Onsite symmetries for models with arbitrary valency

Transcription

1 Onsite symmetries for models with arbitrary valency Ethan Lake (Dated: December 1, 2015) We look at onsite symmetries for string-net models defined on graphs of arbitrary valency. We derive expressions for the action of the onsite symmetry on vertices of any valency, and establish a connection between this action and the topological observables in the model. We also work out examples for finite groups, which serves as good way to compare our results with the partial classification for trivalent graphs obtained earlier. In particular, we show that the action of the onsite symmetry on arbitrary graphs takes a very simple form when the input data is based on the twisted quantum double of a cyclic group. I. INTRODUCTION In previous notes, we ve seen that Z 3 and Z 2 onsite symmetries emerge naturally in string-net models. When the input category is V ec G ω for some ω Z 3 (G, U(1)), we showed that for G = Z N1 Z Nk, one of the N i s must be divisible by 3 in order for the Z 3 onsite symmetry to be nontrivial, and that at least two of the Z i s must be even for the Z 2 onsite symmetry to be realized. One might guess that Z n symmetries are realized for n-valent graphs the goal of these notes is to prove that this guess is correct. First of all, how do we even define string-net models on graphs that aren t trivalent? The natural approach is to decompose vertices with valency > 3 into a small cluster of trivalent vertices and then trace over the internal edges created during the decomposition. The first problem that arises is that there are many different ways of decomposing a given vertex into a collection of trivalent vertices do we stick with one given configuration or try to sum over all possible decompositions? However, it turns out that these subtleties won t really matter in the end. We will work fairly mathematically, and simply associate an n-valent vertex whose legs are labeled by 1,..., n with the space hom(1, 1... n ) without being explicit about the internal structure of the vertex decomposition. First of all, let s set the notation. We ll be using category theory language a lot, and not drawing very many graphs. This is because what we re doing will be applied to graphs of arbitrary valence, which are hard to draw. Throughout, we work in a monoidal category C over C. We ll follow category theory conventions and write the associator as a,y,z : ( Y ) Z (Y Z). (1) We write the unit object as 1, which for us satisfies 1 = = 1. The duality morphisms are written as b : 1, d : 1 (2) which stand for birth and death respectively. In the notation of the last set of notes, birth is identified with a cup + blue dot and death with a cap + red dot. This convention ensures that rigidity looks like (id d ) (b id ) = id (3) for all C. Onsite symmetries arise naturally when we consider iterated duality maps. When C is rigid (which it always is for us), the map extends naturally to a contravariant endofunctor, meaning that the map is a natural monodial endofunctor. When we re doing things diagrammatically, we trace a morphism f : V V by putting it on a coupon and closing it with a loop labeled by V. In our algebraic notation, this corresponds to Tr(f) = d V (f id V ) b V. (4) If g is a similar morphism, we can trace f g by composing the trace of f and g. Now we come to pivotal structure, which is very important for us. We define a pivotal structure p on C by an isomorphism id ( ). It s well-know to math people (and definitely plausible) that p V = p 1 V for all V C. The quantum dimension of a morphism f is Tr(p V f), which we denote as dim V. If f is the morphism in C op corresponding to f, then Tr(p V f ) = Tr(p V f ). If furthermore Tr(p V f ) = Tr(p V f), we call C spherical. The presence of a twist will force C to be spherical (to be discussed later).

