In most textbooks (Georgi, for example) a scalar product on the generators of a Lie Algebra is introduced (the Cartan-Killing form) as $$tr[T^{a}T^{b}]$$ which is promptly diagonalised (for compact algebras) and the generators scaled such that $$tr[T^{a}T^{b}] = \delta^{ab}.$$ In this basis we get that, for example, $$f_{abc} = -i\, tr ([T^{a}, T^{b}]T^{c})$$ that are fully antisymmetric.

Yet I have seen the these relations used for arbitrary (it particular the fundamental) representation as matter of course (maybe up to some normalisation). Is this because $tr[T^{a}T^{b}]$ defines a symmetric matrix in any rep that can thus be diagonalised? Is it a general truth? Or does the diagonalisation in the adjoint imply a diagonal for in any other rep?

I know that the structure constants are essentially fixed for all reps by smoothness and the group product -- is this why fixing the form in one basis for one rep fixes it for that basis in all reps?

For a concrete example, let's suppose I look at SU(2). The adjoint rep is 3 dimensional and I can linearly transform and scale my generators (i.e. the structure constants) so that I get the trace to be diagonal and normalised. This fixes once and for all that the structure constants of SU(2) are $f_{ijk} = \epsilon_{ijk}$, say.

Now I ask someone to construct the fundamental rep; they look for 2x2 matrices satisfying the Lie algebra with these structure constants. They find the Pauli matrices. Why do these come out such that the trace $tr [\sigma^{a} \sigma^{b}] \propto \delta_{ab}$ automatically? It's a different rep...why is it guaranteed?

  • $\begingroup$ Dynkin index. $\endgroup$ Oct 29, 2018 at 13:55
  • $\begingroup$ Thank you for the link. It is potentially helpful, but lacking in the key detail. Why is it guaranteed that if the trace takes the form $g_{ab}$ in some rep then in any other rep it is just a scaling of this same form? That is mñth purpose of my post. $\endgroup$
    – nox
    Oct 29, 2018 at 16:42
  • $\begingroup$ Euxaristw Kosma, but this requires various properties about the ability to construct higher dimensional representations through tensor products. I guess I am looking for something more fundamental about the Lie Algebra. It should follow from something more general. $\endgroup$
    – nox
    Oct 29, 2018 at 19:33
  • $\begingroup$ Yes it does; you might need to replace your text with a superior one. $\endgroup$ Oct 29, 2018 at 22:56
  • $\begingroup$ Ch IV, p 29, on the uniqueness of bilinear invariants up to a number. $\endgroup$ Oct 30, 2018 at 0:26

3 Answers 3


More generally, let there be given an $n$-dimensional complex Lie algebra $(L, [\cdot,\cdot])$.

  1. The adjoint representation ${\rm ad}:L\to {\rm End}(L)$ is defined as $$({\rm ad}x)y~:=~[x,y], \qquad x,y~\in~L.\tag{1}$$

  2. The Killing form $\kappa:L\times L\to \mathbb{C}$ is defined as $$ \kappa(x,y)~:={\rm tr}({\rm ad}x \circ {\rm ad}y), \qquad x,y~\in~L, \tag{2}$$ is bilinear, symmetric, associative. It is non-degenerate iff $L$ is semisimple.

  3. One may show that any bilinear, symmetric, associative form $L\times L\to \mathbb{C}$ is proportional to the Killing form $\kappa$ if $L$ is simple, cf. e.g. this Math.SE post.

  4. Given an arbitrary basis $(t_a)_{a=1,\ldots, n}$ for $L$, define the structure constants $f_{ab}{}^c\in\mathbb{C}$ via $$ [t_a,t_b]~=~\sum_{c=1}^nf_{ab}{}^ct_c , \qquad a,b~\in~\{1,\ldots, n\}. \tag{3}$$

  5. Define Killing metric $$ \kappa_{ab}~:=~\kappa(t_a,t_b) , \qquad a,b~\in~\{1,\ldots, n\}.\tag{4}$$

  6. One may show that the lowered structure constants $$ f_{abc}~:=~\sum_{d=1}^nf_{ab}{}^d\kappa_{dc}\, \qquad a,b,c~\in~\{1,\ldots, n\},\tag{5}$$ are always totally antisymmetric in the indices $abc$.

  7. It is possible to choose the basis $(t_a)_{a=1,\ldots, n}$ such that the Killing metric $\kappa_{ab}$ is diagonal; and in the semisimple case, such that $\kappa_{ab}$ is proportional to the identity $\delta_{ab}$.

  • $\begingroup$ Is there a general method to show that any bilinear symmetric associative form $L \times L\rightarrow \mathbb{C}$ is proportional to $\kappa$ ? $\endgroup$
    – John
    Jan 28, 2023 at 2:05
  • $\begingroup$ Only if $L$ is simple. $\endgroup$
    – Qmechanic
    Jan 28, 2023 at 3:31
  • $\begingroup$ In the case that $L$ is simple, could you tell me where to find that proof? $\endgroup$
    – John
    Jan 28, 2023 at 3:32
  • 1
    $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Jan 29, 2023 at 13:52

Below is a proof outline.

