# 2-nd quantized TQFT formalism?

Suppose that we have a certain TQFT in the Atiyah-Singer sense. It is given by a functor $Z$ which associates:

• To connected oriented $n-1$-manifolds $a, b, \dots$ (in what follows called compact objects) – vector spaces $Z(a), Z(b), \dots$, such as $Z(a^*) = Z(a)^*$ where $a^*$ is an orientation-reversal of $a$.
• To disjoint unions of compact objects – tensor products of their spaces: $Z(a \cup b) = Z(a) \otimes Z(b)$.
• To $n$-cobordism $f$ from $a$ to $b$ – linear operator $Z(f): Z(a) \rightarrow Z(b)$, such that the gluing axiom holds.

For simplicity, let's look at the simplest possible example – a 1d TQFT. Its objects are connected oriented $0$-manifolds or oriented points, which will be denoted by $+$ (future-oriented) and $-$ (past-oriented) respectively. Cobordisms are curves connecting pairs of points, the four possible cobordisms shown on the figure below:

Say we choose to associate a 1-dimensional vector space to objects. Cobordisms are given by (from left to right)

1. "Annihilation": $a \left| e \otimes e^* \right> \rightarrow a$ (here $\left| e \right>$ and $\left| e^* \right>$ are basis element in the 1-dim Hilbert spaces associated to future-oriented and past-oriented points respectively, and $a \in \mathbb{C}$ is an arbitrary complex number).
2. "Pair creation": $a \rightarrow a \left| e \otimes e^* \right>$.
3. Just an identity operator: $a \left| e \right> \rightarrow a \left| e \right>$.
4. Just an identity operator: $a \left| e^* \right> \rightarrow a \left| e^* \right>$.

Now comes a trick which seems natural to me, but I've never seen it described rigorously in the TQFT literature.

The functorial picture sketched above is analogous to the first-quantized formalism, where different Feynman diagrams associate amplitude operators to particle scattering processes. We know that Feynman diagrams find a natural explanation in the language of second quantization – instead of associating an amplitude to each diagram, we sum them all together. The 2nd quantized Hilbert space has the Fock structure:

$$\mathcal{H}_{\text{Fock}} = \oplus_{i=0}^{\infty} \mathcal{H}^{\otimes_{\text{sym}} i}.$$

We are allowed to have many indistinguishable particles in a single quantum state.

Taking this analogy to TQFT would result in the Fock space structure with $\mathcal{H}$ now corresponding to a direct sum of the Hilbert spaces associated to each of the compact objects in the theory. In what follows I will call this "2-nd quantized TQFT" just for the sake of clarity.

Note that the tensor product structure is already there by Atiyah-Singer axioms, but it is not quite equivalent to the Fock structure, because of:

• Symmetrization is absent in Atiyah-Singer, which makes sense in the TQFT context, because a cobordism might not be symmetric under permutations of the objects. However: if we consider a sum of all cobordisms of different topologies instead of just a single cobordism, this asymmetry goes away. To each asymmetric cobordism corresponds its mirror image under the permutation, and the sum is always symmetric. Thus, it makes sense to consider compact objects to be indistinguishable and embrace the Fock structure.
• The overall direct sum over $k$-particle subspaces is absent. In TQFT, the collection of compact objects is among the initial data and is classical. We can never be in a superposition of having a point and not having anything. However: from ordinary QFT we know that those superpositions exist and play a role in the theory. Maybe we could try enlarging our description of TQFT by switching from classical boundary data to quantum?

The resulting "2-nd quantized TQFT" looks quite different from the ordinary formalism of TQFT. Specifically,

• It only has a single Hilbert space to rule them all. This is the Fock space built from the "1-particle" space $\mathcal{H}$ given by a direct sum of spaces associated to types of compact objects in the theory.
• Instead of associating an operator with a cobordism, it has a single operator $\mathcal{P}$ to rule them all. This is (loosely speaking) a sum over all possible cobordisms, or topologies. It has to be idempotent: $$\mathcal{P}^2 = \mathcal{P},$$ because (roughly) applying it twice is equivalent to summing over all topologies, which is also applying it once (manifestation of the gluing axiom in the 2-nd quantized context). Thus I will call it the projection operator.

It is peculiar that what we get is precisely the mathematical structure of timeless quantum mechanics. $\mathcal{P}$ can be thought of as an operator projecting to the "physical degrees of freedom".

