Five-point Toponogov theorem

Toponogov theorem provides an if-and-only-if condition on a metric on four-point space that admits an isometric embedding into a complete nonnegatively curved Riemannian manifold. The only-if part is proved by Victor Toponogov, and the if part follows from a result of Abraham Wald [14, §7].

We show that the so-called Lang–Schroeder–Sturm inequality is the analogous condition for five-point spaces.

The only-if part is well-known, but the if part is new. It was hard to imagine some new restrictions on five-point sets, but now we know there are none.

Let us formulate the Lang–Schroeder–Sturm inequality. Consider an $(n+1)$-point array $(p,x_{1},\dots x_{n})$ in a metric space $X$. We say that the array satisfies Lang–Schroeder–Sturm inequality with center $p$ if for any nonnegative values $\lambda_{1},\dots,\lambda_{n}$ we have

where $a_{ij}=|p-x_{i}|_{X}^{2}+|p-x_{j}|_{X}^{2}-|x_{i}-x_{j}|_{X}^{2}$ and we denote by $|\ -\ |_{X}$ the distance between points in $X$.

Recall that any point array in a complete nonnegatively curved Riemannian manifold (and, more generally, in any nonnegatively curved Alexandrov space) meets the Lang–Schroeder–Sturm inequality [4, 12]. In particular, the Lang–Schroeder–Sturm inequalities for all relabelings of points in a finite metric space $F$ gives a necessary condition for the existence of isometric embedding of $F$ into a complete Riemannian manifold with nonnegative curvature. In this note, we show that this condition is sufficient if $F$ has at most 5 points.

In the next section, we will give a reformulation of the theorem using the so-called (4+1)-point comparison [2, 1] which is also equivalent to graph comparison [8] for the star graph shown on the diagram.

Since we know that Lang–Schroeder–Sturm inequalities are necessary, it remains to construct a complete nonnegatively curved Riemannian manifold that contains an isometric copy of a given 5-point space satisfying the assumptions.

Our proof uses a brute-force search of certain configurations that was originally done on a computer. We present a hand-made proof that was found later. It is still based on brute-force search, and we hope that a more conceptual proof will be found. Our paper is inspired by the note of Vladimir Zolotov and the first author [9]; the results from this note are discussed briefly in the last section.

The (n+1)-comparison is another condition that holds for any $(n+1)$-point array in Alexandrov spaces [2, 1]. It says that given a point array $p,x_{1},\dots,x_{n}$ in a nonnegatively curved Alexandrov space $A$ there is an array $\tilde{p},\tilde{x}_{1},\dots,\tilde{x}_{n}$ in a Hilbert space $\mathbb{H}$ such that

for all $i$ and $j$. Point $p$ will be called the center of comparison.

Let us denote by $S_{n}$ the star graph of order $n$; one central vertex in $S_{n}$ is connected to the remaining $n$. It is easy to see that (n+1)-comparison is equivalent to the $S_{n}$-comparison — a particular type of graph comparison introduced in [8].

For general metric spaces, the (n+1)-comparison implies the Lang–Schroeder–Sturm inequality, briefly $\mathrm{LSS}(n)$. For $n\geqslant 5$ the converse does not hold [8, Section 8]. In this section, we will show that these two conditions are equivalent for $n\leqslant 4$.

Applying the claim, we get the following reformulation of the main theorem.

The following proof is nearly identical to the proof of Proposition 4.1 in [8].

Choose a smooth function $\varphi\colon\mathbb{R}\to\mathbb{R}$ such that $\varphi(x)=0$ if $x\geqslant 0$ and $\varphi(x)>0$, $\varphi^{\prime}(x)<0$ if $x<0$. Consider a configuration of points $\tilde{p},\tilde{x}_{1},\dots,\tilde{x}_{4}\in\mathbb{H}$ such that

and the value

is minimal. Note that $s\geqslant 0$; if $s=0$, then we get the required configuration.

