Methods of determining when Lie algebras are isomorphic I'm just getting started in Lie algebras (and abstract algebra in general) and I was hoping to get some insight into how people generally determine if two Lie algebras are homo/isomorphic of each other. Are there common tests or logic that you employ? I've done some googling and textbook reading but it would be helpful to get some more base level insights from people with experience.
 A: In general, one can explicitly compute this by solving polynomial equations in the entries of the linear map $f:L\rightarrow L'$ arising from the condition
$$
[f(e_i),f(e_j)]_{L'}=f([e_i,e_j]_L)
$$
for a basis $(e_1,\ldots ,e_n)$ of $L$, together with the condition $\det(f)\neq 0$.
In low dimensions, it is indeed possible to solve such equations. If all obvious invariants are equal and we don't see an obvious isomorphism, then we have to do such a computation. 
For example given two $7$-dimensional nilpotent complex Lie algebras $L$ and $L'$ of the same nilpotency class, we usually do not find easy invariants distinguishing them (often the adjoint cohomology $H^p(L,L)$ is a  good invariant, but it is also not so easy to compute it). Then
we can quickly decide whether or not they are isomorphic by the above computation.
A: To show that two Lie algebras are not isomorphic, one usually finds an invariant (roughly, a property that is invariant under isomorphisms) that is different for two the Lie algebras. These include (including some Captain Llama suggested in the comments):


*

*the dimension;

*the dimension of the center;

*the isomorphism type (and dimension) of the derived subalgebra;

*whether the Lie algebra is nilpotent or solvable or semisimple or simple (nilpotence implies solvability, simplicity implies semisimplicity, but the converses of those statements arenot true; solvability and semisimplicity are mutually exclusive);

*the signature of the Killing form (Killing forms are nondegenerate precisely for semisimple Lie algebras, and such algebras of dimension (IIRC) $< \dim A_{17} = 323$ are determined up to isomorphism by this signature);

*isomorphism types (and in particular the dimensions) of the subalgebras in the lower and upper central series.


To show that two Lie algebras $\mathfrak{g}, \mathfrak{h}$ (say, of dimension $n$) are isomorphic, often the most straightforward method is just to construct an explicit isomorphism $\phi : \mathfrak{g} \to \mathfrak{h}$. As Dietrich Burde points out in his good answer, if we choose bases of $\mathfrak{g, h}$, the homomorphism conditions $$\phi([e_i, e_j]) = [\phi(e_i), \phi(e_j)]$$ defines $\frac{1}{2} n (n - 1)$ quadratic conditions in the entries of the matrix representation of $\phi$, and often this system can be solved explicitly.
One can refine this method by adapting the choice of bases to the structure of the Lie algebras. For example, (at least over fields of characteristic $0$) if $\mathfrak{g, h}$ are not semisimple then their respective derived subalgebras, $\mathfrak{[g, g], [h, h]}$ are proper subalgebras. (Supposing the derived subalgebras have the same dimension, say $m$, since if not the Lie algebras are not isomorphic,) we can choose bases of $\mathfrak{g, h}$ adapated in the sense that $\mathfrak{[g, g]} = \operatorname{span}\{e_1, \ldots, e_m\}$ and $\mathfrak{[h, h]} = \operatorname{span}\{f_1, \ldots, f_m\}$. Then, since $\phi(\mathfrak{[g, g]}) = \mathfrak{[h, h]}$, the matrix representation of $\phi$ must have the form $$[\phi] = \pmatrix{\ast&\ast\\&\ast} ,$$ where the block sizes are $m, n - m$. In particular, this forces the $m (n - m)$ of coefficients $a_{ij}$ with $i > m, j \leq m$, to vanish. This strategy and others like it can dramatically simplify the quadratic system in $a_{ij}$ and hence make it easier to construct an explicit isomorphism.
Finally, in small dimensions the Lie algebras (say, over $\Bbb R$ and $\Bbb C$) have been classified up to isomorphism, and often you can quickly determine to which a given Lie algebra is isomorphic by eliminating all of the other possibilities.
