Unitary Poincaré Representations and Little Groups: Examples

The standard $SU(2)$ action on ${\mathbb{C}}^2$ is unitary and smooth; its tensor-square action is also unitary and smooth, and its symmetric subspace $W_2={\mathrm{Sym}}^2({\mathbb{C}}^2)$ is invariant. Hence both restricted actions ${\theta }_1$ and ${\theta }_2$ are strongly continuous unitary representations. The standard action on $W_1={\mathbb{C}}^2$ is irreducible: for every unit vector $(u,v)$, the matrix $\left(\begin{array}{cc}u& -\bar{v}\\ v& \bar{u}\end{array}\right)\in SU(2)$ sends $e_1$ to $(u,v)$, so $SU(2)$ is transitive on the unit sphere and any nonzero invariant subspace is all of $W_1$. For $W_2$, use the basis $e_1\odot e_1,e_1\odot e_2,e_2\odot e_2$. The torus element $t=\mathrm{diag}(e^{i\pi \mathrm{/}6},e^{-i\pi \mathrm{/}6})$ acts with distinct eigenvalues $e^{i\pi \mathrm{/}3},1,e^{-i\pi \mathrm{/}3}$. If a nonzero invariant subspace contains a vector with several weight terms, Lagrange interpolation polynomials in ${\theta }_2(t)$ project onto its nonzero weight components, so it contains at least one basis vector. A group-invariant subspace is invariant under the differentiated $\mathfrak{s}\mathfrak{u}(2)$ action because it is closed and contains each group orbit; complex-linear extension makes it invariant under $E_{12}$ and $E_{21}$ in ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$. These operators send $e_2\odot e_2\mapsto 2e_1\odot e_2\mapsto 2e_1\odot e_1$ and $e_1\odot e_1\mapsto 2e_1\odot e_2\mapsto 2e_2\odot e_2$ (with nonzero coefficients), and from the middle vector they reach both endpoints. Thus every nonzero invariant subspace is all of $W_2$. The torus weights are $z,z^{-1}$ for $W_1$ and $z^2,1,z^{-2}$ for $W_2$, which give the spin labels $1\mathrm{/}2$ and $1$ under the convention in the Example.

On $W_1={\mathbb{C}}^2$, $z=-I$ acts as $-I$. On the whole tensor square, it acts as $(-I)\otimes (-I)=I$, so its restriction to $W_2$ is also $I$. The kernel of $SU(2)\to SO(3)$ is exactly $\{I,z\}$ by [F3].

The momentum $p_m$ is future timelike, hence is in $C_{+}$. By [F1], each ${\theta }_{m_0}$ gives an irreducible positive-energy induction. Applying [F4] to the central kernel $\{I,z\}$, the spin-$\frac{1}{2}$ induction fails the identity condition and cannot descend; the spin-$1$ induction satisfies it and descends to the ordinary Poincaré group.

The circles $SO(2)=\mathbb{R}\mathrm{/}(2\pi \mathbb{Z})$ and $\mathrm{S}\mathrm{p}\mathrm{i}\mathrm{n}(2)=\mathbb{R}\mathrm{/}(4\pi \mathbb{Z})$ are abelian. In an irreducible unitary representation of either circle, each operator commutes with the entire representation, so [F3] makes every operator scalar; the representation is therefore a one-dimensional continuous character. Pull it back to $\mathbb{R}$. By [F4] it is $e^{isr}$ for a unique real $s$. It descends to $SO(2)$ exactly when $e^{2\pi is}=1$, i.e. $s\in \mathbb{Z}$, and to $\mathrm{S}\mathrm{p}\mathrm{i}\mathrm{n}(2)$ exactly when $e^{4\pi is}=1$, i.e. $s\in \frac{1}{2}\mathbb{Z}$. Thus the zero-translation little-group labels in the Example are exhaustive.

