# How unique are extensions of TQFTs to lower dimension?

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!

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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.

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This is a great answer- thanks! – Andy Manion May 15 '14 at 18:27

For your $n=2$ case, with vector spaces assigned to 1-manifolds, the extension to 0-manifolds is unique up to Morita equivalence (of linear 1-categories). This assumes that the extensions assign a semisimple 1-category to a 0-manifold, which might be implied by the extended TQFT axioms depending on which version of those axioms you choose.

For $n-1>1$, there is a categorified version of Morita equivalence for $(n-1)$-categories, and Morita equivalent $(n-1)$-categories lead to TQFTs which are isomorphic at the $(n-1,n)$ level. More generally, the $k$-categories assigned by the two TQFTs to closed $(n-1-k)$-manifolds are Morita equivalent. Morita equivalent $(n-1)$-categories can look somewhat different. For example, Morita equivalent tensor categories (2-categories with one 0-morphism) can have different numbers of (isomorphism classes of) simple objects.

(Warning: There is more than one version of categorified Morita equivalence in the literature. I have in mind the version that uses bimodules rather than functors "all the way up", as described here.)

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Cool! So, to make sure I understand what you're saying, if I have an $(n-1, n)$ TQFT $F$ and two extensions of $F$ all the way down to points, and furthermore if the extensions assign Morita equivalent $(n-1)$-categories to the point, then they assign equivalent categories at every level, right? What happens if you have two fully-extended TQFTs which assign non-equivalent $(n-1)$-categories to a point (and more generally, non-equivalent categories to codim-2 manifolds). Do they have a chance of still being isomorphic at the $(n-1,n)$-level? – Andy Manion Apr 22 '13 at 18:38
In answer to your first question: yes, right. I'm not sure about the second question (non-Morita-equivalent $(n-1)$-categories giving rise to same $(n-1, n)$ structure). I'm not even sure which way I would bet, if I had to bet. – Kevin Walker Apr 22 '13 at 21:25