SL2R as anti de Sitter space cont.

Posted on by arkajad

We continue Becoming anti de Sitter.

Every matrix \Xi in the Lie algebra o(2,2) generates one-parameter group e^{\Xi t} of linear transformations of \mathbf{R}^4. Vectors tangent to orbits of this group form a vector field. Let us find the formula for the vector field generated by \Xi. The orbit through y\in \mathbf{R}^4 is

(1)   \begin{equation*}y(t)=e^{\Xi t}y.\end{equation*}

Differentiating at t=0 we find the vector field \Xi(y)

(2)   \begin{equation*}\Xi(y)=\Xi y.\end{equation*}

If \Xi is a matrix with components \Xi^{\mu}_{\phantom{\mu}\nu}, then \Xi(y) has components

(3)   \begin{equation*}\Xi^{\mu}(y)=\Xi^{\mu}_{\phantom{\mu}\nu}y^{\nu}.\end{equation*}

Vectors tangent to coordinate lines are often denoted as \partial_\mu. Therefore we can write the last formula as:

(4)   \begin{equation*}\Xi(y)=\Xi^{\mu}_{\phantom{\mu}\nu}y^{\nu}\partial_\mu.\end{equation*}

In the last post Becoming anti de Sitter we have constructed six generators \Xi_{(\mu\nu))}. Their vector fields now become

(5)   \begin{equation*}\Xi_{(1,2)}=y^2\partial_1-y^1\partial_2,\Xi_{(1,3)}=y^3\partial_1+y^1\partial_3,\Xi_{(1,4)}=y^4\partial_1+y^1\partial_4,\end{equation*}

(6)   \begin{equation*}\Xi_{(2,3)}=y^3\partial_2+y^2\partial_3,\Xi_{(2,4)}=y^4\partial_2+y^2\partial_4,\Xi_{(3,4)}=-y^4\partial_3+y^3\partial_4.\end{equation*}

Bengtsson and Sandin in their paper “Anti de Sitter space, squashed and stretched” discussed in the previous note use coordinates y^1=X,y^2=Y,y^3=U,y^4=V. Our vector field \Xi_{(1,2)} is the same as their J_{XY}, our \Xi_{(1,3)} is the same as their J_{XU} etc.

In SL(2,R) Killing vector fields in coordinates we introduced six Killing vector fields acting on the group manifold SL(2,R). How they relate to the above six generators of the group O(2,2)?

Vectors from the fields \xi_{iL},\xi_{iR} are tangent to SL(2,R). We have expressed them in coordinates of the group SL(2,R) x^1=\theta,x^2=r,x^3=u. The manifold of SL(2,R) is a hipersurface of dimension 3 in \mathbf{R}^4 endowed with coordinates y^1,y^2,y^3,y^4. What is the relation between components of the same vector in different coordinate systems? The formula is easy to derive and is very simple. If \xi^{i}, (i=1,2,3) are coordinates of the vector in SL(2,R) and \xi^{\mu},\, (\mu=1,2,3,4) are coordinates of the same vector in \mathbf{R}^4, then

(7)   \begin{equation*}\xi^\mu=\frac{\partial y^\mu}{\partial x^{i}}\xi^{i}.\end{equation*}

How y^\mu depend on x^{i}? That is simple. In SL(2,R) vector fields in coordinates we have represented each matrix A from SL(2,R) as

(8)   \begin{equation*} A=\begin{bmatrix} r \cos (\theta )+\frac{u \sin (\theta )}{r} & \frac{\cos (\theta ) u}{r}-r \sin (\theta ) \\ \frac{\sin (\theta )}{r} & \frac{\cos (\theta )}{r}\end{bmatrix}. \end{equation*}

On the other hand, Becoming Anti-de Sitter, we represented it as

(9)   \begin{equation*}A=\begin{bmatrix} V+X & Y+U \\ Y-U & V-X \end{bmatrix}.\end{equation*}

Therefore coordinates y^\mu are easily expressed in terms of x^{i}. It remains to do the calculations. I have used computer algebra software to make these calculations for me. My Mathematica notebook doing all calculations can be downloaded from here. The result of all these calculations is the expression of vector fields \xi_{iL},\xi_{iR} in terms of the generators of O(2,2) used in the paper on anti de Sitter spaces. Here is what I have obtained:

(10)   \begin{eqnarray*} \xi_{1R}&=&-J_1=J_{XU}+J_{YV},\\ \xi_{2R}&=&J_2=J_{YU}-J_{XV},\\ \xi_{3R}&=&J_0=-J_{XY}-J_{UV},\\ \xi_{1L}&=&\tilde{J}_1=J_{YV}-J_{XU},\\ \xi_{2L}&=&\tilde{J}_2=-J_{XV}-J_{YU},\\ \xi_{3L}&=&\tilde{J}_0=J_{XY}-J_{UV}. \end{eqnarray*}

Bengtsson and Sandin introduce then their own parametrization of SL(2,R) and study the invariant metric on the group. We will find the connection between ours and their approaches in the next posts. We came to our problems starting from T-handles spinning freely in zero gravity. They are studying spinning black holes. It is interesting to see and to research similarities.

