10-01 Notes

\section{Topology of the Classifying Space of the Gauge Group}

Our goal is to understand the algebraic topology of the solutions to the Yang Mills equations on a closed, oriented surface $\Sigma$. Following Atiyah and Bott, our strategy will be to do Morse theory for the Yang-Mills functional $\mathcal{E}(A) = \int_M |F_A|^2 dvol$. This will allow us to obtain information about the critical points, i.e. Yang-Mills solutions, from the algebraic topology of the domain of $\mathcal{E}$, which is either $\mathcal{A}/\mathcal{G}$ or simply $\mathcal{A}$ with the note that $\mathcal{E}$ is invariant under the $\mathcal{G}$ action. Because $\mathcal{A}$ is contractible and the action of $\mathcal{G}$ is almost free, the space $\mathcal{A}/\mathcal{G}$ is similar to $B\mathcal{G}$. Thus a first step is to understand the topology of $B\mathcal{G}$.

Let $M$ be a closed, oriented manifold (which later we will specialize to be of dimension $2$). Let $P$ be a principal $U(n)$-bundle over $M$; that is, the frame bundle of a rank $n$ vector bundle $E$ over $M$. Let $\mathcal{G}$ denote the gauge group of $P$ (sometimes denoted $\mathcal{G}_P$ for clarity). We study the classifying space $B\mathcal{G}$.

\begin{prop} (Last Time) $B\mathcal{G} \simeq [M,BG]_P$ \end{prop}

Note that we use $[X,Y]$ to denote the space of maps from $X$ to $Y$, not (as usual) the set of homotopy classes of maps from $X$ to $Y$, which is for us $H_0([X,Y])$, the set of connected components of $[X,Y]$. The subscript $P$ denotes the component of $[M,BG]$ whose elements $\phi$ satsify $\phi^*EG \cong P$ (that is, pull back $EG \rightarrow BG$ to a bundle isomorphic to $P \rightarrow M$). Because $BG$ is the classifying space for $G$, there is a unique component for which this is the case.

Thus we wish to understand the topology of $[M,BG]$. When $BG$ is an Eilenberg-Maclane space $K(\pi,\ell)$, this is given by a theorem of Thom:

\begin{thm} (Thom) $[M,K(\pi,\ell)] \simeq \Pi_{i=0}^\ell K(H^{\ell-i}(M,\pi),i)$ \end{thm}

For example, when $G = U(1)$, we have $BU(1) \simeq K(\mathbb{Z},2) \simeq \mathbb{C} P^\infty$. Thus $[M,BG] \simeq [M,K(\mathbb{Z},2)] \simeq K(H^2(M,\mathbb{Z}),0) \times K(H^1(M,\mathbb{Z}),1)\times K(H^0(M,\mathbb{Z}),2)$

$\simeq H^2(M,\mathbb{Z}) \times (S^1)^{b_1(M)} \times (\mathbb{C} P^\infty)^{b_0(M)}.$

The first term is simply the discrete group $H^2(M,\mathbb{Z})$ itself. The second is a torus of dimension $b_1(M)$ because $H^1(M,\mathbb{Z})$ is always free and $S^1$ is a $K(\mathbb{Z},1)$. Because $\mathbb{C} P^\infty$ is a $K(\mathbb{Z},2)$, we have that the third term is $\mathbb{C} P^\infty$ to the number of path components of $M$.

Thus if we consider a principal $U(1)$ bundle $P \rightarrow M$ over a connected manifold $M$, then $$B\mathcal{G} \simeq [M,BU(1)]_P \simeq (S^1)^{b_1(M)}\times \mathbb{C} P^\infty.$$ This follows from the above with the remark that in Theorem \ref{thom} it is evident that all components are homotopy equivalent and we have simply taken a component from our above description.

\section{Geometric description in the $U(1)$ Case}

When $G = U(1)$, we have that $\mathcal{G} = [M,S^1]$ Endow $M$ with a Riemannian metric. We consider

$\mathcal{G}^h = \left\{ u:M\rightarrow S^1\ |\ d^*(u^{-1} du) = 0\right\}.$

That is, we consider the space of harmonic maps from $M$ to $S^1$.