Let's attempt to build this 2-nd quantized TQFT structure for the familiar example from earlier – the 1d TQFT. A moment of reflection will convince the reader that

• The Hilbert space to rule them all is given by a linear span of $\left| n, m \right>$ for non-negative integers $n, m$ encoding the number of future-oriented and past-oriented points.
• The projection operator maps $\left| n, m \right>$ to $$\mathcal{P} \left| n, m \right> = \left| n - m \right),$$ where the round-bracket ket denotes a vector in the physical Hilbert space, which is a linear span of integers. It can be realized as a self-adjoint operator on the angular-bracket ket Hilbert space to satisfy $\mathcal{P}^2 = \mathcal{P}$ e.g. like this: $$\mathcal{P} \left| n, m \right> = \left| n - m, 0 \right>,$$ or in many other equivalent ways.

In this particular case, the projection operator preserves "charge" conservation.

Finally, it might appear at first sight that this modification breaks the most important mathematical property of TQFT, which is – ability to compute invariants of manifolds. However, the info about those invariants is still encoded in the 2-nd quantized version of the TQFT. We just need different means of extracting information out of it. Instead of using the gluing axiom to calculate invariants of manifolds, we now calculate physical inner-products $$\left< a | b \right> _{\text{physical}} = \left< a | \mathcal{P} | b \right>$$ between states corresponding to initial and final, now fully quantum, data.

Intuitively, "2-nd quantized QFT" could be understood as "quantum topology", since it sums over cobordisms just like path integral sums over fields.

Has this line of thought ever been pursued further? What models have been constructed? Does it indeed give mathematical results, or is there a caveat which I missed? I would love to read a paper about this.

UPDATE:

I will describe here a certain toy model of 2D 2-nd quantized TQFT. At this point, I am not sure if it is of any mathematical value, but if I had to guess – my money is on that it isn't. Whether it is because the model is oversimplified or because the whole programme is flawed is also unknown to me.

It is a well-known fact that 2D TQFTs are equivalent, in the category-theoretic sense, to commutative Frobenius algebras. Such an algebra has an almost-Hopf bialgebra structure (the difference from Hopf being that it lacks the antipode mapping):

$$\mu: \mathcal{A} \otimes \mathcal{A} \rightarrow \mathcal{A} \quad \text{algebraic product}$$ $$\eta: \mathbb{C} \rightarrow A \quad \text{unit element}$$ $$\Delta: \mathcal{A} \rightarrow \mathcal{A} \otimes \mathcal{A} \quad \text{coproduct mapping}$$ $$\varepsilon: \mathcal{A} \rightarrow \mathbb{C} \quad \text{"trace" (or Frobenius form)}$$

Such that certain algebraic properties hold. In our case $\mathcal{A}$ is commutative and symmetric, and the requirement for $\mathcal{A}$ to be a Frobenius algebra is equivalent to a requirement for it to describe a topologically invariant calculus on 2d surfaces. Specifically, the operations above map to the following surfaces:

The following is true: commutative symmetric Frobenius algebras are in 1-to-1 correspondence with 2d TQFTs in the sense of Atiyah-Singer. It can be checked explicitly by constructing a functor from the category of 2-cobordisms to the category of linear maps over Frobenius algebras.

We are looking for the simplest 2d model of the "2-nd quantized TQFT". Let's choose the simplest nontrivial model of "1-st quantized" 2d TQFT first, and then try to 2-nd quantize it.

The Frobenius algebra I am considering here is 2-dimensional. Its basis elements are labeled $$\left| 0 \right>, \left| 1 \right>.$$

We will also consider tensor products of $n$ copies of this Frobenius algebra. The basis on such tensor products is given by $n$ bits (each one is either a zero or a one), and we will denote the basis elements with bitmasks, e.g. $$\left| 1011 \right> \in \mathcal{A}^{\otimes 4}.$$

The bialgebraic structure is given by:

$$\mu \left| ab \right> = \left| (a + b)~\text{mod}~2 \right>$$ $$\eta = \left| 0 \right>$$ $$\Delta \left| 0 \right> = \frac{1}{2} \left| 00 \right> + \frac{1}{2} \left| 11 \right>$$ $$\Delta \left| 1 \right> = \frac{1}{2} \left| 01 \right> + \frac{1}{2} \left| 10 \right>$$ $$\varepsilon \left| 0 \right> = 1, \; \varepsilon \left| 1 \right> = 0$$

In simple words, the basis of $\mathcal{A}$ is the $\mathbb{Z}_2$ group under addition, with the bialgebraic completion given by $\Delta$ and $\varepsilon$.

