# 10-06 Notes

Our goal is to understand the cohomology of the space $\textrm{Map}(\Sigma,BU(n))$ for a closed surface $\Sigma$. Since we have already shown it to be rationally homotopic to a product of Eilenberg-Maclane spaces, we have two remaining goals:

\begin{enumerate} \item Show that $H^{*}(\mathrm{Map}(\Sigma,BU(n));\mathbb{Z})$ is torsion-free. \item Identify integral generators of $H^{*}(\mathrm{Map}(\Sigma,BU(n));\mathbb{Z})$. \end{enumerate}

\subsection{The case $\Sigma=S^{2}$}

Let us suppose for now that $\Sigma=S^{2}$. For a fixed point $p\in S^{2}$, the evaluation map at $p$ gives a fibration $\xymatrix{\Omega^{2}BU(n)\ar[r] & \mathrm{Map}(S^{2},BU(n))\ar[d]^{\mathrm{ev}_{p}}\\ & BU(n)} $ whose fibers are homotopic to $\Omega U(n)$. From Bott's proof of Bott periodicity we know that the energy functional on $\Omega U(n)$ is a Morse function which only has critical points of even index, so from Morse theory it follows that $\Omega U(n)$ has torsion-free cohomology supported entirely in even dimensions. In fact, this is still true if we replace $U(n)$ with any compact Lie group.

Now, we've already seen that this fibration is rationally a product in cohomology: $H^{*}(\mathrm{Map}(S^{2},BU(n));\mathbb{Q})\cong H^{*}(\Omega^{2}BU(n);\mathbb{Q})\otimes_{\mathbb{Q}}H^{*}(BU(n);\mathbb{Q}).$ But both $\Omega^{2}BU(n)$ and $BU(n)$ are torsion-free, so if the Serre spectral sequence has any nontrivial differentials over $\mathbb{Z}$ then their effect should be visible over $\mathbb{Q}$. Thus the fibration must also be a cohomological product over $\mathbb{Z}$, or $H^{*}(\mathrm{Map}(S^{2},BU(n));\mathbb{Z})\cong H^{*}(\Omega^{2}BU(n);\mathbb{Z})\otimes_{\mathbb{Z}}H^{*}(BU(n);\mathbb{Z}).$

\subsection{The general case}

Let $\Sigma=\Sigma_{g}$ be the surface of genus $g$. The quotient of $\Sigma$ by its $1$-skeleton $\bigvee_{i=1}^{2g}S^{1}$ is the sphere $S^{2}$, giving a fibration $\xymatrix{\Omega^{2}BU(n)\ar[r] & \mathrm{Map}(\Sigma,BU(n))\ar[d]\\ & \mathrm{Map}(\bigvee^{2g}S^{1},BU(n))} $ whose last term is $\left(\prod^{2g}\mathrm{Map}_{*}(S^{1},BU(n))\right)\times BU(n)=\left(\prod^{2g}U(n)\right)\times BU(n)$, where $\mathrm{Map}_{*}$ denotes the space of pointed maps. Once again the base and fiber are torsion-free in cohomology, and the total space is rationally their cohomological product, so we can conclude that it is an integral cohomology product as well.

\begin{rem*} In dimensions higher than $2$ this argument can fail if the fiber is not torsion-free. For example, considering $\mathrm{Map}(S^{4},BU(2))$ we get a fibration with fiber $\Omega^{4}BU(2)\simeq\Omega^{3}U(2)$. Now topologically we have a homeomorphism $S^{1}\times SU(2)\cong U(2)$ sending $(e^{i\theta},A)$ to the matrix $\left(\begin{array}{cc} e^{i\theta}\\ & 1\end{array}\right)A$, and $SU(2)\cong S^{3}$, so we see that $\pi_{1}(\Omega^{3}U(2))=\pi_{4}(S^{1}\times S^{3})=\mathbb{Z}/2\mathbb{Z}$.