Suppose $s>0$. Consider the graph $\Gamma$ with 4 vertices labeled by $\tilde{x}_{1},\tilde{x}_{2},\tilde{x}_{3},\tilde{x}_{4}$ such that $(\tilde{x}_{i},\tilde{x}_{j})$ is an edge if and only if $|\tilde{x}_{i}-\tilde{x}_{j}|_{\mathbb{H}}<|x_{i}-x_{j}|_{A}$. Assume $\tilde{x}_{1}$ has a single incident edge, say $(\tilde{x}_{1},\tilde{x}_{2})$. Since $s$ takes minimal value, we have

— a contradiction. It follows that $\Gamma$ contains no end-vertices. Therefore it is isomorphic to one of the four graphs on the diagram.

Without loss of generality, we can assume that $\tilde{p}=0$. Note that any point $\tilde{x}_{i}$ cannot lie in an open half-space with all its adjacent points. Indeed, assume it does, then $\tilde{x}_{i}$ and all its adjacent points lie in a finite-dimensional half-space; denote by $\Pi$ its boundary plane. Then rotating $\tilde{x}_{i}$ slightly around $\Pi$ will increase the distances from $\tilde{x}_{i}$ to all its adjacent vertices; so the value $s$ will decrease — a contradiction.

In the 6-edge case, $\tilde{p}=0$ lies in the convex hull of $\{\tilde{x}_{1},\tilde{x}_{2},\tilde{x}_{3},\tilde{x}_{4}\}$. In particular, $0=\lambda_{1}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\tilde{x}_{1}+\dots+\lambda_{4}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\tilde{x}_{4}$ for some $\lambda_{i}\geqslant 0$ such that $\lambda_{1}+\lambda_{2}+\lambda_{3}+\lambda_{4}=1$. The latter contradicts $\mathrm{LSS}(4)$.

Similarly, in the 5- and 3-edge cases, we can assume that $\tilde{x}_{1}\tilde{x}_{2}\tilde{x}_{3}$ is a 3-cycle of $\Gamma$. In this case, $0=\lambda_{1}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\tilde{x}_{1}+\lambda_{2}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\tilde{x}_{2}+\lambda_{3}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\tilde{x}_{3}$ for some $\lambda_{i}\geqslant 0$ such that $\lambda_{1}+\lambda_{2}+\lambda_{3}=1$, and we arrive at a contradiction with $\mathrm{LSS}(3)$.

Finally, the 4-edge graph (the 4-cycle) cannot occur. In this case, we may think that $\tilde{x}_{1},\tilde{x}_{2},\tilde{x}_{3},\tilde{x}_{4}$ is the 4-cycle. Note that the points $\tilde{x}_{1},\tilde{x}_{2},\tilde{x}_{3},\tilde{x}_{4}$ lie in one plane so that the direction of $\tilde{x}_{1}$ is opposite to $\tilde{x}_{3}$, and the direction of $\tilde{x}_{2}$ is opposite to $\tilde{x}_{4}$. Let us think that this is the horizontal plane in $\mathbb{R}^{3}$. Then rotating the pair $\tilde{x}_{1}$, $\tilde{x}_{3}$ slightly up and the pair $\tilde{x}_{2}$, $\tilde{x}_{4}$ slightly down, decreases $s$ — a contradiction. ∎

In this section, we recall a construction from [11]. Let $\bm{x}=(x_{1},\dots,x_{n})$ be a point array in a metric space $X$.

Choose a simplex $\triangle$ in $\mathbb{R}^{n-1}$; for example, we can take the standard simplex with the first $(n-1)$ of its vertices $v_{1},\dots,v_{n-1}$ form the standard basis on $\mathbb{R}^{n-1}$, and $v_{n}=0$.

