Chern-Osserman type equality for complete surfaces in R^n

TL;DR

This paper establishes a Chern-Osserman type equality for complete surfaces in Euclidean space with finite second fundamental form norm and shows finite mean curvature norm implies quadratic area growth.

Contribution

It introduces a new equality relating geometric quantities of surfaces and links finite mean curvature norm to quadratic area growth.

Findings

Chern-Osserman type equality derived for surfaces with finite second fundamental form

Finite L^2-norm of mean curvature implies at least quadratic area growth

Monotonicity formula used to establish area growth behavior

Abstract

We obtain a Chern-Osserman type equality of a complete properly immersed surface in Euclidean space, provided the L^2-norm of the second fundamental form is finite. Also, by using a monotonicity formula, we prove that if the L^2-norm of mean curvature of a noncompact surface is finite, then it has at least quadratic area growth.

Equations

$$\begin{array}{rl}({\overline{\nabla }}^{2}r)(X,Y)& =XY(r)-{\overline{\nabla }}_{X}Y(r)\\ & =({\nabla }^{2}r)(X,Y)-⟨A(X,Y),\overline{\nabla }r⟩.\end{array}$$

$$({\nabla }^{2}r)(e,e)=\frac{1}{r}(1-⟨e,\nabla r{⟩}^2)+⟨A(e,e),{\nabla }^{\perp }r⟩,$$

$$2\pi \chi (M_t)-{\int }_{M_t}K=\frac{1}{t}\left(v^{\prime }(t)+{\int }_{\partial M_t}\frac{⟨x^{\perp },H⟩}{\mid \nabla r\mid }\right)-{\int }_{\partial M_t}⟨A(\frac{\nabla r}{\mid \nabla r\mid },\frac{\nabla r}{\mid \nabla r\mid }),\frac{{\nabla }^{\perp }r}{\mid \nabla r\mid }⟩,$$

$$\begin{array}{rl}{k}_g& =-⟨{\nabla }_{e}e,\frac{\nabla r}{\mid \nabla r\mid }⟩\\ & =\frac{1}{\mid \nabla r\mid }({\nabla }^{2}r)(e,e)\\ & =\frac{1}{\mid \nabla r\mid }(\frac{1}{r}+⟨A(e,e),{\nabla }^{\perp }r⟩)\\ & =\frac{1}{\mid \nabla r\mid }(\frac{1}{r}+⟨H-A(\frac{\nabla r}{\mid \nabla r\mid },\frac{\nabla r}{\mid \nabla r\mid }),{\nabla }^{\perp }r⟩),\end{array}$$

$$t{v}^{\prime }(t)=t{\int }_{\partial M_t}\frac{\mid {\nabla }^{\perp }r{\mid }^2}{\mid \nabla r\mid }+2v(t)+{\int }_{M_t}⟨x^{\perp },H⟩.$$

$$v^{\prime }(t)={\int }_{\partial M_t}\frac{1}{\mid \nabla r\mid }.$$

$$t{v}^{\prime }(t)$$

$$=t{\int }_{\partial M_t}\frac{\mid {\nabla }^{\perp }r{\mid }^2}{\mid \nabla r\mid }+2v(t)+{\int }_{M_t}⟨x^{\perp },H⟩.\text{qed}$$

$$u_1(t)\triangleq \frac{v(t)}{t^2}-\frac{1}{2t^2}{\int }_{M_t}\mid (x-x_0{)}^{\perp }\mid \mid H\mid +\frac{1}{16}{\int }_{M_t}\mid H{\mid }^2$$

$$u_2(t)\triangleq \frac{v(t)}{t^2}-\frac{1}{t^2}{\int }_{M_t}\mid (x-x_0{)}^{\perp }\mid \mid H\mid +\frac{1}{4}{\int }_{M_t}\mid H{\mid }^2$$