2 2 II. ONSITE SYMMETRIES As before, let C be a pivotal monoidal category. Define the raising and lowering isomorphisms by L Y Z : hom(, Y Z) hom(y, Z), R Z Y : hom( Y, Z) hom(, Z Y ). (5) Graphically, these represent lifting up a leg on a fusion vertex to make it a splitting vertex, or lowering a leg on a splitting vertex to make it fusion, accompanied by taking the dual of the changed leg. We had both left and right L and R in the last notes, but this was a bit redundant. We only need (in this case) the left lowering and right raising operators, since the others can be obtained through conjugation. Now define the spin isomorphism S as S,Y = R Y 1, L,Y 1 : hom(1, Y ) hom(1, Y ). (6) We see that the name spin isomorphism is appropriate since the action of S is to take the splitting vertex 1;, Y and spin the leg around so that it ends up double-conjugated on the other side of Y. Note that in this way of doing things we are free to write 1 = 1, which we couldn t do in Levin and Lin s construction. Graphically, the spin isomorphism acts on morphisms f as Y S,Y (f) = f (7) T n The last definition to make is the linear automorphism that generates the onsite symmetry, which we denote by. It acts on splitting vertices that have n identical objects (and nothing else) coming out of them: T n : hom(1, n ) hom(1, n ). (8) We will call the space n an n-bouquet, since graphically it looks like a bouquet of n -lines joined together. The T n operator acts on morphisms f by where T(f) n = A n (id (n 1) p 1 ) S, (n 1)(f) (9) A n : (n 1) n (10) is a nested associator, which does every possible association operation on the one leg and the n 1 bouquet, given inductively by A 1 = id, A n = (id A n 1 ) a, (n 2), (11) The nested associator is needed to rearrange the internal branching structure of the bouquet so it matches the branching structure of the original splitting space. Graphically, the T n operator looks schematically like: (n 1) (n 1) (n 1) (n 1) T n : (12) where the three lines symbolize a bouquet. The eigenvalue of the onsite rotation operator is then computed as the trace of the T operator: α n () = Tr T n (13) Thinking about things graphically, it s clear that we require (α n ()) n = 1 for any C, since we can think of α n () as describing the action of Z n on hom(1, n ). Also, we should point out that we will define α n () = 0 if n isn t isomorphic to 1, since dim hom(1, n ) = 1 if n = 1 and is 0 otherwise.

3 3 III. COMPUTING THE ONSITE SYMMETRIES This is all well and good, but our formula (13) for the α n () is awful in particular, the nested associator A n is a real nightmare to deal with in computations. We can do a bit better using the same line of reasoning used to derive α 3 () in the last few sets of notes. This comes from the realization that if α n () 0, then an n 1 bouquet of s must be isomorphic to (since α n () is an n-th root of unity). Let B (the bouquet map) be the map exhibiting this isomorphism: B : (n 1) (14) To begin, we note that the following diagram commutes (writing (, Y Z) for hom(, Y Z) when it won t cause confusion) (1, id B) (1, ) (1, (n 1) ) S, S, (n 1) (1, ) (1, (n 1) ) (1, B id ) (15) which graphically, is just (omitting vacuum strings) (1, id (n 1) B) S, S, (n 1) (16) (n 1) (1, B id ) Now, define the maps Υ : d (B 1 id )(A n ) 1 : (n 1) 1 (17) and Γ : (id B)b : 1 (n 1). (18) Υ takes a splitting space with a bouquet on the right, associates it over to the left, and then caps off the diagram with a death map. Γ births a - pair, and then turns the right edge into a bouquet. Γ is easy to draw, and is given graphically by (n 1) Γ = (19) Note that their composition Υ Γ is a map from 1 1. α n () is constructed by sandwiching the T n operator in the middle of this composition. To see this, consider acting on Γ with the spin operator. Looking at the previous commutative diagram and the graphical representation of Γ, we see that S, (n 1)(Γ) = (B id )b (20)

