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Equation of motion and quantization of the system defined by interpreting Hopf invariant as an action functional

+ 3 like - 0 dislike
1058 views

Consider a field theory defined on the circle $S^1$, with one field taking values also in $S^1$. Let the action equal to the winding number of the field. This can be thought of special case of the Hopf invariant for higher dimensional sigma models, for instance, you can analogously take a purely topological Hopf-invariant action for a field theory on $S^3$ where the field values lie in $S^2$.

  • How does one calculate the equation of motion of \(\theta\) from this action functional?
  • How does the above action get quantized?
  • Is this a quantum field theory at all?
asked Apr 6, 2014 in Theoretical Physics by SDevalapurkar (285 points) [ revision history ]
reshown Apr 7, 2014 by dimension10
Most voted comments show all comments

I believe the Hopf invariant for $S^1 \to S^1$ is just the winding number.

"I edited this infernal question to be clear, removing the mistaken identification of the spaces, removing the high-falutin' language, and asking whether the winding number action makes sense. Also downvoted it, because it's a bad question, and probably insincere, designed to test whether we can detect gibberish."

Hi Ron, I'm not sure if either insulting the OP or accusing him of trying to 'trick' the community is a nice way to treat people on this site, let alone attract new users, even when you don't think the question makes a lot of sense. 

@dilaton Yep I'm physicslover on PSE ;)

yes, calling a question "bad" or "infernal" is completely justified if it is really bad, but we probably shouldn't put our speculations about others' intention or sincerity here(of course unless it's an obvious troll), so -1 to Ron's comment. 

@Danu: I am sorry. I retract my comments. Anonymous is totally welcome, and if anonymous would like the original wording back, that's ok too. The "Infernal" was just a jokey way of expressing my own frustration, not a slight on the OP.

Most recent comments show all comments

@Danu Well, I don't really consider Ron's comments to be particularly offensive, and I don't think that we should really have rules preventing rudeness.  

Of course, I disagree with him about the poster's intentions.  

@physicsnewbie I really don't understand the purpose of your "lol" here...   

@dimension10 the "lol" is to lighten the atmosphere a bit, and not give the impression that I see it as serious you and the other admins being banned elsewhere in the past for not "being nice".

1 Answer

+ 1 like - 0 dislike

The action

$\int d\theta$

has a gauge symmetry $d\theta \mapsto d\theta + df$. The gauge equivalence classes are discrete (labelled by the winding number), so the equations of motion don't say anything (and the gauge field is automatically flat for dimensional reasons).

The action is integer-valued, so you get a different "TQFT" for every parameter $\alpha \in U(1)$ by

$S_\alpha = \alpha \int d\theta$.

answered Apr 6, 2014 by Ryan Thorngren (1,605 points) [ revision history ]
edited Apr 7, 2014 by Ron Maimon

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