Convergence of Harmonic Maps

Abstract

In this paper we prove a compactness theorem for sequences of harmonic maps which are defined on converging sequences of Riemannian manifolds.

Appendix: Convergence of Tension Field

In this section we study convergence of the tension fields of the maps $f_i, \tau (f_i)$, under the assumptions of Proposition 3.6.

Assume $(M_i,g_i), f_i, N$ to be as in Proposition 3.6. Moreover consider the following assumption

Assumption 2

The section $s_{i,j}$ is almost harmonic,

$$\begin{aligned} |\tau (s_{i,j})|\le C\cdot \epsilon ''_i, \end{aligned}$$

(21)

and also

$$\begin{aligned} |\nabla _{\bar{X}}ds_{i,j}(X)|\le C\cdot \epsilon ''_i, \end{aligned}$$

(22)

where $X$ is a smooth vector field on $M$ and $\bar{X}$ is its horizontal lift and $\epsilon ''_i$ is a sequence which converges to zero.

Using Assumption 1 and by Theorem 2.4 we have

$$\begin{aligned} \tau (f_i)&= ({\nabla _{e_k}}df_i)e_k+({\nabla _{e_t}}df_i)e_t\\&= ({\nabla _{e_k}}df_i)e_k+\nabla _{{f_i}_*(e_t)}{{f_i}_*(e_t)}{\nonumber }\\&-{f_i}_*(\nabla _{e_t}{e_t})^{H}- {f_i}_*(\nabla _{e_t}{e_t})^V{\nonumber }\\&= ({\nabla _{e_k}}df_i)e_k-{f_i}_*({{\mathrm{H}}}_i)+\tau ({f_i}^{\bot }){\nonumber } \end{aligned}$$

(23)

where $\{e_k,e_t\}$ and $\bar{e}_k$ are as in the proof of Proposition 3.6, ${f_i}^{\bot }$ denotes the restriction of $f_i$ to the fibers $F_i$, and ${{\mathrm{H}}}_i$ is the mean curvature vector of the submanifold $F_i$.

We investigate how each term of the equation above behaves as $f_i$ converges to $f$.

Lemma 3.11

We have

$$\begin{aligned} \lim _{i\rightarrow \infty }\left| d I({\nabla _{e_k}}d f_i)e_k(p)- \left( \Delta _{g_i^M} \tilde{f_i}-\Pi (\tilde{f_i})(d\tilde{f_i},d\tilde{f_i})\right) (\psi _i(p))\right| =0. \end{aligned}$$

(24)

Proof

By the discussion in the proof of Proposition 3.6, we know that $\tilde{f_i}$ converges to $f$ in the $C^1$-topology. Using the composition formula we have

$$\begin{aligned} dI(Bf_i(X_1,X_2))=B(I\circ f_i)(X_1,X_2)-B(\pi _N)(d (I\circ f_i)(X_1),d (I\circ f_i)(X_2)), \end{aligned}$$

and so for $k=1,\ldots ,n$,

$$\begin{aligned} dI(({\nabla _{e_k}}df_i)e_k)=({\nabla _{e_k}}d( I\circ f_i))e_k-B(\pi _N)(d(I\circ f_i)(e_k),d(I\circ f_i)(e_k)). \end{aligned}$$

First we show that

$$\begin{aligned} \lim _{i\rightarrow \infty }|{\nabla _{e_k}}d(I\circ f_i)e_k(p)- \Delta _{g_i^M} \tilde{f_i}(\psi _i(p))|=0. \end{aligned}$$

By definition of $\tilde{f_i}$,

$$\begin{aligned} (\nabla _{{\bar{e}_k}} d\tilde{f_i}){\bar{e}_k}&= \sum \big (d\beta _j{({\bar{e}_k})}\cdot d f_i({s_{i,j}}_*({\bar{e}_k}))\\&\qquad +\beta _j\cdot (\nabla _{{\bar{e}_k}}d(f_i\circ s_{i,j})){\bar{e}_k}+\triangle \beta _j\cdot f_i\circ s_{i,j}\big ). \end{aligned}$$

and again by the composition formula

$$\begin{aligned} \tau (f_i\circ s_{i,j})=B_{{s_{i,j}}_*(\bar{e}_k),{s_{i,j}}_*(\bar{e}_k)} {f_i}+df_i(\tau (s_{i,j})). \end{aligned}$$

(25)

Since $f_i\circ s_{i,j}$ converges in $C^1$ to $f$

$$\begin{aligned}&\lim \limits _{i\rightarrow \infty }|\sum d\beta _j{({\bar{e}_k})}\cdot d f_i({s_{i,j}}_*({\bar{e}_k}))|=0,\\&\lim \limits _{i\rightarrow \infty }\sum \Delta \beta _j\cdot f_i\circ s_{i,j}(x)= \sum \triangle \beta _j\cdot f(x)=0. \end{aligned}$$

