Universal central extension of Lie algebras In the literature, there is the notion of the universal central extension of a Lie algebra. My question is: is there also a notion of universal extensions that are not central? If yes, can you provide a reference please and if not, why is there none?
 A: In my opinion, it's probably not so interesting to consider all extensions. 
If you consider the Lie subalgebra $\mathfrak{n}_n\subset \mathfrak{gl}_n$ of all strictly upper triangular matrices, then $\mathfrak{n}_n$ is an extension of $\mathbf{C}$ because it has a quotient Lie algebra of dimension 1. It follows that even for nilpotent extensions the nilpotency can grow arbitrarily large.
Of course extensions can be even more chaotic than that. 
Edit
Example of a non-central perfect group extension: Consider block matrices over $\mathrm{C}$
$$
 H = \begin{pmatrix}\mathrm{SL_2} & * \\ 0 & \mathrm{SL_2}\end{pmatrix},\;
 K = \begin{pmatrix}I_2 & * \\ 0 & I_2\end{pmatrix},\; G = H/K \cong \mathrm{SL}_2\times \mathrm{SL}_2
$$
The group $H$ is perfect because its Lie algebra is perfect: The diagonal part $\mathfrak{sl}_2\times \mathfrak{sl}_2$ lies in the commutator because $\mathfrak{sl}_2$ is perfect. On the other hand, we have
$$
X = \begin{pmatrix}0 & 0 & b & a \\ 0 & 0 & d & c \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}, Y = \begin{pmatrix}0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0\end{pmatrix},\; [X, Y] = \begin{pmatrix}0 & 0 & a & b \\ 0 & 0 & c & d \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}
$$
so the off-diagonal part is also in the commutator.
Consider the conjugation by the matrix $\mathrm{diag}(1, 1, -1, -1)$. It acts non-trivially on $H$ but trivially on $G$.
It means that non-central extensions can in general have non-trivial automorphisms even if the extension is a perfect group. 
A: Let $K$ be the ground field. An extension of a Lie $K$-algebra $\mathfrak{g}$ means a pair $(\mathfrak{h},p)$ where $p$ is a surjective homomorphism $\mathfrak{h}\to\mathfrak{g}$. A homomorphism between extensions $(\mathfrak{h}_1,p_1)$, $(\mathfrak{h}_2,p_2)$ is a $K$-algebra homomorphism $f:\mathfrak{h}_1\to\mathfrak{h}_2$ such that $p_2\circ f=p_1$. A universal extension of $\mathfrak{g}$, is naturally defined as an initial object in this category.
In the full subcategory of central extensions, an initial object is called a universal central extension (and unlike the claim in a comment to this answer, is a well-adapted definition). This is classical, as well as the fact that a universal central extension of $\mathfrak{g}$ exists if and only if $\mathfrak{g}$ is perfect.
In the whole category of extensions of $\mathfrak{g}$, one has:

Then there is no universal extension of $\mathfrak{g}$ except for $\mathfrak{g}=\{0\}$.

This holds in the category of Lie $K$-algebras, or also, if $\mathfrak{g}$ is finite-dimensional, in the category of finite-dimensional Lie $K$-algebras.
Since $\{0\}$ is an initial object in the category of Lie algebras, $\{0\}\to\{0\}$ is a universal extension. Hence suppose $\mathfrak{g}\neq\{0\}$. By contradiction, let $(\mathfrak{h},p)$ be a universal extension, so $\mathfrak{h}\neq\{0\}$. Consider the first projection $q:\mathfrak{g}\times\mathfrak{h}\to\mathfrak{g}$. Then both homomorphisms $(p,0)$ and $(p,\mathrm{id})$ from $\mathfrak{h}$ to $\mathfrak{g}\times\mathfrak{h}$ lift $p$, and this contradicts the uniqueness in the universal property.
(Note: this is a variation of the argument to show that if $\mathfrak{g}$ is not perfect then it has no universal central extension.)

Edit:

Let $\mathcal{C}$ be a full subcategory of the category of abelian extensions of $\mathfrak{g}$, not consisting of central extensions. Then there is a no universal $\mathcal{C}$-extension (i.e., no initial object in $\mathcal{C}$).
In particular, for $\mathfrak{g}\neq\{0\}$, there is no universal abelian extension of $\mathfrak{g}$.

Proof: let $(\mathfrak{h},p)$ be a universal extension in $\mathcal{C}$. Let $(\mathfrak{s},q)$ be non-central extension in $\mathcal{C}$, with kernel $Z$. Let $f:\mathfrak{h}\to\mathfrak{s}$ be given by the universal property. Choose $z\in Z$ not central; then $\tau:s\mapsto s+[s,z]$ is an automorphism of $\mathfrak{s}$, inducing the identity of $\mathfrak{s}$ modulo $s$. By the uniqueness in the universal property, it follows that $s\circ f=f$. Hence $f$ is valued in the set of points fixed by $s$, which is the centralizer of $z$. Since $q\circ f$ is surjective, we also have $f(\mathfrak{h})+Z=\mathfrak{s}$. Combining, we deduce that $z$ is central, contradiction.
(The second assertion follows since every nontrivial $\mathfrak{g}$ has a nontrivial $\mathfrak{g}$-module, hence has a non-central abelian extension.)
