09-24 Notes

Injectivity

Suppose that \(A\in \mathcal{A}^p_k\) and \((k+1)p>n\). For \(\forall \epsilon>0\), define \[S_{A,\epsilon}=\{A+a|\ d_A^*a=0,\ \left\|{a}\right\|_{L^p_k}<\epsilon\}\] and the map \[m:S_{A,\epsilon}\times \mathcal{G}^p_{k+1}\longrightarrow\mathcal{A}_k^p,\ m(A+a,u)=A+u^{-1}d_Au+u^{-1}au.\] This induces a map \[\bar{m}: S_{A,\epsilon}\times\mathcal{G}^p_{k+1}/Stab_A\longrightarrow \mathcal{A}^p_k.\] Thus we have the following property:

Proposition 1. There exists an \(\epsilon_0>0\), such that \(\forall \epsilon<\epsilon_0\), the map \(\bar{m}\) is a diffeomorphism onto a neighborhood of \(A\).

Proof. Due to the discussion in the former lectures, it remains to show that \(\bar{m}\) is injective. Suppose \(A+a,\ A+b\in S_{A,\epsilon}\) with the same image under the map \(m\). Thus \[u^{-1}au+u^{-1}d_Au=b\Rightarrow d_Au=ub-au.\] Composing both sides with \(d_A^*\), we have \[\begin{aligned} d_A^*d_Au&= -*d_A*(ub-au)\\ &= -*d_A(u*b-*au)\\ &= -*(d_Au\wedge*b+u\wedge d_A*b-d_A*a\wedge u+*a\wedge d_Au)\\ &= -*(d_Au\wedge*b)-*(*a\wedge d_Au). \end{aligned}\] Take the inner product with \(u^0\), the component of \(u\) which is \(L^2\)-orthogonal to \(Ker d_A\). Thus Hölder inequality, \[\left\|{d_Au^0}\right\|_{L^2}^2\leq \left\|{d_Au^0}\right\|_{L^2} (\left\|{a}\right\|_{L^n}+\left\|{b}\right\|_{L^n}) \left\|{u^0}\right\|_{L^{\frac{2n}{n-2}}}.\] Denote the norm of the Sobolev embedding \(L^2_1\hookrightarrow L^{\frac{2n}{n-2}}\) by \(C_{sob}\), thus we have \[\left\|{d_Au^0}\right\|_{L^2}\leq C_{sob} (\left\|{a}\right\|_{L^n}+\left\|{b}\right\|_{L^n})\left\|{u^0}\right\|_{L^2_1}.\] Since \(d_A: L^2_1(X, End(E))\longrightarrow L^2(X,End(E))\) has closed range with finite dimenstional kernel, thus \[\left\|{u^0}\right\|_{L^2_{1,A}}\leq C(\lambda_1)\left\|{d_Au^0}\right\|_{L^2},\] where the constant depends only on the fist positive eigenvalue \(\lambda_1\) of \(d_A\), that is \[\lambda_1=inf\frac{\left\|{d_Au^0}\right\|_{L^2}} {\left\|{u^0}\right\|_L^2}.\] Recall the Kato inequality, that is, if \(A\) is the metric connection on the bundle \(E\longrightarrow X\) and \(s\) is a section of \(E\), then \(|d|s||\leq|\nabla_As|,\ a.e.\). Hence \(\left\|{u^0}\right\|_{L^2_1}\leq\left\|{u^0}\right\|_{L^2_{1,A}}\) and then \[\left\|{d_Au^0}\right\|_{L^2}\leq C_{sob}\cdot C(\lambda_1)\left\|{d_Au^0}\right\|_{L^2}(\left\|{a}\right\|_{L^n} +\left\|{b}\right\|_{L^n}).\] Thus if we choose \(\epsilon>0\) sufficiently small, then we could get contradiction from above unless \(u\in Stab_A\). Therefore the proposition holds for sufficient small positive \(\epsilon\). ◻

