Proving Isomorphism between Lie Algebras

[Ted123]

Homework Statement

Let \mathfrak{g} , \mathfrak{h} be Lie algebras over \mathbb{C}.
(i) When is a mapping \varphi : \mathfrak{g} \to \mathfrak{h} a homomorphism?

(ii) When are the Lie algebras \mathfrak{g} and \mathfrak{h} isomorphic?

(iii) Let \mathfrak{g} be the Lie algebra with basis vectors E,F,G such that the following relations for Lie brackets are satisfied: [E,F]=G,\;\;[E,G]=0,\;\;[F,G]=0. Let \mathfrak{h} be the Lie algebra consisting of 3x3 matrices of the form \begin{bmatrix} 0 & a & c \\ 0 & 0 & b \\ 0 & 0 & 0 \end{bmatrix} where a,b,c are any complex numbers. The vector addition and scalar multiplication on \mathfrak{h} are the usual operations on matrices. The Lie bracket on \mathfrak{h} is defined as the matrix commutator: [X,Y] = XY - YX for any X,Y \in \mathfrak{h}. Prove that the Lie algebras \mathfrak{g} and \mathfrak{h} are isomorphic.

The Attempt at a Solution

Firstly, is this the definition for (i):

\varphi is a homomorphism if \varphi [x,y] = [\varphi (x) , \varphi (y) ] for all x,y\in\mathfrak{g}\,?
What is the definition for (ii)?

For (iii) presumably I first have to show that a mapping \varphi : \mathfrak{g} \to \mathfrak{h} is a homomorphism? If so how do I show \varphi [x,y] = [\varphi (x) , \varphi (y) ]\,?

 

[micromass]

For (i), don't you also want to demand that \phi is linear and stuff?? That is, phi should at least be an algebra morphism.

The definition for (ii) is of course that there exists an isomorphism \phi:\mathfrak{g}\rightarrow \mathfrak{h}. That is: a bijective homomorphism such that its inverse is also a homomorphism.

For (iii), let's not do too much first. Let's first try to figure out what the isomorphism is exactly. Any suggestions??

 

[Ted123]

So we need to define some \varphi which is an isomorphism. Could you help me here?

 

[micromass]

Well, you need to send a basis to a basis. So you need to send E,F and G to a basis. Do you know a basis of those 3x3-matrices?

 

[Ted123]

Well, you need to send a basis to a basis. So you need to send E,F and G to a basis. Do you know a basis of those 3x3-matrices?

\left\{ \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} , \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} , \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \right\}

 

[micromass]

That is a nice proposal. So we set

\phi(aE+bF+cG)=\left(\begin{array}{ccc} 0 & a & b\\ 0 & 0 & c\\ 0 & 0 & 0 \end{array}\right)

So, some questions remain:
- is it bijective?
- is it linear?
- Does it respect the commutator?

 

[Ted123]

That is a nice proposal. So we set

\phi(aE+bF+cG)=\left(\begin{array}{ccc} 0 & a & b\\ 0 & 0 & c\\ 0 & 0 & 0 \end{array}\right)

So, some questions remain:
- is it bijective?
- is it linear?
- Does it respect the commutator?

Let

E = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}
F = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}
G = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}

Then [E,F] = EF - FE = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = G

[E,G] = EG - GE = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}

[F,G] = FG - GF = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}

Does this show that the commutator is satisfied?

How do you show that \varphi is a homomorphism? How do you show \varphi [x,y] = [\varphi (x) , \varphi (y) ] for all x,y\in\mathfrak{g}\,?

 

[micromass]

Well, take x=aE+bF+cG and take y=a'E+b'F+c'G. Try to work out both

\varphi [aE+bF+cG,a^\prime E+b^\prime F+c^\prime G]

and

[\varphi(aE+bF+cG),\varphi(a^\prime E+ b^\prime F+c^\prime G]

Use the properties of the commutator and the definition of \varphi.

 

[Ted123]

Well, take x=aE+bF+cG and take y=a'E+b'F+c'G. Try to work out both

\varphi [aE+bF+cG,a^\prime E+b^\prime F+c^\prime G]

and

[\varphi(aE+bF+cG),\varphi(a^\prime E+ b^\prime F+c^\prime G]

Use the properties of the commutator and the definition of \varphi.

OK, so:

\varphi ( [x,y] ) = \varphi ( xy - yx)
= \varphi \left( \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} - \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} \right)
= \varphi \left( \begin{bmatrix} 0 & 0 & ab' - a'b \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \right) = \begin{bmatrix} 0 & 0 & ab' - a'b \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}
= \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} - \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}
= \varphi (x) \varphi (y) - \varphi (y) \varphi (x) = [\varphi (x) , \varphi (y) ]

Therefore \varphi is a homomorphism?