Consider a quadratic form $W_{\bm{x}}$ on $\mathbb{R}^{n-1}$ that is uniquely defined by

for all $i$ and $j$. It will be called the associated form to the point array $\bm{x}$. The following claim is self-evident:

In particular, the condition $W_{\bm{x}}\geqslant 0$ for a triple $\bm{x}=(x_{1},x_{2},x_{3})$ is equivalent to the three triangle inequalities for the distances between $x_{1}$, $x_{2}$, and $x_{3}$. For an $n$-point array, it implies that $W_{\bm{x}}(v)\geqslant 0$ for any vector $v$ in a plane spanned by a triple of vertices of $\triangle$.

The following claim is a reformulation of Lang–Schroeder–Sturm inequalities for all relabeling of $\bm{x}$:

Assume $(p,x_{1},\dots,x_{n})$ is an array of points with a metric that satisfies $\mathrm{LSS}(n)$ with center at $p$. Suppose $n\leqslant 4$; by Claim 2, we have a comparison configuration $(\tilde{p},\tilde{x}_{1},\dots,\tilde{x}_{n})$ with center $p$.

We say that an array $(p,x_{1},\dots,x_{n})$ is tense with center $p$ if the comparison configuration $(\tilde{p},\tilde{x}_{1},\dots,\tilde{x}_{n})$ is unique up to congruence and isometric to the original array.

Note that if $(p,x_{1},\dots,x_{n})$ is tense, then in its comparison configuration $\tilde{p}$ lies in the convex hull of the remaining points $\tilde{x}_{1},\dots,\tilde{x}_{n}$. In particular,

for some nonnegative coefficients $\lambda_{1},\dots,\lambda_{n}$ such that $\lambda_{1}+\dots+\lambda_{n}=1$. If we can choose all positive $\lambda_{i}$ in $({*})$, then we say that $(p,x_{1},\dots,x_{n})$ is a nondegenerate tense array. The following statement describes nondegenerate tense arrays.

Note that any 2-point array is a degenerate tense array; the center can be chosen arbitrarily.

A 3-point array $(p,x,y)$ is tense with center $p$ if we have equality

Note that any tense 3-point array with distinct points is nondegenerate.

Let $(p,x,y,z)$ be a tense 4-point array. Then there is an isometric comparison configuration $(\tilde{p},\tilde{x},\tilde{y},\tilde{z})$ with $\tilde{p}$ lying in the solid triangle $\tilde{x}\tilde{y}\tilde{z}$. Suppose all points $p$, $x$, $y$, and $z$ are distinct. If $\tilde{p}$ lies in the interior of the solid triangle or the triangle is degenerate, then the array $(\tilde{p},\tilde{x},\tilde{y},\tilde{z})$ is nondegenerate. Otherwise, if $\tilde{p}$ lies on a side, say $[\tilde{x},\tilde{y}]$, and the triangle $[\tilde{x}\tilde{y}\tilde{z}]$ is nondegenerate, then $(\tilde{p},\tilde{x},\tilde{y},\tilde{z})$ is degenerate. In the latter case, the 3-point array $(\tilde{p},\tilde{x},\tilde{y})$ is tense and nondegenerate.

It remains to show that $S$ meets the claim — assume not. That is, for arbitrary small $U\in S$ the metric on $\bm{x}$ defined by the associated quadratic form $W_{\bm{x}}+U$ does not satisfy the $\mathrm{LSS}(4)$-inequality. It means that there is a vector $w\in K_{5}$ such that

Choose a positive quadratic form $I$ and minimal $t>0$ such that

for any $w\in K_{5}$.

Note that for some $w\in K_{5}$, we have equality in $({*}{*})$. The metric on $\bm{x}$ that corresponds to the form $W_{\bm{x}}+U+t{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}I$ satisfies all $\mathrm{LSS}(4)$ inequalities and by 4, it has a tense array with at least 3 points.