\end{rem*}

\subsection{Generators for $H^{*}(B\mathcal{G};\mathbb{Z})$}

In order to find generators of $H^{*}(B\mathcal{G};\mathbb{Z})$, we must find integral generators of $H^{*}(\Omega^{2}BU(n))\cong H^{*}(\Omega U(n))$. For the infinite unitary group $U=\underset{n\to\infty}{\lim}U(n)$, we know that $H^{*}(BU;\mathbb{Z})$ is generated by the Chern classes $c_{1},c_{2},\dots$ of the universal bundle $EU\to BU$. Since the inclusion $U(n)\hookrightarrow U$ induces a map $\Omega^{2}BU(n)\to\Omega^{2}BU\simeq BU,$ where the homotopy equivalence follows from Bott periodicity, we can hope for a similar construction for $\Omega^{2}BU(n)$, but this will require a more detailed look at $BU$.

The space $BU$ can be modeled as the limit of Grassmanians $\underset{n\to\infty}{\lim}\mathrm{Gr}_{n}(\mathbb{C}^{2n})$, but it will be more useful to consider another identification:

\begin{thm*} [Atiyah]\textup{ There is a homotopy equivalence $\mathbb{Z}\times BU\simeq\mathrm{Fred}(H)$, where $\mathrm{Fred}(H)$ is the space of Fredholm operators on the infinite dimensional separable Hilbert space $H$ and the $\mathbb{Z}$ factor captures the indices of these operators.} \end{thm*} \begin{proof} First, we note that $\mathbb{Z}\times BU$ is a classifying space for complex K-theory, i.e. $K(X)\cong[X,BU]$ where $K(X)$ is the Grothendieck group of vector bundles over a compact space $X$. Given a map $f:X\to\mathrm{Fred}(H)$, and a point $x_{0}\in X$ with neighborhood $\mathcal{U}$ small enough so that the orthogonal projection $\mathrm{Im}(f(x_{0}))^{\perp}\to\mathrm{Im}(f(x))^{\perp}$ is injective for all $x\in\mathcal{U}$, we can construct a virtual bundle as follows: Choose a finite cover $\{\mathcal{U}_{i}\}$ of $X$, together with points $x_{i}\in\mathcal{U}_{i}$ having this same property, and let $\{\varphi_{i}\}$ be a partition of unity subordinate to $\{\mathcal{U}_{i}\}$. We can define a map \begin{eqnarray*} \bar{f}:H\oplus\left(\bigoplus_{i=1}^{n}\mathrm{Im}(f(x_{i}))^{\perp}\right) & \to & H\\ (h,h_{1},\dots,h_{n}) & \mapsto & f(x)\cdot h+\sum_{i=1}^{n}\varphi_{i}(x)\cdot h_{i}\end{eqnarray*} which is now surjective, and associate to $f$ the difference $\ker(\bar{f})-\oplus_{i=1}^{n}\mathrm{Im}(f(x_{i}))^{\perp}\in K(X)$. We leave it as an exercise to check that this gives a well-defined map $[X,\mathrm{Fred}(H)]\to K(X)$.

Conversely, we need to associate to each virtual bundle $[V]-[W]\in K(X)$ a map $f:X\to\mathrm{Fred}(H)$. The bundle $[V]-[W]$ can be written equivalently as $[V']-\underline{\mathbb{C}}^{n}$, where $\underline{\mathbb{C}}$ denotes a trivial line bundle, so we construct a map $f:V'\oplus H\to\underline{\mathbb{C}}^{n}\oplus H$ defined by $f(v,h)=(0,h)$. This clearly has kernel $V'$ and cokernel $\underline{\mathbb{C}}^{n}$. But by Kuiper's theorem, which states that $U(H)$ is contractible, the bundle $V'\oplus H$ is a trivial bundle over $H$ and so $V'\oplus H\cong X\times H$. In particular we can express $f$ equivalently as a Fredholm operator on $H$, parametrized by $x\in X$, as desired. \end{proof} Having described a model for $BU$, we would like to explicitly describe an identification $[X,\Omega^{2}BU]\simeq[X,BU]$; note that $[X,\Omega^{2}BU]\simeq[S^{2}\wedge X,BU]$. In fact, there is a natural map $[S^{2}\wedge X,BU]\to[X,BU]$: A given map $f:S^{2}\wedge X\to BU$ corresponds to a vector bundle $V-\underline{\mathbb{C}}^{n}\to S^{2}\wedge X$. Constructing a family of $\bar{\partial}$ operators on the $S^{2}$ factor at each point, $\bar{\partial}:\Omega^{0,0}(V|_{S^{2}\times\{x\}})\to\Omega^{0,1}(V|_{S^{2}\times\{x\}}),$ gives us the desired map $X\to\mathrm{Fred}(H)$:

\begin{thm*} This construction gives an isomorphism $[X,\Omega^{2}BU]\simeq[S^{2}\wedge X,BU]\stackrel{\sim}{\longrightarrow}[X,BU].$

