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Riemannian metric on SL(2,R)- explicit formula

Posted on 2017/05/082017/05/09 by arkajad

Riemannian metric is usually expressed through its metric tensor. For instance in Conformally Euclidean geometry of the upper half-plane we were discussing the SL(2,R) invariant Riemannian metric on the upper half-plane and came out with the formula:

(1) $\begin{equation*}g=\frac{1}{y^2}\begin{bmatrix}1&0\\0&1\end{bmatrix}.\end{equation*}$

Riemannian metric in $n$ dimensions is given by $n\times n$ symmetric matrix $g_{ij}=g_{ji},\,(i,j=1,...,n).$ The components of this matrix are, in general, functions of coordinates of the points of the space under consideration. Knowing the metric, we can calculate scalar product of any two tangent vectors at the same point. Knowing the scalar product of any two vectors we can calculate the metric tensor. It goes as follows. Suppose we have coordinates $x^1,...,x^n$ . Sometimes the coordinates have their names, like $\theta,r,u,$ but then it is often convenient to number them, for instance $x^1=\theta,x^2=r,x^3=u$ , and to refer to them through their indices. When we have coordinates, we have vectors tangent to the coordinate lines. They are usually denoted as $\frac{\partial}{\partial_i}$ or even simply $\partial_i$ . This is a standard notation used in differential geometry texts. So, for instance, $\partial_1$ is vector tangent to the coordinate line $\theta$ , that is the line when $\theta$ is varying while $r,u$ are kept constant. More precisely: $\partial_1$ is a vector field, since we can draw the coordinate of $\theta$ through every point. Therefore when I write $\partial_i$ I mean the vector field or I mean one particular vector at one particular point that should be evident from the context. When it needs to be specified, we can write $(\partial_i)_p$ – which means that we are specifying particular point $p.$

If we have scalar product $(\xi,\eta)_p$ defined at every point $p$ of our space, then the metric tensor at a point $p$ is given by the matrix

(2) $\begin{equation*}g_{ij}(p)= (\partial_i,\partial_j)_p.\end{equation*}$

Vectors $\partial_i$ tangent to the coordinate lines form a basis in the tangent space at every point. When we say that a given $\xi$ has components $\xi^i$ , that means that $\xi=\xi^{i}\partial_i$ – where we use Einstein convention implying summation over the dummy index. Thus we may write:

(3) $\begin{equation*}(\xi,\eta)=(\xi^{i}\partial_i,\eta^{j}\partial_j)=\xi^{i}\eta^{j}(\partial_{i},\partial_j)=g_{ij}\xi^{i}\eta^j=\xi^Tg\eta.\end{equation*}$

We will now calculate explicitly the coordinate expression for the metric on SL(2,R) described in the last post Riemannian metric on SL(2,R).
The coordinates $x^1=\theta,x^2=r,x^3=u$ on the group manifold were introduced in Parametrization of SL(2,R) through

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

Let us now find the vector fields tangent to the coordinate lines:

(5) $\begin{equation*}\partial_1=\frac{\partial A(\theta,r,u)}{\partial \theta}= \begin{bmatrix} \frac{u \cos (\theta )}{r}-r \sin (\theta ) & -r \cos (\theta )-\frac{u \sin (\theta )}{r} \\ \frac{\cos (\theta )}{r} & -\frac{\sin (\theta )}{r} \end{bmatrix}, \end{equation*}$

(6) $\begin{equation*}\partial_2=\frac{\partial A(\theta,r,u)}{\partial r}=\begin{bmatrix} \cos (\theta )-\frac{u \sin (\theta )}{r^2} & -\frac{u \cos (\theta )}{r^2}-\sin (\theta ) \\ -\frac{\sin (\theta )}{r^2} & -\frac{\cos (\theta )}{r^2} \end{bmatrix},\end{equation*}$

(7) $\begin{equation*}\partial_3=\frac{\partial A(\theta,r,u)}{\partial u}=\begin{bmatrix} \frac{\sin (\theta )}{r} & \frac{\cos (\theta )}{r} \\ 0 & 0 \end{bmatrix}.\end{equation*}$

These are tangent vectors at $A$ . According to the prescription described in Riemannian metric on SL(2,R) we need to shift them to the identity, that is we need to calculate $A^{-1}\partial_i$ , and then take trace of their products:

(8) $\begin{equation*}g_{ij}=\frac{1}{2}\mathrm{Tr}(A^{-1}\partial_iA^{-1}\partial_j).\end{equation*}$

I did these calculations using computer. Here are the results:

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

(10) $\begin{equation*}A^{-1}\partial_1=\begin{bmatrix}0&-1\\1&0\end{bmatrix},\end{equation*}$

(11) $\begin{equation*}A^{-1}\partial_2=\begin{bmatrix} \frac{\cos (2 \theta )}{r} & -\frac{\sin (2 \theta )}{r} \\ -\frac{\sin (2 \theta )}{r} & -\frac{\cos (2 \theta )}{r}\end{bmatrix},\end{equation*}$

(12) $\begin{equation*}A^{-1}\partial_3=\begin{bmatrix} \frac{\cos (\theta ) \sin (\theta )}{r^2} & \frac{\cos ^2(\theta )}{r^2} \\ -\frac{\sin ^2(\theta )}{r^2} & -\frac{\cos (\theta ) \sin (\theta )}{r^2} \end{bmatrix},\end{equation*}$

(13) $\begin{equation*}g=\begin{bmatrix} -1 & 0 & \frac{1}{2 r^2} \\ 0 & \frac{1}{r^2} & 0 \\ \frac{1}{2 r^2} & 0 & 0\end{bmatrix}\end{equation*}$

The metric is not diagonal. That means the coordinate lines are not all perpendicular to each other. Moreover, the third line is “light-like”. We have $(\partial_3,\partial_3)=0.$ In Minkowski space-time it would mean that it is a trajectory of an object moving with the speed of light. At the group identity we have $\theta=u=0, r=1.$ At this point we have

(14) $\begin{equation*}(\partial_3)_e= \begin{bmatrix}0&1\\0&0\end{bmatrix}.\end{equation*}$

In SL(2,R) generators and vector fields on the half-plane we have denoted this generator as $Y_-$ .

Once we have the metric, we can now calculate its geodesics and curvature. So, we have a plan for the following notes.

Riemannian metric on SL(2,R)

Posted on 2017/05/072017/05/08 by arkajad

Every Lie group is like a Universe. Now it is time for us to play with the cosmology of SL(2,R). In Real magic – space-time in Lie algebra we have already started this game – we have defined a natural “metric” on the Lie algebra sl(2,R). We have defined scalar product of any two vectors tangent to the group at the group identity. And we have seen that this scalar product is similar to that of Minkowski space-time, except that with only two space (and one time) dimensions.

Riemannian metric, on the other hand, is defined when we have scalar product of tangent vectors at every point. In recent posts we were playing with Riemannian metric on the upper half-plane, which is a homogeneous space for the group, of only two dimensions – the cross section of the torus. Now we want to define Riemannian metric on the whole torus – an interesting extension.

We already have the metric at one point – at the group identity. Can we extend this definition to the whole torus, and do it in a natural way?

Of course we can. Because we are on the group. Here it is how it is being done. Suppose we have two vectors tangent at some group point $g$ . I am using the letter $g$ now, but $g$ here is another notation for a real matrix $A$ of determinant one, an element of SL(2,R). That is we have two paths $\gamma_1(t),\gamma_2(t)$ with $\gamma_1(0)=\gamma_2(0)=g.$ Our two tangent vectors, say $\xi_1,\xi_2$ , are vectors tangent to $\gamma_1(t)$ and $\gamma_2(t)$ at $g$ , they are represented by matrices

(1) $\begin{equation*}\xi_i=\frac{d\gamma_i(t)}{dt}|_{t=0}, \quad (i=1,2).\end{equation*}$

The matrices $\xi_1,\xi_2$ are not in the Lie algebra. For instance, in our case, they will not be (in general) of trace zero. That is because at $t=0$ the two paths are not at the identity. But we can shift them to the identity. The paths $g^{-1}\gamma_i(t)$ at $t=0$ are at the identity. Thus even if $\xi_i$ are not in the Lie algebra, $g^{-1}\xi_i$ are. We can use this fact and define the scalar product at $g$ as follows:

(2) $\begin{equation*}(\xi_1,\xi_2)_g=(g^{-1}\xi_1,g^{-1}\xi_2)_e.\end{equation*}$

I am using the symbol $e$ rather than the unit matrix $I$ to denote the group identity here, because the construction above is quite general, is being used for any Lie group, not just for SL(2,R).

