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To what extent have people classified $n$-dimensional TQFTs that assign a 1-dimensional vector space to every compact oriented $(n-1)$-manifold?

I have some vague reasons to suspect that the classification depends heavily on $n \bmod 8$, but I'm having trouble finding information on this.

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2 Answers 2

They are called "invertible field theories".

See, lectures 17-24 for the relation to Madsen-Tillmann spectra.

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See also: –  Qiaochu Yuan Aug 19 '14 at 15:27

I recently developed a non-linear $\sigma$ model approach ( ) to address this issue. Let $\text{iTO}_L^n$ be the set of the fully extended TQFTs in $n$-dimensions, that assign a 1-dimensional vector space to every compact oriented $(n−1)$-manifold. (Physically, $\text{iTO}_L^n$ is the set of L-type topologically-ordered phases in $n$-dimensional space-time that have no topological excitations.) Such a set, $\text{iTO}_L^n$, actually from an Abelain group.

We find that $\text{iTO}_L^1=\text{iTO}_L^2=\text{iTO}_L^4=\text{iTO}_L^6=0$, $\text{iTO}_L^3=\mathbb{Z}$, $\text{iTO}_L^5=\mathbb{Z}_2$, $\text{iTO}_L^7=2\mathbb{Z}$.

Here $\text{iTO}_L^3=\mathbb{Z}$ is generated by $\omega_3$ with d$\omega_3=p_1$ and $p_1$ the first Pontryagin class.

Similar results are also obtained in using cobordism approach. But we disagree on the generators of $\text{iTO}_L^3=\mathbb{Z}$ and $\text{iTO}_L^7=2\mathbb{Z}$. For example, in , the generator of $\text{iTO}_L^3=\mathbb{Z}$ is $\frac13 \omega_3$ instead of $\omega_3$.

In , the generator of $\text{iTO}_L^3=\mathbb{Z}$ is $\frac16 \omega_3$, instead of $\omega_3$ or $\frac13 \omega_3$.

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Could you maybe add a comment on what that sequence of groups is: $0,0,\mathbb Z,0,\mathbb Z/2,0,2\mathbb Z$ (in particular, how does it continue?). Mathematicians prefer conceptual descriptions without computations to computations without conceptual descriptions. –  André Henriques Dec 9 '14 at 7:01

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