When the target is $S^1$, harmonic maps are reasonably simply understood. Harmonic is a local condition. Thinking of $S^1$ as $\mathbb{R}/2\pi\mathbb{Z}$, if we locally in an open set $U =subset M$ lift $u|_U$ to a map $\log(u|_U): U \rightarrow \mathbb{R}$, the map $u|_U$ is harmonic if and only if $\log(u|_U)$ is. To see this, note that $\Delta \log(u|_U) = d^*d \log(u|_U) = d^*(u^{-1}du)$. (Alternatively if we think of $S^1$ as the unit disk in $\mathbb{C}$, the lift to $\mathbb{R}$ is a function $f$ such that $u|_U = e^{if}$.)

\begin{prop} $\mathcal{G}^h \hookrightarrow \mathcal{G}$ is a homotopy equivalence. \end{prop}

\begin{proof} We define a deformation retraction. Let $u \in \mathcal{G}$. Note that $u^{-1}du \in i\Omega^1(M)$ is closed as locally it is given by $d \log{u|_U}$. Let $h \in \Omega^1(M)$ be the unique harmonic representative of $[-iu^{-1}du]$. Then there is a unique function $f: M \rightarrow \mathbb{R}$ such that $u^{-1}du = ih + idf$ and $\int_M f \ dvol = 0$.

Note that $f$ is smoothly determined by $u$. To see this, note first that $h$ is smoothly determined by $u$ because we have the composition $u \mapsto u^{-1}du \mapsto [u^{-1}du] \mapsto h$ from $\mathcal{G}$ to $H^1(M;\mathbb{R})$ and then to $\mathcal{H}^1(M)$. To see that $f$ is smoothly determined, we use the short exact sequence $0 \rightarrow \mathbb{R} \rightarrow [M,\mathbb{R}] \rightarrow B^1(M;\mathbb{R}) \rightarrow 0$. This has a splitting given by $\frac{1}{vol(M)}\int_M dvol : [M,\mathbb{R}] \rightarrow \mathbb{R}$ and thus we get $f$ smoothly from $h$ and $u$ by the map $B^1(M;\mathbb{R}) \rightarrow [M,\mathbb{R}]$ taking $-iu^{-1}du - h$ to $f$.

We now consider the retraction $r_t: \mathcal{G} \rightarrow \mathcal{G}$ given by $r_t(u) = ue^{-itf}$. Note that $r_0(u) = u$ and $r_1(u) = ue^{-if}$. We have $$r_1(u)^{-1} d r_1(u) = u^{-1} e^{if} d \left(u e^{-if}\right) = u^{-1} e^{if} \left(e^{-if} du - idf e^{-if}\right) = u^{-1} du - idf = h,$$ as desired. \end{proof}

The space $\mathcal{G}^h$ is more readily understood.

\begin{prop} $\mathcal{G}^h \cong U(1) \times H^1(M,\mathbb{Z})$ as topological groups. \end{prop}

\begin{proof} We have $\Phi: \mathcal{G}^h \rightarrow i\mathcal{H}(M) \cong iH^1(M,\mathbb{R})$ by $u \mapsto u^{-1}du = ih \mapsto i[h] = [u^{-1}du]$. This has image $2\pi i H^1(M;\mathbb{Z})$. To see this, note that this has the same image as $\mathcal{G} \rightarrow iH^1(M,\mathbb{R})$ given by $u \mapsto [u^{-1}du]$. If we have a loop $\gamma: S^1 \rightarrow M$ then $\int_{S^1} \gamma^*(u^{-1}du) = \int_{S^1} \left(u \circ\gamma\right)^*(id\theta) = 2\pi i\ \deg(u|_\gamma)$, so the image lies inside $2\pi iH^1(M;\mathbb{Z})$. To show surjectivity, we obtain an element of $\mathcal{G}$ from an element of $2\pi i H^1(M;\mathbb{Z})$ as follows: We choose a basepoint $m \in M$. Given $\alpha \in 2\pi iH^1(M;\mathbb{Z})$ and $x \in M$, we let $\gamma$ be a path from $m$ to $x$ and let $u(x) = e^{\int_\gamma \alpha}$. This is well-defined because the integral of $\alpha$ around any closed loop is in $2\pi i \mathbb{Z}$.

The kernel of the $\Phi$ is evidently the constant maps. Thus we have a short exact sequence $$0 \rightarrow U(1) \rightarrow \mathcal{G}^h \rightarrow 2\pi i H^1(M,\mathbb{Z}) \rightarrow 0.$$ We have already given in the previous paragraph a splitting of this exact sequence $2\pi i H^1(M,\mathbb{Z}) \rightarrow \mathcal{G}^h$. Thus we have a semi-direct product. The conjugation action of $U(1)$ on $2\pi iH^1(M,\mathbb{R})$ must be trivial, however, because the latter is discrete. Thus we have $\mathcal{G}^h \cong U(1) \times H^1(M,\mathbb{Z})$. \end{proof}

This gives us $B\mathcal{G} \simeq B\mathcal{G}^h \simeq BU(1) \times B(\mathbb{Z}^{b_1(M)}) \simeq \mathbb{C} P^\infty \times (S^1)^{b_1(M)},$ reproducing the result we obtained in the previous section for $U(1)$.

