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