$$\begin{array}{rl}\frac{d}{dt}({\int }_{M_t}\mid x^{\perp }\mid \mid H\mid )& ={\int }_{\partial M_t}\frac{\mid x^{\perp }\mid \mid H\mid }{\mid \nabla r\mid }\\ & =t{\int }_{\partial M_t}\frac{\mid {\nabla }^{\perp }r\mid \mid H\mid }{\mid \nabla r\mid }\\ & \le 2{\int }_{\partial M_t}\frac{\mid {\nabla }^{\perp }r{\mid }^2}{\mid \nabla r\mid }+\frac{t^2}{8}{\int }_{\partial M_t}\frac{\mid H{\mid }^2}{\mid \nabla r\mid }.\end{array}$$

$${\int }_{M\setminus B_0(R)}\mid H{\mid }^2<\epsilon .$$

$$Vol(M\cap B_0(t))$$

$$\ge \frac{(t-R{)}^2}{4}\left(u_1(\frac{t-R}{2})-\frac{1}{16}{\int }_{M_{\frac{t-R}{2}}}\mid H{\mid }^2\right)$$

$$\ge \frac{\pi -\epsilon }{4}(t-R{)}^2.$$

$$\begin{array}{rl}2\pi \chi (M_t)-\underset{M_t}{\int }K& \ge \frac{1}{t}\left(v^{\prime }(t)-\mid {\int }_{\partial M_t}\frac{⟨x^{\perp },H⟩}{\mid \nabla r\mid }\mid \right)-\underset{\partial M_t}{\int }\mid A\mid \frac{\mid {\nabla }^{\perp }r\mid }{\mid \nabla r\mid }\\ & \ge \frac{v^{\prime }}{2t}-\frac{t}{2}{\int }_{\partial M_t}\frac{\mid H{\mid }^2}{\mid \nabla r\mid }-\underset{\partial M_t}{\int }(\frac{t\mid A{\mid }^2}{2\mid \nabla r\mid }+\frac{\mid {\nabla }^{\perp }r{\mid }^2}{2t\mid \nabla r\mid })\\ (byProposition2.3)& =\frac{v^{\prime }}{2t}-\frac{t}{2}{\int }_{\partial M_t}\frac{\mid H{\mid }^2}{\mid \nabla r\mid }-\frac{t}{2}{\int }_{\partial M_t}\frac{\mid A{\mid }^2}{\mid \nabla r\mid }-\frac{t{v}^{\prime }-(2v(t)+{\int }_{M_t}⟨x^{\perp },H⟩)}{2t^2}\\ & =\frac{v}{t^2}-\frac{t}{2}{\int }_{\partial M_t}\frac{\mid H{\mid }^2}{\mid \nabla r\mid }-\frac{t}{2}{\int }_{\partial M_t}\frac{\mid A{\mid }^2}{\mid \nabla r\mid }+\frac{{\int }_{M_t}⟨x^{\perp },H⟩}{2t^2}\\ & \ge \frac{v}{t^2}-\frac{t}{2}{\int }_{\partial M_t}\frac{\mid H{\mid }^2}{\mid \nabla r\mid }-\frac{t}{2}{\int }_{\partial M_t}\frac{\mid A{\mid }^2}{\mid \nabla r\mid }-\frac{\sqrt{v{\int }_{M_t}\mid H{\mid }^2}}{2t}\\ & \ge \frac{v}{2t^2}-\frac{t}{2}{\int }_{\partial M_t}\frac{\mid H{\mid }^2}{\mid \nabla r\mid }-\frac{t}{2}{\int }_{\partial M_t}\frac{\mid A{\mid }^2}{\mid \nabla r\mid }-\frac{1}{8}{\int }_{M_t}\mid H{\mid }^2,\end{array}$$

Full text

1. Introduction

Let M be a complete minimal surface in $\mathds{R}^{n}$ with finite total curvature, Chern and Osserman [2], [7] proved that

[TABLE]

where K is the Gauss curvature of M, $\chi(M)$ is the Euler characteristic of M and $k$ is the number of ends of M. Further results were obtained by Jorge and Meeks [5] that

[TABLE]

where $B(t)$ is the extrinsic ball of radius t.

When M is a general surface properly immersed in $\mathds{R}^{n}$ with $\displaystyle\int_{M}\mid A\mid^{2}<\infty$, where A is the second fundamental form of the immersion, White [9] proved that $\displaystyle\frac{1}{2\pi}\int_{M}K$ must be an integer. In this paper, we present a general version of (1.2), where M is a general surface properly immersed in $\mathds{R}^{n}$ with the $L^{2}$-norm of the second fundamental form is finite.

