We use the eigenvalues of the Cartan generators (=diagonal generators) of a given gauge group as quantum numbers in physics. Are these numbers somehow fixed and if not, what transformations are allowed?

The easiest example is $SU(2)$ with just one Cartan generator $H_1$, which is commonly written in terms of the Pauli matrix $\sigma_3= \begin{pmatrix} 1&0\\0&-1 \end{pmatrix}$: $H_1 = \frac{1}{2} \sigma_3$ and therefore

$$H_1= \begin{pmatrix} \frac{1}{2} &0\\0&-\frac{1}{2} \end{pmatrix} $$

Would $H_1 = \begin{pmatrix} \frac{1}{7} &0\\0&-\frac{1}{7} \end{pmatrix} $ or $H_1= \begin{pmatrix} -\frac{1}{2} &0\\0&\frac{1}{2} \end{pmatrix}$ equally "work"?

A bit more involved example would be $SU(3)$, which has two Cartan generators $H_1=\frac{1}{2} \lambda_3$ and $H_2=\frac{1}{2} \lambda_8$, where $\lambda_3$ and $\lambda_3$ denote Gell-Mann matrices.

How unique are the diagonal entries of these matrices? In what ways are we allowed to transform the Cartan generators (and with them of course the corresponding quantum numbers)?

(One allowed transformation is certainly which one we call $H_1$ and which one $H_2$, i.e. permuations. $H_1 \leftrightarrow H_2$ )

  • $\begingroup$ Since the weights of a rep are in the Cartan dual, they specify the eigenvalues of the Cartan generators. Hence, the eigenvalues are invariants of a representation, that is, they are unchanged by any transformation. Am I missing something, or is that what you are asking? $\endgroup$
    – ACuriousMind
    Jul 20, 2015 at 11:37
  • $\begingroup$ @ACuriousMind I don't think that is correct. For example in this paper arxiv.org/abs/1502.06929 the authors explicitly say: "We choose the five linearly independent Cartan generators as follows:..." There seems to be some freedom and I'm trying to understand what exactly we are allowed to change, i.e. in what ways the Cartan generator eigenvalues that we use to label weights can be modified. One obviously allowed transformation, for example is simple permutations: $H_1 \leftrightarrow H_2$ $\endgroup$
    – jak
    Jul 20, 2015 at 11:44
  • $\begingroup$ @ACuriousMind They copied this "choice" from this paper arxiv.org/abs/hep-ph/9309312 , where they explain that they use linear combinations of the basis elements of the Cartan subalgebra in order to define their "choice" of Cartan subalgebra. $\endgroup$
    – jak
    Jul 20, 2015 at 11:55
  • $\begingroup$ Ah, I see, you are talking about the freedom in the choice of Cartan basis, not about transforming the chosen basis. I think your answer lies in using the Casimirs computed from the eigenvalues instead of the eigenvalues themselves, but I need to check that. $\endgroup$
    – ACuriousMind
    Jul 20, 2015 at 11:57

1 Answer 1


Here we will for simplicity just consider an arbitrary finite-dimensional complex$^1$ semisimple Lie algebra $\mathfrak{g}$.

I) One may show that the CSAs are precisely the maximal toral Lie subalgebras of $\mathfrak{g}$. In particular CSAs are abelian.

Also the Killing form $\kappa:\mathfrak{g}\times \mathfrak{g}\to \mathbb{C}$ (which is non-degenerate) has a non-degenerate restriction to a CSA, so a CSA is canonically an inner product space, and canonically isomorphic to its dual vector space.

Moreover, all CSAs of $\mathfrak{g}$ have the same dimension (called the rank $r$), and are conjugated with each other, i.e. related via inner automorphisms of the Lie algebra $\mathfrak{g}$. So in this sense all choices of CSA are equivalent.

II) Consider from now on an arbitrary but fixed given choice of CSA $\mathfrak{h}\subset \mathfrak{g}$.

Obviously, one may pick an arbitrary basis $(H_1, \ldots, H_r)$ for $\mathfrak{h}$.

A root $\alpha\in \mathfrak{h}^{\ast}$ belongs to the dual vector space of $\mathfrak{h}$. Its defining property is $$\tag{1}\exists x\in \mathfrak{g}\backslash\{0\}~ \forall H\in\mathfrak{h} :~~[H,x]~=~\alpha(H)x.$$

From a physics perspective, the Lie algebra element $x\in \mathfrak{g}$ plays the role of a generalized raising/lowering creation/annihilation operator and the root $\alpha\in \mathfrak{h}^{\ast}$ plays the role of a generalized quantum number.

Note in particular that the definition (1) in principle does not depend on a choice of basis $(H_1, \ldots, H_r)$.

NB: Be aware that authors often use other associative/invariant bilinear forms than the canonical Killing form $\kappa$. This may induce a non-canonical isomorphism $\mathfrak{h}^{\ast}\cong \mathfrak{h}$ and non-canonical normalizations of roots.


$^1$ Many results and properties for complex Lie algebras continues to hold for real Lie algebras, although sometimes in modified form.

  • $\begingroup$ thanks for your answer. The key point, I think, as already mentioned by @ACuriousMind is that we can pick a basis for the CSA. I'm still unsure what this means in concrete terms, say, for the the two examples I mentioned above $SU(2)$ and $SU(3)$. I've never seen different matrices as Cartan algebra elements for the corresponding algebras (= never different diagonal entries). Can you give another example of a possible basis choice for the CSA, specifically for $SU(2)$ and $SU(3)$? $\endgroup$
    – jak
    Jul 21, 2015 at 6:13

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