09-10 Notes
Let \(X\) be a compact Riemanian manifold, and \(P\) be a principal bundle over \(X\) with structure group \(G\), associated to a vector bundle \(E \rightarrow X\). We denote the space of \(L_k^p\) connections by \(\mathcal{A}_k^p\), more precisely, \(\mathcal{A}_k^p\) = \(\{\) A connection \(|\) if \(\tau: \pi^{-1}(U) \rightarrow U \times P\) is a \(C^\infty\) trivialization, \(\varphi \in C_o^\infty(U)\), and \(a_\tau \in \Omega^1(U,\mathfrak{g})\), then \(\varphi\alpha_\tau \in L_k^p(U, T^*U \otimes \mathfrak{g}) \}\). We denote the \(L_{k+1}^p\) gauge transformations by \(\mathcal{G}_{k+1}^p\), more precisely \(\mathcal{G}_{k+1}^p\) = \(\{ u \in L_{k+1}^p\)(\(X\), End \(E\)) \(| u^*u = 1\) a.e. \(\}\). Last time, we noted that we have a smooth action \(\mathcal{G}_{k+1}^p \times \mathcal{A}_k^p \rightarrow \mathcal{A}_k^p\) provided \(p(k+1) > n\), we denote the quotient of this action by \(\mathcal{B}_k^p\) = \(\mathcal{A}_k^p / \mathcal{G}_{k+1}^p\).
In light of the Sobolev Embedding theorems, we can ask the following question: does an embedding \(\mathcal{A}_k^p \hookrightarrow \mathcal{A}_l^q\) yield a map \(\mathcal{B}_k^p \hookrightarrow \mathcal{B}_l^q\)?
Proposition 1. Suppose \(u \in \mathcal{G}_0^\infty = \{ u \in L^\infty(X, End E) | u^*u = 1\) a.e. \(\}\) and \(\exists A, \tilde{A} \in \mathcal{A}_k^p\) such that \(u\cdot A = \tilde{A}\), then \(u \in \mathcal{G}_{k+1}^p\).
Proof. We assume \(u \in \mathcal{G}_0^\infty\) because that is the weakest possible restriction. In a local trivialization, we have \[u\cdot A = \tilde{A} \Rightarrow u^{-1}du + u^{-1}a^{\tau}u = \tilde{a}^\tau \Rightarrow du = u\tilde{a}^\tau - a^\tau u\] If \(\tilde{a}^\tau, a^\tau \in L^p\) then the above implies that \[du \in L^p \Rightarrow u \in L_1^p .\] Next, if we have \(\tilde{a}^\tau, a^\tau \in L_1^p\) then this implies, by the Leibnitz rule, \[\nabla du = u \nabla \tilde{a}^\tau - \nabla a^\tau u + \nabla u \tilde{a}^\tau - a^\tau \nabla u\] The first two terms in the above, \(u \nabla \tilde{a}^\tau, \nabla a^\tau u \in L^p\). For the third and fourth terms, we know that \(\nabla u \in L^p\). We also have \(\tilde{a}^\tau, a^\tau \in L_1^p,\) but by the Sobolev embedding theorems, we have \(L_1^p \hookrightarrow L^q\) for some q. This implies that \(\nabla u \tilde{a}^\tau, a^\tau \nabla u \in L^n,\) where \(\frac{1}{n} = \frac{1}{p} + \frac{1}{q}.\) If \(p \leq n\), then \(L^n \hookrightarrow L^p\), which means \(\nabla du \in L^p\) and we are done. If \(p > n\), then \(L^p \hookrightarrow L^n\), and so \(\nabla du \in L^n .\) Then: \[\nabla du \in L^n \Rightarrow u \in L_2^n \Rightarrow \nabla u \in L_1^n \hookrightarrow L^r \forall r \Rightarrow \nabla u \tilde{a}^\tau, a^\tau \nabla u \in L^p \Rightarrow \nabla du \in L^p .\] We repeat this "bootstrapping" argument and obtain \(u \in \mathcal{G}^p_{k+1}\), for any k we desire. ◻
Recall that if \(p(k+1) > n\) then \(\mathcal{B}_k^p\) is Hausdorff. Consider the borderline case, with \(\mathcal{A}_0^n, \mathcal{G}_1^n,\) in this case \(\mathcal{G}_1^n\) is a group, but not a topological group.
