10-15 Notes

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 \(\left\{ A\in\mathcal{A}_{E}|d_{A}\! *\! F_{A}=0\right\} / \mathcal{G}_{E}\), where \(\mathcal{G}_E\) 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 \(\det=1\).

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 \(\det(E\otimes L,A\otimes \lambda)=(\Delta\otimes L^{\otimes n},\delta\otimes\lambda^n)\).

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

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.

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. # \(\lambda_1=\lambda_2\). 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. # \(\lambda_1\neq\lambda_2\). 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.

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 \(\mathcal{A}_{E,\delta}\) 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 \(\mathfrak{su}(E)=\mathbb{R}\oplus V\), where \(V=E_{1}\otimes E_{2}^{*}\). with a connection \(A\) having parallel curvature⁴.

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

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

Lemma 1. 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 \(\Omega^{0,1}\), \(\square_{A}=\bar{\square}_{A}=\frac{1}{2}\Delta_{A}\).

Additional notes

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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 \[\nabla_A F_A=(1\otimes**)\nabla_A F_A=(1\otimes*)\nabla_A* F_A=(\nabla_A*F_A)\otimes *1=(d_A*F_A)\otimes *1\in\Omega^1(\Sigma)\otimes_{\Omega^0}\Omega^2(\Sigma;\mathrm{ad}(\mathfrak{g})).\]↩︎