Let $M$ be a complex manifold, $R^{\nabla}$ be the curvature operator for connections $\nabla$. Consider a polynomial function $f:\operatorname M_n(\mathbb{C})\to\mathbb{C}$. For the gauge group $\operatorname{GL}_n(\mathbb{C})$, if $f(A)=f(gAg^{-1})$ where $A \in \operatorname M_n(\mathbb{C})$ and $g \in \operatorname{GL}_n(\mathbb{C})$, then $f$ is said to be an invariant polynomial function. Let $I^k(\operatorname M_n(\mathbb{C}))$ be the set of all such polynomials of degree $k$. Also, $\bigoplus_{k\geq0}I^{k}(\operatorname M_n(\mathbb{C}))=I(\operatorname M_n(\mathbb{C}))$.

We shall need $\phi_n(A)=det(A)$, $n>1$, and $\phi_1(A)=Tr(A)$.

Define a global differential form ($2k$-form) $f(R^{\nabla}) \in \mathbb{A}^{2k}(M)$. If we have the de Rham cohomology group $H^{2k}(M)$, then the Weil homomorphism is defined as the map $\omega:I(\operatorname M_n(\mathbb{C}))\to \bigoplus_{k\geq 0}H^{2k}(M)$.

The Chern forms $c_{i}(R^{\nabla})=\phi_{i}(\frac{\sqrt{-1}}{2\pi}R^{\nabla})$.

For the complex vector bundle $(\mathbb{E},\pi,M)$, where $\mathbb{E}$ is the total space, the Chern classes are defined as $c_{i}(\mathbb{E}) \in H^{2k}(M)$.

Therefore $c_{i}(\mathbb{E})\mathrel{:=}\omega(c_{i}(R^{\nabla}))=[c_{i}(R^{\nabla})]$ (de Rham cohomology class).

The Chern forms $c_{i}(R^{\nabla}) \in \mathbb{A}^{2i}(\mathbb{E})$.

$\mathbb{A}^{2i}(\mathbb{E})$ is sheaf of smooth $\mathbb{E}$-valued $2i$ forms on $M$.

Cohomology groups are very important in geometry for understanding the invariants that can be defined on manifolds. That is, the transformations that keep some special properties of the manifold and which analogous to the gauge transformations in physics.

Chern classes are special type of cohomology classes. If the first Chern class vanishes for a particular manifold, then it must be a Ricci-flat manifold. For example the Calabi–Yau manifolds (they have lots of other special properties, e.g., trivial canonical bundle, etc.).

But what do the higher Chern classes mean? What uses are those higher cohomology classes corresponding to the higher Chern classes of?

  • $\begingroup$ One answer is that if $\mathbb E$ is a trivial vector bundle, then all of its Chern classes vanish. Therefore higher Chern classes are obstructions to a bundle being trivial. $\endgroup$ Apr 20 '19 at 13:21

This is a big topic, which should be covered in the union of many standard texts (Chern, Griffiths-Harris, Milnor-Stasheff...). I'll list a few answers off the top of my head.

  1. Suppose that $L$ is the tautological line bundle on the complex projective space plane, then it's clearly not trivial, and neither is $V= L\oplus L^{-1}$. But $c_1(V)=0$, because the trace of curvature is zero for an induced connection. Of course, $V$ is certainly not trivial. So it's natural to look for higher cohomological obstructions (as Arun suggested), e.g. $c_2(V) = -c_1(L)^2\not=0$ would work.
  2. For universal bundles on Grassmanians, Chern classes have natural geometric interpretations involving Schubert cycles.
  3. Chern classes come up in formulas expressing answers to natural geometric questions: Gauss-Bonnet, Riemann-Roch, or more general index theorems.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.