Are the Structure Constants $c^a_{bc}$ of a Lie Algebra always totally antisymmetric so,

$$ c_{abc} = c_{bca} = c_{cab} $$

Or is this just the case for semi-simple algebras?

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    $\begingroup$ If $x,y\in L$ then $[x,y] = -[y,x]$. $\endgroup$ Commented Oct 21, 2015 at 7:39
  • $\begingroup$ hey @NicolasBourbaki I'm a physicist rather than a mathematician so I don't really know what $L$ is here! Are you calling $L$ a Lie Algebra? $\endgroup$ Commented Oct 21, 2015 at 7:44
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    $\begingroup$ I added more details for you. $\endgroup$ Commented Oct 21, 2015 at 7:49
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    $\begingroup$ The answer you choose is only prove the antisymmetric in two indices. It's not the proof for totally antisymmetric. Only Idempotent's answer is correct. Totally antisymmetric is only right for compact and semisimple Lie algebra. $\endgroup$
    – 346699
    Commented May 2, 2017 at 21:52

4 Answers 4


Previous answers have made the point that total antisymmetry would be more than just $c^K_{IJ} = - c^K_{JI}$, but additionally invariance of the structure constants under cyclic permutations of the indices. You get this if your Lie algebra is compact and semisimple, but you have to choose the basis elements strategically for it to work.

If the Lie algebra $\mathfrak{g}$ is compact and semisimple, you can prove that the Killing form on $\mathfrak{g}$, namely the bilinear form

$ K(X,Y) \equiv \mbox{Tr} \, (XY) $

is negative definite (for a proof, see these notes by Peter Woit). This means that the negative of the Killing form gives you a positive definite inner product on your Lie algebra (recall that a Lie algebra is just a vector space with some extra structure, so it makes sense that one could use that extra structure to define a natural inner product on the Lie algebra. Note that even if you are not in the compact semisimple case, you can define the closely related trace inner product, or Hilbert-Schmidt inner product, $\langle X,Y \rangle_{Tr} \equiv \mbox{Tr} \, (X Y^\dagger)$).

Suppose that $\mathfrak{g}$ is compact and semisimple, and denote the inner product given by the negative of the Killing form by $\langle X,Y \rangle_K \equiv - \mbox{Tr} \, (XY)$. Choose a basis $\{ T_I \}$ for $\mathfrak{g}$ which is normalized with respect to $\langle \cdot , \cdot \rangle_K$, that is, $\langle T_I , T_J \rangle_K = \delta_{IJ}$. Now the structure constants in terms of this basis are defined by the relation

$[ T_J , T_K] = c^I_{JK} T_I$.

Multiplying both sides by another basis element $T_L$ and taking the trace yields

$\mbox{Tr} \, \left( T_L [T_J , T_K] \right) = c^I_{JK} \mbox{Tr} \, (T_L T_I) = c^I_{JK} (-\delta_{LI}) = - c^L_{JK}$.

On the other hand, you have

$\mbox{Tr} \, \left( T_L [T_I , T_J] \right) = \mbox{Tr} \, (T_L T_J T_K) - \mbox{Tr} \, (T_L T_K T_J) = \mbox{Tr} \, (T_L T_J T_K) - \mbox{Tr} \, (T_J T_L T_K) = \mbox{Tr} \, ([T_L,T_J] T_K) = \mbox{Tr} \, (c^M_{LJ} T_M T_K) = c^M_{LJ} (-\delta_{MK}) = - c^K_{LJ}$,

where the cyclic property of the trace has been used. So

$c^L_{JK} = c^K_{LJ}$.


They are, if you clarify what you mean by structure contants with three lower indices.

You can always use the Killing form to do it. Let $\mathfrak{g}$ denote your Lie algebra. Killing form $K$ is a symmetric bilinear form defined by

$K(x,y) = Tr(ad_{x} \cdot ad_{y})$,

where $ad_{x} \in End(\mathfrak{g})$ is the operator of the adjoint representation $ad_{x}(y) = [x,y]_{\mathfrak{g}}$.

