# How many degrees of freedom in a massless $2$-form field?

Consider the Kalb-Ramond field $$B_{\mu\nu}$$ which is basically a massless $$2$$-form field with the Lagrangian $$\mathcal L = \frac{1}{2}P_{\alpha\mu\nu}P^{\alpha\mu\nu}\,,$$ where $$P_{\alpha\mu\nu} \equiv \partial_{[\alpha}B_{\mu\nu]}$$ is the field strength, invariant under the gauge transformation $$B_{\mu\nu} \to B_{\mu\nu} + \partial_{[\mu}\epsilon_{\nu]}\,.$$ I am trying to calculate the number of degrees of freedom the theory has.

A general $$4\times4$$ antisymmetric matrix has $$6$$ independent entries. Let us try to fix the redundancy by choosing a gauge $$\epsilon_\mu$$ such that the gauge-fixed field is divergence free. \begin{align} \partial^\alpha\left( B_{\alpha\beta} + \partial_{[\alpha}\epsilon_{\beta]} \right) &= 0 \\ \Rightarrow \left(\delta^\alpha_\beta\square- \partial^\alpha\partial_\beta\right)\epsilon_\alpha &= -\partial^\alpha B_{\alpha\beta}\,. \end{align} The above is nothing but Maxwell's equations, and hence $$\epsilon_\alpha$$ has $$3$$ off-shell degrees of freedom. This means we are left with $$6-3=3$$ degrees of freedom for $$B_{\mu\nu}$$.

However, we should be able to kill $$2$$ more degrees of freedom because we know that a massless $$2$$-form field is physically equivalent to a massless scalar field which has only $$1$$ degree of freedom.

Do you see where the remaining gauge redundancy is?

It is natural to generalize to an Abelian $$p$$-form gauge field $$A~=~\frac{1}{p!} A_{\mu_1\mu_2\ldots\mu_p} \mathrm{d}x^{\mu_1}\wedge\ldots\wedge \mathrm{d}x^{\mu_p}\tag{1}$$ with $$\begin{pmatrix} D \cr p \end{pmatrix}$$ real component fields $$A_{\mu_1\mu_2\ldots\mu_p}$$ in a $$D$$-dimensional spacetime.

I) Massless case:

1. There is a gauge symmetry $$\delta A ~=~\mathrm{d}\Lambda , \qquad \Lambda~=~\frac{1}{(p\!-\!1)!} \Lambda_{\mu_1\mu_2\ldots\mu_{p-1}} \mathrm{d}x^{\mu_1}\wedge\ldots\wedge \mathrm{d}x^{\mu_{p-1}}, \tag{2}$$ with $$\begin{pmatrix} D \cr p\!-\!1 \end{pmatrix}$$ gauge parameters $$\Lambda_{\mu_1\mu_2\ldots\mu_{p-1}}$$; and a gauge-for-gauge symmetry $$\delta \Lambda ~=~\mathrm{d}\xi , \qquad \xi~=~\frac{1}{(r\!-\!2)!} \xi_{\mu_1\mu_2\ldots\mu_{p-2}} \mathrm{d}x^{\mu_1}\wedge\ldots\wedge \mathrm{d}x^{\mu_{p-2}}, \tag{3}$$ with $$\begin{pmatrix} D \cr p\!-\!2 \end{pmatrix}$$ gauge-for-gauge parameters $$\xi_{\mu_1\mu_2\ldots\mu_{p-1}}$$; and a gauge-for-gauge-for-gauge symmetry $$\ldots$$; and so forth.

2. Lemma: There are $$\begin{pmatrix} D\!-\!1 \cr p\!-\!1 \end{pmatrix}$$ independent gauge symmetries; there are $$\begin{pmatrix} D\!-\!1 \cr p\!-\!2 \end{pmatrix}$$ independent gauge-for-gauge symmetries; there are $$\begin{pmatrix} D\!-\!1 \cr p\!-\!3 \end{pmatrix}$$ independent gauge-for-gauge-for-gauge symmetries; and so forth.

Sketched proof: This is correct for $$p=1$$. Now use induction $$\begin{pmatrix} D\!-\!1 \cr p\!-\!1 \end{pmatrix}=\begin{pmatrix} D \cr p\!-\!1 \end{pmatrix}-\begin{pmatrix} D\!-\!1 \cr p\!-\!2 \end{pmatrix}$$ for $$p\geq 2$$ while keeping $$D$$ fixed. $$\Box$$

3. From the EL equations $$\sum_{\mu_0=0}^{D-1}d_{\mu_0}F^{\mu_0\mu_1\ldots\mu_p}~= ~0,\tag{4}$$ we see that the temporal gauge fields $$A^{0i_1i_2\ldots i_{p-1}},\qquad i_1, i_2, \ldots, i_{p-1}~\in~ \{1,\ldots,D\!-\!1\},\tag{5}$$ are not propagating, i.e. their time derivatives don't appear. They are fixed by boundary conditions (up to non-trivial topology).

