Linear Lie algebra isomorphic to two dimensional algebra

Question

Find a linear Lie algebra isomorphic to the nonabelian two dimensional algebra with basis $x,y$ such that $[x,y]=x$. (Hint: Look at the adjoint representation.)

$\DeclareMathOperator{\ad}{ad}$The adjoint representation takes $a\in L$ to $\ad a$, which is the map from $L$ to itself taking $b\mapsto [a,b]$. So I'm looking at $\ad x$, which is represented by the matrix $$\phi(x)=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$ (since $[x,x]=0, [x,y]=x$). On the other hand, $\ad y$ is represented by $$\phi(y)=\begin{pmatrix} -1 & 0 \\ 0 & 0 \end{pmatrix}$$ (since $[y,x]=-x, [y,y]=0$). So the element in the isomorphic linear Lie algebra corresponding to $z=cx+dy$ is $$\phi(z)=\begin{pmatrix} -d & c \\ 0 & 0 \end{pmatrix}.$$

I can verify by hand that $\phi([z_1,z_2])=[\phi(z_1),\phi(z_2)]$ by using $z_1=ax+by, z_2=cx+dy$ and expanding. (Is there an easier way to see?) Do I need to check that the linear algebra satisfies the Lie algebra axioms?

Answer 1

If $\def\g{\mathfrak{g}}\g$ is a Lie algebra and $\def\End{\mathfrak{end}}\End(\g)$ is its Lie algebra of $k$-linear endomorphisms, there is a map $\def\ad{\operatorname{ad}}\ad:\g\to\End(\g)$ such that $\ad(x)(y)=[x,y]$.

• You can immediately check that $\ad$ is a $k$-linear map;

• you can check that $\ad$ is a morphism of Lie algebras: indeed, the Jacobi identity for $\g$ says precisely this, once you arrange it properly; and, finally

• the kernel of the map $\ad$ is precisely the center of $\g$.

Finally, you can check two additional facts:

• The image of a morphism of Lie algebras $f:\mathfrak a\to\mathfrak b$ is always a Lie subalgebra of its codomain.

• The center of your $2$-dimensional algebra $\mathfrak s$ is trivial.

Putting all this things together, we get that the map $\ad:\mathfrak s\to\End(\mathfrak s)$ is an injective morphism of Lie algebras, so an isomorphism of Lie algebras from its domain to its image. Since its image is obviously a linear Lie algebra, we get what you want.

Answer 2

Take a look at the two dimensional lie subalgebra of $\mathfrak{gl}_2(\mathbb{R})$ spanned by the two vectors $$\begin{pmatrix} 1& 0 \\ 0 & 0 \end{pmatrix}\text{ and }\begin{pmatrix} 0& 1 \\ 0 & 0 \end{pmatrix}.$$ This does the job (even if you are not talking about real lie algebras, this does it for arbitary fields)

Just as a remark, this lie algebra is in some sense the smallest lie algebra, which is solvable, but not nilpotent and the corresponding connected lie subgroup of $\operatorname{GL}_2(\mathbb{R})$ is $$\{\begin{pmatrix} e^x& y \\ 0 & 1 \end{pmatrix}|x,y\in\mathbb{R}\}$$

Answer 3

We map an element $z=cx+dy\in L$ to the matrix $$\phi(z)=\begin{pmatrix} -d & c \\ 0 & 0 \end{pmatrix}.$$ The resulting linear algebra is a subalgebra of $gl(2,F)$, so it satisfies the Lie algebra axioms. Moreover, it is closed, because the product and difference of two $2\times 2$ matrices with bottom row being zero is another matrix of the same form, so if two matrices $A,B$ are of this form, then $AB-BA$ is also of this form. So the linear algebra is a Lie algebra.

Now we only need to show that $[\phi(z_1)\phi(z_2)]=\phi([z_1z_2])$ for $z_1,z_2\in L$. Suppose $z_1=ax+by$ and $z_2=cx+dy$. Then $$[z_1z_2] = [ax,cx]+[ax,dy]+[bx,cy]+[bx,dy]=ac[x,x]+ad[x,y]+bc[y,x]+bd[y,y]=(ad-bc)[x,y]=(ad-bc)x.$$ So $\phi([z_1z_2]) = (ad-bc)\phi(x).$

On the other hand, $[\phi(z_1)\phi(z_2)] = \phi(z_1)\phi(z_2)-\phi(z_2)\phi(z_1)$. Substituting the matrices for $\phi(z_1)$ and $\phi(z_2)$, we get the result.

Is there a more intuitive way to see that $[\phi(z_1)\phi(z_2)]=\phi([z_1z_2])$?