A polyhedron comparison theorem for 3-manifolds with positive scalar curvature

Abstract

The study of comparison theorems in geometry has a rich history. In this paper, we establish a comparison theorem for polyhedra in 3-manifolds with nonnegative scalar curvature, answering affirmatively a dihedral rigidity conjecture by Gromov. For a large collections of polyhedra with interior non-negative scalar curvature and mean convex faces, we prove the dihedral angles along its edges cannot be everywhere less or equal than those of the corresponding Euclidean model, unless it is isometric to a flat polyhedron.

Appendix A

We provide some general calculation for infinitesimal variations of geometric quantities of properly immersed hypersurfaces under variations of the ambient manifold $(M^{n+1},g)$ that leave the boundary of the hypersurface inside $\partial M$. We also refer the readers to the thorough treatment in [26] and [4] (warning: the choice of orientation for the unit normal vector field N in [4] is the opposite to ours).

We keep the notations used in Sect. 2.1 and for each $t\in (-\varepsilon ,\varepsilon )$, we use the subscript t for the terms related to $\Sigma _t$. Recall that $Y=\frac{\partial \Psi (t,\cdot )}{\partial t}$ is the deformation vector field. Denote $Y_0$ the tangent part of Y on $\Sigma$, $Y_0$ the tangent part of Y on $\partial \Sigma$. Let $v=\left\langle Y,N \right\rangle$. For $q\in \Sigma$, let $e_1,\ldots ,e_n$ be an orthonormal basis of $T_q\Sigma$, and let $e_i(t)=d\Psi _t(e_i)$. Let $S_0,S_1$ be the shape operators of $\Sigma \subset M$ and $\partial M\subset M$. Precisely, $S_0(Z_1)=-\nabla _{Z_1}N$, $S_1(Z_2)=\nabla _{Z_2}X$. We have:

Lemma A.1

(Lemma 4.1(1) of [26], Proposition 15 of [4])

$$\begin{aligned} \nabla _Y N=-\nabla ^\Sigma v-S_0(Y_0). \end{aligned}$$

(A.1)

We use Lemma A.1 to calculate the evolution of the contact angle along the boundary.

Lemma A.2

Let $\gamma$ denote the contact angle between $\Sigma$ and $F_j$. Then

$$\begin{aligned} \frac{d }{d t}\bigg \vert _{t=0}\left\langle N_t,X_t \right\rangle =-\sin \gamma \frac{\partial v}{\partial \nu }+(\cos \gamma ) A(\nu ,\nu )v+{{\,\mathrm{II}\,}}({\overline{\nu }},{\overline{\nu }})v+\left\langle L,\nabla ^{\partial \Sigma }\gamma _j \right\rangle v, \end{aligned}$$

(A.2)

where L is a bounded vector field on $\partial \Sigma$.

In particular, if each $\Sigma _t$ meets $F_j$ at constant angle $\gamma _j$, then on $F_j$,

$$\begin{aligned} \frac{\partial v_t}{\partial \nu _t}=\left[ (\cot \gamma _j) A_t(\nu _t,\nu _t)+\frac{1}{\sin \gamma _j}{{\,\mathrm{II}\,}}(\overline{\nu _t},\overline{\nu _t})\right] v_t. \end{aligned}$$

Proof

Let us fix one boundary face $F_j$ and denote $\gamma _j$ by $\gamma$. By Lemma A.1,

$$\begin{aligned} \begin{aligned} \frac{d }{d t}\bigg \vert _{t=0}\left\langle N_t,X_t \right\rangle&=\left\langle \nabla _Y N,X \right\rangle +\left\langle N,\nabla _Y X \right\rangle \\&=-\left\langle \nabla ^\Sigma v,X \right\rangle -\left\langle S_0(Y_0),X \right\rangle +\left\langle N,\nabla _Y X \right\rangle .\end{aligned} \end{aligned}$$

On $\partial M$, Y decomposes into $Y=Y_1-\frac{v}{\sin \gamma }{\overline{\nu }}$. Notice that since $X=\cos \gamma N+\sin \gamma N$,

$$\begin{aligned} \left\langle S_0(Y_0),X \right\rangle =\left\langle S_0(Y_0),\cos \gamma N+\sin \gamma \nu \right\rangle =\sin \gamma A(Y_0,\nu ). \end{aligned}$$

