The Effective Dynamics of the Volume Preserving Mean Curvature Flow

Abstract

We consider the dynamics of small closed submanifolds (‘bubbles’) under the volume preserving mean curvature flow. We construct a map from ($\text {n}+1$)-dimensional Euclidean space into a given ($\text {n}+1$)-dimensional Riemannian manifold which characterizes the existence, stability and dynamics of constant mean curvature submanifolds. This is done in terms of a reduced area function on the Euclidean space, which is given constructively and can be computed perturbatively. This allows us to derive adiabatic and effective dynamics of the bubbles. The results can be mapped by rescaling to the dynamics of fixed size bubbles in almost Euclidean Riemannian manifolds.

Appendices

Appendix A: Expansions of the Induced Metric and Mean Curvature

We Taylor expand the induced metric and mean curvature of $\theta _{\rho ,z}$ and estimate the non-linear terms in $\phi$ in $H^k$, $k>\frac{n}{2}+2$. One may consult [18] for a thorough study on such classical expansions, we mostly follow the notation in [24]. We abuse slightly notation by denoting the pushforward of $\omega$ through $d\theta _{\rho ,z}$ by the same symbol $\omega$. For instance, given $x^i$,$\partial _i$ a set of coordinates and the associated vector fields, we write the corresponding basis of tangent vector fields on $S^n_z$: $\zeta _i:=\lambda (1+\phi )\partial _i\omega +\lambda \partial _i\phi \, \omega \in TS^n_z$.

Lemma A.1

(see [24], Lemmas 2.1 and 2.4) The following expansions are valid:

$$\begin{aligned} \lambda ^{-2}(1+\phi )^{-2}g(\zeta _i,\zeta _j)&=\, g(\partial _i\omega ,\partial _j\omega )+\frac{1}{3}\text {R}(\omega ,\partial _i\omega ,\omega ,\partial _j\omega )\cdot \lambda ^2(1+\phi )^2\nonumber \\&\quad +\frac{\partial _i\phi \partial _j\phi }{(1+\phi )^2}+\lambda ^3 \mathrm{R}_{\lambda , z}^{(0)}( \phi ) +\lambda ^2 \mathrm{R}_{\lambda , z}^{(2)}( \phi ) \end{aligned}$$

(A.1)

where R is the Riemann curvature tensor and the remainders $\mathrm R_{\lambda , z}^{(j)}( \phi ), j=0, 2,$ are local terms depending on $\phi$ and $\partial \phi$ and satisfying the estimates

$$\begin{aligned}&\Vert \partial _{\lambda }^i\partial _z^\alpha \mathrm R_{\lambda , z}^{(r)}(\phi ,\partial \phi )\Vert _{H^{k-1}}\lesssim \sum _{j\le i,\beta \le \alpha }\Vert \partial _{\lambda }^j\partial _z^\beta \phi \Vert ^r_{H^k} \end{aligned}$$

(A.2)

provided $\Vert \phi \Vert _{H^k} \lesssim 1$ and $i+|\alpha |\le 2$, and

$$\begin{aligned} \lambda H(\theta _{\rho ,z})=\;n-\frac{1}{3}\text {Ric}_z\lambda ^2 - (\Delta _{\mathbb {S}^n}+n)\phi +\lambda ^2M'_{\lambda , z} \phi +\lambda ^2 N'_{\lambda , z}(\phi )+\lambda ^3 H_{\lambda , z}, \end{aligned}$$

(A.3)

where $\text {Ric}_z: \omega \rightarrow \text {Ric}_z(\omega ,\omega )$ (Ric$(\cdot ,\cdot )$ is the Ricci curvature tensor of M), $M'_{\lambda , z}\phi$, $N'_{\lambda , z}(\phi )$ and $H_{\lambda , z}$ are linear non-linear terms in $\phi$ and in its derivatives up to order two, respectively, and independent of $\phi$ term, satisfying the estimates:

$$\begin{aligned} \Vert \partial _{\lambda }^i\partial _z^\alpha M'_{\lambda , z}(\phi )\Vert _{H^{k-2}}&\lesssim \sum _{j\le i, \beta \le \alpha }\Vert \partial _{\lambda }^j\partial _z^\beta \phi \Vert _{H^k}, \end{aligned}$$