It can be checked that this definition indeed gives a topologically-invariant calculus on 2d surfaces and disjoint unions of circles. It comes down to checking the following relation:

Now let's attempt to 2-nd quantize it in the sense described in the question.

The Fock space structure is easy to infer. It is a direct sum of $k$-circle subspaces, and each subspace is a tensor product of $k$ copies of $\mathcal{A}$. However, since circles are indistinguishable, we have to restrict this structure to completely symmetric states. The basis of the $k$-circle subspace of the Fock space is given by $$\left| n, m \right>$$ for $n, m$ non-negative integers satisfying $n + m = k$. The physical meaning of $n$ and $m$ is clear – they correspond to the number of circles in "0" and "1" states respectively.

We are still missing the projector $\mathcal{P}$. Let's calculate the matrix element $$\left< p, q \right| \mathcal{P} \left| n, m \right>.$$

We are loosely associating with it a sum over all possible bulk topologies connecting $p+q$ circles with $n+m$ circles.

Let's start by considering a connection between two circles. Here's a list of possible bulk topologies:

There's a dedicated disconnected topology, to which our Frobenius algebra associates a projector map to the eigenspace spanned by $\left| 0 \right>$, and there's an infinite series of Riemann surfaces with increasing genus, all of which correspond to the identity operator on $\mathcal{A}$ (our TQFT is oversimplified – it doesn't distinguish between Riemann surfaces of different genus). We are interested in the some of these, appropriately normalized. It follows that in the limit where all intermediate topologies are summed over, the first (disconnected) one gives an infinitesimally small contribution:

$$S = \lim_{g_{\max} \rightarrow \infty} \frac{1}{g_{\max} + 1} \left( \left| 0 \right> \left< 0 \right| + g_{\max} \cdot \mathbb{1} \right) = \mathbb{1}.$$

We thus conjecture that only connected topologies contribute significantly to the overall sum.

The other complication that we would like to get rid of is a possibility of "bubbles" – freely floating Riemann surfaces, disconnected from the circles on the boundary. We conjecture that those contribute a (possibly, diverging) overall normalization constant to $\mathcal{P}$. Thus, we are free to not consider them for now, and correct for this later by rescaling $\mathcal{P}$ to satisfy

$$\mathcal{P}^2 = \mathcal{P}.$$

(Note that it is still a highly nontrivial requirement that the resulting bare $\mathcal{P}_0$ can be rescaled in this way, so we should expect nontrivial structure in the 2-nd quantized TQFT).

These observations significantly simplify further analysis. Indeed, it follows that the matrix element comes solely from the connected Riemann surface with any genus (for simplicity, chose it to be a punctured sphere).

It is straightforward to show that the amplitude is given by $$\left< p, q \right| \mathcal{P} \left| n, m \right> \sim \frac{(p + q)!}{p! q! 2^{p+q-1}} \cdot \Theta(m, q),$$

where $\Theta(m, q)$ is equal to $1$ when $m$ and $q$ are both even or both odd, and $0$ otherwise.

This theory needs renormalization though. $\mathcal{P}$ as it stands is not a projection operator, however, it is easy to make it into a projection operator by introducing a cut-off $\Lambda$.

In this toy model, the physical meaning of $\Lambda$ is straightforward – it is the maximal value that $q$ and $m$ can take (we are cutting our Hilbert space off by throwing away states with a significantly large number of "1" bits in their bitmask).