4 4 Now consider the T n operator acting on Γ. Referring to the definition of T n that and using the above equation, we see Now compose on the left with Υ, which kills of the nested associator: T n (Γ) = A n (id (n 1) p 1 )(B id )b (21) Υ T n (Γ) = d (B 1 p 1 )(B id )b = d (id p 1 )b = dim (22) On the other hand, n = 1, and so the LHS is also α n ()(Υ Γ). So we finally have an explicit form for α n (), namely α n ()(Υ Γ) = dim. (23) This still involves the nested associator, but at least it gives us a simple graphical interpretation of α n (). We use α n () = dim (Γ 1 Υ 1 ) = dim d (id B 1 ) A n (B id )b (24) to draw α n () (rather schematically) as 1 α n () = dim ev (25) where the wiggly lines in the middle represent the nested associator transferring an n 2 bouquet from the right side to the left side. Of course, the nested associator also acts on the bouquet s internal structure as well. Note that our final expression for α n () only involves associators and the bouquet map, and not the raising / lowering / spin operators. This is important because these operators all involve deforming the graph in a way that would affect neighboring vertices. This means that our expression for α n () is well-defined regardless of what kind of graphical environment the vertex in question is located in. Our latest expression for α n () is still a bit ugly. We can beautify it quite a bit if the input category is V ec G ω for finite G and ω Z 3 (G, U(1)). In this case, we can choose a gauge in which all the quantum dimensions and bouquet maps are trivial. Therefore, the only factors contributing to α n (g) are those that come from the nested associator A n g. Unpacking the definition of the nested associator, we see that A n g gives us a product over all associators whose middle argument runs over every possible bouquet size and whose first and third arguments are g. With the simple fusion rules that the twisted quantum double gives us, we can write a k-bouquet of g s as g k. Finally, we require that α n (g) 0 only when g n is the identity. Putting all this together, we have α n (g) = δ gn,1 n 1 k=1 ω(g, g k, g) (26) This is a really awesome formula. Before, we were figuring out α 2 and α 3 in twisted Z n essentially by exhaustion we would play around with gauge transformations until we found a product which transformed nontrivially under the onsite symmetry in a gauge-invariant way, and if we couldn t find any such terms, assume that α was trivial. By contrast, the above formula instantly tells us when we ll have nontrivial onsite symmetry, and it does it for all α n and arbitrary finite G! Very cool. IV. CONNECTION WITH TOPOLOGICAL SPIN Given that onsite symmetries are related to the eigenvalues of processes that move a bunch of identical labels around each other, we might wonder if they can be connected to the topological observables of the model. Indeed this is the case, as we will sketch in this section.

5 We are interested in the cases when the input category C is a ribbon category, equipped with a half-braiding commutativity constraint c,y : Y Y, which needs to satisfy (Y c,z ) a Y,,Z (c,y Z) = a Y,Z, c,y Z a,y,z (27) Ribbon categories also come with a twist θ : id C id C, which is an isomorphism of functors whose tensor structure is given by θ Y = (θ θ Y )c,y c Y,, (θ ) = θ (28) If we know the twist, the pivotal structure on C, p : id C ( ), is constrained to be p = (d id )a 1,, (id c 1, )a,,(b id )θ (29) which diagrammatically looks like the twisted loop we re familiar with. As usual, we let Z(C) denote the (left) Drinfeld center of C. It consists of pairs (, c ) where is an object of C and c : ( ) ( ) is a half-braiding constraint. Consider the canonical forgetful functor A : (, c ) that forgets about the braiding structure of Z(C). The adjoint of this functor is A : C Z(C), which sends objects to A() = (Y, c Y ) dim C hom(y,) (30) (Y,c Y ) Z(C) (see e.g. Luger The quantum double of tensor categories and subfactors ). We can use another result of Luger s to relate the trace of the T n operator to the trace of the twists of the images of A: 5 where as usual, TrT n = 1 dim C Tr(θ n A() ) (31) dim C = C 2 (32) is the global dimension of C. Showing this takes a while, and has already been done in the mathematics community, so we won t reproduce the proof here. It should seem reasonable, though. When performing the nested association we re associating a graph that s been acted on by the spin isomorphism, and looking at the picture in (7) tells us that a sum over twist powers might be expected. We then connect this with α n () and use the explicit expression for the adjoint A() given by equation 31, obtaining α n () = 1 dim C (Y,c Y ) Z(C) θ n Y dim Y (dim C hom(, Y )) (33) which is a pretty nice way to connect the onsite symmetries with the topological observables in the model. This formula becomes particularly simple when the input category is based on the twisted quantum double of some finite group G. In this case, we know that we can parametrize the different eigenvectors of T by pairs (A, µ), where A, µ = 1 G g C A,h Z g χ g µ(h) g, h (34) with C A denoting a conjugacy class of G, Z g denoting the centralizer of g, and χ g µ(h) = Tr(ρ g µ(h)), with ρ a projective representation ρ g : Z g GL(Z g ) whose factor set is given by γ g (x, y), which is the defining 2-cocycle of the associator we re working with. The eigenvalues of these basis vectors are of course the twists: Following Yuting s twisted quantum double paper, we arrive at (θ A µ ) g = θµ A = χg A µ (g A ) (35) dim µ g 1 k=1 ω(g, g k, g) (36)

