Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 22 / 2

Use the convention $\nabla =d+A$ with anti-Hermitian connection one-forms. For a unitary complex line bundle, $A$ is locally an $i\mathbb{R}$-valued one-form and $F_{\nabla }=dA$, since the Lie algebra is abelian. Under a change of unitary frame, the vector-bundle curvature transforms by conjugation, which is trivial for $U(1)$. Hence $F_{\nabla }$ is a globally defined imaginary-valued two-form, and $i{F}_{\nabla }$ is an ordinary real two-form. Equivalently, $\mathrm{End}(L)$ has its canonical scalar trivialization, so no nontrivial coefficient bundle remains.

Locally $d{F}_{\nabla }=d^{2}A=0$, and these equalities patch globally. If another unitary connection is ${\nabla }^{\prime }=\nabla +i\beta$ for a global real one-form $\beta$, thenThus the real vector-bundle curvature is closed and its de Rham cohomology class is independent of the unitary connection.

With the First Chern class normalization $c_1(L{)}_{\mathbb{R}}=[i{F}_{\nabla }/(2\pi )]$, the transverse section also identifies this class as $\mathrm{PD}[\Sigma ]$. Indeed, the complex orientation of each normal disk makes the section wind positively once around a transverse zero. The Euler class of the underlying oriented plane bundle counts these zeros with their induced orientation, and equals $c_1(L)$. This records the role of the section and orientation hypotheses; connection independence itself did not require choosing a section.

Start with any unitary connection ${\nabla }_0$. The Hodge decomposition theorem gives a unique real harmonic differential form $\eta$ representing $[i{F}_{{\nabla }_0}]$. Since they have the same de Rham cohomology class, $\eta -i{F}_{{\nabla }_0}=d\beta$ for a smooth real one-form $\beta$. Set $\nabla ={\nabla }_0-i\beta$. The vector-bundle curvature difference formula givesThis proves existence of a harmonic-curvature unitary line connection for any metric and any complex line bundle.

Two such unitary connections have the same vector-bundle curvature because harmonic representatives are unique. Their difference is therefore $i\gamma$ with $d\gamma =0$. When $b_1(X)=0$, write $\gamma =d\phi$ globally. The bundle gauge transformation $u=e^{i\phi }$ changes $\nabla$ to $u^{-1}\nabla u=\nabla +id\phi$. Thus the harmonic-curvature connection is unique up to gauge. On a general base, closed forms modulo the integral-period forms coming from circle-valued gauge transformations leave the torus $H^1(X;\mathbb{R})/(2\pi H^1(X;\mathbb{Z}))$ of possible gauge classes; the hypothesis removes this freedom.

The reverse-oriented Complex projective plane has intersection form $(-1)$, so $b^{+}=0$, while $b_1=0$. The Hodge star splits harmonic two-forms into self-dual and anti-self-dual subspaces, and their dimensions are the positive and negative indices of the intersection form. Therefore every harmonic two-form is an anti-self-dual two-form for every chosen metric.

By part (b), every complex line bundle has a unitary connection with harmonic real vector-bundle curvature, so this connection is an ASD connection. Conversely, an ASD connection on a complex line bundle has $dF=0$ and $d\ast F=-dF=0$, so its vector-bundle curvature is harmonic. The uniqueness in part (b) therefore applies to all ASD connections on that fixed complex line bundle. Each complex line bundle has exactly one gauge orbit of ASD connections.

Let $a,b\in H^2(S^2\times S^2;\mathbb{Z})$ be the two factor classes, normalized by $a^2=b^2=0$ and $⟨ab,[X]⟩=1$. Choose the complex line bundle $L$ with $c_1(L)=a+b$, for example the tensor product of pullbacks of degree-one complex line bundles from the two factors. ThenIf $L$ had an ASD connection, the real closed form $\alpha =iF/(2\pi )$ would represent this class and satisfy $\ast \alpha =-\alpha$. Hencea contradiction. This complex line bundle admits no ASD connection for any metric of the given product orientation. For the reverse product orientation, take $c_1(L)=a-b$ instead, which then has square $+2$. This is the positive-square obstruction to ASD line connections.

The suggested rank-two construction gives the same obstruction. The split bundle $E=L\oplus L^{\ast }$ has trivial determinant andAn ASD connection on $L$ would induce an ASD connection on $E$, but its instanton number would be $\Vert F_E{\Vert }_2^2/(8{\pi }^2)\ge 0$, contradicting $c_2(E)[X]=-2$.