On Gromov’s rigidity theorem for polytopes with acute angles

For each 0 ≤ k ≤ q , the set

is non-empty.

For each 0 ≤ k ≤ q , the gradient of the function u k with respect to the Euclidean metric is given by a unit vector N k ∈ S n − 1 .

Theorem 1.3 (cf. M. Gromov [4, Section 3.18])

Suppose that n ≥ 3 is an integer, and

is a compact, convex polytope in R n with non-empty interior. Suppose that Assumptions 1.1 and 1.2 are satisfied. Let 𝑔 be a Riemannian metric on R n . For each k ∈ { 0 , 1 , … , q } , we denote by ν k = ∇ u k / | ∇ u k | the unit normal vector to the level sets of u k with respect to the metric 𝑔. We assume that the following conditions are satisfied.

• The scalar curvature of 𝑔 is nonnegative at each point in Ω.

• For each k ∈ { 0 , 1 , … , q } , the mean curvature of the hypersurface { u k = 0 } with respect to 𝑔 is nonnegative at each point in Ω ∩ { u k = 0 } .

• If 𝑗 and 𝑘 are integers with 0 ≤ j < k ≤ q and 𝑥 is a point in Ω with u j ⁢ ( x ) = u k ⁢ ( x ) = 0 , then ⟨ ν j , ν k ⟩ ≤ ⟨ N j , N k ⟩ ≤ 0 at the point 𝑥.

• the Riemann curvature tensor of 𝑔 vanishes at each point in Ω.

• For each k ∈ { 0 , 1 , … , q } , the second fundamental form of the hypersurface { u k = 0 } with respect to 𝑔 vanishes at each point in Ω ∩ { u k = 0 } .

• If 𝑗 and 𝑘 are integers with 0 ≤ j < k ≤ q and 𝑥 is a point in Ω with u j ⁢ ( x ) = u k ⁢ ( x ) = 0 , then ⟨ ν j , ν k ⟩ = ⟨ N j , N k ⟩ at the point 𝑥.

Let 0 ≤ j < k ≤ q . If there exists a point x ∈ Ω satisfying

then ⟨ N j , N k ⟩ ≤ 0 .

Given ε ∈ ( 0 , 1 ) , we can find a small positive real number 𝛿 (depending on 𝜀) with the following property. Suppose that 0 ≤ j < k ≤ q . If there exists a point x ∈ Ω with − δ ≤ u j ⁢ ( x ) ≤ 0 and − δ ≤ u k ⁢ ( x ) ≤ 0 , then ⟨ N j , N k ⟩ ≤ ε .

Proof

Suppose that the assertion is false. We consider a sequence of counterexamples and pass to the limit. In the limit, we obtain a pair of integers 0 ≤ j < k ≤ q and a point x ∈ Ω such that u j ⁢ ( x ) = u k ⁢ ( x ) = 0 and ⟨ N j , N k ⟩ ≥ ε . This contradicts Assumption 2.1. ∎

We can find a large constant Λ > 1 with the following property. Suppose that 𝑥 is a point in Ω and a 0 , … , a q are nonnegative real numbers such that a i = 0 for all 0 ≤ i ≤ q satisfying u i ⁢ ( x ) ≤ − 2 ⁢ Λ − 1 . Then ∑ i = 0 q a i ≤ Λ ⁢ | ∑ i = 0 q a i ⁢ N i | .

Proof

Suppose that the assertion is false. We consider a sequence of counterexamples and pass to the limit. In the limit, we obtain a point x ∈ Ω and a collection of nonnegative real numbers a 0 , … , a q with the following properties.

• ∑ i = 0 q a i = 1 .

• ∑ i = 0 q a i ⁢ N i = 0 .

• a i = 0 for all 0 ≤ i ≤ q satisfying u i ⁢ ( x ) < 0 .