Proposition 2. The \(G_1^n\) orbits are weakly (hence strogly) closed in \(A_0^n.\)
Proof. Suppose \(A \in \mathcal{A}_0^n\) and \(u_i \in \mathcal{G}_1^n\) such that \(u_i \cdot A \rightharpoonup \tilde{A}\) in \(L^n\), then we want to show that \(\exists u \in \mathcal{G}_1^n\) such that \(u \cdot A = \tilde{A}\). (Note: \(\rightharpoonup\) denotes weak convergence.) Locally, for a trivialization \(\tau\), we have: \(u_i^{-1}du_i + u_1^{-1}a^{\tau}u_i = \tilde{a}_i^{\tau}\) and by assumption \(\tilde{a}_i^{\tau} \stackrel{L^n}\rightharpoonup a^{\tau}\). Then we have \(du_i = u_i\tilde{a}_i^{\tau} - a^{\tau}u_i\), and weak convergence in \(L^n\) implies that the right hand side is uniformly bounded in \(L^n\), so in particular, \(u_i\) is uniformly bounded in \(L_1^n\). This implies that we can find a subsequence such that \(u_{i_k} \stackrel{L^n_1}\rightharpoonup u\). From the Sobolev Embedding Theorem, we have that \(L_1^n \hookrightarrow L^p\) for all \(p\), so we have a subsequence \(u_i \stackrel{L^p}\rightarrow u\), which implies \(u_i \rightarrow u\) a.e., thus \(u^*u = 1\) a.e. and so \(u \in \mathcal{G}_1^n\). Now, in the equation \(du_i = u_i\tilde{a}_i^{\tau} - a^{\tau}u_i\), we have \(du_i \rightharpoonup du\), and \(a^{\tau}u_i \rightharpoonup a^{\tau}u\) (because \(u_i \rightharpoonup u\)). We want to show that \(u\tilde{a}^{\tau} = \tilde{\tilde{a}}\) in \(L^{n-\epsilon}\) for small \(\epsilon\), where \(\tilde{\tilde{a}}\) is such that we have a subsequence \(u_{i}\tilde{a}_i^{\tau} \rightharpoonup \tilde{\tilde{a}}\). Since we have \(u_i \stackrel{L^p}\rightarrow u\) for all \(p\), we can find a large \(N\) for which \(u_i \in L^N\) and \(\tilde{a}_i^{\tau} \in L^n\), which would imply by the Lemma below that \(u_i\tilde{a}_i^{\tau} \rightharpoonup u\tilde{a}^{\tau} = \tilde{\tilde{a}}\) in \(L^{n-\epsilon}\). ◻
Lemma 1. If \(a_i \in L^p, b_i \in L^q, a_i \stackrel{L^p}\rightharpoonup a, b_i \stackrel{L^q}\rightharpoonup b\), then \(a_ib_i \stackrel{L^r}\rightharpoonup ab\) where \(\frac{1}{r}=\frac{1}{p} + \frac{1}{q}\).
We will now use the above proposition to prove the following:
Proposition 3. \(\mathcal{B}_0^n\) is a metric space (and hence Hausdorff).
Proof. Define a metric \[d([A],[\tilde{A}])= \inf_{u, \tilde{u} \in \mathcal{G}_1^n} ||uA - \tilde{u}\tilde{A}||_{L^n(X)}\]
Since we can rewrite \(uA-\tilde{u}\tilde{A} = u(A-u^{-1}\tilde{u}A)\), so we have the following \(\left\|{uA - \tilde{u}\tilde{A}}\right\|_{L^n(X)} = ||u(A-u^{-1}\tilde{u}A)||_{L^n(X)} = ||A-u^{-1}\tilde{u}A||_{L^n(X)}\). Hence we can rewrite the metric as follows: \[d([A],[\tilde{A}])= \inf_{u \in \mathcal{G}_1^n} ||A - u\tilde{A}||_{L^n(X)}\]
We check that \(d([A],[\tilde{A}])\) satisfies the conditions of a metric space:
Symmetry. Clearly, \[d([A],[\tilde{A}]) = \inf_{u \in \mathcal{G}_1^n} ||A - u\tilde{A}||_{L^n(X)} = \inf_{u \in \mathcal{G}_1^n} ||\tilde{A} - uA||_{L^n(X)} = d([\tilde{A}],[A])\]
Triangle inequality. In the \(L^n\) metric, we have: \[||A-u\tilde{A}||_{L^n} \leq ||A-v\tilde{\tilde{A}}||_{L^n} + ||v\tilde{\tilde{A}}-u\tilde{A}||_{L^n}\] We take \(\inf_{u \in \mathcal{G}_1^n}\) of both sides, and get; \[d([A],[\tilde{A}]) \leq d([A], [\tilde{\tilde{A}}]) + d([\tilde{\tilde{A}}],[A])\]
\(d([A],[\tilde{A}]) = 0 \Rightarrow [A]= [\tilde{A}]\). If \(d([A],[\tilde{A}]) = 0\), then there exists a sequence \(u_i\) such that \(\lim_{i \rightarrow \infty} ||u_i A - \tilde{A}||_{L^n} = 0\). So the points \(u-i A\) lie in some \(L^n\) ball about \(\tilde{A}\), this implies that, after possibly passing to some subsequence, \(u_i A \rightharpoonup \tilde{\tilde{A}}\) for some \(\tilde{\tilde{A}}\). However, by the previous proposition, \(\mathcal{G}_i^n\) orbits are weakly closed in \(\mathcal{A}_0^n\), so \(\tilde{\tilde{A}} = u\tilde{A}\) for some \(u \in \mathcal{G}_i^n\). Since we have a subsequence \(u_i\) for which \(\lim_{i \rightarrow \infty} ||u_i A - \tilde{A}||_{L^n} = 0\), we have \(\tilde{\tilde{A}} = \tilde{A} = uA\). Hence, \(A\) and \(\tilde{A}\) are in same \(\mathcal{G}_1^n\) orbit, so \([A]=[\tilde{A}]\).
