# Homework Help: 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

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

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

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

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

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

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

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

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

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

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