If a triangle can be displaced without distortion, must the surface have constant curvature?

Question

Suppose $S$ is a Riemannian 2-manifold (e.g. a surface in $\mathbb{R}^3$). Let $T$ be a geodesic triangle on $S$: a triangle whose edges are geodesics. If $T$ can be moved around arbitrarily on $S$ while remaining congruent (edge lengths the same, vertex angles the same), does this imply that $S$ has constant curvature?

I realize this is a naive question. If $S$ has constant curvature, then $T$ can be moved around without distortion. I would like to see reasoning for the reverse: If $T$ can be moved around while maintaining congruence, then $S$ must have constant curvature. What is not clear to me is how to formalize "moved around."

Answer 1

Already Riemann in his famous "On the Hypotheses Which Lie at the Bases of Geometry" concludes that the spaces of constant curvature are precisely those in which figures can move without distortion. However, as you correctly remarked, a free mobility of rigid bodies is a rather subtle notion. For historical perspective on this problem see https://arxiv.org/abs/math/0305023 (From quaternions to cosmology: spaces of constant curvature, ca. 1873-1925, by Moritz Epple) and https://arxiv.org/abs/1310.7334 (The problem of space in the light of relativity: the views of H. Weyl and E. Cartan, by Erhard Scholz). Also Jürgen Jost provides a detailed and interesting commentaries on Riemann's paper in the book https://www.springer.com/us/book/9783319260402

Answer 2

This is sort of an answer and sort of not. I'll let you be the judge:

Suppose you formulate the question, not in terms of 'motion' (which you left vague) but terms of 'freely copying' a triangle $T$, as follows:

Let $(S,g)$ be a Riemannian surface, and let $T$ be a triangle, i.e., a triple of points $(p_1,p_2,p_3)$ on the surface together with three geodesic segments $\gamma_{ij}=\gamma_{ji}$ for $i\not=j$ where $p_i$ and $p_j$ are the endpoints of $\gamma_{ij}$. Let $\ell_{ij}>0$ be the length of the geodesic segment $\gamma_{ij}$.

Now, suppose that, side-angle-side holds for this specific $T$ in the following sense: Whenever $(q_1,q_2,q_3)$ are three points of $S$ and $\eta_{12}$, respectively $\eta_{13}$, are geodesic segments of length $\ell_{12}$, respectively $\ell_{13}$, with endpoints $q_1$ and $q_2$, respectively $q_1$ and $q_3$, so that the angle between these geodesic segements at $q_1$ is the same as the angle between the geodesic segments $\gamma_{12}$ and $\gamma_{13}$ at $p_1$, then there exists a geodesic segment $\eta_{23}$ of length $\ell_{23}$ with endpoints $q_2$ and $q_3$, such that for all $i,j,k$ distinct, the angle between $\eta_{ij}$ and $\eta_{ik}$ is the same as the angle between $\gamma_{ij}$ and $\gamma_{ik}$.

Does it then follow that $S$ has constant Gauss curvature?

The answer is 'no', even if $S$ is a sphere: Let $(S,g)$ be a Zoll sphere all of whose geodesics are closed and of length $L$. Fix a point $p$ and let $T$ be a triangle with vertices $(p_1,p_2,p_3) = (p,p,p)$ and let the sides be any three geodesic segments $\gamma_{ij}$ of length $L$ with endpoints $p_i=p$. This 'triangle' can be copied to any triangle $T'$ with vertices $q_i=q$ by choosing the geodesic segments $\eta_{ij}$ so that the angles between sides are equal to the angles between the corresponding sides of $T$.

Answer 3

It is more convenient to talk about hinges, i.e., pair of sides issuing from a vertex. In constant curvature spaces hinges can be moved around without distortion so long as the angle at the vertex is preserved. In rank-1 symmetric spaces hinges can be moved around without distortion so long as a pair of angles at the vertex are preserved. Thus, in the context of complex projective spaces, one has a well-developed trigonometry, including a theorem of cosines expressing the third side in terms of the edges of the hinge and the two angles. This seems to have been first written about by Shirokov in the 1950s. For a study of complex projective trigonometry see e.g., this 1991 article in Geometriae Dedicata.

Answer 4

Here's one approach to formalizing "moved around". Let $G=(V,E)$ be a graph. Let $p:V\rightarrow S$ be a placement of $G$, that is, a map from the vertex set of $G$ to $S$. Let us call the data $(G,p)$ a framework on $S$. Let us define a motion of $(G,p)$ to be a continuous family of placements $f:V\times[0,1]\rightarrow S$ such that:

1. $f(v,0) = p(v)$ for all $v\in V$, that is, the motion begins at $p$.
2. $d_S(f(u,t),f(v,t))=d_S(p(u),p(v))$ for all $uv\in E$ and all $t\in[0,1]$, where $d_S(\cdot,\cdot)$ is the distance function on $S$. This condition just states that the motion preserves the lengths of all edges of $G$.
3. $\alpha_t(u,v,w)=\alpha_0(u,v,w)$ for all triples of vertices $uvw$ such that $uv\in E$ and $vw\in E$, where $\alpha_t(u,v,w)$ is the angle between the geodesic segments $uv$ and $vw$ at $v$. This condition ensures that the motion preserves all angles between adjacent pairs of edges of $G$.

