Mass cocycle for translations and boosts

The two parenthesizations of a triple product in $G_d$ have the same central coordinate, since both give $\theta +{\theta }^{\mathrm{\prime }}+{\theta }^{\mathrm{\prime }\mathrm{\prime }}-v^{\mathrm{\prime }}\cdot a-v^{\mathrm{\prime }\mathrm{\prime }}\cdot a-v^{\mathrm{\prime }\mathrm{\prime }}\cdot a^{\mathrm{\prime }}$. The identity is $(0,0,0)$ and the inverse of $(a,v,\theta )$ is $(-a,-v,-\theta -v\cdot a)$. Multiplying by $(0,0,\eta )$ on either side adds $\eta$ to the central coordinate and leaves $(a,v)$ fixed, so $Z$ is central; the quotient forgets $\theta$ and is the additive group ${\mathbb{R}}^{2d}$. The product and inverse maps are continuous in the product Euclidean topology: their coordinates are sums and the polynomial expression $\theta +{\theta }^{\mathrm{\prime }}-v^{\mathrm{\prime }}\cdot a$ or $-\theta -v\cdot a$. Thus $G_d$ is a topological group.

By [A2] and [F1], $\mathcal{H}=L^2({\mathbb{R}}^d,dx;\mathbb{C})$ with the stated pairing is a complex Hilbert space, so the operator definitions act on the required Hilbert space.

Translation is well-defined on almost-everywhere classes by [F2], and $\mathrm{\parallel }T(a)f{\mathrm{\parallel }}_2^2=\int \mathrm{\mid }f(x-a){\mathrm{\mid }}^{2}dx=\int \mathrm{\mid }f(x){\mathrm{\mid }}^{2}dx$ by [F3]. The map is linear with inverse $T(-a)$; hence it is unitary by [F5] and step 1.2.

The multiplier $e^{imv\cdot x\mathrm{/}\mathrm{\hslash }}=\exp (imv\cdot x\mathrm{/}\mathrm{\hslash })$ is continuous, hence measurable, by [F15], and has modulus one by [F16]. Therefore $B_m(v)$ preserves the norm. It is linear, and the addition law [F14] gives $B_m(v)B_m(-v)=I$; hence $B_m(-v)$ is its inverse and it is unitary by [F5] and step 1.2.

For almost every $x$, $(B_m(v)T(a)f)(x)=e^{imv\cdot x\mathrm{/}\mathrm{\hslash }}f(x-a)$ while $(T(a)B_m(v)f)(x)=e^{imv\cdot (x-a)\mathrm{/}\mathrm{\hslash }}f(x-a)$. By the addition law [F14], $e^{imv\cdot x\mathrm{/}\mathrm{\hslash }}=e^{imv\cdot (x-a)\mathrm{/}\mathrm{\hslash }}{e}^{imv\cdot a\mathrm{/}\mathrm{\hslash }}$, proving the commutation identity. Multiplying it on the right by $B_m(v{)}^{\ast }T(a{)}^{\ast }$ gives the displayed loop identity.

Let $g=u+iw\in C_c^{\mathrm{\infty }}({\mathbb{R}}^d;\mathbb{C})$. For the unitary families established in steps 2.1 and 2.2, [F10] gives $u,w\in C_c({\mathbb{R}}^d)$; applying [F9] to each component and using $\mathrm{\parallel }h+ik{\mathrm{\parallel }}_2^2=\mathrm{\parallel }h{\mathrm{\parallel }}_2^2+\mathrm{\parallel }k{\mathrm{\parallel }}_2^2$ yields $\mathrm{\parallel }T(a)g-T(a_0)g{\mathrm{\parallel }}_2\to 0$ as $a\to a_0$. If $v_n\to v_0$, continuity of the scalar exponential [F15] gives pointwise convergence $\mathrm{\mid }{e}^{im(v_n-v_0)\cdot x\mathrm{/}\mathrm{\hslash }}-1{\mathrm{\mid }}^2\mathrm{\mid }g(x){\mathrm{\mid }}^2\to 0$; by [F16] and the triangle inequality [F4], this is bounded by $4\mathrm{\mid }g(x){\mathrm{\mid }}^2$, which is integrable. Dominated convergence [F12] gives $\mathrm{\parallel }{B}_m(v_n)g-B_m(v_0)g{\mathrm{\parallel }}_2\to 0$. Since Euclidean parameter spaces are metric, both orbit maps are continuous on this dense smooth subspace.

