Mass is constant on an irreducible positive super-Poincare representation

The function $s(p)=-{\eta }^{\mu \nu }{p}_{\mu }{p}_{\nu }=(p^0{)}^2-\mathrm{\mid }\vec{p}{\mathrm{\mid }}^2$ is continuous and nonnegative on the closed future cone in [F1]. For each Borel $A\subseteq [0,\mathrm{\infty })$, $\{p\in {\mathcal{C}}_{+}^{\ast }:s(p)\in A\}$ is Borel in the full covector space because ${\mathcal{C}}_{+}^{\ast }$ is closed and $s$ is continuous there. Inverse images preserve disjoint unions, and the preimage of $[0,\mathrm{\infty })$ is ${\mathcal{C}}_{+}^{\ast }$, whose projection is $I$ by [F1]; therefore the pushforward $F(A):=E(\{p\in {\mathcal{C}}_{+}^{\ast }:s(p)\in A\})$ is a normalized PVM on $[0,\mathrm{\infty })$ by strong countable additivity of $E$.

For each Borel $A$, $F(A)$ is the projection $E(\{p\in {\mathcal{C}}_{+}^{\ast }:s(p)\in A\})$. The charge-domain reduction in [F1] therefore gives $F(A)D(Q)\subseteq D(Q)$ and $QF(A)\psi =F(A)Q\psi$ for every $\psi \in D(Q)$; the same argument applies to $Q^{\ast }$. This proves the spectral-subspace assertion.

The determinant formula in [F1] and $\det g=1$ give $\det \mathrm{\Phi }({\mathrm{\Lambda }}_{g}x)=\det (g\mathrm{\Phi }(x)g^{†})=\det \mathrm{\Phi }(x)$, hence ${\eta }_{\mu \nu }({\mathrm{\Lambda }}_{g}x{)}^{\mu }({\mathrm{\Lambda }}_{g}x{)}^{\nu }={\eta }_{\mu \nu }{x}^{\mu }{x}^{\nu }$. Polarization of this quadratic identity gives ${\mathrm{\Lambda }}_g^{\mathsf{T}}\eta {\mathrm{\Lambda }}_g=\eta$. By the dual-action definition in [F1], a covector column transforms as $({\mathrm{\Lambda }}_g^{-1}{)}^{\mathsf{T}}p$; the inverse-metric identity ${\mathrm{\Lambda }}_g^{-1}{\eta }^{-1}({\mathrm{\Lambda }}_g^{-1}{)}^{\mathsf{T}}={\eta }^{-1}$ therefore gives $s({\mathrm{\Lambda }}_g^{\vee }p)=s(p)$. Under the metric identification $p^{\mathrm{♯}}={\eta }^{-1}p$, this dual action satisfies $({\mathrm{\Lambda }}_g^{\vee }p{)}^{\mathrm{♯}}={\mathrm{\Lambda }}_g{p}^{\mathrm{♯}}$; preservation of the future vector cone therefore gives preservation of ${\mathcal{C}}_{+}^{\ast }$. Lorentz covariance of $E$ now implies $U(g,0)F(A)U(g,0{)}^{\ast }=F(A)$. Translation unitaries are bounded Borel functions of $E$, so they commute with every $F(A)$; $\mathrm{\Gamma }$ commutes with $E$ by [F1]. Hence $\mathrm{ran}F(A)$ is closed and graded, invariant under the integrated even group, and invariant under every charge and adjoint with dense restricted domains: density follows by applying the bounded projection $F(A)$ to a dense operator domain. Irreducibility as defined in [F1] forces $F(A)=0$ or $F(A)=I$.

