have:
where on the right the multiplication is that of a cohomology ring: cup product. For the normalization property, one computes the first Chern class of the complex projective line; see Chern class#Example: the complex tangent bundle of the Riemann sphere.
Since Ω E ⊗ E ′ = Ω E ⊗ I E ′ + I E ⊗ Ω E ′ {\displaystyle \Omega _{E\otimes E'}=\Omega _{E}\otimes I_{E'}+I_{E}\otimes \Omega _{E'}} ,4 we also have:
Finally, the Chern character of E is given by
where Ω {\displaystyle \Omega } is the curvature form of some connection on E (since Ω {\displaystyle \Omega } is nilpotent, it is a polynomial in Ω {\displaystyle \Omega } .) Then ch is a ring homomorphism:
Now suppose, in some ring R containing the cohomology ring H ∗ ( M , C ) {\displaystyle H^{*}(M,\mathbb {C} )} , there is the factorization of the polynomial in t:
where λ j {\displaystyle \lambda _{j}} are in R (they are sometimes called Chern roots.) Then ch ( E ) = e λ j {\displaystyle \operatorname {ch} (E)=e^{\lambda _{j}}} .
If E is a smooth real vector bundle on a manifold M, then the k-th Pontrjagin class of E is given as:
where we wrote E ⊗ C {\displaystyle E\otimes \mathbb {C} } for the complexification of E. Equivalently, it is the image under the Chern–Weil homomorphism of the invariant polynomial g 2 k {\displaystyle g_{2k}} on g l n ( R ) {\displaystyle {\mathfrak {gl}}_{n}(\mathbb {R} )} given by:
Let E be a holomorphic (complex-)vector bundle on a complex manifold M. The curvature form Ω {\displaystyle \Omega } of E, with respect to some hermitian metric, is not just a 2-form, but is in fact a (1, 1)-form (see holomorphic vector bundle#Hermitian metrics on a holomorphic vector bundle). Hence, the Chern–Weil homomorphism assumes the form: with G = GL n ( C ) {\displaystyle G=\operatorname {GL} _{n}(\mathbb {C} )} ,
Kobayashi & Nomizu 1969, Ch. XII. - Kobayashi, Shoshichi; Nomizu, Katsumi (1969), Foundations of Differential Geometry, vol. 2 (new ed.), Wiley-Interscience (published 2004), MR 0152974 https://mathscinet.ams.org/mathscinet-getitem?mr=0152974 ↩
The argument for the independent of a choice of connection here is taken from: Akhil Mathew, Notes on Kodaira vanishing "Archived copy" (PDF). Archived from the original (PDF) on 2014-12-17. Retrieved 2014-12-11.{{cite web}}: CS1 maint: archived copy as title (link). Kobayashi-Nomizu, the main reference, gives a more concrete argument. https://web.archive.org/web/20141217025038/https://math.berkeley.edu/~amathew/kodaira.pdf ↩
Editorial note: This definition is consistent with the reference except we have t, which is t −1 there. Our choice seems more standard and is consistent with our "Chern class" article. /wiki/Chern_class ↩
Proof: By definition, ∇ E ⊗ E ′ ( s ⊗ s ′ ) = ∇ E s ⊗ s ′ + s ⊗ ∇ E ′ s ′ {\displaystyle \nabla ^{E\otimes E'}(s\otimes s')=\nabla ^{E}s\otimes s'+s\otimes \nabla ^{E'}s'} . Now compute the square of ∇ E ⊗ E ′ {\displaystyle \nabla ^{E\otimes E'}} using Leibniz's rule. ↩