6 where g is the order of g, which is chosen to be a representative element of the conjugacy class C A. It s easy to check that the RHS is the same for every g C A, and so the power of the twist on the LHS is a class function, as it should be. Note the similarity between (36) and (26)! We can then use this result and (33) to compute the action of the onsite symmetries directly in terms of topological observables. Although this might have been expected based on the twistiness of the T n isomorphism we traced over to get αn (), it s by no means obvious. We note that this also tells us that the onsite symmetries are class functions. 6 V. EAMPLES A. Cyclic twisted quantum double Now it s time to actually apply this and calculate the onsite symmetries for a few models. The most accessible choices are phases derived from the twisted quantum double. The twisted quantum double is also good to look at because we already have been able to partially classify onsite symmetries for trivalent graphs. We can use our knowledge of cohomology to simplify equation 26 a bit. We know that in the twisted quantum double of Z N, pf ω(f, g, h) = exp N 2 (g + h [g + h] N) (37) where p Z N parametrizes the different solutions and the brackets denote addition mod N. If we want to compute α n (g), we see that if [ng] N 0, α n (g) = 0. So then assume that [ng] N = 0. We have α n (g) = n 1 k=1 = exp ) N 2 g([gk] N + g [gk + g] N ) ) n 1 N 2 g ([kg] N + g [(k + 1)g] N ) k=1 { g 2 N 2 (n 1) + (g 2 [ng] N ) } ) ( p exp ( p ( p = exp p = exp N 2 ng2 which is very nice and simple. What s more, it has built-in gauge invariance. Under transformation by a coboundary b B 3 (Z N, U(1)), we have (switching to multiplicative notation) which implies that b(i, jk)b(j, k) ω(i, j, k) ω(i, j, k) b(i, j)b(ij, k) α n (g) n 1 k ω(g, g k, g) b(g, gk+1 )b(g k, g) b(g, g k )b(g k+1, g) The first terms product together to give the usual α n (g). For k 1 and k n 1, every coboundary term is canceled by its neighbors on the left and right. This leaves only half of the first and last terms. The first term is completely killed, and we end up with (38) (39) (40) α n (g) α n (g) b(g, gn ) b(g n, g) = αn (g) (41) since α n (g) δ gn,1. Thus, when given in this form every onsite symmetry is manifestly gauge-invariant under coboundary transformations. This also tells us that the onsite symmetries are class functions. To see this, we note that ω(xfx 1, xgx 1, xhx 1 ) = ω x (f, g, h) is a 3-cocycle cohomologous to ω(f, g, h), where x G. Therefore α n (g) = α n (xgx 1 ), up to a transformation of α n by some b B 3 (G, U(1)). But we just showed that α n is invariant under such transformations, and so α n is a class function as claimed.