Choose an array $Q$ that remains to be tense as $U\to 0$. Note that $Q$ is isometric to an array in Euclidean space. Since $W_{\bm{x}}+U+t{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}I\to W_{\bm{x}}$ as $U\to 0$, the array $Q$ must contain one of the three-point tense arrays for the original metric. The latter is impossible since $t>0$ — a contradiction. ∎

Denote by $\mathcal{A}_{5}$ the space of metrics on a 5-point set $F=\{a,b,c,d,e\}$ that admits an embedding into a Riemannian manifold with nonnegative curvature. The associated quadratic forms for spaces in $\mathcal{A}_{5}$ form a convex cone in the space of all quadratic forms on $\mathbb{R}^{4}$. The latter follows since nonnegative curvature survives after rescaling and passing to a product space.

Denote by $\mathcal{B}_{5}$ the space of metrics on $F$ that satisfies all Lang–Schroeder–Sturm inequalities for all relabelings. As well as for $\mathcal{A}_{5}$, the associated forms for spaces in $\mathcal{B}_{5}$ form a convex cone in the space of all quadratic forms on $\mathbb{R}^{4}$.

Since the associated quadratic form describes its metric completely, we may identify $\mathcal{A}_{5}$ and $\mathcal{B}_{5}$ with subsets in $\mathbb{R}^{10}$ — the space of quadratic forms on $\mathbb{R}^{4}$. This way we can think that $\mathcal{A}_{5}$ and $\mathcal{B}_{5}$ are convex cones in $\mathbb{R}^{10}$.

The set $\mathcal{B}_{5}$ is a cone so it does not have extremal points except the origin. The origin corresponds to degenerate metric with all zero distances. But $\mathcal{B}_{5}$ is a cone over a convex compact set $\mathcal{B}_{5}^{\prime}$ in the sphere $\mathbb{S}^{9}\subset\mathbb{R}^{10}$. The extremal points of $\mathcal{B}_{5}^{\prime}$ correspond to extremal rays of $\mathcal{B}_{5}$; metrics on extremal rays will be called extremal. Note that if an extremal metric $\rho$ lies in the interior of a line segment between metrics $\rho^{\prime}$ and $\rho^{\prime\prime}$ in $\mathcal{B}_{5}$, then both metrics $\rho^{\prime}$ and $\rho^{\prime\prime}$ are proportional to $\rho$.

Since Lang–Schroeder–Sturm inequalities are necessary for the existence of isometric embedding into a complete nonnegatively curved Riemannian manifold, we have that

To prove the theorem we need to show that the opposite inclusion holds as well. Since $\mathcal{B}_{5}$ is the convex hull of its extremal metrics, it is sufficient to prove the following:

The remaining part of the proof is broken into cases:

• $\diamond$

  $F$ contains a 5-point tense set. In this case, $F$ admits an isometric embedding into Euclidean space — the problem is solved.

• $\diamond$

  $F$ contains a 4-point tense set. This case follows from Proposition 6 below.

• $\diamond$

  $F$ contains only 3-point tense sets. This is the hardest part of the proof; it follows from Proposition 7. ∎

In the following proof, we first construct a nonnegatively curved Alexandrov space with an isometric copy of $F$ and then smooth it. The space will be a doubling of a convex polyhedral set in $\mathbb{R}^{3}$.

We use notations

for hinge, its angle measure, and the model angle respectively.

By the definition of a tense array, we can choose an array $(\tilde{p},\tilde{x}_{1},\tilde{x}_{2},\tilde{x}_{3})$ in $\mathbb{R}^{2}$ that is isometric to $(p,x_{1},x_{2},x_{3})$. Consider $\mathbb{R}^{2}$ as a plane in $\mathbb{R}^{3}$.

By 2, we can apply the (4+1)-comparison. It implies the existence of point $\tilde{q}\in\mathbb{R}^{3}$ such that

for any $i$.

Further, let us show that there are points $\tilde{q}_{1}$, $\tilde{q}_{2}$, $\tilde{q}_{3}$ in the plane thru $\tilde{p}$, $\tilde{x}_{1}$, $\tilde{x}_{2}$, and $\tilde{x}_{3}$ such that the following four conditions

hold for all $i$ and $j$.

