two Lie algebras: Isomorphic or not

Question

I saw following examples of Lie algebras in a book of physics:

$L_1=\langle x,y,z\rangle$ $[x,y]=z$, $[y,z]=x$ and $[z,x]=y$.

$L_2=\langle x,y,z\rangle$, $[x,y]=z$, $[y,z]=-x$, $[z,x]=y$.

These Lie algebras are $3$-dimensional and the book considers over $\mathbb{R}$.

I want to know whether they are isomorphic and also consider field to be of arbitrary characteristic. Can one state whether they are isomorphic or not. I will try then for proof.

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My attempt: if $-1$ is a square in $F$, then let $a^2=-1$. In $L_2$, $$[x, ay]=az, [ay,az]=-a^2x=x, [az,x]=ay.$$ This means $x\mapsto x, y\mapsto ay, z\mapsto az$ gives isomorphism between $L_1$ and $L_2$.

What happens if $-1$ is not square in $F$? Of course, above map is not isomorphism; can there be other isomorphism? How to tackle this problem?

Answer 1

The first Lie algebra $L_1$ is the cross product.

https://en.wikipedia.org/wiki/Cross_product#Coordinate_notation

You can write the second like so:

$[2y,x+z]=[2y,x]+[2y,z]=-2(x+z)$. $[2y,x-z]=-2z+2x=2(x-z)$. $[x-z,x+z]=2[x,z]=2y$. This is the presentation of $SL2$

https://en.wikipedia.org/wiki/Special_linear_Lie_algebra

Over the field of real number they are not isomorphic.

https://mathoverflow.net/questions/165656/how-many-three-dimensional-real-lie-algebras-are-there

Answer 2

Over the real numbers, the first Lie algebra is isomorphic to $\mathfrak{so}_3(\mathbb{R})$, and the second is isomorphic to $\mathfrak{sl}_2(\mathbb{R})$. These two Lie algebras are not isomorphic, as we already know at MSE - see here. An easy argument is, in addition, that the first algebra has no $2$-dimensional subalgebra, but the second one has.