0
$\begingroup$

Is it possible that in a lie algebra one may have a generator that does not appear on any of the commutators?

If the above line is too physicisty for you, maybe I can try to translate to a more mathematicians question:

A lie algebra is a vector space $\mathfrak{g}$ equipped with an commutator map from $\mathfrak{g}\times\mathfrak{g}\longrightarrow\mathfrak{g}$ that satisfies the Jacobi identity. Is it possible that the range of this map to be a proper subset of $\mathfrak{g}$?

I hope the last question conveys the same message. Any reference or link will be of help too.

$\endgroup$
3
  • 1
    $\begingroup$ If I understand your question correctly, that happens quite often, an extreme case being that the Lie algebra is commutative, i.e. $[x,y]=0$ for all $x,y$. $\endgroup$ Jan 4, 2019 at 17:34
  • $\begingroup$ @TorstenSchoeneberg I was about to post that as an answer. Why don't you do that? $\endgroup$ Jan 4, 2019 at 17:34
  • $\begingroup$ @JoséCarlosSantos: I wanted to leave the possibility open that I misunderstood or the question was misphrased and something more profound was asked. But I've upgraded it to an answer now. $\endgroup$ Jan 4, 2019 at 17:55

1 Answer 1

4
$\begingroup$

It happens quite often that the image of

$$ [ \cdot, \cdot]: \mathfrak{g} \times \mathfrak{g} \rightarrow \mathfrak{g}$$

is a proper subset of $\mathfrak{g}$. An extreme case is given by commutative (also called abelian) Lie algebras, where the Lie bracket is constant $=0$, in other words the above image is just $\lbrace 0 \rbrace$ (but $\mathfrak{g}$ can be any vector space you like).

Other examples are all nilpotent and even all solvable Lie algebras.

It is worth noting that in general, the image of the Lie bracket is not even a vector subspace (scalar multiples are obviously fine, but the sum of commutators has no reason to be a commutator itself). The case that the space generated by the image (which by abuse of notation is usually denoted $[\mathfrak{g}, \mathfrak{g}]$, whereas formally it should be something like "$\langle[\mathfrak{g}, \mathfrak{g}]\rangle$") is all of $\mathfrak{g}$ has a special name: Such a Lie algebra is called perfect.

Now note that your condition is even more restrictive than being perfect! (For examples of perfect Lie algebras where not every element is a commutator, cf. YCor's answer here: https://math.stackexchange.com/a/2365525/96384.)

As Tobias Kildetoft points out in a comment, on the other hand, a big class of Lie algebras for which indeed each element is a commutator are split semisimple Lie algebras, at least if the cardinality of your ground field is big enough (certainly over all infinite fields, and since you say you're a physicist, I assume you are interested only in $\Bbb R$ or $\Bbb C$ anyway). This is a non-trivial theorem. See the reference in Dietrich Burde's answer here: https://math.stackexchange.com/a/771802/96384. For an elementary but by no means trivial proof just for the case of $\mathfrak{sl}_n(\Bbb R)$ and $\mathfrak{sl}_n(\Bbb C)$, see user1551's answer here: https://math.stackexchange.com/a/252324/96384

$\endgroup$
1
  • $\begingroup$ Though it is worth noting that the condition does hold for semisimple Lie algebras (at least as far as I recall, and I don't recall why it holds). $\endgroup$ Jan 4, 2019 at 19:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .