As the title suggests, I am wondering about the dimensionality of state spaces in $d$-dimensional TQFTs. As of yet I have mostly been concerned with the mathematical, functorial definition of TQFTs as functors $\mathcal{Z}:\mathrm{Cob}_d\rightarrow \mathcal{C}$ in which one requires the target category $\mathcal{C}$ to be monoidal, left and right rigid (meaning it exhibits a distinguished left and right duality on the level of objects), amongst other things. If one chooses $\mathrm{Vect}_\mathbb{k}$ as the target category this literally translates to the requirement that the state space attributed to closed ($d-1$)-dimensional manifolds is finite dimensional.

My Physical understanding is very patchy and I would really love to fill the gaps. Currently my understanding is that TQFTs arise as topological subsectors of QFTs with SUSY. To be more precise I think that given a SUSY QFT, I can look at the cohomology ring defined by the differential $Q$, where $Q$ is a generator of the SUSY transformation. Now if I restrict to $Q$-closed observables, I am thinking that I should somehow end up with finite dimensional state spaces.

Most likely a lot of the things I said in this second paragraph are wrong and I would really appreciate if someone could point out my mistakes as well as correct them.

Ultimately the question I would like to ask is what the physical reason for the finite dimensionality of the state spaces is.

Also, I would be delighted if someone could point me to a source that could help me clarify some of my other misconceptions.

I would greatly appreciate any answer!


1 Answer 1


The theories you described, are the so-called Witten-type TQFTs. These are not the most general TQFTs, nay the most common. The most common ones are instead of the Schwarz type. Moreover, it is the Schwarz type TQFTs that can be described in the functorial way that you are familiar with. On a related note, according to Urs Schreiber, in this Phys.SE question, Witten type TQFTs can also be seen as Schwarz type (and hence given a functorial interpretation) from a very highbrow perspective. For all these reasons I will focus on Schwarz type TQFTs and to save myself some time, I will imply that whenever I mention the word TQFT.

Now, TQFTs arise most naturally in physics when one looks at the very low energy limit of a regular QFT. Starting with a regular QFT, you can ride the renormalisation group, flowing all the way down to the deep infrared. This means that you path-integrate out every massive degree of freedom possible, leaving yourself only with an effective action comprised of massless fields. This leaves you with a low-energy effective field theory which falls in one of the following three classes.

  1. It is is gapless. There is a continuum of excitations which you can reach without spending energy. It is described either by a free or by an interacting CFT, depending on the details.
  2. It is trivially gapped. There is a unique ground state, and then an energy gap. If you go below that energy gap, the theory is empty. It is the trivial theory containing only the identity operator, corresponding to the unique vacuum state.
  3. It is non-trivially gapped. This is a theory of countably many degenerate vacua. Its spectrum is a discrete set of operators corresponding to the various vacua.

It is this last case that is in fact what a TQFT is.

I will explain why in a moment, but assuming this, you already have your answer. The state space of the TQFT is that set of vacua. It is a set of at most countably infinite dimension. Note that also on the mathematics side which you explained in your first paragraph, choosing the category of vector spaces as a target category does not immediately imply that the state space is finite dimensional. Choosing the category of finite dimensional vector spaces does. Indeed, this choice is implicit in the mathematics literature, and indeed, this reflects most TQFTs arising from the above procedure. $\newcommand{\ket}[1]{\left|#1\right>}$

Now, to convince you why this non-trivially gapped low-energy theory is a TQFT, note that if I start with one of my vacua, say I call it $\ket{0}$, I can construct another one, by acting with one of the operators I have in my disposal, say $V\ket{0}$. I can measure the energy of that state by acting with the Hamiltonian. Since this is also a vacuum, it has the same energy as $\ket{0}$. Therefore, $$HV\ket{0} = E_0 V\ket{0} = V E_0\ket{0} = V H\ket{0},$$ i.e. $V$ commutes with the Hamiltonian. In fact it follows (remember, that an implicit assumption in the functorial definition of TQFTs is that they are defined in Euclidean signature, therefore my choice of time direction, and hence Hamiltonian is arbitrary) that $V$ commutes with the whole stress tensor, $T_{\mu\nu}$. Quantum mechanically, the stress tensor is defined by $$\left<T_{\mu\nu}\right>:= \frac{\delta Z}{\delta g^{\mu\nu}}.$$ Therefore the fact that $V$ commutes with the stress tensor translates to the following statement in the language of Euclidean path integrals: $$\frac{\delta}{\delta g^{\mu\nu}}\left<\cdots V \cdots\right> = 0,$$ where the $\cdots$ represent arbitrary insertions. In other words, $V$ carries no dependence on the metric! It is topological! All in all, this shows that every operator in this non-trivially gapped low-energy effective theory is completely topological. And that's what a TQFT is.

  • $\begingroup$ First of all thank you so much for your detailed answer! I see that my wording was not ideal. What I was trying to say was that if we have in addition a distinguished duality, the vector spaces have to be finite dimensional. This however is not true, the existence of a nondegenerate paring does not imply that. $\endgroup$ Commented Sep 14, 2023 at 19:59
  • $\begingroup$ If I follow your explanation correctly, ultimately we conclude that TQFTs have countably infinite state spaces. Can you perhaps give an explanation for when and why they turn out to be of finite dimension? $\endgroup$ Commented Sep 14, 2023 at 20:08

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