Lie Algebras

1. Sep 14, 2011

Ted123

1. The problem statement, all variables and given/known data

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.

3. 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 $$\varphi [x,y] = [\varphi (x) , \varphi (y) ]\,?$$

Last edited: Sep 14, 2011
2. Sep 14, 2011

micromass

Staff Emeritus
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??

3. Sep 14, 2011

Ted123

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

4. Sep 14, 2011

micromass

Staff Emeritus
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?

5. Sep 14, 2011

Ted123

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

Last edited: Sep 14, 2011
6. Sep 14, 2011

micromass

Staff Emeritus
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?

7. Sep 14, 2011

Ted123

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 $$x,y\in\mathfrak{g}\,?$$

Last edited: Sep 14, 2011
8. Sep 14, 2011

micromass

Staff Emeritus
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 $\varphi$.

9. Sep 14, 2011

Ted123

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?

10. Sep 14, 2011

micromass

Staff Emeritus
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

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

11. Sep 14, 2011

Ted123

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?

12. Sep 14, 2011

micromass

Staff Emeritus
Yes, this is ok!

13. Sep 14, 2011

Ted123

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?

14. Sep 14, 2011

micromass

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

15. Sep 14, 2011

Ted123

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?

16. Sep 14, 2011

micromass

Staff Emeritus
Ah yes! But isn't that trivial?

17. Sep 14, 2011

Ted123

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.

18. Sep 14, 2011

micromass

Staff Emeritus
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!!

19. Sep 14, 2011

Ted123

Isn't $\varphi$ the identity function?

20. Sep 14, 2011

micromass

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