According to the Baez-Dolan cobordism hypothesis, an extended TQFT is determined by its value on a single point. This value a fully dualizable object of a symmetric monoidal $n$ category (a fully dualizable object is a higher categorical analogue of a finite dimensional vector space). The Alexander polynomial is a quantum invariant, and comes from a TQFT. 

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> How can an "Alexander polynomial" TQFT be put into an extended TQFT, and what is its value at a single point?

The question I just asked is closely related to [this question](https://mathoverflow.net/questions/9847/why-is-the-alexander-polynomial-a-quantum-invariant). I also asked the question on the ldtopology blog [here](http://ldtopology.wordpress.com/2010/02/02/the-alexander-polynomial-as-a-quantum-invariant-part-2/), and Theo Johnson-Freyd suggested that MO might be the place to ask it.  

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Briefly, I will summarize what an extended TQFT is. A TQFT as a symmetric monoidal functor Z:**Cob**(n)-> **Vect**(k) from the tensor category of $n-1$ dimensional manifolds and cobordisms between them to the tensor category of vector spaces over a field k. An extended TQFT is a symmetric monoidal functor Z:**Cob**k(n)->**C** from the n-category of cobordisms to a symmetric monoidal n-category **C**. I vote for the [introduction to Lurie's expository account of his work proving the Baez-Dolan cobordism hypothesis](http://www.math.harvard.edu/~lurie/papers/cobordism.pdf) as the best place to read about why extended TQFT's are natural objects, to understand their motivation, and to understand why people are so excited about them. An extended TQFT assigns a fully dualizable object to a point, and a higher “trace” on this object to a closed n-dimensional manifolds.