Area of convex surface

This is the sequel of the post about Bean’s theorem on convex curve.

I will state the smooth version of Bean’s theorem in the general dimension.

Theorem 1 (Bean’s theorem, 2011, Smooth version) Suppose {\Sigma} is a ((n-1)-dimensional) smooth convex hypersurface contained in the interior of another smooth compact hypersurface S in {\mathbb{R}^n}, then the area of {\Sigma} is not greater than the area of S.

First of all let us make some remarks. By a smooth convex hypersurface we mean {\Sigma} is compact, connected and such that for all {p}, we can find a (outward) normal vector {\nu(p)} such that all {q\in \Sigma}, {\langle q-p, \nu(p)\rangle\leq0}. i.e. {\Sigma } lies on the non-positive side of {\nu}.

Since {\Sigma} (and also S) is a compact hypersurface in {\mathbb{R}^n}, it is orientable (see also here). Thus we can find a global outward normal field {\nu} of {\Sigma}, and thus we can globally define its second fundamental form {h(X, Y):=\langle \nabla^{\mathbb{R}^n}_X \nu, Y \rangle} without any ambiguity, once {\nu} is fixed (here \nabla^{\mathbb{R}^n}_X \nu is only the ordinary directional derivative of the vector field \nu in \mathbb{R}^n in direction X). In our convention, the second fundamental form of the unit sphere with respect to its outward normal is {h(X, X)=|X|^2}.

We claim that

Lemma 2 If {\Sigma} is convex, then its second fundamental form with respect to the outward normal {\nu} is always non-negative definite.

Proof: Since {\Sigma} is convex, it is locally convex in the sense that at any point {p}, {\Sigma} can be locally represented by a graph of function {f:T_p\Sigma\cong \mathbb{R}^{n-1} \rightarrow \mathbb{R}} with {f(0)=\nabla f(0)=0} and {f\geq 0}. (What does that mean precisely, and how is {f} related to {\nu}? )

Since {f\geq 0}, by Taylor’s theorem, at {0} (corresponding to the point {p}), for any tangent vector {X\in T_p\Sigma\cong\mathbb{R}^{n-1}}

\displaystyle 0\leq \text{Hess} \, f(X, X)

But it is easy to see that the hessian of f is equal to the second fundamental form:

\displaystyle 0\leq \text{Hess} \, f(X, X)=h(X, X).


Remark: Under some additional assumptions, the converse is also true (but much harder) by a result of Sacksteder.

Lemma 3 For any hypersurface {S} in {\mathbb{R}^n}, suppose {V} is a unit vector field defined on a neighborhood of {S} such that {V|_S} is a normal vector field to {S}. Then at each point {p} of {S}, we have

\displaystyle div(V)=H(p)

where {H} is the mean curvature of {S\subset \mathbb{R}^n} w.r.t. {V|_S} (i.e. sum of eigenvalues of the second fundamental form w.r.t. {V|_S}, which is symmetric). In our notation, the mean curvature of the unit sphere in {\mathbb{R}^3} is 2.

Proof: Since {\langle V, V\rangle\equiv1}, differentiating it gives {\nabla ^{\mathbb{R}^n}_V V\perp V} and thus at {p}

\displaystyle H(p)=\sum_{i=1}^{n-1}\langle \nabla ^{\mathbb{R}^n}_{e_i}V, e_i\rangle=\sum_{i=1}^{n-1}\langle \nabla ^{\mathbb{R}^n}_{e_i}V, e_i\rangle+\langle \nabla ^{\mathbb{R}^n}_V V, V\rangle=div(V)

where {e_i} forms an orthonormal basis of T_p\Sigma. \Box

Now for convex {\Sigma} and {\rho \geq 0}, we define the surface {\Sigma ^\rho} by

\displaystyle \Sigma ^\rho:=\{x+\rho \nu(x): x\in \Sigma\}.

By our previously result (Lemma 2 in here), {\Sigma^\rho} is a regular surface (i.e. without any self-intersection). We claim

Lemma 4 The mean curvature of {\Sigma^\rho} is non-negative if {\Sigma} is convex.

Proof: Let {X=X(u^1, \cdots, u^{n-1})} be a local parametrization of {\Sigma } around {p=X(0)}, let us for simplicity denote {\nu(X(u^1, \cdots, u^{n-1}))} simply by {\nu(u^1, \cdots, u^{n-1})}. Then for fixed {\rho}

\displaystyle \tilde X(u^1, \cdots, u^{n-1}):=X(u^1, \cdots, u^{n-1})+\rho\vec\nu(u^1, \cdots, u^{n-1})

gives a natural parametrization for {\Sigma ^\rho}. It is easy to see that {\nu(x)} is also the normal vector of {\Sigma^\rho} at {\tilde X(x)} and that {\frac{\partial \nu}{\partial u^i}\in T_p\Sigma}. Let us compute the second fundamental form {\tilde h} of {\Sigma^\rho} (denote {\frac{\partial X}{\partial u^i}} by {X_i} etc) at {\tilde X(0)}:

\displaystyle  \tilde h(\tilde X_i, \tilde X_i)=\langle \nabla^{\mathbb{R}^n} _{\tilde X_i}\nu, \tilde X_i\rangle=\langle \nabla^{\mathbb{R}^n} _{X_i+\rho \nu_i}\nu, X_i+\rho\nu_i\rangle=h(X_i+\rho \nu_i, X_i+\rho \nu_i)\geq0.

Thus the second fundamental form of {\Sigma^\rho} is also non-negative and so its mean curvature is non-negative. \Box

Proof:(Proof of Bean’s theorem)
Extend the outward normal {\nu} of {\Sigma} to the region {\Omega} bounded between {\Sigma} and {S} by parallel translating {\nu} along its outward direction. Then by our previous result (Lemma 2 in here), this defines a vector field {X} on {\Omega}, extending {\nu}.

It is easy to see that at each point {x\in \Omega}, {X(x)} is the normal to the surface {\Sigma^{\rho}} containing {x} where {\rho=d(x, \Sigma)}. Thus by Lemma 4 and Lemma 3,

\displaystyle div X\geq 0.

Now, we apply divergence theorem to {X} in {\Omega},

\displaystyle  \int_{S}\langle X, \nu_S\rangle dS- \int_{\Sigma}\langle X, \nu\rangle dS=\int_\Omega div X dx\geq 0

where {\nu_S} is the unit outward normal of {S}. But clearly {\langle X, \nu_S\rangle \leq 1}, and {\langle X, \nu \rangle=1}, so

\displaystyle Area(S)- Area (\Sigma)\geq 0.


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2 Responses to Area of convex surface

  1. Yin Tat Lee says:

    Is it related to Bishop volume comparison theorem?
    (I didn’t even read your article nor remember what’s Bishop volume comparison theorem)

  2. KKK says:

    No. At least I don’t see the relation. The Bishop comparison theorem compares the volume of geodesic balls when we have a comparison of the Ricci curvatures between a certain Riemannian manifold and a space form.

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