Chern–Weil homomorphism

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

Given a principal G-bundle P on M, there is an associated homomorphism of ${\displaystyle \mathbb{C}}$-algebras,

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

called the Chern–Weil homomorphism, where on the right cohomology is de Rham cohomology. This homomorphism is obtained by taking invariant polynomials in the curvature of any connection on the given bundle. If G is either compact or semi-simple, then the cohomology ring of the classifying space for G-bundles, ${\displaystyle BG}$, is isomorphic to the algebra ${\displaystyle \mathbb{C}[\mathfrak{g}{]}^G}$ of invariant polynomials:

${\displaystyle H^{\ast }(BG;\mathbb{C})\cong \mathbb{C}[\mathfrak{g}{]}^G.}$

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

${\displaystyle H^k(BG;\mathbb{C})=\underrightarrow{\lim }\ker (d:{\mathrm{\Omega }}^k(B_{j}G)\to {\mathrm{\Omega }}^{k+1}(B_{j}G))/\mathrm{im}d.}$

when ${\displaystyle BG=\underrightarrow{\lim }{B}_{j}G}$ and ${\displaystyle B_{j}G}$ are manifolds.)

Definition of the homomorphism

Choose any connection form ω in P, and let Ω be the associated curvature form; i.e., ${\displaystyle \mathrm{\Omega }=D\omega }$, the exterior covariant derivative of ω. If ${\displaystyle f\in \mathbb{C}[\mathfrak{g}{]}^G}$ is a homogeneous polynomial function of degree k; i.e., ${\displaystyle f(ax)=a^{k}f(x)}$ for any complex number a and x in ${\displaystyle \mathfrak{g}}$, then, viewing f as a symmetric multilinear functional on $\prod _1^k\mathfrak{g}$ (see the ring of polynomial functions), let ${\displaystyle f(\mathrm{\Omega })}$ be the (scalar-valued) 2k-form on P given by

${\displaystyle f(\mathrm{\Omega })(v_1,\dots ,v_{2k})=\frac{1}{(2k)!}\sum _{\sigma \in {\mathfrak{S}}_{2k}}{ϵ}_{\sigma }f(\mathrm{\Omega }(v_{\sigma (1)},v_{\sigma (2)}),\dots ,\mathrm{\Omega }(v_{\sigma (2k-1)},v_{\sigma (2k)}))}$

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

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

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

Lemma 2: If a form of ${\displaystyle \phi }$ on P descends to a form on M, then ${\displaystyle d\phi =D\phi }$.

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

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

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

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

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

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

Next, we show that the de Rham cohomology class of ${\displaystyle \overline{f}(\mathrm{\Omega })}$ on M is independent of a choice of connection. Let ${\displaystyle {\omega }_0,{\omega }_1}$ be arbitrary connection forms on P and let ${\displaystyle p:P\times \mathbb{R}\to P}$ be the projection. Put

${\displaystyle {\omega }^{\prime }=t{p}^{\ast }{\omega }_1+(1-t)p^{\ast }{\omega }_0}$

where t is a smooth function on ${\displaystyle P\times \mathbb{R}}$ given by ${\displaystyle (x,s)\mapsto s}$. Let ${\displaystyle {\mathrm{\Omega }}^{\prime },{\mathrm{\Omega }}_0,{\mathrm{\Omega }}_1}$ be the curvature forms of ${\displaystyle {\omega }^{\prime },{\omega }_0,{\omega }_1}$. Let ${\displaystyle i_s:M\to M\times \mathbb{R},x\mapsto (x,s)}$ be the inclusions. Then ${\displaystyle i_0}$ is homotopic to ${\displaystyle i_1}$. Thus, ${\displaystyle i_0^{\ast }\overline{f}({\mathrm{\Omega }}^{\prime })}$ and ${\displaystyle i_1^{\ast }\overline{f}({\mathrm{\Omega }}^{\prime })}$ belong to the same de Rham cohomology class by the homotopy invariance of de Rham cohomology. Finally, by naturality and by uniqueness of descending,

${\displaystyle i_0^{\ast }\overline{f}({\mathrm{\Omega }}^{\prime })=\overline{f}({\mathrm{\Omega }}_0)}$

and the same for ${\displaystyle {\mathrm{\Omega }}_1}$. Hence, ${\displaystyle \overline{f}({\mathrm{\Omega }}_0),\overline{f}({\mathrm{\Omega }}_1)}$ belong to the same cohomology class.

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

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

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

${\displaystyle \mathbb{C}[\mathfrak{g}{]}^G\to H^{\ast }(M;\mathbb{C})}$

is an algebra homomorphism.