A group presentation is defined as a collection of generators (such that every element is a product of powers of those) and relations between the generators such that the group is uniquely defined by those generators and their relations.

My questions addresses the proper definition of presentations of Lie algebras. Are they defined completely analoguously such that we have

  • A set of generators $X_{i}$ (such that every element is a linear combination of the $X_i$ and can be constructed from the Lie bracket of the $X_i$?)
  • The Lie bracket relations between the $X_i$?

Would this suffice as a definition of a Lie algebra presentation or would there be missing something?

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    $\begingroup$ That's pretty much it: see section 18 of Humphreys's book to see a presentation of each complex semisimple Lie algebra. $\endgroup$ – Lord Shark the Unknown Apr 20 '17 at 7:39
  • $\begingroup$ Is complex semisimplicity necessary or does it work for an arbitray Lie algebra? $\endgroup$ – Jakob Elias Apr 20 '17 at 9:23
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    $\begingroup$ No, it works in general. Humphreys is just doing a very important example. $\endgroup$ – Lord Shark the Unknown Apr 20 '17 at 9:25
  • $\begingroup$ An important example is the class of quadratically presented Lie algebras, for which the relators have the form $\sum\lambda_{i,j}[x_i,x_j]$ for generators $x_i$. $\endgroup$ – YCor Apr 20 '17 at 12:58
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    $\begingroup$ Btw presentation just means that we have a free Lie algebra and generators of some ideal of relations, called relators. It works in other contexts, associative algebras, etc. Also there are semigroups presentations, except that in this case relators are not elements, but rather a list of equalities between given elements, e.g., $\langle x,y|x^2=y^3\rangle$. $\endgroup$ – YCor Apr 20 '17 at 13:01

That's correct, but I would make what the $X_i$ are more clear. A Lie algebra is a vector space $V$, with Lie bracket $[\cdot, \cdot]: V \times V \to V$, so I would say a presentation is

  1. A basis $\{X_i\}$ of $V$; along with
  2. A collection of scalars $a_{ij}^k$ such that $[X_i, X_j] = \sum_k a_{ij}^k X_k$.

Which informally, is just some basis $\{X_i\}$ of $V$ that you know the Lie bracket on. The scalars $a_{ij}^k$ are called the structure constants of the Lie algebra, and they must satsify certain relations, due to the fact that they come from a Lie bracket, not any scalars will do. For example, we must have $a_{ii}^k = 0$, and $a_{ij}^k = -a_{ji}^k$, amongst others.

There is a different notion of just bare generators and relations, which is closer to a group presentation. But, like group presentations, they can be quite difficult to work with, and you might end up having to send some of your generators $X_i$ to $0$ in order to satisfy the relations.

  • $\begingroup$ Also worth pointing out that those scalars are called the structure constants. $\endgroup$ – Tobias Kildetoft Apr 20 '17 at 8:26
  • $\begingroup$ Though this is not quite analoguous to a group presentation unless you drop the requirement of the elements forming a basis, since otherwise there are choices of constants not resulting in a valid Lie algebra, while all choices of relations result in a valid group. $\endgroup$ – Tobias Kildetoft Apr 20 '17 at 8:27
  • $\begingroup$ @TobiasKildetoft: I'll add in that they are called the structure constants. As for your second comment, I agree that it's not the same style as a group presentation, but it's what I most often hear called a "lie algebra presentation", and what I interpreted as being closest to the question asked. $\endgroup$ – Joppy Apr 20 '17 at 9:01

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