Theorem 1.1. *Let M be a complete properly immersed noncompact oriented surface in $\mathds{R}^{n}$, A the second fundamental form of the immersion, r the distance of  $\mathds{R}^{n}$ from a fixed point and $M_{t}=\{x\in M:r(x)<t\}$, $\chi(M)$ the Euler characteristic of M. Suppose $\displaystyle\int_{M}\mid A\mid^{2}<\infty$, then

1.$\displaystyle\lim\limits_{t\rightarrow\infty}\frac{areaM_{t}}{\pi t^{2}}$ exists and is a positive integer;

2.$\displaystyle\lim\limits_{t\rightarrow\infty}\frac{areaM_{t}}{\pi t^{2}}=\chi(M)-\frac{1}{2\pi}\int_{M}K$.*

Since $\displaystyle\int_{M}\mid A\mid^{2}<+\infty$, then $\displaystyle\int_{M}\mid K\mid<+\infty$ by Gauss equation. When M is a complete surface with finite total Gaussian curvature, Huber [4] proved that M has finite topological type. And Cohn-Vossen [3] obtained:

[TABLE]

The explicit equality was obtained by Shiohama [8]:

[TABLE]

where $D(t)$ denote the area of geodesic balls of radius t at a fixed point. Our theorem shows that (1.4) also holds with extrinsic balls instead of geodesic balls if M is properly immersed in $\mathds{R}^{n}$.

The proof of Theorem 1.1 is based on two monotonicity formulas (Theorem 2.4). The monotonicity formulas also have an interesting application, namely, if the $L^{2}$-norm of mean curvature $H$ of the surface is finite, then it has at least quadratic area growth.

Corollary 1.2. (see also Corollary 2.5) Let M be a complete properly immersed noncompact surface in $\mathds{R}^{n}$ with $\displaystyle\int_{M}\mid H\mid^{2}<\infty$, then the volume of the intersection of M and the extrinsic balls has at least quadratic area growth.

2. Preliminaries

Let $x:M\rightarrow\mathds{R}^{n}$ be a complete properly immersed surface in $\mathds{R}^{n}$, $r$ the distance function of $\mathds{R}^{n}$ from a fixed point. For simplicity, we always assume the fixed point to be 0, unless otherwise specifie. Denote the covariant derivatve of $\mathds{R}^{n}$ and M by $\overline{{\nabla}}$ and $\nabla$ respectively. Let X, Y be two tangent vector fields of M, then

[TABLE]

The equality (2.1), together with the fact that $\displaystyle\overline{\nabla}^{2}r=\frac{1}{r}(g_{st}-dr\otimes dr)$, where $g_{st}$ denotes the standard metric of $\mathds{R}^{n}$, implies

Proposition 2.1. For any unit tangent vector $e$ of M,

[TABLE]

where $\nabla^{\perp}r$ is the projection of $\overline{\nabla}$ onto the normal of M.

By Sard’s theorem, for $a.e.t>0$, $M_{t}=\{x\in M:r(x)<t\}$ is a related compact open subset of M with the boundary $\partial M_{t}$ being a closed immersed curve of M. Let $v(t)=areaM_{t}$ , A the second fundamental form of M, and $H=trA$ the mean curvature vector.

Proposition 2.2. Suppose M is a complete properly immersed surface in $\mathds{R}^{n}$. Then for $a.e.\ t>0$,

[TABLE]

where $x^{\perp}$ is the projection of position vector $x$ onto the normal of M.

Proof.

By the Gauss-Bonnet formula, it’s sufficient to verify

[TABLE]

where $k_{g}$ denote the geodesic curvature of $\partial M_{t}$ in M.

Suppose $e$ is the unit tangent vector of $\partial M_{t}$. Since the normal of $\partial M_{t}$ is $\frac{\nabla r}{\mid\nabla r\mid}$,

[TABLE]

where the third equality follows by Proposition 2.1. Then by using co-area formula, $\displaystyle v^{\prime}(t)=\int_{\partial M_{t}}\frac{1}{\mid\nabla r\mid}$ and the fact that $\displaystyle\nabla^{\bot}r=\frac{x^{\bot}}{r}$, we obtain (2.2). ∎

Proposition 2.3. Let M be a complete properly immersed surface in $\mathds{R}^{n}$, then

[TABLE]

Proof.

Since $\displaystyle\frac{1}{2}\Delta r^{2}=2+\langle x,H\rangle$, integrating over $M_{t}$ and using the Green’s formula,

[TABLE]

By the co-area formula ,

[TABLE]

So we have,

[TABLE]

Theorem 2.4. Let M be a complete properly immersed surface in $\mathds{R}^{n}$, r the distance of $\mathds{R}^{n}$ from a fixed point $x_{0}$, H the mean curvature of M, $M_{t}=\{x\in M:r(x)<t\}$, $v(t)=area{M_{t}}$, then both

[TABLE]

and

[TABLE]

are monotone nondecreasing in t.

Proof.

For simplicity, we assume $x_{0}=0$. By Proposition 2.3, we have

[TABLE]

By co-area formula and the weighted mean value inequalities,

[TABLE]

Combining (2.5) and (2.6), we have

[TABLE]

or equivalently,

[TABLE]

Dividing both sides of (2.8) by $t^{3}$ yields

[TABLE]

this proves that $u_{1}(t)$ is monotone nondecreasing in t.

If we make slight modifications to (2.5) and (2.6), we have

[TABLE]

and

[TABLE]

Combining $(2.5)^{\prime}$ and $(2.6)^{\prime}$, we obtain

[TABLE]

i.e. $u_{2}(t)$ is monotone nondecreasing in t. ∎

Remark 2.4. From the poof, we can see that the theorem is also valid for noncomplete surface, for t with $\partial M\cap B_{x_{0}}(t)=\emptyset$, where $B_{x_{0}}(t)$ is the ball in $\mathds{R}^{n}$ of radius t and centered at $x_{0}$.

By Theorem 2.4, we can get various volume estimates under suitable restrictions on mean curvature $H$.

Corollary 2.5. Let M be a complete properly immersed noncompact surface in $\mathds{R}^{n}$ with $\displaystyle\int_{M}\mid H\mid^{2}<\infty$, then the volume of the intersection of M and the extrinsic balls has at least quadratic area growth.

Proof.

Without loss of generality, we assume the center of the extrinsic balls to be 0. Since $\displaystyle\int_{M}\mid H\mid^{2}<\infty$, for a given $\varepsilon>0$, there exists $R>0$, such that

[TABLE]

Now for $t>R$ large enough, choosing a point $p\in M\cap\partial B_{0}(\frac{t+R}{2})$, then $B_{p}(\frac{t-R}{2})\subset B_{0}(t)\setminus B_{0}(R)$, so we have

[TABLE]

Taking $x_{0}=p$ in Theorem 2.4, then we have

[TABLE]

Combining (2.10) and (2.11), we obtain

[TABLE]

The conclusion follows by chosing $\varepsilon$ small. ∎

3. Proof of Theorem 1.1

Lemma 3.1. Let M be as in Theorem 1.1, then both $\displaystyle\lim\limits_{t\rightarrow\infty}\frac{v(t)}{t^{2}}$ and $\displaystyle\lim\limits_{t\rightarrow\infty}\frac{\int_{M_{t}}\mid x^{\perp}\mid\mid H\mid}{t^{2}}$ exist.

Proof.

First we prove:

Claim: $\displaystyle\liminf\limits_{t\rightarrow\infty}\frac{v}{t^{2}}<+\infty.$

Proof of the claim: Since by the weighted mean value inequality,

[TABLE]

by Proposition 2.2, we have

[TABLE]

where we use the weighted mean value inequalities in the second and the last equality, while the second equality count backwards follows from Cauchy’s inequality.

