Reducing massive representation of the Poincare group to the massless one

I want to ask about the connection for massive and massless representation of the Poincare group. Sorry for the awkwardness.

First I must to represent the formalism for both of cases.

Massive representation.

There is the formalism for an arbitrary spin-$s$ and massive representation of the Poincare group: $$(\partial^{2} + m^{2})\psi_{a_{1}...a_{n}\dot {b}_{1}...\dot {b}_{k}}(x) = 0, \qquad (1)$$ $$\partial^{a \dot b}\psi_{a...a_{n - 1}\dot {b}...\dot {b}_{k - 1}}(x) = 0, \qquad (2)$$ where $n + k = 2s$; field $\psi$ is symmetric by the all of indices and has the following transformation law under the Lorentz group $\left(\frac{n}{2}, \frac{k}{2}\right)$: $$\quad \psi_{a_{1}...a_{n}\dot {b}_{1}...\dot {b}_{k}}' = N_{a_{1}}^{\quad c_{1}}...N_{a_{n}}^{\quad c_{n}}{N^{*}}_{\dot {b}_{1}}^{\quad \dot {d}_{1}}...{N^{*}}_{\dot {b}_{k}}^{\quad \dot {d}_{k}}\psi_{c_{1}...c_{n}\dot {d}_{1}...\dot {d}_{k}}.$$

$(1)$ gives the irreducibility by mass (or the statement $P^{2}\psi = m^2\psi$ for first Casimir operator of the Poincare group), while $(2)$ gives the irreducibility by spin (or the statement $W^{2}\psi = -m^{2}s(s + 1)\psi$ for the second Casimir operator). This representation has $2s + 1$ spin degrees of freedom.

For the case of an integer spin-$s$ representation refers to the traceless transverse symmetric tensor of rank s $A_{\mu_{1}...\mu_{s}}$: $$A^{\mu}_{\quad \mu ...\mu_{s - 2}} = 0, \quad \partial_{\mu}A^{\mu ...\mu_{s - 1}} = 0,$$
$$(\partial^{2} + m^{2})A_{\mu_{1}...\mu_{s}} = 0,\quad A_{\mu_{1}...\mu_{i}...\mu_{j}...\mu_{s}} = A_{\mu_{1}...\mu_{j}...\mu_{i}...\mu_{s}}.$$

Massless representation.

There is the formalism for an arbitrary helicity-$\lambda$ and massless representations (for given helicity there is infinite number of the representations!) of the Poincare group: $$\partial^{a \dot {d}}\psi_{a...a_{n - 1}\dot {b}_{1}...\dot {b}_{k}} = 0. \qquad (3)$$ $$\partial^{\dot {b} c}\psi_{a_{1}...a_{n}\dot {b}...\dot {b}_{k - 1}} = 0. \qquad (4)$$ Both of them realize the irreducibility by mass $\partial^{2} \psi = 0$ and irreducibility by helicity, $$W_{c \dot {c}}\psi_{a_{1}...a_{n}\dot {b}_{1}...\dot {b}_{k}} = \frac{n - k}{2}p_{c \dot {c}}\psi_{a_{1}...a_{n}\dot {b}_{1}...\dot {b}_{k}} = \lambda p_{c\dot {c}}\psi_{a_{1}...a_{n}\dot {b}_{1}...\dot {b}_{k}}.$$ The representation has only one independent component, as it must be for massless one. If the representation is real we must take the direct sum of representations $\left( \frac{n}{2}, \frac{k}{2} \right) + \left( \frac{k}{2} , \frac{n}{2} \right)$, so the number of degrees of freedom will increase from 1 to 2.

My question.

I'm interesting only in an integer spin representations.

For spin-1 massive case and field (I choosed one of three possible variants) $$\psi_{a \dot {b}} \to A_{\mu} = \frac{1}{2}\tilde {\sigma}_{\mu}^{\dot {b} a}\psi_{a \dot {b}}$$ there is three degrees of freedom, which refers to the transverse condition $\partial_{\mu}A^{\mu} = 0$.

