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Tag: quivers

quivers versus quilts

We have associated to a subgroup of the modular group $PSL_2(\mathbb{Z}) $ a quiver (that is, an oriented graph). For example, one verifies that the fundamental domain of the subgroup $\Gamma_0(2) $ (an index 3 subgroup) is depicted on the right by the region between the thick lines with the identification of edges as indicated. The associated quiver is then

\[
\xymatrix{i \ar[rr]^a \ar[dd]^b & & 1 \ar@/^/[ld]^h \ar@/_/[ld]_i \\
& \rho \ar@/^/[lu]^d \ar@/_/[lu]_e \ar[rd]^f & \\
0 \ar[ru]^g & & i+1 \ar[uu]^c}
\]

The corresponding “dessin d’enfant” are the green edges in the picture. But, the red dot on the left boundary is identied with the red dot on the lower circular boundary, so the dessin of the modular subgroup $\Gamma_0(2) $ is

\[
\xymatrix{| \ar@{-}[r] & \bullet \ar@{-}@/^8ex/[r] \ar@{-}@/_8ex/[r] & -}
\]

Here, the three red dots (all of them even points in the Dedekind tessellation) give (after the identification) the two points indicated by a $\mid $ whereas the blue dot (an odd point in the tessellation) is depicted by a $\bullet $. There is another ‘quiver-like’ picture associated to this dessin, a quilt of the modular subgroup $\Gamma_0(2) $ as studied by John Conway and Tim Hsu.

On the left, a quilt-diagram copied from Hsu’s book Quilts : central extensions, braid actions, and finite groups, exercise 3.3.9. This ‘quiver’ has also 5 vertices and 7 arrows as our quiver above, so is there a connection?

A quilt is a gadget to study transitive permutation representations of the braid group $B_3 $ (rather than its quotient, the modular group $PSL_2(\mathbb{Z}) = B_3/\langle Z \rangle $ where $\langle Z \rangle $ is the cyclic center of $B_3 $. The $Z $-stabilizer subgroup of all elements in a transitive permutation representation of $B_3 $ is the same and hence of the form $\langle Z^M \rangle $ where M is called the modulus of the representation. The arrow-data of a quilt, that is the direction of certain edges and their labeling with numbers from $\mathbb{Z}/M \mathbb{Z} $ (which have to satisfy some requirements, the flow rules, but more about that another time) encode the Z-action on the permutation representation. The dimension of the representation is $M \times k $ where $k $ is the number of half-edges in the dessin. In the above example, the modulus is 5 and the dessin has 3 (half)edges, so it depicts a 15-dimensional permutation representation of $B_3 $.

If we forget the Z-action (that is, the arrow information), we get a permutation representation of the modular group (that is a dessin). So, if we delete the labels and directions on the edges we get what Hsu calls a modular quilt, that is, a picture consisting of thick edges (the dessin) together with dotted edges which are called the seams of the modular quilt. The modular quilt is merely another way to depict a fundamental domain of the corresponding subgroup of the modular group. For the above example, we have the indicated correspondences between the fundamental domain of $\Gamma_0(2) $ in the upper half-plane (on the left) and as a modular quilt (on the right)

That is, we can also get our quiver (or its opposite quiver) from the modular quilt by fixing the orientation of one 2-cell. For example, if we fix the orientation of the 2-cell $\vec{fch} $ we get our quiver back from the modular quilt


\[
\xymatrix{i \ar[rr]^a \ar[dd]^b & & 1 \ar@/^/[ld]^h \ar@/_/[ld]_i \\
& \rho \ar@/^/[lu]^d \ar@/_/[lu]_e \ar[rd]^f & \\
0 \ar[ru]^g & & i+1 \ar[uu]^c}
\]

This shows that the quiver (or its opposite) associated to a (conjugacy class of a) subgroup of $PSL_2(\mathbb{Z}) $ does not depend on the choice of embedding of the dessin (or associated cuboid tree diagram) in the upper half-plane. For, one can get the modular quilt from the dessin by adding one extra vertex for every connected component of the complement of the dessin (in the example, the two vertices corresponding to 0 and 1) and drawing a triangulation from them (the dotted lines or ‘seams’).

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the modular group and superpotentials (2)

Last time we have that that one can represent (the conjugacy class of) a finite index subgroup of the modular group $\Gamma = PSL_2(\mathbb{Z}) $ by a Farey symbol or by a dessin or by its fundamental domain. Today we will associate a quiver to it.

