# Harmonic connections on line bundles

Let $M$ be a manifold equipped with a complex line bundle $L$, and $\lambda$ a unitary connection on $L$. Let $P$ be the associated $U_1$ principal bundle. Since $U_1$ is abelian, the bundle $\mathrm{Ad}P := P\times_{\mathrm Ad}U_1$ is trivial. Thus $\mathcal{G}_L:=\Gamma(\mathrm{Ad} P)=\mathrm{Map}(M,U_1)$. Similarly, $\mathcal{A}_L=\lambda+\Omega^1(M;\mathrm{ad}P)=\lambda+\Omega^1(M;i\mathbb{R})$.

**Definition:** A connection $\lambda$ on a line bundle is said to be *harmonic* if its curvature is harmonic: that is, if it satisfies the Yang-Mills equation $d^* F_\lambda=0$.

Note that $d_A=d$ and $d_A^*=d^*$ on $\Omega^\bullet(M;\mathrm{ad}P)\cong i\Omega^\bullet(M)$. If $M$ is a surface, then being harmonic is equivalent to constant $*\!F_\lambda$.

**Theorem:** For any complex line bundle $L$, there is a connection $\lambda_0$ which is harmonic and has zero periods. Such a connection is unique up to gauge transformation.

**Proof:** An arbitrary connection on $L$ can be written as $\lambda+a$ for $a\in\Omega^1(M;i\mathbb{R})$. The curvature is $F_{\lambda+a}=F_\lambda+da$.

__1) Uniqueness:__ We first show that any two such connections must be gauge equivalent.

By standard Chern-Weil theory, $iF_{\lambda_0}$ is a harmonic representative for $2\pi c_1(L)\in H^1(M;\mathbb{R})$. By Hodge theory, this harmonic representative is uniquely determined. Thus any two harmonic connections on the same line bundle must have identical curvatures.

Now suppose $\lambda+a_1$ and $\lambda+a_2$ are two connections on $L$ with identical curvature and holonomy. Identical curvatures means that $a_1-a_2$ is closed, and identical holonomy means $[a_1-a_2]\in H^1(M;2\pi\mathbb{Z})\otimes\mathbb{R}$. (This ensures that $\exp\int_\gamma a_1=\exp\int_\gamma a_2$ for any loop $\gamma$).

Since $U_1$ is a $K(\mathbb{Z},1)$, we can represent $[a_1-a_2]$ as a (smooth) map $u:M\longrightarrow U_1$. This map satisfies $[u^{-1}\, du]=[a_1-a_2]$. Thus $u^{-1}\, du-a_1+a_2$ is exact, and is $i\,df$ for some function $f:M\longrightarrow\mathbb{R}$. The $U_1$-valued function $u \exp(if)$ therefore defines a gauge transformation taking $\lambda+a_2$ to $\lambda+a_1$, proving uniqueness.

__2) Existence:__

By the Hodge decomposition theorem, we can write $iF_\lambda=h_2-d\alpha-d^*\beta$ for some harmonic two-form $h_2$, $\alpha\in\Omega^1(M)$, and $\beta\in\Omega^3(M)$.

The Bianchi identity tells us that $0=dF_\lambda \implies 0=dd^*\beta \implies 0=\langle \beta,dd^*\beta\rangle \implies 0=d^*\beta$. Therefore, $iF_{\lambda+i\alpha}=h_2$, so $\lambda+i\alpha$ is a harmonic connection. (Note that $[h_2]=2\pi c_1(L)$.)

It remains to modify $\lambda+i\alpha$ to have zero holonomy. Since $U_1$ is abelian, the holonomy of any connection is an element of $\mathrm{Hom}(\pi_1(M),U_1)\cong\mathrm{Hom}(H_1(M;\mathbb{Z}),U_1)$. If we use the exponential map to identify $\frac{\mathbb{R}}{2\pi \mathbb{Z}}$ with $U_1$, then holonomy may be thought of as an element of $\mathrm{Pic}^0(M):=\frac{H^1(M;\mathbb{R})}{H^1(M;2\pi \mathbb{Z})}$.

Now $\mathrm{hol}(\lambda+i\alpha)\in \mathrm{Pic}^0(M)$. Let $\nu\in H^1(M;\mathbb{R})$ be a representative of $\mathrm{hol}(\lambda+i\alpha)$. Then $\mathrm{hol}(\lambda+i\alpha-i\nu)=0$, and $F_{\lambda+i\alpha-i\nu}=F_{\lambda+i\alpha}+i\,d\nu=F_{\lambda+i\alpha}$, so $\lambda_0:=\lambda+i\alpha-i\nu$ is a harmonic connection on $L$ with zero period, as desired.

$\blacksquare$

The structure of connections over a line bundle now becomes transparent. Any connection can be written as $\lambda=\lambda_0+a=\lambda_0+i\,h_1+i\,df+i\,d^*\beta$, where $a\in i\Omega^1(M;\mathbb{R})$, and $a=h_1+df+d^*\beta$ is the Hodge decomposition of $a$. The curvature is $i F_\lambda=h_2-dd^*\beta$. (Recall that $h_2$ was the unique harmonic representative of $2\pi c_1(L)$). By the Hodge decomposition theorem, $h_2 \perp \mathrm{Im}d$, so $\mathcal{E}(\lambda)=\Vert h_2\Vert^2+\Vert dd^*\beta\Vert^2$. Thus the Yang-Mills functional is the norm of the (fixed) harmonic part of the curvature, plus the norm of the non-harmonic part of the curvature.

Now consider the connections modulo gauge. We saw in the proof of uniqueness modulo gauge that a connection is determined by its curvature and holonomy. Thus the Hodge decomposition of $\lambda-\lambda_0$ gives an isomorphism $\mathcal{B}_L\cong\mathrm{Pic}^0(M)\times\mathrm{Im}(\Omega^1\overset{d}{\rightarrow}\Omega^2)$.

Now $\mathcal{E}$ is independent of the $\mathrm{Pic}^0(M)$ part, and returns a constant plus the norm (squared) of the image of $d$. Thus the Yang-Mills connections are precisely the harmonic connections, they are parameterized by $\mathrm{Pic}^0(M)$, and all have index zero.