# Anomalies in the definition of Turaev's TQFT

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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Can you add the definition of anomaly (if it is short)? Thanks. –  Tim Porter May 16 '13 at 5:56
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. –  Primo May 16 '13 at 6:46
@Tim Porter Does this help? –  Primo May 16 '13 at 7:46
Could you clarify a bit further? Are $\tau(M_i)$ Hilbert spaces? Of what algebra is $k$ an invertible element? –  Chris Heunen May 16 '13 at 9:46
@Chris Heunen $\tau(M_i)$ is not Hilbert space. Neither $\tau(\partial M)$, in general. –  Primo May 16 '13 at 10:13