Regularity and compactness of harmonic-Einstein equations

Abstract

Let (M, x, g) be pointed Riemannian manifold with $\mathrm{Vol}(B_1(x))\ge \mathrm{v}>0$ and satisfy harmonic-Einstein equation $\mathrm{Ric}_g-\nabla u\otimes \nabla u=\lambda g$ with $|\lambda |\le n-1$, where $u:(M,g)\rightarrow (N,h)$ is a harmonic map to a fixed compact Riemannian manifold (N, h). Then for any $p<1$, we prove the $L^p$ curvature estimate . As a consequence, if (N, h) has nonpositive sectional curvature, we have $|\mathrm{Ric}|\le C(n,\mathrm{v},N)$. That means harmonic-Einstein equation automatically implies bounded Ricci curvature provided nonpositive sectional curvature of (N, h). Let $(X,x_\infty , d,u_\infty )$ be the limit of a sequence of harmonic-Einstein manifolds $(M_i,x_i,g_i,u_i)$. We show that the singular set of X is closed and the convergence is smooth on the regular part. We also prove an orbifold type compactness theorem of harmonic-Einstein equations with bounded $\int _M|\mathrm{Rm}|^{n/2}$ without assuming nonpositive sectional curvature of (N, h), which generalizes Xu’s compactness result in Yiyan (Adv. Math. 231(2):680–708, 2012).

Appendix: Kato inequality

In this appendix, we prove the improved Kato inequality in Lemma 5.8. The proof is the same as in [26]. For the reader’s convenience, we give a proof here. Let us recall Lemma 5.8

Lemma 6.1

(Kato inequality) Let (M, g) be a Riemmanian manifold with dimensional n. Then we have the follow Kato inequality:

$$\begin{aligned} (1+4\eta (n))\left| \nabla \right| \mathrm{Rm}\left| \right| ^2\le \left| \nabla \mathrm{Rm}\right| ^2+C(n)\left| \nabla \mathrm{Ric}\right| ^2, \end{aligned}$$

(6.1)

where $\eta (n), C(n)$ are two universal constants.

Proof

The proof is essentially just a repeated application of the second Bianchi identity. For any fixed $y\in M$, choose a normal coordinate such that $|\nabla |\mathrm{Rm}||(y)=\partial _1 |\mathrm{Rm}|(y)$. Then at y, we have

$$\begin{aligned} \left| \nabla \left| \mathrm{Rm}\right| ^2\right| =\left| \partial _1\left| \mathrm{Rm}\right| ^2\right| =2\left| \langle \nabla _1\mathrm{Rm},\mathrm{Rm}\rangle \right| \le 2\left( \sum _{ijkl}R_{ijkl,1}^2\right) ^{1/2}\left( \sum _{ijkl}R_{ijkl}^2\right) ^{1/2}\,. \end{aligned}$$

(6.2)

To prove the lemma, it suffices to show $(1+\sigma (n))\sum _{ijkl}R_{ijkl,1}^2\le \sum _{ijklp}R_{ijkl,p}^2+C(n)\sum _{ijp}Rc_{ij,p}^2$ for some dimensional constant $\sigma (n)$ and C(n). In fact, it suffices to show $\sum _{ijkl}R_{ijkl,1}^2\le C(n)\left( \sum _{ijkl}\sum _{p\ge 2}R_{ijkl,p}^2+\sum _{ijp}Rc_{ij,p}^2 \right) :=C(n)\Xi$. Hence we only need to prove $R_{\alpha \beta \gamma \delta ,1}^2\le C(n)\Xi$ for any fixed $\alpha ,\beta ,\gamma ,\delta =1,\ldots ,n$.

Case 1: For $\alpha ,\beta ,\gamma ,\delta \ge 2$, then by second Bianchi identity and Cauchy inequality, we have

$$\begin{aligned} R_{\alpha \beta \gamma \delta ,1}^2=\left( R_{\alpha \beta \delta 1,\gamma }+R_{\alpha \beta 1\gamma ,\delta }\right) ^2\le 2\left( R_{\alpha \beta \delta 1,\gamma }^2+R_{\alpha \beta 1\gamma ,\delta }^2\right) \le 2\Xi \, . \end{aligned}$$

(6.3)

Case 2: For $\alpha =1$ and $\beta ,\gamma ,\delta \ge 2$, then by second Bianchi identity and Cauchy inequality, we have

$$\begin{aligned} R_{1\beta \gamma \delta ,1}^2=\left( (R_{1\beta \delta 1,\gamma }+R_{1\beta 1\gamma ,\delta }\right) ^2\le 2\left( R_{1\beta \delta 1,\gamma }^2+R_{1\beta 1\gamma ,\delta }^2\right) \le 2\Xi \, . \end{aligned}$$

(6.4)

Case 3: For $\alpha =\delta =1$ and $\beta ,\gamma \ge 2$, then by Cauchy inequality, we have

$$\begin{aligned} R_{1\beta \gamma 1,1}^2=\left( Rc_{\beta \gamma ,1}-\sum _{\alpha \ge 2}R_{\alpha \beta \gamma \alpha ,1}\right) ^2\le n\left( Rc_{\beta \gamma ,1}^2+\sum _{\alpha \ge 2}R_{\alpha \beta \gamma \alpha ,1}^2\right) \le n\left( \Xi +2\sum _{\alpha \ge 2}\Xi \right) \le n(2n-1)\Xi \, , \end{aligned}$$

(6.5)

where we have used the estimates in Case 1 to the second inequality.

Thus, by the symmetric relations of curvature tensor, we have proved $R_{\alpha \beta \gamma \delta ,1}^2\le C(n)\Xi$. Hence, we prove $\sum _{ijkl}R_{ijkl,1}^2\le C(n)\left( \sum _{ijkl}\sum _{p\ge 2}R_{ijkl,p}^2+\sum _{ijp}Rc_{ij,p}^2 \right)$. This finishes the proof of Lemma 5.8. $\square$