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Edit: We thank Vladimir Matveev for his comment on this post which leeds us to revise the question as follows:

Assume that $M_{g}$ is a compact Riemann surface with constant negative cuvature (That is $g>1$).

Assume that $ABC$ is a triangle on $M_{g}$. Is there a triangle $A'B'C'$ such that $AB=A'B',\;\;AC=A'C',\;\;BC=B'C'$ but there is no an isometry which carries the first triangle to the second one. (Note that the sides of triangles are geodesics.We can consider both cases minimizing or not minimizing geodesics)

As another question: Assume that ABC and A'B'C' are two triangles which satisfies the above edges equality and they are homologues to zero. Does this imply that their interiors have the same volume?

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    $\begingroup$ just take a hyperbolic surface admitting a self-isometry, take a small triange ABC on it and its image A'B'C' w.r.t. the self-isometry, and slightly change the metric in a small neightborhood inside the first triangle. You would not change the edges but there will be no isometry that carries the first to the second, and their interiours have different volume. $\endgroup$ Jul 24, 2015 at 20:32
  • $\begingroup$ @VladimirSMatveev Could you please more explain? My question was about $M_{g}$ with its standard metrics whose isometry group is at most 84(g-1). It seems that you change the geometry, yes? $\endgroup$ Jul 24, 2015 at 20:40
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    $\begingroup$ I misunderstood your question in which actually you did not specify that your metric is of constand curvature. Indeed, you are right, in my example I changed the geometry $\endgroup$ Jul 24, 2015 at 21:04
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    $\begingroup$ May be it would help if you give other definitions you are working with: for example you said ``triangle''. I assume that the sides are geodesics. Are they minimal geodesics or simply geodesics? $\endgroup$ Jul 24, 2015 at 21:08
  • $\begingroup$ @VladimirSMatveev I am sorry for some unclear point in my question. Now I revise it. $\endgroup$ Jul 24, 2015 at 21:25

2 Answers 2

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If by isometry, you mean "isometry of the surface", the answer is NO. Indeed, the isometry group of a hyperbolic surface is finite. So, pick a small triangle $\Delta=ABC,$ and let $\Delta(\theta)$ be $\Delta$ rotated by an angle $\theta$ about some point $x_0.$ By finiteness of the isometry group, there is a finite set of $\theta for which there is an isometry.

For the second, the answer is obviously NO. Pick your surface in such a way that there are two separating geodesics of the same length (but which bound different pieces of the surface). Since such a geodesic can be thought of as a triangle, you have your answer.

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  • $\begingroup$ Help me to understand your reasoning. First, by transitivity, you reduce the problem to the case when the two triangles have a vertex in common, say $x_0$: then you are basically asking whether the isotropy subgroup $G_{x_0}$ is 2-transitive (any triangle with one vertex $x_0$ is uniquely determined by two other points). And you say "NO" by using finiteness of $G_{x_0}$: is this enough? What if the surface is the hyperbolic plane: does this finiteness argument works? $\endgroup$ Jul 25, 2015 at 3:25
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    $\begingroup$ The OP's question was a general question, so any counterexample works. In fact, the reasoning works for ANY pair of triangles, and an arbitrary point $x_0.$ Since the isometry group of $M_g$ is finite, a generic pair of congruent triangles will not have an isometry mapping one to another. The OP explicitly specified that the surface was compact. Of course, the isometry group of $\mathbb{H}^2$ is very far from finite, and there the statement is tautologically true. $\endgroup$
    – Igor Rivin
    Jul 25, 2015 at 3:52
  • $\begingroup$ Now I understand. You are absolutely right. $\endgroup$ Jul 25, 2015 at 9:44
  • $\begingroup$ @IgorRivin I am sorry if my question is elementary:What do you mean by rotation? We know that $M_{g}$ is not embedded in R^3, isometrically. $\endgroup$ Jul 25, 2015 at 18:35
  • $\begingroup$ Rotation is understood in a universal cover (pick your triangles small enough so that they both lie in a disk). $\endgroup$
    – Igor Rivin
    Jul 25, 2015 at 18:48
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Am I right to assume that you wanted to ask weather the two triangles on the same surface are congruent after lifting them to the universal cover (the hyperbolic plane)? Then the answer is again no, in the case when we don't know what kind of subsurface of the surface each triangle bounds (on both sides). For instance one triangle could bound a disk, so it lifts to a standard hyperbolic triangle, but the other could bound a hyperbolic torus with polygonal (triangular) boundary on one side and a surface of even higher genus on the other (the boundary being the triangle). By the way, it doesn't even have to bound anything specific, i.e. it might have a non-zero homology class. As Igor Rivin pointed out, a special case is a simple closed geodesic whose length is equal to the perimeter of a triangle on the surface that lifts to a standard triangle on the hyperbolic plane via the universal covering map. Then we can choose three points on the geodesic in question so that it gets split into three segments, each equal in length to the corresponding edge from the other triangle. Analogous arguments (more or less similar to Igor Rivin's above) give also an answer to your second question, which is again no. There are even more non-trivial examples where both triangles are strictly "convex". i.e. the angles at their vertices (on one side) are less than $\pi/2$. In the closed geodesic example, draw three very very small geodesic segments of the same length $\varepsilon$ perpendicular to the closed geodesic at the three points we have already fixed. All segments should be on "one side" of the geodesic. Then the other three endpoints would form such a "convex triangle. Then perturb the other, "trivial" triangle so that it has the same corresponding edge-lengths. You can carry our this construction even on a pair of pants without the need to visualize the whole surface. It is much easier. The bottom line is, that homology is not strict enough. Probably you want the two triangles to be freely homotopic to zero on the surface, i.e. they both bound disks? Then the answer in this case should be yes, they are congruent, after lifting them the the hyperbolic plane (please, correct me if I am wrong here).