By these conditions, $\tilde{q}_{1}$ must lie on the circle with the center at $x_{1}$ and radius $|x_{1}-q|_{F}$. Denote by $\Gamma_{1}$ the intersection of this circle with the angle vertical to the hinge $[\tilde{x}_{1}\,{}^{\tilde{x}_{2}}_{\tilde{x}_{3}}]$. Let us show that $\tilde{q}_{1}$ can be chosen on $\Gamma_{1}$. The point $\tilde{q}_{1}$ has to satisfy additional three conditions:

for $i\neq 1$. Each condition describes a subarc of $\Gamma_{1}$, say $\breve{X}_{1,2}$, $\breve{X}_{1,3}$, and $\breve{P}_{1}$.

By the construction and comparison we have

for $i\neq 1$. These inequalities and identities imply that each pair of arcs $\breve{X}_{1,2}$, $\breve{X}_{1,3}$, and $\breve{P}_{1}$ have a nonempty intersection. By 1-dimensional Helly’s theorem, all three arcs intersect; so we can choose $\tilde{q}_{1}$ in this intersection. The same way we construct $\tilde{q}_{2}$ and $\tilde{q}_{3}$.

Now let us show that there is a point $\tilde{s}\in\mathbb{R}^{2}$ such that

for all $i$. In other words, the following four closed balls have a nonempty intersection: $\bar{B}[\tilde{p},|p-q|_{F}]$ and $\bar{B}[\tilde{x}_{i},|x_{i}-q|_{F}]$ for all $i$. Indeed, by the overlap lemma [2], any 3 of these balls have a nonempty intersection; it remains to apply Helly’s theorem. Note that we can assume that $\tilde{s}$ lies in the convex hull of $\tilde{x}_{1}$, $\tilde{x}_{2}$, and $\tilde{x}_{3}$.

The four perpendicular bisectors to $[\tilde{s},\tilde{q}]$, $[\tilde{s},\tilde{q}_{1}]$, $[\tilde{s},\tilde{q}_{2}]$, $[\tilde{s},\tilde{q}_{3}]$ cut from $\mathbb{R}^{3}$ a closed convex set $V$ that contains $\tilde{s}$. (It might be a one-sided infinite triangular prism or, if $\tilde{q}$ lies in the plane of the triangle, a two-sided infinite quadrangular prism.) Note that the inequalities $({*})$ and $({*}{*})$ imply that $V$ contains the points $\tilde{p}$, $\tilde{x}_{1}$, $\tilde{x}_{2}$, and $\tilde{x}_{3}$.

Consider the doubling $W$ of $V$ with respect to its boundary; it is an Alexandrov space with nonnegative curvature [10, 5.2]. Denote by $\iota_{1}$ and $\iota_{2}$ the two isometric embeddings $V\to W$. By construction, the array $\hat{p}=\iota_{1}(\tilde{p})$, $\hat{x}_{1}=\iota_{1}(\tilde{x}_{1})$, $\hat{x}_{2}=\iota_{1}(\tilde{x}_{2})$, $\hat{x}_{3}=\iota_{1}(\tilde{x}_{3})$, $\hat{s}=\iota_{2}(\tilde{s})$ in $W$ is isometric to the array $(p,q,x_{1},x_{2},x_{3})$ in $F$.

Finally, we need to show that the obtained space can be smoothed into a Riemannian manifold that still has an isometric copy of $F$. This part is divided into two steps; first, we show that the construction above can be made so that the points $\tilde{s}$, $\tilde{p}$, $\tilde{x}_{1}$, $\tilde{x}_{2}$, $\tilde{x}_{3}$ do not lie on the edges of $V$. In this case, there is a compliment, say $U$, of a neighborhood of the singular set in $W$ that contains the 5-point set together with all the geodesics between them. After that, we construct a smooth Riemannian manifold with nonnegative curvature that contains an isometric copy of $U$.