Becoming anti de Sitter

Posted on by arkajad

In the last post we were discussing Killing vector fields of the group SL(2,R). It was done without specifying any reason for doing it – except that it somehow came in our way naturally. But now there is an opportunity to relate our theme to something that is fashionable in theoretical physics: holographic principle and AdS/CFT correspondence

We were playing with AdS without knowing it. Here AdS stands for “anti-de-Sitter” space. Let us therefore look into the content of one pedagogical paper dealing with the subject: “Anti de Sitter space, squashed and stretched” by Ingemar Beengtsson and Patrik Sandrin . We will not be squashing and stretching – not yet. Our task is “to connect” to what other people are doing. Let us start reading Section 2 of the paper “Geodetic congruence in anti-de Sitter space“. There we read:

For the 2+1 dimensional case the definition can be reformulated in an interesting way. Anti-de Sitter space can be regarded as the group manifold of SL(2,{\bf R}), that is as the set of matrices

(1)   \begin{equation*} A = \left[ \begin{array}{cc} V+X & Y+U \\ Y-U & V-X \end{array} \right] \ , \hspace{10mm} \mbox{det}A = U^2 + V^2 - X^2 - Y^2 = 1 \ . \end{equation*}

It is clear that every SL(2,R) matrix A=\left[\begin{smallmatrix}\alpha&\beta\\ \gamma&\delta\end{smallmatrix}\right] can be uniquely written in the above form.

But Section 2 starts with something else:

\noindent Anti-de Sitter space is defined as a quadric surface embedded in a flat space of signature (+ \dots +--). Thus 2+1 dimensional anti-de Sitter space is defined as the hypersurface

(2)   \begin{equation*} X^2 + Y^2 - U^2 - V^2 = - 1 \end{equation*}

\noindent embedded in a 4 dimensional flat space with the metric

(3)   \begin{equation*} ds^2 = dX^2 + dY^2 - dU^2 - dV^2 \ . \end{equation*}

\noindent The Killing vectors are denoted J_{XY} = X\partial_Y - Y\partial_X,
J_{XU} = X\partial_U + U\partial_X, and so on. The topology is now
{\bf R}^2 \times {\bf S}^1, and one may wish to go to the covering
space in order to remove the closed timelike curves. Our arguments
will mostly not depend on whether this final step is taken.

For the 2+1 dimensional case the definition can be reformulated in an interesting way. Anti-de Sitter space can be regarded as the group manifold of SL(2,{\bf R}), that is as the set of matrices

(4)   \begin{equation*} g = \left[ \begin{array}{cc} V+X & Y+U \\ Y-U & V-X \end{array} \right] \ , \hspace{10mm} \mbox{det}g = U^2 + V^2 - X^2 - Y^2 = 1 \ . \end{equation*}

\noindent The group manifold is equipped with its natural metric, which is invariant under transformations g \rightarrow g_1gg_2^{-1}, g_1, g_2 \in SL(2, {\bf R}). The Killing vectors can now be organized into two orthonormal and mutually commuting sets,

(5)   \begin{eqnarray*} & J_1 = - J_{XU} - J_{YV} \hspace{15mm} & \tilde{J}_1 = - J_{XU} + J_{YV} \\ & J_2 = - J_{XV} + J_{YU} \hspace{15mm} & \tilde{J}_2 = - J_{XV} - J_{YU} \\ & J_0 = - J_{XY} - J_{UV} \hspace{15mm} & \tilde{J}_0 = J_{XY} - J_{UV} \ . \end{eqnarray*}

\noindent They obey

(6)   \begin{equation*} ||J_1||^2 = ||J_2||^2 = - ||J_0||^2 = 1 \ , \hspace{3mm} ||\tilde{J}_1||^2 = ||\tilde{J}_2||^2 = - ||\tilde{J}_0||^2 = 1 \ . \end{equation*}

The story here is this: 2\times 2 real matrices form a four-dimensional real vector space. We can use \alpha,\beta,\gamma,\delta or V,U,X,Y as coordinates y^1,y^2,y^3,y^4 there. The condition of being of determinant one defines a three-dimensional hypersurface in \mathbf{R}^4. We can endow \mathbf{R}^4 with scalar product determined by the matrix G defined by:

(7)   \begin{equation*}G=\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}.\end{equation*}

The scalar product is then defined as

(8)   \begin{equation*}(y,y')=y^TGy'=G_{ij}y^{i}y'^{j}=y^1y'^1+y^2y'^2-y^3y'^3-y^4y'^4.\end{equation*}

This scalar product is invariant with respect to the group SO(2,2) of 4\times 4 real matrices A satisfying:

(9)   \begin{equation*}A^TGA=G.\end{equation*}

That is, if A\in O(2,2) then (Ay,Ay')=(y,y') for all y,y' in \mathbf{R}^4.

