I'm studying Weinberg's QFT books, and regarding symmetries I'm quite confused about the distinction between a group and its representations in Weinberg's presentation.
First of all, Weinberg states that symmetry transformations are ray transformations preserving the probabilities $P(\mathscr{R}\to \mathscr{R}_n)$ defined for a ray $\mathscr{R}$ and a family of mutualy orthogonal rays $\mathscr{R}_n$ to be
$$P(\mathscr{R}\to \mathscr{R}_n)=|(\Psi,\Psi_n)|^2,\quad \forall\Psi\in \mathscr{R},\Psi_n\in \mathscr{R}_n.\tag{2.1.7}$$
He then points out (p. 52) that:
The set of symmetry transformations has certain properties that defines it as a group. If $T_1$ is a transformation that takes rays $\mathscr{R}_n$ into $\mathscr{R}_n'$ and $T_2$ is another transformation that takes $\mathscr{R}_n'$ into $\mathscr{R}_n''$, then the result of performing both transformations is another symmetry transformation, which we write $T_2T_1$, that takes $\mathscr{R}_n$ into $\mathscr{R}_n''$. Also, a symmmetry transofrmation $T$ which takes rays $\mathscr{R}_n$ into $\mathscr{R}_n'$ has an inverse written $T^{-1}$ which takes $\mathscr{R}_n'$ into $\mathscr{R}_n$ and there is an identity transformation $T =1$ which leaves rays unchanged.
He also states Wigner's theorem that states that symmetry transformations defined as above can be realized either by unitary linear or antiunitary and antiliear operators on the Hilbert space $\mathscr{H}$. In his notation, for every symmetry transformation $T$ one gets a unitary operator $U(T)$. Then Weinberg proves that
$$U(T_2)U(T_1)=e^{i\phi(T_2,T_1)}U(T_2T_1), \tag{2.2.14}$$
saying that $U(T)$ is a projective representation of the symmetry transformations.
After this recap, here are my questions:
So by (1) above, to deal with symmetryes, Weinberg is actualy implictly considering that there is a group $G$ such that we have one homomorphism $T$ mapping $G$ into the group of ray transformations?
In other words, for every $g \in G$ we have $T(g)$ a ray transformation. Upon imposing the symmetry requirement by (2) above, we have that all $T(g)$ descends to one $U(T(g))$ and these $U(T(g))$ form a projective representation of $G$ .
Because of that in the end he forgets $T$ altogether and directly works with projective representations of $G$ on the Hilbert space of states. Is that it?
The main difference between what I'm writing and Weinberg is that I'm trying to abstract a group from its representations.
So I'm guessing there is one underlying group $G$ of symmetries, which give rise to the ray transformations and then to the projective representations, while Weinberg seems to identify $G$ with the ray transformations themselves.
Is my point of view correct of considering there is one abstract group behind all of this? Or there is actually no group behind the ray transformations, and Weinberg is actually defining a group with the ray transformations themselves, instead?