It is well known that the theta term

$\int d^4x\frac{\theta}{4\pi}Tr[F\wedge F]=\int d^4x\frac{\theta}{4\pi}\epsilon_{\mu\nu\sigma\lambda}Tr[F^{\mu\nu}F^{\sigma\lambda}]$

is a topological term, since the integral is instanton number on a 4-d manifold (mathematically, the integral is the 2nd Chern-Character).

Besides, it is definitely metric independent. As far as I know, metric independent is at least a necessary condition for a theory to be a topological quantum field theory (TQFT). For example, Chern-Simons and B-F are metric independent.

So, now, the questions come:

  1. Is metric independent a sufficient condition for a theory to be a TQFT?

  2. Is theta term above a TQFT?

My guess it is both are not, but I am not sure. I checked out some reference and found the mathematical definition of a TQFT, it has to satisfy Atiyah-Segal axioms(see [1] or [2]). But it is clear to be how to prove theta term is or isn't a TQFT, because category is too abstract to me. Could someone give helps?

Thanks in advance.

[1] http://en.wikipedia.org/wiki/Topological_quantum_field_theory

[2] http://webusers.imj-prg.fr/~christian.blanchet/Articles/EMP_axiomatic_TQFT.pdf

  • 2
    $\begingroup$ A "term" is not a QFT. You want to know whether the action here defines a TQFT, but you need to define/find out the spaces of states you are looking at on the manifold in order to see whether this is a TQFT, since the TQFT axioms are formulated mostly in terms of the finite-dimensional spaces of states. So, what is the space of state you would associate to a spatial slice of a theory with that action? $\endgroup$
    – ACuriousMind
    Mar 28, 2015 at 19:12
  • 2
    $\begingroup$ As @ACuriousMind has pointed out, you need to figure out the space of states on a spatial manifold. One way to proceed is the following: first, find the classical phase space by solving equations of motion. Then do canonical quantization. For Chern-Simons theory or BF theory, the equation of motion sets the field strength to zero (in the absence of sources), so there are only flat connections (up to gauge transformations). However, $F\wedge F$ term does not have any effect on the equation of motion on a closed manifold. So this is already a sign that it is not a TQFT. $\endgroup$
    – Meng Cheng
    Mar 28, 2015 at 19:33
  • $\begingroup$ Hi, @MengCheng and AcuriousMind, Thank you for your answer. Right, $F\wedge F$ term has trivial eom so it is not a TFT. Another question is: Is the following TFT, $l=\psi d\psi\wedge d\psi\wedge d\psi$, where $\psi$ are grass-man fields on 3-manifold and the theory has non-zero eom. Since all TFT I know are bosonic, and are gauge theories, it will be surprising to have a fermionic, non-gauge TFT. $\endgroup$
    – Blue
    Apr 1, 2015 at 15:54
  • $\begingroup$ @ACuriousMind thank you for answer and the same question as above. $\endgroup$
    – Blue
    Apr 1, 2015 at 15:55
  • 1
    $\begingroup$ @Blue You can write many non-gauge topological terms, but it does not mean they are TQFT. The reason that usually one needs gauge theories is because gauge theories naturally have non-local observables, like Wilson loops and higher-dimensional generalizations. Otherwise, like in your theory, one can easily write down many local observables (e.g. $\psi d\psi$), unless your EOF sets all these to zero, but I doubt that is the case. $\endgroup$
    – Meng Cheng
    Apr 1, 2015 at 16:14


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