We can find a large constant Ξ ∈ ( 2 ⁢ Λ , ∞ ) with the following property. Suppose that 1 ≤ k ≤ q , and suppose that 𝑥 is a point in Ω with

Moreover, suppose that a 0 , … , a k − 1 are nonnegative real numbers such that a i = 0 for all 0 ≤ i ≤ k − 1 satisfying u i ⁢ ( x ) ≤ − 2 ⁢ Ξ − 1 . Then

Proof

Let Λ denote the constant in Lemma 2.3. Let us fix a positive real number δ ∈ ( 0 , Λ − 1 ) so that the conclusion of Lemma 2.2 holds with ε = 1 2 ⁢ Λ − 1 . Let us define

We claim that Ξ has the desired property. To prove this, suppose that 1 ≤ k ≤ q , and suppose that 𝑥 is a point in Ω with − δ ≤ u k ⁢ ( x ) ≤ 0 . Moreover, suppose that a 0 , … , a k − 1 are nonnegative real numbers such that a i = 0 for all 0 ≤ i ≤ k − 1 satisfying u i ⁢ ( x ) ≤ − δ . We define

Clearly,

To estimate the term on the right-hand side, we distinguish two cases.

Case 1.  Suppose first that b ≥ 0 . Lemma 2.3 implies

Case 2.  Suppose next that b ≤ 0 . Using Lemma 2.3, we obtain

Moreover, it follows from Lemma 2.2 that ⟨ N i , N k ⟩ ≤ 1 2 ⁢ Λ − 1 for all 0 ≤ i ≤ k − 1 satisfying − δ ≤ u i ⁢ ( x ) ≤ 0 . This implies

Consequently,

This completes the proof of Lemma 2.4. ∎

We define a collection of smooth functions u ̂ 0 , … , u ̂ q on R n so that u ̂ 0 = u 0 and

for 1 ≤ k ≤ q . Moreover, we define Ω ̂ : = { u ̂ q ≤ 0 } and Σ ̂ : = { u ̂ q = 0 } .

We have

for 1 ≤ k ≤ q . In particular, u ̂ 0 ≤ u ̂ 1 ≤ ⋯ ≤ u ̂ q .

Proof

This follows from the fact that | t | ≤ η ⁢ ( t ) ≤ | t | + 1 for all t ∈ R . ∎

We have

In particular, ⋂ m = 0 q { u m ≤ − λ 0 − 1 } ⊂ Ω ̂ ⊂ Ω .

Proof

This follows immediately from Lemma 2.6. ∎

Let 0 ≤ k ≤ q . Then | d ⁢ u ̂ k | ≤ C at each point in R n . Moreover, u ̂ k is a convex function and the Hessian of u ̂ k is bounded by C ⁢ λ k at each point in R n . Here, 𝐶 is independent of 𝛾 and λ 0 .

Proof

The proof is by induction on 𝑘. The assertion is obvious for k = 0 . Suppose next that 1 ≤ k ≤ q and that the assertion is true for u ̂ k − 1 . The differential of u k is given by

Moreover,

where D 2 ⁢ u ̂ k denotes the Hessian of u ̂ k with respect to the Euclidean metric and D 2 ⁢ u ̂ k − 1 denotes the Hessian of u ̂ k − 1 with respect to the Euclidean metric. From this, we deduce that the assertion is true for u ̂ k . This completes the proof of Lemma 2.8. ∎

Let 0 ≤ k ≤ q and let 𝑥 be a point in Ω. Then we can find nonnegative real numbers a 0 , … , a k such that ∑ i = 0 k a i = 1 and d ⁢ u ̂ k = ∑ i = 0 k a i ⁢ d ⁢ u i at the point 𝑥. Moreover, a i = 0 for all 0 ≤ i ≤ k satisfying u i ⁢ ( x ) < u ̂ k ⁢ ( x ) − 2 ⁢ k ⁢ λ i − 1 .

Proof

We argue by induction on 𝑘. The assertion is clearly true for k = 0 . We next assume that 1 ≤ k ≤ q , and the assertion is true for k − 1 . To prove the assertion for 𝑘, we fix an arbitrary point x ∈ Ω .

Case 1.  Suppose that u ̂ k − 1 ⁢ ( x ) − u k ⁢ ( x ) > λ k − 1 . In this case, u ̂ k = u ̂ k − 1 in an open neighborhood of 𝑥. In view of the induction hypothesis, we can find nonnegative real numbers b 0 , … , b k − 1 such that ∑ i = 0 k − 1 b i = 1 and d ⁢ u ̂ k − 1 = ∑ i = 0 k − 1 b i ⁢ d ⁢ u i at the point 𝑥. Moreover, b i = 0 for all 0 ≤ i ≤ k − 1 satisfying u i ⁢ ( x ) < u ̂ k − 1 ⁢ ( x ) − 2 ⁢ ( k − 1 ) ⁢ λ i − 1 . We define nonnegative real numbers a 0 , … , a k by a i = b i for 0 ≤ i ≤ k − 1 and a k = 0 . Then

at the point 𝑥. Moreover, a i = 0 for all 0 ≤ i ≤ k satisfying u i ⁢ ( x ) < u ̂ k ⁢ ( x ) − 2 ⁢ k ⁢ λ i − 1 .

