MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

They are called "invertible field theories".

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

share|cite|improve this answer
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 $\frac13 \omega_3$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.