# Topological theta term and topological quantum field theory

+ 3 like - 0 dislike
1688 views

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 the 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 independence 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 independence a sufficient condition for a theory to be a TQFT?

2. Does the theta term above give rise to a TQFT?

My guess is that both answers are negative, but I am not sure. I checked out some reference and found the mathematical definition of a TQFT, it has to satisfy the Atiyah-Segal axioms (see  or ). But I do not know how to prove whether or not the theta term gives a TQFT, because category is too abstract to me. Could someone give help?

This post imported from StackExchange Physics at 2015-04-02 13:06 (UTC), posted by SE-user Blue

edited Apr 13, 2015
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?

This post imported from StackExchange Physics at 2015-04-02 13:06 (UTC), posted by SE-user ACuriousMind
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.

This post imported from StackExchange Physics at 2015-04-02 13:06 (UTC), posted by SE-user Meng Cheng
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.

This post imported from StackExchange Physics at 2015-04-02 13:06 (UTC), posted by SE-user Blue
@ACuriousMind thank you for answer and the same question as above.

This post imported from StackExchange Physics at 2015-04-02 13:06 (UTC), posted by SE-user Blue
@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.

This post imported from StackExchange Physics at 2015-04-02 13:06 (UTC), posted by SE-user Meng Cheng

By the way, Blue, there are simple examples of fermionic TQFTs with no gauge fields starting in 1+1d. The theories I have in mind are theories where the only degree of freedom is a spin structure. A good action principle in 2d is given by the Arf invariant of the spin structure. See my paper with Kapustin, Turzillo, and Wang: http://arxiv.org/abs/1406.7329 . Coupling to these sorts of TQFTs is like inserting factors of $(-1)^F$ in the path integral, or imposing certain boundary conditions for fermions.

+ 6 like - 0 dislike

There is a (canonical) way to make a TQFT out of topological terms for a gauge field. First of all, note that a TQFT has no local degrees of freedom, so we cannot allow non-flat configurations. For a $U(1)$ gauge field, this means that the Chern class of the gauge bundle (the thing which in real-valued cohomology equals $F/2\pi$) has to be torsion as an integral cohomology class (though it still may be nontrivial!). If for a spacetime $X$ we have $H^2(X,\mathbb{Z}) =$free$\oplus$finite, then this is enough restriction to define a path integral (it will be a finite sum over different gauge sectors). This TQFT for your topological term is a sort of BF theory.

Note you can think of the allowable topological terms as the allowable ways to couple your QFT to a TQFT. See http://arxiv.org/abs/1401.0740 for a nice introduction. These two ideas are in some sense inverse maps to each other.

answered Apr 2, 2015 by (1,895 points)
edited Apr 3, 2015 by 40227

Could you please edit the original question slightly so that it becomes clearer?

Sorry Arnold, I don't know what kind of edit you have in mind.

For example that hinted to in the comment by ACuriousMind. Just improve the language so that the statements make proper sense on the technical level. (If needed, I'll improve then any remaining purely grammatical problems.)

I see. ACuriousMind is referencing the Atiyah-Segal axioms, but these should maybe not be the gold standard for definition what a TQFT is. The Atiyah-Segal data is something one should try to produce from a QFT that happens to be topological. There are interesting questions like what is required to produce an Atiyah-Segal TQFT? Is an action enough? A Lagrangian? Do we need to make more choices?

I edited the question myself. Please re-edit it if I introduced problems while cleaning up the language.

+ 4 like - 0 dislike

The question seems to consider TQFT of "Schwarz-type", i.e. defined by path integral with a metric independent action.  Ryan Thorngren has given an answer in this context.

But another standard way to construct a TQFT is to consider a TQFT of "Witten-type", defined by topological twist of a non-topological quantum field theory. Topological twist means to change the Lorentz transformation properties of the fields of the non-topological QFT to construct from them a BRST like charge $Q$, i.e. a scalar operator satisfying $Q^2=0$, such that the energy-momentum tensor $T_{\mu \nu}$ is $Q$-exact. Restricting the twisted theory to the $Q$-closed states and operators, $Q$-exact states and operators decouple and so we obtain a TQFT.

One standard example is the Donaldson-Witten theory, obtained by topological twisting of the pure $N=2$ super Yang-Mills four dimensional theory. After twisting, the action is precisely equal to the theta term up to a $Q$-exact term. Fields of the TQFT are antiselfdual connections, i.e. instantons, and the partition function can be expanded over the instanton configurations weighted by the action, i.e. by the theta term, i.e. by the instanton number.

answered Apr 12, 2015 by (5,120 points)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar\varnothing$sicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.