Proving Isomorphism between Lie Algebras

In summary: \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & ab \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} - \begin{bmatrix} 0 & 0 & ab' \\ 0 & 0 &... \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & a & c \\ 0 & 0 & b \\ 0 & 0 & 0 \end{bmatrix} \right)
  • #36
Ted123 said:
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.
 
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  • #37
micromass said:
Yes, but you won't be able to do this with Lie algebra's which are isomorphic.

So if I define [itex]\varphi : \mathfrak{f} \to \mathfrak{h}[/itex] by: [tex]\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}[/tex] how do I show the lie bracket [itex][X,Y]=0[/itex] for all [itex]X,Y \in \mathfrak{f}[/itex] isn't satisfied?
 
  • #38
No, you can't take a specific [itex]\varphi[/itex].

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

You know nothing about [itex]\varphi[/itex] except that it's an isomorphism.
 
  • #39
micromass said:
No, you can't take a specific [itex]\varphi[/itex].

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

You know nothing about [itex]\varphi[/itex] except that it's an isomorphism.

So if we assume [itex]\varphi[/itex] is an isomorphism.

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

I'm not seeing how to use the fact that [itex]\varphi[/itex] is 1-1:

[itex]x=y \Rightarrow \varphi (x) = \varphi (y)[/itex]
 
  • #40
Try to calculate


[tex]\varphi [E,F],~\varphi [F,G],~\varphi [E,G][/tex]

in several ways. See if you can get a contradiction.
 
  • #41
micromass said:
Try to calculate


[tex]\varphi [E,F],~\varphi [F,G],~\varphi [E,G][/tex]

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

Are we talking about the same [itex]E,F,G \in\mathfrak{g}[/itex] as before?

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

I thought I had to do something with diagonal matrices in [itex]\mathfrak{f}[/itex] and show that the lie bracket [x,y]=0 can't be satisfied?
 
  • #42
Ted123 said:
Are we talking about the same [itex]E,F,G \in\mathfrak{g}[/itex] as before?

Yes.

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

That works as well.
 
  • #43
micromass said:
Yes.



That works as well.

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

Also since [itex]\mathfrak{g}\cong\mathfrak{h}[/itex], if we show [itex]\mathfrak{f}\ncong\mathfrak{g}[/itex] then does that automatically imply that [itex]\mathfrak{f}\ncong\mathfrak{h}[/itex] ?
 
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  • #44
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 [itex]\mathfrak{g}\cong\mathfrak{h}[/itex].

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

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

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

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

[itex]\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}[/itex]

[itex]\Rightarrow aE +bF+cG = a'E + b'F + c'G[/itex] i.e. [itex]x=y[/itex]

How would I show [itex]\varphi[/itex] is onto?
 
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  • #45
Is this an exam??
 
  • #46
micromass said:
Is this an exam??

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