What I will be writing now is “elementary”in the sense that “everybody in the business” knows it, and if asked will often not be able to tell where and when she/he learned it. But this is a blog, and the subject is so pretty that it would be a pity if people “not in the business” would miss it.

The equation (2) can then be written as (y,y)=-1. It determines a “generalized hyperboloid” in \mathbf{R}^4 that is invariant with respect to the action of O(2,2). Thus the situation is analogous to the one we have seen in The disk and the hyperbolic model. There we had the Poincaré disk realized as a two-dimensional hyperboloid in a three-dimensional space with signature (2,1), here we have SL(2,R) realized as a generalized hyperboloid in four-dimensional space with signature (2,2). Before it was the group O(2,1) that was acting on the hyperboloid, now is the group O(2,2). Let us look at the vector fields of the generators of this group. By differentiating Eq. (9) at group identity we find that each generator \Xi must satisfy the equation:

(10)   \begin{equation*}\Xi^TG+G\Xi=0.\end{equation*}

This equation can be also written as

(11)   \begin{equation*}(G\Xi)^T+G\Xi=0.\end{equation*}

Thus G\Xi must be antisymmetric. In n dimensions the space of antisymmetric matrices is n(n-1)/2-dimensional. For us n=4, therefore the Lie algebra so(2,2) is 6-dimensional, like the Lie algebra so(4) – they are simply related by matrix multiplication \Xi\mapsto G\Xi. We need a basis in so(2,2), so let us start with a basis in so(4). Let M_{(\mu\nu)} denote the elementary antisymmetric matrix that has 1 in row \mu, column \nu and -1 in row \nu column \mu for \mu\neq \nu, and zeros everywhere else. In a formula

    \[ (M_{(\mu\nu)})_{\alpha\beta}=\delta_{\alpha\mu}\delta_{\beta\nu}-\delta_{\alpha\nu}\delta_{\beta\mu},\]

where \delta_{\mu\nu} is the Kronecker delta symbol: \delta_{\mu\nu}=1 for \mu=\nu, and =0 for \mu\neq\nu.

As we have mentioned above, the matrices \Xi_{(\mu\nu)}=G^{-1}M_{(\mu\nu)} form then the basis in the Lie algebra so(2,2). We can list them as follows

(12)   \begin{equation*}\Xi_{(12)}=\left[\begin{smallmatrix}0&1&0&0\\-1&0&0&0\\0&0&0&0\\0&0&0&0\end{smallmatrix}\right], \Xi_{(13)}=\left[\begin{smallmatrix}0&0&1&0\\0&0&0&0\\1&0&0&0\\0&0&0&0\end{smallmatrix}\right], \Xi_{(14)}=\left[\begin{smallmatrix}0&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&0\end{smallmatrix}\right].\end{equation*}

(13)   \begin{equation*}\Xi_{(23)}=\left[\begin{smallmatrix}0&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&0\end{smallmatrix}\right], \Xi_{(24)}=\left[\begin{smallmatrix}0&0&0&0\\0&0&0&1\\0&0&0&0\\0&1&0&0\end{smallmatrix}\right], \Xi_{(34)}=\left[\begin{smallmatrix}0&0&0&0\\0&0&0&0\\0&0&0&-1\\0&0&1&0\end{smallmatrix}\right].\end{equation*}

On the path to AdS

In the next post we will relate these generators to J_i,\tilde{J}_i from the Anti de Sitter paper by Bengtsson et al and to our Killing vector fields
\xi_{iL},\xi_{iR} from the last note

SL(2,R) Killing vector fields in coordinates

Posted on by arkajad

In Parametrization of SL(2,R) we introduced global coordinates x^1=\theta,x^2=r,x^3=u on the group SL(2,R). Any matrix A in SL(2,R) can be uniquely written as

(1)   \begin{equation*} A(\theta,r,u)=\begin{bmatrix} r \cos (\theta )+\frac{u \sin (\theta )}{r} & \frac{\cos (\theta ) u}{r}-r \sin (\theta ) \\ \frac{\sin (\theta )}{r} & \frac{\cos (\theta )}{r}\end{bmatrix}. \end{equation*}

