Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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.)

share|improve this answer
Hi, thanks for your answer. I have some comments and doubts still left over. It would be great if you would clarify them. I am given to believe that a Lie algebra is a vector space along with a bilinear form which has satisfies some properties. Now, does Weinberg mean to say that we can take a representation of "the" Lie algebra on a vector space and view it as adjoint representation of "another" Lie algebra? So, that means, according to him a Lie group can have many Lie algebras? Is this notion widely followed across physics and mathematics communities? ... –  user4235 Mar 3 '12 at 11:45
... I think that it is more widely accepted to regard these "a" Lie algebras as the representations of "the" Lie algebras. Am I correct? Or is it that this (confusing) convention followed much across QFT and string theory circles? Please share as much as know regarding this so that when I see such things elsewhere, I can try to solve my confusion by the context of the matter. –  user4235 Mar 3 '12 at 11:47
Moreover, I didn't quite follow your answer to point numbered 3. I think I have an idea that you are trying to hint at projective representations or something like that. But, I think Weinberg already takes care of that and makes an assumption about them in the middle paragraph of Page 53 just before discussion about connected Lie groups. Plus, please answer the specific question of point number 4. Can Lie algebras be defined for groups which are not connected? In the end, I would like to thank you again for a very precise answer which removed quite of troubling confusion from my mind. –  user4235 Mar 3 '12 at 11:51
I've updated my answer. Weinberg doesn't say much, but the commonly accepted terminology is that a Lie group has one single Lie group associated to it. The notion of "a" Lie algebra is independent of any group, you can study representations of Lie algebras without ever having heard of what a Lie group is. In particular, the concept of "a" Lie algebra is independent of "the" Lie algebra associated to a group or any representation thereof. –  Greg Graviton Mar 4 '12 at 9:18
I probably don't know enough about particle physics to answer that one. In general, I recommend that you seek other books besides Weinberg. (To be honest, I find his discussion of Lie groups to be unintelligible to anyone who doesn't know what they are already.) Reading books in parallel has often helped me tremendously as one book explains what the other doesn't and vice versa. For an introduction to relativistic quantum field theory, I can recommend Sakurai's "Advanced Quantum Mechanics" as he explains some concepts most clearly. –  Greg Graviton Mar 8 '12 at 11:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.