For example, the modular group itself is represented by the Farey symbol
[tex]\xymatrix{\infty \ar@{-}[r]_{\circ} & 0 \ar@{-}[r]_{\bullet} & \infty}[/tex] or by its dessin (the green circle-edge) or by its fundamental domain which is the region of the upper halfplane bounded by the red and blue vertical boundaries. Both the red and blue boundary consist of TWO edges which are identified with each other and are therefore called a and b. These edges carry a natural orientation given by circling counter-clockwise along the boundary of the marked triangle (or clockwise along the boundary of the upper unmarked triangle having $\infty $ as its third vertex). That is the edge a is oriented from $i $ to $0 $ (or from $i $ to $\infty $) and the edge b is oriented from $0 $ to $\rho $ (or from $\infty $ to $\rho $) and the green edge c (which is an inner edge so carries no identifications) from $\rho $ to $i $. That is, the fundamental region consists of two triangles, glued together along their boundary which is the oriented cycle $\vec{abc} $ consistent with the fact that the compactification of $\mathcal{H}/\Gamma $ is the 2-sphere $S^2 = \mathbb{P}^1_{\mathbb{C}} $. Under this identification the triangle-boundary abc can be seen to circle the equator whereas the top triangle gives the upper half sphere and the lower triangle the lower half sphere. Emphasizing the orientation we can depict the triangle-boundary as the quiver

[tex]\xymatrix{i \ar[rd]_a & & \rho \ar[ll]_c \\ & 0 \ar[ru]_b}[/tex]

embedded in the 2-sphere. Note that quiver is just a fancy name for an oriented graph…

Okay, let’s look at the next case, that of the unique index 2 subgroup $\Gamma_2 $ represented by the Farey symbol [tex]\xymatrix{\infty \ar@{-}[r]_{\bullet} & 0 \ar@{-}[r]_{\bullet} & \infty}[/tex] or the dessin (the two green edges) or by its fundamental domain consisting of the 4 triangles where again the left and right vertical boundaries are to be identified in parts.

That is we have 6 edges on the 2-sphere $\mathcal{H}/\Gamma_2 = S^2 $ all of them oriented by the above rule. So, for example the lower-right triangle is oriented as $\vec{cfb} $. To see how this oriented graph (the quiver) is embedded in $S^2 $ view the big lower region (cdab) as the under hemisphere and the big upper region (abcd) as the upper hemisphere. So, the two green edges together with a and b are the equator and the remaining two yellow edges form the two parts of a bigcircle connecting the north and south pole. That is, the graph are the cut-lines if we cut the sphere in 4 equal parts. The corresponding quiver-picture is

[tex]\xymatrix{& i \ar@/^/[dd]^f \ar@/_/[dd]_e & \\
\rho^2 \ar[ru]^d & & \rho \ar[lu]_c \\
& 0 \ar[lu]^a \ar[ru]_b &}[/tex]

As a mental check, verify that the index 3 subgroup determined by the Farey symbol [tex]\xymatrix{\infty \ar@{-}[r]_{\circ} & 0 \ar@{-}[r]_{\circ} & 1 \ar@{-}[r]_{\circ} & \infty}[/tex] , whose fundamental domain with identifications is given on the left, has as its associated quiver picture

[tex]\xymatrix{& & \rho \ar[lld]_d \ar[ld]^f \ar[rd]^e & \\
i \ar[rrd]_a & i+1 \ar[rd]^b & & \omega \ar[ld]^c \\
& & 0 \ar[uu]^h \ar@/^/[uu]^g \ar@/_/[uu]_i &}[/tex]

whereas the index 3 subgroup determined by the Farey symbol [tex]\xymatrix{\infty \ar@{-}[r]_{1} & 0 \ar@{-}[r]_{1} & 1 \ar@{-}[r]_{\circ} & \infty}[/tex], whose fundamental domain with identifications is depicted on the right, has as its associated quiver

[tex]\xymatrix{i \ar[rr]^a \ar[dd]^b & & 1 \ar@/^/[ld]^h \ar@/_/[ld]_i \\
& \rho \ar@/^/[lu]^d \ar@/_/[lu]_e \ar[rd]^f & \\
0 \ar[ru]^g & & i+1 \ar[uu]^c}[/tex]

Next time, we will use these quivers to define superpotentials…

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Superpotentials and Calabi-Yaus

Yesterday, Jan Stienstra gave a talk at theARTS entitled “Quivers, superpotentials and Dimer Models”. He started off by telling that the talk was based on a paper he put on the arXiv Hypergeometric Systems in two Variables, Quivers, Dimers and Dessins d’Enfants but that he was not going to say a thing about dessins but would rather focuss on the connection with superpotentials instead…pleasing some members of the public, while driving others to utter despair.

