Lower volume growth estimates for Self-shrinkers of mean curvature flow

We obtain a Calabi-Yau type lower volume growth estimates for complete noncompact self-shrinkers of the mean curvature flow, more precisely, every complete noncompact properly immersed self-shrinker has at least linear volume growth.


Introduction
On a complete noncompact Riemannian manifold M n with nonnegative Ricci curvature, there are two well known theorems on volume growth estimates of geodesic balls. One is the classic Bishop volume comparison theorem (see [12], [16]) which says the geodesic balls have at most Euclidean growth, i.e., there exists some positive constant C such that Vol(B x 0 (r)) ≤ Cr n (1.1) holds for r > 0 sufficiently large. The other is a theorem proved by Calabi [1] and Yau [18] independently, which says the geodesic balls of such manifolds have at least linear volume growth, that is Vol(B x 0 (r)) ≥ Cr (1.2) holds for some positive constant C.
In this paper, we consider the volume growth estimates on self-shrinkers. Note that there are many similarities between self-shrinkers and gradient shrinking solitons. Self-shrinkers give homothetically self-shrinking solutions to mean curvature flow, and describe possible blow ups at a given singularity of the mean curvature flow. While gradient shrinking Ricci solitons also correspond to the self-similar solutions to Hamilton's Ricci flow, and often arise as Type I singularity models.
Before we state our main theorem, we would like to give a roughly brief review about the already known results on volume growth of gradient shrinking Ricci solitons and self-shrinkers.
For an n-dimensional complete noncompact gradient shrinking Ricci soliton (M, g, f ) satisfying 3) The first author was supported by NSFC No. 10971110 and Tsinghua University-K. U. Leuven Bilateral Scientific Cooperation Fund.
H. -D. Cao and D. Zhou [4] proved that it has at most Euclidean volume growth (see also [7], [19]). On the lower volume growth estimate, H.-D. Cao and X.P. Zhu [2] proved that any complete noncompact gradient shrinking Ricci soliton must have infinite volume. In fact, they showed that there is some positive constant C such that Vol(B x 0 (r)) ≥ C ln ln r for r sufficiently large. If the Ricci curvature is bounded, Carillo-Ni [6] showed that the volume grows at least linearly. If the average scalar curvature satisfies Rdv ≤ δ for δ < n/2 and r sufficiently large, then Cao-Zhou [4] showed that there exists some positive constant C such that Vol(B x 0 (r)) ≥ Cr n−2δ . In [14] O. Munteanu and J. Wang proved the sharp result that every complete noncompact gradient shrinking Ricci soliton has at least linear volume growth, which answered the question asked by , [2]) and Lei Ni that if a Calabi-Yau type lower volume growth estimate holds complete noncompact gradient shrinking Ricci solitons.
Theorem A (Munteanu-Wang [14]) Let (M, g, f ) be a complete noncompact gradient shrinking Ricci soliton, then for any x 0 ∈ M there exists a constant C > 0 such that where B x 0 (r) is the geodesic ball of M of radius r centered at x 0 ∈ M .
For a complete noncompact self-shrinker X : M n → R n+m satisfying Lu Wang [17] proved that every entire graphical self-shrinker has polynomial volume growth. Then Q. Ding and Y. L. Xin [9] generalized it and showed that if the immersion is proper, then the self-shrinker has at most Euclidean volume growth. After that, Cheng and Zhou [7] improved Ding-Xin's result and gave a sharp volume growth estimate, they showed that Vol(B x 0 (r)) ≤ Cr n−2β , with β ≤ inf |H| 2 , where the ball B x 0 (r) is defined by In this paper, we consider the lower volume growth estimates for complete noncompact self-shrinkers, an analogue Munteanu-Wang's result will be proved. where the ball B x 0 (r) is defined as (1.5).
Remark 1.2. Note that this is sharp because the volume of the cylinder self-shrinker X : S n−1 ( 2(n − 1)) × R → R n+1 grows linearly.
Acknowledgment. We would like to thank Prof. Huai-Dong Cao for his help and comments, thank Robert Haslhofer for informing us the paper [14] of Ovidiu Munteanu and Jiaping Wang. Special thanks go to Ovidiu Munteanu for particularly valuable comments and pointing out an error in (2.14) in the draft of this paper. We also express our thanks to Prof. Detang Zhou for his interests in our work.

Preliminary
For a complete immersed self-shrinker X : M n → R n+m satisfies (1.4), we have Note that for a gradient shrinking Ricci soliton which satisfies (1.3), we take the trace in (1.3) and get The main idea of this paper is comparing the two equations (2.1) and (2.3), in fact, we can correspond |H| 2 to R, and 1 4 |X| 2 to f , then exploring the similarities between self-shrinker and gradient shrinking Ricci soliton.
Denote ρ(x) = |X|, we have Then by the co-area formula (cf. [16]), we have Now we state the following Lemma: Remark 2.1. From the fourth equality in the above proof, we can get that is, the average of |H| 2 is bounded by n/2. Lemma 2.2. Let X : M n → R n+m be a complete noncompact properly immersed self-shrinker, then Integrating the above equation from r 2 to r 1 , we get Choose r 0 = 2(n + 2), and let r 1 > r 2 ≥ r 0 . Since η(r) is nonnegative and nondecreasing in r, we have where we used (2.12) in the last inequality. This completes the proof of Lemma 2.2.
Remark 2.2. Let r 2 = r 0 and r = r 1 sufficiently large in Lemma 2.2, we can obtain that for r ≥ |X 0 |. This recovers Ding-Xin's result [9], which states that every complete noncompact properly immersed self-shrinker has at most Euclidean volume growth.
In the last of this section, we recall the Logarithmic Sobolev inequality for submanifolds in Euclidean space, this was shown by K. Ecker in [10].
Proposition 2.1 (LSI). Let X : M n → R n+m be an n-dimensional submanifold with measure dv, then the following inequality holds for any nonnegative function f for which all integrals are well-defined and finite, where C(n) is a positive constant depending on n.
On self-shrinker which satisfies (1.4), the Logarithmic Sobolev inequalities (2.14) implies the following two inequalities: (1) For any nonnegative function f which satisfies the normalization holds.
(2) By substituting f = ue In order to prove Theorem 1.1, we need the following Lemma which holds for any complete properly immersed submanifold in Euclidean space.
Proof. In B x 0 (r) we have Integrating the above equation over B x 0 (s) for s ≤ r x 0 (s), where the last equality is due to the co-area formula. This implies Integrating from ǫ > 0 to r, we have Let ǫ → 0, by lim ǫ→0 + Vx 0 (ǫ) ǫ n = ω n , we have V x 0 (r) ≥ κr n , (κ = ω n e −C ) (3.4) Remark 3.1. As pointed out to us by Ovidiu Munteanu, Lemma 3.1 also follows from Michael-Simon Sobolev inequality, see page 377 in [13].
