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  APS $\eta$-invariant and spin-Ising TQFT

+ 3 like - 0 dislike

I am interested in the relation between the Atiyah-Patodi-Singer-$\eta$ invariant and spin topological quantum field theory. In the paper "Gapped Boundary Phases of Topological Insulators via Weak Coupling"


by Seiberg and Witten, they presented such a relation between the two. 

Let the $3+1$ dimensional manifold $X$ be the world volume of a topological insulator. Its bounday $W$ is a $2+1$ dimensional manifold. Let $\chi$ be a massless Dirac fermion on the bounday manifold $W$. Then, integrating out the boundary fermion 

$$\int_{W}d^{3}x\,i\bar{\chi}D \!\!\!\!/\,\chi\,, $$

where $D_{\mu}=\partial_{\mu}+iA_{\mu}$, one has the partition function


So far, this is just the standard parity anomaly in odd dimensions. 

On page 35, the authors claimed that the factor $\Phi=e^{-i\pi\eta/2}$ is actually the partition function of a topological quantum field theory, called spin-Ising TQFT. The name comes from the fact that it is related with the 2D Ising model of CFT. The authors explained that this is due to the "Free-Dai theorem".


I don't really understand much from the paper of Freed-Dai theorem because of its heavy mathematics. But from my understanding, it is saying that the $\eta$ invariant is some kind of topological invariant and cobordism invariant, which satisfies the gluing and surgery axioms of TQFT. Thus, the factor $\Phi=e^{-i\pi\eta/2}$ can be treated as the partition function of some TQFT.

Now the question is why this TQFT is the so-called spin-Ising TQFT. The authors claim that the partition function of the spin-Ising TQFT should be of modulus $1$ because the $\mathbb{Z}_{2}$ chiral ring generated by the field $\psi$ (from the 2D Ising model $\left\{1,\sigma,\psi\right\}$) has only one representation.

Question 1: Why does the fact that the chiral algebra has one representation makes its partition function being of modulus $1$?

The authors then showed an example, taking the manifold $W$ to be $S^{2}\times S^{1}$, that the partition function of the corresponding spin-Ising TQFT is $\pm 1$, which is indeed a phase. Then, by Freed-Dai theorem, they claimed that in general the partition function of a spin-Ising TQFT should be $\Phi=e^{-i\pi\eta/2}$. 

I don't really understand much from the paper of Freed-Dai theorem. Could anyone please enlighten me on how one should apply that theorem to this case?

The authors explained in the following that if $W$ has a boundary $\Sigma$, then the product of the path integral of the chiral fermion $\psi$ on $\Sigma$ and the factor $\Phi$ is smooth and well-defined because of the Freed-Dai theorem.

However, in our case, the manifold $W$ itself is the boundary of the $3+1$-manifold $X$, and so $W$ has no bounday at all. How should one understand the explanation provided by the authors?

I also posted my question at 


New Edition: Suppose the $2+1$ dimensional manifold $W^{\prime}$ has a boundary $\Sigma$. There is a fermionic field $\psi$ defined on the boundary $\Sigma$. From the Freed-Dai theorem explained in the paper by Seiberg and Witten, the following path-integral is smooth and well-defined.


where $\mathcal{Z}_{\phi}[\Sigma]$ is the partition function of the $2D$ fermion $\psi$ on $\Sigma$. The authors claimed that this fermion are related with the spin-Ising model of $2D$ CFT. 

Is that true that this is just the holomorphic sector of the free Majorana fermion in $2D$? 

The motivation behind the above question is that I found that the Ising model in RCFT really looks like a sum over even spin structures of the $2D$ free majorana fermion. If this were true, does it mean that 


asked Oct 22, 2018 in Theoretical Physics by Libertarian Feudalist Bot (270 points) [ revision history ]
edited Oct 23, 2018 by Libertarian Feudalist Bot

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