# A class of Lie groups with $f^{abc} \neq -f^{acb}$ (not fully anti-symmetrized) or $f^{abc} \neq f^{bca}$ (not-cyclic)

With the motivation to understand the Lie group structure constraint on a non-Abelian Chern-Simons theory, could some experts give a class of Lie groups with structure constants cannot fully anti-symmetrized in any choice of basis into

$$f^{abc} \neq -f^{acb}$$

or cannot obey cycic condition in any choice of basis

$$f^{abc} \neq f^{bca}$$

, please also with some explicit examples?

I read a note from a QFT class of Barton Zwiebach, learned that this condition $f^{abc} =-f^{acb}$ where compact semi-simple Lie algebra can always be taken to satisfy it.

(ps. Here the commutator of generators is $$[T^a,T^b]=if^{abc}T^c$$ So, $f^{abc}=-f^{bac}$ by definition.)

How this relates to real/complex, compactness, simple or not, simply-connected or not of Lie groups? Many thanks.

• Structure constants are not a property of a Lie group or even of a Lie algebra, they are a property of a basis of a Lie algebra. Also, I don't know what convention you're using for indices because you've written them all as having the same variance. Do you mean $[x_a, x_b] = \sum f^{abc} x_c$ where $x_i$ is a basis, and if so, shouldn't the RHS be written $\sum f_{ab}^c x_c$? Or are you also implicitly using a nondegenerate Killing form somewhere? – Qiaochu Yuan Dec 21 '13 at 5:45
• I had clarified. I am asking when a Lie group is NOT possible to choose a Lie algebra basis such that $f^{abc}=f^{acb}$. Could you provide explicit examples? – cycles Dec 21 '13 at 5:55
• This condition confuses me. For a compact Lie algebra the natural condition is $f^{abc} = -f^{acb}$ as far as I can tell (it holds whenever you take the basis to be orthonormal with respect to the negation of the Killing form). Can you check that this is the condition you care about? – Qiaochu Yuan Dec 21 '13 at 6:21
• @cycles: Qiaochu is right. Actually, there is no homomorphism from the permutation group on $\{1,2,3\}$ to $\{-1,1\}$ mapping $(12)$ to $-1$ and $(23)$ to $+1$. As a consequence, a Lie algebra with basis satisfying your condition $f^{abc}=f^{acb}$ is necessarily abelian (because $[T^a,T^b]_c=[T^a,T^c]_b=-[T^c,T^a]_b=-[T^c,T^b]_a=[T^b,T^c]_a=[T^b,T^a]_c=-[T^a,T^b]_c$.) My answer actually concerns the much more relevant condition $f^{abc}=-f^{acb}$. – YCor Dec 21 '13 at 10:40
• Yes, I totally mess up last night (was too sleepy), what I look for is exactly $f^{abc} \neq - f^{acb}$ or $f^{abc} \neq f^{bca}$. Thank you for pointing out. – cycles Dec 21 '13 at 18:10

(I assume the ground field is an algebraically closed field of characteristic zero. It would have been useful to specify which field you have in mind.) Your condition amounts to asking whether there exists a invariant nondegenerate quadratic form, i.e. a nondegenerate bracket $\langle\cdot,\cdot\rangle$ satisfying $$\langle x,[y,z]\rangle=\langle [x,y],z\rangle,\quad\forall x,y,z,$$ or equivalently such that the trilinear form $(x,y,z)\mapsto \langle x,[y,z]\rangle$ is alternating. Such Lie algebras have several names in the literature, including "quadratic Lie algebras", "metric Lie algebras", "self-dual Lie algebras". (Beware that there is sometimes some ambiguity on whether this denotes a Lie algebra endowed with such a bilinear form, or a Lie algebra admitting at least such a form.) Anyway there is a lot of available information.

Here is now a few examples of Lie algebra with no such structure: the first two are based on the observation that, given a quadratic Lie algebra, the orthogonal of the derived subalgebra is necessarily equal to the center. Thus it is enough to find a finite-dimensional Lie algebra such that the dimension of the center and the codimension of the derived subalgebra do not coincide.

1) For instance, the 2-dimensional non-abelian Lie algebra cannot be made (non-degenerate) quadratic.

2) A nilpotent example is the Heisenberg Lie algebra with basis $(e_1,e_2,e_3)$ with $[e_1,e_2]=e_3$ and all other brackets zero: the derived subalgebra has codimension 2, while the center has dimension 1. More generally, if $d\ge 3$, if we consider the standard filiform Lie algebra with basis $(e_1,\dots e_d)$ with $[e_1,e_i]=e_{i+1}$ for $i=2\le i<d$ and all other brackets zero, the center has derived subalgebra has codimension 2 while the center has dimension 1. Also in the $(2m+1)$-dimensional Heisenberg algebra ($2m+1\ge 3$), the center and the derived subalgebra coincide and are 1-dimensional, so the codimension is $2m\neq 1$ and again there is no nondegenerate invariant bilinear form.

3) Also, here is a third counterexample, in which the above argument does not work (the Lie algebra is perfect and the center is zero): the semidirect product $\mathfrak{g}=\mathfrak{sl}_2\ltimes \mathfrak{v}$, where $\mathfrak{v}$ is a finite-dimensional irreducible representation of dimension $\neq 1,3$ (for instance, the standard 2-dimensional representation). Then the only ideals of $\mathfrak{g}$ are $\mathfrak{g}$, $\mathfrak{v}$ and $\{0\}$. But if there were a nondegenerate invariant quadratic form, the orthogonal of $\mathfrak{v}$ would be an ideal of the complementary dimension, which does not exist since $\dim(\mathfrak{v})\neq 3$.

In case you are working over an arbitrary field (e.g., the reals), you are asking whether there is a nondegenerate quadratic form which is in addition equivalent to the standard scalar product, but this is even more restrictive.

• Dear Yves, do you know whether those examples given by you can form a required [group structure in Chern-Simons theory]( physics.stackexchange.com/questions/90946/…)? – wonderich Dec 23 '13 at 1:07
• PS: I checked on the web: metric Lie algebra also appears in the literature as a real Lie algebra endowed with a scalar product, not assumed invariant (this is natural in the study of left-invariant Riemannian structures on Lie groups). – YCor Dec 27 '13 at 12:32