Choose any nonzero vector and normalize it to a unit vector $\psi \in \mathcal{H}$; set $\mu (A)=⟨F(A)\psi ,\psi ⟩$. By [F3] this is a countably additive probability measure, and by step 2.2 every value is $0$ or $1$. For any measurable sets $A_n$, define disjoint $C_n=A_n\setminus {\bigcup }_{k<n}{A}_k$. If $C\subseteq A$, countable additivity on $A=C\bigsqcup (A\setminus C)$ gives $\mu (C)\le \mu (A)$; hence $\mu ({\bigcup }_n{A}_n)={\sum }_n\mu (C_n)\le {\sum }_n\mu (A_n)$. Thus countable subadditivity is available here from countable additivity and nonnegativity, without an additional supplier. The set $R=\{r\ge 0:\mu ([0,r])=1\}$ is nonempty: otherwise countable subadditivity applied to $[0,\mathrm{\infty })={\bigcup }_{n\ge 1}[0,n]$ would give $\mu ([0,\mathrm{\infty }))=0$, contradicting that $\mu$ is a probability measure. It is bounded below by $0$, so [F5] applied to $-R$ gives its finite greatest lower bound $m^2=\inf R\ge 0$.

For every $\epsilon >0$, the greatest-lower-bound property in [F5] gives $r\in R$ with $r<m^2+\epsilon$, so $\mu ([0,m^2+\epsilon ])=1$. If $m^2>\epsilon$, then $m^2-\epsilon <m^2$ is not in $R$, and the zero-or-one property gives $\mu ([0,m^2-\epsilon ])=0$. Thus for all sufficiently large $n$, the nested intervals $I_n=(\max (0,m^2-1\mathrm{/}n),m^2+1\mathrm{/}n]$ have measure one; when $m^2=0$, use $I_n=[0,1\mathrm{/}n]$. Here is the needed continuity-from-above argument. If $(D_n{)}_{n\ge n_0}$ is any decreasing sequence of measurable sets with $\mu (D_{n_0})<\mathrm{\infty }$, put $C_n=D_n\setminus D_{n+1}$. Then the $C_n$ are disjoint and $D_k=({\bigcap }_{n\ge n_0}{D}_n)\bigsqcup {\bigcup }_{n\ge k}{C}_n$ for each $k\ge n_0$. Countable additivity gives $\mu (D_k)=\mu ({\bigcap }_{n\ge n_0}{D}_n)+{\sum }_{n\ge k}\mu (C_n)$. Since ${\bigcup }_{n\ge n_0}{C}_n\subseteq D_{n_0}$, monotonicity of $\mu$ from step 3.1 bounds the series by $\mu (D_{n_0})<\mathrm{\infty }$; its tails therefore tend to zero. Thus ${\lim }_k\mu (D_k)=\mu ({\bigcap }_{n\ge n_0}{D}_n)$. Apply this to the tail $(I_n{)}_{n\ge n_0}$ on which $\mu (I_n)=1$; its intersection is $\{m^2\}$, so $\mu (\{m^2\})={\lim }_{n\ge n_0}\mu (I_n)=1$. Therefore $F(\{m^2\})\mathrm{\ne }0$, and step 2.2 forces $F(\{m^2\})=I$. If $A$ contains $m^2$, monotonicity gives $F(A)=I$; if not, then $\mu (A)=0$ because $A$ is disjoint from $\{m^2\}$, so the zero-or-one property gives $F(A)=0$. This rule also proves uniqueness: any second value with projection $I$ would have to equal $m^2$, since the singleton projections for distinct values are orthogonal.

By [F4], $m^2$ has a unique nonnegative square root $m$. Step 2.1 shows that each charge maps vectors in a fixed mass-squared subspace into that same subspace; step 4.1 shows there is only one such spectral value in the irreducible representation. The argument uses the pushforward PVM rather than a normalizable sharp-momentum vector, and it includes $m=0$ without introducing a rest frame.

If an additional unitary group $V(g)$ preserves $D(Q)$ and satisfies $V(g)Q\psi =QV(g)\psi$ there, then for every $\psi \in D(Q)$ one has $V(g)(Q\psi )=Q(V(g)\psi )$; this is exactly the intertwining relation. Without that additional commutation hypothesis no gauge representation equality follows.