# Are there indecomposable unsolvable four and five dimensional Lie algebras?

In many places one can find an information that real Lie algebras are classified up to dimension 5. The only reference containing a classification I was able to find is Invariants of real low dimensional Lie algebras. The list presented there contains indecomposable Lie algebras of dimension up to five, yet every for and five dimensional algebra presented there is solvable. Is it true that there are no indecomposable, unsolvable algebras in that dimensions or the article very implicitly deals in fact only with solvable algebras.

An answer or a solid reference for a classification will be very appreciated.

Every finite dimensional real Lie algebra has the form $\mathfrak g=\mathfrak l\ltimes\mathfrak r$ where $\mathfrak l$ is semisimple and $\mathfrak r$ is solvable (Levi decomposition). The smallest semisimple Lie algebras are $\mathfrak{su}(1,1)=\mathfrak{sl}(2,\mathbb R)$ and $\mathfrak{su}(2)$, both of dimension $3$. The next larger ones have dimension $6$. So if $\mathfrak g$ is non-solvable of dimension $\le5$ then $\mathfrak l$ is one of the two $3$-dimensional ones. The indecomposability implies that $\mathfrak l$ acts non-trivially on $\mathfrak r$. From this, one gets easily that $\mathfrak{su}(1,1)$, $\mathfrak{su}(2)$, and $\mathfrak{sl}(2,\mathbb R)\ltimes\mathbb R^2$ are all the non-solvable indecomposable Lie algebras of dimension $\le5$. The last algebra is item $A_{5,40}$ of Table II of loc. cit. which is incorrectly labeled as solvable.

As already mentioned, an easy argument using the Levi decomposition shows that there are three distinct indecomposable non-solvable Lie algebras of dimension $5$.
However, I think it is worth to point out in this context that a full classification of all real $5$-dimensional Lie algebras has been given by G.M. Mubarakzyanov in $1963$, Izv. Vyssh. Uchebn. Zaved. Mat. 34. Besides the indecomposable non-solvable algebras we also have of course the decomposable non-solvable Lie algebras $\mathfrak{sl}(2,\mathbb R)\times \mathfrak{r}_2(\mathbb{R})$ and $\mathfrak{sl}(2,\mathbb R)\times \mathbb{R}^2$, where $\mathfrak{r}_2(\mathbb{R})$ denotes the non-abelian Lie algebra of dimension $2$.
The Lie brackets for $\mathfrak{sl}(2,\mathbb R)\ltimes\mathbb R^2$, in a basis $e_1,\ldots e_5$ are given by $$[e_1,e_2]=e_3,\; [e_1,e_3]=-2e_1,\; [e_1,e_5]=e_4, \; [e_2,e_3]=2e_2,\; [e_2,e_4]=e_5,$$

$$[e_3,e_4]=e_4,\; [e_3,e_5]=-e_5.$$

A further reference is the paper of Roman Popovych (2003).

• Of course you don't need the classification for the problem given here. It's an easy exercise solved in Friedrich's answer, only relying on Levi decomposition and classification of simple real Lie algebras of dimension $\le 5$. – YCor Apr 10 '17 at 15:28
• It may be worth pointing out that the 2003 paper by Popovych and others was published in that year in a math physics journal: ams.org/mathscinet-getitem?mr=2004893 Even though Lie himself had classified 4-dimensional real Lie algebras, the authors seem to give a more usable account. – Jim Humphreys Apr 10 '17 at 15:30

The classification of indecomposable non-solvable Lie algebras in dimension $\le 7$ over a field of characteristic zero is a simple exercise.

First, the semisimple ones have dimension 3 or 6: in dimension 3, $\mathfrak{so}(q)$ for quadratic forms in 3 variables, in dimension 6, the same over a quadratic extension, or products of two simple ones of dimension 3.

Next we use a Levi decomposition $\mathfrak{s}\ltimes\mathfrak{r}$. If $\mathfrak{s}$ has dimension 6, $\mathfrak{r}$ has dimension $\le 1$ and is a direct factor.

If $\mathfrak{s}$ has dimension 3, one has to discuss on $\mathfrak{r}$. If $\mathfrak{r}$ is abelian, it comes with a representation of $\mathfrak{r}$, which splits into irreducibles. If a 1-dimensional rep occurs, then it is a direct factor, so in the indecomposable cases the only possibilities are 2,3,4 and 2+2. (Note that if $\mathfrak{s}$ has a nontrivial 2-dimensional rep, then it is isomorphic to $\mathfrak{sl}_2$; the only 3-dimensional irreducible is the adjoint rep. In dim 4, I'm not sure in general fields, but in the real case both the split form and the non-split form have a unique 4-dimensional irreducible rep, which in the case of $\mathfrak{su}_2$ is not absolutely irreducible).

If $\mathfrak{r}$ is nilpotent and not abelian then it is (by the classification of nilpotent Lie algebras of dimension $\le 4$, either $\mathfrak{h}_3$ (Heisenberg), or $\mathfrak{h}_3\times\mathfrak{a}_1$ (its product with a 1-dimensional abelian) or $\mathfrak{f}_4$ (filiform of dimension 4). In the last case, the derivation algebra is solvable and hence $\mathfrak{r}$ is a direct factor. In the first two cases, either $\mathfrak{r}$ is a direct factor, or we have the following: modulo the center, we have an irreducible 2-dimensional $\mathfrak{s}$-module (hence $\mathfrak{s}$ is $\mathfrak{sl}_2$); then this lifts to an irreducible 2-dimensional submodule in $\mathfrak{r}$, and we see we have the (unique) nontrivial semidirect product $\mathfrak{sl}_2\ltimes\mathfrak{h}_3$, or its direct product with $\mathfrak{a}_1$.

The last case is when $\mathfrak{r}$ is not nilpotent. In a solvable Lie algebra $\mathfrak{r}$ with nilpotent radical, it is easy to check that $\dim(\mathfrak{n})\ge\dim(\mathfrak{r})/2$, with equality only for powers of the 2-dimensional nonabelian Lie algebra. The latter has a solvable derivation algebra and hence only occurs as direct factor. Otherwise in our case the pair $(\dim(\mathfrak{r}),\dim(\mathfrak{n}))$ is $(4,3)$ or $(3,2)$. Since in these cases $\mathfrak{n}$ has codimension 1, it has a $\mathfrak{s}$-invariant complement. So the Lie algebra has the form $(\mathfrak{s}\times\mathfrak{a}_1)\ltimes\mathfrak{n}$, where $\mathfrak{a}_1$ does not act on $\mathfrak{n}$ in a nilpotent way, and acts in the centralizer of the $\mathfrak{s}$-action. Excluding the case with abelian direct factors, we find $\mathfrak{gl}_2\ltimes\mathfrak{h}_3$, $(\mathfrak{s}\times\mathfrak{a}_1)\ltimes\mathfrak{v}$ where $\mathfrak{v}$ is an irreducible of dimension 2 or 3 and $\mathfrak{a}_1$ acts by scalar multiplication.

So the list (dimension in parentheses):

• (3) $\mathfrak{s}$ simple of dimension 3 ($\mathfrak{so}$ of some 3-dimensional quadratic form, unique up to scalar multiplication and equivalence: in the real case, $\mathfrak{sl}_2=\mathfrak{so}(2,1)$ or $\mathfrak{su}_2=\mathfrak{so}(3)$)
• (6) $\mathfrak{s}$ simple of dimension 6 (idem over a quadratic extension of the ground field; in the real case: $\mathfrak{sl}_2(\mathbf{C})=\mathfrak{so}_3(\mathbf{C})$)
• (5) $\mathfrak{sl}_2(K)\ltimes K^2$
• (7) $\mathfrak{sl}_2(K)\ltimes (K^2\oplus K^2)$
• (6) $\mathfrak{sl}_2\ltimes\mathfrak{h}_3$
• (6) $\mathfrak{gl}_2(K)\ltimes K^2$
• (7) $\mathfrak{gl}_2\ltimes\mathfrak{h}_3$
• (6) $\mathfrak{s}\ltimes V_\mathfrak{s}$ where $\mathfrak{s}$ is simple of dimension 3 and $V_\mathfrak{s}$ is the adjoint representation of $\mathfrak{s}$
• (7) $(\mathfrak{s}\times\mathfrak{a}_1)\ltimes V_\mathfrak{s}$ where $\mathfrak{s}$ is 3-dimensional simple, $\mathfrak{a}_1$ acts by scalar multiplication
• (7) $\mathfrak{s}\ltimes V$, $V$ irreducible of dimension 4. To be more precise, in the real case, the only case beyond the split case $\mathfrak{sl}_2(K)\ltimes\mathrm{Sym}^3(K^2)$ is $\mathfrak{su}_2\ltimes\mathbf{C}^2$.
• "The classification of indecomposable non-solvable Lie algebras in dimension $≤7$ over a field of characteristic zero is a simple exercise" that puzzles me because from this one aip.scitation.org/doi/abs/10.1063/1.528721 we have that the classification of six dimensional solvable Lie algebras is available, since You argue that non-solvable indecomposable Lie algebras in dimension six are also easy to classifie than by adding decomposable 6 dimensional non-solvable (we have classification for ≤5) we would obtain classification of six dimensional ones, am I wrong? – J.E.M.S Apr 10 '17 at 18:52
• @J.E.M.S Yes you're right. But it's well-known that for fixed small dimension, the most difficult part is the solvable case. – YCor Apr 11 '17 at 17:15
• My point was that if we could obtain a list of in-decomposable non-solvable lie algebras in dimension 6 than we would have then full classifications of Lie algebras in dimension six (since we have classification of solvable ones) but (due to my knowledge, please correct me if I am wrong) there is no classification of real Lie algebras in dimension three. – J.E.M.S Apr 19 '17 at 14:07
• In dimension 3 of course the classification of real Lie algebras is known, and it's not hard (google can easily find references). Beyond the two simple ones, there are the solvable ones, all of the form $\mathbf{R}\ltimes\mathbf{R}^2$, whose classification is an exercise (= classify $2\times 2$ matrices up to nonzero scalar multiplication and conjugation). – YCor Apr 19 '17 at 14:59