Remark 1. Observe that the map \(\bar{m}\) is also injective on a small \(L^n\)-ball in the slice not just the \(L^p_k\)-ball, which follows from the Multipication theorem. And the \(L^n\) norm of \(1\)-form is scaling invariant. Indeed, let \(\delta_\lambda: \mathbb{R}^n\longrightarrow\mathbb{R}^n\) be \(\delta_\lambda(x)=\lambda x\). For \(a\in\Omega^1_c(\mathbb{R}^n)\), \[\begin{aligned} \int_{\mathbb{R}^n} |\delta_\lambda^*a|^ndx^1\wedge\cdots\wedge dx^n&= \int_{\mathbb{R}^n}\lambda^n|a(\lambda x)|^ndx^1\wedge\cdots\wedge dx^n\\ &= \int_{\mathbb{R}^n}|a|^ndx^1\wedge\cdots\wedge dx^n. \end{aligned}\] Thus for \(\forall\epsilon>0\), define \[\tilde{S}_{A,\epsilon}=\{A+a|\ d_A^*a=0,\ a\in L^p_k,\ \left\|{a}\right\|_{L^n}\leq\epsilon\}.\] Then we can define \(m:\tilde{S}_{A,\epsilon}\times\mathcal{G}^p_{k+1} \longrightarrow\mathcal{A}^p_k\) and \(\bar{m}:\tilde{S}_{A,\epsilon}\times \mathcal{G}^p_{k+1}/Stab_A \longrightarrow\mathcal{A}^p_k\) identically with the former maps, and have the following property,

Proposition 2. \(\bar{m}\) is a diffeomorphism onto an \(L^n\) ball about \(A\) in \(\mathcal{A}^p_k\).

Yang-Mills Functional

We now consider the equation involving curvature. First define the Yang-Mills functional. Given \(A\in \mathcal{A}^p_k\) a metric connection on the bundle \(E\longrightarrow X\), \((k+1)p>n\), and the curvature \(F_A=dA+A\wedge A\).

Definition 1. Yang-Mills functional \(\mathcal{E}(A)=\int_X|F_A|^2dvol\).

Especially, in dimension \(2\) and \(4\), the Yang-Mills functional has nice properties: in dimension \(4\), it is scaling invariant, since the \(L^2\) norm of \(2\)-form is scaling invariant and is invarant under the conformal change of metric.
Now we would like to derive the Euler-Lagrangian equation for Yang-Mills functional. By definition, for \(t\in \mathbb{R}\) and \(a\in L^p_k\), \[\begin{aligned} \frac{d}{dt}\mathcal{E}(A+ta)|_{t=0} &= \frac{d}{dt}\int_X-tr(F_{A+ta}\wedge *F_{A+ta})dvol|_{t=0}\\ &= -2\int_Xtr(F_A\wedge*d_Aa)dvol\\ &= <F_A, d_Aa>_{L^2}.\\ \end{aligned}\] Therefore the Euler-Lagrangian equation of \(\mathcal{E}\) is \(d_A^*F_A=0\), which is also called Yang-Mills equation.
Observe that the Bianchi identity gives that \(d_AF_A=0\), and thus the solution \(F_A\) of the Yang-Mills equation is harmonic \(2\)-form. Locally the Yang-Mills equation is \[d^*_{a^\tau}(da^\tau+a^\tau\wedge a^\tau)=0,\] where \(A=d+a^\tau\). Thus \[d^*da^\tau+Q(a^\tau,\nabla a^\tau)+C(a^\tau,a^\tau,a^\tau)=0,\] where \(Q\) is some quardratic term and \(C\) is some cubic term. Note that in dimension greater or equal to \(2\), the linearized equation of the Yang-Mills equation is not elliptic.
In dimension \(2\), the solution \(F_A\) of Yang-Mills equation satisfies that \(*F_A\in\Omega^0(X, End(E))\) is parallel. Suppose that \(\xi\) is a paralle section of skew Hermitian bundle \(U(E)\longrightarrow X\). Decompose \(E\) into direct sum of the eigenspace \(E_j\), which corresponds to the eigenvalue \(i\lambda_j\) and \(\lambda_j\in\mathbb{R}\). Thus \(\xi|_{E_j}=i\lambda_j\). If \(e_\lambda\) is a local eigensection of \(E\), \(\xi e_\lambda=i\lambda e_\lambda\), then \[\nabla_X^A(\xi e_\lambda)=\nabla^A_X\xi\cdot e_\lambda+\xi\nabla^A_Xe_\lambda =i\lambda(\nabla^A_Xe_\lambda).\] Hence \(\nabla^A\) preserves the eigenspace. It follows that \(*F_A\) is a scaling endomorphism. And the eigenvalues of a parallel endomorphisms are constants and the eigenspaces are vector bundle which is projections of flat bundles.