Case 2.  Suppose that u ̂ k − 1 ⁢ ( x ) − u k ⁢ ( x ) < − λ k − 1 . In this case, u ̂ k = u k in an open neighborhood of 𝑥. We define nonnegative real numbers a 0 , … , a k by a i = 0 for 0 ≤ i ≤ k − 1 and a k = 1 . Then ∑ i = 0 k a i = 1 and d ⁢ u ̂ k = d ⁢ u k = ∑ i = 0 k a i ⁢ d ⁢ u i . Moreover, a i = 0 for all 0 ≤ i ≤ k satisfying u i ⁢ ( x ) < u ̂ k ⁢ ( x ) − 2 ⁢ k ⁢ λ i − 1 .

Case 3.  Suppose that | u ̂ k − 1 ⁢ ( x ) − u k ⁢ ( x ) | ≤ λ k − 1 . In this case, Lemma 2.6 implies

In view of the induction hypothesis, we can find nonnegative real numbers b 0 , … , b k − 1 such that ∑ i = 0 k − 1 b i = 1 and d ⁢ u ̂ k − 1 = ∑ i = 0 k − 1 b i ⁢ d ⁢ u i at the point 𝑥. Moreover, we have b i = 0 for all 0 ≤ i ≤ k − 1 satisfying u i ⁢ ( x ) < u ̂ k − 1 ⁢ ( x ) − 2 ⁢ ( k − 1 ) ⁢ λ i − 1 . We define nonnegative real numbers a 0 , … , a k by

for 0 ≤ i ≤ k − 1 , and

Then

If λ 0 is sufficiently large, then the following holds. Let 0 ≤ k ≤ q . If 𝑥 is a point in Ω satisfying − Λ − 1 ≤ u ̂ k ⁢ ( x ) ≤ 0 , then | d ⁢ u ̂ k | ≥ Λ − 1 at the point 𝑥. In particular, Σ ̂ is a smooth hypersurface.

Proof

This follows by combining Lemma 2.3 and Lemma 2.9. ∎

If λ 0 is sufficiently large, then the following holds. Let 0 ≤ j < k ≤ q . If 𝑥 is a point in Ω satisfying − Ξ − 1 ≤ u ̂ j ⁢ ( x ) ≤ 0 , − Ξ − 1 ≤ u k ⁢ ( x ) ≤ 0 , then | d ⁢ u ̂ j ∧ d ⁢ u k | ≥ Ξ − 1 at the point 𝑥.

Proof

Consider a point x ∈ Ω satisfying − Ξ − 1 ≤ u ̂ j ⁢ ( x ) ≤ 0 and − Ξ − 1 ≤ u k ⁢ ( x ) ≤ 0 . By Lemma 2.9, we can find nonnegative real numbers a 0 , … , a j such that ∑ i = 0 j a i = 1 and d ⁢ u ̂ j = ∑ i = 0 j a i ⁢ d ⁢ u i at the point 𝑥. Moreover, a i = 0 for all i ∈ { 0 , 1 , … , j } satisfying u i ⁢ ( x ) < u ̂ j ⁢ ( x ) − 2 ⁢ j ⁢ λ i − 1 . As − Ξ − 1 ≤ u ̂ j ⁢ ( x ) ≤ 0 , it follows that a i = 0 for all i ∈ { 0 , 1 , … , j } satisfying u i ⁢ ( x ) ≤ − 2 ⁢ Ξ − 1 . Using Assumption 1.2 and Lemma 2.4, we obtain

at the point 𝑥. This completes the proof of Lemma 2.11. ∎

Let 1 ≤ k ≤ q . Then

for all 0 < r ≤ 1 . Here, the expression on the left-hand side represents the ( n − 1 ) -dimensional measure. Moreover, B r ⁢ ( p ) denotes a Euclidean ball of radius 𝑟, and 𝐶 is independent of 𝛾 and λ 0 .