[Such frameworks are related to the point-line frameworks of Jackson and Owen, and also work of Tay, Whiteley, Jackson and Jordán and others on 2D molecular graphs and frameworks (see e.g. this paper of Jackson and Jordán.]

Your question is essentially: Let $(G,p)$ be the framework constructed from a geodesic triangle $T$ on $S$. Suppose there exists a motion from $(G,p)$ to any other congruent geodesic triangle (i.e. one with the same lengths, angles and orientation as $T$). Does this imply that $S$ has constant curvature?

I suspect the answer is no, for a possibly silly reason. It's not obvious to me that generic embedded surfaces need to have any pairs of distinct congruent geodesic triangles at all; the condition would be vacuous on such surfaces, which also don't have constant curvature. So let us add an additional condition on $S$ that such pairs exist.

edit:

Thanks to the answer of Zurab Silagadze, I can see that a positive answer to a related question was claimed by Riemann in §II.4 of his famous paper "Ueber die Hypothesen welche der Geometrie zu Grunde liegen", (see also this English translation by Clifford). Here is an edited version of Clifford's translation of the passage in question:

The common character of manifolds with constant curvature may also be expressed thus, that figures may be moved in them without stretching. For clearly figures could not be arbitrarily shifted and turned round in them if the curvature at each point were not the same in all directions. On the other hand, however, the measure-relations of the manifold are entirely determined by the curvature; they are therefore exactly the same in all directions at one point as at another, and consequently the same constructions can be made from it: whence it follows that in manifolds with constant curvature figures may be given any arbitrary position.

It is not clear to me what "figures" are being considered here, and I admit to not understanding what exactly is proved here, if anything. According to §2.2 of Hans Freudenthal's "Lie groups in the foundations of geometry", the first proof was given by Rudolf Lipschitz in the 1870 paper Fortgesetzte Untersuchungen in Betreff der ganzen homogenen Functionen von n Differentialen. Unfortunately, my German is not up to the task of finding the precise statement in this paper. Any takers?

Answer 5

Suppose that for any two triangles, (pairwise) equal edge sizes imply (pairwise) equal angles between geodesics. Then the curvature has to be constant.

By contradiction, suppose there exist $\epsilon > 0$, $k_1, k_2$ such that $k_1 < k_2 - 3\epsilon$ and points $p_1, p_2$ such that in the (geodesic) ball $B_\epsilon(p_1)$ Gauss curvature is $\le k_1 + \epsilon$ and in $B_\epsilon(p_2)$ Gauss curvature is $ \ge k_2 - \epsilon$. This implies that in $B_\epsilon(p_1)$ all non-degenerate triangles have all angles smaller than the corresponding angles of the comparison triangle in $M_{k_1 + \epsilon}$, where $M_{k_1 + \epsilon}$ is a surface of a constant curvature $k_1 + \epsilon$, and "the comparison triangle" means the triangle with same side lengths. Similarly in $B_\epsilon(p_2)$ all non-degenerate triangles have all angles bigger than the corresponding angles of the comparison triangle in $M_{k_2 - \epsilon}$.

Now if you take any triangle $T$ with the diameter $<\epsilon/10$ in our surface then you can find triangles with the same lengths as $T$ in both $B_\epsilon(p_1)$ and $B_\epsilon(p_2)$. And since we have that "equal edge sizes imply equal angles" then we have: [the angle in $T$] $=$[the angle in comparison triangle in $B_\epsilon(p_1)]$ $<$ [the angle in comparison triangle in $M_{k_1 + \epsilon}]$ $<$
[the angle in comparison triangle in $M_{k_2 - \epsilon}$] $<$ [the angle in comparison triangle in $B_\epsilon(p_2)$] $=$ [the angle in $T$]. And that's a contradiction.

PS: The same argument works for "if two triangles have two pairs of pairwise equal lengths and also the angle between them is the same then the remaining lengths are equal too". This seems even more reasonable formalization of "T can be moved around while maintaining congruence".

Answer 6

The integral over a triangle of the curvature is equal to the difference between the sum of angles for a triangle in a flat surface (that is, $\pi$) and the actual sum of angles for that triangle. $\sum \theta = \pi + \int \kappa $. For two triangles to be congruent, their sum of angles must be equal. So it suffices to show that moving the triangle does not change $\int \kappa $.