8
$\begingroup$

The paper arXiv:hep-th/9308043 proves that the partition function of an arbitrary (unitary) two-dimensional topological theory is given by $$ Z(\lambda)=\sum_{i=1}^n\lambda_i^{g-1}\tag1 $$ where $g$ is the genus of the surface, and $\{\lambda_i\}$ are some non-negative real numbers characterizing the topological order. The paper claims that this latter set defines the TQFT uniquely.

I'd like to understand this last statement better. In particular, I'd like to understand whether the map $\{\lambda_i\}\leftrightarrow \text{TQFT}$ is one-to-one and onto.

  • It is clear that if $Z(\lambda)=Z(\lambda')$ for all $g$, then $\{\lambda_i\}=\{\lambda'_i\}$ (as sets, i.e., modulo permutations), so TQFTs with different $\{\lambda_i\}$ are necessarily different.
  • The converse is less clear. Is it possible that there is more than one TQFT for a given $\{\lambda_i\}$? In other words, is it possible to have different TQFTs that share the same $\{\lambda_i\}$?

The partition function is not the only object relevant to a QFT, we also need observables. If all expectation values agree, then the two theories are identical; but, as a matter of principle, it is possible to have two QFTs with the same vacuum expectation value (partition function), but different higher-order moments. So it is not clear to me that the set $\{\lambda_i\}$ characterizes the order uniquely. It does determine the partition function, but does it determine the full theory?

$\endgroup$

2 Answers 2

7
$\begingroup$

I don't think there is more than one TQFT, given a set $\lambda_i$. I might be wrong, but I'll try explain my reasoning.

The space of observables in the theory is the space of boundary states. Inserting an operator into a correlation function is simply inserting a boundary with specified boundary conditions into the surface. We can write something like $$ \langle\mathcal{O}_{i_1}\cdots\mathcal{O}_{i_n}\rangle=Z[M_{g,n};\ i_1,i_2,\ldots,i_n] $$ where the RHS is the path integral over a manifold with genus $g$ and $n$ circle boundary components with boundary conditions labeled respectively by the $i_k$.

This index $i$ running over boundary conditions is the same index summed over in the expression you wrote \begin{equation} Z[M_g]=\sum_i\lambda_i^{-\frac{1}{2}(2-2g)}, \end{equation} up to a change in basis, of course. If we choose the appropriate basis for our observables, and scale the observables appropriately we can get \begin{equation} \langle\mathcal{O}_{i_1}\cdots\mathcal{O}_{i_n}\rangle=Z[M_{g,n};\ i_1,i_2,\ldots,i_n]=\delta_{i_1i_2\cdots i_n}\lambda_{i_1}^{-\frac{1}{2}(2-2g-n)} \end{equation} which is the generalization of the expression for $Z[M_{g}]$ to the case of manifolds with boundaries. You can check that the above expression is right by tracing over boundary conditions on two boundaries. This should be the same as gluing the two boundaries together. Given the above expression, it seems that not even the correlation functions contain additional information beyond the $\lambda_i$.

In his answer Ryan Thorngren gives a possible counterexample where two seemingly different TQFTs have the same $\lambda_i$, namely $\lambda_i=1$. The two algebras in question are group algebras $\mathbb{C}[G]$ and $\mathbb{C}[G']$ where $G\neq G'$ are abelian groups with $|G|=|G'|$. Construct a basis for $\mathbb{C}[G]$ by $$ e_q\equiv \sum_{g\in G}\chi_q(g)g\in \mathbb{C}[G] $$ where $\chi_q$ are the irreducible characters of $G$, labeled by $q$. It's easy to show that $e_q\star e_p = \left|G\right|\delta_{qp}\delta_q^r e_r$. So the only information about the group retained by the algebra multiplication is $|G|$. The counit $\epsilon$ (see the link Ryan provided) in this basis is given by $\epsilon(e_q)=\chi_q(1_G)/|G|=1/|G|$, which likewise contains no group data other than $|G|$. So the seemingly different Frobenius algebras $\mathbb{C}[G]$ and $\mathbb{C}[G']$ are in fact the same.

Edit in response to AccidentalFourierTransform's comment: Take the following with a grain of salt, as I'm in the process of learning much of this stuff myself, a fact I probably should have made clear at the top.

So a line operator would be something that acts on the circle Hilbert space, so something that mixes the $\mathcal{O}_i$ operators. You can describe its action on the Hilbert space by a matrix $W_{ji}$. Nontrivially inserting a line operator $\hat{W}$, with associated matrix $W_{ji}$, into a genus $g$ partition function would give $$ \langle \hat{W}\rangle_{g} = \sum_{ij}W_{ji}\langle \mathcal{O}_j\mathcal{O}_i \rangle_{g-1}. $$ Again, I'm not sure, but I suppose there are two perspectives you could take here. If you allow your algebra of line operators to include all $N\times N$ matrices $W_{ij}$, (where $N$ is the Hilbert space dimension), then there is no distinction between TQFTs beyond the $\lambda_i$. On the other hand if you take the algebra of line operators to be some subalgebra of all $N\times N$ matrices, in other words make some choice of line operator content, that choice will distinguish between theories that are otherwise the same.

As an example of the second option, say you have a gauge group $G$ and take only the corresponding Wilson lines as your line operators. Take the states with definite holonomy as a basis for your Hilbert space. The Wilson lines are all diagonal in this basis, and so make up just a subalgebra of the algebra of matrices $W_{ij}$, namely the subalgebra of diagonal matrices.

I don't see why you couldn't include all possible matrices $W_{ij}$ if you wanted to. After all, the bilocal operator $\sum_{ij}W_{ji}\mathcal{O}_j\mathcal{O}_i$ seems well-defined, and defining $\hat{W}$ by the above rule seems to make sense and to give an object that acts like a line operator. There could be something I'm missing here, though.

I don't know when/if this distinction is important. So it may very well be right to say that line operator content can distinguish TQFTs with the same $\lambda_i$. That said, when two TQFTs with the same $\lambda_i$ have a line operator in common (in the sense that it has the same action on the Hilbert space in either theory), you'll get the same values for its insertions, by the above rule.

$\endgroup$
4
  • $\begingroup$ Nice! Wow I didn't realize the group algebra was so lame. I've been spoiled thinking only about the category $Vec_G$. :) I guess special Frobenius algebras over C are indeed boring, as I think this paper suggests pdfs.semanticscholar.org/f3b1/… $\endgroup$ Aug 27, 2020 at 1:55
  • $\begingroup$ Wonderful argument John Gardiner! $\endgroup$ Aug 27, 2020 at 3:34
  • 1
    $\begingroup$ Very interesting, thank you! May I ask whether $\mathcal O_i$ are local operators or line operators? $\endgroup$ Aug 28, 2020 at 17:33
  • $\begingroup$ The $\mathcal{O}_i$ are just local operators. I've edited my answer to say something about line operators. $\endgroup$ Aug 29, 2020 at 2:59
6
$\begingroup$

I believe the answer is no. EDIT: I defer to John Gardiner, my original answer is below

A 2d TQFT is known to be classified by its associated Frobenius algebra, which is generated by the cup/cap and the pair of pants. It is an algebra $A$ with a multiplication $\mu:A \otimes A \to A$, a co-multiplication $\delta: A \to A \otimes A$, and some other stuff.

What the paper you link discusses is that the $\lambda_i$ are the eigenvalues of the endomorphism $\mu \circ \delta$, which is the map associated with the twice-punctured torus (this makes it clear where the formula for the partition function comes from also). The reason these eigenvalues, or even this map, does not determine the Frobenius algebra is because there are so-called special Frobenius algebras where $\mu \circ \delta = id$. See https://ncatlab.org/nlab/show/Frobenius+algebra#special_frobenius_algebras .

I think physically what the example on that nlab page means is that if we have some spontaneously broken finite symmetry $G$, then we can define a 2d TQFT whose partition function on every surface is 1, but the TQFT remembers the group law of $G$.

$\endgroup$
2
  • $\begingroup$ Thanks! Frobenius was actually going to be "part two" tomorrow :-P I guess the question is, now, whether special Frobenius defines a non-trivial TQFT. Is a theory with $\lambda=1$ non-trivial? The partition function is $Z\equiv 1$ for any surface. Is there any observable with non-trivial expectation value in this theory? Or is it really the empty theory? $\endgroup$ Aug 22, 2020 at 18:41
  • $\begingroup$ @AccidentalFourierTransform right... I'm a little confused about it too. The example on the nlab page is certainly a nontrivial special Frobenius algebra, but the torus partition function doesn't compute its ordinary vector space dimension... $\endgroup$ Aug 22, 2020 at 18:45

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.