Restricting the Lie bracket If we have a Lie algebra $\mathfrak{g}$ over a field $k$ with a Lie bracket $[-,-]$, then is there a way how we can restrict/change the Lie bracket?
For example, if I think about Lie algebra $\mathfrak{g}$ as a $k$-algebra $\mathfrak{g}$ with a $k$-linear map $[-,-]:\mathfrak{g}\otimes \mathfrak{g}\to \mathfrak{g}$ that factors through $\wedge^2 \mathfrak{g}$ i.e. we have that
$$\mathfrak{g}\otimes \mathfrak{g}\to\wedge^2 \mathfrak{g}\to \mathfrak{g}.$$
Then can I reassign the Lie bracket by just doing
$$\mathfrak{g}\otimes \mathfrak{g}\to\wedge^2 \mathfrak{g}\to \mathfrak{g}\to \mathfrak{g}$$
where the last map is the linear map $R:\mathfrak{g}\to \mathfrak{g}$ with certain conditions.
I hope that makes sense.
 A: I'll not literally answer the question.
Start from a Lie bracket $[\cdot,\cdot]$ on some, say, $n$-dimensional space, defining a Lie algebra $\mathfrak{g}$. Question: what is a Lie bracket $[\cdot,\cdot]'$ "very near" $[\cdot,\cdot]$?
If we forget the Jacobi identity, i.e., only consider skew-symmetric brackets, there is nothing mysterious: the set of skew-symmetric brackets is a linear space (of dimension $n^2(n-1)/2$).
Let's thus write $[\cdot,\cdot]'=[\cdot,\cdot]+tb(\cdot,\cdot)$. (This is some kind of "physicist" heuristics.) And think of $t$ as a very small scalar.
Now assume that $[\cdot,\cdot]'$ also satisfies Jacobi. Then expanding, the constant terms in $t$ vanish, the main term is in $t$ and we remove the term in $t^2$: the condition is
$$\sum_{i=0}^2 b(x_i,[x_{i+1},x_{i+1}])+[x_0,b(x_{i+1},x_{i+2})]=0$$
for all $x_0,x_1,x_2$, where indices are meant modulo 3. This is a linear condition on $b$. The set of $b$ satisfying this conditions is denoted $B^2(\mathfrak{g},\mathfrak{g})$. Call these "cocycles".
Another heuristic: what is a "stupid" way to deform $\mathfrak{g}$. Namely, we can conjugate the bracket by a linear automorphism $u$ (which yields another bracket, unless the linear automorphism is a Lie algebra automorphisms). That is, one considers the new bracket $(x,y)\mapsto u^{-1}([u(x),u(y)]$. Let's do this with an automorphism close to the identity, say, $\mathrm{Id}+tM$. Clearing terms in $O(t^2)$, one gets $$(x,y)\mapsto (\mathrm{Id}-tM)[x+tMx,y+tMy])=[x,y]+t([Mx,y]+[x,MY]-M[x,y]).$$
And one sees that $(x,y)\mapsto [Mx,y]+[x,MY]-M[x,y]$ is indeed a cocycle. The cocycles of this special form (for some linear endomorphism $M$) are called coboundaries, and form a subspace of the space of cocycles. The quotient space cocycles/coboundaries is a cohomology space called $H^2(\mathfrak{g},\mathfrak{g})$. It can be quite efficiently computed by a computer from the structure constants if $n$ is reasonable (e.g., $n=50$).
Heuristically, this space is the tangent space of the space of brackets modulo linear automorphism group. There are results in the direction of making this rigorous, e.g. in Vergne's PhD (this paper, in French https://eudml.org/doc/87158).
Things are not as nice as we could: given a cocycle, it is not always true that we can "integrate" it into a deformation: notably when $\mathfrak{g}$ is abelian (in which being a cocycle is an empty condition) this often fails. Still there are positive results. For instance, when $\mathfrak{g}$ is semisimple, this cohomology space is zero, and indeed every small deformation is "trivial".