Edit: Maybe I am trying to fit the question to the answers given so far, but it seems to me the question is roughly meant to sound something like that: Let $M$ be a compact hyperbolic surface and assume we have three geodesic segments $AB, BC, CA$ on the surface forming one simple triangular piece-wise geodesic curve $\Delta$ and three more $A'B', B'C', C'A'$ forming another such curve $\Delta'$. It is given that $|AB|=|A'B'|, \, |BC|=|B'C'|, \, |CA|=|C'A'|$, where $|AB|$ is the hyperbolic length of the geodesic segment $AB$, etc.

Is there an isometry $h : N(\Delta) \to N(\Delta')$ between (small enough) tubular neighborhoods $N(\Delta)$ and $N(\Delta')$ of $\Delta$ and $\Delta'$ respectively (collars, if you prefer this terminology), which maps $\Delta$ to $\Delta'$, i.e. $h(\Delta) = \Delta'$?

The answer is in general no, but in some cases, (for example when both triangular curves can be freely homotoped to points) there is. But there are other, more non-trivial cases, in which such isometry exists. For instance, since $\Delta$ is a simple closed loop on $M$ there exists a unique simple closed geodesic $\delta$ freely homotopic to $\Delta$ (in the non-trivial case). The same is true for $\Delta'$. Denote by $\delta'$ the simiple closed geodesic freely homotopic to $\Delta'$. Assume for simplicity that $\Delta$ and $\delta$ bound a cylinder embedded in $M$. Assume the same is true for $\Delta'$ and $\delta$. Then, if you take the points $A, B, C$ and draw on the cylinder bounded by $\Delta$ and $\delta$ the shortest geodesic perpendiculars from each of these three points to the geodesic $\delta$, you obtain three geodesic segments of length $h_A, h_B, h_C$ orthogonal to $\delta$. Do the same for $\Delta'$ and $\delta'$. Then $|AB|=|A'B'|, \, |BC|=|B'C'|, \, |CA|=|C'A'|$ and $h_A=h_{A'}, \, h_B=h_{B'}, \, h_C=h_{C'}$ if and only if there are two tubular neighborhoods that are isometric together with $\Delta$ and $\Delta'$. In this case, $|\delta|=|\delta'|$ (this is a necessary condition). To generalize to non embedded cylinders, consider oriented distances to between the vertices of the triangles and the corresponding simple closed geodesics.

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  • $\begingroup$ I try to underestand the details of your answer. But in my question I did not concern with lifting triangles and universal covering. In fact my question is near to your material after the "EDIT" but my question was global. In fact my question: We have a triangle on Mg and we search for another triangle congruent to it but no global isometry can send the initial triangle to this triangle. I confess that I need times to understand your and the other answer. $\endgroup$ Jul 25, 2015 at 18:44
  • $\begingroup$ Yeah, it seems to me that your comment confirms my assumption that what I wrote after the edit is what you are probably interested in. I basically tried to put it in a more ''formal" way. I am sorry if I have been a bit sloppy and not too precise. $\endgroup$ Jul 26, 2015 at 1:21
  • $\begingroup$ I think when people say congruent they mean global isometry, like a ''motion" of the whole space (e.g. hyperbolic plane) that moves one triangle onto the other, so that they match precisely. It is a notion related to the isometry group of the whole space (hence Igor's answer). The term isometry, however, I think means something more general and can be used in a more flexible way: a map that preserves lengths (and angles when angles are well defined). Hence my definition of isometric triangular piece-wise geodesic curves (we call triangles) in terms of annular/tubular/collar/ neighborhoods. $\endgroup$ Jul 26, 2015 at 1:26
  • $\begingroup$ I am sorry. I was incorrect to use congruent for "two triangle with equal sides". $\endgroup$ Jul 26, 2015 at 3:49
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    $\begingroup$ By the way, the definition with tubular neighborhoods, allows for two isometric triangles to have interiors with arbitrary topology, and so two isometric triangles homologous to zero can bound interiors with completely different area. $\endgroup$ Jul 26, 2015 at 9:10

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