(A.4)

$$\begin{aligned} \Vert \partial _{\lambda }^i\partial _z^\alpha N'_{\lambda , z}(\phi )\Vert _{H^{k-2}}&\lesssim \sum _{j\le i,\beta \le \alpha }\Vert \partial _{\lambda }^j\partial _z^\beta \phi \Vert ^2_{H^k}, \end{aligned}$$

(A.5)

$$\begin{aligned} \Vert \partial _{\lambda }^i\partial _z^\alpha H_{\lambda , z}\Vert _{H^{k-2}}&\lesssim 1, \end{aligned}$$

(A.6)

and similarly for $N'_{\lambda , z}(\phi ')-N'_{\lambda , z}(\phi )$, for $i + |\alpha |\le 2$, $k>\frac{n}{2}+2$. (Above, if $\phi$ is independent of $\lambda$ and z, then only the term with $j= |\beta |=0$ survives on the r.h.s..)

Proof

Both expressions (A.1) and (A.3) can be read from [24]. Bounds (A.4) are immediate by definition. In order to derive estimate (A.6) for the non-linearity $N_{\lambda , z}(\phi )$ we have to first examine its structure. Recall that the mean curvature of $\theta _{\rho ,z}$ is given by⁴

$$\begin{aligned} H(\theta _{\rho ,z})=\sum _{i,j}g^{i j}g(\nabla _{\zeta _i}\nu ,\zeta _j)=\text {div}_{\theta _{\rho ,z}}\nu \end{aligned}$$

(A.8)

where $g^{ij}$ is the inverse matrix to the metric $g_{ij}:=g(\zeta _i,\zeta _j)$ and $\nu$ the outward unit normal vector field on $\theta _{\rho ,z}$:

$$\begin{aligned} \nu =\frac{-\omega +\lambda \sum _{i,j}g^{ij}\partial _i\phi \zeta _j}{\sqrt{1-\lambda ^2g(\nabla _{\mathbb {S}^n}\phi ,\nabla _{\mathbb {S}^n}\phi )}} \end{aligned}$$

(A.9)

Note that $\nu$ is well defined for $\lambda$ and $\Vert \phi \Vert _{H^k}$ appropriately small. Hence, we observe that the non-linear terms in $\phi$ arising in the expansion of $H(\theta _{\rho ,z})$ are of the form $b(\lambda \phi (\omega ),\lambda \partial \phi (\omega ))\lambda \partial ^2\phi (\omega )$, where b(s, t) is a simple function, uniformly bounded together with its derivatives, provided $|s|\ll 1$ and $|t|\ll 1$. Using these estimates it is not hard (but somewhat tedious) to show that $\Vert b(\lambda \phi ,\lambda \partial \phi )\lambda \partial ^2\phi \Vert _{H^{k-2}}\lesssim \lambda ^r\Vert \phi \Vert ^r_{H^k}$ for some $r\ge 2$, provided $\Vert \phi \Vert _{H^k}\ll 1$ (here we use the condition $k>\frac{n}{2}+2$). This completes the proof of the lemma. $\square$

Eq (A.1) implies

$$\begin{aligned} \lambda ^{-n}(1+\phi )^{-n}\sqrt{\text {det}g_{\theta _{\rho ,z}}}&=\,1+\frac{1}{6}\text {Ric}(\omega ,\omega )\lambda ^2(1+\phi )^2+\frac{1}{2}\frac{|\nabla _{\mathbb {S}^n}\phi |^2}{(1+\phi )^2}\nonumber \\&\quad +\lambda ^3\mathrm{S_{\lambda , z}^{(0)}}( \phi )+\lambda ^2\text {S}_{\lambda , z}^{(2)}(\phi ) \end{aligned}$$

