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Tag: superpotential

the modular group and superpotentials (1)

Here I will go over the last post at a more leisurely pace, focussing on a couple of far more trivial examples. Here’s the goal : we want to assign a quiver-superpotential to any subgroup of finite index of the modular group. So fix such a subgroup $\Gamma’ $ of the modular group $\Gamma=PSL_2(\mathbb{Z}) $ and consider the associated permutation representation of $\Gamma $ on the left-cosets $\Gamma/\Gamma’ $. As $\Gamma \simeq C_2 \ast C_3 $ this representation is determined by the action of the order 2 and order 3 generators of the modular group. There are a number of combinatorial gadgets to control the subgroup $\Gamma’ $ and the associated permutation representation : (generalized) Farey symbols and dessins d’enfants.

Recall that the modular group acts on the upper-halfplane (the ‘hyperbolic plane’) by Moebius transformations, so to any subgroup $\Gamma’ $ we can associate a fundamental domain for its restricted action. The dessins and the Farey symbols give us a particular choice of these fundamental domains. Let us consider the two most trivial subgroups of all : the modular group itself (so $\Gamma/\Gamma $ is just one element and therefore the associated permutation representation is just the trivial representation) and the unique index two subgroup $\Gamma_2 $ (so there are two cosets $\Gamma/\Gamma_2 $ and the order 2 generator interchanges these two while the order 3 generator acts trivially on them). The fundamental domains of $\Gamma $ (left) and $\Gamma_2 $ (right) are depicted below

In both cases the fundamental domain is bounded by the thick black (hyperbolic) edges. The left-domain consists of two hyperbolic triangles (the upper domain has $\infty $ as the third vertex) and the right-domain has 4 triangles. In general, if the subgroup $\Gamma’ $ has index n, then its fundamental domain will consist of $2n $ hyperbolic triangles. Note that these triangles are part of the Dedekind tessellation so really depict the action of $PGL_2(\mathbb{Z} $ and any $\Gamma $-hyperbolic triangle consists of one black and one white triangle in Dedekind’s coloring. We will indicate the color of a triangle by a black circle if the corresponding triangle is black. Of course, the bounding edges of the fundamental domain need to be identified and the Farey symbol is a notation device to clarify this. The Farey symbols of the above domains are
[tex]\xymatrix{\infty \ar@{-}[r]_{\circ} & 0 \ar@{-}[r]_{\bullet} & \infty}[/tex] and [tex]\xymatrix{\infty \ar@{-}[r]_{\bullet} & 0 \ar@{-}[r]_{\bullet} & \infty}[/tex] respectively. In both cases this indicates that the two bounding edges on the left are to be identified as are the two bounding edges on the right (so, in particular, after identification $\infty $ coincides with $0 $). Hence, after identification, the $\Gamma $ domain consists of two triangles on the vertices ${ 0,i,\rho } $ (where $\rho=e^{2 \pi i}{6} $) (the blue dots) sharing all three edges, the $\Gamma_2 $ domain consists of 4 triangles on the 4 vertices ${ 0,i,\rho,\rho^2 } $ (the blue dots). In general we have three types of vertices : cusps (such as 0 or $\infty $), even vertices (such as $i $ where there are 4 hyperbolic edges in the Dedekind tessellation) and odd vertices (such as $\rho $ and $\rho^2 $ where there are 6 hyperbolic edges in the tessellation).

Another combinatorial gadget assigned to the fundamental domain is the cuboid tree diagram or dessin. It consists of all odd and even vertices on the boundary of the domain, together with all odd and even vertices in the interior. These vertices are then connected with the hyperbolic edges connecting them. If we color the even vertices red and the odds blue we have the indicated dessins for our two examples (the green pictures). An half-edge is an edge connecting a red and a blue vertex in the dessin and we number all half-edges. So, the $\Gamma $-dessin has 1 half-edge whereas the $\Gamma_2 $-dessin has two (in general, the number of these half-edges is equal to the index of the subgroup). Observe also that every triangle has exactly one half-edge as one of its three edges. The dessin gives all information to calculate the permutation representation on the coset-set $\Gamma/\Gamma’ $ : the action of the order 2 generator of $\Gamma $ is given by taking for each internal red vertex the two-cycle $~(a,b) $ where a and b are the numbers of the two half-edges connected to the red vertex and the action of the order 3 generator is given by taking for every internal blue vertex the three cycle $~(c,d,e) $ where c, d and e are the numbers of the three half-edges connected to the blue vertex in counter-clockwise ordering. Our two examples above are a bit too simplistic to view this in action. There are no internal blue vertices, so the action of the order 3 generator is trivial in both cases. For $\Gamma $ there is also no red internal vertex, whence this is indeed the trivial representation whereas for $\Gamma_2 $ there is one internal red vertex, so the action of the order 2 generator is given by $~(1,2) $, which is indeed the representation representation on $\Gamma/\Gamma_2 $. In general, if the index of the subgroup $\Gamma’ $ is n, then we call the subgroup of the symmetric group on n letters $S_n $ generated by the action-elements of the order 2 and order 3 generator the monodromy group of the permutation representation (or of the subgroup). In the trivial cases here, the monodromy groups are the trivial group (for $\Gamma $) and $C_2 $ (for $\Gamma_2 $).

