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Homework Help: Lie Algebras

  1. Sep 14, 2011 #1
    1. The problem statement, all variables and given/known data

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

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

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

    3. The attempt at a solution

    Firstly, is this the definition for (i):

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

    For (iii) presumably I first have to show that a mapping [tex]\varphi : \mathfrak{g} \to \mathfrak{h}[/tex] is a homomorphism? If so how do I show [tex]\varphi [x,y] = [\varphi (x) , \varphi (y) ]\,?[/tex]
     
    Last edited: Sep 14, 2011
  2. jcsd
  3. Sep 14, 2011 #2
    For (i), don't you also want to demand that [itex]\phi[/itex] 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 [itex]\phi:\mathfrak{g}\rightarrow \mathfrak{h}[/itex]. 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??
     
  4. Sep 14, 2011 #3
    So we need to define some [tex]\varphi[/tex] which is an isomorphism. Could you help me here?
     
  5. Sep 14, 2011 #4
    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?
     
  6. Sep 14, 2011 #5
    [tex]\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\}[/tex]
     
    Last edited: Sep 14, 2011
  7. Sep 14, 2011 #6
    That is a nice proposal. So we set

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

    So, some questions remain:
    - is it bijective?
    - is it linear?
    - Does it respect the commutator?
     
  8. Sep 14, 2011 #7
    Let

    [tex]E = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}[/tex]
    [tex]F = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}[/tex]
    [tex]G = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}[/tex]

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

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

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

    Does this show that the commutator is satisfied?

    How do you show that [tex]\varphi[/tex] is a homomorphism? How do you show [tex]\varphi [x,y] = [\varphi (x) , \varphi (y) ][/tex] for all [tex]x,y\in\mathfrak{g}\,?[/tex]
     
    Last edited: Sep 14, 2011
  9. Sep 14, 2011 #8
    Well, take x=aE+bF+cG and take y=a'E+b'F+c'G. Try to work out both

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

    and

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

    Use the properties of the commutator and the definition of [itex]\varphi[/itex].
     
  10. Sep 14, 2011 #9
    OK, so:

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

    Therefore [tex]\varphi[/tex] is a homomorphism?
     
  11. Sep 14, 2011 #10
    No, since you don't know that [itex][x,y]=xy-yx[/itex]. You know nothing about the Lie bracket in [itex]\mathfrak{g}[/itex], except [E,F]=G, [E,G]=[F,G]=0. So use these relations to calculate

    [tex][aE+bF+cG,a^\prime E+b^\prime F+c^\prime G][/tex]
     
  12. Sep 14, 2011 #11
    OK, so using the bilinearity property of the Lie bracket and the property that [x,x] = 0 for all [tex]x\in\mathfrak{g}[/tex] together with those 3 relations I get:

    [tex]\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)[/tex] and from here the calculation continues like it did in my last attempt.

    Does this now show that [tex]\varphi[/tex] is a homomorphism?
     
  13. Sep 14, 2011 #12
    Yes, this is ok!
     
  14. Sep 14, 2011 #13
    So do I have to show that [itex]\varphi[/itex] is also a linear transformation?

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

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

    Then this shows [itex]\varphi[/itex] is a homomorphism. Then how do I show that it is a bijective map to show [itex]\varphi[/itex] is an isomorphism? After this the question is answered is it not?
     
  15. Sep 14, 2011 #14
    Well, try to show that the function is injective and surjective. This is the last thing you need to do.
     
  16. Sep 14, 2011 #15
    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?
     
  17. Sep 14, 2011 #16
    Ah yes! But isn't that trivial?
     
  18. Sep 14, 2011 #17
    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?)

    [tex]E = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}[/tex]
    [tex]F = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}[/tex]
    [tex]G = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}[/tex]

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

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

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


    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.
     
  19. Sep 14, 2011 #18
    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!!
     
  20. Sep 14, 2011 #19
    Isn't [itex]\varphi[/itex] the identity function?
     
  21. Sep 14, 2011 #20
    No: E,F and G are not matrices in general.
     
  22. Sep 14, 2011 #21
    OK.

    There is one further question.

    Give an example of a Lie algebra [itex]\mathfrak{f}[/itex] over the field [itex]\mathbb{C}[/itex] such that the dimension of [itex]\mathfrak{f}[/itex] is 3 but [itex]\mathfrak{f}[/itex] is not isomorphic to [itex]\mathfrak{g}[/itex] and [itex]\mathfrak{h}[/itex]. Show that your example meets the required conditions.
     
  23. Sep 14, 2011 #22
    And, what do you think?
     
  24. Sep 14, 2011 #23
    Special linear lie algebra [itex]\mathfrak{sl}_3(\mathbb{C})\,?[/itex]
     
  25. Sep 14, 2011 #24
    That doesn't have dimension 3 if I recall correctly.

    Try some trivial Lie bracket.
     
  26. Sep 14, 2011 #25
    How about [itex]\mathbb{C}^3[/itex], the Lie algebra consisting of 3x1 column vectors with entries in [itex]\mathbb{C}[/itex] ?
     
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