7 In addition to coboundaries, there s another gauge degree of freedom we need to keep in mind. Looking back to our derivation of α n (g), we see that we actually have α n (g) dim g we just haven t been writing it since normally the quantum dimensions in the twisted quantum double are trivial for every element. We observe that a gauge transformation by a homomorphism ξ(g) : Z N U(1) that acts as α n (g) αn (g) ξ(g), dim g dim g ξ(g) (42) leaves the input fusion data invariant and satisfies all the self-consistency equations. Thus we have some freedom to transfer the action of onsite symmetries onto the quantum dimensions of the labels in the theory. Firstly, suppose we want to kill as many onsite symmetries as possible. Can we kill all onsite symmetries? Importantly, no this is because of the g 2 in the exponential in our result for α n (g), and the requirement that ξ be a homomorphism, In general, the best we can do is to choose ξ(g) = exp(pg/n) (note that ξ cannot be a function of n). With this choice, ( { ng α n 2 (g) = δ gn,1 exp p N 2 g }) g, dim g = exp (43) N N which ensures that α 2 (g) = 0 for all N, p and g, since α 2 (g) can only be nonzero if 2g = N. The Z 2 onsite symmetry is the only one we can completely kill in this way. Of course, if N is odd the onsite Z 2 symmetry is trivial anyway, and we don t need the gauge transformation. Using this choice for the quantum dimensions, we look to see what types of onsite symmetries are realized for twisted Z N. In the table below, a means that a nontrivial onsite symmetry is realized. The rows list the group used, the columns list the valence of vertices that have nontrivial symmetry, and we let p = 1. 7 n = 2 n = 3 n = 4 n = 5 n = 6 Z 2 Z 3 Z 4 Z 5 Z 6 Z 7 Z 8 Z 9 Z 10 (44) In retrospect, the pattern is clear: we get nontrivial onsite symmetry when the valency of the vertex divides the order of the input group, except for 2-valent vertices, whose symmetries we ve already gauged away. Interestingly, we see that tetravalent Z 2 graphs still have a nontrivial onsite symmetry. If we extend the table further, we see that Z 2 models have a nontrivial onsite symmetry of eigenvalue 1 for any graph whose valency is a multiple of 4. The table above is for the choice p = 1. When p N, some onsite symmetries are lost. For example, when p = 1 then the Z 6 model has nontrivial symmetries for n = 3, 4, 6. When p = ±2 nontrivial symmetries occur only for n = 3 and n = 6, and when p = ±3 only n = 4 is nontrivial. B. Non-cyclic twisted quantum double Now we look at the twisted quantum double picture for more general finite abelian groups. To get the associators, we can use, for an arbitrary finite abelian group G = Z N1 Z NM, H 3 Z Ni, U(1) = Z N Z gcd(ni,n j) Z gcd(ni,n j,n k ) (45) i i i<j The first case to consider is Z N Z N. We will write an element g Z 2 N as g = (g 1, g 2 ) T. Applying the Kunneth formula, we see that H 3 (Z 2 N, U(1)) = Z3 N, and so the different associators will be generated by three different elements. i<j<k

8 8 The generators of the cohomology group are v 1 (f, g, h) = exp N 2 f 1(g 1 + h 1 [g 1 + h 1 ] N ) v 2 (f, g, h) = exp N 2 f 2(g 2 + h 2 [g 2 + h 2 ] N ) v 3 (f, g, h) = exp N 2 f 1(g 2 + h 2 [g 2 + h 2 ] N ) (46) The first and second generators represent self-couplings between the magnetic fluxes in each Z N gauge group, while the third establishes a pairwise coupling between the fluxes in the different gauge groups. We should point out switching the 1 and 2 subscripts in the equation for the third generator gives us something cohomologous to v 3, so three generators are sufficient. A general cocycle in H 3 (Z 2 N, U(1)) can be obtained by taking product of powers of the v i s. As such, an element ω H 3 (Z 2 N, U(1)) can be written as ω(f, g, h) = exp N 2 f T a b P (g + h [g + h] N ), P =, a, b, c, Z N (47) 0 c Using the same procedure as before, we compute the onsite symmetries in the model as n n α n (g) = δ g n 1,1δ g n 2,1 exp N 2 (ag2 1 + bg 1 g 2 + cg2) 2 = δ g n 1,1δ g n 2,1 exp N 2 gt P g (48) As before, this formula is manifestly invariant under coboundary transformations, and admits ξ : Z N Z N U(1) gauge transformations with the same constraints involving the quantum dimensions as in the cyclic case. The coupling between the two Z N components parametrized by b is very important. When b = 0, the onsite symmetries are determined simply as a product of two Z N models. When b 0, more symmetries emerge, even in the simplest case of Z 2 Z 2. In this case, let us take the matrix P that yields a coupled semion semion model (a = b = c = 1). In this model, α 2 ((1, 1) T ) = 1 is actually impossible to gauge away with ξ. One can see this by computing α 2 ((1, 0) T )α 2 ((0, 1) T )α 2 ((1, 1) T ) = 1, and realizing that the quantity α n (f)α n (g)α n (h) is ξ-gauge invariant if fgh = 1, since ξ must be a homomorphism. Let a = b = c = 1. We can then construct a table similar to the one before showing when onsite symmetries are realized: n = 2 n = 3 n = 4 n = 5 n = 6 Z 2 Z 2 Z 3 Z 3 Z 4 Z 4 Z 5 Z 5 Z 6 Z 6 (49) We see that as long as the valency n of the graph satisfies gcd(n, n) 1, an onsite symmetry is realized. Now we turn to the slightly more general case of Z M N, where M > 2. The third cohomology group in this case is H 3 (Z M N, U(1)) = Z M+M(M 1)/2+M(M 1)(M 2)/6 N (50) which we can derive using the Kunneth formula. There are three generators, given by (here i, j, k Z N, with 1 i j k N assumed on each line separately) v1(f, i g, h) = exp N 2 f i(g i + h i [g i + h i ] N ) v ij 2 (f, g, h) = exp N 2 f i(g j + h j [g j + h j ] N ) (51) v ijk 3 (f, g, h) = exp N 2 f ig j h k

