Chern–Weil homomorphism

In mathematics, the Chern–Weil homomorphism is a basic construction in the Chern–Weil theory that computes topological invariants of vector bundles and principal bundles on a smooth manifold M in terms of connections and curvature representing classes in the de Rham cohomology rings of M. That is, the theory forms a bridge between the areas of algebraic topology and differential geometry. It was developed in the late 1940s by Shiing-Shen Chern and André Weil, in the wake of proofs of the generalized Gauss–Bonnet theorem. This theory was an important step in the theory of characteristic classes.

Let G be a real or complex Lie group with Lie algebra $\mathfrak g$; and let $\mathbb{C}[\mathfrak g]$ denote the algebra of $\mathbb{C}$-valued polynomials on $\mathfrak g$ (exactly the same argument works if we used $\mathbb{R}$ instead of $\mathbb{C}$.) Let $\mathbb{C}[\mathfrak g]^G$ be the subalgebra of fixed points in $\mathbb C[\mathfrak g]$ under the adjoint action of G; that is, it consists of all polynomials f such that for any g in G and x in $\mathfrak{g}$, $f(\operatorname{Ad}_g x) = f(x).$

The Chern–Weil homomorphism is a homomorphism of $\mathbb C$-algebras

$\mathbb C[\mathfrak g]^{G} \to H^*(M,\mathbb C)$

where on the right cohomology is de Rham cohomology. Such a homomorphism exists and is uniquely defined for every principal G-bundle P on M. If G is compact, then under the homomorphism, the cohomology ring of the classifying space for G-bundles BG is isomorphic to the algebra $\mathbb C[\mathfrak g]^{G}$ of invariant polynomials:

$H^*(BG, \mathbb{C}) \cong \mathbb C[\mathfrak g]^{G}.$

(The cohomology ring of BG can still be given in the de Rham sense:

$H^k(BG, \mathbb{C}) = \varinjlim \operatorname{ker} (d: \Omega^k(B_jG) \to \Omega^{k+1}(B_jG))/\operatorname{im} d.$

when $BG = \varinjlim B_jG$ and $B_jG$ are manifolds.) For non-compact groups like SL(n,R), there may be cohomology classes that are not represented by invariant polynomials.

Definition of the homomorphism

Choose any connection form ω in P, and let Ω be the associated curvature 2-form; i.e., Ω = Dω, the exterior covariant derivative of ω. If $f\in\mathbb C[\mathfrak g]^G$ is a homogeneous polynomial function of degree k; i.e., $f(a x) = a^k f(x)$ for any complex number a and x in $\mathfrak g$, then, viewing f as a symmetric multilinear functional on $\prod_1^k \mathfrak{g}$ (see the ring of polynomial functions), let

$f(\Omega)$

be the (scalar-valued) 2k-form on P given by

$f(\Omega)(v_1,\dots,v_{2k})=\frac{1}{(2k)!}\sum_{\sigma\in\mathfrak S_{2k}}\epsilon_\sigma f(\Omega(v_{\sigma(1)},v_{\sigma(2)}),\dots,\Omega(v_{\sigma(2k-1)}, v_{\sigma(2k)}))$

where v_i are tangent vectors to P, $\epsilon_\sigma$ is the sign of the permutation $\sigma$ in the symmetric group on 2k numbers $\mathfrak S_{2k}$ (see Lie algebra-valued forms#Operations as well as Pfaffian).

If, moreover, f is invariant; i.e., $f(\operatorname{Ad}_g x) = f(x)$, then one can show that $f(\Omega)$ 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. First, that $f(\Omega)$ is a closed form follows from the next two lemmas:

Lemma 1: The form $f(\Omega)$ on P descends to a (unique) form $\overline{f}(\Omega)$ on M; i.e., there is a form on M that pulls-back to $f(\Omega)$.

Lemma 2: If a form φ on P descends to a form on M, then dφ = Dφ.

Indeed, Bianchi's second identity says $D \Omega = 0$ and, since D is a graded derivation, $D f(\Omega) = 0.$ Finally, Lemma 1 says $f(\Omega)$ satisfies the hypothesis of Lemma 2.

To see Lemma 2, let $\pi: P \to M$ be the projection and h be the projection of $T_u P$ onto the horizontal subspace. Then Lemma 2 is a consequence of the fact that $d \pi(h v) = d \pi(v)$ (the kernel of $d \pi$ is precisely the vertical subspace.) As for Lemma 1, first note

$f(\Omega)(d R_g(v_1), \dots, d R_g(v_{2k})) = f(\Omega)(v_1, \dots, v_{2k}), \, R_g(u) = ug;$

which is because $R_g^* \Omega = \operatorname{Ad}_{g^{-1}} \Omega$ and f is invariant. Thus, one can define $\overline{f}(\Omega)$ by the formula:

$\overline{f}(\Omega)(\overline{v_1}, \dots, \overline{v_{2k}}) = f(\Omega)(v_1, \dots, v_{2k})$

where $v_i$ are any lifts of $\overline{v_i}$: $d \pi(v_i) = \overline{v}_i$.

Next, we show that the de Rham cohomology class of $\overline{f}(\Omega)$ on M is independent of a choice of connection. Let $\omega_0, \omega_1$ be arbitrary connection forms on P and let $p: P \times \mathbb{R} \to P$ be the projection. Put

$\omega' = t \, p^* \omega_1 + (1 - t) \, p^* \omega_0$

where t is a smooth function on $P \times \mathbb{R}$ given by $(x, s) \mapsto s$. Let $\Omega', \Omega_0, \Omega_1$ be the curvature forms of $\omega', \omega_0, \omega_1$. Let $i_s: M \to M \times \mathbb{R}, \, x \mapsto (x, s)$ be the inclusions. Then $i_0$ is homotopic to $i_1$. Thus, $i_0^* \overline{f}(\Omega')$ and $i_1^* \overline{f}(\Omega')$ belong to the same de Rham cohomology class by the homotopy invariance of de Rham cohomology. Finally, by naturality and by uniqueness of descending,

$i_0^* \overline{f}(\Omega') = \overline{f}(\Omega_0)$

and the same for $\Omega_1$. Hence, $\overline{f}(\Omega_0), \overline{f}(\Omega_1)$ belong to the same cohomology class.

