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In his book Quantum invariants of knots and 3-manifolds page 124, Turaev defined a TQFT $\tau$ axiomatically.

For a cobordism $(M, \partial_{-}M, \partial_{+}M)$, a TQFT assignes a $k$-homomorhism $\tau(M)$ from the projective $k$-module $\tau(\partial_{-}M)$ to the projective $k$-module $\tau(\partial_{+}M)$. Here $k$ is a (ground) ring.

One of the axioms is a functoriality. Let $M_1$ and $M_2$ be a cobordisms and let $M$ be a cobordism obtained by gluing $M_1$ and $M_2$ along a homeomorphism $f$ from $\partial_+(M_1)$ to $\partial_{-}(M_2)$. Then the functoriality says that

$\tau(M)=k\tau(M_2)\circ f_{*} \circ \tau(M_1)$, where $\tau$ is a TQFT and $k$ is an invertible element called anomaly for the pair $(M_1, M_2,f)$.

My question is whether this is associative or not. Namely, if we have two pairs $(M_1, M_2,f)$ and $(M_2, M_3,g)$, then we can calculate the anomaly in two way, gluing first along $f$ and then $g$, or gluing along $g$ first then $f$.

Are anomalies obtained in these ways same? If so, how do we prove it from the axioms of Turaev TQFT?

If we regard $(M_1 \cup_f M_2)\cup_g M_3$ and $M_1 \cup_f (M_2\cup_g M_3)$ are the same, then we should have the anomalies the same. Or if we think there is a homeomophism between these spaces, then the anomalies should be the same up to invertible factor. Right?

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  • $\begingroup$ Can you add the definition of anomaly (if it is short)? Thanks. $\endgroup$
    – Tim Porter
    Commented May 16, 2013 at 5:56
  • $\begingroup$ An anomaly for a pair $(M_1, M-2, f)$ is an invertible element satisfying $\tau(M)=k\tau(M_2)\circ f_{*} \circ \tau(M_1)$. If we can take $k=1$ for all pairs, we say that the tqft is anomaly free. $\endgroup$
    – user22741
    Commented May 16, 2013 at 6:46
  • $\begingroup$ @Tim Porter Does this help? $\endgroup$
    – user22741
    Commented May 16, 2013 at 7:46
  • $\begingroup$ Could you clarify a bit further? Are $\tau(M_i)$ Hilbert spaces? Of what algebra is $k$ an invertible element? $\endgroup$
    – Chris Heunen
    Commented May 16, 2013 at 9:46
  • $\begingroup$ @Chris Heunen $\tau(M_i)$ is not Hilbert space. Neither $\tau(\partial M)$, in general. $\endgroup$
    – user22741
    Commented May 16, 2013 at 10:13

1 Answer 1

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I think in the section you refer to Turaev is just defining what one might call a "projective TQFT" or a "TQFT up to phase factors". In later sections he discusses how to remove the anomaly by tweaking the cobordism category.

Turaev's book is probably not the best place to learn about Reshetikhin-Turaev TQFTs for a beginner. You might want to supplement your reading with one or more of

  • The original Reshetikhin-Turaev papers
  • The book by Kauffman and Lins
  • The paper by Blanchet, Habeggar, Masbaum and Vogel
  • My TQFT notes from 1991
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