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I'm looking into $SL(2,\mathbb{R})$ group and it's algebra. I found on line that the $sl(2,\mathbb{R})$ algebra is given by the two by two real matrices of trace zero. This Lie algebra has dimension three; a standard basis is given as

$$X=\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}, Y=\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}, Z=\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix}$$

with commutation relations $[X,Y]=2Y$, $[X,Z]=-2Z$, $[Y,Z]=X$, and Jacobi identity is satisfied (I calculated via Mathematica).

Now, in an article about Kerr/CFT correspondence, the Near-Horizon Extreme Kerr (NHEK) geometry, has an enhanced $SL(2,\mathbb{R})\times U(1)$ isometry group, with Killing vectors that generate $SL(2,\mathbb{R})$ group

\begin{equation} \tilde{J}_0=2\partial_\tau \end{equation}

\begin{equation} \tilde{J}_1=\frac{2r\sin\tau}{\sqrt{1+r^2}}\partial_\tau-2\sqrt{1+r^2}\cos\tau\partial_r+\frac{2\sin\tau}{\sqrt{1+r^2}}\partial_\varphi \end{equation}

\begin{equation} \tilde{J}_2=-\frac{2r\cos\tau}{\sqrt{1+r^2}}\partial_\tau-2\sqrt{1+r^2}\sin\tau\partial_r-\frac{2\cos\tau}{\sqrt{1+r^2}}\partial_\varphi \end{equation}

with algebra that satisfies

\begin{equation} [\tilde{J}_0,\tilde{J}_1]=-2\tilde{J}_2,\quad [\tilde{J}_0,\tilde{J}_2]=2\tilde{J}_1,\quad [\tilde{J}_1,\tilde{J}_2]=2\tilde{J}_0 \end{equation}

and Jacobi identity is also satisfied.

Now, my question is, since this is $SL(2,\mathbb{R})$ group, shouldn't the algebra be the same? That is, shouldn't Lie brackets be identical in the first and second case? Why is there a difference?

In one case I have $[X,Y]=2Y$, and in other $[X,Y]=-2Z$ basically. Is this because of how we defined the generators?

I'm kinda confused because how do I know the latter are indeed generators of $SL(2,\mathbb{R})$. I mean, all I need for Lie algebra is to have the basic axioms fullfilled, and that's it, right?

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  • $\begingroup$ Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files, e.g., arxiv.org/abs/0809.4266 $\endgroup$ – Qmechanic Jan 10 '14 at 20:57
  • $\begingroup$ Sorry, forgot about that, corrected it now. $\endgroup$ – dingo_d Jan 10 '14 at 20:59
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Given Lie Algebras $\mathfrak{g}$ and $\mathfrak{g'}$ structure constants need not to be the same in order for $\mathfrak{g}$ and $\mathfrak{g'}$ to define the same Lie Algebra. Since any Lie Algebra is by definition a vector space with a product (the commutator) that satisfies certain properties, it is indeed a linear space. Then it is all up to a change of base.

Probably the two algebras you are facing are indeed the same, but written in with a different bases for the vector space.

My suggestion is that you should find a matrix $M$ that brings you from one base to the other.

Basically if you call $T_i$ the generators of $\mathfrak{g}$ and $\tilde{T}_i$ the generators of $\mathfrak{g}$, as long as you find a matrix that does $$MT_iM^{-1}=\tilde{T}_i$$ then $\mathfrak{g}$ and $\mathfrak{g'}$ deifne the same Lie Algebra.

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  • $\begingroup$ I was just reading something about structure constants and thought that that might be the answer! So it is only difference in the given representation after all. $\endgroup$ – dingo_d Jan 10 '14 at 21:07
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Often in physics versus math the group generators will throw in a factor of 2 or i in there for convenience. Same with the Pauli matrices versus the actual SU(2) matrices.

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