The construction thus gives the linear map: (cf. Lemma 1)

$\mathbb C[\mathfrak g]^{G}_k \rightarrow H^{2k}(M,\mathbb C), \, f \mapsto \left[\overline{f}(\Omega)\right].$

In fact, one can check that the map thus obtained:

$\mathbb C[\mathfrak g]^{G} \rightarrow H^*(M,\mathbb C)$

is an algebra homomorphism.

Example: Chern classes and Chern character

Let $G = GL_n(\mathbb{C})$ and $\mathfrak{g} = \mathfrak{gl}_n(\mathbb{C})$ its Lie algebra. For each x in $\mathfrak{g}$, we can consider its characteristic polynomial in t:

$\det \left( I - t{x \over 2 \pi i} \right) = \sum_{k=0}^n f_k(x) t^k,$

where i is the square root of -1. Then $f_k$ are invariant polynomials on $\mathfrak{g}$, 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:

$c_k(E) \in H^{2k}(M, \mathbb{Z})$

is given as the image of f_k under the Chern–Weil homomorphism defined by E (or more precisely the frame bundle of E). If t = 1, then $\det \left(I - {x \over 2 \pi i} \right) = 1 + f_1(x) + \cdots + f_n(x)$ is an invariant polynomial. The total Chern class of E is the image of this polynomial; that is,

$c(E) = 1 + c_1(E) + \cdots + c_n(E).$

Directly from the definition, one can show c_j, c given above satisfy the axioms of Chern classes. For example, for the Whitney sum formula, we consider

$c_t(E) = [\det \left( I - t {\Omega / 2 \pi i} \right)]$

where we wrote Ω 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. Now, suppose E is a direct sum of vector bundles E_i's and Ω_i the curvature form of E_i so that, in the matrix term, Ω is the block diagonal matrix with Ω_I's on the diagonal. Then, since $\det(I - t\Omega/2\pi i) = \det(I - t\Omega_1/2\pi i) \wedge \dots \wedge \det(I - t\Omega_m/2\pi i)$, we have:

$c_t(E) = c_t(E_1) \cdots c_t(E_m)$

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 $\Omega_{E \otimes E'} = \Omega_E \otimes I_{E'} + I_{E} \otimes \Omega_{E'}$, we also have:

$c_1(E \otimes E') = c_1(E) \operatorname{rk}(E') + \operatorname{rk}(E) c_1(E').$

Finally, the Chern character of E is given by

$\operatorname{ch}(E) = [\operatorname{tr}(e^{-\Omega/2\pi i})] \in H^*(M, \mathbb{Q})$

where Ω is the curvature form of some connection on E (since Ω is nilpotent, it is a polynomial in Ω.) Then ch is a ring homomorphism:

$\operatorname{ch}(E \oplus F) = \operatorname{ch}(E) + \operatorname{ch}(F), \, \operatorname{ch}(E \otimes F) = \operatorname{ch}(E) \operatorname{ch}(F).$

Now suppose, in some ring R containing the cohomology ring H(M, C), there is the factorization of the polynomial in t:

$c_t(E) = \prod_{j=0}^n (1 + \lambda_j t)$

where λ_j are in R (they are sometimes called Chern roots.) Then $\operatorname{ch}(E) = e^{\lambda_j}$.

Example: Pontrjagin classes

If E is a smooth real vector bundle on a manifold M, then the k-th Pontrjagin class of E is given as:

$p_k(E) = (-1)^k c_{2k}(E \otimes \mathbb{C}) \in H^{4k}(M, \mathbb{Z})$

where we wrote $E \otimes \mathbb{C}$ for the complexification of E. Equivalently, it is the image under the Chern–Weil homomorphism of the invariant polynomial $g_{2k}$ on $\mathfrak{gl}_n(\mathbb{R})$ given by:

$\operatorname{det}\left(I - t {x \over 2 \pi}\right) = \sum_{k = 0}^n g_k(x) t^k.$

The homomorphism for holomorphic vector bundles

Let E be a holomorphic (complex-)vector bundle on a complex manifold M. The curvature form Ω 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 $G = GL_n(\mathbb{C})$,

$\mathbb{C}[\mathfrak{g}]_k \to H^{k, k}(M, \mathbb{C}), f \mapsto [f(\Omega)].$

References

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• Shiing-Shen Chern, Complex Manifolds Without Potential Theory (Springer-Verlag Press, 1995) ISBN 0-387-90422-0, ISBN 3-540-90422-0.

  The appendix of this book: "Geometry of Characteristic Classes" is a very neat and profound introduction to the development of the ideas of characteristic classes.

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Further reading

• Kobayashi-Nomizu 1969, Ch. XII.
• The argument for the independent of a choice of connection here is taken from: Akhil Mathew, Notes on Kodaira vanishing [1]. Kobayashi-Nomizu, the main reference, gives a more concrete argument.
• 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.
• Proof: By definition, $\nabla^{E \otimes E'}(s \otimes s') = \nabla^{E} s \otimes s' + s \otimes\nabla^{E'} s'$. Now compute the square of $\nabla^{E \otimes E'}$ using Leibniz's rule.