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.

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    $\begingroup$ Do you want to assume furthermore that the Lie bracket is the usual commutator, $[a, b] := a b - b a$? You may be interested in the notion of universal enveloping algebra: en.wikipedia.org/wiki/Universal_enveloping_algebra $\endgroup$ – Travis Feb 19 '18 at 15:51
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    $\begingroup$ What examples of L did you consider? It is somewhat not trivial to find them, in fact! $\endgroup$ – Mariano Suárez-Álvarez Feb 19 '18 at 16:39
  • $\begingroup$ Are you sure that your Lie algebras are also rings? Your Lie algebra is certainly closed under the multiplication by the underlying field of scalars by definition (it's a vector space). But the Lie bracket fails associativity, which is usually required of the multiplication operation of a ring. Maybe this answer helps? math.stackexchange.com/questions/380177/… $\endgroup$ – Odds-Bodkins Feb 19 '18 at 17:09
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    $\begingroup$ I'm puzzled by your assertion that all your examples were multiplicatively closed. For example $sl_2(\Bbb R)$ is not multiplicatively closed in $M_2(\Bbb R)$ as it contains a fourth root of the identity... $\endgroup$ – Arnaud D. Feb 19 '18 at 17:42
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    $\begingroup$ @Odds-Bodkins Indeed I did not know that most of Lie algebras are not associative. I generate $L$ and then construct $G = e^L$, what made me thought the opposite. $\endgroup$ – Cassius Manuel Feb 21 '18 at 15:06

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:


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.

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    $\begingroup$ The last sentence does not make any sense, as Lie algebras a pre-Lie algebras... Also, the sentence «L is endowed with a biinvariant left symmetric algebras» seems incomplete, or something. $\endgroup$ – Mariano Suárez-Álvarez Feb 19 '18 at 16:39
  • $\begingroup$ why it does not make sense ? $\endgroup$ – Tsemo Aristide Feb 19 '18 at 16:41
  • $\begingroup$ Left symmetric algebras have been rediscovered by Loday or one of its fellows who were working in the theory of operads and they renamed it pre-Lie algebras. $\endgroup$ – Tsemo Aristide Feb 19 '18 at 16:44
  • $\begingroup$ @Mariano Suárez-Álvarez the theory of operads create by May ( I believe) was on the spotlight in the 90's thanks to Manin who has published a memoire in Montreal on quantum algebras an quadratic algebras, inspired by this work, Ginzburg and Kapranov have published a paper on quadratic operads which are renew the interest on operads. I remember that at that time i was working on that subject and I have generalized the tensor product defined by Maning in the category of quadratic algebras and by Ginzburg and Kapranov in the category of quadratic operads to the notion of quadratic categories. $\endgroup$ – Tsemo Aristide Feb 19 '18 at 18:24
  • $\begingroup$ This result was published in my D.E.A (diplôme d'études approfondies) During that time, Loday has published the notes of one of his Bourbaki seminar called La renaissance des opérades. It is this new interest of operads which lead Loday and his colleagues to rediscover the notion of left symmetric algebras that they called pre Lie algebras. $\endgroup$ – Tsemo Aristide Feb 19 '18 at 18:25

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