For arbitrary $f\in \mathcal{H}$, approximate it in $L^2$ by $g\in C_c^{\mathrm{\infty }}({\mathbb{R}}^d;\mathbb{C})$ using [F8]. Unitarity in steps 2.1 and 2.2 gives $\mathrm{\parallel }T(a)f-T(a_0)f{\mathrm{\parallel }}_2\le 2\mathrm{\parallel }f-g{\mathrm{\parallel }}_2+\mathrm{\parallel }T(a)g-T(a_0)g{\mathrm{\parallel }}_2$, and the analogous estimate for $B_m(v)$; letting the approximation error be arbitrarily small extends both continuities to every $f$. The scalar phase $e^{im\theta \mathrm{/}\mathrm{\hslash }}$ is continuous in $\theta$ by [F15]. Moreover, $\mathrm{\parallel }{\mathrm{\Pi }}_m(a,v,\theta )f-{\mathrm{\Pi }}_m(a_0,v_0,{\theta }_0)f{\mathrm{\parallel }}_2\le \mathrm{\mid }{e}^{im\theta \mathrm{/}\mathrm{\hslash }}-e^{im{\theta }_0\mathrm{/}\mathrm{\hslash }}\mathrm{\mid }\mathrm{\parallel }f{\mathrm{\parallel }}_2+\mathrm{\parallel }T(a)f-T(a_0)f{\mathrm{\parallel }}_2+\mathrm{\parallel }(B_m(v)-B_m(v_0))T(a_0)f{\mathrm{\parallel }}_2,$ so the orbit map of ${\mathrm{\Pi }}_m$ is norm-continuous on the product Euclidean group.

Using $T(a)B_m(v^{\mathrm{\prime }})=e^{-im{v}^{\mathrm{\prime }}\cdot a\mathrm{/}\mathrm{\hslash }}{B}_m(v^{\mathrm{\prime }})T(a)$ from step 3.1 and the translation law together with the exponential addition law [F14], multiplication gives ${\mathrm{\Pi }}_m(a,v,\theta ){\mathrm{\Pi }}_m(a^{\mathrm{\prime }},v^{\mathrm{\prime }},{\theta }^{\mathrm{\prime }})={\mathrm{\Pi }}_m(a+a^{\mathrm{\prime }},v+v^{\mathrm{\prime }},\theta +{\theta }^{\mathrm{\prime }}-v^{\mathrm{\prime }}\cdot a)$. By step 1.1 this is the group law; the operators are unitary by steps 2.1 and 2.2, and their orbit maps are continuous by step 4.1. Thus ${\mathrm{\Pi }}_m$ is a unitary representation by [F5], and ${\mathrm{\Pi }}_m(0,0,\eta )=e^{im\eta \mathrm{/}\mathrm{\hslash }}I$ is its central character.

For a finite direct sum the central subgroup acts as ${⨁}_j{e}^{i{m}_j\eta \mathrm{/}\mathrm{\hslash }}{I}_j$. If $m_j\mathrm{\ne }{m}_k$, equality of the two central characters as functions would imply $\exp (i{m}_j\eta \mathrm{/}\mathrm{\hslash })=\exp (i{m}_k\eta \mathrm{/}\mathrm{\hslash })$ for every real $\eta$. By [F13], $\exp (0)=1$; by the derivative at zero from [F15], for every nonzero real $\alpha$ we have ${\lim }_{\eta \to 0}(\exp (i\alpha \eta )-1)\mathrm{/}\eta =i\alpha$, since factoring $i\alpha$ reduces this limit to the complex derivative of $\exp$ at $0$. Applying this with $\alpha =m_j\mathrm{/}\mathrm{\hslash }$ and $\alpha =m_k\mathrm{/}\mathrm{\hslash }$ would give $i{m}_j\mathrm{/}\mathrm{\hslash }=i{m}_k\mathrm{/}\mathrm{\hslash }$, contrary to $m_j\mathrm{\ne }{m}_k$. Therefore the central characters are pairwise distinct.

The displayed form on $V={\mathbb{R}}^d\oplus {\mathbb{R}}^d$ is bilinear and alternating. If $\omega ((a,p),(a^{\mathrm{\prime }},p^{\mathrm{\prime }}))=0$ for every $(a^{\mathrm{\prime }},p^{\mathrm{\prime }})$, choosing $a^{\mathrm{\prime }}=0$ and arbitrary $p^{\mathrm{\prime }}$ forces $a=0$, while choosing $p^{\mathrm{\prime }}=0$ and arbitrary $a^{\mathrm{\prime }}$ forces $p=0$; hence it is nondegenerate on the $2d$-dimensional real space. Since $\mathrm{\hslash }>0$, the parameter $\lambda =1\mathrm{/}\mathrm{\hslash }$ is nonzero. Substituting $p=mv$ into the displayed Weyl formula gives $W(a,mv)=e^{-imv\cdot a\mathrm{/}(2\mathrm{\hslash })}{B}_m(v)T(a)$. Direct multiplication, using the exponential addition law [F14], gives $W(a,p)W(a^{\mathrm{\prime }},p^{\mathrm{\prime }})=e^{-i\omega ((a,p),(a^{\mathrm{\prime }},p^{\mathrm{\prime }}))\mathrm{/}(2\mathrm{\hslash })}W(a+a^{\mathrm{\prime }},p+p^{\mathrm{\prime }})$. The published model result [F7] verifies this system is irreducible and regular under Countable Choice [A2], and [F6] gives unitary equivalence for every other system with the same finite phase space and nonzero parameter $1\mathrm{/}\mathrm{\hslash }$ under AC [A1].