There is a "folk theorem" (alternatively, a fun and easy exercise) which asserts that a 2D TQFT is the same as a commutative Frobenius algebra. Now, to every compact oriented manifold $X$ we can associate a natural Frobenius algebra, namely the cohomology ring $H^\ast(X)$ with the Poincare duality pairing. Thus to every compact oriented manifold $X$ we can associate a 2D TQFT.

Is this a coincidence? Is there any reason we might have expected this TQFT to pop up?

When $X$ is a compact symplectic manifold, perhaps the appearance of the Frobenius algebra can be explained by the fact that the quantum cohomology of $X$, which comes from the A-twisted sigma-model with target $X$, becomes the ordinary cohomology of $X$ upon passing to the "large volume limit".

But for a general compact oriented $X$? I don't see how we might interpret the appearance of the Frobenius algebra in some quantum-field-theoretic way. Maybe there is an explanation via Morse homology?


2 Answers 2


These 2D TQFTs do not come from extended theories (unless X is discrete). I interpret this as saying that these theories are non-local (in the 2D bordism) and so you will have trouble interpreting them in a traditional QFT framework. You will have to do something funny and non-local, like squashing your circles to points and surfaces to graphs, as in the Cohen work mentioned by Tim.

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    $\begingroup$ Hi Chris: That is very interesting. To be precise, are you saying that the 2-1 TQFT coming from any non-discrete manifold cannot come from a 2-1-0 TQFT? Is this perhaps somewhere in your thesis? $\endgroup$ Dec 30, 2009 at 15:01
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    $\begingroup$ Hi Kevin. Yes, that's right. This is Corollary 4.6.15 in my thesis (on page 211). For any 0-1-2 TQFT over a ring the commutative algebra associated to the circle must be separable and projective. Over a field this implies it must be semi-simple, and over Z it is equivalent to being a finite direct sum of copies of Z (as algebras!). In particular, this can only happen if the cohomology is concentrated in degree zero. This statement holds in the framed TQFT setting, and so is true for all variations of 2D TQFTs: oriented, unoriented, spin, with G-bunlde, etc. $\endgroup$ Dec 30, 2009 at 15:24
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    $\begingroup$ That's super awesome. $\endgroup$ Dec 30, 2009 at 16:10

There is indeed a Morse homology explanation; and, in the symplectic case, it's a degeneration of the Hamiltonian Floer cohomology picture. In a nutshell, you degenerate the surfaces to graphs, and then use a different Morse function for each edge. This has been explored (e.g.) by Ralph Cohen, initially in a paper with Betz and more recently with Norbury:



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