Graviton polarization in higher dimensions

Question

It's not difficult to see that the graviton in $D$ spacetime dimensions has $(D-3)D/2$ polarizations. In $D=4$ there are two $\epsilon^{\pm}_{\mu\nu}$. What I find curious is that in $D=4$ I can actually pick $\epsilon^{\pm}_{\mu\nu}=\epsilon^{\pm}_{\mu}\epsilon^{\pm}_{\nu}$ where $\epsilon^{\pm}_{\mu}$ are the two polarizations (of definite elicity $\pm1$) for a massless spin-1 particle like the photon. In higher dimension this doesn't seem possible since the photon has $D-2$ polarizations so that the number $(D-2)(D-1)/2$ of $\epsilon^{\lambda}_{\mu}\epsilon^{\lambda^\prime}_{\mu}$ pairs doesn't match the number $(D-3)D/2$ of graviton polarization. Well, unless somehow I consider only a smaller subset of them, say adding a constraint or removing one of them $$(D-2)(D-1)/2-1=(D-3)D/2$$ as in $D=4$ where $\epsilon^{+}_{\mu}\epsilon^{-}_{\mu}$ is discarded having zero elicity.

Is there an analogous constructions for $\epsilon^{\lambda}_{\mu\nu}$ in terms of the spin-1 polarizations $\epsilon^\sigma_\mu$ in higher dimensions? I suspect that something similar may happen only if I involve polarizations of higher elicity as well.

Answer 1

IMHO, the current use of the word helicities happens only when one is looking at some representation of $SU(2)$.

1) Now, a first point of view is to try to go back to representations of $ \otimes^n SU(2)$, when working with representations of $SO(D-2)$. In the best case, you will have different kind of "helicities".

Suppose we work with $D=6$, so spin-$1$ massless particles are in the fundamental representation of $SO(4)$, which I write $4$. In term of $SU(2) \otimes SU(2)$ representations, this gives : $4 \to (2,2)$

[here I write the number of states in the representations]

So, multiplying photon representations gives $4 \times 4 \to (2,2) \times (2,2) = (3,3) + (1,3) + (3,1) + (1,1)$

$(3,3)$ is the graviton traceless symmetric representation that we are looking for, with $9 = \dfrac{6 (6-3)}{2}$

So here photons have "helicities" $(\pm 1, \pm 1)$, while gravitons have "helicities" $(0 \pm 1, 0 \pm 1)$

Gravitons states could be written from photons states, for instance :

$(+1,+1) = (+1,+1) (+1,+1)$

$(-1,-1) = (-1,-1) (-1,-1)$

$(+1,-1) = (+1,-1) (+1,-1)$

$(-1,+1) = (-1,+1) (-1,+1)$

$(+1,0) = \frac{1}{\sqrt 2} [ (+1,+1) (+1,-1) + (+1,-1) (+1,+1)]$

$(0,0) = \frac{1}{ 2} [ (+1,-1) (+1,-1) + (+1,-1) (-1,+1) + (-1,+1) (+1,-1) + (-1,+1) (-1,+1)]$

and so on.

2) A second point of view is to work directly with the representations of $SO(D-2)$

Let us use this (french) Lie group on-line tool (Université de Poitiers). Choose $D3 (SO(6))$, "Tensor product decomposition" (then "proceed"). Let's type $(1,0,0) \times (1,0,0)$, (then "start"), and you get $(2,0,0) + (0,1,1)+(0,0,0)$. Here we are working with Dynkin indices.

So $(2,0,0)$ is the graviton symmetric traceless representation, and it is also the highest weight state of the representation. You may get the other states of the representation by substracting with the simple roots you may directly from the Cartan matrix of $D3= SO(6) = SU(4)$ (they are the lines of the Cartan matrix) until you get no positive number. Here the simple roots are $(2,-1,0), (-1,2,-1), (0,-1,2)$. So, for instance, substracting the first root, you get the state $(2,0,0) - (2,-1,0) = (0,1,0)$, and so on.

So each state for the gravitons (or the photons) could be represented by $3$ integers, so it is an alternative way to classify the states into a given representation.