The construction of the facets of $V$ above implies that each ellipsoid has a tangent plane that contains all the ellipsoids on one side — these planes are the perpendicular bisectors to $[\tilde{s},\tilde{q}]$ and $[\tilde{s},\tilde{q}_{i}]$ for all $i$. Note that any choice of such planes does the trick — they can be used instead of the perpendicular bisectors discussed above. These planes will be called $p$-plane and $x_{i}$-planes respectively. We need to choose them so that no pair of these planes pass thru $\tilde{p}$, $\tilde{x}_{1}$, $\tilde{x}_{2}$, or $\tilde{x}_{3}$. The latter is only possible if the corresponding ellipsoid degenerates to a line segment.

We can assume that one of the ellipsoids is nondegenerate; otherwise, the array $\tilde{p}$, $\tilde{q}$, $\tilde{x}_{1}$, $\tilde{x}_{2}$, $\tilde{x}_{3}$ is isometric to $F$; in this case, $F$ is isometric to a subset of Euclidean space. Further, if only one ellipsoid, say $X_{1}$ is degenerate, then we can move $\tilde{s}$ slightly making this ellipsoid nondegenerate and keeping the rest of its properties. So we can assume that three or two ellipsoids are degenerate.

Suppose $X_{1}$, $X_{2}$, $X_{3}$ are degenerate (picture on the left), then it is easy to choose $x_{i}$-planes tangent to $P$; it solves our problem. Another triple of ellipsoids, say $P$, $X_{1}$, $X_{2}$ might be degenerate only if $\tilde{p}\in[\tilde{x}_{1},\tilde{x}_{2}]$. This case is even simpler — we can choose one plane that contains $\tilde{p}$, $\tilde{x}_{1}$, $\tilde{x}_{2}$, and tangent to $X_{3}$.

Now, suppose exactly two ellipsoids are degenerate; note that in this case $P$ is nondegenerate. Therefore we can assume that $X_{1}$ and $X_{2}$ are degenerate. Further, we can assume that $\tilde{s}\in[\tilde{x}_{1},\tilde{x}_{2}]$; if not we can slightly move $\tilde{s}$ toward $\tilde{x}_{1}$ and $\tilde{x}_{2}$ making $X_{1}$ and $X_{2}$ nondegenerate and keep the rest properties of $\tilde{s}$. Since a focus of $P$ lies on $[\tilde{x}_{1},\tilde{x}_{2}]$, we have that $x_{1}$-plane cannot be $x_{2}$-plane and the other way around.

Suppose that $P$ does not lie in the convex hull of the remaining three ellipsoids and the same holds for $X_{3}$ (middle picture). Then it is easy to make the required choice of planes.

In the remaining case (see picture on the right), either $P$ or $X_{3}$ lies in the convex hull of the remaining three ellipsoids. Suppose it is $P$, draw a $p$-plane; note that it is also an $x_{3}$-plane; it might be also $x_{1}$- or $x_{2}$-plane, but cannot be both. It remains to add $x_{1}$-plane and/or $x_{2}$-plane as needed. Since an $x_{1}$-plane cannot be $x_{2}$-plane and the other way around, we will not get two planes passing thru $x_{1}$ or $x_{2}$.

The case when $X_{3}$ lies in the convex hull of the rest is identical.

A three-point tense set $\{a,b,c\}$ with center $b$ will be briefly denoted by $abc$. Observe that $F$ has tense set $abc$ if and only if

On the diagrams, we will connect three-point tense sets by a smooth curve so that the center is in the middle. For example, the given diagram corresponds to a metric on $\{a,b,c,d,e\}$ with five tense sets $abc$, $bcd$, $cda$, $dae$, $aec$.

In other words, the points in $F$ can be labeled by $\{a,b,c,d,e\}$ so that it has one of the following three tense-set configurations:

In the following proof, we use only a small part of this classification. Namely, we use that it is either the first case (the cycle) or there are two tense sets with a shared center ($bed$ and $cea$). However, the proof of this small part takes nearly as long as the complete classification. (We could exclude cases 11, 12 and 16 on page 7, but we decided to keep them.)