\section{Equivariant Topology and the Borel Construction}

Instead of considering the quotient space $\mathcal{A}/\mathcal{G}$, we consider the equivariant topology of the space $\mathcal{A}$ with its $\mathcal{G}$ action. The idea is to take the homotopy quotient of $\mathcal{A}$ by $\mathcal{G}$. That is, we find a $G$-space homotopy equivalent to $\mathcal{A}$ (via maps respecting the $G$-action) on which the action of $\mathcal{G}$ is free, and take the quotient.

We have the Borel construction. Let $X$ be a topological space with a $G$-action, for $G$ a topological group. Let $X_G = X\times_G EG = (X\times EG)/G$. The action of $G$ on $X\times EG$ is free because it is free on $EG$, and the inclusion of $X$ into $X\times EG$ is a $G$-map which is a homotopy equivalence, because $EG$ is contractible. Note that if $X$ and $EG$ are (e.g. Hilbert) manifolds (as is the case when $G$ is a Lie group), then so is $X_G$.

We define $H_G^*(X) = H^*(X_G)$, the equivariant cohomology. This is much better behaved than $H^*(X/G)$. Equivariant homology and homotopy groups are defined analogously.

We have maps $\alpha: X_G \rightarrow X/G$ and $\beta: X_G \rightarrow BG$ induced by the projections. The fibers of $\alpha$ are $\alpha^{-1}(x) = EG/\textrm{Stab}(x) \simeq B\textrm{Stab}(x)$. The fibers of $\beta$ are simply $X$, so $\beta$ expresses $X_G$ as an $X$ bundle over $BG$.

This gives the equivariant cohomology $H_G^*(X)$ the structure of a bi-module over $H^*(X/G)$ and $H^*(BG)$ via cupping with $\alpha^*$ and with $\beta^*$. We note that when the action of $G$ is free, $X_G \simeq X/G$ and the action of $H^*(BG)$ is trivial.

\section{Equivariant Cohomology of the Gauge Group Action}

We consider the gauge group action of $\mathcal{G}$ on $\mathcal{A}$. We have the fibration $\mathcal{A} \rightarrow \mathcal{A}_\mathcal{G} \rightarrow B\mathcal{G}$ as in the previous section. But note that $\mathcal{A}$ is an affine space and thus contractible. Thus $\mathcal{A}_\mathcal{G} \simeq B\mathcal{G}$ and so we have $H^*_\mathcal{G}(\mathcal{A}) = H^*(B\mathcal{G})$. We wish to compute this. Restricting to the case $M = \Sigma$, a closed, oriented surface, we have that $B\mathcal{G} = [\Sigma, BU(n)]_P$. For $n > 1$, the space $BU(n)$ is not an Eilenberg-Maclane space, but we will see that it has the same rational homotopy type as a product of Eilenberg-Maclane spaces.

The cohomology of $BU(n)$ is generated by the Chern classes: $H^*(BU(n),\mathbb{Z}) \cong \mathbb{Z}[c_1,\ldots, c_n]$, where $c_i$ is of degree $2i$. Thus we have $\mathbf{c} = \prod_{i=1}^n c_i: BU(n) \rightarrow \prod_{i=1}^n K(\mathbb{Z},2i)$. The cohomology of $K(\mathbb{Z},n)$ is complicated over $\mathbb{Z}$, but over $\mathbb{Q}$ it is given by $H^*(K(\mathbb{Z},n), \mathbb{Q}) = \mathbb{Q}[x_n]$, where this is understood as the graded commutative polynomial ring; that is, the usual polynomial ring if $n$ is even and $\mathbb{Q}[x_n]/x_n^2$ if $n$ is odd. Thus $H^*(\prod_{i=1}^n K(\mathbb{Z},2i),\mathbb{Q}) \cong \mathbb{Q}[x_2,x_4,\ldots,x_{2n}]$ and so $\mathbf{c}$ is an isomorphism on rational cohomology.

Thus the map $[\Sigma,BU(n)] \stackrel{\mathbf{c}}{\rightarrow} [\Sigma,\prod_{i=1}^n K(\mathbb{Z},2i)]$ is also an isomorphism on rational cohomology. Letting $g$ be the genus of $\Sigma$, we have $[\Sigma,K(\mathbb{Z},2i)] \simeq K(\mathbb{Z},2i) \times K(\mathbb{Z},2i-1)^{2g}\times K(\mathbb{Z},2i-2)$. Thus $B\mathcal{G}$ is rationally homotopy equivalent to the appropriate product of $K(\mathbb{Z},2k)$'s (leaving off $K(\mathbb{Z},0)$ because $B\mathcal{G} = [\Sigma,BU(n)]_P$, a component of $\Sigma,BU(n)$).

\section{References}