Also, ${\psi _i}_*(e_k-{s_{i,j}}_*(\bar{e}_k))=0$ and so $e_k-{s_{i,j}}_*(\bar{e}_k)$ is vertical. On the other hand

$$\begin{aligned} |e_k-{s_{i,j}}_*(\bar{e}_k)|\le \epsilon _i. \end{aligned}$$

By inequality (11) and almost harmonicity of $s_{i,j}$ (21), the second term on the right hand side of (25) converges to zero. Again by inequality (12) and (22), we have

$$\begin{aligned}&\lim _{i\rightarrow \infty }|(\nabla _{e_k}{d f_i})(e_k-{s_{i,j}}_*(\bar{e}_k))|=0,\\&\lim _{i\rightarrow \infty }|(\nabla _ {(e_k-{s_{i,j}}_*(\bar{e}_k))}d{f_i})e_k|=0. \end{aligned}$$

Finally

$$\begin{aligned} \lim _{i\rightarrow \infty }|(\nabla _{e_k} d(I\circ f_i))e_k(p)-(\nabla _{\bar{e}_k}d\tilde{f_i})\bar{e}_k(\psi _(p))|=0. \end{aligned}$$

We have the same for the second term

$$\begin{aligned} \lim _{i\rightarrow \infty }|\Pi (f_i)(p)(df_i,df_i)-\Pi (\tilde{f_i})(\psi _i(p))(d\tilde{f_i},d\tilde{f_i})|=0. \end{aligned}$$

$\square$

By the above lemma and ${\psi _i}_*(\tfrac{{{\mathrm{dvol}}}_{M_i}}{{{\mathrm{vol}}}(M_i)})=\Phi _i{{\mathrm{dvol}}}_{M}$, we have

$$\begin{aligned}&\lim \limits _{i\rightarrow \infty }\left| \int _{M_i}\langle dI(({\nabla _{e_k}}d f_i)e_k),\eta _i\rangle ~\tfrac{{{\mathrm{dvol}}}_{M_i}}{{{\mathrm{vol}}}(M_i)}\right. \\&\quad \left. - \int _M \langle \Delta ^{g_i^M} \tilde{f_i}-\Pi (\tilde{f_i})(d \tilde{f_i},d\tilde{f_i}),\eta \rangle ~\Phi _i{{\mathrm{dvol}}}_{g_i^M}\right| =0, \end{aligned}$$

and we conclude

$$\begin{aligned}&\lim \limits _{i\rightarrow \infty }\int _{M_i}\langle dI(({\nabla _{e_k}}d f_i)e_k),\eta _i\rangle ~\tfrac{{{\mathrm{dvol}}}_{M_i}}{{{\mathrm{vol}}}(M_i)}\nonumber \\&=\int _{M}\left[ \langle df,d\eta \rangle +\langle df(\nabla \ln \Phi )-\Pi (f)(df,df),\eta \rangle \right] ~ \Phi {{\mathrm{dvol}}}_M. \end{aligned}$$

(26)

Here $\eta$ is a test map on $M$ and $\eta _i=\eta \circ \psi _i$. Now we will consider the second and third terms in the decomposition of $\tau (f_i)$.

Lemma 3.12

With the same assumptions as above

• i. $\lim \limits _{i\rightarrow \infty } \int _{M_i}\langle df_i({{\mathrm{H}}}_i),\eta _i\rangle ~ \tfrac{{{\mathrm{dvol}}}_{M_i}}{{{\mathrm{vol}}}(M_i)}=-\int _{M}\langle df(\nabla \ln \Phi ),\eta \rangle ~ \Phi {{\mathrm{dvol}}}_M$.

• ii. $\lim \limits _{i\rightarrow \infty }\Vert \tau ({f_i}^{\bot })\Vert =0$.

Here ${{\mathrm{H}}}_i$ denotes the mean curvature vector of the fibers $F_i^x=\psi ^{-1}_i(x)$.

Before we prove Lemma 3.12, we prove the following lemma which we need for the proof of part i.

