Proving Chern forms are closed via Bianchi identity

Question

Let $M$ be a smooth real manifold and $E \rightarrow M$ an Hermitean vector bundle over it. Define Chern classes as

$$c(M)=\sum_{i=0}^m c_i(E)t^i=\det \left( \textrm{Id} +\frac{i}{2\pi}F \right) = \prod_i (1 + \lambda_i t),$$

Letting $F \triangleq D^2$ denote the curvature two-form associated with the connection $D$ on $M$ – a ($\mathfrak{u}(E)$-valued two-form?), and $\lambda_i$ denote the eigenvalues of the normalization $i/2\pi F$. I have seen it asserted that one may use the Bianchi identity to show that the Chern forms are closed and hence represent cohomology classes, i.e.

$$ DF = 0 \implies dc_i(E) = 0 $$

How does one see this from the definition of the Chern classes as above?

Answer 1

In the book "Differential Geometry: Connections, Curvature and Characteristic Classes" by L. Tu (which is an excellent book), the following is proved.

• The ring of invariant polynomials on $\mathfrak{gl}(n,\mathbb{C})$ is generated as a ring by the trace polynomials $f_k(F)=\text{tr}(F^k)$, where $F\in\mathfrak{gl}(n,\mathbb{C})$. In particular, every polynomial on $\mathfrak{gl}(n,\mathbb{C})$ is a polynomial in the trace polynomials.
• $d\text{tr}(F)=\text{tr}(dF)$ for any matrix $F$ of $k$-forms

Now, the Chern class is defined via an invariant polynomial on $\mathfrak{gl}(n,\mathbb{C})$. Let $A$ be a connection $1$-form (or rather its local representative), and $F$ its curvature $2$-form. Then $$d\text{tr}(F^k)=\text{tr}(d(F^k))=\text{tr}(F^k\wedge A-A\wedge F^k)=\text{tr}(F^k\wedge A)-\text{tr}(A\wedge F^k)=0$$ Now, we have established the result for the curvature $2$-form on the trace polynomials, and therefore $dP(F)=0$ for any invariant polynomial on $\mathfrak{gl}(n,\mathbb{C})$. In particular, for the polynomial which defines the Chern class $c(E)$. The fact that $d(F^k)=F^k\wedge A-A\wedge F^k$ is, in the book, referred to as the "generalised second Bianchi identity", for obvious reasons. The ordinary second Bianchi identity is the case $k=1$.