# 10-15 Notes

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\section{The Setup}

We have a complex rank $n$ vector bundle over a Riemann surface: $\mathbb{C}^{n}\rightarrow E\rightarrow\Sigma^{2}$.

Denote the Chern number by $k:=\left\langle c_{1}(E),\left[\Sigma\right]\right\rangle$.

The moduli space of Yang Mills connections is defined to be $\left\{ A\in\mathcal{A}_{E}|d_{A}\! *\! F_{A}=0\right\} / \G_{E}$, where $\G_E$ is the group of unitary gauge transformations. Let $N(n,k)$ denote the subspace of connections for which the Yang-Mills energy achieves its absolute minimum.

Given a connection $\delta$ for $\Delta=\det E$, we consider the corresponding $SU_n$ moduli space $N_{\delta}$ of minimal energy Yang-Mills connections inside $\left\{ A\in\mathcal{A}_{E}|d_{A}\!*\!F_{A}=0,\,\,\det A=\delta\right\} /\mathcal{G}_{E}^{1}$, where $\mathcal{G}_{E}^{1}$ denotes the unitary endomorphisms of $E$ with $\det=1$.

\section{Twisting by a line bundle}

If $F_{\delta}$ is not harmonic, then $N_{\delta}(n,k)=\emptyset$. [1] If it is harmonic, then $N_{\delta}(n,k) \subset N(n,k)$. [2]

Given $\lambda$ a harmonic connection on $L$, we can twist by mapping $(E,A)\mapsto(E\otimes L,A\otimes\lambda)$.

If $A$ is Yang-Mills if and only if $A\otimes\lambda$ is Yang-Mills. Furthermore, if $\det(E,A)=(\Delta,\delta)$, then $\det(E\otimes L,A\otimes \lambda)=(\Delta\otimes L^{\otimes n},\delta\otimes\lambda^n)$.

\begin{thm} $N_{\delta}(n,k) \cong N_{\delta\otimes\lambda^n}(n,k+nl)$, where $l:=\left\langle c_{1}(L),\left[\Sigma\right]\right\rangle$. \end{thm}

Define $\mbox{Pic}^{0}(\Sigma):=H^{1}(\Sigma;\mathbb{R})/H^{1}(\Sigma;\mathbb{Z})$. This torus is equivalent to the moduli space of flat connections on the trivial bundle, modulo gauge.[3]

If $L$ is the trivial line bundle, then the twist gives a map $N_{\delta}(n,k)\times\mbox{Pic}^{0}(\Sigma)\rightarrow N(n,k)$.

This is surjective: Consider some $\left[(E,\tilde{A})\right]\in N(n,k)$. Then for some imaginary 1-form $\tilde{\lambda}$, we have $\det \tilde{A}=\delta+\tilde{\lambda}=\delta\otimes\tilde{\lambda}$. (In writing $\delta\otimes\tilde{\lambda}$, we are thinking of $\tilde{\lambda}$ as a flat connection in trivial bundle). We have a short exact sequence of abelian groups $0\longrightarrow (\mathbb{Z}_n)^{2g}\longrightarrow \mathrm{Pic}^0(\Sigma)\overset{m_n}{\longrightarrow} \mathrm{Pic}^0(\Sigma)\longrightarrow 0$, where $m_n$ denotes multiplication by $n$. Pick some $\mu\in (m_n)^{-1}(\tilde{\lambda})$. Then $\left[(E,A\otimes\mu^*)\right]\in N_{\delta}(n,k)$, and $(A\otimes\mu^*)\otimes\mu=A$.

Fact: If $n$ and $k$ are coprime, then $N_{\delta}(n,k)\times(S^{1})^{2g}\rightarrow N(n,k)$ is a $n^{2g}$-covering space.

\section{The case $n=2$, $k=1$}

Consider $E$ a rank two vector bundle with $A\in\mathcal{A}_\delta$. Recall that $*\!F_A$, as a covariantly constant endomorphism of $E$, has eigenvalues $-i\lambda_1$ and $-i\lambda_2$, and decomposes $E$ into eigenspaces. We set $\mathrm{vol}(\Sigma)=2\pi$, so that $k=\left\langle c_{1}(E),\left[\Sigma\right]\right\rangle =\lambda_1+\lambda_2$. The energy is $\mathcal{E}(A)=(2\pi)^{-1}((\lambda_1)^2+(\lambda_2)^2)$. Note that since $k=\lambda_1+\lambda_2$, energy is smallest when $\lambda_1\approx\lambda_2$.

There are two cases.

1. $\lambda_1=\lambda_2$. It follows that $\lambda_i=\frac{1}{2}k$, and energy is $(2\pi)^{-1}\frac{1}{2}k^2$. In the case $k=1$, we have $\lambda_2=1-\lambda_1$, so the absolute minimum of energy occurs when the $\lambda_i$ are each one-half.
2. $\lambda_1\neq\lambda_2$. We get distinct eigenspaces $E=E_{1}\oplus E_{2}$. Each line bundle $E_i$ has its own connection $A_i$, and

$*\!F_{A}=\left(\begin{matrix} -i\lambda_{1}\\ & -i\lambda_{2}\end{matrix}\right)$.  Thus, $\lambda_i=\left\langle c_{1}(E_i),\left[\Sigma\right]\right\rangle\in\mathbb{Z}$.  If $k=1$, then the energy becomes $2\lambda_1^2-2\lambda_1+1$.  The connection $A_1$ determines $A_2$ since $\det A=A_{1}\otimes A_{2}=\delta$ is fixed.  Since $A$ is Yang-Mills, we know that the $A_i$ are harmonic.  Thus connections of this type correspond to splittings of $E$, together with a harmonic connection on one of the subbundles.