9 9 The last one makes the Z M N case a little weird. We can thus write an arbitrary cocycle as { ω(f, g, h) = exp f T N 2 P (g + h [g + h] N ) + Q ijk } f i g j h k (52) where P is a matrix in Z M N and Q is a tensor in ZM M M N. If Q is non-diagonal, the quasiparticles that emerge have non-abelian statistics, even though the input data is abelian. The onsite symmetries are ( { }) n 1 α n n (g) = exp N 2 g T P g + Q ijk g i [kg j ] N g k (53) For simplicity s sake, the examples we ll write down will come from Z 3 N. In the table below, we have assumed both P and Q have 1 s above the diagonal and 0 s elsewhere, which gives us a prototypical model with non-abelian quasiparticles. With this choice, we have k=1 n = 2 n = 3 n = 4 n = 5 n = 6 Z 2 Z 2 Z 2 Z 3 Z 3 Z 3 Z 4 Z 4 Z 4 Z 5 Z 5 Z 5 Z 6 Z 6 Z 6 Finally, we can consider the dihedral groups D N. If we want to write ω(f, g, h)ω(g, h, (gh) (fgh)) ω(g, g fg, h) (54) = δγ f (g, h) (55) where γ f is a twisted 1-cochain, we must choose N odd. In this case we use D N = ZN Z 2 to write each element as a tuple (F, f) where F Z 2 and f Z N. The associator is ( [ ω((f, f), (G, g), (H, h)) = exp ( 1) G+H N 2 f { ( 1) H g + h [( 1) H } g + h] N + N 2 F GH/2 ] ) (56) The onsite symmetries are easy to compute, but the actual expressions are fairly long so we won t write them down. We note that if (G, g) D N is even (i.e, if G = 0), the associator reduces to that of the regular twisted Z N. With the gauge choice ξ((g, g)) = ( 1) G, we can eliminate all the Z 2 onsite symmetries. Similarly to the abelian case, we have n = 2 n = 3 n = 4 n = 5 n = 6 n = 7 n = 8 n = 9 n = 10 n = 11 n = 12 D 3 D 5 D 7 D 9 D 11 which is fairly predictable except for D 3 lacking any onsite symmetries for 9-valent graphs. In general, when the valency of the graph is a power of N, D N has no onsite symmetries. (57) C. General abelian group Now we let G = Z N1 Z Nk. We obtain H 3 (G, U(1)) through the Kunneth formula as usual. The generators are v1(f, i g, h) = exp Ni 2 f i (g i + h i [g i + h i ] G ) v ij 2 (f, g, h) = exp f i (g j + h j [g j + h j ] G ) (58) N i N j ( ) v ijk 3 (f, g, h) = exp gcd(n i, N j, N k ) f ig j h k

10 That s all for now. In the next set of notes, I ll carry out the same analysis for a few non-abelian models. I also have an idea concerning how to make the onsite symmetry / topological spin connection a bit more transparent by using the tube algebra that I m still figuring out. 10