When is a TQFT the dimensional reduction of a higher dimensional TQFT?

Question

In Lurie's framework for TQFT's, a TQFT is a symmetric mondoial functor from $Cob_n(n)$ to some symmetric monoidal $n$-category $\mathcal{C}$. One can construct an $(n-1)$-dimensional TQFT from an $n$-dimensional TQFT by taking the product of a cobordism with a circle first: $Red(F)(M) = F(M \times S^1)$.

My first question is: what are necessary or sufficient conditions for a TQFT to be the dimensional reduction of a higher-dimensional TQFT?

As an illustration, consider the following. For example, the 2-dimensional TQFT $K^\tau_G(G)$ for a compact simple Lie group $G$ arises as the dimensional reduction of Chern-Simons theory. In Consistent Orientation of Moduli Spaces, Freed-Hopkins-Teleman remark that the fact that $K^\tau_G(G)$ can be refined to be defined over $\mathbb{Z}$ instead of $\mathbb{C}$ "reflects that the theory is a dimensional reduction". Can this statement be made precise?

Answer 1

Asking for sufficient conditions that an $n$-dimensional TQFT lifts to $n+1$ dimensions is like asking for sufficient conditions that an egg will hatch into a bird. You might think of necessary conditions. The only reasonable sufficient condition is to see the egg hatch, i.e., to see the lift or categorification of the TQFT into the next dimension. Lest that sound like a fatuous pronouncement, the reason is that an $(n+1)$-dimensional TQFT is often vastly more complicated than its $n$-dimensional predecessor. Certainly a Chern-Simons TQFT in 3 dimensions is vastly more complicated than its dimensional reduction.

As for necessary conditions, that depends on what kind of bird you expect. Suppose that you are set on a specific target category $\mathcal{C}$. Then you know that $F(M \times S^1)$ is the trace of the identity functor on $M$, and that gives you a lot of information. For example, in the dimensional reduction of a Chern-Simons TQFT, you always assign a non-negative integer to a closed surface. You have to, because its lift is vector space.

The problem is that $\mathcal{C}$ might be negotiable. Suppose that you have a 2D TQFT in which you assign a negative integer to a surface. If know that that should lift to a vector space, that's impossible. But it could instead lift to a supervector space (a graded vector space with its signed dimension), or a chain complex, or something else. If the value is not even an integer, that might not be a show-stopper either: Its lift could be a graded space with a Poincaré-Hilbert series, and there could be a formal sum of that series with a non-integer answer. One important moral example is Khovanov homology, even though it is not a complete TQFT. Khovanov homology categorifies the Jones polynomial. A Laurent polynomial with both signs is an inauspicious candidate for categorification, but the categorification exists.