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

Question

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?

Answer 1

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).

Answer 2

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.