I am looking for a counter example which shows, that a full rational 2D CFT (with respect to a given chiral subtheory) is not characterized by its modular invariant partition function. People tell me such an example exists, but noone could point out a concrete example or reference to me.

In the framework of Fuchs, Runkel, Schweigert etc. full CFTs are classified by Morita equivalence classes of Frobenius algebras in a modular tensor category $\mathcal C$ (the representation category of the chiral CFT), so my question more concretely:

I am looking for two special symmetric haploid Frobenius algebra objects $A,A'$ in a modular tensor category $\mathcal C$, which are not Morita equivalent but whose full centres $Z(A),Z(A')$ are equivalent as objects (they cannot be equivalent as algebra objects due to a result of Kong and Runkel), i.e. have the same modular invariant partition function $Z(A)_{ij}=Z(A')_{ij}$.

I am also fine with an example in any other formulation of CFT any reference in physics literature, non-degenerate braided subfactors or similar...


I apologize since my answer will involve shameless self-promotion. You can find one example of this kind on page 36 of my slides. In this example one Frobenius algebra is commutative and another is not Morita equivalent to a commutative algebra.

  • $\begingroup$ It is no problem I actually looked into this slides before and it's a shame that I overlooked this before. Is this example unitary? I remember that it follows from some result of Evans that for such a modular invariant always exist a commutative Frobenius algebra (subfactor). $\endgroup$ – Marcel Bischoff May 26 '13 at 22:40
  • $\begingroup$ I think this is the counterexample some people were telling me about. Do you have a reference to this? $\endgroup$ – Marcel Bischoff May 27 '13 at 6:46
  • $\begingroup$ So you are talking about the $\mathcal E_3$ modular invariant of $G_2$ at level 3, I guess. One way to obtain it is the conformal inclusion $G_{2,3}\subset E_{6,1}$, this should be the commutative Frobenius algebra you are talking about, how is the inequivalent one obtained? $\endgroup$ – Marcel Bischoff May 27 '13 at 7:02
  • $\begingroup$ Yes, the first algebra $A$ is constructed via conformal embedding as you suggested. The second one is an extension of $A$. Namely, the category of dyslectic $A-$modules is pointed with underlying group ${\mathbb Z}/3$ (since it is equivalent to $E_{6,1}$), so it contains a unique non-commutative algebra which is a sum of all elements of ${\mathbb Z}/3$ as an object. This algebra (considered as algebra in $G_{2,3}$) is what we want. The reason this example works is that two non-vacuum representations of $E_{6,1}$ restrict to the same representation of $G_{2,3}$. $\endgroup$ – Victor Ostrik May 27 '13 at 15:03
  • $\begingroup$ Thank you very much, is this written down somewhere? $\endgroup$ – Marcel Bischoff May 27 '13 at 19:55

I have no idea what a ``symmetric haploid Frobenius algebra" is, so my answer may be terribly naive, but given an even-self dual lattice of rank $d$ there is a well known lattice construction of a rational (actually holomorphic) conformal field theory with central charge $c=d$. There are two even self-dual lattices of rank $16$ ($E_8 \times E_8$ and $Spin(32)/(Z/2)$) and these give rise to two inequivalent holomorphic Conformal Field Theories with $c=16$. The partition function of these theories is $\Theta_{\Gamma}/\eta^{16}$ where $\eta$ is the Dedekind eta function and $\Theta_\Gamma$ is the theta function of the associated lattice. The theta functions for these two lattices are the same because they are modular forms of weight $8$ and there is a unique such modular form. This is related to the famous "can one hear the shape of a drum" problem. Thus this pair of lattices gives an example of two inequivalent CFT's with the same partition function.