(A.10)

where Ric is the Ricci curvature tensor of M and $\mathrm S_{\lambda , z}^{(0)}, r=0, 2,$ are local terms depending on $\phi$ and $\partial \phi$ satisfying the estimates

$$\begin{aligned}&\Vert \partial _{\lambda }^i\partial _z^\alpha \mathrm S_{\lambda , z}^{(r)}(\phi ,\partial \phi )\Vert _{H^{k-1}}\lesssim \Vert \phi \Vert _{H^k}^r \end{aligned}$$

(A.11)

provided $\Vert \phi \Vert _{H^k} \lesssim 1$ and $i+|\alpha |\le 2$. Furthermore, multiplying (A.10) by $\lambda ^{n}(1+\phi )^{n}$, integrating over $\mathbb {S}^n$ and taking into account that $\int _{\mathbb {S}^n}\phi =0$, and using the co-area formula and polar coordinates, we obtain:

Corollary A.2

We have the following expansions

$$\begin{aligned} A(\Phi _\lambda (z))&=\;\lambda ^n\bigg (a_n\big [1-\frac{1}{6(n+1)}R(z)\lambda ^2\big ]+\lambda ^3\mathrm{Q_{\lambda , z}^{(0)}}( \phi )+\lambda ^2\text {Q}_{\lambda , z}^{(2)}(\phi ) \end{aligned}$$

(A.12)

where $a_n$ denotes the area of the Euclidean unit sphere $\mathbb {S}^n$ and R is the scalar curvature of M, and

$$\begin{aligned} V_{enc}(\Phi _\lambda (z))&=\;\lambda ^{n+1}\bigg ( \frac{a_n}{n+1}\bigg [1-\frac{1}{6(n+3)}R(z)\lambda ^2 \bigg ]+\lambda ^3\mathrm{T_{\lambda , z}^{(0)}}( \phi )+\lambda ^2\text {T}_{\lambda , z}^{(2)}(\phi ) \end{aligned}$$

(A.13)

The remainders $\mathrm Q_{\lambda , z}^{(r)}$ and $\mathrm T_{\lambda , z}^{(r)}$ above are local expression in $\phi ,\partial \phi$ satisfying the estimates of the type of (4.2).

Appendix B: Existence of Barycenter

In this section, we show that barycenter exists. That is, we show that Eq. (3.4) has a solution. To begin, we state some useful estimates.

Lemma B.1

For the terms defined in equations (3.2)–(3.3), we have the estimate

$$\begin{aligned}&\Vert \sigma -1\Vert _{H^{k-1}} \lesssim \lambda \Vert \phi _{\lambda ,z}\Vert _{H^k} \end{aligned}$$

(B.1)

$$\begin{aligned}&\Vert \sigma (\xi )-\sigma \Vert _{H^{k-1}} \lesssim \lambda \Vert \xi \Vert _{H^k} \end{aligned}$$

(B.2)

$$\begin{aligned}&\Vert f_i - \omega _i \Vert _{H^{k-1}} \lesssim \lambda (1+\Vert \phi _{\lambda ,z}\Vert _{H^k} + \Vert \partial _{z^i} \phi _{\lambda ,z} \Vert _{H^k} ) \text { for } i=1,\ldots ,n+1 \end{aligned}$$

(B.3)

$$\begin{aligned}&\Vert f_i(\xi ) - f_i\Vert _{H^{k-1}} \lesssim \lambda \Vert \xi \Vert _{H^k} \text { for } i=1,\ldots ,n+1 \end{aligned}$$

(B.4)

$$\begin{aligned}&\Vert f_0 - 1 \Vert _{H^{k-1}} \lesssim \lambda \Vert \partial _\lambda \phi _{\lambda ,z} \Vert _{H^k} + \Vert \phi _{\lambda ,z} \Vert _{H^k} \end{aligned}$$