We also have the vector decomposition on $\partial M$ with respect to the orthonormal basis ${\overline{\nu }},X$:

$$\begin{aligned} N=\cos \gamma X-\sin \gamma {\overline{\nu }},\qquad \nu =\cos \gamma {\overline{\nu }}+\sin \gamma X. \end{aligned}$$

(A.3)

Since $\left\langle X,X \right\rangle =1$ along $\partial M$, we have $\left\langle X,\nabla _Z X \right\rangle =0$ for any vector Z on $\partial M$. We have

$$\begin{aligned} \begin{aligned} \frac{d }{d t}\bigg \vert _{t=0}\left\langle N_t,X_t \right\rangle&=-\sin \gamma \frac{\partial v}{\partial \nu }-\left\langle S_0(Y_0),X \right\rangle \\&\qquad \qquad +\left\langle \cos \gamma X-\sin \gamma {\overline{\nu }},\nabla _{Y_1-\frac{v}{\sin \gamma }{\overline{\nu }}}X \right\rangle \\&=-\sin \gamma \frac{\partial v}{\partial \nu }-\sin \gamma A(Y_0,\nu )-\sin \gamma \left\langle {\overline{\nu }},\nabla _{Y_1}X \right\rangle +\left\langle {\overline{\nu }},\nabla _{{\overline{\nu }}}X \right\rangle v. \end{aligned} \end{aligned}$$

Now we deal with the second and the third terms above. Notice that on $\partial \Sigma \cap F_j$,

$$\begin{aligned} Y_0=Y_1-(\cot \gamma ) v\nu . \end{aligned}$$

Thus $A(Y_0,\nu )=A(Y_1,\nu )-(\cot \gamma ) v A(\nu ,\nu )=-\left\langle \nabla _{Y_1}N,\nu \right\rangle -(\cot \gamma ) A(\nu ,\nu )v$. On the other hand, using the vector decomposition (A.3), we find

$$\begin{aligned} \begin{aligned} \left\langle \nabla _{Y_1}N,\nu \right\rangle&=\left\langle \nabla _{Y_1}(\cos \gamma X-\sin \gamma {\overline{\nu }}),\cos \gamma {\overline{\nu }}+\sin \gamma X \right\rangle \\&=\cos ^2\gamma \left\langle \nabla _{Y_1}X,{\overline{\nu }} \right\rangle -\sin ^2\gamma \left\langle \nabla _{Y_1}{\overline{\nu }},X \right\rangle +\left\langle L,\nabla ^{\partial \Sigma } \gamma \right\rangle .\\&=\left\langle \nabla _{Y_1}X,{\overline{\nu }} \right\rangle +\left\langle L,\nabla ^{\partial \Sigma } \gamma \right\rangle . \end{aligned} \end{aligned}$$

Here L is a vector field along $\partial \Sigma$, and $|L|\le C=C(Y,X,\nu )$. Thus we conclude that

$$\begin{aligned} \frac{d }{d t}\bigg \vert _{t=0}\left\langle N_t,X_t \right\rangle =-\sin \gamma \frac{\partial v}{\partial \nu }+(\cos \gamma )A(\nu ,\nu )v+{{\,\mathrm{II}\,}}({\overline{\nu }},{\overline{\nu }})v+\left\langle L,\nabla ^{\partial \Sigma }\gamma \right\rangle , \end{aligned}$$

as desired. $\square$

The evolution equation of the mean curvature has been studied in many circumstances. We refer the readers to the thorough calculation in Proposition 16, [4]:

Lemma A.3

(Proposition 16 of [4]) Let $H_t$ be the mean curvature of $\Sigma _t$. Then

$$\begin{aligned} \frac{d }{d t}\bigg \vert _{t=0} H_t=\Delta _\Sigma v+({{\,\mathrm{Ric}\,}}(N,N)+|A|^2)v-\left\langle \nabla _\Sigma H,Y_0 \right\rangle . \end{aligned}$$

In particular, if each $\Sigma _t$ has constant mean curvature, then

$$\begin{aligned} \frac{d }{d t} H_t=\Delta _{\Sigma _t} v_t+({{\,\mathrm{Ric}\,}}(N_t,N_t)+|A_t|^2)v_t. \end{aligned}$$