  • $\begingroup$ Very nice! I think this answers the question. I'll explain how. Let $\Lambda_1$, $\Lambda_2$ be the unimodular lattices. The chiral CFT with 2 distinct full CFTs (with same partition function) is the one corresponding to the lattice $\Lambda_1 \cap \Lambda_2$. The unique full CFT with chiral symmetry given by the lattice VOA $V_{\Lambda_1}$ is in particular a full CFT with chiral symmetry $V_{\Lambda_1\cap \Lambda_2}$. Similarly for the unique full CFT with chiral symmetry $V_{\Lambda_2}$. Moreover, these 2 full CFTs are non-isomorphic because their maximal chiral algebras are not isomorphic. $\endgroup$ – André Henriques May 26 '13 at 21:33
  • $\begingroup$ I am afraid this concerns just chiral CFTs, but it provides a counterexample for something other I am thinking of. Do this to lattices have a common sublattice of rank 16, then maybe it could be useful (not sure, though) to built a counterexample for my original question... $\endgroup$ – Marcel Bischoff May 26 '13 at 22:44
  • $\begingroup$ André, how do you see that $V_{\Lambda_1\cap\Lambda_2}$ has full rank and if this is true how do you see that this two "conformal inclusions" give the same modular invariant? $\endgroup$ – Marcel Bischoff May 26 '13 at 22:46
  • $\begingroup$ Because, the chiral symmetry is rational iff the $V_{\Lambda_1\cap\Lambda_2}$ has full rank. $\endgroup$ – Marcel Bischoff May 26 '13 at 22:49
  • $\begingroup$ @Marcel can you define what you mean by a "full rational CFT?" I suspect I must know what this is but am not familiar with this terminology. $\endgroup$ – Jeff Harvey May 26 '13 at 22:58

There is an SU(3) example here in arXiv:math/0008056 pages 21 and 22 for $SU(3)_9$ in $E_6$ and also in $E_6 \times Z_3$ - or see also pages 11 and 12 of arXiv:1002.2348, page 6 of arXiv:0906.4252

  • 1
    $\begingroup$ Dear David, welcome to MO. Your answer is informative. However, it might be more reader-friendly if you could make it somewhat more self-contained. $\endgroup$ – André Henriques Jun 5 '13 at 19:22
  • $\begingroup$ Thank you very much David. Again shame on me because I also have this paper printed and partialy read. I see this example is in the same spirit as Victor Ostrik's, again the two non-trivial reps of $E_{6,1}$ restrict to the same rep of $\mathrm{SU}(3)_9$ (or $G_2,3$ in Victor Ostrik's example). $\endgroup$ – Marcel Bischoff Jun 6 '13 at 17:56

I'm going to try to connect the language of Jeff Harvey's answer to the language of your question. It seems to give an answer to your title question, but not to the question you asked in the main text.

If you are given an even unimodular lattice $L$, the resulting vertex algebra $V_L$ has trivial module category, i.e., equivalent to complex vector spaces. From this datum, you can make a CFT in the Fuchs-Runkel-Schweigert sense by taking the trivial unital algebra in this category, analogous to taking the algebra $\mathbb{C}$ in the category of $\mathbb{C}$-vector spaces. The partition function is equal to $\frac{\theta_L}{\eta^{rank(L)}}$, which is modular-invariant up to an order 3 character.

If you have two even unimodular lattices, then you get two equivalent categories (both equivalent to the category of complex vector spaces), and if you take the trivial algebras, then the full CFTs are considered Morita equivalent in this definition. In other words, I suspect this is not an answer to the precise question you asked. However, in the case of the two rank 16 examples with equal theta functions (hence equal partition functions), there is a very meaningful sense in which the CFTs are not equivalent, since the vertex algebras have nonisomorphic automorphism groups. In particular, the CFTs have different sets of twining characters, and the corresponding orbifold theories are also quite different.

Regarding the question of intersections in the comments, it follows from the $I\!I_{m,n}$ construction (see Wikipedia) that one can choose embeddings in Euclidean space so that the intersection of lattices is finite index, hence full rank. In particular, both $I\!I_{8,0} \times I\!I_{8,0}$ and $I\!I_{16,0}$ contain $(2\mathbb{Z})^{\oplus 16}$.