Since $\displaystyle\int_{M}\mid H\mid^{2}+\int_{M}\mid A\mid^{2}<+\infty$, there exists a sequence $\{\tau_{i}\}$ diverging to infinity such that

[TABLE]

Otherwise, we must have $\displaystyle\liminf\limits_{t\rightarrow\infty}t(\int_{\partial M_{t}}\frac{\mid H\mid^{2}}{\mid\nabla r\mid}+\int_{\partial M_{t}}\frac{\mid A\mid^{2}}{\mid\nabla r\mid})=\delta>0$.  So for sufficient large t, we have

[TABLE]

i.e.

[TABLE]

When you integrate t, by the co-area formula, it is as bounded on the left as it is diverging on the right, a contradiction.

Then taking $t=\tau_{i}$ in (3.2), together with the fact that

[TABLE]

we have $\displaystyle\limsup\limits_{i\rightarrow\infty}\frac{v(t_{i})}{t_{i}^{2}}<+\infty$, which implies $\displaystyle\liminf\limits_{t\rightarrow\infty}\frac{v}{t^{2}}\leq\limsup\limits_{i\rightarrow\infty}\frac{v(t_{i})}{t_{i}^{2}}<+\infty$. This proves the claim.

Let $u_{1}(t)$ and $u_{2}(t)$ be as in Theorem 2.4 with $x_{0}=0$. By the claim, we have

[TABLE]

Combining (3.4) and Theorem 2.4, we know that both $u_{1}(t)$ and $u_{2}(t)$ have finite limit as $t\rightarrow\infty$.

Since

[TABLE]

we conclude that both $\displaystyle\lim\limits_{t\rightarrow\infty}\frac{v(t)}{t^{2}}$ and $\displaystyle\lim\limits_{t\rightarrow\infty}\frac{\int_{M_{t}}\mid x^{\perp}\mid\mid H\mid}{t^{2}}$ exist. ∎

Lemma 3.2. There exists a sequence $\{t_{k}\}$ diverging to infinity such that

[TABLE]

Proof.

Let $u_{1}(t)$, $u_{2}(t)$ be as in Lemma 3.1. Since $u_{1}(t)+u_{2}(t)+\int_{M_{t}}\mid H\mid^{2}+\int_{M_{t}}\mid A\mid^{2}$ is bounded, arguing as in the proof of the claim in Lemma 3.1, we know that there is a sequence $\{t_{k}\}$ diverging to infinity such that

[TABLE]

Since derivative of each function in left side of (3.6) is nonnegative, we have

[TABLE]

as   $k\rightarrow\infty.$ Combining (3.5) and (3.7), we get

[TABLE]

as   $k\rightarrow\infty.$ So we obtain

[TABLE]

where we use the fact that $\lim\limits_{t\rightarrow\infty}\frac{v(t)}{t^{2}}$ and $\lim\limits_{t\rightarrow\infty}\frac{\int_{M_{t}}\mid x^{\perp}\mid\mid H\mid}{t^{2}}$ exist by Lemma 3.1, this proves (ii).

By co-area formula, when $k\longrightarrow\infty$,

[TABLE]

Combining (3.9) and (3.10), we have

[TABLE]

Then (i) follows from (3.9) and (3.11). ∎

Proof of Theorem 1.1   By Proposition 2.3, we have

[TABLE]

then Lemma 3.2 implies

[TABLE]

Taking $t=t_{k}$ in Proposition 2.2 and letting $k\rightarrow\infty$, together with (3.13) and Lemma 3.2, we get

[TABLE]

which implies

[TABLE]

Since the extrinsic distance is smaller than intrinsic distance, we clearly have

[TABLE]

where D(t) is the area of geodesic balls of radius t at a fixed point.

Combining (1.4), (3.15) and (3.16), we conclude that

[TABLE]

Furthermore, by the main theorem of White [9], we know that $\frac{1}{2\pi}\int_{M}K$ is an integer, so is $\lim\limits_{t\rightarrow\infty}\frac{v(t)}{\pi t^{2}}$, and this limit must be positive by Corollary 2.5. This completes the proof of Theorem 1.1.

Corollary 3.3 Let M be a complete properly immersed noncompact oriented surface in $\mathds{R}^{n}$ with $\displaystyle\int_{M}\mid A\mid^{2}<4\pi$, then $\chi(M)=1$.**