# Why the characters of the Minkowski spacetime translation group involve the Minkowski metric?

The translation group of Minkowski spacetime is just the additive group $$\mathbb{R}^4$$. Indeed, if $$x\in \mathbb{R}^4$$ is a point in Minkowski spacetime, the translation $$T_v$$ acts on $$x$$ by

$$T_vx=x+v.$$

This is one isometry of said spacetime metric $$\eta = \operatorname{diag}(-1,+1,+1,+1)$$.

Now, I've read that the characters of the translation group are just labelled by $$p\in \mathbb{R}^4$$ and read

$$\chi_p(v)=e^{i\eta(p,v)}=e^{i\eta_{\mu\nu} p^\mu v^\mu}.$$

My question here is: how did the Minkowski metric inner product ended up in the character?

I mean, a character is a continuous group homomorphism from a group $$G$$ to the circle group $$S^1$$.

In this case, it would be $$\chi : \mathbb{R}^4\to S^1$$ with $$\chi(v+w)=\chi(v)\chi(w)$$.

Further since it belongs to the circle $$\chi(v)=e^{i\theta(v)}$$ with $$\theta : \mathbb{R}^4\to \mathbb{R}$$ and $$\theta(v+w)=\theta(v)+\theta(w)$$.

Since $$\chi$$ must be continuous, $$\theta$$ must be continuous. Further, since $$\theta$$ is continuous and additive it is linear. Hence $$\theta$$ is a covector. In that case, if $$v = \sum v^\mu e_\mu$$ in some basis it follows that if $$\theta_\mu = \theta(e_\mu)$$ then

$$\theta(v)=v^\mu \theta_\mu$$

And hence

$$\chi(v)=e^{iv^\mu \theta_\mu}$$

Now, the above is not the Minkowski inner product. Is just the action of a covector in a vector. In other words: it is

$$\theta_\mu v^\mu = \theta_0 v^0+\theta_1v^1+\theta_2v^2+\theta_3v^3$$

But the usual form is

$$\eta(p,v)=-p_0v^0+p_1v^1+p_2v^2+p_3v^3$$

Why is that? How does the Minkowski metric ends up in the characters of the Minkowski translation group? What am I missing here?

One obvious answer is that the metric induces one isomorphism between $$\mathbb{R}^4$$ and its dual so any covector $$\theta$$ as above is $$\theta(v)=\eta(p,v)$$ for some $$p\in \mathbb{R}^4$$.

But this is just one representation that is always available. It is one option, not something forced upon us.

This has something to do with the adjoint action of the Lorentz group on the translation group which comes about when we realize the translation group as a normal subgroup of the full Poincare group?