For $\rho >0$, [F2] gives a single induced little-group representation for $E$, because its point stabilizer in $SO(2)$ is trivial. For the double cover the stabilizer is $\{\pm 1\}$, whose two one-dimensional characters have values $ϵ=+1$ and $ϵ=-1$ on $z$. Thus the two double-cover little representations are inequivalent and exhaust this orbit's unitary dual.

In $\mathrm{S}\mathrm{p}\mathrm{i}\mathrm{n}(2)=\mathbb{R}\mathrm{/}(4\pi \mathbb{Z})$ the element $z=[2\pi ]$ has order two. The character ${\theta }_s$ sends it to $e^{2\pi is}$, which is $+1$ for integer $s$ and $-1$ for noninteger half-integer $s$. Applying [F6] to the full Poincaré induction gives precisely the stated helicity descent outcomes.

Each little representation in step 1.2 is irreducible by [F2]. In a double-cover branch it is induced from $H={\mathbb{R}}^2⋊\{1,z\}$ with $\sigma ={\chi }_{\xi }\otimes ϵ$. The element $z$ is central in $\tilde{E}$ and belongs to $H$, so it fixes every quotient point; its Radon--Nikodym factor is $1$ almost everywhere. By [F7], $({\theta }_{\rho ,ϵ}(z)F)(x)=F(z^{-1}x)=F(x{z}^{-1})=\sigma (z^{-1}{)}^{-1}F(x)=ϵF(x)$, so $z$ acts on this intermediate little-group representation as $ϵI$ on its Hilbert completion. Now its induction to the full Poincare group is irreducible and positive-energy by [F5], and [F6] transfers this computed central value to that induction. Thus the $+1$ branch factors through the base group and the $-1$ branch fails the necessary identity condition. The base group has no cover kernel. This verifies both continuous-spin central parities and their descent.

The zero translation orbit ($\rho =0$) and the nonzero spheres ($\rho >0$) are disjoint and exhaust the dual by [F2]. Steps 1.1–2.2 therefore give the stated helicity and continuous-spin branches, including the distinct base and double-cover central actions.

The vector ${\xi }_{\rho }\mathrm{\ne }0$, so the dual orbit $S_{\rho }=\{\eta \in {\mathbb{R}}^2:\mathrm{\mid }\eta \mathrm{\mid }=\rho \}$ does not contain zero. By [F1], ${\theta }_{\rho }$ is the irreducible induction from ${\chi }_{{\xi }_{\rho }}$ and, in the induced model, translation by $a\in {\mathbb{R}}^2$ multiplies over $\eta \in S_{\rho }$ by $e^{i\eta \cdot a}$.

Choose $a=(\pi \mathrm{/}\rho )e_1$. At $\eta ={\xi }_{\rho }$ the multiplier is $e^{i{\xi }_{\rho }\cdot a}=e^{i\pi }=-1$. It remains different from $1$ on an open arc around ${\xi }_{\rho }$ by continuity. The orbit base is the circle $SO(2)$, and the full-support measure in [F2] assigns positive measure to this arc. Thus the translation operator is not the identity, proving that ${\mathbb{R}}^2$ acts nontrivially.

Choose any nonzero future null $p_{+}$ and the little-group representation ${\theta }_{\rho }$. It is irreducible by [F1], so [F3] gives an irreducible positive-energy Poincaré induction. Its little-group translations remain nontrivial by step 2.1. Hence the example lies in the continuous-spin branch with nonzero null-little-group translations; it cannot be replaced by a helicity representation, for which those translations are trivial.

The map $r\mapsto e^{i\sqrt{2}r}$ is a continuous one-dimensional unitary character of the additive group $\mathbb{R}$, so it is irreducible. The momentum $p_m$ is future timelike and lies in $C_{+}$. Applying [F2] gives an irreducible positive-energy induced representation on the universal cover.

By [F1], $z^N$ has little-group coordinate $2\pi N$. Its character value is ${\theta }_{\sqrt{2}}(z^N)=e^{2\pi iN\sqrt{2}}$. By [F3], descent to $K_N$ would require $N\sqrt{2}\in \mathbb{Z}$.

