Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 22 / 2

A connection on a vector bundle is a $\mathbb{C}$-linear map $D:A^0(E)\to A^1(E)$ satisfying $D(fs)=df\otimes s+fDs$ for smooth functions. Its extension is the covariant exterior derivativeSquaring gives a $C^{\infty }$-linear operator $D^{2}s=\Theta s$, defining the vector-bundle curvature $\Theta \in A^2(\mathrm{End}E)$.

For a line bundle, $\mathrm{End}E$ is canonically the trivial line bundle, so this vector-bundle curvature is an ordinary global complex two-form. In a local frame with $D=d+A$, the scalar identity $A\wedge A=0$ gives $\Theta =dA$. Therefore $d\Theta =0$. The calculations in the two subparts also explain the gluing directly.

Use a row of local frame sections $e=(e_1,\dots ,e_m)$ and columns of coefficients: $D(eu)=e(du+Au)$. If $e^{\prime }=eg$ for a smooth invertible matrix-valued function $g$, thenHence the change of frame of a vector-bundle connection isThis convention specifies both the placement of the frame matrix and the sign of the inhomogeneous term. The vector-bundle curvature transforms by ${\Theta }^{\prime }=g^{-1}\Theta g$, either by applying $D^2$ to the new frame or by substituting in the formula of the next subpart.

For a section represented by a column $u$, expand the covariant exterior derivative using the graded Leibniz rule:Therefore the Cartan curvature matrix equation isThe product includes endomorphism composition; it need not vanish for a matrix-valued one-form. For a scalar frame change $g$, $A^{\prime }=A+g^{-1}dg$ and $d(g^{-1}dg)=0$, confirming that the line-bundle two-forms $dA$ patch globally.

The smooth exponential sequence isLocal logarithms make it exact as a sequence of sheaves. A line bundle has a transition class in $H^1(M,{\mathcal{C}}^{\infty }({\mathbb{C}}^{\times }))$; its image under the connecting map defines the First Chern class. For a holomorphic line bundle the holomorphic exponential sequence gives the same class by naturality.

Here is the explicit Čech-de Rham curvature descent. On a good cover, write $e_j=e_i{g}_{ij}$ and choose logarithms $g_{ij}=\exp (2\pi i{f}_{ij})$. The integer cocyclerepresents $c_1(E)$. By the proved frame formula,Put $B_i=-A_i/(2\pi i)$ and $F=i\Theta /(2\pi )=d{B}_i$. Then $\delta B=-df$ and $\delta f=c$. In the Čech-de Rham double complex, with total differential $D_{\mathrm{tot}}=\delta +(-1{)}^{p}d$ on Čech degree $p$,Thus the global closed differential form and the integer cocycle represent the same class under de Rham theorem:This proves integrality, including the sign and normalization. It identifies the image of the integral class; vector-bundle curvature alone cannot recover torsion classes lost in passage to complex coefficients.

Two connections on the same line bundle differ by a global scalar one-form $a$. Their vector-bundle curvatures satisfy ${\Theta }_1-\Theta =da$, so the normalized representatives differ by an exact differential form. Hence the class is independent of the connection.

Use the integral normalization of the Fubini-Study form. On the chart $Z_j\ne 0$, define the integrally normalized Fubini-Study form byThe potentials on overlaps differ by the logarithm of the squared modulus of a nowhere-zero holomorphic function, whose $\partial \bar{\partial }$ is zero. Thus the forms patch and are closed. The dual of the tautological metric on $\mathcal{O}(1)$ has local squared norm $h=(1+\sum |z_k{|}^2{)}^{-1}$. Its Chern connection has $A=\partial \log h$ and vector-bundle curvature $\Theta =\bar{\partial }\partial \log h=\partial \bar{\partial }\log (1+\sum |z_k{|}^2)$. Therefore $[{\omega }_{\mathrm{int}}]=c_1(\mathcal{O}(1))$, proving the requested integrality. With the unnormalized convention ${\omega }_{FS}=i\partial \bar{\partial }\log (1+\sum |z_k{|}^2)$, this is ${\omega }_{\mathrm{int}}={\omega }_{FS}/(2\pi )$.