# 09-29 Notes

## Review of Yang-Mills functional

$P\to M$ unitary frame bundle of a vector bundle $E\to M$.

Yang-Mills functional: $\mathcal{E}(A)=\int_M |F_A|^2=\int_{M} -\textrm{tr}(F_A \wedge *F_A)$.

Euler-Lagrange equations for $\mathcal{E}$: $d_A^*F_A=0$.

First Bianchi identity: $d_A F_A=0$.

So $F_A$ is a harmonic two-form.

## Yang-Mills functional in dimension 2

From now on $n=2$ and $M=\Sigma$. The Euler-Lagrange equations translate to:

$d_A^*F_A=0 \iff d_A *F_A=0 \iff *F_A$ is a parallel endomorphism.

LEMMA. The eigenvalues of a parallel endomorphism $\xi$ are constant. \begin{proof} For each $m\geq 0$, consider the trace $\textrm{tr}(\xi^m)\in C^{\infty}(\Sigma)$ of $\xi^m\in \Gamma(\Sigma,\textrm{End}\, E)$. Since $A$ is a metric connection, $d\, \textrm{tr}(\xi^m) = \textrm{tr}(\sum_{i=0}^{m-1} \xi^i (d_A \xi) \xi^{m-i-1})$ which vanishes since $d_A \xi=0$. Therefore $\textrm{tr}(\xi^m)$ is constant, that is the sum of all $m-$th powers of the eigenvalues is constant. The elementary symmetric polynomials in the eigenvalues $\lambda_j$ are algebraically generated by the sums $\sum_j \lambda_j^m$, therefore the coefficients of the characteristic polynomial of $\xi$ are constant. Hence all eigenvalues are constant. \end{proof}

Let $A$ be a Yang-Mills connection on $E \to \Sigma$. The lemma applied to $\xi=*F_A$ shows that $E$ splits as a sum of eigenvector bundles,

$$E = \bigoplus_{i\lambda \in \textrm{Spec}(*F_A)} E_{\lambda},$$

where $*F_A|_{E_\lambda} = i \lambda$.

Question. What $\lambda$ values can we have?

Answer. $\lambda$ only depends on the topology of $E_{\lambda}$:

$$\left[\frac{i}{2\pi} \textrm{tr} F_A \right] = c_1(E_{\lambda})_{\R} \in H^2(\Sigma,\R).$$

On $E_{\lambda}$, $*F_A=i\lambda \, \textrm{Id}$, so $F_A= i \lambda \, \textrm{Id} *1$, so $\textrm{tr} F_A = i\lambda \,\textrm{rk} (E_{\lambda})*1$. Thus the eigenvalues $\lambda$ can be recovered from topological data:

$$<c_1(E_{\lambda}),[\Sigma]> = -\frac{\lambda}{2\pi} \textrm{rk}(E_{\lambda}) \textrm{vol}(\Sigma).$$

The Yang-Mills functional is also topologically determined:

$$\begin{array}{lll} \mathcal{E}(A) & = & \displaystyle \int -\textrm{tr}(F_A \wedge *F_A)\\ & = & \displaystyle \int - \textrm{tr}((i\lambda\, \textrm{Id}_{E_{\lambda}} • 1) (i\lambda\, \textrm{Id}_{E_{\lambda}})) \\ & = & \displaystyle \sum \lambda^2 \,\textrm{rk}(E_{\lambda})\\ & = & \displaystyle (2\pi)^2 \sum \frac{c_1(E_{\lambda})^2}{\textrm{rk}(E_{\lambda})} \end{array}$$

where we abbreviated $c_1(E_\lambda)=<c_1(E_{\lambda}),[\Sigma]>$ and we normalized $\textrm{vol}(\Sigma)=1$.

## The Classifying Space of $\mathcal{G}$

Idea 1. The Morse theory for the functional $\mathcal{E}$ relates the topology of $\mathcal{A}$ and $\mathcal{G}$.

Idea 2. $\mathcal{A}$ is contractible since it's an affine space, $\mathcal{G}$ almost acts freely, therefore $\mathcal{B}=\mathcal{A}/\mathcal{G}$ is almost the classifying space $B\mathcal{G}$ for $\mathcal{G}$.

For a topological group $G$ there exists a classifying space $BG$ together with a principal $G-$bundle $EG \to BG$, such that \begin{enumerate} \item $EG$ is contractible;

\item $G$ acts freely on $EG$;

\item $BG \cong EG/G$;

\item every principal $G-$bundle $P \to X$ is isomorphic to the pullback of $EG$ via a map $X\to BG$, called classifying map, which is unique up to homotopy;

\item $BG$ is unique up to homotopy. \end{enumerate}

THEOREM (Atiyah-Bott) Let $P\to M$ be a principal $G-$bundle with group of gauge transformations $\mathcal{G}_P.$ Then there is a homotopy equivalence $$B\mathcal{G}_P \simeq [M,BG]_P.$$

Remark. Here $[M,BG]_P$ is the collection of classifying maps $M\to BG$ for $P$. Convention: $[X,Y]$ is the space of maps $X\to Y$, not the homotopy classes of such maps.