No positive integer $N$ satisfies this condition. If $N\sqrt{2}=k\in \mathbb{Z}$, then $\sqrt{2}=k\mathrm{/}N$ is rational. To recall the elementary contradiction, write $\sqrt{2}=a\mathrm{/}b$ in lowest positive terms; $a^2=2b^2$ makes $a$ even, then makes $b$ even, contradicting coprimality. Thus the representation fails the descent test for every finite $N$.

The displayed formula is multiplicative because $\mathbb{R}\times \mathbb{Z}$ is a direct product: ${\theta }_{t,\alpha }((u,j)+(v,k))=e^{it(u+v)}{e}^{2\pi i\alpha (j+k)}={\theta }_{t,\alpha }(u,j){\theta }_{t,\alpha }(v,k)$. It is continuous in $u$, and $\mathbb{Z}$ is discrete. A one-dimensional unitary representation is irreducible.

For $\rho >0$, evaluating the characters with $t=\rho$ and $t=-\rho$ at $(u,j)=(\pi \mathrm{/}(2\rho ),0)$ gives respectively $i$ and $-i$, so they are inequivalent. This is consistent with the orbit classification because the connected rotation factor is $SO(1)=\{1\}$ and cannot reverse the real translation character.

The deck generator is $(0,1)$, so ${\theta }_{t,\alpha }(z)=e^{2\pi i\alpha }$ and the value is independent of $t$. As $\alpha$ ranges over $\mathbb{R}\mathrm{/}\mathbb{Z}$, these are all of $\mathbb{T}$. On $K_N$, the relation $z^N=1$ forces and is forced by $e^{2\pi iN\alpha }=1$, equivalently $N\alpha \in \mathbb{Z}$, giving precisely the $N$ possible phases.

For every $t=\pm \rho$ and allowed $\alpha$, [F4] gives a positive-energy irreducible Poincaré induction. The independent central-phase calculation in step 2.2 and the sign calculation in step 2.1 show that neither the finite-cover descent condition [F3] nor the phase parameter identifies the two signs. This proves the stated D=3 distinction.

For $q=(p^0,p^1)$, the connected boost of rapidity $s$ acts on light-cone coordinates by $(p^0+p^1,p^0-p^1)\mapsto (e^s(p^0+p^1),e^{-s}(p^0-p^1))$. For $p_{+}$ the second coordinate is zero and the first positive; for $p_{-}$ the first coordinate is zero and the second positive. These zero-coordinate patterns are preserved by every finite boost, so no element of $L$ carries $p_{+}$ to $p_{-}$. Both have $p^0>0$ and $\eta (p,p)=0$, hence are future null.

Their little groups are trivial by [F2], and the unique representation of the trivial group is its one-dimensional identity representation. Applying [F3] to the two distinct future null orbits yields irreducible positive-energy inductions, and its orbit-equivalence criterion makes ${\pi }_{+}$ and ${\pi }_{-}$ inequivalent. Their squared masses are both zero by direct substitution into $\eta (p,p)=-(p^0{)}^2+(p^1{)}^2$.

The function ${\theta }_s(r)=e^{isr}$ is a continuous one-dimensional unitary character of the additive group $K=\mathbb{R}$, hence is irreducible. Since $p=0$, [F1] gives $K_p=K$, and [F2] yields its induced representation over the zero orbit. Induction from the whole group stabilizer identifies with the inducing representation itself, so the formula is ${\mathrm{\Pi }}_s(a,r)={\chi }_0(a){\theta }_s(r)=e^{isr}$.

The representation is nontrivial: at $r=\pi \mathrm{/}s$ it has value $e^{i\pi }=-1$. Its translation subgroup acts identically because ${\chi }_0=1$, and [F2] guarantees positive energy since the inducing momentum $0$ lies in $C_{+}$. This proves that a nontrivial Lorentz character lies over zero momentum.