In some discussions of bosonization, it is stressed that the duality between free bosons and free fermions requires the use of a compact boson. For example, in a review article by Senechal, the following statement is made:

In order for bosonization to be rigorously defined, the field $\phi$ must have an angular character. In other words, the target space of the field must be a circle of radius $R$ (to be kept general for the moment), so that $\phi$ and $\phi + 2\pi R$ are identified. We say that the boson is compactified on a circle.

In other sources, however, no mention of the compactness of $\phi$ is made. For example, in the review article by von-Delft and Schoeller, there is no mention of the word "compact". Moreover, books like Fradkin's and Shankar's respective condensed matter field theory books seem to make no mention of this issue, to the best of my knowledge.

As far as I understand, theories of compact and non-compact bosons are quite nontrivially different. At first glance, both theories appear to have the same action: $$ S[\phi] = \frac{K}{2} \int d^2x (\partial_{\mu} \phi)^2. $$ However, a massless non-compact boson is a trivial free theory, and the coefficient $K$ can always be rescaled away by changing our normalization of $\phi$. On the other hand, rescaling a compact boson nontrivially changes the theory, and the theory can undergo the famous KT transition at a critical value of $K$.

Other purely circumstantial evidence that the boson obtained by bosonizing fermions is compact includes:

  • In the central bosonization formula, $\psi \sim e^{-i \phi}$, $\phi$ appears inside an exponential. As a result, all the observables one considers in the bosonized side of the theory seem to respect the periodicity of $\phi$. This is transparent from the form of the bosonized action as well: it's easy to find terms in the fermionic theory which translate to objects like $\cos \phi$ in the action of the bosonized theory, but one can never generate a mass term like $m \phi^2$ to my knowledge.

  • In books on the Luttinger liquid, the parameter $K$ above is said to be related to the compactification radius of the boson. In Fradkin's book, for example, $K$ is referred to as the compactification radius, but then the compactness of $\phi$ is never used thereafter.

  • There is a self-duality which trades $\phi$ for its dual field $\theta$, and correspondingly $K$ to $1/K$. As far as I know, this is again related to the fact that $\phi$ is compact; if $\phi$ were non-compact, we could rescale $K$ to whatever we want without changing the physics. It's only when $\phi$ is compact that this duality becomes nontrivial.

However, the above are all only circumstantial evidence, and don't explain why a compactified boson is necessary. Moreover, books like Fradkin's and review articles like von-Delft's, which put quite a bit of work into proving the bosonization formulae, never seem to run into trouble by not using a compact boson.

So to summarize, here is the question: in abelian bosonization, where a Dirac fermion is equivalent to a massless boson, is the boson compact or noncompact? Is there ever freedom to use one or the other, or must it be compact (or noncompact)? And most importantly, how can I see why the boson is compact or not?

  • $\begingroup$ I don't have Shankar or Fradkin readily to hand, but von-Delft and Shoeller certain discuss the importance of the system having a finite length $L$ when setting up bosonization, which amounts to saying the system is compact even if they don't use that word. They then go on to consider the limit that $L\rightarrow \infty$, but are clear that in principle $L$ should always be finite even if it becomes large. I would be surprised if the other sources don't do something similar $\endgroup$ Commented Oct 8, 2021 at 10:00
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    $\begingroup$ @BySymmetry The question is about the compactness of the space of values of $\phi(x)$, not about the compactness of the space of values of $x$. $\endgroup$ Commented Oct 8, 2021 at 13:33
  • $\begingroup$ For a compact boson $\phi\simeq \phi+2\pi$, it is incorrect to claim that there is a duality under $\phi\to\theta$ and $K\to1/K$, and at $K=1$ the theory is self dual. In fact one can show that for $K=1$, the compactification radius of $\theta$ is $\pi$. This is an error in Fradkin's book. See my question and answer here physics.stackexchange.com/questions/422416/…. $\endgroup$ Commented Jan 31 at 3:04

2 Answers 2


It is compact.

The simplest approach is to look at symmetries. The most obvious symmetry has algebra $\mathfrak{u}_1$, given by rephasing the fermion or shifting the boson. What is the global form of this symmetry? (i.e., what is the group for this algebra?)

It is clear that the fermion has group $U(1)$, as the group element is a phase.

For the boson, the group is $\mathbb R$ for the non-compact case and $U(1)$ for the compact case. The reason is obvious: the set of transformations $\phi\mapsto\phi+\alpha$ has the structure of $\mathbb R$ if $\alpha$ is arbitrary, but colapses to $U(1)$ if $\alpha\cong\alpha+2\pi$.

Hence, if these symmetries are to be identical (which is a necessary condition for duality), the boson must be compact.

(You can also convince yourself that these theories are CFTs and their chiral algebra is precisely the affine extension of $\mathfrak u_1$; both are rational with respect to this algebra and in fact holomorphic. For the fermions this is obvious but for the bosons this is only true at $R=1$ (in some normalization).)

