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Of course I'm talking about the algebraic notion of field.

In a few words, if a TQFT consists of a functor $Z\colon Cob(n)\to \mathbf{Vec}_k$, I'm wondering if there are sensible relations among different choices of $k$... Some example coming to my mind:

  1. Fix a real valued TQFT. Can I turn the functor sending objects to $Z(M)\otimes_\mathbb R\mathbb C$ (the "complexification" of the real vector space associated to $M$) into a complex valued TQFT?

  2. What if I consider, in general, a $k$-valued TQFT and a $K$-valued TQFT (with $K$ an extension of $k$)? What if $K$ is finite, infinite, algebraic, Galois or not, positive charateristic or not, etc etc.?

(Feel free to retag the question).

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Yes it makes sense to talk about TQFTs over different fields and yes you can ask the usual kind of questions about base extensions and descent. You may find this recent preprint of Etingof-Gelaki interesting (it's about the same question for fusion categories, but via Turaev-Viro those give TQFTs).

The answer to your first question is trivial: you have a functor Cob(n)-> Vec_k and you compose it with the base extension functor Vec_k -> Vec_K. I'm not sure what the question is in your second question.

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  • $\begingroup$ The answer to your first question is trivial: you have a functor Cob(n)-> Vec_k and you compose it with the base extension functor Vec_k -> Vec_K. I don't think it is so trivial: a TQFT must have some special properties the simple composition $F=(-\otimes_k K)\circ Z_k\colon Cob(n)\to Vec_k\to Vec_K$ does not have; for example what's happen if i want to compute $F$ on the disjoint union of two manifolds? $\endgroup$
    – fosco
    Commented May 21, 2011 at 7:22
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    $\begingroup$ @tetrapharmakon: I assume the special properties you mention are that the tqft must be a symmetric monoidal functor. The base change functor is symmetric monoidal so these properties (really structures) are preserved. $\endgroup$
    – Chris Schommer-Pries
    Commented May 21, 2011 at 14:44

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