Consider Poincare group $\mathrm{ISO}(1,d-1)$, given by $\mathbb R^{1,d-1}\rtimes SO(1,d-1)$ in signature $(1,d-1)$, for some odd $d \geq 3$. Denote the universal cover of the component connected to identity by $G \ltimes \mathbb R^{1,d-1}$. Denote the generators of $\mathbb R$ part by $P^\mu$. I'm trying to understand the action of an element of its maximal torus on some irreducible representations.

Look at some particular unitary irreducible representations defined as follows: call $\Psi_{p,\sigma}$ an improper basis of Hilbert space $H$, which we take to be an eigenvector of the operators $P^\mu$ with eigenvalue $p^\mu$. There are also some continuity hypotheses that I'm omitting, as well as trace-class conditions. If we denote the quadratic Casimir by $p^2$, we focus on the case $p^2=0$ with $p^0>0$. Using the method of induced representation, we can classify representations by looking at little group representations, and by requiring the latter to be finite dimensional we can take as little group $\mathrm{Spin}(d-2)$ in this case (more precisely we are requiring the noncompact part of little group $ISO(d-2)$ to be represented trivially), therefore $\sigma$ is a discrete label spanning one such finite dimensional representation. (If $d=3$, the universal cover is not Spin, but let me be sloppy.)

We can represent the action of the abstract generator associated to $\Lambda\in\mathrm{Spin}(1,d-1)$ on a state $\Psi$ as

$$ U(\Lambda) \Psi_{p,\sigma} = N \sum_{\sigma'} D(W)_{\sigma,\sigma'} \Psi_{\Lambda p,\sigma'} $$


  • $N$ is a numerical normalization factor
  • $k=(1,0 \ldots 0,1)$
  • $L$ is a Lorentz transformation such that $p = Lk$
  • the little group element $W=W(\Lambda,p)$ such that $Wk=k$ is given by $W=L^{-1}(\Lambda p) \Lambda L(p)$
  • $D$ is a matrix that furnishes the representation we are looking for.

Now I would like to know whether there is a choice for the Cartan subalgebra of $\mathfrak{so}(d-2)$ and of $\mathfrak{so}(1,d-1)$, and possibly a choice of $L$, such that for $\Lambda \in Spin(d-1)\subset\mathrm{Spin}(1,d-1)$ the above matrix $D$ takes a simple form, possibly a bunch of phases (related to the parametrization of the Cartan of $\mathfrak{so}(1,d-1)$).

The goal I have in mind is to compute traces of the form

$$ tr_H e^{i \beta P^0} \, U(\Lambda) $$

for $\Lambda \in Spin(d-1)$, in terms of $Spin(d-1)$ characters, so I guess my true question is whether it's possible in this setup to use the induced representation method to compute characters of the larger group $Spin(d-1)$ using little group $Spin(d-2)$ representations, namely has some relation of type $$tr_H U(\Lambda)= \int tr D(W) \, \delta^{(d-1)}(\Lambda p-p) \, d\mu(p) = \frac{char (\Lambda)}{\det (1-\Lambda)}$$ where $d\mu=\delta(p^2) \theta(p^0) d^d p$, a chance to be true? The point is that naively the character would be for a larger group than the little group, whose elements can move us from a little group orbit, and therefore amke it vanish.

Also, does quartic Casimir play any role similar to what happens for $d=4$?

I understand the relevant theory was studied by Mackey in the 1950s, so if any expert could help pedagogically translate it to this particular problem it would be helpful. Some related sources are

  • related although inequivalent question
  • question in lower dimension
  • paper1; paper2; paper3 where they extend Frobenius formula for characters of Poincare in dimension 3+1
  • 2
    $\begingroup$ What do you call $\mathrm{ISO}(1,d-1)$? never seen this notation. $\endgroup$
    – YCor
    Apr 13, 2017 at 15:00
  • 4
    $\begingroup$ ISO is short for isometry. $\endgroup$
    – user1504
    Apr 13, 2017 at 15:02
  • 6
    $\begingroup$ Then the standard notation is $\mathrm{O}(1,d-1)$. ISO sounds weird, recall that SO means "Special Orthogonal", not "SOmetry", and standard abbreviation for Isometry groups in general are $\mathrm{Isom}$ or $\mathrm{Is}$ (not just capital such as ISOM or IS). $\endgroup$
    – YCor
    Apr 13, 2017 at 16:50
  • 4
    $\begingroup$ o is orthogonal, so special orhtogonal, iso the full isometry $\endgroup$
    – jj_p
    Apr 13, 2017 at 19:55
  • 6
    $\begingroup$ I had always thought the I stands for "inhomogeneous" as opposed to "homogeneous". Therefore inhomogeneous is pseudo-orthogonal semdirect with translations as the first line of the question suggests. $\endgroup$ Apr 14, 2017 at 7:50

2 Answers 2


I believe that you are asking for the analogue of the Frobenius formula in the theory of induced representations. The paper "On the primitive characters of the Poincaré group" (PDF file) by Joos and Schrader does this for the physically interesting representations of the 4-dimensional Poincaré group; although it is a delicate issue due to the infinite dimensionality of the representations. I am not aware of any paper doing this in higher dimension.

  • $\begingroup$ Thank you for the reference, that paper indeed computes the most general answer for Poincare in $d=4$. Since I'm only interested in a somehow simpler case, namely spatial rotations, although I'm looking at $d>4$, perhaps is there a reference discussing that? $\endgroup$
    – jj_p
    Apr 20, 2017 at 23:20

I am not sure this is the right answer (and this is too long for comments). Let $\chi : \mathbb{R}^d\rightarrow \mathbb{C}^*$ be a unitary character (by duality, this may be viewed as an element of the dual vector space $\mathbb{R}^d$). The assumptions seem to mean that the isotropy $M$ of the character $\chi$ in $SO(1,d-1)$ is compact. Hence we take a finite dimensional representation $\sigma$ of $M$ (could be trivial) and consider the induced representation $\pi =Ind _{\mathbb{R}^d \rtimes M} ^{\mathbb{R}^d \rtimes SO(1,d-1)} (\chi \otimes \sigma)$. If this is your representation, the restriction to the maximal compact $K=O(d-1)$ is simply the induced representation $Ind _M^K (\sigma)$. Its restriction to the compact maximal torus $T\subset K$ can easily be determined, since this has nothing to do with the representation $\pi $; this is simply an induced representation on $K$ restricted to $T$.

  • $\begingroup$ Hi, I'm not sure I fully understand your answer; could you please write down the first one or two SO(10) characters that you'd get for d=11? I also expect that after the first few small reps, one is not able to write it in the form above, unless using virtual characters. $\endgroup$
    – jj_p
    Apr 17, 2017 at 20:00

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