Is L isomorphic to sl(2,C)?

In summary, the conversation involves a discussion of a Lie Algebra L with complex numbers a, b, and c. The basis for L is found to be non-abelian and simple, and it is shown to be isomorphic to sl(2,C). The conversation then moves on to discussing how to show L is an isomorphism, but it is concluded that it is already established as an isomorphism due to its sub-algebra being isomorphic to sl(2,C).
  • #1
ElDavidas
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0

Homework Statement



Take

[tex] L = \left(\begin{array}{ccc}0 & -a & -b \\b & c & 0 \\a & 0 & -c\end{array}\right) [/tex]

where a,b,c are complex numbers.

Homework Equations



I find that a basis for the above Lie Algebra is

[tex]e_1 = \left(\begin{array}{ccc}0 & -1 & 0 \\0 & 0 & 0 \\1 & 0 & 0 \end{array}\right) [/tex]

[tex]e_2 = \left(\begin{array}{ccc}0 & 0 & -1 \\1 & 0 & 0 \\0 & 0 & 0 \end{array}\right) [/tex]

[tex]e_3 = \left(\begin{array}{ccc}0 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & -1 \end{array}\right) [/tex]

I then calculate all the products [itex] [e_i,e_j] [/itex] and see that L is non-abelian and simple

The Attempt at a Solution



The question then asks show L is isomorphic to sl(2,C). I have found [itex] e,f,h \in L [/itex] such that [itex] [h,e] = 2e, [h,f] = -2f, [e,f] = h[/itex]
where,

[tex]h = \left(\begin{array}{ccc}0 & 0 & 0 \\0 & 2 & 0 \\0 & 0 & -2 \end{array}\right) [/tex]

[tex]e = \left(\begin{array}{ccc}0 & 0 & -\sqrt{2} \\ \sqrt{2} & 0 & 0 \\0 & 0 & 0 \end{array}\right) [/tex]

[tex]f = \left(\begin{array}{ccc}0 & -\sqrt{2} & 0 \\0 & 0 & 0 \\ \sqrt{2} & 0 & 0 \end{array}\right) [/tex]

if I haven't made any mistakes. I see L is homomorphic to sl(2,C) but how do I show it's an isomorphism? (i.e show the injection and surjection). I know the definitions for an injection map and a surjection map but don't know how to apply it in this case.

Thanks in advance for any help.
 
Last edited:
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  • #2
You've just written down a sub-algebra isomorphic to sl_2, and clearly it is all of the space (just by dimension arguments). There is nothing more to show.

You haven't actually written down a map so you can't apply the notion of injection or surjection. If you want to put in a map - there is an obvious one - then it is trivially an injection (and a surjection).
 

1. What is a Lie Algebra isomorphism?

A Lie Algebra isomorphism is a mapping between two Lie algebras that preserves the algebraic structure, such as commutators and the Lie bracket. This means that the two Lie algebras are essentially the same, just represented in different ways.

2. What is the significance of Lie Algebra isomorphism?

Lie Algebra isomorphism is significant because it allows us to study and understand different Lie algebras by relating them to each other. It also helps us to identify similarities and differences between different Lie algebras, and to find connections between seemingly unrelated mathematical concepts.

3. How is Lie Algebra isomorphism different from group isomorphism?

Lie Algebra isomorphism is a specific type of isomorphism that deals with Lie algebras, which are linear algebraic structures. Group isomorphism, on the other hand, deals with groups, which are sets with a binary operation. While both involve preserving algebraic structure, the specific structures being preserved are different.

4. Can two Lie algebras be isomorphic but not identical?

Yes, two Lie algebras can be isomorphic but not identical. This means that they have the same algebraic structure, but may be represented in different ways. For example, a Lie algebra can be isomorphic to its dual space, but the two are not identical.

5. How is Lie Algebra isomorphism used in physics?

Lie Algebra isomorphism is used in physics to study the symmetries and conservation laws of physical systems. By relating different Lie algebras, we can uncover hidden connections between different physical theories and gain a deeper understanding of the fundamental principles of physics.

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