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.
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. ◻
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 \(E_{\lambda}\):
\[\left[\frac{i}{2\pi} \textrm{tr} F_A \right] = c_1(E_{\lambda})_{\mathbb{R}} \in H^2(\Sigma,\mathbb{R}).\]
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
\(EG\) is contractible;
\(G\) acts freely on \(EG\);
\(BG \cong EG/G\);
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;
\(BG\) is unique up to homotopy.
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, \(\pi_0 (L^p_k(X,Y)) \cong \pi_0 (C^0_{\textrm{CO}}(X,Y))\).
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. ◻
Remarks.
\([M,BG]=\bigcup_{[P]} [M,BG]_P\) is the union of the connected components of \([M,BG]\), where \([P]\) denote the isomorphism classes of principal bundles.
\(\pi_n(BG)=\pi_{n-1}(G)\) by the homotopy long exact sequence for the fibration \(G\to EG \to BG\).
Example \(G=U(1)\) is a \(K(\mathbb{Z},1)\), therefore \(BG = K(\mathbb{Z},2) = \mathbb{C}P^{\infty}\) by the previous remark. We want to understand \([M,BU(1)]=[M,K(\mathbb{Z},2)]\) 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.
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 \(BK(H^{k-1-q}(M,\pi),q)=K(H^{k-(q+1)}(M,\pi),q+1)\). ◻
Example: \([M,K(\mathbb{Z},2)]\)
For compact connected \(M\),
\[\begin{array}{lll} [M,K(\mathbb{Z},2)] & \simeq & K(H^2(M,\mathbb{Z}),0) \times K(H^1(M,\mathbb{Z}),1) \times K(H^0(M,\mathbb{Z}),2) \\ & \simeq & H^2(M,\mathbb{Z}) \times (S^1)^{b_1(M)} \times \mathbb{C}P^{\infty} \end{array}\]
where \(H^2(M,\mathbb{Z})\) 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 \(i\Omega^1(M,\mathbb{R})\). 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 \(a\in \Omega^1(M,\mathbb{R})\).
CLAIM. There is a \(u\in \mathcal{G}\) such that \(u^*A \in S_{A_0}\).
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 \(d^*\tilde{a}=0\). ◻
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 \([M,K(\mathbb{Z},2)]\).
