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

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.

• My guess is the answer is something like: we can use any product between $a$ and $P$ that we like, since we are just talking about the translation group, which doesn't know anything about the Minkowski product. To make things convenient later we take the convention that positive time translations are mapped to $e^{itP^0}$ while positive space translations are mapped to $e^{-ixP^1}$ etc. Any physical meaning for the Minkowski product there comes later when we restrict our representation to positive energies only. – Luke Pritchett Jan 28 at 2:24
• The SNAG theorem is a neat generalization of the Stone's theorem and thus one can use the spectral theorem for both operator types (unitary and self-adjoint) to arrive at the result via the "exponential of an unbounded self-adjoint operator", a quite delicate mathematical concept. @Valter Moretti. – DanielC Jan 28 at 17:34
• I think I made some progress, but there is one last bit which is exactly what @LukePritchett talks about in his comment. How the Minkowski inner product ends up in the exponent if the translation group "knows nothing about said product"? I think the answer is in the fact that if $U(a,\Lambda)$ is a unitary representation of the Poincare group, then $$U(b,\Lambda)U(a,1)U(b,\Lambda)^\dagger = U(\Lambda a,1)$$. This implies in particular that $$U(b,\Lambda)P^\mu U(b,\Lambda)^\dagger = \Lambda^\mu_\nu P^\nu$$ so I think that somehow the answer comes from this. I just don't know how. – user1620696 Jan 28 at 18:05

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 $$P^0=Y_1$$, $$P^1=-Y_2$$, $$P^2=-Y_3$$, $$P^3=-Y_4$$.
• Thanks for the answer ! I've notice one could make such choice, but why is it convenient to do so? I mean, what's the motivation behind it? Has it to do with the fact that when the full Poincare group is considered we have the relation $$U(b,\Lambda)U(a,1)U(b,\Lambda)^\dagger = U(\Lambda a,1)$$ – user1620696 Feb 6 at 20:55
• This has nothing to do with the full Poincare group. Even if all we got is the Lorentz group, it is convenient to write everything in terms of Lorentz invariant objects like $\eta$ so one can easily track how things transform by Lorentz even in the midst of complicated computations (think, e.g., 5 loop Feynman diagram). So we do this because it is convenient and because we can. – Abdelmalek Abdesselam Feb 6 at 21:27