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A formal group $F$ is an intermediate object between Lie group $G$ and Lie algebra $\mathfrak{g}$.

A formal group is called simple if it has no nontrivial sub-formal group. On the other hand, a simple Lie group is more or less defined in terms of whether the associated Lie algebra is simple (e.g., section 4).

More precisely, let $F$ be a formal group over a ring $R$ and $$F(X,Y)=X+Y+\underset{\text{quadratic terms}}{F_2(X,Y)}+\text{higher degree terms}$$ be a formal group law representing $F$, the Lie bracket is defined by $$[x,y]=F_2(x,y)-F_2(y,x).$$ This is how a formal group (law) gives rise to a Lie algebra.

As Wikipedia says, over characteristic $0$ field, formal group laws are essentially the same as finite-dimensional Lie algebra, my question is:

Does a simple formal group over a characteristic $0$ field or ring give rise to simple Lie algebra?

Thanks

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    $\begingroup$ I don't think simplicity of a Lie group is defined by whether the Lie algebra is simple. I think that most people wouldn't call $\operatorname{SL}_2(\mathbb R)$ simple, since it has $\{\pm1\}$ as a non-trivial normal subgroup, but it does have simple Lie algebra $\mathfrak{sl}_2(\mathbb R)$. (Despite this we would call it semisimple, or, to emphasise that it's semisimple with only one factor, sometimes ‘almost simple’.) $\endgroup$
    – LSpice
    Commented Oct 29 at 10:46
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    $\begingroup$ @LSpice, thanks for the comment. I found something in the answer here. You may be right about it. Anyway, if you keep aside the Lie group here, my question is what is the relationship between a simple formal group and it's Lie algebra. Does the property of a formal group of being "simple" transfer to its Lie algebra? I think there might not be a direct relationship but I may be wrong because how can the coefficients of quadratic terms of a formal group law determine if it's Lie algebra is simple. $\endgroup$
    – Learner
    Commented Oct 29 at 11:54
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    $\begingroup$ @LSpice This is the exact opposite of my experience, which is that most people would call $SL_2(\mathbb R)$ simple. The Wikipedia page on simple Lie groups gives the definition that "a simple Lie group is a connected non-abelian Lie group $G$ which does not have nontrivial connected normal subgroups" which fits with this. $\endgroup$
    – Will Sawin
    Commented Oct 30 at 11:53
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    $\begingroup$ I believe over $\mathbb Z_p$ it's possible to have a simple formal group of dimension $>1$ whose underlying Lie algebra is commutative, and therefore certainly not simple. For a concrete example if you take an abelian variety of genus $3$ with good reduction and Newton polygon of Frobenius with slopes $1/3, 2/3$ then the formal group law will split at most into $1$-dimensional and $2$-dimensional pieces, but will certainly be commutative. $\endgroup$
    – Will Sawin
    Commented Oct 30 at 12:33
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    $\begingroup$ I was definitely too hasty in saying what most people would do! I think I should instead have said that, in the context of algebraic groups, I think that it is reasonably common to call the algebraic group $\operatorname{SL}_2$ both semisimple and almost-simple, and just to avoid the term simple as somewhere between confusing and useless. Historically, Borel–Tits do this; and, contemporaneously, Milne does. (I don't regard Wiki's definition as saying much about mathematical practice, but I do attach weight to the experience of my colleagues here!) $\endgroup$
    – LSpice
    Commented Oct 30 at 13:10

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