I would like to clarify the interpretation of the notion of "charge" in therms of theory of Lie algebras. There it is stated that

So, for example, when the symmetry group is a Lie group $G$, then the charge operators correspond to the simple roots of the root system of the (associated) Lie algebra $\mathfrak{g}$; the discreteness of the root system accounting for the quantization of the charge. The simple roots are used, as all the other roots can be obtained as linear combinations of these. The general roots are often called raising and lowering operators, or ladder operators.

I'm not sure if I got it and would like to clarify if I understood the used terminology correctly. Let recall that the root decomposition of the Lie algebra $\mathfrak{g}$ comes from it's decomposition $\mathfrak{g}= \mathfrak{h} \oplus \bigoplus_{\lambda \in \mathfrak{h}^*} V_{\lambda}$ with respect to the adjoint representation $\text{ad}: \mathfrak{g} \to \mathfrak{gl}(\mathfrak{g}), X \mapsto [X,-]$, where $ V_{\lambda}:= \{ X \in \mathfrak{g} \ \vert \ [X,h] = \lambda(h)\cdot X \text{ for all } h \in \mathfrak{h} \}$ is the root space with respect to root $\lambda \in \mathfrak{h}^*$, the space of functionals $\mathfrak{h} \to \mathbb{C}$.

Question: I would like to clarify if I understand it correctly. Do they mean that the charge operators are precisely given as the "pure" elements from the root spaces $V_{\lambda}$? And the so called "central charges" are precisely elements from $\mathfrak{h}$? In turn, arbitrary linear combinations of these elements (ie the "not pure ones") cannot be called "charges")?

(besides: is this also exactly what a physicist would synonymously call "generators" of the corresponding Lie algebra? )

Another point is - if so far what I wrote is correct - what is then in terms of this terminology the "conserved charge number", which is given as stated in the quoted text as eigenvalue of the charge operator well defined? Cannot it happen that the charge operator have several eigenvalues? Are they then all called "conserved charge numbers"? In other words, does to every "charge operator" belong exactly one unique "conserved charge number" (given as a unique eigenvalue; but then why there should be exactly one unique one?) - what one should physically expect - or could it happen that there are more?

  • $\begingroup$ Well, just a few clarifying points: (1) $\lambda\in\mathfrak{h}^{*}$ and (2) $V_{\lambda}:= \{ X \in \mathfrak{g} \ \vert \ [X,h] = \lambda(h)\cdot X\mbox{ for every }h\in\mathfrak{h} \}$. $\endgroup$ Commented Jul 10, 2023 at 16:17
  • $\begingroup$ @AlexNelson: thanks, the "for all" should be added $\endgroup$
    – user267839
    Commented Jul 10, 2023 at 16:25

2 Answers 2


Note that the Wikipedia article lacks any citations for this claim, and the claim is indeed wrong - this is not some sort of universally accepted definition of "charge" (and the claim about the "discreteness of the root system" being related to "quantization" is so non-sensical that I will dismiss it simply with the observation that both classical and quantum mechanics have largely the same symmetry algebras but only one of these shows "quantization"). It's easy to prove this wrong when looking at how people talk about the color charge of the strong interaction:

Quarks carry the strong "color charge" because they transform in the fundamental representation $\mathbf{3}$ of $\mathfrak{su}(3)$ and that representation is three-dimensional so you can assign the names of three colors "red, green, blue" to a basis and talk about that. The anti-quarks transform in the anti-fundamental representation (the conjugate representation of the fundamental) $\bar{\mathbf{3}}$ and hence have "anti-color charges": Anti-red, anti-green, anti-blue.

Finally, gluons carry a "color-anticolor" charge because they transform in the adjoint representation $\mathbf{8}$ and you can get that by $\mathbf{3}\otimes \bar{\mathbf{3}} = \mathbf{8} \oplus\mathbf{1}$ so every gluon in the adjoint can be thought of as being the combination of a color and an anticolor (with the combination corresponding to the singlet $\mathbf{1}$ disallowed): Green-antired, etc.

None of this has anything to do with (simple) roots, as $\mathfrak{su}(3)$ has 2 simple roots (the Dynkin diagrams of $\mathfrak{su}(n)$ is $A_{n-1}$) and 6 roots in total, but what we observe in the language above is that people assign 1 charge value to a quark and a pair of two charge values to a gluon.

