Very confused by Lie Algrbera

1. Feb 6, 2010

latentcorpse

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?

2. Feb 6, 2010

Dick

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: Feb 6, 2010
3. Feb 6, 2010

latentcorpse

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 $[T_i , T_j]= \epsilon_{ijk} T_k$?

4. Feb 6, 2010

Dick

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.

5. Feb 6, 2010

latentcorpse

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.

6. Feb 6, 2010

Dick

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.

7. Feb 6, 2010

latentcorpse

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?

8. Feb 6, 2010

Dick

They look ok to me. By complexification I mean multiplying some of your generators by 'i'.

9. Feb 6, 2010

latentcorpse

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. Feb 6, 2010

latentcorpse

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. Feb 6, 2010

Dick

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