Is a formal deformation of a Lie algebra an example of a formal group law?

Question

I stumbled across the following definition of the formal deformation of a Lie algebra, and it looks like a group object in the category of formal schemes (not necessarily commutative or 1-dimensional). I would really like to know if this is a correct interpretation.

Consider a Lie algebra $\mathfrak{g}$ of dimension $N$ over an arbitrary field $k$. Let us denote the basis elements of $\mathfrak{g}$ by $\{x_1,..., x_N\}$, and write the Lie bracket as $$[x_i, x_j] = C^k_{ij}x_k$$

where the coefficients $C^k_{ij}$ are the structure constants. We denote by $L_N(k)$ the space of structural tensors of N-dimensional Lie algebras. Then a one-parameter deformation of a Lie algebra g, whose structure constants belong to $L_N (k)$, is a continuous curve over $L_N (k)$. (...)

A formal one-parameter deformation is defined by the Lie brackets: $$[a,b]_t = F_0(a,b) + tF_1(a,b) + ... +t^mF_m(a,b)$$

where $F_0$ denotes the original Lie bracket $[-, -]$. Jacobi identity implies relations between the tensors $F_m$. The first such deformation relation is that $F_1$ must be a two-cocycle of $\mathfrak{g}$. We call $[-,-]_t$ a first-order, or infinitesimal, deformation if it satisfies the Jacobi identity up to $t^2$. It follows that first-order deformations correspond to elements of the space of two-cocycles $Z^2(\mathfrak{g},\mathfrak{g})$. A deformation is called of order $n$ if it is defined modulo $t^{n+1}$.

Consider now a deformation $\mathfrak{g}_t = [-,-]_t$ not as a family of Lie algebras, but as a Lie algebra over the ring $k[[t]]$ of formal power series over $k$.

A natural generalization is to allow more parameters, which amounts to consider $k[[t_1, . . . , t_k]]$ as the base (or, even more generally, to take an arbitrary commutative algebra $A$ over $k$, with unit as the base).

Assume that A admits an augmentation $\epsilon: A \to k$, such that $\epsilon$ is a $k$-algebra homomorphism. The ideal $m_\epsilon := ker(\epsilon)$ is a maximal ideal of $A$, and, given a maximal ideal $m$ of $A$ with $A/m \simeq k$, the natural quotient map defines an augmentation. If $A$ has a unique maximal ideal, the deformation with base $A$ is called local. If $A$ is the colimit of local algebras, the deformation is called formal.

-- Deformations and Contractions of Lie algebras, Fialowski and Montigny

Here's my question: Does a formal power series define a formal deformation of a Lie algebra?

Answer 1

Yes, it does define a formal deformation of a Lie algebra in the following sense: let $\mathfrak{g}_0$ be a Lie algebra over a field $k$ and $A$ be a $k$-algebra with specified point $t_0\in {\rm Spec}(A)$ and residue field $k_{t_0}=k$. A deformation of $\mathfrak{g}_0$ is a Lie algebra $\mathfrak{a}$ over $A$ together with an isomorphism of Lie algebras over $k$, $$ \phi\colon \mathfrak{g}_0\rightarrow \mathfrak{a}\otimes_A k. $$ The Lie algebra $\mathfrak{a}_k=\mathfrak{a}\otimes_A k$ is called the limit algebra or the special fibre of the deformation $\mathfrak{a}$.

So a formal deformation of $\mathfrak{g}_0$ can be a deformation over, say, the ring $A=k[[t]]$ of formal power series. This ring is uniquely determined as a complete regular $1$-dimensional local $k$-algebra.

For references on deformations, contractions and degenerations of Lie algebras and algebraic groups, see here.