(B.5)

$$\begin{aligned}&\Vert f_i(\xi ) - f_i\Vert _{H^{k-1}} \lesssim \lambda \Vert \xi \Vert _{H^k} \end{aligned}$$

(B.6)

Proof

Recall that $\psi (\omega ) = \exp _z(\rho (\xi )\omega )$, where $\rho (\xi ) := \lambda (1+\phi _{\lambda ,z} +\xi )$. Using the definitions (3.2), we compute the difference, $\sigma (\xi )-\sigma$. To this end, it suffices to compute the differences of each factor in (3.2):

$$\begin{aligned} \sigma (\xi ) - \sigma&= g_\psi (\nu (\theta _{\rho (\xi ), z}(\omega )), \partial _s \theta _{s, z}(\omega ) \mid _{s=\rho (\xi )}) - g_{\theta _{\lambda ,z})} (\nu (\theta _{\rho (0), z}(\omega )), \partial _s \theta _{s, z}(\omega ) \mid _{s=\rho (0)}) \\&= g_\psi (\nu (\theta _{\rho (\xi ), z}(\omega )), \partial _s \theta _{s, z}(\omega ) \mid _{s=\rho (\xi )}) - g_\psi (T\nu (\theta _{\rho (0), z}(\omega )) , T\partial _s \theta _{s, z}(\omega ) \mid _{s=\rho (0)}) \end{aligned}$$

where T is the parallel transport from $\theta _{\lambda ,z}(\omega )$ to $\psi (\omega )$. Continuing the estimate, we have

$$\begin{aligned}&= g_\psi (\nu (\theta _{\rho (\xi )}) - T\nu (\theta _{\rho (0), z}(\omega )), \partial _s \theta _{s, z}(\omega ) \mid _{s=\rho (\xi )}) \\&\quad + g_\psi (T\nu (\theta _{\rho (0), z}(\omega )), \partial _s \theta _{s, z}(\omega ) \mid _{s=\rho (\xi )} - T\partial _s \theta _{s, z}(\omega ) \mid _{s=\rho (0)}) . \end{aligned}$$

Recalling that $\theta _{\lambda ,z} = \exp _z(\rho (0)\omega )$ and letting $\omega ^0 := 1$, we find furthermore

$$\begin{aligned}&g(\nu (\theta _{\rho (0), z}(\omega )), \partial _s \theta _{s, z}(\omega ) \mid _{s=\rho (0)}) \mid _{\lambda = 0} = g(\nu (\theta _{\lambda , z}(\omega )) \mid _{\lambda = 0}, \partial _s \theta _{s, z}(\omega ) \mid _{s=0})= 1 \\&\quad g(\nu (\theta _{\rho (0), z}(\omega )), \partial _{z^i} \theta _{\rho (0), z}(\omega )) \mid _{\lambda = 0} = \omega ^i,\quad i = 0, 1,\ldots ,n+1. \end{aligned}$$

Similarly, we write $f_i(\xi ) - f_i$. Hence, using this, we only need to estimate the following items to complete the proof:

$$\begin{aligned}&g_p - g_{p'} \\&T(s)-T(s') \\&\partial _s \exp _z(s\omega ) \mid _{s=s_1} -T(s_1-s_2, \exp _z(s_2\omega ))\partial _s \exp _z(s\omega ) \mid _{s=s_2} \\&\partial _z \exp _z(s_1 \omega ) - T(s_1-s_2, \exp _z(s_2\omega ))\partial _z \exp _z(s_2 \omega ) \\&\rho (\xi ) - \rho (0) \\&\partial _\lambda \rho - \partial _\lambda \rho \mid _{\phi _{\lambda ,z}=0} \\&\rho (\xi ) \\&\nu (\psi ) -T(\lambda , \theta _{\lambda ,z})\nu (\theta _{\lambda ,z}) \end{aligned}$$

