10-20 Notes
Review of the notation
\(\mathbb{C}^2 \to E \to \Sigma\) with \(\det E = \Delta\) and \(<c_1(\Delta),[\Sigma]> =1\)
\(A \in \mathcal{A}_{\delta} = \{ A \in \mathcal{A}_E: \det A = \delta \}\)
\(\mathcal{G}' = \{ u: P \to U_2: u(pg) = gu(p) g^{-1}, \det u =1 \}\)
\(\mathcal{E}(A)= \int_{\Sigma} |F_A|^2\)
\(N(2,1) = \{\) set of minima of \(\mathcal{E}\) modulo \(\mathcal{G}'\}\)
If \(\delta\) is not harmonic, then \(N(2,1)=\emptyset\).
If \(\delta\) is harmonic, then there are two cases: the eigenvalues of \(*F_A\) are equal or they are distinct. In this lecture, we will compute the Morse indices of the critical points of \(\mathcal{E}\) when the eigenvalues are distinct.
Notation: the eigenvalues of \(*F_A\) will be denoted \(-i\lambda_1\) and \(-i\lambda_2\), so that normalizing vol\((\Sigma)=2\pi\), the decomposition into eigenbundles \(E=E_1\oplus E_2\) have Chern classes \(c_1(E_1)= \lambda_1\) and \(c_2(E_2)=\lambda_2\) (Lecture 7).
\(\textrm{Hess}_A \mathcal{E} = d_A^* d_A + *[*F_A,\cdot]\) as a map \(\Omega^1(\Sigma,\mathfrak{su}(E)) \to \Omega^1(\Sigma,\mathfrak{su}(E))\).
\(\ker \textrm{Hess}_A \mathcal{E} = T_{[A]} \mathcal{B}\)
We now add \(d_Ad_A^*\) to \(\textrm{Hess}_A \mathcal{E}\) to get rid of this kernel. This affects only the positive eigenspace, it does not affect the Morse index.
\(Q_A \mathcal{E} = d_A^* d_A + d_A d_A^* + *[*F_A,\cdot] = \Delta_A + *[*F_A,\cdot]\)
Morse index calculation
Consider \(E= E_1 \oplus E_2\).
An element of \(\mathfrak{su}(E)\) has the form
\[\left( \begin{array}{ll} it & z \\ -\overline{z} & -it \end{array} \right).\]
So \(\mathfrak{su}(E)\), as a subset of \(\textrm{Hom}(E_1\oplus E_2, E_1\oplus E_2\)), breaks up into two bundles: a trivial line bundle \(i\mathbb{R}= \varepsilon{e}_{\mathbb{R}}\) which corresponds to multiplication by \(it\) on \(E_1\) and by \(-it\) on \(E_2\), and a complex line bundle \(\textrm{Hom}(E_2,E_1) = E_1 \otimes E_2^*\) which is multiplication by \(-\overline{z}\) (and dualizing, a homomorphism \(E_1 \to E_2\) given by multiplication by \(z\)). Briefly:
\(\mathfrak{su}(E) = i\mathbb{R}\oplus (E_1 \otimes E_2^*) = \mathcal{\varepsilon}_{\mathbb{R}} \oplus L\).
\(\Omega^1(\Sigma,\mathfrak{su}(E)) = \Omega^1(\Sigma,\varepsilon_{\mathbb{R}}) \oplus \Omega^1(\Sigma,L)\)
Now the Hodge \(*\) acts by \(*^2=-1\) on \(\Omega^1(\Sigma,L)\). This breaks up \(\Omega^1\) into \(\mp i\) eigenspaces:
\(\Omega^1(\Sigma,L) = \Omega^{1,0}(\Sigma,L) \oplus \Omega^{0,1}(\Sigma,L)\)
Conventions: 1.) \(*(dx+idy)=dy-idx=-i(dx+idy)\) so \(*dz=-idz\). 2.) From now on, we omit \((\Sigma,L)\) from the above notation.
\(*F_A\) acts by \(-i\lambda_1\) on \(E_1\) and by \(-i\lambda_2\) on \(E_2\). Therefore it acts by \(0\) on \(\varepsilon_{\mathbb{R}}\) and by \(-i(\lambda_1-\lambda_2)\) on \(L\). Therefore, \(Q_A\) acts by \(\Delta_A\), \(\Delta_A-(\lambda_1-\lambda_2)\) and \(\Delta_A+(\lambda_1-\lambda_2)\) respectively on the three summands of \(\Omega^1(\Sigma,\mathfrak{su}(E)) = \Omega^1(\Sigma,\varepsilon_{\mathbb{R}}) \oplus \Omega^{1,0} \oplus \Omega^{0,1}\).
