Say I have an "ordinary" TQFT $F$ of dimension $n$, assigning groups or vector spaces to closed $(n-1)$-manifolds and linear maps to cobordisms. Consider the different ways $F$ can be obtained from a TQFT "extended one step" which assigns categories to manifolds of dimension $n-2$ (often derived categories of algebras or dg / $A\_{\infty}$ algebras).


Is there expected to be any uniqueness to these extensions of $F$? For example, are there cases where you can extend the same $F$ two ways, but the corresponding (derived) categories associated to a codimension-2 manifold aren't equivalent?


This question is motivated partly by the situation in bordered Heegaard Floer homology; the dg algebra associated to a surface depends on a choice of parametrization, but these distinct algebras end up having equivalent derived categories (of type D or type A modules).


I'd be happy with a toy example in lower dimensions or anything illustrating this uniqueness holding or not holding- maybe in some restricted context. I'd also be curious to know if there's an $F$ which doesn't extend at all (maybe this is basic knowledge for the experts...)


Thanks!



How unique are extensions of TQFTs to lower dimension?
The question of which tqfts extend is a very interesting one. To make the question more mathematically precise, we can fix the target n-categories and ask for the tqfts to extend with respect to those targets. Then I can give precise answers. 


In general there are both existence and uniqueness issues, even in the n=2 case. That case is pretty instructive, and it doesn't get much simpler by increasing dimensions. 


It is well known that a 2D non-extended (oriented) tqft in vector spaces is the same thing as a commutative Frobenius algebra. Now we can ask that this "extends to points" using our favorite target 2-category, like linear categories or the 2-category of algebras, bimodules, and maps. There turns out to not be much difference between these two choices, so I will work with the latter. If you prefer the former, than you should just think of the category of modules associated to the algebra. It doesn't really effect what I am going to say.


Now a 2D extended (oriented) tqft in algebras, bimodules, and maps is the same thing as a *non-commutative* symmetric Frobenius algebra which is **fully-dualizable**. Over a perfect field fully-dualizable is the same as finite dimensional and semisimple. 


Now if A is such a fully-dualizable Frobenius algebra, then the center Z(A) will be a commutative Frobenius algebra and this is the algebra corresponding the the non-extended part of the 2D TQFTs. Since A is semisimple, this commutative algebra is semisimple. 


Thus, **not every 2D TQFT extends to points** (at least with the usual target categories). An explicit counter example is given by the non-semisimple commutative Frobenius algebra $k[x]/x^{n+1}$, where the trace is given by picking off the $x^n$-coefficient. 


Moreover we also see that uniqueness is an issue. For example consider working over the real numbers. Then the algebra $\mathbb{R}$, (with trivial trace) is a semisimple Frobenius algebra. The center is, of course, also $\mathbb{R}$. But we also have the quaternion algebra $\mathbb{H}$, which is also a semisimple Frobenius algebra (the trace is projection onto the real line in $\mathbb{H}$). The center of $\mathbb{H}$ is also $\mathbb{R}$, and so this gives an example of two extended tqfts which have the same underlying non-extended 2D tqft. Note that $\mathbb{H}$ and $\mathbb{R}$ are not Morita equivalent, and so in particular these are genuinely different extended tqfts. 


**Different extended 2D tqfts can have the same underlying non-extended 2D tqft.**


