Lengthening a tetrahedron

Abstract

We give rigorous, computer assisted proofs of a number of statements about the effect on the volume of lengthening various edges of a tetrahedron. Our results give new and sharp polynomial inequalities concerning the Cayley–Menger determinant and its partial derivatives.

Appendix: Existence in all dimensions

Here I’ll give the proof of Theorem 1.2. I learned all the arguments here from Peter Doyle and Igor Rivin. Let $D$ stand for a list $\{d_{ij}\}$. We will perform operations componentwise, so that $D+t=\{d_{ij}+t\}$, etc. We define tetrahedral lists in all dimensions just as in the $3$ dimensional case. Let $T$ denote the space of tetrahedral lists.

Theorem 6.1

If $D \in T$ then $\sqrt{D} \in T$.

See [7, Corollary 4.8]. In [7] this result is attributed to Von Neumann, though Rivin calls it Schoenberg’s result.

Theorem 6.2

If $A,B \in T$ then $\sqrt{A^2+B^2} \in T$. The simplex represented by $\sqrt{A^2+B^2}$ has larger volume than the simplex represented by $A$.

Proof

See [4] for a proof. See also [1, Lemma 1]. Here is a self-contained proof.

Given a quadratic form $Q$ and a linear isomorphism $L$, we have the new quadratic form

$$\begin{aligned} L^*Q(v,w)=Q(L^{-1}(v),L^{-1}(w)). \end{aligned}$$

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Let $\Delta \subset {{\varvec{R}}}^n$ denote some copy of the regular simplex. Let $L_A$ denote the linear transformation which carries $\Delta$ to a simplex $\Delta _A$ whose lengths are realized by $A$. Let $Q_A$ be the quadratic form such that $L_A^*(Q_A)$ is the standard quadratic form—i.e. the dot product. By construction, $Q_A$ assigns the length list $A$ to the sides of $\Delta$. Likewise define $Q_B$. The sum $Q_C=Q_A+Q_B$ is also positive definite. Given an edge $e$ of $\Delta$. Let $C$ denote the list of lengths that $Q_C$ assigns to $\Delta$. We compute

$$\begin{aligned} C_e=\sqrt{Q_C(e,e)}=\sqrt{Q_A(e,e)+Q_B(e,e)}=\sqrt{A_e^2+B_e^2}. \end{aligned}$$

This shows that $Q_C$ assigns the corresponding number on the list $\sqrt{A^2+B^2}$ to the edge $e$. But then let $L_C$ be the linear transformation which pushes $L_C$ forward to the standard quadratic form. By construction $L_C(\Delta )$ is the simplex realizing the list $\sqrt{A^2+B^2}$.

Note that $L_C(e)>L_A(e)$ for all $e \in {{\varvec{R}}}^n$. But then the linear map carrying $\Delta _A$ to $\Delta _C$ strictly increases all distances, and hence also increases volume. $\square$

Let $\phi _t$ denote the lengthening flow. That is, $\phi _t(D)=D+t$. Let $D_t=\phi _t(D)$.

Lemma 6.3

$$\begin{aligned} \frac{d}{dt}\mathrm{vol}(D_t) \ge 0 \end{aligned}$$

for all $D \in T$.

Proof

Choose some point $D \in T$. By Theorem 6.1 and scaling, $\sqrt{2tD} \in T$. By Theorem 6.2. $\sqrt{D^2+2tD} \in T$. But we can write out

$$\begin{aligned} \sqrt{D^2+2tD}=D(\sqrt{1+2t/D}) = D+t + \mathrm{higher\ order\ terms} \end{aligned}$$

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Theorem 6.2 shows that

$$\begin{aligned} \mathrm{vol}(D+t+\mathrm{higher\ order\ terms})>\mathrm{vol}(D_0). \end{aligned}$$

Taking the limit as $t \rightarrow 0$ we get the result of this lemma. $\square$

Lemma 6.3 implies that $D_t \in T$ for all $t$, because $\mathrm{vol}(D_t)$ cannot converge to $0$. This takes part of the existence statement in Theorem 1.2.

To see that $\mathrm{vol}(D_t)$ is strictly increasing, we look more carefully at the proof of Theorem 6.2. When all the terms on the list $B$ have size $O(t)$, and $t$ is much smaller than the minimum length on the list $A$, the quadratic form $Q_C$ assigns the length $A_e+O(t)$ to each unit length vector $e \in {{\varvec{R}}}^n$. But then the linear map $\Delta _A$ to $\Delta _C$ increases all unit distances by $O(t)$, and hence increases volume by $O(t)$ as well. This gives

$$\begin{aligned} \mathrm{vol}(D+t+\mathrm{higher\ order\ terms})>\mathrm{vol}(D_0)+C_Dt, \end{aligned}$$

for some positive constant $C_D$. Taking the limit, we see that $\mathrm{vol}(D_t)$ is strictly increasing, and in fact has positive derivative. This completes the proof of Theorem 1.2.