Anyway, it gave me the opportunity to figure out for myself what dessins might have to do with dimers, whathever these beasts are. Soon enough he put on a slide containing the definition of a dimer and from that moment on I was lost in my own thoughts… realizing that a dessin d’enfant had to be a dimer for the Dedekind tessellation of its associated Riemann surface!
and a few minutes later I could slap myself on the head for not having thought of this before :

There is a natural way to associate to a Farey symbol (aka a permutation representation of the modular group) a quiver and a superpotential (aka a necklace) defining (conjecturally) a Calabi-Yau algebra! Moreover, different embeddings of the cuboid tree diagrams in the hyperbolic plane may (again conjecturally) give rise to all sorts of arty-farty fanshi-wanshi dualities…

I’ll give here the details of the simplest example I worked out during the talk and will come back to general procedure later, when I’ve done a reference check. I don’t claim any originality here and probably all of this is contained in Stienstra’s paper or in some physics-paper, so if you know of a reference, please leave a comment. Okay, remember the Dedekind tessellation ?

So, all hyperbolic triangles we will encounter below are colored black or white. Now, take a Farey symbol and consider its associated special polygon in the hyperbolic plane. If we start with the Farey symbol

[tex]\xymatrix{\infty \ar@{-}_{(1)}[r] & 0 \ar@{-}_{\bullet}[r] & 1 \ar@{-}_{(1)}[r] & \infty} [/tex]

we get the special polygonal region bounded by the thick edges, the vertical edges are identified as are the two bottom edges. Hence, this fundamental domain has 6 vertices (the 5 blue dots and the point at $i \infty $) and 8 hyperbolic triangles (4 colored black, indicated by a black dot, and 4 white ones).

Right, now let us associate a quiver to this triangulation (which embeds the quiver in the corresponding Riemann surface). The vertices of the triangulation are also the vertices of the quiver (so in our case we are going for a quiver with 6 vertices). Every hyperbolic edge in the triangulation gives one arrow in the quiver between the corresponding vertices. The orientation of the arrow is determined by the color of a triangle of which it is an edge : if the triangle is black, we run around its edges counter-clockwise and if the triangle is white we run over its edges clockwise (that is, the orientation of the arrow is independent of the choice of triangles to determine it). In our example, there is one arrows directed from the vertex at $i $ to the vertex at $0 $, whether you use the black triangle on the left to determine the orientation or the white triangle on the right. If we do this for all edges in the triangulation we arrive at the quiver below

where x,y and z are the three finite vertices on the $\frac{1}{2} $-axis from bottom to top and where I’ve used the physics-convention for double arrows, that is there are two F-arrows, two G-arrows and two H-arrows. Observe that the quiver is of Calabi-Yau type meaning that there are as much arrows coming into a vertex as there are arrows leaving the vertex.

Now that we have our quiver we determine the superpotential as follows. Fix an orientation on the Riemann surface (for example counter-clockwise) and sum over all black triangles the product of the edge-arrows counterclockwise MINUS sum over all white triangles
the product of the edge arrows counterclockwise. So, in our example we have the cubic superpotential

$IH’B+HAG+G’DF+FEC-BHI-H’G’A-GFD-CEF’ $

From this we get the associated noncommutative algebra, which is the quotient of the path algebra of the above quiver modulo the following ‘commutativity relations’

$\begin{cases} GH &=G’H’ \\ IH’ &= IH \\ FE &= F’E \\ F’G’ &= FG \\ CF &= CF’ \\ EC &= GD \\ G’D &= EC \\ HA &= DF \\ DF’ &= H’A \\ AG &= BI \\ BI &= AG’ \end{cases} $

and morally this should be a Calabi-Yau algebra (( can someone who knows more about CYs verify this? )). This concludes the walk through of the procedure. Summarizing : to every Farey-symbol one associates a Calabi-Yau quiver and superpotential, possibly giving a Calabi-Yau algebra!

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