Remark. As long as the maps we consider are continuous, the algebraic topology is the same no matter what class of maps $M\to BG$ we are looking at (smooth maps, Sobolev maps, etc.). The idea is that by applying a heat flow on the space of maps or on families of maps, these can be turned into smooth maps/families (one could also try to use convolutions to smoothen on patches, but it's messy). A useful general statement along these lines:

Theorem (Palais, Foundations of Global Non-linear Analysis, W. A. Benjamin 1968) Above the Sobolev borderline, so $L_k^p(X,Y)$ is a subset of $C^0(X,Y)$, there is a homotopy equivalence

$$L^p_k(X,Y) \simeq C^0_{\textrm{CO}}(X,Y)$$

with the space of continuous maps with respect to the compact-open topology. Borderline case: the components are the same, $\pi_0 (L^p_k(X,Y)) \cong \pi_0 (C^0_{\textrm{CO}}(X,Y))$.

\begin{proof}(Sketch) Let $E\mathcal{G}_P = [P,EG]^G$ be the collection of $G-$equivariant maps $P\to EG$. We will show that $E\mathcal{G}_P$ is contractible.

Suppose first that $P=M\times G$ is trivial. Then $[P,EG]^G = [M,EG]$ is contractible since $EG$ is.

In the general case, we assume $M$ is compact. Pick a metric on $M$. Cover $M$ by finitely many balls $B(x_i,r_i)$, such that also $B(x_i,r_i/2)$ would cover $M$. We now show that we can contract the maps $P\to EG$ down to a constant on each ball, one at a time.

By the trivial case, $[P|_{B(x_1,r_1)},EG]^G$ is contractible so let $r_t$ be a contraction. Pick $\varphi: B(x_1,r_1) \to [0,1]$ with

$$\begin{array}{ll} \varphi(B(x_1,r_1/2))=1\\ \varphi(\partial B(x_1,r_1))=0. \end{array}$$ Given $f:P \to EG$, we can contract $f$ down to a constant on $B(x_1,r_1/2)$ without changing $f$ outside $B(x_1,r_1)$:

$$\tilde{r}_t f(p) = \left\{ \begin{array}{ll} f(p) & \textrm{if } p \notin B(x_1,r_1)\\ r_{t\varphi(\pi p)} \circ f(p) & \textrm{if } p\in B(x_1,r_1) \end{array}\right.$$

Proceed to $B(x_2,r_2)$ and repeat the procedure for the map $\tilde{r}_1f:P \to EG$. Since the $B(x_i,r_i/2)$ cover $M$, at the end we get a constant map to the basepoint. \end{proof}

Remarks. \begin{enumerate} \item $[M,BG]=\bigcup_{[P]} [M,BG]_P$ is the union of the connected components of $[M,BG]$, where $[P]$ denote the isomorphism classes of principal bundles.

\item $\pi_n(BG)=\pi_{n-1}(G)$ by the homotopy long exact sequence for the fibration $G\to EG \to BG$. \end{enumerate}

Example $G=U(1)$ is a $K(\Z,1)$, therefore $BG = K(\Z,2) = \C P^{\infty}$ by the previous remark. We want to understand $[M,BU(1)]=[M,K(\Z,2)]$ better.

THEOREM (Thom) Let $M$ be a CW complex. Then

$$[M,K(\pi,k)] = \prod_{q=0}^k K(H^{k-q}(M,\pi),q).$$

Remark. Recall that the reduced suspension $\Sigma M = S^1 \wedge M$ is left-adjoint to $\Omega$ for based maps: $[\Sigma M, K]_* \cong [M,\Omega K]_*$. Also recall that $H^q(M,\pi)$ can be computed as the homotopy classes of maps $M \to K(\pi,q)$. Let $K_q= K(\pi,q)$ then one can show that the homotopy groups of the two sides of the theorem are isomorphic:

$$\begin{array}{l} \pi_0 [M,K_k] \cong H^k(M,\pi)\\ \pi_1 [M,K_k] \cong \pi_0 [\Sigma M,K_k] \cong H^k(\Sigma M,\pi) \cong H^{k-1}(M,\pi)\\ \pi_q [M,K_k] \cong \pi_0 [\Sigma^q M,K_k] \cong H^k(\Sigma^q M,\pi) \cong H^{k-q}(M,\pi) \end{array}$$

So by Whitehead's theorem, it would be sufficient to find a map between the two spaces in the theorem which induces an isomorphism on homotopy groups.

\begin{proof} Let $K_q=K(\pi,q)$. Recall that the path-space fibration $\Omega K_q \to P K_q \to K_q$ gives a classifying space $BK_{q-1}=K_q$ for principal $(K_{q-1}=\Omega K_q)$-bundles. Thus

$$[M,K_k] = [M,BK_{k-1}] = \bigcup_{[P]} [M,BK_{k-1}]_P,$$

where $P$ are the principal $K_{k-1}$-bundles.

