Chern–Weil homomorphism explained
In mathematics, the Chern–Weil homomorphism is a basic construction in 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 and let
denote the algebra of
-valued
polynomials on
(exactly the same argument works if we used
instead of Let
be the
subalgebra of fixed points in
under the
adjoint action of
G; that is, the subalgebra consisting of all polynomials
f such that
f(\operatorname{Ad}gx)=f(x)
, for all
g in
G and
x in
,
Given a principal G-bundle P on M, there is an associated homomorphism of
-algebras,
\Complex[akg]G\toH*(M;\Complex)
,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,
, is isomorphic to the algebra
of invariant polynomials:
H*(BG;\Complex)\cong\Complex[akg]G.
(The cohomology ring of
BG can still be given in the de Rham sense:
Hk(BG;\Complex)=\varinjlim\operatorname{ker}(d\colon
\to\Omegak+1(BjG))/\operatorname{im}d.
when
and
are manifolds.)
Definition of the homomorphism
Choose any connection form ω in P, and let Ω be the associated curvature form; i.e., the exterior covariant derivative of ω. If
is a homogeneous polynomial function of degree
k; i.e.,
for any complex number
a and
x in then, viewing
f as a symmetric multilinear functional on
(see the
ring of polynomial functions), let
be the (scalar-valued) 2
k-form on
P given by
f(\Omega)(v1,...,v2k)=
\epsilon\sigmaf(\Omega(v\sigma(1),v\sigma(2)),...,\Omega(v\sigma(2k-1),v\sigma(2k)))
where
vi are tangent vectors to
P,
is the sign of the permutation
in the symmetric group on 2
k numbers
(see Lie algebra-valued forms#Operations as well as
Pfaffian).
If, moreover, f is invariant; i.e.,
f(\operatorname{Ad}gx)=f(x)
, then one can show that
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
. First, that
is a closed form follows from the next two lemmas:
Lemma 1: The form
on
P descends to a (unique) form
on
M; i.e., there is a form on
M that pulls-back to
.
Lemma 2: If a form of
on
P descends to a form on
M, then
.
Indeed, Bianchi's second identity says
and, since
D is a graded derivation,
Finally, Lemma 1 says
satisfies the hypothesis of Lemma 2.
To see Lemma 2, let
be the projection and
h be the projection of
onto the horizontal subspace. Then Lemma 2 is a consequence of the fact that
(the kernel of
is precisely the vertical subspace.) As for Lemma 1, first note
f(\Omega)(dRg(v1),...,dRg(v2k))=f(\Omega)(v1,...,v2k),Rg(u)=ug;
which is because
and
f is invariant. Thus, one can define
by the formula:
\overline{f}(\Omega)(\overline{v1},...,\overline{v2k
}) = f(\Omega)(v_1, \dots, v_),where
are any lifts of
:
.
Next, we show that the de Rham cohomology class of
on
M is independent of a choice of connection.
[1] Let
be arbitrary connection forms on
P and let
be the projection. Put
\omega'=tp*\omega1+(1-t)p*\omega0
where
t is a smooth function on
given by
. Let
be the curvature forms of
. Let
is:M\toM x R,x\mapsto(x,s)
be the inclusions. Then
is homotopic to
. Thus,
and
belong to the same de Rham cohomology class by the homotopy invariance of de Rham cohomology. Finally, by naturality and by uniqueness of descending,
\overline{f}(\Omega')=\overline{f}(\Omega0)
and the same for
. Hence,
\overline{f}(\Omega0),\overline{f}(\Omega1)
belong to the same cohomology class.
The construction thus gives the linear map: (cf. Lemma 1)
\toH2k(M;\Complex),f\mapsto\left[\overline{f}(\Omega)\right].
In fact, one can check that the map thus obtained:
\Complex[akg]G\toH*(M;\Complex)
is an algebra homomorphism.
Example: Chern classes and Chern character
Let
G=\operatorname{GL}n(\Complex)
and
its Lie algebra. For each
x in
, we can consider its
characteristic polynomial in
t:
[2] \det\left(I-t{x\over2\pii}\right)=
fk(x)tk,
where
i is the square root of -1. Then
are invariant polynomials on
, 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
under the Chern–Weil homomorphism defined by
E (or more precisely the frame bundle of
E). If
t = 1, then
\det\left(I-{x\over2\pii}\right)=1+f1(x)+ … +fn(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
and
c given above satisfy the axioms of Chern classes. For example, for the Whitney sum formula, we consider
ct(E)=[\det\left(I-t{\Omega/2\pii}\right)],
where we wrote
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
. Now, suppose
E is a direct sum of vector bundles
's and
the curvature form of
so that, in the matrix term,
is the block diagonal matrix with Ω
I's on the diagonal. Then, since 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 .
Since
\OmegaE=\OmegaE ⊗ IE'+IE ⊗ \OmegaE'
,
[3] we also have:
c1(E ⊗ E')=c1(E)\operatorname{rank}(E')+\operatorname{rank}(E)c1(E').
Finally, the Chern character of E is given by
\operatorname{ch}(E)=[\operatorname{tr}(e-\Omega/2\pi)]\inH*(M,\Q)
where
is the curvature form of some connection on
E (since
is nilpotent, it is a polynomial in
.) Then ch is a
ring homomorphism:
\operatorname{ch}(E ⊕ F)=\operatorname{ch}(E)+\operatorname{ch}(F),\operatorname{ch}(E ⊗ F)=\operatorname{ch}(E)\operatorname{ch}(F).
Now suppose, in some ring
R containing the cohomology ring
, there is the factorization of the polynomial in
t:
where
are in
R (they are sometimes called Chern roots.) Then
.
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:
pk(E)=(-1)kc2k(E ⊗ \Complex)\inH4k(M;\Z)
where we wrote
for the
complexification of
E. Equivalently, it is the image under the Chern–Weil homomorphism of the invariant polynomial
on
given by:
\operatorname{det}\left(I-t{x\over2\pi}\right)=
gk(x)tk.
The homomorphism for holomorphic vector bundles
Let E be a holomorphic (complex-)vector bundle on a complex manifold M. The curvature form
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=\operatorname{GL}n(\Complex)
,
\Complex[ak{g}]k\toHk,(M;\Complex),f\mapsto[f(\Omega)].
References
- .
- .
- , . (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.)
- .
- .
- .
- .
Further reading
- Chern-Weil forms and abstract homotopy theory. 1301.5959. 2013. Daniel S.. Freed. Dan Freed. Michael J.. Hopkins. Michael J. Hopkins. Bulletin of the American Mathematical Society. (N.S.) . 50 . 3. 431–468. 10.1090/S0273-0979-2013-01415-0. 3049871. 51755613.
Notes and References
- The argument for the independent of a choice of connection here is taken from: Akhil Mathew, Notes on Kodaira vanishing Web site: Archived copy . 2014-12-11 . dead . https://web.archive.org/web/20141217025038/https://math.berkeley.edu/~amathew/kodaira.pdf . 2014-12-17 . . 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,
\nablaE(s ⊗ s')=\nablaEs ⊗ s'+s ⊗ \nablaE's'
. Now compute the square of
using Leibniz's rule.