Now, by the classification lemma, we can assume that two tense triples in $F$ have a common center. Let us relabel $F$ by $x,v_{1},v_{2},w_{1},w_{2}$ so that $F$ has tense triples $v_{1}xv_{2}$ and $w_{1}xw_{2}$.

First, we will construct an Alexandrov space $L$ — a flat disc with at most four singular points. The disc $L$ will be triangulated by four triangles with vertices $\hat{x}$, $\hat{v}_{1}$, $\hat{v}_{2}$, $\hat{w}_{1}$, $\hat{w}_{2}$ as shown on the diagram. Each of the four triangles has at most one singular point; in other words, each triangle is a solid geodesic triangle in a cone. The sides of the triangles are the same as in $F$.

Note that the metric on the obtained disc is completely determined by the 12 angles of the triangles. It remains to choose these angles in such a way that $L$ has nonnegative curvature and the map $\iota\colon F\to L$ defined by $x\mapsto\hat{x}$, $v_{i}\mapsto\hat{v}_{i}$, $w_{i}\mapsto\hat{w}_{i}$ is distance-preserving. By construction, $\iota$ is distance-nonexpanding; therefore we only need to show that $\iota$ is distance-noncontracting.

This part is divided into two steps.

First, we need to assume that the 12 angles of the triangles are at least as large as the corresponding model angles; that is,

for all $i$ and $j$.

Further, choose a three-edge path in the triangulation connecting $v_{1}$ to $v_{2}$ (or $w_{1}$ to $w_{2}$), say $v_{1}w_{1}xv_{2}$. Consider the plane polygonal line $\tilde{v}_{1}\tilde{w}_{1}\tilde{x}\tilde{v}_{2}$ with the same angles and sides as in $L$ such that $\tilde{v}_{1}$ and $\tilde{v}_{2}$ lie on the opposite sides from the line $\tilde{w}_{1}\tilde{x}$. Set

The next group of conditions has eight comparisons:

for any $i$. Finally, we need a group of eight identities:

for any $i$.

Now, let us show that these conditions imply that $\iota$ distance-noncontracting. Suppose $\gamma$ is a curve from $\hat{x}$ to $\hat{v}_{i}$ that lies completely in one of the triangles adjacent to the edge $\hat{x}\hat{v}_{i}$. Note that the inequalities in $({*})$ imply that

The same holds for any pair $(\hat{x},\hat{w}_{i})$ and $(\hat{v}_{i},\hat{w}_{j})$. It implies that minimizing geodesic from $x$ to any point on four edges $\hat{x}\hat{v}_{i}$ or $\hat{x}\hat{w}_{i}$ runs in the corresponding edge; in particular, we have

for each $i$. We also get that each of the four edges $\hat{x}\hat{v}_{i}$ or $\hat{x}\hat{w}_{i}$ is a convex set in $L$; in particular, each of these edges can be crossed at most once by a shortest path in $L$.

Suppose that there is a curve $\gamma$ from $\hat{v}_{1}$ to $\hat{v}_{2}$ that is shorter than $|v_{1}-v_{2}|_{F}$. Since two edges $\hat{v}_{1}\hat{x}$ and $\hat{x}\hat{v}_{2}$ have total length $|v_{1}-v_{2}|_{F}$, we can assume that $\gamma$ runs in a pair of two adjacent triangles, say $[\hat{v}_{1}\hat{x}\hat{w}_{1}]$ and $[\hat{v}_{2}\hat{x}\hat{w}_{1}]$. From above, $\gamma$ crosses the edge $\hat{x}\hat{w}_{1}$ once. Denote by $\hat{z}_{1}$ and $\hat{z}_{2}$ the singular points in the triangles $[\hat{v}_{1}\hat{x}\hat{w}_{1}]$ and $[\hat{v}_{2}\hat{x}\hat{w}_{1}]$. We have the following 4 options: