Is there any one dimensional subalgebra which is an Ideal of the two dimensional non-abelian Lie Algebra?

i.e. is it invariant as a subalgebra of the 2D non-abelian algebra

I read that "all the subalgebras of an abelian algebra are automatically invariant"~Mathematical Methods for Physicists - George B. Arfken

I would assume this includes the trivial case of the group itself.

However, it then occurred to me that this would only hold if I consider the algebra as a subalgebra of itself and not necessarily in its own right.

Furthermore, can you prove with with the structure constants, generators or other methods of Lie Algebras that this is the case.


What motivated this is a tutorial in which, I proved there are no 2 dimensional semi-simple Lie Algebras by showing the metric was not invertible.

Another definition of semisimple is that it cannot contain any invariant abelian subalgebras. The abelian 2D algebra is clearly not semisimple. However it becomes a bit less trivial with the non-abelian 2D algebra. Clearly the only non trivial subalgebra of the non abelian 2D is the 1D subalgebra. The 1D algebra ticks the box for being abelian but can I also show it is an invariant subalgebra of the non-abelian 2D Lie algebra.

If so I can then get peace of mind and show there are no semi-simple non-abelian 2D lie Algebras!


  • $\begingroup$ Sorry I thought the abelian case was trivial. I edited it to be the non-abelian case in the question $\endgroup$ – Alexander McFarlane Oct 22 '15 at 19:03
  • $\begingroup$ What do you mean by "the" one-dimensional subalgebra? Every one-dimensional vector subspace is a subalgebra. $\endgroup$ – Eric Wofsey Oct 22 '15 at 22:56
  • $\begingroup$ In the two dimensional case the only non trivial subalgebra is the one dimensional case. I want to know if as a subalgebra given the conditions it is an ideal $\endgroup$ – Alexander McFarlane Oct 23 '15 at 0:28
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    $\begingroup$ Er, I meant one-dimensional subalgebras of the 2-dimensional non-abelian algebra. The issue is your use of the word "the": any Lie algebra of dimension $>1$ has many different 1-dimensional subalgebras (uncountably many, even), and it is possible that some of them are ideals and others are not. $\endgroup$ – Eric Wofsey Oct 23 '15 at 0:40
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    $\begingroup$ Incidentally, there is a priori the same issue with saying "the" 2-dimensional nonabelian Lie algebra, but it turns out that all of them are isomorphic, so there is essentially only one. $\endgroup$ – Eric Wofsey Oct 23 '15 at 0:47

You can get peace of mind because there are no semisimple non-abelian 2D Lie algebras. Suppose that you have a Lie algebra $L$ of dimension $2$ with basis $\{e_1,e_2\}$. Then its commutator subalgebra $[L,L]$ has dimension $1$ (or dimension $0$, which would mean that $L$ is abelian which you have excluded). Indeed, $[e_1,e_2]=\alpha e_1+\beta e_2$, hence $[L,L]=\langle \alpha e_1+\beta e_2\rangle$ is $1$-dimensional. Every $1$-dimensional Lie algebra has trivial commutator, hence we have $[[L,L],[L,L]]=0$. This says that $L$ is $2$-step solvable and non-abelian. So it cannot be semisimple.

  • $\begingroup$ What does 2 step solvable mean with relevance to the semi-simple definition? I haven't come across this yet $\endgroup$ – Alexander McFarlane Oct 23 '15 at 18:27
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    $\begingroup$ If $L$ is semisimple then $[L,L]=L$, and $[[L,L],[L,L]]=L$, and not $0$. $\endgroup$ – Dietrich Burde Oct 23 '15 at 18:35

Ok so I have an answer which isn't very rigorous (standard physics) but satisfied me.

The (non-trivial) generators of the sub algebra of the 2D non-abelian algebra give,

$$ \tag{1} [T_a,T_b] = T_a $$ $$ \tag{2} [T_a,T_b] = T_b $$

This means we have two groups of the 1D subalgebra if we choose one (say for example $T_a$ here) of the generators to be the generator of the 1D subalgebra, we can see we have two groups of the 1D subalgebra.

  • Equation 1, which is invariant
  • Equation 2, which is non-invariant

Obviously, there are two other cases if you pick $T_b$ as the 1D subalgebra generator.

Let me know if anyone has any discrepancies with this argument. Love feedback cheers!


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