Proof

Lemma 2.11 implies that | d ⁢ u ̂ k − 1 ∧ d ⁢ u k | ≥ Ξ − 1 at each point in

Let us consider two arbitrary real numbers a , b ∈ [ − 2 ⁢ λ k − 1 , 0 ] . We apply Proposition A.1 with m = n − 1 and with 𝑓 defined as the restriction of the function u ̂ k − 1 − a to the hyperplane { u k = b } . This gives

for all 0 < r ≤ 1 , where the expression on the left-hand side represents the ( n − 2 ) -dimensional measure. Since d ⁢ u ̂ k − 1 ∧ d ⁢ u k ≠ 0 at each point in

it follows that

for all a , b ∈ [ − 2 ⁢ λ k − 1 , 0 ] and all 0 < r ≤ 1 , where the expression on the left-hand side represents the ( n − 2 ) -dimensional measure.

For abbreviation, let

Moreover, we put S k + = S k ∩ { u ̂ k − 1 − u k ≥ 0 } and S k − = S k ∩ { u ̂ k − 1 − u k ≤ 0 } .

For each b ∈ [ − 2 ⁢ λ k − 1 , 0 ] , we can find a real number a ∈ [ b , 0 ] (depending on 𝑏) such that S k + ∩ { u k = b } ⊂ { u ̂ k − 1 = a } . Using (2.1), we obtain

for each b ∈ [ − 2 ⁢ λ k − 1 , 0 ] and all 0 < r ≤ 1 , where the expression on the left-hand side represents the ( n − 2 ) -dimensional measure. Similarly, for each a ∈ [ − 2 ⁢ λ k − 1 , 0 ] , we can find a real number b ∈ [ a , 0 ] (depending on 𝑎) such that S k − ∩ { u ̂ k − 1 = a } ⊂ { u k = b } . Using (2.1), we obtain

for each a ∈ [ − 2 ⁢ λ k − 1 , 0 ] and all 0 < r ≤ 1 , where the expression on the left-hand side represents the ( n − 2 ) -dimensional measure. In the next step, we integrate inequality (2.2) over b ∈ [ − 2 ⁢ λ k − 1 , 0 ] , and we integrate inequality (2.3) over a ∈ [ − 2 ⁢ λ k − 1 , 0 ] . Using the co-area formula, we obtain

for all 0 < r ≤ 1 . It follows from Lemma 2.8 that | d ⁢ u ̂ k | ≤ C at each point in Ω. Using the identity

we obtain

at each point in S k + and

at each point in S k − . Putting these facts together, we conclude that

for all 0 < r ≤ 1 . Thus, | S k ∩ B r ⁢ ( p ) | ≤ C ⁢ λ k − 1 ⁢ r n − 2 for all 0 < r ≤ 1 . This completes the proof of Lemma 2.12. ∎

Suppose that i , j , k ∈ { 0 , 1 , … , q } are pairwise distinct. If λ 0 is sufficiently large, then

for all λ 0 − 1 ≤ r ≤ 1 . Here, the expression on the left-hand side represents the ( n − 1 ) -dimensional measure. Moreover, B r ⁢ ( p ) denotes a Euclidean ball of radius 𝑟, and 𝐶 is independent of 𝛾 and λ 0 .

Proof

As above, we define

We distinguish two cases.

Case 1.  Suppose that Ω ∩ { u i = 0 } ∩ { u j = 0 } ∩ { u k = 0 } = ∅ . Then

if λ 0 is sufficiently large. Consequently,

if λ 0 is sufficiently large. Thus, the assertion is trivially true in this case.

Case 2.  Suppose that Ω ∩ { u i = 0 } ∩ { u j = 0 } ∩ { u k = 0 } ≠ ∅ . It follows from Assumption 1.1 that the hyperplanes { u i = 0 } , { u j = 0 } , { u k = 0 } must intersect transversally. Suppose now that λ 0 − 1 ≤ r ≤ 1 . We can cover the set

by C ⁢ ( λ 0 ⁢ r ) n − 3 balls of radius λ 0 − 1 . By Lemma 2.12, the intersection of S k with each such ball has area at most C ⁢ λ k − 1 ⁢ λ 0 2 − n . Therefore, the set