What is the Definition of Lie Algebra \mathfrak{so}(2,1)?

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Homework Help Overview

The discussion revolves around the definition and properties of the Lie algebra \mathfrak{so}(2,1), particularly focusing on its structure and the conditions that define its elements. Participants explore the relationship between the Lie algebra and the special orthogonal group SO(2,1).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants examine the definition of \mathfrak{so}(2,1) and discuss the implications of the conditions for its elements, including the trace and symmetry properties. There is inquiry into deriving properties from the group SO(2,1) and how to find a basis that satisfies specific commutation relations.

Discussion Status

The conversation is active, with participants sharing their reasoning and checking assumptions about the structure of the Lie algebra. Some guidance has been offered regarding the use of complexification to achieve desired commutation relations, and there is an ongoing exploration of the implications of different choices for the generators.

Contextual Notes

Participants are working under the constraints of defining a Lie algebra and ensuring that the generators meet specific algebraic properties. There is mention of commutation relations that may involve factors that complicate the relationships between the generators.

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What is the definition of the lie algebra \mathfrak{so}(2,1)?

i want to say that it is the set

\{ X \in GL(3,\mathbb{R}) | X^t=-X \text{ and } \text{trace}(X)=0 \}

is that true?
 
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You derive the properties of so(2,1) from the definition of SO(2,1), as in your other post. a is in SO(2,1) if a^T(eta)a=eta, right? Set a(t)=exp(t*X) and differentiate at t=0 to get the properties of X. X^T=(-X) isn't quite right.
 
Last edited:
i get

\frac{d}{dt} \left( e^{t X^T} \eta e^{tX} \right)_{t=0}= \left(X^T e^{tX^T} \eta e^{tX} + e^{tX^T} \eta X e^{tX} \right)_{t=0} = X^T \eta + \eta X

and since \frac{d}{dt} \left( \eta \right)_{t=0}

we get

X^T \eta + \eta X = 0 \Rightarrow X = \eta^{-1} X^T \eta

which would mean

\mathfrak{so}(2,1)= \{ X \in \text{ Mat}(2,\mathbb{R}) | \text{ trace}(X)=0 \quad \text{and } X=-\eta^{-1} X^T \eta

is that right?

how would one go about finding a basis for \mathfrak{so}(2,1) that satisifes [T_i , T_j]= \epsilon_{ijk} T_k?
 
That looks right. I haven't looked at Lie algebra stuff for a while. But since so(2,1) is only 3 dimensional, I'd look for an explicit basis and see what the commutators look like, to start.
 
so the condition X=-\eta^{-1} X^T \eta tells us X_{11}=X_{22}=X_{33}=0 as well as X_{12}=-X_{21},X_{13}=X_{31},X_{23}=X_{32}

so we could have something like

T_1= \left( \begin{array}{ccc} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right), T_2=\left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{array} \right), T_3=\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right)

these work except some of the commutator relationships have factors of -1 in front of them that i can't get rid of.

e.g. [T_2,T_3]=-T_1. swapping the entry in the matrix that is negative doesn't make a difference (as it shouldn't since the basis isn't affected by scalar multiplication). any ideas?
thanks.
 
Not really. Like I said, this isn't something I do everyday. I think you've got the Lie algebra right. You've got one rotation generator T1 and two boost generators T2 and T3. Do you have to go the complexification of the Lie algebra to get the epsilon in the commutation relation? Don't know offhand.
 
i assume my T1,T2,T3 are ok as they satisfy the properties required, right?

sorry but i don't understand what you mean by the complexification of the lie algebra to get an epsilon?
 
latentcorpse said:
i assume my T1,T2,T3 are ok as they satisfy the properties required, right?

sorry but i don't understand what you mean by the complexification of the lie algebra to get an epsilon?

They look ok to me. By complexification I mean multiplying some of your generators by 'i'.
 
yes but regardless of whether i complexify 1,2 or all 3 of them, that means the commutation relation will have to change to something like

[T_l,T_m]=i \epsilon_{lmn}T_n and i don't want that factor of i at the front.
 
  • #10
actually, scratch that i multiplied T_1 and T_2 by i and left T_3 alone and it's working fine. thanks for the tip!
 
  • #11
latentcorpse said:
actually, scratch that i multiplied T_1 and T_2 by i and left T_3 alone and it's working fine. thanks for the tip!

Hmm. Really? I had to put i's on T2 and T3 and flip their order. But you've got the idea.
 

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