Define $(a,b)_X \equiv \mathrm{Tr}(X^a X^b)$ where $a$ and $b$ are generators of a Lie algebra and $X$ is an arbitrary representation. $(a,b)_{\textbf{ad}}$, where $\textbf{ad}$ stands for the adjoint representation, is also called the Killing form, and often denoted as $\langle a,b \rangle$.

There are a few definitions that are useful.

  • Bilinear form: See this link for its definition. Notice that $(\cdot,\cdot)_X$ in general, and $\langle \cdot, \cdot \rangle$ in particular, are bilinear forms.
  • Nondegenerate bilinear form: See this link for its definition. There is a theorem due to Cartan that $\langle \cdot,\cdot \rangle$ is nondegenerate if and only if the algebra is semi-simple.
  • Invariant bilinear form: A bilinear form $B$ of a Lie algebra is invariant if $B([x,y], z) = B([x,[y, z])$ for every $x$, $y$, and $z$ in the algebra. Again, notice that $(\cdot,\cdot)_X$ in general, and $\langle \cdot,\cdot \rangle$ in particular, are invariant bilinear forms. This can be shown through direct calculation.

($(a,b)_X$ is also symmetric in $a$ and $b$. But we don't really care about this property for this proof.)

Now comes the critical part. There is a theorem that any invariant bilinear form of a simple Lie algebra is proportional to the Killing form. Its proof can be found here.

In Georgi's book (Sectoin 2.4), it is explained in detail that for any compact Lie algebra, we can choose a basis such that $\langle a, b \rangle = \lambda \delta_{ab}$. Once this is done, we have automatically $(a, b)_X = \lambda_X \delta_{ab}$ for every representation $X$, assuming the algebra is also simple.


As you say, it is clear that \begin{equation} \text{Tr} [T^aT^b] = \text{Tr} [T^bT^a] \end{equation} Therefore, this is clearly defines some symmetric matrix, which we can diagonalize. Through the linearity of the trace, this procedure also allows us to construct a new basis for which the desired $\text{Tr}[T^aT^b] = \delta^{ab}$ holds. Note we have no specialized to any representation, so this is generally allowed.

Another way of thinking about this is to treat the Cartan-Killing form as a dot product for Lie algebras. The fact that it is a symmetric, bilinear form allows you to diagonalize your vector space such that the dot product becomes identity.

I believe the statement that "given a choice of structure constants in one basis (eg adjoint) such that the trace is $\delta^{ab}$, for all other representations which have the same set of structure constants their generators also satisfy the trace condition" is false. This can be intuitively seen by the fact that the structure constants fix the commutators, while the trace fixes the anti-commutators; generically these shouldn't be related (though I don't have an explicit counterexample).

In the special case of the fundamental and the adjoint, however, they are in fact related with the same choice of structure constants, as Cosmas Zachos points out. This is due to the special fact that the adjoint is defined off of the fundamental, ie. the basis choice between the two have good reason to be nicely related to each other.

  • $\begingroup$ Yes I agree with this in principle. However, the Cartan-Killing form is defined in the adjoint. The thing is: once I fix the basis for the gens in the adjoint I have fixed the structure constants (and their scale) and these are rep independent. So it's not obvious to me that I can then go to another rep (fundamental, say) and expect to find the orthonormal form. If I then similarity transform or scale, I will probably break the form in the adjoint as my structure constants will have been modified. Do you have any comment on this? $\endgroup$
    – nox
    Oct 29, 2018 at 15:31
  • $\begingroup$ Perhaps calling it a Cartan-Killing form is the point of confusion. The only necessarily ingredient is the trace, which when acting on two operators is symmetric and bilinear. This statement is representation independent; there is no need to specify the adjoint rep for the operators by this argument. As you say, generally speaking I don't expect there to be a relation between the bases in different representations. $\endgroup$
    – Aaron
    Oct 29, 2018 at 15:42
  • $\begingroup$ Well this is my point: once I fix the trace in one rep, this selects the gens of that rep but also fixes the structure constants for all reps. Then in another rep if I want to maintain the same structure constants (supposedly rep independent) then I can no longer play with the trace I'm that other rep. Do you see my quandary? $\endgroup$
    – nox
    Oct 29, 2018 at 15:50
  • $\begingroup$ Are you sure that fixing the trace in one form (eg adjoint) fixes it for all reps if you preserve the structure constants? I could be wrong, but generically I don't believe that the trace will have the same form in other representations if you preserve the structure constants; preserving commutation doesn't preserve multiplication. $\endgroup$
    – Aaron
    Oct 29, 2018 at 16:17
  • $\begingroup$ That's precisely my question: I don't believe that it does fix the trace in all reps. Yet I must maintain the structure constants, because they are rep independient (fixed by smoothness and the group product) which leads me to believe that once I fix the form one rep, I have no further freedom to fix it in another rep. It is induced by the requirement of satisfying the Lie Algebra with the structure constants fixed by diagonalising the form in the adjoint. Your doubt in the comment above is essentially a rewording of my question! $\endgroup$
    – nox
    Oct 29, 2018 at 16:37

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