Lemma 3.13

We have

$$\begin{aligned} \int _M\eta d\ln \Phi (X)~\Phi {{\mathrm{dvol}}}_M=-\lim _{i\rightarrow \infty }\int _{M_i} \eta \langle X,{{\mathrm{H}}}_i\rangle ~ \tfrac{{{\mathrm{dvol}}}_{M_i}}{{{\mathrm{vol}}}(M_i)}. \end{aligned}$$

(27)

Proof

Suppose $X$ is a smooth vector field on $M$ and $X_i$ its horizontal lift on $M_i$. The flow $\theta _t^i$ of $X_i$ sends fibers to fibers diffeomorphically. By the first variation formula

$$\begin{aligned} \left. \frac{d}{dt} \right| _{t = 0} {\theta _t^i}^*({{\mathrm{dvol}}}_{F^x_i})=-\int _{F^x_i}\langle X_i,{{\mathrm{H}}}^x_i \rangle ~ {{\mathrm{dvol}}}_{F^x_i}. \end{aligned}$$

(28)

Also

$$\begin{aligned} \Phi _i(x)=\tfrac{{{\mathrm{vol}}}(\psi _i^{-1}(x))}{{{\mathrm{vol}}}(M_i)}. \end{aligned}$$

and by (28),

$$\begin{aligned} d\Phi _i(X)(x)=-\int _{F^x_i}\langle X_i,{{\mathrm{H}}}^x_i \rangle ~ \tfrac{{{\mathrm{dvol}}}_{F^x_i}}{{{\mathrm{vol}}}(M_i)}, \end{aligned}$$

For an arbitrary $\eta$ in $C^{\infty }({M})$, we prove

$$\begin{aligned} \int _{M} \eta d\Phi _i(X)~ {{\mathrm{dvol}}}_{g_i^M}=-\int _{M_i}\eta _i\langle X_i,{{\mathrm{H}}}_i \rangle ~ \tfrac{{{\mathrm{dvol}}}_{M_i}}{{{\mathrm{vol}}}(M_i)}. \end{aligned}$$

(29)

If we consider $(U_{\gamma },h_{\gamma })$ as a local trivialization of the fibration $\psi _i$, then

$$\begin{aligned} \int _{M} \chi _{U_{\gamma }} d\Phi _i(X)~ {{\mathrm{dvol}}}^{g_i^M}=-\int _{U_{\gamma }}\int _{F^x_i}\chi _{U_{\gamma }}\langle X_i,{{\mathrm{H}}}^x_i\rangle ~ \tfrac{{{\mathrm{dvol}}}_{F^x_i}}{{{\mathrm{vol}}}(M_i)}{{\mathrm{dvol}}}_{g_i^M}, \end{aligned}$$

and so

$$\begin{aligned} \int _{M} \chi _{U_{\gamma }} d\Phi _i(X)~ {{\mathrm{dvol}}}^{g_i^M}_{M}=-\int _{\psi _i^{-1}( U_{\gamma })}\langle X_i,{{\mathrm{H}}}_i \rangle ~ \tfrac{{{\mathrm{dvol}}}_{M_i}}{{{\mathrm{vol}}}(M_i)}, \end{aligned}$$

where $\chi _{U_{\gamma }}$ denotes the characteristic function on $U_{\gamma }$ and so we have (29). The functions $\Phi _i$ goes to $\Phi$ in $C^{\infty }$ and also ${{\mathrm{dvol}}}^{g_i^M}$ goes to ${{\mathrm{dvol}}}_M$ as $i$ goes to infinity. Letting $i$ go to $\infty$ on the both sides of (29) and by the definition of weak derivatives

$$\begin{aligned} \int _M\eta d\ln \Phi (X)~\Phi {{\mathrm{dvol}}}_M=-\lim _{i\rightarrow \infty }\int _{M_i} \eta \langle X,{{\mathrm{H}}}_i\rangle ~ \tfrac{{{\mathrm{dvol}}}_{M_i}}{{{\mathrm{vol}}}(M_i)}. \end{aligned}$$

$\square$

Proof of Lemma 3.12

Part i follows directly from Lemma 3.13.

To prove part ii consider

$$\begin{aligned} \tau ({f_i}^{\bot })=\nabla _{{f_i}_*(e_t)}{{f_i}_*(e_t)}-{f_i}_*( \nabla _{e_t}{e_t})^{V}. \end{aligned}$$

From (11) and (12)

$$\begin{aligned} |\nabla _{{f_i}_*(e_t)}{{f_i}_*(e_t)}|&< C\cdot {\epsilon '_i},\\ \Vert {f_i}_*(\nabla _{e_t}{e_t})^{V}\Vert _{L^{\infty }}&< C\cdot { \epsilon '_i}|(\nabla _{e_t}{e_t})^V|, \end{aligned}$$

where $C$ is a constant independent of $i$. It follows that

$$\begin{aligned} \lim _{i\rightarrow \infty }\Vert \tau ({f_i}^{\bot })\Vert =0. \end{aligned}$$

$\square$