Differential Geometry – First Chern Class of E and det E

Question

Let $\pi:E\to M$ be a vector bundle, and $\nabla$ a connection. My definition of the first Chern class is $$c_1(E)=\left[tr\left(\frac{i}{2\pi}F^\nabla\right)\right],$$ where $F^\nabla$ is the curvature tensor. Let det $E$ be the top exterior power, det$(E)=\Lambda^n(E)$. Why does $c_1(E)=c_1(\text{det}(E))$? If $\widetilde\nabla$ is the induced connection on det $E$, then I believe $$F^{\widetilde\nabla}(X,Y)(s_1\wedge\dots\wedge s_n)=F^\nabla(X,Y)s_1\wedge\dots\wedge F^\nabla(X,Y)s_n.$$Does the answer follow from this? I was told that $F^{\widetilde\nabla}=\text{tr}\ (F^\nabla)$. If this is true, then I see why they have the same Chern class, but I don't see why that's true.

Answer 1

Here is the simplest way I know to see this. Consider the case where $E = L_1\oplus\dots\oplus L_n$. Then with the direct sum connection on $E$, the curvature $2$-form $F^\nabla$ will be the matrix with the curvature $2$-forms $F^{\nabla_1},\dots,F^{\nabla_n}$ of the respective line bundles on the diagonal. So $\text{tr}\,F^{\nabla} = \sum\limits_{j=1}^n F^{\nabla_j}$. On the other hand, $\Lambda^n(E) \cong L_1\otimes\dots\otimes L_n$. But with the induced connection on the tensor product, the curvature forms of the tensor product line bundle is the sum of the curvature forms. So we get the same result.

An alternative approach (see, for example, pp. 411-413 of Griffiths and Harris) is to think of the Chern class $c_j(E)$ as the Poincaré dual of the locus where $n-j+1$ generic sections of $E$ become linearly dependent. So $c_1(E)$ is the Poincaré dual of the locus where $n$ generic sections $s_1,\dots,s_n$ become linearly dependent. This is the set where $s_1\wedge\dots\wedge s_n = 0$, so it's the zero-locus of a generic section of $\Lambda^n(E)$, which is what $c_1(\Lambda^n(E))$ represents.