By the previous theorem, $[M,BK_{k-1}]_P = B\mathcal{G}_P$. We need to understand $\mathcal{G}_P$.

Because $\pi$ is abelian and $BK_k=K_{k+1}$, it is possible to inductively construct the $K_k$ as abelian groups. Therefore the gauge group is

$$\mathcal{G}_P = [P,Ad(K_{k-1})]^{K_{k-1}} = [M,K_{k-1}].$$

The $[P]$ are indexed by $\pi_0[M,K_k] =H^k(M,\pi)=K(H^k(M,\pi),0)$, and by induction on $k$ we may assume the result for $k-1$:

$$\mathcal{G}_P =[M,K_{k-1}] = \prod_{q=0}^{k-1} K(H^{k-1-q}(M,\pi),q).$$

Therefore, $[M,K_k] = K(H^k(M,\pi),0) \times B \mathcal{G}_P$ becomes

$$[M,K_k] = K(H^k(M,\pi),0) \times \prod_{q=0}^{k-1} BK(H^{k-1-q}(M,\pi),q).$$

The result follows on using $BK(H^{k-1-q}(M,\pi),q)=K(H^{k-(q+1)}(M,\pi),q+1)$. \end{proof}

## Example: $[M,K(\Z,2)]$

For compact connected $M$,

$$\begin{array}{lll} [M,K(\Z,2)] & \simeq & K(H^2(M,\Z),0) \times K(H^1(M,\Z),1) \times K(H^0(M,\Z),2) \\ & \simeq & H^2(M,\Z) \times (S^1)^{b_1(M)} \times \C P^{\infty} \end{array}$$

where $H^2(M,\Z)$ classifies the isomorphism classes of complex line bundles on $M$.

The space of connections $\mathcal{A}_L$ for a line bundle $L\to M$ is an affine space for $i\Omega^1(M,\R)$. The gauge group $\mathcal{G} = [M,S^1]$ does not act freely on $\mathcal{A}_L$ since the constant maps fix the connection.

Fix a basepoint $x_0 \in M$. Let $\mathcal{G}^0 = \{ u \in \mathcal{G}: u(x_0)=1 \}$. The fibration $\mathcal{G}^0 \to \mathcal{G} \stackrel{ev}{\to} S^1$ comes with a section which gives the constant maps.

Now $\mathcal{G}^0$ acts freely on connections, therefore

$$E\mathcal{G}^0 = \mathcal{A}_L \qquad\textrm{and}\qquad B\mathcal{G}^0 = \mathcal{A}_L/\mathcal{G}^0.$$

Consider a connection $A=A_0 + a$, where $a\in \Omega^1(M,\R)$.

CLAIM. There is a $u\in \mathcal{G}$ such that $u^*A \in S_{A_0}$.

\begin{proof} By the Hodge decomposition,

$$a = i(df + \textrm{harmonic} + d^*\textrm{two-form})$$

Let $\tilde{a}=a-idf$. Try $u=e^{-if}$:

$$u^*A = A+u^{-1}du = A_0 + a + u^{-1}du = A_0 + \tilde{a}$$ Thus $u^*A \in S_{A_0}$ since $d^*\tilde{a}=0$.\end{proof}

Therefore the orbit of $A$ always hits the slice, so $\mathcal{A}/\mathcal{G} = S_{A_0}/\tilde{\mathcal{G}}$ where

$$\tilde{\mathcal{G}}=\{ u: u^*S_{A_0} = S_{A_0} \}.$$

The orbit of $A=A_0+\tilde{a} \in S_{A_0}$ under $u \in \tilde{\mathcal{G}}$ gives $u^*(A_0+\tilde{a}) = A_0 + \tilde{a} + u^{-1}du \in S_{A_0}$. This means precisely that

$$d^*(u^{-1}du)=0$$

or equivalently, since $u^{-1}du = d\log u$, the local function $\log u$ is harmonic.

Idea. these harmonic maps $M \to S^1$ give the unique harmonic representatives of homotopy classes $M \to S^1$. The latter calculates $H^1(M)$ and these harmonic maps are responsible for the torus part $(S^1)^{b_1(M)}$ in the above description of $[M,K(\Z,2)]$.