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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?

Branimir Ćaćić
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PJ Miller
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3 Answers3

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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.

Mariano Suárez-Álvarez
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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 $[\phi(z_1)\phi(z_2)]=\phi([z_1z_2])$?

PJ Miller
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  • You say «The resulting linear algebra is a subalgebra of $gl(2,F)$, so it satisfies the Lie algebra axioms» and immediately after that «Moreover, it is closed»: that cannot be done that way, if it were not closed the statements of your first sentence would not even make sense! The morevover, in particular, is quite not well-chosen. – Mariano Suárez-Álvarez Aug 10 '13 at 15:49
  • You do not need to give an ismorphism. Your question assumes, that there IS a lie algebra with this relation. Obviously this relation determines the lie algebra completely, so if you find one, any other is isomorphic to the one you found. At least if you fix the field. – archipelago Aug 10 '13 at 15:50
  • @archipelago, well, the image of the map could well be quotient of the original algebra. I do not think your comment applies here. – Mariano Suárez-Álvarez Aug 10 '13 at 15:55
  • If everything is two dimensional there is no need for taking care of quotients or anything else. It is really obvious, that there is only one two dimensional lie algebra with this relation, since the relation determines the whole bracket, if the lie algebra is two dimensional. – archipelago Aug 10 '13 at 16:05
  • And as his given algebra is clearly two dimensional there is no need for giving an isomorphism. He only has to check, that it is a lie algebra and it fulfills the relation. Nothing else. – archipelago Aug 10 '13 at 16:06
  • Nothing else, assuming all the observations you are using to justify the "nothing else" are made! One has to at least notice that those observations are to be made. «Obviousness» is a relative property: what is obvious to you might not be obvious to others, and that should be obvious! (And when one is at the point in one's education about lie algebras where, for example, the fact that the image of a Lie algebra map is a Lie algebra is not obvious, then being told that things are obvious in all likelyhood does not help) – Mariano Suárez-Álvarez Aug 10 '13 at 16:13
  • Take $\mathfrak{a}$ and $\mathfrak{g}$ be two two-dimensional lie algebras both with a basis, which fulfills the given realtion. Mapping the one basis to the other in the way the relation suggests is an isomorphism of vector spaces as it is injective and both have the same dimension. Since the relation is the only one,determines the bracket completely and the map is exactly defined in a way, that it preserves the relation, it is an isomoprhism of lie algebras. – archipelago Aug 10 '13 at 16:24
  • I don't feel bad if I call this "obvious" to someone, who takes a course about lie algebras and had one of linear algebra before, what I assumed. – archipelago Aug 10 '13 at 16:26
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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 $$\{\begin{pmatrix} e^x& y \\ 0 & 1 \end{pmatrix}|x,y\in\mathbb{R}\}$$

archipelago
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