Proving Isomorphism between Lie Algebras

[Ted123]

Try a very trivial Lie bracket. What's the easiest Lie bracket around??

The one that's identically 0

 

[micromass]

The one that's identically 0

Indeed!

Note that this is exactly the Lie algebra of the diagonal 3x3 matrices!

 

[Ted123]

Indeed!

Note that this is exactly the Lie algebra of the diagonal 3x3 matrices!

So $D_3(\mathbb{C})$, the space of all 3x3 diagonal matrices with entries in $\mathbb{C}$ with lie bracket $[X,Y]= 0$ for all $X,Y\in D_3(\mathbb{C})$ is not isomorphic to $\mathfrak{g}$ and $\mathfrak{h}$?

When the question says 'show that your example meets the required conditions' does this mean I have to prove that it has dimension 3 and prove that it is not isomorphic? Would showing that a map from $\mathfrak{f} \to \mathfrak{g}$ and $\mathfrak{f} \to \mathfrak{h}$ defined by something aren't homomorphisms suffice?

 

[micromass]

So $D_3(\mathbb{C})$, the space of all 3x3 diagonal matrices with entries in $\mathbb{C}$ with lie bracket $[X,Y]= 0$ for all $X,Y\in D_3(\mathbb{C})$ is not isomorphic to $\mathfrak{g}$ and $\mathfrak{h}$?

When the question says 'show that your example meets the required conditions' does this mean I have to prove that it has dimension 3 and prove that it is not isomorphic?

Yes.

Would showing that a map from $\mathfrak{f} \to \mathfrak{g}$ and $\mathfrak{f} \to \mathfrak{h}$ defined by something aren't homomorphisms suffice?

Won't work. There will certainly exist homormorphisms between those Lie algebra's. But there might not exist any isomorphism.

So, take $\varphi$ an isomorphism and try to deduce a contradiction.

 

[Ted123]

Yes.


Won't work. There will certainly exist homormorphisms between those Lie algebra's. But there might not exist any isomorphism.

So, take $\varphi$ an isomorphism and try to deduce a contradiction.

So can I just define mappings $$\varphi : \mathfrak{f} \to \mathfrak{g}$$ and $$\varphi : \mathfrak{f} \to \mathfrak{h}$$ in any way, assume they are isomorphisms and contradict this by showing the lie bracket isn't preseved?

But if I did this, I could do this with lie algebras which are isomorphic, assume I have a mapping that is an isomorphism (even though it isn't) and contradict the assumption. I need to show that there exist no isomorphisms at all between the lie algebras don't I?

 

[micromass]

assume I have a mapping that is an isomorphism (even though it isn't) and contradict the assumption.

Yes, but you won't be able to do this with Lie algebra's which are isomorphic.

 

[Ted123]

Yes, but you won't be able to do this with Lie algebra's which are isomorphic.

So if I define $\varphi : \mathfrak{f} \to \mathfrak{h}$ by: $$\varphi \left( \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} \right) = \begin{bmatrix} 0 & a & c \\ 0 & 0 & b \\ 0 & 0 & 0 \end{bmatrix}$$ how do I show the lie bracket $[X,Y]=0$ for all $X,Y \in \mathfrak{f}$ isn't satisfied?

 

[micromass]

No, you can't take a specific $\varphi$.

What you do is take an arbitrary homomorphism $\varphi$ and assume that it is an isomorphism. You then show a contradiction.

You know nothing about $\varphi$ except that it's an isomorphism.

 

[Ted123]

No, you can't take a specific $\varphi$.

What you do is take an arbitrary homomorphism $\varphi$ and assume that it is an isomorphism. You then show a contradiction.

You know nothing about $\varphi$ except that it's an isomorphism.

So if we assume $\varphi$ is an isomorphism.

Then $\varphi ([x,y]) = [\varphi (x) , \varphi (y)]$ for all $x,y \in \mathfrak{f}$ since $\varphi$ is a homomorphism.

I'm not seeing how to use the fact that $\varphi$ is 1-1:

$x=y \Rightarrow \varphi (x) = \varphi (y)$

 

[micromass]

Try to calculate


$$\varphi [E,F],~\varphi [F,G],~\varphi [E,G]$$

in several ways. See if you can get a contradiction.

 

[Ted123]

Try to calculate


$$\varphi [E,F],~\varphi [F,G],~\varphi [E,G]$$

in several ways. See if you can get a contradiction.

Are we talking about the same $E,F,G \in\mathfrak{g}$ as before?

From the fact that $\varphi$ is a homomorphism, $\varphi ([E,F]) = [\varphi (E),\varphi (F)]$ etc.

I thought I had to do something with diagonal matrices in $\mathfrak{f}$ and show that the lie bracket [x,y]=0 can't be satisfied?

 

[micromass]

Are we talking about the same $E,F,G \in\mathfrak{g}$ as before?

Yes.

I thought I had to do something with diagonal matrices in $\mathfrak{f}$ and show that the lie bracket [x,y]=0 can't be satisfied?

That works as well.

 

[Ted123]

Yes.



That works as well.

How am I supposed to calculate $\varphi ([E,F])$ etc. at all when I don't know what $\varphi$ is?

Also since $\mathfrak{g}\cong\mathfrak{h}$, if we show $\mathfrak{f}\ncong\mathfrak{g}$ then does that automatically imply that $\mathfrak{f}\ncong\mathfrak{h}$ ?

 

[Ted123]

The question on giving an example and showing that the example meets the required conditions is worth very few marks in this exam paper compared to the question on showing $\mathfrak{g}\cong\mathfrak{h}$.

Isn't there some obvious property that $D_3(\mathbb{C})$ endowed with lie bracket $[X,Y]=0$ has that means it cannot be isomorphic to $\mathfrak{g}$ or $\mathfrak{h}$ ?

Can't I just say that since there is a non-trivial Lie bracket in both $\mathfrak{g}$ and $\mathfrak{h}$, they cannot be isomorphic to $\mathfrak{f}$ since all Lie brackets are trivial in $\mathfrak{f}$ and isomorphisms preserve Lie brackets?

Also for showing that $\varphi (aE + bF + cG) = \begin{bmatrix} 0 & a & c \\ 0 & 0 & b \\ 0 & 0 & 0 \end{bmatrix}$ is 1-1 is this proof OK:

Letting $x = aE+bF+cG \in \mathfrak{g}$ and $y = a'E+b'F+c'G \in \mathfrak{g}$,

$\varphi (x) = \varphi (y) \Rightarrow \begin{bmatrix} 0 & a & c \\ 0 & 0 & b \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & a' & c' \\ 0 & 0 & b' \\ 0 & 0 & 0 \end{bmatrix}$

$\Rightarrow aE +bF+cG = a'E + b'F + c'G$ i.e. $x=y$

How would I show $\varphi$ is onto?

 

[micromass]

Is this an exam??

 

[Ted123]

Is this an exam??

Just a past paper but we don't get the solutions you see