One can show that the Killing form $K$ is ad-invariant, that is it satisfies the identity

$(\ast) \hspace{4mm} K([x,y]_{\mathfrak{g}},z) + K(y, [x,z]_{\mathfrak{g}}) = 0$,

for all $x,y,z \in \mathfrak{g}$. Now, choose a fixed basis $\{ t_{i} \}_{i=1}^{\dim{\mathfrak{g}}}$ of the vector space $\mathfrak{g}$. As already noted in the comments above, the structure constants in this given basis are defined by the equation

$[t_{i},t_{j}]_{\mathfrak{g}} = {c_{ij}}^{k} t_{k}$

However, having the Killing form, you can use it to lower one of the indices, that is define them using the equation

$c_{ijk} = K( [t_{i},t_{j}]_{\mathfrak{g}}, t_{k})$

These numbers indeed form a components of the completely skew-symmetric tensor on $\mathfrak{g}$. It is obviously skew-symmetric in the first two indices due to the skew-symmetry of $[\cdot,\cdot]_{\mathfrak{g}}$. It is easy excercise to see that the skew-symmetry in the other two indices (say the second and third) follows from the ad-invariance $(\ast)$.

Note that only for semisimple Lie groups, the set of numbers $c_{ijk}$ is equivalent to the structure constants ${c_{ij}}^{k}$. It can even happen that all of $c_{ijk}$ are zero.

  • $\begingroup$ +1. This should be the accepted answer because it points out that total anti-symmetry only holds for the structure constants with the index lowered, and that this follows immediately from the assumption of ad-invariance of the Killing form, making no reference at all to specific Lie algebras or the realization of the Killing form as a tr(XY) which only holds for some Lie algebras (as the accepted answer annoyingly does). $\endgroup$
    – dennis
    Commented Nov 9, 2023 at 21:58

Consider the unique non-abelian two-dimensional Lie algebra $L$. Give it a basis $x, y$ such that $[x, y] = y$.

$c^y_{xy} = 1$, but $c^x_{yy} = 0$.

Edit: I think it is important to note that this is not the standard definition of antisymmetry. See the other answer(s).

  • $\begingroup$ out of curiosity, why do you give it the basis $[x, y] = y$? Why not $[x,y]= c$? $\endgroup$ Commented Oct 21, 2015 at 7:53
  • $\begingroup$ For convenience alone ^_^. $\endgroup$ Commented Oct 21, 2015 at 7:54

Let $L$ be a Lie algebra, say over the real numbers $\mathbb{R}$. We assume that $L$ is a finite dimensional vector space. Let $x_1,...,x_n\in L$, be a basis for this Lie algebra.

The Lie bracket $[x_i,x_j]\in L$, and so we can write it as a combination of the basis vectors. That is, $$ [x_i,x_j] = \sum_k c^k_{i,j} x_k $$

Since the Lie bracket is anti-commutative, $$ [x_j,x_i] = -[x_i,x_j] \implies \sum_k c_{j,i}^k x_k = \sum_k -c_{i,j}^k x_k $$ Thus, $c^k_{i,j} = -c^k_{j,i}$

Note: I am assuming that is what you mean by "anti-symmetric", if not tell me, and I will delete this reply.

  • $\begingroup$ no this is great. I just wasn't sure if the rule only applied to semi-simple algebras or over all Lie algebras $\endgroup$ Commented Oct 21, 2015 at 7:52
  • $\begingroup$ @AlexanderMcFarlane As a physicist what book do you use to learn Lie algebra? $\endgroup$ Commented Oct 21, 2015 at 7:53
  • $\begingroup$ about 500 of them that's the problem! 85-90 hr a week course so no time to sit and think deeply about everything :( I found Group Theory for Physicists pretty good. I wish I had time to actually properly go through it rather than just lecture notes $\endgroup$ Commented Oct 21, 2015 at 7:56
  • $\begingroup$ This only shows that the structure constants $c^k_{ij}=:c_{ijk}$ are antiymmetric w.r.t. to the permutation of its first two indices, which is trivial. Totally antisymmetric means antisymmetric w.r.t. any permutation of two indices, so the proof is missing the non trivial part: proving $c_{ijk} = -c_{kji} = c_{kij}$. $\endgroup$
    – Albert
    Commented May 9, 2023 at 15:29

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