  • $\begingroup$ This is compelling, but it raises further questions. First, the Dirac fermions have an additional $U(1)$ chiral symmetry, which doesn't seem to be present on the boson side. Is this because the chiral anomaly breaks the symmetry in the Dirac theory anyway? Second, when people try to give evidence of the duality, they usually compute correlation functions on the Bose and Fermi sides and show that they are equal. Why is it sufficient to compute correlation functions of non-compact bosons? Indeed, it's not clear to me that one can simply compute correlation functions of the compact boson anyway. $\endgroup$
    – Zack
    Commented Oct 8, 2021 at 15:38
  • $\begingroup$ @Zack There is no chiral anomaly, the fermion is free. The boson has two $\mathfrak u_1$'s too. In the old String Theory literature these are known as momentum and winding symmetries, and indeed (in a given T-duality frame) only one of them is obvious. Anyway, in checking the duality you need to compute correlators of the compact boson, not those of the non-compact one. Your best bet is to check ST books, the correlators of compact bosons are what they call "vertex operators", essentially $e^{i\phi}$ (which manifestly respects the $2\pi$ identification). $\endgroup$ Commented Oct 8, 2021 at 16:02
  • $\begingroup$ Thank you, one more question: in the condensed matter literature I'm familiar with, it seems extremely common to treat $\phi$ as if it were an ordinary noncompact field with a Gaussian action. For example, it's commonly said that the Luttinger liquid without Umklapp scattering is mapped to a free boson theory. With Umklapp, one gets a Sine-Gordon theory instead, and one then performs RG treating $\phi$ as Gaussian to obtain the phase diagram. So, how should I understand when it's okay and when it's dangerous to neglect the compactness of $\phi$? $\endgroup$
    – Zack
    Commented Oct 8, 2021 at 16:10
  • $\begingroup$ @Zack I am not all that familiar with the cond-mat literature, so I cannot really judge whether they are just being careless or not. But as a rule of thumb I'd say: you should never neglect compactness, unless perhaps if you are just doing some simple like looking at a local observable in perturbation theory. (Such observables are insensitive to global issues anyway). $\endgroup$ Commented Oct 8, 2021 at 17:20
  • $\begingroup$ Maybe I can ask a specific question, instead of alluding vaguely to cond-mat literature: if I use bosonization to map the massive Thirring model to the Sine-Gordon model, I would be tempted to conclude that the MT model undergoes a KT transition at certain parameters. But as far as I understand, the usual RG analysis of the SG model leading to the KT transition assumes that $\phi$ is noncompact. Am I thus wrong to conclude that the MT model has a KT transition? $\endgroup$
    – Zack
    Commented Oct 8, 2021 at 17:41

When in doubt, you can go back to the torus partition function. For the free Dirac fermion, it is \begin{equation} Z = \frac{1}{2} \left ( \left | \frac{\vartheta_2(\tau)}{\eta(\tau)} \right |^2 + \left | \frac{\vartheta_3(\tau)}{\eta(\tau)} \right |^2 + \left | \frac{\vartheta_4(\tau)}{\eta(\tau)} \right |^2 \right ). \end{equation} This agrees with the partition function of the free boson at $\sqrt{2}$ times the self dual radius. Some techniques for evaluating related partition functions (along with this specific example) are given in Chapter 8 of Ginsparg.

This can be used as a starting point to find the fermionic description of compact free bosons at other radii. The point is that the radius can be tuned by adding the current-curent deformation $\partial \phi \bar{\partial} \phi$. After applying the duality, this becomes a four-fermi interaction which defines the massless Thirring model. You can then find another pair of theories related by abelian bosonization if you add a fermion mass. Since this breaks conformal symmetry, it leads to a non-trivial S-matrix. This massive Thirring model S-matrix is the same as the one for the sine-Gordon model as shown by Coleman.

  • $\begingroup$ Thanks for your comments -- I'm still learning CFT, so I don't yet understand the meaning of the partition function you wrote. But your last point does raise another confusion of mine: although the Sine-Gordon action seems to respect the periodicity of $\phi$, my understanding was that the SG theory was a theory of a non-compact boson -- see, ex, the mode expansion in Coleman's Eq. 2.1. This is also important in the traditional discussion of the KT transition in the SG theory, where the scaling dimension of $\cos \phi$ is evaluated using the Gaussian action. $\endgroup$
    – Zack
    Commented Oct 8, 2021 at 17:31
  • $\begingroup$ So, should I regard SG as a theory of a compact or non-compact boson, and does the RG analysis change between the two cases? $\endgroup$
    – Zack
    Commented Oct 8, 2021 at 17:31
  • $\begingroup$ I haven't read Coleman's steps in awhile but I thought SG described a compact boson. In $\cos(\beta \phi)$, you can shift $\phi$ by $\frac{2\pi}{\beta}$ meaning the radius should be $\frac{1}{\beta}$. If this is true, (1.9) agrees with what the current-current term is supposed to do. You get an infinite radius at $g = \infty$ but a finite radius at $g = 0$. $\endgroup$ Commented Oct 8, 2021 at 19:12
  • $\begingroup$ Also, compact and non-compact free bosons are both Gaussian if you just look at the action. The radius is important for determining which local operators are admissible in the theory. But if someone just hands you an operator built from $\phi$, you can read off its scaling dimension without knowing the radius. $\endgroup$ Commented Oct 8, 2021 at 19:16
  • $\begingroup$ Perhaps it's just terminological, but by "Gaussian" I mean that the path integral and correlation functions can be evaluated by a simple Gaussian integral. This is not the case for the compact boson, where a Gaussian integral amounts to neglecting the compactness of $\phi$ and "misses" vortices. Perhaps the confusion is because Coleman's paper only treats the zero charge sector of the Thirring model, which means no winding of the boson. But I'm not totally positive, I would think even with no winding there could still be vortices so long as the total charge is zero. $\endgroup$
    – Zack
    Commented Oct 8, 2021 at 22:35

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