Next we will prove that every complete noncompact properly immersed self-shrinker has infinite volume, the argument in the following proof is an adoption of Cao-Zhu's [2] proof on that complete noncompact shrinking Ricci solitons have infinite volume. Lemma 3.2. Every complete noncompact properly immersed self-shrinker X : M n → R n+m has infinite volume Proof. We are going to show that if M has finite volume, then we shall obtain a contradiction to the Logarithmic Sobolev inequality (2.15). We denote the annulus region here ρ(x) = |X|. Since X : M n → R n+m is complete noncompact properly immersed, X(M ) cannot be contained in a compact Euclidean ballB(R) with radius R < +∞. Then for k large enough, A(k, k + 1) contains at least 2 2k−1 disjoint balls Noting that on self-shrinker thus by Lemma 3.1, each ball B x i (r) has at least volume κ2 −kn , here κ = ω n e −1 . So we have Suppose that Vol(M ) < +∞, then for every ǫ > 0, there exists a large constants k 0 > 0 such that if k 2 > k 1 > k 0 , we have and we can also choose k 1 , k 2 in such a way that In deed, we may first choose K > 0 sufficiently large, and let k 1 ≈ K/2, k 2 ≈ 3K/2, suppose (3.8) does not hold, i.e., then we are done, otherwise we can repeat this process, after j steps we get But we have already assumed Vol(M ) is finite, so after finitely many steps (3.8) must hold for some k 2 > k 1 . Thus for any ǫ > 0 we can choose k 1 and k 2 ≈ 3k 1 such that both (3.7) and (3.8) are valid. Now we are going to derive a contradiction to the Logarithmic Sobolev inequality (2.15). We define a smooth cut-off function ψ(t) by Then let we can choose L such that By the Logarithmic Sobolev inequality (2.15) we have where we have used |∇ρ(x)| ≤ 1 and the elementary inequality t ln t ≥ − 1 e for 0 ≤ t ≤ 1. Then (3.8) implies, where the last equality is due to (3.9). On the other hand, by (3.7) (3.9) and 0 ≤ ψ ≤ 1, we have 1 ≤ e 2L ǫ. So we can make L arbitrary large by letting ǫ > 0 sufficiently small, this contradicts with (3.10) because C(n) is just a universal positive constant depending on n. Therefore M must have infinite volume.
Remark 3.2. In the paper [7], Xu Cheng and Detang Zhou proved that if the self-shrinker is not properly immersed, then it must also have infinite volume. Now we are ready to prove Theorem 1.1.
Proof of Theorem 1.1. We use the similar arguments of Munteanu-Wang's in their proof of Theorem A. First we can choose c > 0 such that V (r) > 0 for r ≥ c. To prove Theorem 1.1, it suffices to show there exists a constant C > 0 depending only on n such that (3.12) V (r) ≥ Cr hold for all r ≥ c. Indeed, if (3.12) holds, then for ∀x 0 ∈ M , since for r sufficiently large, Now we are going to prove (3.12) by contradiction. Assume that for any ǫ > 0, there exits r ≥ c such that V (r) ≤ ǫr (3.14) Without loss of generality, we can assume r ∈ N and consider the following set: Obviously D = Ø because r ∈ D, we want to prove that any integer k ≥ r is in D.
For t ≥ c, we define a function u by Substituting u(x) into the Logarithmic Sobolev inequaltiy (2.16), we obtain with C 0 = C(n) + 4 + 1 e , here we have used |∇ρ(x)| ≤ 1 and the elementary inequality t ln t ≥ − 1 e for 0 ≤ t ≤ 1. From Lemma 2.2, we have This implies for t sufficiently large, t So there exists some constant C 1 (n) such that for all t ≥ C 1 (n), whereC 1 depending only on n. Combining (3.18) and (3.19) gives that for all t ≥ C 1 (n) + 1, where C 2 depending only on n. Note that we can assume r ≥ C 1 (n) + 1 for the r satisfying (3.14). In fact, if for any give ǫ > 0, all the r which satisfies (3.14) is bounded above by C 1 (n) + 1, then V (r) ≥ ǫr holds for any r > C 1 (n) + 1, this implies M has at least linear volume growth.
Then for all integers r ≤ t ≤ k, we have t ∈ D, (3.20) implies If we choose ǫ such that 2C 2 ǫ < 1, and noting that Iterating from t = r to t = k and summing up give that where we used (3.19) in the last inequality. Therefore Noting that r ≤ k + 1, therefore (3.27) implies k + 1 ∈ D. Then by induction we conclude that D contains all the integers k ≥ r. However (3.27) implies V (k) ≤ 2ǫr, for any integer k ≥ r This implies that M has finite volume, which contradicts with Lemma 3.2. So there exists no such r > c such that V (r) ≤ ǫr with ǫ > 0 chosen in (3.26). That is V (r) ≥ ǫr for r > c, and this completes the proof of Theorem 1.1.