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?


3 Answers 3


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.

  • $\begingroup$ 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, 2014 at 21:07

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.


I know that this is quite late, but in the paper by Bardeen and Horowitz, where NHEK originated, they give the following Killing vectors, which after an easy rescalation, fulfill the Lie algebra of $\mathfrak{sl\left(2,\mathbb{R}\right)}$: \begin{align} \xi_1 = \partial_t, \quad \xi_2 = \partial_\phi, \quad \xi_3 = t\partial_t - r \partial_r, \quad \xi_4 = \left(\frac{t^2}{2} + \frac{1}{2r^2}\right)\partial_t - tr\partial_r - \frac{1}{r}\partial_\phi\,. \end{align}


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