As a safety-check let us work out all these concepts in the next simplest examples, those of some subgroups of index 3. Consider the Farey symbols

[tex]\xymatrix{\infty \ar@{-}[r]_{\circ} & 0 \ar@{-}[r]_{\circ} & 1 \ar@{-}[r]_{\circ} & \infty}[/tex] and
[tex]\xymatrix{\infty \ar@{-}[r]_{\circ} & 0 \ar@{-}[r]_{1} & 1 \ar@{-}[r]_{1} & \infty}[/tex]

In these cases the fundamental domain consists of 6 triangles with the indicated vertices (the blue dots). The distinction between the two is that in the first case, one identifies the two edges of the left, resp. bottom, resp. right boundary (so, in particular, 0,1 and $\infty $ are identified) whereas in the second one identifies the two edges of the left boundary and identifies the edges of the bottom with those of the right boundary (here, 0 is identified only with $\infty $ but also $1+i $ is indetified with $\frac{1}{2}+\frac{1}{2}i $).

In both cases the dessin seems to be the same (and given by the picture on the right). However, in the first case all three red vertices are distinct hence there are no internal red vertices in this case whereas in the second case we should identify the bottom and right-hand red vertex which then becomes an internal red vertex of the dessin!

Hence, if we order the three green half-edges 1,2,3 starting with the bottom one and counting counter-clockwise we see that in both cases the action of the order 3-generator of $\Gamma $ is given by the 3-cycle $~(1,2,3) $. The action of the order 2-generator is trivial in the first case, while given by the 2-cycle $~(1,2) $ in the second case. Therefore, the monodromy group is the cylic group $C_3 $ in the first case and is the symmetric group $S_3 $ in the second case.

Next time we will associate a quiver to these vertices and triangles as well as a cubic superpotential which will then allow us to define a noncommutative algebra associated to any subgroup of the modular group. The monodromy group of the situation will then reappear as a group of algebra-automorphisms of this noncommutative algebra!

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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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Segal’s formal neighbourhood result

Yesterday, Ed Segal gave a talk at the Arts. His title “Superpotential algebras from 3-fold singularities” didnt look too promising to me. And sure enough it was all there again : stringtheory, D-branes, Calabi-Yaus, superpotentials, all the pseudo-physics babble that spreads virally among the youngest generation of algebraists and geometers.

Fortunately, his talk did contain a general ringtheoretic gem. After a bit of polishing up this gem, contained in his paper The A-infinity Deformation Theory of a Point and the Derived Categories of Local Calabi-Yaus, can be stated as follows.

Let $A $ be a $\mathbb{C} $-algebra and let $M = S_1 \oplus \ldots \oplus S_k $ be a finite dimensional semi-simple representation with distinct simple components. Let $\mathfrak{m} $ be the kernel of the algebra epimorphism $A \rightarrow S $ to the semi-simple algebra $S=End(M) $. Then, the $\mathfrak{m} $-adic completion of $A $ is Morita-equivalent to the completion of a quiver-algebra with relations. The nice thing is that both the quiver and relations come in a canonical way from the $A_{\infty} $-structure on the Ext-algebra $Ext^{\bullet}_A(M,M) $. More precisely, there is an isomorphism

$\hat{A}_{\mathfrak{m}} \simeq \frac{\hat{T}_S(Ext^1_A(M,M)^{\ast})}{(Im(HMC)^{\ast})} $

where the homotopy Maurer-Cartan map comes from the $A_{\infty} $ structure maps

$HMC = \oplus_i m_i~:~T_S(Ext_A^1(M,M)) \rightarrow Ext^2_A(M,M) $

and hence the defining relations of the completion are given by the image of the dual of this map.

For ages, Ive known this result in the trivial case of formally smooth algebras (where $Ext^2_A(M,M)=0 $ and hence there are no relations to divide out) and where it is a consequence of a special case of the Cuntz-Quillen “tubular neighborhood” result. Completions of formally smooth algebras at semi-simples are Morita equivalent to completions of path algebras. This fact motivated all the local-quiver technology that was developed here in Antwerp over the last decade (see my book if you want to know the details).

Also for 3-dimensional Calabi-Yau algebras it states that the completions at semi-simples are Morita equivalent to completions of quotients of path algebras by the relations coming from a superpotential (aka a necklace) by taking partial noncommutative derivatives. Here the essential ingredient is that $Ext^2_A(M,M)^{\ast} \simeq Ext^1_A(M,M) $ in this case.

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