4. This leaves us with the spatial gauge fields $$A^{i_1i_2\ldots i_p},\qquad i_1, i_2, \ldots, i_p~\in~ \{1,\ldots,D\!-\!1\},\tag{6}$$ which are $$\fbox{\begin{pmatrix} D\!-\!1 \cr p \end{pmatrix} \text{ massless propagating off-shell DOF,}}\tag{7}$$ which have $$\begin{pmatrix} D\!-\!2 \cr p\!-\!1 \end{pmatrix}$$ remaining independent gauge symmetries, cf. the Lemma.

5. The Lorenz gauge conditions$$^1$$ $$\sum_{\mu_0=0}^{D-1}d_{\mu_0}A^{\mu_0i_1\ldots i_{p-1}}~=~ 0, \qquad i_1, i_2, \ldots, i_{p-1}~\in~ \{1,\ldots,D\!-\!1\},\tag{8}$$ or equivalently (since there are no temporal gauge fields left), the $$\begin{pmatrix} D\!-\!2 \cr p\!-\!1 \end{pmatrix}$$ Coulomb gauge conditions $$\sum_{i_0=1}^{D-1}d_{i_0}A^{i_0a_1\ldots a_{p-1}}~=~0,\qquad a_1, \ldots, a_{p-1}~\in~ \{1,\ldots,D\!-\!2\},\tag{9}$$ (which match the number of remaining independent gauge symmetries) can be used to eliminate polarizations along one spatial direction, say $$x^{D-1}$$. Therefore there are only $$\begin{pmatrix} D\!-\!1 \cr p \end{pmatrix}-\begin{pmatrix} D\!-\!2 \cr p\!-\!1 \end{pmatrix}~=~\fbox{\begin{pmatrix} D\!-\!2 \cr p \end{pmatrix} \text{ massless on-shell DOF,}} \tag{10}$$ given by transversal component fields $$A^{a_1a_2\ldots a_p},\qquad a_1, a_2, \ldots, a_p\in \{1,\ldots,D\!-\!2\},\tag{11}$$ which each satisfies a decoupled wave eq. $$\Box A^{a_1a_2\ldots a_p}~=~0.\tag{12}$$ For the 4D Kalb-Ramond 2-form field, this leaves just 1 component, cf. OP's question.

II) Massive case:

1. There is no gauge-symmetry, so all field components are
$$\fbox{\begin{pmatrix} D \cr p \end{pmatrix} \text{ massive propagating off-shell DOF.}}\tag{13}$$

2. The massive EL equations imply $$\begin{pmatrix} D \cr p\!-\!1 \end{pmatrix}$$ Lorenz conditions $$\sum_{\mu_0=0}^{D-1}d_{\mu_0}A^{\mu_0\mu_1\ldots \mu_{p-1}}~=~ 0, \qquad \mu_1, \mu_2, \ldots, \mu_{p-1}~\in~ \{0,\ldots,D\!-\!1\}.\tag{14}$$ They follow from $$\begin{pmatrix} D\!-\!1 \cr p\!-\!1 \end{pmatrix}$$ spatial Lorenz conditions $$\sum_{\mu_0=0}^{D-1}d_{\mu_0}A^{i_0i_1\ldots i_{p-1}}~=~ 0, \qquad i_1, i_2, \ldots, i_{p-1}~\in~ \{1,\ldots,D\!-\!1\},\tag{15}$$ which can be used to eliminate polarizations along the temporal direction $$x^{0}$$. Therefore there are only $$\begin{pmatrix} D \cr p \end{pmatrix}-\begin{pmatrix} D\!-\!1 \cr p\!-\!1 \end{pmatrix}~=~\fbox{\begin{pmatrix} D\!-\!1 \cr p \end{pmatrix} \text{ massive on-shell DOF,}} \tag{16}$$ given by spatial component fields $$A^{i_1i_2\ldots i_p},\qquad i_1, i_2, \ldots, i_p\in \{1,\ldots,D\!-\!1\},\tag{17}$$ which each satisfies a decoupled wave eq. $$\Box A^{i_1i_2\ldots i_p}~=~0.\tag{18}$$

III) Alternatively, the massive $$p$$-form in $$D$$ spacetime dimensions can be gotten from dimensional reduction of the massless $$p$$-form in $$D\!+\!1$$ spacetime dimensions by eliminating $$x^D$$-components of the gauge field $$A$$ via gauge symmetry, and identifying momentum $$p^D=m$$.

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$$^1$$ One may show that the Lorenz conditions (9) not directly listed in eq. (8) still follow indirectly from eq. (8).

• Thanks. If you don't mind me asking, what is the answer if $r$ is larger than $D-2$? For example, how many degrees of freedom are there in a massless $3$-form in $4$ dimensions? Thanks, again. – Nanashi No Gombe Jan 29 '19 at 14:03
• The latter example is 1 off-shell DOF & 0 on-shell DOF. – Qmechanic Jan 30 '19 at 11:39

All field entries with time component are non-dynamical; you can see this using the field equations :

$$\partial^{\alpha} \partial_{[\alpha} B_{\beta \gamma]} = 0$$

Therefore, use temporal gauge :

$$B_{0i} = 0$$

The dynamical variables are : $$B_{ij}$$.

Similarly the gauge transformations :

$$B_{ij} \rightarrow B_{ij} +\partial_{[i} \epsilon_{j]}$$

Hence dof = 3 - 2 = 1.

One can use the exact same argument to show that a photon has 2 dof.