Snake Graphs from Triangulated Orbifolds

Abstract

We give an explicit combinatorial formula for the Laurent expansion of any arc or closed curve on an unpunctured triangulated orbifold. We do this by extending the snake graph construction of Musiker, Schiffler, and Williams to unpunctured orbifolds. In the case of an ordinary arc, this gives a combinatorial proof of positivity to the generalized cluster algebra from this orbifold.

Cluster scattering diagrams and theta functions for reciprocal generalized cluster algebras

“…Note that in the above theorem, the Laurent polynomial coefficients are in ZP. Although these coefficients are in fact strictly non-negative for certain subclasses of generalized cluster algebras [1,10], there is currently no proof of positivity for arbitrary generalized cluster algebras. Because ordinary cluster scattering diagrams were used to prove positivity for arbitrary ordinary cluster algebras, the conjectural positivity of arbitrary generalized cluster algebras is a powerful motivator for defining generalized cluster scattering diagrams.…”

“…We can see that D in,s is not consistent by computing p γ,Din,s (z (0,1) ) as: 1) 1 + z (−1,0) (0,1),(0,1) 1 + z (0,1) 1 + z (−1,0) (0,1),(0,1)…”

“…Whereas, here p γ,Din,s z (0,1) = z (0,1) . Making the diagram D in,s consistent requires adding the single wall 1) . Following the same loop γ as above, our calculation now has the additional step z (0,1) 1 + z (−1,1) d3 − → z (0,1) 1 + z (−1,1) (0,1),(−1,−1) • 1) and now p γ,Ds z (0,1) = z (0,1) .…”

“…Making the diagram D in,s consistent requires adding the single wall 1) . Following the same loop γ as above, our calculation now has the additional step z (0,1) 1 + z (−1,1) d3 − → z (0,1) 1 + z (−1,1) (0,1),(−1,−1) • 1) and now p γ,Ds z (0,1) = z (0,1) . In this example, the necessary wall-crossing automorphism can be seen by inspection of p γ,Din,s z (0,1) .…”

“…Similarly, we compute p γ ′ ,Din,s as 1) 1 + az (0,1) + az (0,2) + z (0,3) (1,1),(1,0) 1 + z (−1,0) 1 + az (0,1) + az (0,2) + z (0,3) (−1,0), (1,0) = z (1,1) 1 + az (0,1) + az (0,2) + z (0,3) 1 + z (−1,0) 1 + az (0,1) + az (0,2) + z (0,3) = z (1,1) 1 + az (0,1) + az (0,2) + z (0,3) + z (−1,0)…”