where $g_p$ is the value of the metric at p, T(s, z) is parallel transport from z along $\omega$ for time s. Then the difference of quantities in the statement of the question are composition and smooth functions of the above with at most two derivatives of $\phi _{\lambda ,z}$ and $\xi$. Note that the first four quantities only depends on the ambient geometry of $M^{n+1}$. Since we are working lcoally, we may assume that we are working on a compact subset of M. Thus, the first four are smooth functions, the first four quantity exhibits Lipschitz estimates in the difference in their argument. For example,

$$\begin{aligned} |\partial _z \exp _z(s_1 \omega ) - T(s_1-s_2, \exp _z(s_2\omega ))\partial _z \exp _z(s_2 \omega )| \lesssim |s_1-s_2| \end{aligned}$$

uniformly on $M^{n+1}$. The next 3 expressions have the obvious estimate

$$\begin{aligned}&\Vert \rho (\xi ) - \rho (0) \Vert _{H^k} = \lambda \Vert \xi \Vert _{H^k} \\&\Vert \partial _\lambda \rho - \partial _\lambda \rho \mid _{\phi _{\lambda ,z}=0}\Vert _{H^k} \lesssim \lambda \Vert \partial _\lambda \phi _{\lambda ,z} \Vert _{H^k} + \Vert \phi _{\lambda ,z} \Vert _{H^k}\\&\Vert \partial _z \rho - \partial _z \rho \mid _{\phi _{\lambda ,z}=0}\Vert _{H^k} \lesssim \lambda (\Vert \partial _z \phi _{\lambda ,z} \Vert _{H^k} + \Vert \phi _{\lambda ,z} \Vert _{H^k}) \\&\Vert \rho (\xi )\Vert _{H^k} \lesssim \lambda (\Vert \xi \Vert _{H^k}+\Vert \phi _{\lambda ,z}\Vert _{H^k}). \end{aligned}$$

Finally, the last one follows from the fact $\nu (\psi )$ is a none singular rational function of $\psi$. $\square$

Now, let

$$\begin{aligned} P(\lambda , z \psi ) : \mathbb {R}\times \mathbb {R}^{n+1} \times H^k \rightarrow \mathbb {R}\times \mathbb {R}^{n+1} = (\langle f_0\sigma , \xi \rangle , \cdots , \langle f_{n+1} \sigma , \xi \rangle ) \end{aligned}$$

where $\xi$ is defined by Eq. (3.1).

Lemma B.2

Any $\psi$ sufficiently close to a CMC $\theta _{\lambda _0,z_0} = \exp _{z_0}(\lambda _0(1+\phi _{\lambda _0,z_0}))$ can be written in the form (2.2), with (3.1) and $\lambda$ and z satisfying the equation

$$\begin{aligned} P(\lambda , z, \psi ) = 0 \end{aligned}$$

Proof

We note that $P(\lambda _0,z_0, \theta _{\lambda _0,z_0}) = 0$. By the implicit function theorem, it suffices to show that

(i):

P is $C^1$, and

(ii):

$(\partial _{\lambda ,z} P)(\lambda _0,z_0,\theta _{\lambda _0,z_0})$ is an invertible matrix.

(i) We check, by definition of $f_i$ and $\xi$, that they are real functions of $\lambda$, z and $\phi _{\lambda ,z}$ up to 1 derivative. Since $\phi _{\lambda ,z}$ is $C^1$ in $\lambda$ and z (cf. Proposition 2.4), we see that $f_i$ and $\sigma$ are $C^1$ in $\lambda$ and z. Since P is linear in $\xi$, using the estimate of Proposition 2.4 again, we see that P is $C^1$ in $\psi$ as well.