◻
Before we try to understand the the orbits of the gauge group action and what the quotient spaces \(\mathcal{B}_k^p\) = \(\mathcal{A}_k^p / \mathcal{G}_{k+1}^p\) look like, we will describe the local structure of \(X/G\) where \(X\) is a smooth finite dimensional space, and \(G\) is a compact Lie group acting smoothly on \(X\). First a simple example:
Example 1. \(\mathbb{C}/ S^1 = \mathbb{R}^+\) and \(\mathbb{C}^n / (S^1)^n = (\mathbb{R}^+)^n\) but with some singularity, so we ask: how bad is this singularity?
In general, for \(x \in X\) we want to investigate the neighborhood of the orbit \(\O_{x_0} = Gx_0\). As usual, we denote the stabilizer of the action to be \(Stab_{x_0} = \{g \in G | gx_0 = x_0 \}\) (note: this is actually a closed Lie subgroup of G), and then the orbit is \(\O_{x_0} = G/{Stab_{x_0}}\), which is a maniold in its own right. We take a \(G\)-invariant Riemannian metric on \(X\), then the tangent space of \(X\) at \(x_0\) is \(T_{x_0}=T\O_{x_0} \oplus (T\O_{x_0})^{\bot}\), where \(^\bot\) denotes the orthoganal complement with respect to the chosen Riemannian metric. The map \(exp_{x_0} : T_{x_0} \rightarrow X\) allows us to define a slice \(\S_{x_0}\) of the action: where \(\S_{x_0}\) is a submanifold that is the embeddeding of a \(Stab_{x_0}\)-invariant neighborhood of \(O \in (T\O_{x_0})^{\bot}\) into \(X\).
Proposition 4. The map \(\S_{x_0} \times G \stackrel{i}{\rightarrow}X\), given by \((x,g) \mapsto xg\) and the \(Stab_{x_0}\) action on \(\S_{x_0} \times G\), given by \(h \cdot (x,g) \mapsto (xh, h^{-1}g)\), yields a map \(\bar{i} :(\S_{x_0} \times G)/Stab_{x_0} \rightarrow X\) which (after possibly shrinking \(\S_{x_0}\)) is a diffeomorphism of onto a neighborhood of \(\O_{x_0}\).
The proposition above basically says that a neighborhood of \([x_0] \in X/G\) is homeomorhpic to \(\S_{x_0}/Stab_{x_0}\).
To apply the above discussion to the setting of the gauge group acting on the space of connections, we first must determine what is the stabilizer of a given connection \(A\) in \(\mathcal{G}\). In terms of the covariant derivative, \(u \in Stab_A\) if: \(d_u \cdot A = u^{-1}d_A u = d_A + u^{-1} d_A u\), so \(u\) must be such that \(d_A + u^{-1} d_A u = d_A\), meaning \(u^{-1} d_A u = 0\). Assuming that \(u^{-1}\) is an invertible endomorphism, this is equivalent to \(d_A u = 0\).
Example 2. Given the following fiber bundle \(P\) over \(X\), \(S^1 \rightarrow P \rightarrow X\), we have an associated frame bundle: a complex line bundle \(E \rightarrow X\). The gauge group in this case is \(Map(X, S^1)\), and \(End(E) = E^* \otimes E = \mathbb{C} \times X\), hence connections are trivial. For a \(u: X \rightarrow S^1\), we have \(d_A u = du = 0\) implying that \(u\) is constant. Thus, \(Stab_A = S^1\).
Example 3. Given a Hermitian vector bundle \(\mathbb{C} \rightarrow E \rightarrow X\), we construct a frame bundle \(P=Fr_{U(n)}E\). We note that \(End(E) = End_0(E) \otimes \mathbb{C}\cdot Id\), where \(End_0(E)\) are the traceless endomorphisms and \(\mathbb{C}\cdot Id\) are multiples of the identity. Since if \(A\) is a connection in \(E\) then \(\nabla\) is \(\nabla_A \otimes Id + Id \otimes \nabla_A\) and constant maps to \(\mathbb{C}\cdot Id = Z(End(E))\) stabilize \(A\).
Example 4. More generally, for a principal bundle, \(G \rightarrow P \rightarrow X\), where \(G\) is the structure group, we have \(Aut(P) = \{ \hat{u}: P \rightarrow G | \hat{u}(pg) = \hat{u}(p)g \}\), so if \(z \in Z(G)\), then \(\hat{u}_z: P \rightarrow G\) which sends \(\hat{u}_z(p) = z\) is always parallel.
So gauge transformations are determined by the value at a single point of \(P\) plus a parallel transport, thus we can think of \(Stab_A \subset G\) as a closed Lie subgroup of G.