 

[micromass]

OK, so:

\varphi ( [x,y] ) = \varphi ( xy - yx)

No, since you don't know that [x,y]=xy-yx. You know nothing about the Lie bracket in \mathfrak{g}, except [E,F]=G, [E,G]=[F,G]=0. So use these relations to calculate

[aE+bF+cG,a^\prime E+b^\prime F+c^\prime G]

 

[Ted123]

No, since you don't know that [x,y]=xy-yx. You know nothing about the Lie bracket in \mathfrak{g}, except [E,F]=G, [E,G]=[F,G]=0. So use these relations to calculate

[aE+bF+cG,a^\prime E+b^\prime F+c^\prime G]

OK, so using the bilinearity property of the Lie bracket and the property that [x,x] = 0 for all x\in\mathfrak{g} together with those 3 relations I get:

\varphi ( [x,y] ) = \varphi ( (ab' - a'b)G ) = \varphi \left( \begin{bmatrix} 0 & 0 & ab' - a'b \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \right) and from here the calculation continues like it did in my last attempt.

Does this now show that \varphi is a homomorphism?

 

[micromass]

Yes, this is ok!

 

[Ted123]

Yes, this is ok!

So do I have to show that \varphi is also a linear transformation?

i.e. \varphi ( x + y) = \varphi (x) + \varphi (y) for all x,y\in\mathfrak{g}

and \varphi (\alpha x) = \alpha \varphi (x) for all \alpha \in\mathbb{C} , x\in\mathfrak{g}\,?

Then this shows \varphi is a homomorphism. Then how do I show that it is a bijective map to show \varphi is an isomorphism? After this the question is answered is it not?

 

[micromass]

Well, try to show that the function is injective and surjective. This is the last thing you need to do.

 

[Ted123]

Well, try to show that the function is injective and surjective. This is the last thing you need to do.

Don't I have to show that it is linear (i.e. a linear transformation) to complete the proof of it being a homomorphism or is this clear already?

 

[micromass]

Don't I have to show that it is linear (i.e. a linear transformation) to complete the proof of it being a homomorphism or is this clear already?

Ah yes! But isn't that trivial?

 

[Ted123]

Ah yes! But isn't that trivial?

Yeah it is.

Did I have to include the bit below in the argument? (Is it correct that I'm using the Lie bracket commutator in h to show that it satisifes the relations in g?)

E = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}
F = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}
G = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}

[E,F] = EF - FE = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = G

[E,G] = EG - GE = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}

[F,G] = FG - GF = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}


Also are there any theorems that I can appeal to to show that a homomorphism is an isomorphism rather than go through the injective/surjective proofs from first principles? Things like the Group and Ring theory theorems.

 

[micromass]

Yeah it is.

Did I have to include the bit below in the argument? (Is it correct that I'm using the Lie bracket commutator in h to show that it satisifes the relations in g?)

E = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}
F = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}
G = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}

[E,F] = EF - FE = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = G

[E,G] = EG - GE = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}

[F,G] = FG - GF = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}

Yes, this is good.

Also are there any theorems that I can appeal to to show that a homomorphism is an isomorphism rather than go through the injective/surjective proofs from first principles? Things like the Group and Ring theory theorems.

I suppose there are also isomorphism theorems for Lie algebra's, but they're of no use here. It's very easy to show injective/surjective in this case!

 

[Ted123]

Yes, this is good.



I suppose there are also isomorphism theorems for Lie algebra's, but they're of no use here. It's very easy to show injective/surjective in this case!

Isn't \varphi the identity function?

 

[micromass]

Isn't \varphi the identity function?

No: E,F and G are not matrices in general.

 

[Ted123]

No: E,F and G are not matrices in general.

OK.

There is one further question.

Give an example of a Lie algebra \mathfrak{f} over the field \mathbb{C} such that the dimension of \mathfrak{f} is 3 but \mathfrak{f} is not isomorphic to \mathfrak{g} and \mathfrak{h}. Show that your example meets the required conditions.

 

[micromass]

OK.

There is one further question.

Give an example of a Lie algebra \mathfrak{f} over the field \mathbb{C} such that the dimension of \mathfrak{f} is 3 but \mathfrak{f} is not isomorphic to \mathfrak{g} and \mathfrak{h}. Show that your example meets the required conditions.

And, what do you think?

 

[Ted123]

And, what do you think?

Special linear lie algebra \mathfrak{sl}_3(\mathbb{C})\,?

 

[micromass]

General linear lie algebra \mathfrak{gl}_3(\mathbb{C})\,?

That doesn't have dimension 3 if I recall correctly.

Try some trivial Lie bracket.

 

[Ted123]

That doesn't have dimension 3 if I recall correctly.

Try some trivial Lie bracket.

How about \mathbb{C}^3, the Lie algebra consisting of 3x1 column vectors with entries in \mathbb{C} ?

 

[micromass]

How about \mathbb{C}^3, the Lie algebra consisting of 3x1 column vectors with entries in \mathbb{C} ?

With which Lie bracket?

 

[Ted123]

With which Lie bracket?

Commutator: [X,Y] = XY - YX for all X,Y \in \mathbb{C}^3

 

[micromass]

Commutator: [X,Y] = XY - YX for all X,Y \in \mathbb{C}^3

OK, but how is multiplication defined in \mathbb{C}^3??

 

[Ted123]

OK, but how is multiplication defined in \mathbb{C}^3??

Oh yeah, it's not!

Can you suggest a lie bracket that would work?

 

[micromass]

Oh yeah, it's not!

Can you suggest a lie bracket that would work?

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

 

[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