2
$\begingroup$

There is a kind of duality transformations between antisymmetric tensor fields which I learnt from a series of lectures by Gia Dvali on quantum field theory. I have not been able to locate a source for further reading. So, I am not sure if it is mainstream physics and with how much grain of salt I should take its seriousness.

The algorithm goes like this.

  1. One starts with a Lagrangian, say that of a massless scalar field. $$ \mathcal L = \partial_\mu \phi \partial^\mu \phi \equiv Z_\mu(\phi) Z^\mu(\phi) \,,\tag{1} $$ where $Z_\mu(\phi) \equiv \partial_\mu \phi$ is the “field strength”.

  2. The field strength satisfies a Bianchi identity, $\partial_{[\nu}Z_{\mu]} = 0$. Adding this identity as a constraint to the Lagrangian using a Lagrange multiplier does not change the theory. The following Lagrangian is physically equivalent to the previous one. $$ \mathcal L = Z_\mu Z^\mu + B_{\alpha\beta}\epsilon^{\alpha\beta\nu\mu}\partial_\nu Z_\mu\,,\tag{2} $$ where a two-form $B_{\alpha\beta}$ is introduced as a Lagrange multiplier, known as the Kalb-Ramond field.

  3. Treating $Z_\mu$ as a fundamental field and “integrating out” this field through its equation of motion leaves us with the following dual description. $$ \mathcal L \sim F_{\nu\alpha\beta}F^{\nu\alpha\beta}\,,\tag{3} $$ where $F^{\nu\alpha\beta} \equiv \partial^{[\nu} B^{\alpha\beta]}$ is the field strength of the Kalb-Ramond field $B_{\mu\nu}$.

  4. One can verify that because of gauge symmetry, $B_{\mu\nu} \to B_{\mu\nu} + \partial_{[\mu}\epsilon_{\nu]}$, the number of on-shell degrees of freedom is exactly the same as that of the scalar field.

One can carry out this algorithm to establish duality between several antisymmetric fields.

\begin{array}{cccc} 1 \text{ DOF:}& \text{Massless }\phi&\leftrightarrow &\text{Massless }B_{\mu\nu}\\ 2 \text{ DOF:}& \text{Massless }A_\mu&\leftrightarrow &\text{Massless }Y_{\mu}\\ 3 \text{ DOF:}& \text{Massive }\textbf{A}_\mu&\leftrightarrow &\text{Massive }\textbf{B}_{\mu\nu}\\ \end{array}

I would like you to kindly comment on the sensibility of the above duality transformations, namely whether they truly establish physical equivalence between the two theories, or there are complications such as due to representation theory or the simple fact that the theories do not have matching off-shell degrees of freedom.

$\endgroup$
1
$\begingroup$
  1. By integrating out the 1-form $Z$ in the parent action $$\begin{align}S_{\rm parent}&~=~\int\!\left( -\frac{1}{2}Z\wedge\star Z+Z\wedge\mathrm{d}B\right)\cr &~=~-\frac{1}{2}\int \!\left(Z- \star^{-1}\mathrm{d}B\right)\wedge\left( \star Z- \mathrm{d}B\right) -\frac{1}{2}\int \!\mathrm{d}B\wedge\star \mathrm{d}B,\end{align}\tag{2} $$ one directly gets the action $$S~=~ -\frac{1}{2}\int \!\mathrm{d}B\wedge\star \mathrm{d}B \tag{3}.$$ So the actions (2) & (3) are quantum mechanically equivalent as long as one doesn't ask questions about the $Z$-sector.

  2. On the other hand, it is possible to gauge the shift-symmetry $\mathrm{d}B\leadsto DB:=\mathrm{d}B+V$, $\mathrm{d}V=0$, of the 2-form $B$ in the action (3). This leads to the dual parent action $$ \widetilde{S}_{\rm parent}~=~\int\!\left( -\frac{1}{2}\int V\wedge \star V + V\wedge \mathrm{d}\phi\right),\tag{4}$$ cf. Ref. 1. By integrating the last term by parts, the 0-form $\phi$ becomes a Lagrange multiplier. By integrating out the 0-form $\phi$, the dual parent action (4) is classically equivalent to the action (3), up to topological obstructions (which among other things means that $V$ is assumed to be exact).

  3. By integrating out the 3-form $V$ in the dual parent action (4), one directly gets the dual action $$\widetilde{S}~=~-\frac{1}{2}\int \!\mathrm{d}\phi\wedge\star \mathrm{d}\phi.\tag{1}$$ So the actions (1) & (4) are quantum mechanically equivalent as long as one doesn't ask questions about the $V$-sector.

  4. Alternatively, integrating out the 2-form $B$ in the parent action (2) also establishes classical equivalence with the dual action (1).

  5. Remarkably, the quartet mechanism of the actions (1)-(4) works for any Lorentzian spacetime dimension, and for fields of arbitrary form-degrees. (Of course the form degree of the field strength & the dual field strength have to add up to the spacetime dimension.) One may check that the kinetic terms have standard normalization and signs. In particularly, dynamical variables (with positive kinetic terms) don't turn into ghost variables (with negative kinetic terms) under the duality, and vice-versa.

  6. The counting of on-shell & off-shell DOF of massless & massive $p$-forms is explained in my Phys.SE answer here.

References:

  1. S. E. Hjelmeland & U. Lindström, Duality for the Non-Specialist, arXiv:hep-th/9705122; Chapter 1.
$\endgroup$
0
$\begingroup$

Dualizing a gauge potential in vacuum to its dual potential is completely standard, although there can be global obstructions the existence of the dual. Here's how it works in arbitrary dimensions and with arbitrary fields:

Given a potential $A$ and a field strength $F = \mathrm{d}A$, the vacuum equations of motion are $\mathrm{d}F = 0$ and $\mathrm{d}{\star}F = 0$. By the Poincaré lemma, at least on contractible open patches we can therefore find a potential for ${\star}F$ as $\mathrm{d}B = {\star}F$. You can express the Lagrangian as a function of $A$ or of $B$, the equations of motion don't change.

Note that this is purely geometric. We don't need to "integrate out" auxiliary fields or anything. The symmetry of the equations of motion between the field strength and its dual is all we need. Note that therefore dualizing becomes impossible once there is a current we couple to, since then we have $\mathrm{d}F = 0$ and $\mathrm{d}{\star}F = j$ and we can no longer find a potential for the dual. That is, as soon as you couple the free theory to something else, you have to decide which of the two dual potentials is the "real" one.

$\endgroup$
  • $\begingroup$ What can you say about the spins of the two dual fields? $\endgroup$ – Nanashi No Gombe Feb 3 at 14:01
  • $\begingroup$ @NanashiNoGombe It is not hard to deduce that the prescription for dualization I gave maps a $p$-form to a $d-p-2$-form in $d$ dimensions. $\endgroup$ – ACuriousMind Feb 3 at 14:14
  • $\begingroup$ I understand that. But if they represent the same thing, shouldn't they have the same spins? For example, a scalar has spin 0, whereas Kalb-Ramond has spin 1. $\endgroup$ – Nanashi No Gombe Feb 3 at 14:20
  • $\begingroup$ @NanashiNoGombe Why should they have the same spin? Being "equivalent physical theories" means leading to the same equations of motions, the Lorentz indices of some objects we invent to get some Lagrangian to do that for us are completely irrelevant for that. $\endgroup$ – ACuriousMind Feb 3 at 14:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.