TITLE: Classifying TQFTs with 1d vector spaces
QUESTION [13 upvotes]: 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.

REPLY [4 votes]: I recently developed a non-linear $\sigma$ model approach ( http://arxiv.org/abs/1410.8477 ) 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
http://arxiv.org/abs/1403.1467 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 http://arxiv.org/abs/1403.1467 , the generator of $\text{iTO}_L^3=\mathbb{Z}$
is $\frac13 \omega_3$ instead of $\omega_3$.
In http://arxiv.org/abs/1406.7278 , the generator of $\text{iTO}_L^3=\mathbb{Z}$
is $\frac16 \omega_3$, instead of $\omega_3$ or $\frac13 \omega_3$.