\end{thm*} The evaluation map $\mathrm{ev}:S^{2}\times\mathrm{Map}_{*}(S^{2},BU(n))\to BU(n)$ pulls back to a bundle $V=\mathrm{ev}^{*}(EU(n))$: $\xymatrix{\mathrm{ev}^{*}(EU(n))\ar[r]\ar[d] & EU(n)\ar[d]\\ S^{2}\times\mathrm{Map}_{*}(S^{2},BU(n))\ar[r] & BU(n)}$ and if we identify $V$ with the corresponding $\sigma\in K(\mathrm{Map}_{*}(S^{2},BU(n)))$, the Chern classes $c_{i}(\sigma)$ will be integral generators of $H^{*}(\Omega^{2}BU(n))$ as desired.

Similarly, given a vector bundle $E\to\Sigma$ and a connection $A\in\mathcal{A}$ on $\Sigma$, we have a family of Fredholm operators $\mathcal{A}\to\mathrm{Fred}(H)$ of the form $A\mapsto(\bar{\partial}_{A}:\Omega^{0,0}(E)\to\Omega^{0,1}(E)).$ This family is $\mathcal{G}$-equivariant in the sense that $\bar{\partial}_{g\cdot A}=g^{-1}\circ\bar{\partial}_{A}\circ g$, so it corresponds to an element of the equivariant K-theory $K_{\mathcal{G}}(\mathcal{A})$, whose Chern classes $c_{i}$ are generators of $H_{\mathcal{G}}^{2i}(\mathcal{A})\cong H^{2i}(B\mathcal{G})$.

\subsection{Minima of the Yang-Mills functional}

Ideally, we would like the minima of the Yang-Mills functional to be free orbits of $\mathcal{G}$. Recall that in general, if $d_{A}*F_{A}=0$ then we can write $E$ as a sum $\bigoplus_{\lambda}E_{\lambda}$, where up to a constant factor the $\lambda$ are the eigenvalues of $*F_{A}$ and we have $\lambda=\frac{c_{1}(E_{\lambda})}{\mathrm{rk}(E_{\lambda})}$. The Yang-Mills functional in this case satisfies (again up to a constant) $\mathcal{E}(A)=\sum_{\lambda}\left(\frac{c_{1}(E_{\lambda})}{\mathrm{rk}(E_{\lambda})}\right)^{2}\cdot\mathrm{rk}(E_{\lambda})=\sum_{\lambda}\frac{c_{1}(E_{\lambda})^{2}}{\mathrm{rk}(E_{\lambda})},$ and so to minimize $\mathcal{E}(A)$ when $n=c_{1}(E)=\sum_{\lambda}c_{1}(E_{\lambda})$ is fixed, we need each of the $c_{1}(E_{\lambda})$ to be as close to equal as possible.

\begin{example*} Consider $U(2)$ bundles over $\Sigma$. Either $E=E_{\lambda}$ does not split, so $\mathcal{E}(A)=\frac{n^{2}}{2}$, or $E=L_{1}\oplus L_{2}$ is a sum of line bundles with Chern classes $n_{1}$ and $n_{2}$ and then $\mathcal{E}(A)=n_{1}^{2}+n_{2}^{2}$. If $n$ is odd, then $n_{1}\not=n_{2}$ and so $n_{1}^{2}+n_{2}^{2}>\frac{n^{2}}{2}$, meaning that we cannot have minima which are direct sums of line bundles. On the other hand, if $n$ is even and $|n_{1}|=|n_{2}|=\frac{|n|}{2}$ then minima can be direct sums. This means that when $n$ is odd, the minima $A$ all satisfy $\mathrm{Stab}(A)=U(1)$, consisting of constant gauge transforms, but for $n$ even the stabilizer can jump from $U(1)$ to $U(1)\times U(1)$ among the minimal action solutions. \end{example*} In the next lecture, we'll continue this analysis for higher rank bundles and see that the analogous condition is for $c_{1}(E)$ and $\mathrm{rk}(E)$ to be coprime.