Why there's a Lorentz inner product in the unitary representations of the translation group?

Question

Consider Minkowski spacetime. Its translation group is just the additive group $\mathbb{R}^4$. This is an abelian locally compact group.

Next, consider one unitary representation $T : \mathbb{R}^4\to \mathfrak{U}(\mathscr{H})$ on the Hilbert space $\mathscr{H}$. It is said that the SNAG theorem implies that $$T(a)=\exp\left[i\eta_{\mu\nu}a^\mu P^\nu\right]$$

where $P^\mu$ are four Hermitian commuting observables and $\eta_{\mu\nu}$ is the Minkowski metric.

I want to see how to derive this from the SNAG theorem. The theorem is stated as follows (Barut's group theory book):

SNAG (Stone-Naimark-Ambrose-Godement) Theorem: Let $T$ be an unitary continuous representation of an abelian locally compact group $G$ in a Hilbert space $\mathscr{H}$. Then there exists on the character group $\hat{G}$ a spectral measure $E$ such that $$T(x)=\int_{\hat{G}}\langle \hat{x},x\rangle dE(\hat{x})$$

Now it is possible to show that for the additive group $\mathbb{R}^n$ the SNAG theorem tells us that there are $n$ self-adjoint commuting operators $Y_1,\dots,Y_n$ such that

$$T(x)=\exp\left[i\sum_{k=1}^n x^k Y_k\right].$$

These operators $Y_k$ are defined in terms of the spectral measure $E$ given by the SNAG theorem as

$$Y_k=\int y_k dE(y).$$

Now, the Minkowski spacetime translation group is exactly $\mathbb{R}^4$ so this theorem should apply. Indeed it is almost it, except that for Minkowski spacetime, the operators are $P_0,\dots, P_3$ and

$$T(a)=\exp\left[i \eta_{\mu\nu}a^\mu P^\nu\right]$$

I can't get why. How does the Minkowski inner product ends up there if the translation group is just $\mathbb{R}^4$ which has nothing to do with the metric structure?

This has something to do with the realization of the translation group as a subgroup of the Poincare group so that if $U(\Lambda,a)$ is a unitary representation of the latter one has $U(1,a)$ a unitary representation of the translations satisfying

$$U(\Lambda,b)U(1,a)U(\Lambda,b)^\dagger=U(1,\Lambda a)$$

I think the answer comes from this, but I don't know how to justify it.

Answer 1

I will use the signature $(+,-,-,-)$ for the Minkowski metric $\eta$.

If you got so far as showing that $$ \forall\ {\rm continuous\ unitary\ repesentation}\ T\ {\rm of}\ (\mathbb{R}^4,+), $$ $$ \exists\ {\rm commuting\ self-adjoint\ operators} \ Y_1,\ldots,Y_4, $$ $$ \forall x\in \mathbb{R}^4,\ \ T(x)=\exp[i(x^1T_1+x^2T_2+x^3T_3+x^4T_4)] $$ then the last step to conclude that $$ \forall\ {\rm continuous\ unitary\ repesentation}\ T\ {\rm of}\ (\mathbb{R}^4,+), $$ $$ \exists\ {\rm commuting\ self-adjoint\ operators} \ P^0,\ldots,P^3, $$ $$ \forall a=(a^0,\ldots,a^3)\in \mathbb{R}^4,\ \ T(a)=\exp[i(a^0P^0-a^1P^1-a^2P^2-a^3P^3)] $$ is trivial: just define $P^0=Y_1$, $P^1=-Y_2$, $P^2=-Y_3$, $P^3=-Y_4$.

BTW, as mentioned in DanielC's comment, appealing to the SNAG Theorem is overkill here. The much more elementary Stone-von Neumann Theorem is enough.