Idea. \(\Delta_A\) has non-negative eigenvalues and the zero eigenvectors of \(\Delta_A=\Delta\) on \(\Omega^1(\Sigma,\mathbb{R})\) are the harmonic one-forms. We may assume that \(\lambda_1 - \lambda_2 >0\), so \(\Delta_A + (\lambda_1-\lambda_2)\) has positive eigenvalues. So our aim is to find the spectrum of \(\Delta_A - (\lambda_1-\lambda_2)\).
Let \(\Pi^{p,q}\) be the projection to \(\Omega^{p,q}\), therefore
\[\Pi^{1,0}= \frac{1}{2} (1+i*) \;\;\textrm{ and }\;\; \Pi^{0,1} = \frac{1}{2} (1-i*).\]
Recall \(d_A = \partial_A + \overline{\partial}_A\) on \(\Omega^{0,0}\), with
\[\partial_A = \Pi^{1,0} \circ d_A = \frac{1}{2} (1+i*) d_A \;\; \textrm{ and } \;\; \overline{\partial}_A= \Pi^{0,1}\circ d_A = \frac{1}{2} (1-i*) d_A.\]
Define \(\Box_A = \partial_A \partial^*_A + \partial^*_A \partial_A\) and \(\overline{\Box}_A = \overline{\partial}_A \overline{\partial}^*_A + \overline{\partial}^*_A \overline{\partial}_A\).
Observe that \(\Box_A\), \(\overline{\Box}_A\), and \(\Delta_A\) are self-adjoint operators, and therefore their spectrum is non-negative, real and discrete.
We need an abstract result about positive spectra \(\textrm{Spec}^+(T) = \{ \lambda \in \textrm{Spec(T)}: \lambda>0 \}\) of an operator \(T:H_1 \to H_2\) between Hilbert spaces.
LEMMA. Let \(D:H_1 \to H_2\) be a bounded operator, and let \(T=D^*D\) and \(T'=DD^*\). Then \(\textrm{Spec}^+(T) = \textrm{Spec}^+(T')\).
Proof. \(T\) is self-adjoint, so it has non-negative eigenvalues. Observe that if \(\psi\) is an eigenvector of \(T\) with eigenvalue \(\lambda\), \(D^*D \psi = \lambda \psi\), then \(D\psi\) is an eigenvector of \(T'\) with the same eigenvalue: \(DD^*(D\psi) = D(\lambda \psi) = \lambda D\psi\). Their norms are related by
\[\| D\psi \|^2 = <\psi, D^*D\psi> = \lambda \|\psi\|^2.\]
Therefore, for all such \(\lambda>0\), \(\|\psi\|\neq 0\) iff \(\|D\psi\|\neq 0\). ◻
In our case, \(\textrm{Spec}^+(\Box_A|_{\Omega^{0,0}}) = \textrm{Spec}^+(\Box_A|_{\Omega^{1,0}})\) since \(\Box_A|_{\Omega^{0,0}}\) and \(\Box_A|_{\Omega^{1,0}}\) are respectively \(\partial_A^*\partial_A\) and \(\partial_A \partial_A^*\). Similarly, \(\textrm{Spec}^+(\overline{\Box}_A|_{\Omega^{0,0}}) = \textrm{Spec}^+(\overline{\Box}_A|_{\Omega^{0,1}})\).
LEMMA. On \(\Omega^{0,0}\),
\[\Box_A = \frac{1}{2} (\Delta_A+i*F_A) \;\; \textrm{ and }\;\; \overline{\Box}_A = \frac{1}{2} (\Delta_A - i^*F_A).\]
Moreover \(\Box_A = \frac{1}{2} \Delta_A\) on \(\Omega^{1,0}\), and \(\overline{\Box}_A = \frac{1}{2} \Delta_A\) on \(\Omega^{0,1}\).