  • $\begingroup$ Thanks for pointing out that both examples are "framed", I should have realized this regarding some other question I asked some time ago. Now the inclusions $(2\mathbb Z)^{\oplus 16}\subset \Lambda_i$ gives rise to two different non-equivalent Frobenius algebras in the MTC associated with $(2\mathbb Z)^{\oplus 16}$. If they have the same Modular invariant, i.e. if the full centres are equivalent as objects, this would also answer the question in the box, I guess this is completely equivalent to the remark which André pointed out above $\endgroup$ – Marcel Bischoff May 27 '13 at 5:54
  • $\begingroup$ I am quite sure now, that the two Frob. mentioned in the las comment have different modular invariant coupling matrix $Z_{ij}$. In general I learned from Ingo Runkel that it can be shown that for Schelleken algebras the object $Z(A)$ already fixes the Frobenius algebra. So lattices could not give a counter example to the stronger question in the box. $\endgroup$ – Marcel Bischoff May 27 '13 at 6:19
  • $\begingroup$ I see. So you are saying that it is possible to have two different modules for a given VOA that have the same character. Is that right? $\endgroup$ – André Henriques May 27 '13 at 11:42
  • $\begingroup$ I am not sure if I am saying this, but Jeff Harvey does provide an example, doesn't he? What I am saying is that a commutative Frob algebra which is a Schellekens algebra is fixed by its object, in other words the Izumi--Kosaki cohomology for it vanishes. Because the MTC of a even lattice contains just contains automorphisms, so each Frob algebra is Schellekens. So for the above example should follow that the two conformal inclusions $(2\mathbb Z)^{\oplus 16} \subset \Lambda_1,\Lambda_2$ need to have different modular invariant coupling matrices. $\endgroup$ – Marcel Bischoff May 27 '13 at 20:14

[Edit: my answer is wrong.]

The set of full CFTs with a given modular invariant is cassified by a kind of second cohomology that was introduced in the paper "On a subfactor analogue of the second cohomology" by Izumi and Kosaki.

This "second cohomology" $H^2(N\subset M)$ is a pointed set that classifies those extensions $N\subset \tilde M$ with the property that ${}_NL^2M_N\cong {}_NL^2\tilde M_N$ (isomorphism of $N$-$N$-bimodules). In that same paper, the authors show that when the subfactor is of the form $M^G\subset M$ for the outer action of a finite group $G$, then their $H^2$ is equal to the 2nd group cohomolgoy with $U(1)$ coefficients (also equal to $H^3(BG,\mathbb{Z})$).

Now take a conformal net $\mathcal A$, take a finite group $G$ that acts on it, and consider the inclusion $\mathcal A^G\subset \mathcal A$. Now, you probably know that full CFTs with left and right chiral algebras given by $\mathcal A^G$ correspond bijectively to relatively local extensions of $\mathcal A^G\subset \mathcal B$, which correspond bijectively to $Q$-systems in $Rep(\mathcal A^G)$ (a.k.a. Frobenius algebra objects). The modular invariant remembers only the isomorphism class of the vacuum Hilbert space of $\mathcal B$ as an object of $Rep(\mathcal A^G)$, which is the same as remembering the isomorphism class of the bimodule $\{\}_ \{\mathcal A^G(I)\} L^2\mathcal B(I)_ \{\mathcal A^G(I)\}$.

The extension $\mathcal A^G\subset \mathcal A$ corresponds to a given modular invariant of $\mathcal A^G$, and there will be as many other extensions $\mathcal A^G\subset \mathcal B$ with same modular invariant as there are elements in the group $H^2(G,U(1))$.

  • $\begingroup$ So by crossing out, you say that the modular invariant does not only remember the isomorphism class? Yes I would expect this because the module categories are not isomorphic, though i have no idea how to prove this. $\endgroup$ – Marcel Bischoff May 26 '13 at 22:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.