But in massless case we also can introduce gauge transformations $$A_{\mu} \to A_{\mu}' = A_{\mu} + \partial_{\mu} \varphi$$ which don't change the transverse condition and Klein-Gordon equation: $$\partial_{\mu}A^{\mu}{'} = \partial_{\mu}A^{\mu} + \partial^{2}\varphi = 0 \Rightarrow \partial^{2}\varphi = 0,$$ $$\partial^{2}A_{\mu}' = \partial^{2}A_{\mu} = 0.$$ So we can also satisfy the condition $u_{\mu}A^{\mu} = 0$, where $u_{\mu}$ is some timelike vector. So the number of independent components reduce to two (we have the case when representation is real, so this representation gives one particle with two possible states of helicity).

By the full analogy with spin-1 massive case I got the result with spin-2 massive case.

In the beginning it has $$2s + 1 = 10 (symmetry) - 4 (transverse) - 1(traceless) = 5$$ independent components. When $m = 0$ we have the gauge transformations $$A_{\mu \nu} \to A_{\mu \nu}{'} = A_{\mu \nu} + \partial_{\mu}\varepsilon_{\nu} + \partial_{\nu}\varepsilon_{\mu},$$ by which we can set the condition $u_{\mu}A^{\mu \nu} = 0$ which decreases the number of degrees of freedom by three (the another one is equal to one $k_{\mu}A^{\mu \nu}u_{\nu}$ from $\partial_{\mu} A^{\mu \nu} = 0$). So we also have 2 independent components.

For the arbitrary integer spin representation I also can get two components, because there is gauge transformations $$A_{\mu_{1}...\mu_{s}}{'} = A_{\mu_{1}...\mu_{s}} + \partial_{\mu_{1}}\varepsilon_{\mu_{2}} + ...,$$ by which I also can reduce the number of degrees of freedom of my field.

Finally, my question (edited): can this reduction from massive to massless case be derived on the language of $(1)-(4)$?

• To make sure I understand the question. Your (2) makes field divergenceless, in vector notation you impose something like $\partial^\nu \phi_{\nu\mu....}=0$. From (3) or (4) one gets (2), but they are stronger than (2), in particular they impose $\square \phi_{\nu\mu...}=0$ while (2) does not. So I do not think one can get (3) and (4) from (2) – John Jan 3 '14 at 20:12
• I've got a feeling that $\psi$ in your sections are the same. In (1)-(2) your $\psi$ is a gauge potential, analog $A_\mu$, while your (3)-(4) is an analog of field stengh $F_{\mu\nu}$. The Bianchi identities $\partial_\mu F_{\nu\sigma}+2\mbox{more}=0$ implies that $F=dA$, i.e. $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$. This how the two approaches are related. The best way to show this is to solve equations by Fourier transform. Say for (3,4) you get $\int \xi_a...\xi_a \bar{\xi}_{\dot{b}}...\bar{\xi}_{\dot{b}} f(\xi) \exp{\xi_c\bar{\xi}_{\dot{d}} x^{c\dot{d}}}$. – John Jan 5 '14 at 8:51
• One more example is the free graviton $g_{\mu\nu}$ that is described by $\psi_{aa,\dot{b}\dot{b}}$ and equivalently by the Weyl tensor $\psi_{aaaa}, \psi_{\dot{a}\dot{a}\dot{a}\dot{a}}$ – John Jan 5 '14 at 8:56
• $\square A_\mu=0$ or $\square\psi_{a\dot{a}}=0$+constraint (3) can be solved analogously. There is nothing special about $4d$ spinor language you are using. The same things can be asked in any dimension. While it is nonlocal to go from (3,4) to (1,2), it is easy to see that $\partial ...\partial \psi$ with indices contracted appropriately obeys (3,4) as a consequence of (1,2). It is obvious for $A_\mu$ and $F_{\mu\nu}$, which both describes massless spin-one. – John Jan 5 '14 at 9:02
• So, to summarize, there are two maps. One maps gauge potential (1,2) to gauge invariant (3,4), it is given by applying derivatives like in $A_\mu$ - $F_{\mu\nu}$ example. The map in the opposite direction is nonlocal, i.e. one can reconstruct $A_\mu$ from $F_{\mu\nu}$, but it is an integral. So the simplest way to relate the two approaches is to show that the solution spaces are isomorphic with the help of Fourier transform – John Jan 5 '14 at 19:09