It is well known that properties of Lie groups can be studied through those of their associated Lie algebra. I am interested here in $\mathbb{C}$-vector spaces and the finite-dimensional case. In that context, assume we have a Lie group $G$ with $L$ being its Lie algebra. For simplicity, let us assume, moreover, that $L$ is a subset of $M_n(\mathbb{C})$, and the Lie bracket is induced by the multiplication of matrices.
Edit: I am referring to $[A,B] = AB - BA$.
In this context, along my late (and short) work on particular examples, I am constructing only $L$'s which were not only Lie algebras, but also multiplicatively closed (under the regular multiplication of matrices), i.e. they all were associatve rings.
Edit: An example of how I generate $L$ would be the following: For a given vector $v$, I consider a set of $m$ matrices $M_i$, $i\in[m]$, satsfying $M_iv = 0$. The $\mathbb{C}$-vector space generated by $\{M_i\}$ will be my $L$, i.e. $L = <\{M_i\}>_{\mathbb{C}}$. Now let us assume the multiplicative closeness, i.e. that they satisfy $M_i M_j \in L $. I think that I could get some advantages on $G$ after all these assumptions.
Therefore I wonder:
- Assuming that $L$ is a (associaitve) ring, what can we say about $G$?.
- In the particular case of the construction described above, in which $L = <\{M_i\}>_{\mathbb{C}}$, $M_iv = 0$ for all $i$ and $M_i M_j \in L $, what could we say about $G = e^L$?
That's all, thanks for your help and sorry for the editing mess.
If $L$ is a Lie subalgera of $M_n(\mathbb{C})$ such that for every $A,B\in L, AB\in L$ where $L$ is endowed with the canonical multiplication of matrices, then $L$ is endowed with a biinvariant left symmetric algebras. Left symmetric algebras (recently renamed) pre-Lie algebra where defined by Koszul and Vinberg in the 50's of the last century.
They are defined by $a(bc)-(ab)c=b(ac)-(ba)c$ and $ab-ba=[a,b]$ as you see associative algebras are particular examples. Geometrically, this is equivalent to endow $G$ with a connection whose curvature and torsion form vanish identically, or with a left invariant atlas whose coordinate change are affine maps. The link between the both notion is like so:
Given a left symmetric algebra, you obtain a representation $L\rightarrow aff(L)$ defined by $a\rightarrow (a,L_a)$ where $L_a(b)=ab$. This representation induces an affine representation of $f:G\rightarrow Aff(L)$ the simply connected Lie group whose Lie algebra is $L$ which has an open orbit (the orbit of $0)$. The map $D:G\rightarrow L$ defined by $D(g)=f(g)(0)$ is called the developing map. The left invariant affine connection of $G$ is the pullback of the standard connection of the vector space $L$ by $D$, and conversely, a representation $D:G\rightarrow Aff(\mathbb{C}^n)$ (where $n$ is the dimension of $L$) which has an open orbit defines a left symmetric algebra.
To see that to define a biinvariant left symmetric algebra, it is equivalent to define an associative algebra, remark that biinvariant means that the left symmetric product is invariant by the adjoint: A Left invariant connection $\nabla$ defined on $G$ induces a product on $L$ defined by $ab=\nabla_ab$. The connection is biinvariant implies that it is invariant by the left and right multiplication, which is equivalent to see that it is invariant by the adjoint: it can be written like so:
Let $a,b,c\in L$ and $exp_t(a)$ the flow of the vector field $a$, write $\phi_t(g)=exp_t(a)gexp_{-t}(a)$, $\phi_t^*\nabla=\nabla$ if $\nabla$ is biinvariant. By differentiating $\phi_t^*\nabla_bc=\nabla_{\phi_t^*b}\phi_t^*c$ at the identity of $G$, you obtain:
$[a,bc]=[a,b]c+b[a,c]$
you can rewrite it:
$a(bc)-(bc)a=(ab-ba)c+b(ac-ca)$ which is equivalent to say
$a(bc)-(ab)c-(b(ac)-(ba)c)=(bc)a-b(ca)$ since $L$ is left symmetric, you deduce that $a(bc)-(ab)c-(b(ac)-(ba)c)=0$ and $b(ca)=(bc)a$ which is equivalent to saying that $L$ is associative.
It has been shown (more than 50 years ago) in the 60's of the last centuries that a non zero left symmetric algebra is not semi-simple. This is a consequence of the Whitehead's lemma which says that $H^1(L,.)=0$ if $L$ is semi-simple, saying that $ab-ba=[a,b]$ is equivalent to saying that the identity map of $L$ is a $1$-cocycle for the representation $a\rightarrow L_a$, since $H^1=0$, it is a boundary, there exits $e\in L$ such that $a=ae$, but we have $[a,e]=ae-ea$, this implies that $Id_L=L_e-ad_e$, but since $L$ is semi-simple its image by $L$ is contained in $sl(L)$ and $ad_e\in sl(L)$, we deduce that $tr(Id_L)=tr(L_e)-tr(ad_e)=0$ which implies that $L=\{0\}$. Contradiction.