The resulting value is: $$\left< p, q \right| \mathcal{P}(\Lambda) \left| n, m \right> = \frac{(p + q)!}{\Lambda p! q! 2^{p+q}} \cdot \Theta(m, q).$$

Did we really just arrived at a 2-nd quantized TQFT? There's only one way to find out, lets check its consistency. Specifically, we want to check that $\mathcal{P}(\Lambda)^2 = \mathcal{P}(\Lambda)$:

$$\left< p, q \right| \mathcal{P}(\Lambda)^2 \left| n, m \right> = \sum_{s, t} \left< p, q \right| \mathcal{P}(\Lambda) \left| s, t \right> \left< s, t \right| \mathcal{P}(\Lambda) \left| n, m \right> =$$ $$\sum_{s,t} \frac{1}{\Lambda^2} \frac{(p + q)!}{p! q! 2^{p+q}} \frac{(s + t)!}{s! t! 2^{s+t}} \cdot \Theta(q, t) \Theta(t, m) =$$ $$\sum_t \Theta(t, m) \cdot \frac{2}{\Lambda^2} \frac{(p + q)!}{p! q! 2^{p+q}} \Theta(q, m) = \frac{\Lambda}{2} \cdot \frac{2}{\Lambda^2} \frac{(p + q)!}{p! q! 2^{p+q}} \Theta(q, m) =$$ $$\frac{(p + q)!}{\Lambda p! q! 2^{p+q}} \cdot \Theta(m, q) = \left< p, q \right| \mathcal{P}(\Lambda) \left| n, m \right>,$$

thus $\mathcal{P}(\Lambda)$ is indeed a projection operator.

Finally, let's compute the "covariant vacuum" of this theory, defined as $$\left| \Omega \right> = \mathcal{P} \left| 0, 0 \right>.$$

A direct computation shows that the theory is never in the "no circles" state. The covariant vacuum is homogeneously distributed between all states with even $m$.

As next steps, 1. The physical Hilbert space could be computed from $\mathcal{P}$, or at least its dimensionality in the $\Lambda \rightarrow \infty$ limit. 2. A less trivial example, distinguishing between different Riemann surfaces can be studied.

UPDATE2: The physical sector of the toy model outlined above is just a 2-dim Hilbert space, with a basis corresponding to "even number of ones" and "odd number of ones". It is easy to see from a direct application of $\mathcal{P}$ to elements of the Fock space basis $\left| n, m \right>$.

• I think what you're describing is just quantum gravity with the metric already integrated out (now just the topological sector remaining to integrate). Above 2 dimensions there are lots of manifolds and it is very hard to regulate the sum over cobordisms. youtube.com/watch?v=VuZWvu2JFKk :P Anyway look up Ponzano-Regge and the many follow-ups which were somewhat successful in deriving Einstein-Hilbert in 2+1D from techniques which became associated with TQFT, like state-sums, later on! – Ryan Thorngren Apr 20 '18 at 6:39
• @RyanThorngren interesting. I have thought about this, and here’s why I think it is not quite the case: Tuarev-Viro (which is basically a regulated Ponzano-Regge) sums are triangulation-independent as long is the triangulation remains on a single manifold... So it is in fact just a TQFT! Its projection operator is not the same as in “2-nd quantized TQFT”, in fact, it is just an ordinary TQFT projection operator applied to a cylinder topology. – Prof. Legolasov Apr 20 '18 at 14:54
• @RyanThorngren it is required by Atiyah-Singer axioms that to a cylinder cobordism corresponds an identity operator. Turaev-Viro lives on an “extended” kinematical space of spin networks, thus, the relevant object associated with a cylinder here is the projection operator to the physical sector, hence the spin foam sum projector. – Prof. Legolasov Apr 20 '18 at 14:58
• @RyanThorngren but in “2-nd quantized TQFT” the sum is over cobordisms themselves. It is only a quantum gravity with metric integrated out if the underlying quantum gravity is not only background independent (read generally covariant) like Turaev-Viro, but also topology independent. In fact, I saw some general ideas about topology independence in LQG literature, but it only suggestions. As to the part where cobordisms are hard to regulate — you may be right, but note that unlike in the path integral case, there’s a countable amount of those in 2d and 3d. Probably impossible in 4+d though. – Prof. Legolasov Apr 20 '18 at 15:03
• I think in Ponzano-Regge to get a proper gravity theory, not just a TQFT, you also have to sum over the spacetimes. Maybe nobody figured out how to do it in that context, but it's very familiar, eg. in AdS/CFT. Typically you're only allowed to specify the asymptotic geometry to get well-defined gravity amplitudes. You should read about the topological string by the way. It's a 2d version of what you're saying, where you take the A or B model and sum over the worldsheets. They're able to do it because string theory is mathemagical. – Ryan Thorngren Apr 20 '18 at 15:54