10-20 Notes

Review of the notation

$\C^2 \to E \to \Sigma$ 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 $i\R= \varepsilon{e}_{\R}$ 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:

$\mathfrak{su}(E) = i\R \oplus (E_1 \otimes E_2^*) = \mathcal{\varepsilon}_{\R} \oplus L$.

$\Omega^1(\Sigma,\mathfrak{su}(E)) = \Omega^1(\Sigma,\varepsilon_{\R}) \oplus \Omega^1(\Sigma,L)$

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 $\varepsilon_{\R}$ 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 $\Omega^1(\Sigma,\mathfrak{su}(E)) = \Omega^1(\Sigma,\varepsilon_{\R}) \oplus \Omega^{1,0} \oplus \Omega^{0,1}$.

Idea. $\Delta_A$ has non-negative eigenvalues and the zero eigenvectors of $\Delta_A=\Delta$ on $\Omega^1(\Sigma,\R)$ 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 $\textrm{Spec}^+(T) = \textrm{Spec}^+(T')$.

\begin{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 $\|D\psi\|\neq 0$. \end{proof}

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 $\Omega^{0,1}$.

\begin{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 $\Omega^{1,0}$,

$$ \begin{array}{ll} \overline{\Box}_A = \overline{\partial}^*_A \overline{\partial}_A & = \frac{1}{2} (1+i*)d_A^*d_A \\ & = \frac{1}{2} [d_A^*d_A+ i*d_A^*d_A] \\ & = \frac{1}{2}[ d_A^*d_A-d_A*d_A*] \\ & = \frac{1}{2} \Delta_A. \end{array}$$ \end{proof}

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 $\lambda_1-\lambda_2 + g -1$.