Weinberg's way of deriving Lie algebra related to a Lie group

Question

I was reading the second chapter of the first volume of Weinberg's books on QFT. I am quite confused by the way he derives the Lie algebra of a connected Lie group.

He starts with a connected Lie group of symmetry transformations and then consider how it would be represented on the Hilbert space. Then, he says that as it is a symmetry transformation and because it is connected to identity continuously (because it is locally path connected), the transformation U(T) corresponding to a particular symmetry transformation T must be unitary by Wigner's theorem. So, he takes a group element close to identity and expand it as a power series and considers their compositions to derive at the commutation relations for the operators representing generators of the Lie algebra.

Much of this can be followed on pages 53-55. My doubts and confusions are as follows:

1. Does this give a Lie algebra of the underlying group of Ts or the group induced on Hilbert space of the U(T)s?

2. Is it that we presuppose a Lie algebra with some generators already out there and think of all this stuff on Hilbert space as its representation and then derive the commutation relations of the representation of generators to get the abstract Lie product relations of Lie algebra? If yes, then are the Lie algebras of Ts and U(T)s same or different? Please be precise on this.

3. If all the above is true, then why do we need specifically representations on Hilbert space? Any representation would do, right? And where does all this talk about unitarity of representations used in the following derivation of the Lie algebra? I can't see any application of Wigner's theorem in what follows. I think that all that follows can be done on the representation on any space. But then this brings us to the question as to if a representation of the group exists on that vector space? Is it why we need Wigner's theorem and Hilbert space? Then, why don't we take the representations on 4D space-time itself where a representation exists trivially by definition? If this is true, how do we go in general about finding about what kind of representations can exist for a Lie algebra?

4. In general, provided with a Lie group, how does a mathematician go on to derive its Lie algebra? Should the Lie group be connected? Can Lie algebras be defined for Lie groups which are not connected?

Answer 1

Here is how I understand Weinberg's discussion:

First of all: following Weinberg's words precisely, he only says that "Such a set of commutation relations is known as a Lie algebra". He motivates the commutation relations by considering a representation $U(T)$ of a Lie group $T$, but he does not clarify the relationship between these notions. The commonly accepted terminology has to be learned from elsewhere; I'll try to expand on it here.

Note that mathematicians distinguish between "a Lie algebra" and "the Lie algebra associated to a Lie group".

Any algebra of operators that fulfills (2.2.22) is "a" Lie algebra.

But for each Lie group $T$, there is "the" "most general" Lie algebra $\mathfrak{t}$ that fulfills (2.2.22). It arises from the structure constants $f^a_{bc}$ that Weinberg mentions. Note that the structure constants do not depend on the representation on some Hilbert space, the only depend on the Lie group $T$.

To find this "most general" Lie algebra associated to a Lie group, you can look at a very special vector space, namely the tangent space at the identity element $1\in T$. The ajdoint representation of the group on this space will give rise to a Lie bracket $[·,·]$ on that vector space and turn it into said "most general" Lie algebra.

As for your particular questions:

1. Weinberg only the mentions "a Lie algebra", though it comes from a group. It's "the" lie algebra of the group $U(T)$.
2. In general, every such Lie algebra is a homomorphic image of "the" Lie algebra associated to $T$. These Lie algebras are different from each other; for instance, the vector spaces may have different dimensions.

3. Weinberg invokes Wigners theorem for the following reason: a priori, the symmetry group $T$ acts on rays. Remember that the vector $|ψ\rangle$ represents the same physical state as the vector $ξ|ψ\rangle$ which arise from multiplication with an arbitrary complex number of mangitude $|ξ|=1$. A ray is the set of all vectors that arise in this fashion; they all represent the same physical state.

   Now, the symmetry maps physical states to physical states, i.e. sets of vectors to sets of vectors. But each symmetry operation is allowed to permutate vectors within the set, after all, they are indistinguishable physically. It is not clear at all that a mapping $T_1$ on rays can be recast as a mapping $U(T_1)$ that acts on individual vectors and is linear. Neither is it clear that $U(T_1T_2) = U(T_1)U(T_2)$, because these operations may map the rays to each other, but they might permute the vectors within a ray rather differently, which would manifest in a phase $U(T_1T_2)|ψ\rangle = ξ·U(T_1)U(T_2)|ψ\rangle$.

   We want the symmetry to act on a Hilbert space, though, because this allows us use the familiar expression for the commutator of two operators $[X,Y] = XY-YX$. After all, it involves the addition (subtraction) of two operators, something that is not available for symmetries merely acting on rays.

   Other than that, Hilbert spaces are completely unnecessary for defining Lie algebras or Lie groups.

   The question of how to find all representations of a Lie group or of a Lie algebra is beyond the scope of this answer. It leads to topics such as semisimple Lie algebras and their classification.

4. Mathematicians obtain the Lie algebra associated to a Lie group by considering the tangent space at the identity element, as mentioned above. The Lie group doesn't need to be connected for this to be well-defined. (But it is only useful for studying the part of the group that is connected to the identity.)