Thus once we have scalar product defined in the Lie algebra, we have it defined everywhere.

Of course one can ask: why not use another definition, with right shifts? What is wrong with

(3) $\begin{equation*}(\xi_1,\xi_2)_g=(\xi_1g^{-1},\xi_2g^{-1})_e.\end{equation*}$

Nothing is wrong. The scalar product at the identity that we have defined in Real magic – space-time in Lie algebra has the property of invariance expressed there in Eq. (11). It is a homework to show that using this property we can prove that the two definitions above lead to the same Riemannian metric!

Parametrization of SL(2,R)

Posted on 2017/05/062017/05/07 by arkajad

We are going to discuss Riemannian metric on SL(2,R), which will be more difficult than Riemannian metric on the Poincare disk $D$ or on the upper half-plane $\mathbb{H}$ , because SL(2,R) is three-dimensional. In fact it will be a pseudo-Riemannian metric, because SL(2,R) with its natural metric looks more like space-time (one minus and two pluses, or one plus and two minuses), while $D$ (and $\mathbb{H}$ ) looks like space alone (all pluses).

In order to introduce the metric, to calculate geodesics and Curvature, we need to introduce coordinates. Well, in fact we don’t have to, but we will. It was the French mathematician Élie Cartan Cartan who invented the technique of doing geometry without coordinates, using the so called “moving frame” instead. That is sometimes very useful. But for now we will follow the old-fashioned straightforward way, we will use coordinates.

The group SL(2,R) is isomorphic to SU(1,1) via Cayley transformation, and we have already parametrized SU(1,1) in SU(1,1) parametrization and SU(1,1) decomposition , so in principle we could use what we already have. But we can also try something new, and then see how it relates to the old. Here I am going to use parametrization suggested by Keith Conrad in his notes on SL(2,R). Here is the beginning of his paper “DECOMPOSING $\mathrm{SL}_2(R)$ ”

We are introducing three one-parameter subgroups of SL(2,R) called $\mathcal{K,A,N}$

(1) $\begin{equation*}\mathcal{K}=\begin{bmatrix}\cos \theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix},\end{equation*}$

(2) $\begin{equation*}\mathcal{A}=\begin{bmatrix}r &0\\0&1/r\end{bmatrix},\end{equation*}$

(3) $\begin{equation*}\mathcal{N}=\begin{bmatrix}1&u\\0&1\end{bmatrix}.\end{equation*}$

Then, as Conrad is showing, every SL(2,R) matrix $A$ decomposes uniquely into the product $\mathcal{KAN}$ . But we will change the order. Instead of writing $A=\mathcal{KAN}$ , we will write $\mathcal{NAK}.$ There is a tradition in group theory to use the KAN order. If you search Google for “KAN decomposition”, you will find a lot of information. If you look for “NAK decomposition” – you will find much less, mainly written by eccentrics.
We prefer NAK, because it better corresponds to the polar decomposition of SU(1,1) that we have already discussed.

So, we write $A$ as $\mathcal{NAK}$ , that is, after multiplying the three matrices, as

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

Now $\theta,r,u$ are coordinates of the element $A$ of the group. Of course we would like to know how they relate to coordinates that we have introduced in the group SU(1,1). For instance in Right circular orbits in SU(1,1) we were representing SU(1,1) as a torus. How our new parameters relate to the parameters of the torus?

To answer this question we need to Cayley transform $A$ back to SU(1,1) and then express $\theta$ and $z$ of the torus in terms of $\theta,r,u$ of the NAK decomposition. I did it, using any algebraic software it is a simple task. The end result is that $\theta$ is the same (that is why I have chosen NAK instead of KAN, while $z$ is an algebraic function of $r,u$ . With $z=x+iy$ we have:

(5) $\begin{equation*} x=-\frac{r^4+u^2-1}{r^4+2 r^2+u^2+1},\quad y=-\frac{2 u}{\left(r^2+1\right)^2+u^2}. \end{equation*}$

Thus after Cayley transform to SU(1,1) $r,u$ become coordinates on the unit disk – the cross-section of the torus. Here are their lines transformed to the disk: