# 10-08 Notes

\section{Generators for the cohomology of $B\mathcal{G}$}

Following the previous lecture the goal is to identify generators for the cohomology of $B\mathcal{G} = \text{Map} [\Sigma , BU_n]$. Consider a principal bundle $SU_n \rightarrow P \rightarrow \Sigma$ which is the bundle of frames $(v_1,..,v_n)$ of a vector bundle $E$ with volume $1$. The volume element is the same as a trivialization of the determinant line bundle $\Delta=\text{det} E$. Let $\delta$ be a unitary connection on $\Delta$ and consider $$\mathcal{A}^{\delta}=\{A | A \text{ a unitary connection on E s.t.} \text{det}(A) = \delta \}.$$ $\mathcal{A}^{\delta}$ is an affine space modeled by $\Omega^1(su(E))$ where $su(E)$ is the traceless skew hermitian endomorphisms of $E$.

The subgroup $\mathcal{G}^\delta$ of the gauge transformations of $E$ preserving $\mathcal{A}^\delta$ is the space of sections of the bundle of unitary endomorphisms of $E$ with determinant $1$. Now let $\mathcal{B}^\delta = \mathcal{A}^\delta/\mathcal{G}^\delta$ and consider the fibration $(\mathcal{A}^\delta \times E)/\mathcal{G}^\delta \stackrel{\pi}{\rightarrow} \mathcal{B}^\delta \times \Sigma$. The fiber above a point is given by $\pi^{-1}([A]\times \sigma)=E|_{\sigma}/Stab(A)$. In trying to make this a vector bundle the stabilizer of $A$ needs to be gotten rid of. As the center $Z(SU_n) \subset \mathcal{G}^{\delta}$ is clearly in the center a first attempt is to get rid of this part. To this end, consider instead the fibration $$(\mathcal{A}^{\delta}\times \text{End}^0(E))/\mathcal{G}^{\delta} \stackrel{\pi}{\rightarrow} \mathcal{B}^{\delta} \times \Sigma,$$ where $End^0(E)$ denotes the bundle of traceless endomorphisms of $E$. Now as $Z(SU_n)$ acts trivially on $\text{End}^0(E)$ that part of the stabilizer has been taken care of. However there is still an issue in the case where $\text{Stab}(A) \neq Z(SU_n)$. Hence it can only be said that $\mathcal{E}|_{\mathcal{B}_*^{\delta} \times \Sigma}$ is a vector bundle, where $\mathcal{E}= (\mathcal{A}^{\delta}\times \text{End}^0(E))/\mathcal{G}^{\delta}$ and $\mathcal{B}_*^{\delta}=\{[A]|\text{Stab}(A)=Z(SU(n))\}$ is the space of irreducible connections.

Now the Chern classes $c_i(\mathcal{E}) \in H^{2i}(\mathcal{B}_*^{\delta} \times \Sigma)$ can be decomposed using Kunneth formula and used to define $\mu_i : H_*(\Sigma) \rightarrow H^*(\mathcal{B}_*^{\delta})$ via $$\mu_i(a) = c_i(\mathcal{E})/a \in H^{2i-\text{deg}(a)}.$$ Here $c_i(\mathcal{E})/a$ denotes the evaluation of the $H^{\text{deg}a}(\Sigma) \otimes H^{2i-\text{deg}a}(\mathcal{B}_*^{\delta})$ component of $c_i(\mathcal{E})$ on $a$. In the case that there is another vector bundle $\mathcal{V}$ such that $\text{End}^0(\mathcal{V}) = \mathcal{E}$, the Chern classes $c_i(\mathcal{E})$ can be written as polynomials in $c_i(\mathcal{V})$. This formula can be inverted in general, regardless of whether $\mathcal{V}$ exists to define $c_i(\mathcal{V})$. The convention is to use $c_i(\mathcal{V})$ rather than $c_i(\mathcal{E})$ to define $\mu_i$. Since the original bundle $E$ was arbitrary and $\mathcal{B}=\mathcal{A}/\mathcal{G}$ is a classifying space for $\mathcal{G}$ the $\mu_i$’s give the required generators for $H^*(B\mathcal{G})$.

\section{Stabilizers in the rank 2 case}

Consider a rank $2$ bundle with a unitary connection $A$ on it. It was shown in lecture 4 that $\text{Stab}_{p_0}(A)=\text{Centralizer of}\text{ hol}_{p_0}(A)$. Hence $\text{Stab}(A) = \{\pm 1\}, U_1 \text{ or } SU_2$. Assuming $\langle c_1(E), [\Sigma] \rangle = 1$ and $\text{vol}\Sigma = 2\pi$ these three cases can be dealt with as follows –

$\mathbf{Case 1}$: $\text{Stab}(A) = SU_2$. In this case the holonomy would be $\{\pm 1\}$ and since it changes continuously it must be locally constant and the connection is flat ($F_A = 0$). Thus it cannot be true that $\langle c_1(E), [\Sigma] \rangle = 1$ and this case cannot happen.

$\mathbf{Case 2}$: $\text{Stab}(A) = U_1$. In this case the bundle splits $E = L_1 \oplus L_2$ and $A$ preserves the splitting. The curvature operator splits as $*F_A = \begin{bmatrix} i\lambda_1 & 0 \\ 0 & i\lambda_2 \end{bmatrix}$, where $\lambda_i = -c_1(L_i)$ are not necessarily distinct. As $\lambda_1+\lambda_2=-c_1(E)=-1$, if $\lambda_1=-h$ the energy is given by $\mathcal{E}(A)=2h^2-2h+1$.

$\mathbf{Case 3}$: $\text{Stab}(A) = \pm 1$. In this case $E$ would have full holonomy $SU_2$ and hence it cannot split. As the eigenspaces of $*F_A$ give a holomorphic splitting of $E$ it must be the case that there is just one eigenspace $*F_A = \begin{bmatrix} i\lambda & 0 \\ 0 & i\lambda \end{bmatrix}$. Here $\lambda = -\frac{1}{2}c_1(E) = -\frac{1}{2}$ and hence the energy is $\mathcal{E}(A)=\frac{1}{2}$.

\section{Hessian of YMF}

Since the goal is to understand the topology of $N^0(n,k)=\{A \in A^{\delta}| \text{rank}(E)=n, c_1(E)=k\}/\mathcal{G}$ using the Yang Mills functional, the Morse index and nullity of the critical points must first be understood. The computation to this end gives $F_{A+ta} = F_A + td_A +\frac{t^2}{2}[a,a]$, where $A$ is a fixed unitary connection, $a \in \Omega^1(su(E))$ and $[a,a](x,y) \doteq [a(x),a(y)]$ by definition. The energy is $$\mathcal{E}(A+ta) = \mathcal(A) + 2t \langle F_A , d_Aa\rangle + t^2(\langle d_Aa, d_Aa \rangle + \langle F_A,[a,a]\rangle) + O(t^3),$$ and the Hessian can be computed as \begin{eqnarray*} \text{Hessian}_A(a) &=& \langle d_Aa,d_Aa \rangle + \langle F_A , [a,a]\rangle \\ &=& \langle d_Aa,d_Aa \rangle + \int (*F_A , [a,a]) \\ &=& \langle d_Aa,d_Aa \rangle + \int ([a,*F_A] , a) \\ &=& \langle d_Aa,d_Aa \rangle + \int (**[*F_A,a] , a) \\ &=& \langle L_Aa , a \rangle, \end{eqnarray*} where $L_A a = d_A^*d_Aa + *[*F_A,a]$. The Hessian can be viewed as an operator on the tangent space $\Omega^1(\Sigma,su(E))$. It will be shown in the next lecture that the tangent space to the gauge orbit $T_A(\mathcal{G}.A)$ is in the nullspace of $L_A$. However $L_A$ can be replaced by $H_A = L_A + d_Ad_A^*$ to get an operator with the same number of negative eigenvalues and the nullity of $L_A$ restricted to a slice ($\text{ker} L_A|_{T_A(\mathcal{G}.A)^{\perp}}$). The operator $H_A$ acts on $\Omega^1(\Sigma,su(E))$. In the case where $E=L_1\oplus L_2$ is a sum of line bundles $su(E) = i \mathbb{R} \oplus L_1^* \otimes L_2 = i\mathbb{R} \oplus V$, and there is a decomposition $\Omega^1(su(E)) = \Omega^1(\Sigma) \oplus \Omega^1(V) = \Omega^1(\Sigma) \oplus \Omega^{0,1}(V) \oplus \Omega^{1,0}(V)$. In the next lecture the nullity and index of $H_A$ will be studied along the lines of the Kodaira-Nakano calculation.