If A is the matrix with components A_{ij}, then its coordinates can be expressed as functions of the matrix components as follows

(2)   \begin{eqnarray*} x^1(A)&=&\mathrm{atan2}(A_{21},A_{22}),\\ x^2(A)&=&\frac{1}{\sqrt{A_{21}^2+A_{22}^2}},\\ x^3(A)&=&\frac{A_{11}A_{21}+A_{12}A_{22}}{A_{21}^2+A_{22}^2}. \end{eqnarray*}

The function \mathrm{atan2}(y,x) returns the angle \theta, 0\leq\theta<2\pi of the complex number x+iy=\sqrt{x^2+y^2}e^{i\theta}.

Once we have coordinates, it is easy to calculate components of tangent vectors to any given path A(t) – they are given by derivatives of the coordinates x^{i}(A(t)). We will calculate now the vector fields resulting from left and right actions of one-parameter subgroups of SL(2,R) that we have already met in SL(2,R) generators and vector fields on the half-plane

We have introduced there the one-parameter groups, that we will denote now as U_1,U_2,U_3, generated by X_1,X_2,X_3:

(3)   \begin{eqnarray*} X_1&=&\begin{bmatrix}0&1\\1&0\end{bmatrix},\\ X_2&=&\begin{bmatrix}-1&0\\0&1\end{bmatrix},\\ X_3&=&\begin{bmatrix}0&-1\\1&0\end{bmatrix}. \end{eqnarray*}

We can take exponentials of the generators and construct one-parameter subgroups

(4)   \begin{eqnarray*} U_1(t)&=&\exp tX_1=\begin{bmatrix}\cosh t&\sinh t\\ \sinh t&\cosh t\end{bmatrix},\\ U_2(t)&=&\exp tX_2=\begin{bmatrix}\exp -t&0\\ 0&\exp t\end{bmatrix},\\ U_3(t)&=&\exp tX_3=\begin{bmatrix}\cos t&-\sin t\\ \sin t&\cos t\end{bmatrix}. \end{eqnarray*}

In SL(2,R) generators and vector fields on the half-plane we were acting with these transformations on the upper half-plane using fractional linear representation. Now we will be acting on SL(2,R) itself via group multiplication, either left or right.

Let us start with U_1 acting from the left. At A we have the trajectory U_1(t)A. Its coordinates are x^{i}(t)=x^{i}(U_1(t)A). Let us denote by \xi_{1,L} the tangent vector field, with components \xi_{1,L}^{i}, \,(i=1,2,3) Then

(5)   \begin{equation*}\xi_{1,L}^{i}=\frac{dx^{i}(U_1(t)A)}{dt}|_{t=0}.\end{equation*}

We can calculate it easily using algebra software. The result is:

(6)   \begin{equation*}\xi_{1,L}=(r^2,-ur,1+r^4-u^2),\end{equation*}

The same way we get

(7)   \begin{equation*}\xi_{2,L}=(0,-r,-2u),\end{equation*}

(8)   \begin{equation*}\xi_{3,L}=(0,-ur,-1+r^4-u^2),\end{equation*}

Then we can calculate vector fields of right shifts

(9)   \begin{equation*}\xi_{j,R}^{i}=\frac{dx^{i}(AU_j(t))}{dt}|_{t=0},\end{equation*}

to obtain:

(10)   \begin{equation*}\xi_{1,R}=(\cos (2 \theta ),-r \sin (2 \theta ),2 r^2 \cos (2 \theta )),\end{equation*}

(11)   \begin{equation*}\xi_{2,R}=(-2 \sin (\theta ) \cos (\theta ),-r \cos (2 \theta ),-4 r^2 \sin (\theta ) \cos (\theta )),\end{equation*}

(12)   \begin{equation*}\xi_{3,R}=(1,0,0),\end{equation*}

We know that our metric is bi-invariant. That means the vector fields of the left and right shifts generate one-parameter group of isometries. They are called Killing fields of the metric. In differential geometry one shows that a vector field \xi is a Killing vector field for metric g if and only if the Lie derivative L_\xi of the metric vanishes. Lie derivative of any symmetric tensor T_{ab} is defined as

(13)   \begin{equation*} (L_\xi T)_{ab}=\xi^c\frac{\partial T_{ab}}{\partial x^c}+\frac{\partial \xi^c}{\partial x^{a}}T_{cb}+\frac{\partial \xi^c}{\partial x^{b}}T_{ac}. \end{equation*}

Using any computer algebra software it is easy to verify that Lie derivatives of our metric with respect to all six vector fields \xi_{j,L},\xi_{j,R} indeed vanish. Mathematica notebook verifying this property can be downloaded here.