The Chern–Weil homomorphism and invariant polynomials
Build closed global forms from curvature, with explicit normalizations and an honest account of lost torsion.
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The question for today
What you’ll understand
- Explain why invariant curvature polynomials descend and are closed.
- Use Chern, Pontryagin, and Euler curvature normalizations consistently.
- Separate connection-independent classes from connection-dependent forms.
Characteristic classes were defined topologically, but a connection can represent their real images by concrete differential forms. The Chern–Weil construction takes an invariant polynomial on the Lie algebra and evaluates it on curvature. Conjugation invariance makes the local expressions glue, while the Bianchi identity makes the resulting form closed. This is not a method for reconstructing every integral class: torsion disappears under the map to real cohomology. We work with smooth bundles and smooth connections over paracompact manifolds, and specify the representation and normalization before identifying a curvature form with a familiar characteristic class.
Invariant polynomials turn covariance into global forms
Let be a symmetric -linear form on the Lie algebra invariant under the adjoint action. Evaluating on curvature factors gives a degree- form. The principal curvature is horizontal and equivariant; applying makes it invariant as well, so it is basic and descends uniquely to the base. In local notation this says because curvature changes by conjugation. An arbitrary polynomial lacking this invariance need not produce an atlas-independent expression.
The Bianchi identity and infinitesimal invariance imply that the descended form is closed. Varying the connection along an affine path shows that its derivative is exact, so the de Rham class depends on the bundle and , rather than on the chosen connection. Multiplication of invariant polynomials becomes wedge product of cohomology classes, yielding the Chern–Weil homomorphism. The exact transgression formula will be derived next; here it justifies why a continuously changing curvature density can represent an unchanged class.
Fixing the normalizations
For a unitary connection with skew-Hermitian matrix potential and , use the total Chern form . Thus the first Chern form is . This equals the coefficient convention in ; forgetting the minus sign in that expression reverses the first class. For a real orthogonal connection the normalized first Pontryagin form is . The representation used for the trace is the defining real representation.
For an oriented real rank- bundle with an orthogonal connection, the Euler form is the Pfaffian of . Fix the standard convention that the Pfaffian of equals . For the oriented plane with complex structure , a potential has Euler form . Under the corresponding complex line convention this agrees with the first Chern form, since its complex curvature is .
Forms do not retain every topological obstruction
A compact oriented base allows integration of top-degree characteristic forms to obtain real characteristic numbers. When the polynomial is integrally normalized, these agree with the corresponding integer cohomological pairings. A local calculation alone does not establish integrality; that statement uses the topological normalization and the bundle's global gluing. An arbitrary real multiple of an invariant polynomial still defines a valid real Chern–Weil class, but its periods need not be integers.
A flat connection makes every positive-degree curvature polynomial vanish, so the corresponding real characteristic classes vanish. Integral torsion classes can nevertheless remain nonzero, and flat holonomy can retain further information. Likewise two different connections may have different local curvature densities while their Chern–Weil forms differ only by an exact form. On a closed base their matching top-degree integrals then agree by Stokes. With a boundary, an exact difference can produce a boundary integral, so closed-base and relative-boundary conclusions must be distinguished.
Work it through
A Chern number from two local potentials
On the unit sphere with area form , choose line potentials and . Compute the first Chern number in our convention.
Each potential is smooth on its own polar chart: the vanishing angular coefficient removes the apparent singularity at its included pole. On the overlap for .
Both yield . The sign agrees across charts because the connection transformation adds an exact local angular term.
The first Chern form is . Its integral is times the sphere area , hence three.
The line has first Chern number 3 for the specified sphere orientation, transition, and curvature convention. The nonzero integral forbids a global smooth potential for this connection on one trivial line.
The living notebook
Explore the model
Interactive lab · matrix
An invariant survives shear
Pause/Reset, set diagonal scale 1 and shear 2. Predict determinant 1. Reset and set both controls to scale 1 and shear −2: the determinant remains 1 while the sheared square changes direction. The selected matrix also has trace 2 in both cases. Readouts describe that final matrix during interpolation, and control changes retain time. This illustrates selective invariance; it does not evaluate an invariant polynomial on a curvature two-form.
- Area scale · det A
- 1.4
- Singular values · σmax, σmin
- 1.597, 0.876
- Condition number · κ₂(A)
- 1.823
The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA scalar invariant deliberately forgets some coordinate data, so its limits must be understood.
Pause & reason
Check your understanding
Your turn
Try it yourself
On oriented , let an orthogonal rank-two connection have curvature , where and . Compute its Euler and first Pontryagin forms and verify .
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Use J²=−I, trace I=2, and the fact that two two-forms commute under wedge product.
Compare with the solution
The Euler form is . Since , its trace is , so . Here , hence . These are forms on a contractible base and represent zero positive-degree de Rham classes; a nonzero local density does not imply a nonzero global cohomology class.
Keep the idea
Take it with you
- Invariant polynomials make curvature expressions global and closed.
- Normalization depends on the declared connection and representation conventions.
- Chern–Weil forms represent real classes and cannot detect integral torsion.
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