Proof. Observe that \(\partial_A = \Pi^{1,0} \circ d_A\) so \(\partial^*_A = d_A^* \circ \Pi^{1,0}\). So on \(\Omega^{0,0}\),
\[\Box_A= \partial_A^*\partial_A = d_A^* \frac{1}{2} (1+i*) d_A = \frac{1}{2} d_A^* d_A - \frac{i}{2}*d_A**d_A = \frac{1}{2} (\Delta_A +i*F_A),\]
where recall \(d_A^* = -*d_A*\). Proceed similarly for \(\overline{\Box}_A\).
On \(\Omega^{1,0}\),
\[\begin{array}{ll} \Box_A = \partial_A \partial^*_A & = -\Pi^{1,0}d_A * d_A * \Pi^{1,0} \\ & = -\frac{1}{2} (1+i*)d_A*d_A*\frac{1}{2}(1+i*) \\ & = -\frac{1}{4} [d_A*d_A*+ i*d_A*d_A*+ d_A*d_A*i* + i*d_A*d_A*i*] \\ & = -\frac{1}{4}[ d_A*d_A*+*d_A*d_A+d_A*d_A* + *d_A*d_A] \\ & = -\frac{1}{4}[ -d_Ad_A^*-d_A^*d_A-d_Ad_A^* - d_A^*d_A] \\ & = \frac{1}{2} \Delta_A. \end{array}\]
On \(\Omega^{1,0}\),
\[\begin{array}{ll} \overline{\Box}_A = \overline{\partial}^*_A \overline{\partial}_A & = \frac{1}{2} (1+i*)d_A^*d_A \\ & = \frac{1}{2} [d_A^*d_A+ i*d_A^*d_A] \\ & = \frac{1}{2}[ d_A^*d_A-d_A*d_A*] \\ & = \frac{1}{2} \Delta_A. \end{array}\] ◻
Denote \(\sigma = \lambda_1 -\lambda_2 >0\). Then on \(\Omega^{0,0}\),
\[\Box_A = \frac{1}{2} (\Delta_A + i(-i\sigma)) = \frac{1}{2}(\Delta_A + \sigma)\]
and similarly \(\overline{\Box}_A=\frac{1}{2} (\Delta_A - \sigma)\). So
\[\Box_A = \overline{\Box}_A + \sigma.\]
Now \(\textrm{Spec}(\overline{\Box}_A|_{\Omega^{0,0}})\geq 0\), so \(\textrm{Spec}^+(\Box_A|_{\Omega^{1,0}})=\textrm{Spec}^+(\Box_A|_{\Omega^{0,0}})\geq \sigma\). Therefore
\[\textrm{Spec}^+(\Delta_A|_{\Omega^{1,0}}) = 2\textrm{Spec}^+(\Box_A|_{\Omega^{1,0}})\geq 2\sigma.\]
The operator \(Q_A-\Delta_A=*[*F_A,\cdot]\) acts by \(-\sigma\) on \(\Omega^{1,0}\). Therefore \(\textrm{Spec}^+(Q_A|_{\Omega^{1,0}}) \geq 2\sigma-\sigma>0\). So the negative eigenspace of \(Q_A|_{\Omega^{1,0}}\) corresponds to \(\ker \Delta_A|_{\Omega^{1,0}}\), which is where \(Q_A\) acts by \(-\sigma\).
By sheaf cohomology,
\[\ker \Delta_A = H^0(\Sigma,\Omega^{1,0} \otimes L),\]
and by Serre duality this is isomorphic to \(H^1(\Sigma,L^*)^*\).
Recall Riemann-Roch for a vector bundle \(V\) over \(\Sigma\): \(h^0(V)-h^1(V) = c_1(V) + \textrm{rk}(V)(1-g)\). In our case,
\[\textrm{dim} H^0(\Sigma,L^*) - \textrm{dim} H^1(\Sigma,L^*) = c_1(L^*)+1-g.\]
The space of sections \(H^0(\Sigma,L^*)=0\) because
\[c_1(L^*)=-c_1(L) = -c_1(E_1\otimes E_2^*)=-\sigma<0.\]
Therefore, we finished calculating the Morse index \(\textrm{dim} \ker(\Delta_A|_{\Omega^{1,0}}) = \textrm{dim}H^1(\Sigma,L^*)\):
THEOREM. The Morse index of a critical point of the Yang-Mills functional in the case \(\lambda_1> \lambda_2\) is \(\lambda_1-\lambda_2 + g -1\).