Really, the only uncontroversial thing we can say about what people mean when they talk about "charge" in this generalized context of non-Abelian algebras is that they mean the charged object transforms in a non-trivial representation of that algebra.

Finally, what is true in general is that if the algebra is a symmetry algebra, then all the Lie algebra generators (not the roots but the generators, i.e. a basis of the Lie algebra) are "charges" in the Noetherian sense: The Hamiltonian formulation of mechanics means that there is a representation of the symmetry algebra on the functions on the state space and the images $f_i$ of the generators under that representation fulfill $\{H,f_i\} = 0$ (where $\{,\}$ is either the classical Poisson bracket or the quantum commutator) by definition of a Hamiltonian symmetry and so they are conserved charges, i.e. the values of these functions are constant along trajectories just like the Hamiltonian itself is.

  • $\begingroup$ let me try to understand the last paraphraph: so in order to talk about charges wrt a symmetry algebra, then it is neccessary to make an explicit choice of the basis element, ie the charges cannot be defined intrinsically? $\endgroup$
    – user267839
    Commented Jul 10, 2023 at 17:46
  • $\begingroup$ @user267839 No, you can of course look at the vector of these conserved charges and phrase the conservation law more abstractly without having to pick. E.g. in the classical formulation you have a momentum map $\mu : M\to \mathfrak{g}^\ast$ and for any $h \in\mathfrak{g}^\ast$ (which is just an assignment of values to the individual generators but abstracted away from a particular basis) you have $\mu^{-1}(h)$ as the set of points with that particular value of the "charges". $\endgroup$
    – ACuriousMind
    Commented Jul 10, 2023 at 17:51

The general idea is that weight vectors for a representation correspond to types of particles, and weights correspond to quantum numbers. The root vectors (being weight vectors for the adjoint representation) correspond to types of gauge bosons, and roots correspond to their quantum numbers. But root vectors always act "in the obvious way" on weight vectors, and roots gain an interpretation as charges in this manner.

Just to give some citations (there are many, but here are the random ones at hand). Jürgen Fuchs and Christoph Schweigert's Symmetries, Lie Algebras and Representations: A Graduate Course for Physicists, chapter 6, notes (pg 86):

The root space decomposition (6.7) has important implications in gauge theories. The gauge bosons of a non-abelian Yang-Mills theory carry the adjoint representation of some simple Lie algebra; the quantum numbers of the gauge bosons are zero if they correspond to generators of the Cartan subalgebra, and otherwise they are given by the roots (see e.g. [O'Raifeartaigh 19861). Since the roots are non-zero, the gauge bosons of a non-abelian gauge theory are charged and therefore self-interact. For example, in the theory of electro-weak interactions the $W^{\pm}$ gauge bosons correspond to the positive and negative root, respectively, of an $\mathfrak{sl}(2)$ algebra and hence are charged, while a linear combination of the photon and the Z boson, which are uncharged, corresponds to the Cartan subalgebra. This is in sharp contrast to the situation in abelian gauge theories like electrodynamics, in which there are no roots at all so that the gauge bosons do not self-interact.

The reference given is to L. O'Raifeartaigh's Group Structure of Gauge Theories, Cambridge University Press (1986). See especially chapter 11, section 5, where O'Raifeartaigh observes that for a rank $\ell$ Lie algebra, "Another important set of elements are the charges $Q_{i}$, $i=1, ..., \ell$ defined (for any Lie algebra) as the elements of the Cartan [subalgebra] which combine with the elements $E_{\pm i}$ associated with primitive roots $\alpha_{i}$ to form $\ell$ [copies of the Lie algebra for] $SU(2)$ subalgebras $\{E_{i},E_{-i},Q_{i}\}$..." (pg 133).

But O'Raifeartaigh's discussion of the Cartan subalgebra in chapter 3, section 5, is also worth reading. There he discusses Cartan subalgebras and charges further.

  • $\begingroup$ "roots gain an interpretation as charges". Hier does it specialize to the naive physical situation where charges appear as honest numbers? Recall, a root is by definition a functional $\mathfrak{h} \to \mathbb{C}$... ie is this notion of charge "backward compatible" with classical terminology? $\endgroup$
    – user267839
    Commented Jul 11, 2023 at 15:50

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