(ii) We compute

$$\begin{aligned} (\partial _{\lambda ,z} P)(\lambda _0,z_0,\theta _{\lambda _0,z_0}) = ( \langle f_i\sigma , \partial _{\lambda ,z} \xi \rangle ) \end{aligned}$$

(B.7)

since $\xi = 0$ for $\psi = \theta _{\lambda _0,z_0}$. To compute $\partial _\lambda \xi$, we use the fact that, by definition of $\xi$,

$$\begin{aligned} \exp _{z_0}(\lambda (1+\phi _{\lambda ,z_0} + \xi )\omega ) = \exp _{z_0}(\lambda _0(1+\phi _{\lambda _0,z_0})\omega ) \end{aligned}$$

(B.8)

for $\lambda$ sufficiently close to $\lambda _0$. (Note that we suppressed the identification $I_z$ between $T_zM$ and $\mathbb {R}^{n+1}$ here as z is not varied.) Taking $\partial _\lambda$ and evaluate at $\lambda = 0$, we get the result of

$$\begin{aligned} \partial _s \exp _{z_0}(s \omega ) \mid _{s=\lambda _0(1+\phi _{\lambda _0,z_0})} (1+ \phi _{\lambda _0,z_0} + \lambda _0 \partial _\lambda \phi _{\lambda ,z_0} \mid _{\lambda =\lambda _0} + \lambda _0 \partial _\lambda \xi \mid _{\lambda = \lambda _0}) = 0 \end{aligned}$$

(B.9)

Contracting with $\nu (\theta _{\lambda _0,\theta _0})$, we get

$$\begin{aligned} 0 = f_0 + \lambda _0 \sigma \partial _\lambda \xi , \end{aligned}$$

(B.10)

where the last line follow from Proposition 2.4. To compute $\partial _z \xi \mid _{\lambda _0,z_0}$, we consider curves $z(t) : (-\epsilon , \epsilon ) \rightarrow M$ with $z(0) = z_0$ and $\dot{z}(0) = v$ for any $v \in T_{z_0}M$ fixed. Then any variation of

$$\begin{aligned} \exp _{z(t)}(\lambda _0(1+\phi _{\lambda _0,z(t)} + \xi ) I_{z(t)}(\omega )) \end{aligned}$$

(B.11)

is tangential. Taking derivative with respect to t and setting $t=0$, we see that the expression (B.11) becomes

$$\begin{aligned}&[ (\partial _x \exp _{x} (\lambda _0(1+\phi _{\lambda _0,z_0}) I_x(\omega ))) \mid _{x=z_0} \\&\quad + \partial _s \exp _{z_0}(s\omega ) \mid _{s = \lambda _0(1+\phi _{\lambda _0,z_0})} (\lambda _0 \partial _z \phi _{\lambda _0,z} \mid _{z=z_0} + \lambda _0 \partial _z \xi \mid _{z=z_0})] \dot{z}(0) \end{aligned}$$

Using the fact that varying z is only tangential, it follows that

$$\begin{aligned}&0 = g(\nu (\theta _{\lambda _0,z_0}), [(\partial _x \exp _{x} (\lambda _0(1+\phi _{\lambda _0,z_0}) I_x(\omega ))) \mid _{x=z_0} \end{aligned}$$

(B.12)

$$\begin{aligned}&\quad + \partial _s \exp _{z_0}(s\omega ) \mid _{s = \lambda _0(1+\phi _{\lambda _0,z_0})} (\lambda _0 \partial _z \phi _{\lambda _0,z} \mid _{z=z_0})]v) \end{aligned}$$

(B.13)

$$\begin{aligned}&\quad + g(\nu (\theta _{\lambda _0,z_0}), \lambda _0 \partial _z \xi \mid _{z=z_0}) v) \end{aligned}$$

(B.14)

for any $v \in T_zM$. By definition of $f_i$,

$$\begin{aligned} f_i + \lambda _0 \sigma \partial _{z^i} \xi = 0 \end{aligned}$$

(B.15)

It follows that

$$\begin{aligned} (\partial _{\lambda ,z} P)(\lambda _0,z_0,\theta _{\lambda _0,z_0}) = -\lambda _0^{-1} \langle f_i,f_j \rangle \end{aligned}$$

(B.16)

is invertible by Lemma B.1. $\square$