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Patera and Winternitz have carried out extensive classification of three and four dimensional Lie algebras. When I tried to look for classification for three dimensional Lie algebra with non-zero commutations given as:

$[e_{1}, e_{2}]=e_{2}, [e_{1}, e_{3}]=2\,e_{3} $

I could not find corresponding classification in table 1, similarly for four dimensional Lie algebra with non-zero commutation given as:

$[e_{1}, e_{3}]=e_{3}, [e_{1}, e_{4}]=e_{4}, [e_{2}, e_{3}]=e_{3}$

there is not any classification given there in table 2. It seems that they have missed these three and four dimensional algebras, I guess this might be due large number of possible Lie algebras of dimension three and four and it is natural for them to miss those algebras.

I wonder is there any limit to number of three and four dimensional Lie algebras with different non-zero commutations?

IgotiT
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  • have you shown, that your lie algebras are not isomorphic to the lie algebras in the tables? – Claudius Jun 06 '16 at 03:06
  • @user218931 I haven't tried that and I really don't know how to check for isomorphism between algebras. Can you help how I can check isomorphism? – IgotiT Jun 06 '16 at 03:29
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    The answer to the title question: There are infinitely many non-isomorphic Lie algebras of dimension $3$ resp. $4$ over a field $K$ of characteristic zero. Your first Lie algebra is $\mathfrak{r}_{3,1/2}$ in the classification list (in the book "Lie groups and Lie algebras III" by Vinberg and Onishchik). – Dietrich Burde Jun 06 '16 at 09:05
  • @DietrichBurde: Many many thanks, I was missing you on another query http://math.stackexchange.com/questions/1801804/does-solvability-of-lie-algebra-have-useful-application-in-study-of-pdes – IgotiT Jun 06 '16 at 10:35
  • @IgotiT The study of PDE's is not really my field. I can tell you why solvability is interesting and useful in general. – Dietrich Burde Jun 06 '16 at 11:23
  • @DietrichBurde: Suppose I add to new elements $e4$ and $e5$ in basis ${e1, e2, e3}$ of three dimensional algebra without altering the non zero commutations $[e_{1}, e_{2}]=e_{2}, [e_{1}, e_{3}]=2,e_{3} $, would that effect classification anyway ? – IgotiT Jun 06 '16 at 12:40
  • Yes, it would affect the classification. Then we would deal with $5$-dimensional Lie algebras, which are much more complicated (but have been classified, too). There are many possibilities so satisfy the Jacobi identity for $e_1,\ldots,e_5$, keeping $[e_1,e_2]=e_2,[e_1,e_3]=2e_3$. – Dietrich Burde Jun 06 '16 at 12:43
  • Let us continue this discussion in chat. – IgotiT Jun 06 '16 at 15:21

1 Answers1

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Your 3-dimensional Lie-algebra is isomorphic to $A_{3,5}^{1/2}$ in the table: Denoting by $(\mathfrak g,[.,.])$ your Lie-algebra (with non-zero relations $[e_1,e_2] = e_2$ and $[e_1,e_3] = 2e_3$) and by $(A_{3,5}^{1/2},(.,.))$ the Lie-algebra with non-zero relations $(e_1,e_3) =e_1$ and $(e_2,e_3) = \frac12 e_2$, then $$ \phi\colon \mathfrak g\longrightarrow A_{3,5}^{1/2},\quad e_1\mapsto -2e_3,\; e_2\mapsto e_2,\; e_3\mapsto e_1 $$ gives an isomorphism of Lie-algebras: Namely \begin{align*} \phi([e_1,e_2]) &= \phi(e_2) = e_2 = (e_2,2e_3) = (-2e_3,e_2) = (\phi(e_1), \phi(e_2))\\ \phi([e_1,e_3]) &= \phi(2e_3) = 2e_1 = (e_1,2e_3) = (-2e_3,e_1) = (\phi(e_1),\phi(e_3))\\ \phi([e_2,e_3]) &= 0 = (e_2,e_1) = (\phi(e_2),\phi(e_3)), \end{align*} i. e. $\phi$ preserves the Lie-brackets.

Your 4-dimensional Lie-algebra is isomorphic to $2A_2$, as can be seen by replacing your $e_1$ by $e_1-e_2$ (the non-zero relation then become $[e_1,e_4]=e_4$, $[e_2,e_3] = e_3$). I leave it to you, to write down the explicit isomorphism.

Claudius
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  • Can you please give me your email id so that I can acknowledge you in my work if needed ? My email id is [email protected], you can also mail me at this address. – IgotiT Jun 10 '16 at 05:48
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    Thank you kindly. But this solution is not enough to be acknowledged in your work. If you really want to give credit, you may of course refer to this answer. Please feel free to use it as you wish. – Claudius Jun 10 '16 at 06:20