\section{The Hessian}

Our goal is to deduce the topology of $N_{\delta}(2,1)$ based on the topology of $\mathcal{B}_{\delta}(2,1)$ and the topology of the higher critical points.

We denote by $\A_{E,\delta}$ the space of connections in $E$ with determinant connection $\delta$. The tangent space is $T_{A}\mathcal{A}_{E,\delta}=\Omega^{1}(\Sigma;\mathfrak{su}(E))$. Recall the definition of energy $\mathcal{E}(A):=\int_{\Sigma}\left|F_{A}\right|^{2}$.

The Hessian of $\mathcal{E}$ at a connection $A$ is an endomorphism $\mbox{Hess}_{A}\mathcal{E}:\Omega^{1}(\Sigma;\mathfrak{su}E)\circlearrowleft$. It is given by $\mbox{Hess}_{A}\mathcal{E}(a)=d_{A}^{*}d_{A}a+*\left[*\!F_{A},a\right]$.

For any Yang-Mills connection $A$, we would like to know 1) number of negative eigenvalues of $\mathrm{Hess}_{A}$, which corresponds to the Morse index of $\mathcal{E}$ at $A$, and 2) number of zero eigenvalues of $\mathrm{Hess}_{A}$, which corresponds to the nullity.

Observation: Nullity is infinite dimensional, since $\ker(\mathrm{Hess}_{A}\mathcal{E})$ contains the image of the gauge orbit $d_{A}\sigma$, where $\sigma\in\Omega^{0}(\Sigma;\mathfrak{su}(E))$.

To verify that the gauge orbit is indeed in the kernel, we compute $$\mathrm{Hess}_{A}\mathcal{E}(d_{A}\sigma)=d_{A}^{*}d_{A}d_{A}\sigma+*\left[*\!F_{A},d_{A}\sigma\right]=-*\!d_{A}*\!\left[F_{A},\sigma\right]+*\!\left[*\!F_{A},d_{A}\sigma\right].$$ Now $$-\!*d_{A}\left[*\!F_{A},\sigma\right]=-\!*\!(\left[d_{A}*\!F_{A}\rightarrow0,\sigma\right]+\left[*\!F_{A},d_{A}\sigma\right])=-*\left[*\!F_{A},d_{A}\sigma\right].$$ Thus the two terms cancel to give zero.

To compensate for this infinite dimensional kernel, we restrict $\mathrm{Hess}_{A}\mathcal{E}|_{\ker d_{A}^{*}=T_{\left[A\right]}\mathcal{B}}$ to the slice. Equivalently, we can insert an extra term to get $Q_A=\mathrm{Hess}_{A}\mathcal{E}+d_{A}d_{A}^{*}$. To see why this works, note that $$T_{A}\mathcal{A}_{0}=(\ker d_{A}^{*}=T_{\left[A\right]}\mathcal{B})\oplus\mbox{Im}d_{A}.$$ Then we get $$Q_{A}=\left(\begin{matrix} \mbox{Hess}_{A}\mathcal{E}|_{\ker d_{A}^{*}} & 0\\ \bullet & d_{A}d_{A}^{*}\end{matrix}\right).$$ We could check by direct computation that $\bullet=0$, but a simpler way is to observe that $\mathrm{Hess}_{A}$ is self-adjoint.

We now have that $\mathrm{Index}(\mathrm{Hess}_{A}\mathcal{E})=\mbox{Index}(Q_{A})$, and $\mathrm{Nullity}(\mathrm{Hess}_{A}\mathcal{E}|_{\ker d_{A}^{*}})=\mathrm{Nullity}(Q_{A})$.

We now wish to study the operator $$Q_{A}=d_{A}^{*}d_{A}+d_{A}d_{A}^{*}+*\left[*\!F_{A},\bullet\right],$$ along the lines of the Kodaira-Nakano vanishing theorem.

Fix a Riemannian metric on $\Sigma$. Then for any complex bundle $E$, the Hodge star operator is an endomorphism $*:\Omega^{1}(\Sigma;E) \circlearrowleft$ satisfying $*^{2}=-I$. This decomposes one-forms into eigenspaces $*=-i$ on $\Omega^{1,0}$ and $*=i$ on $\Omega^{0,1}$. In coordinates, $$*\!dz=*\!(dx+idy)=dy-idx=-i(dx+idy)=-idz.$$ Thus $\Omega^{1}(\Sigma;E)=\Omega^{1,0}(\Sigma;E)\oplus\Omega^{0,1}(\Sigma,E)$.

In the following lecture, we will consider differential forms with values in the adjoint bundle $\mathfrak{su}(E)\rightarrow\Sigma$. For $E=E_{1}\oplus E_{2}$, there is an isomorphism $\mathfrak{su}(E)=\R\oplus V$, where $V=E_{1}\otimes E_{2}^{*}$. with a connection $A$ having parallel curvature[4].

$\Omega^{0,0}(\Sigma,V)\overset{d_{A}}{\rightarrow}\Omega^{1}(\Sigma,V)$, $d_{A}=\partial_{A}+\bar{\partial}_{A}$, $\partial_{A}:\Omega^{0,0}(\Sigma,V)\rightarrow\Omega^{1,0}(\Sigma,V)$.

$\xymatrix{\Omega^{1,0}(\Sigma,V)\ar[r]^{\bar{\partial}_{A}} & \Omega^{1,1}(\Sigma,V)\\ \Omega^{0,0}(\Sigma,V)\ar[r]^{\bar{\partial}_{A}}\ar[u]^{\partial_{A}} & \Omega^{0,1}(\Sigma,V)\ar[u]^{\partial_{A}}}$

We can construct also $\square_{A}=\partial_{A}\partial_{A}^{*}+\partial_{A}^{*}\partial_{A}$, $\bar{\square}_{A}=\bar{\partial}_{A}\bar{\partial}_{A}^{*}+\bar{\partial}_{A}^{*}\bar{\partial}_{A}$. $\Delta_{A}=d_{A}d_{A}^{*}+d_{A}d_{A}^{*}$.

\begin{lem} On $\Omega^{0,0}$, $\square_{A}=\frac{1}{2}\Delta_{A}+\frac{i}{2}*\!F_{A}$, $\bar{\square}_{A}=\frac{1}{2}\Delta_{A}-\frac{i}{2}*\!F_{A}$. On $\Omega^{1,0}$ or $\Omega^{0,1}$, $\square_{A}=\bar{\square}_{A}=\frac{1}{2}\Delta_{A}$. \end{lem}

\section{Additional notes}

1. The determinant map $\mathcal{A}_E\overset{\det}{\longrightarrow}\mathcal{A}_\Delta$ preserves the Yang-Mills condition since $d^*_A F_A=0 \implies d^*F_{\det A}=d^*\mathrm{tr}(F_A)=\mathrm{tr}(d^*_A F_A)=0$. If $\delta=\det A$ isn't harmonic (Yang-Mills), then $A$ isn't Yang-Mills either.
2. This isn't immediately obvious since the gauge group $\mathcal{G}_E$ of $N(n,k)$ is larger than the gauge group $\mathcal{G}_E^1$ of $N_{\delta}(n,k)$. We should verify that the obvious map $N_{\delta}(n,k)\longrightarrow N(n,k)$ is injective. If $A_2=A_1\cdot g$ for some $g\in \mathcal{G}_E$ and $\delta=\det A_2=\det A_1$, then does $A_2=A_1\cdot \tilde{g}$ for some $\tilde{g}\in\mathcal{G}_E^1$? In a local trivialization, $a_2^\tau=(g^\tau)^{-1} a_1 g^\tau+(g^\tau)^{-1} dg^\tau$ and $\delta=\mathrm{tr }\, a_2^\tau=\mathrm{tr}\, a_1^\tau$. It follows that $0=\mathrm{tr}\left((g^\tau)^{-1} dg^\tau\right)$. Recall the formula for the derivative of the determinant $d \det g^\tau=\det g^\tau \mathrm{tr}\left((g^\tau)^{-1} dg^\tau\right)$. Thus $\det g^\tau$ is locally constant. Since $\det$ is conjugation-invariant, $\det g^\tau$ is independent of the trivialization $\tau$, so $\det g$ is a well-defined constant $U_1$-valued function. Upon picking any $n$-th root, we may set $\tilde{g}:=(\det g)^{-\frac{1}{n}} g$.
3. The trivial line bundle has a canonical trivial connection. Relative to this trivial connection, general connections correspond to elements of $i\Omega^1(\Sigma;\mathbb{R})$. Flat connections correspond to closed one-forms. Connections modulo gauge transformations which are of the form $e^{if}$ for $f:\Sigma\longrightarrow\mathbb{R}$ correspond to $H^1(\Sigma;\mathbb{R})$. A general gauge transform $u:\Sigma\longrightarrow U_1$ changes a 1-form $ia$ by $(ia)\cdot u=ia+u^{-1}du$, and $u^{-1}du$ is a closed $1$-form with $2\pi i$ integral periods. Thus flat connections in the trivial line bundle modulo gauge correspond to $\mathrm{Pic}^0(\Sigma)$.
4. Note that over a surface, a connection is Yang-Mills iff it has parallel curvature, since $$\nabla_A F_A=(1\otimes**)\nabla_A F_A=(1\otimes*)\nabla_A* F_A=(\nabla_A*F_A)\otimes *1=(d_A*F_A)\otimes *1\in\Omega^1(\Sigma)\otimes_{\Omega^0}\Omega^2(\Sigma;\mathrm{ad}(\mathfrak{g})).$$