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 $T : \mathbb{R}^4\to \mathrm{U}(\mathcal{H})$ on the Hilbert space $\mathcal{H}$. 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.