Category: SU(1,1)

Real magic – space-time in Lie algebra

Posted on by arkajad

We start with a partial recall of events as they transpired so far. A month ago we became hyperbolic. The post Getting hyperbolic started with this sentence:

Without knowing it, during the last three posts (Our first field expedition, Our second field expedition, The Third Expedition) we became hyperbolic. Hyperbolic and conformal. Conformal and relativistic, relativistic and non-Euclidean.

Then we introduced the group SU(1,1) and its action on the disk. Then, in From SU(1,1) to the Lorentz group we realized that the disk can be considered as a projection of a hyperboloid in 2+1 dimensional space-time, where the hyperboloid is inside the future light cone. We were contemplating the image below

Space-time hyperboloid and the Poincare disk models

and we have shown that there is a group homomorphism from SU(1,1) to the Lorentz group SO(2,1) of 2+1 dimensional special relativity. We have calculated the formula explicitly in Eq. (7) there. If A=\left[\begin{smallmatrix}\lambda&\mu\\ \bar{\mu}&\bar{\lambda}\end{smallmatrix}\right] is a complex matrix from SU(1,1) with \lambda=\lambda_r+i\lambda_i,\,\mu=\mu_r+i\mu_i split into real and imaginary parts, then the real 3\times 3 matrix L(A) from SO(2,1) is given by the formula:

(1)   \begin{equation*}L(A)=\begin{bmatrix} -\lambda_i^2+\lambda_r^2-\mu_i^2+\mu_r^2 & 2 \lambda_i \lambda_r-2 \mu_i \mu_r & 2 \lambda_r \mu_r-2 \lambda_i \mu_i \\ -2 \lambda_i \lambda_r-2 \mu_i \mu_r & -\lambda_i^2+\lambda_r^2+\mu_i^2-\mu_r^2 & -2 \lambda_r \mu_i-2 \lambda_i \mu_r \\ 2 \lambda_i \mu_i+2 \lambda_r \mu_r & 2 \lambda_i \mu_r-2 \lambda_r \mu_i & \lambda_i^2+\lambda_r^2+\mu_i^2+\mu_r^2 \end{bmatrix}.\end{equation*}

The map A\mapsto L(A) is a group homomorphism, that is L(I)=I and L(A_1A_2)=L(A_1)L(A_2). ( Of course in L(I)=I we denote by the same symbol "I" identity matrices of different sizes.)

After that, in Getting real, we used Cayley transform that defines group isomorphism between the complex matrix group SU(1,1) and the real matrix group SL(2,R). With

(2)   \begin{equation*}\mathcal{C}=\frac{1}{1+i}\begin{bmatrix}i&1\\-i&1\end{bmatrix}.\end{equation*}

(3)   \begin{equation*}\mathcal{C}^{-1}=\frac{1}{1-i}\begin{bmatrix}-i&i\\1&1\end{bmatrix}.\end{equation*}

we have that if A'=\left[\begin{smallmatrix}\alpha&\beta\\ \gamma&\delta\end{smallmatrix}\right] is a real matrix from SL(2,R) (i.e. \det A'=\alpha\delta-\beta\gamma=1), then \mathcal{C}A'\mathcal{C}^{-1} is in SU(1,1). If we calculate explicitly \lambda and \mu in terms of \alpha,\beta,\gamma,\delta, the result is:

(4)   \begin{eqnarray*} \lambda_r&=&\frac{\alpha+\delta}{2},\\ \lambda_i&=&\frac{\beta-\gamma}{2},\\ \mu_r&=&\frac{\delta-\alpha}{2},\\ \mu_i&=&\frac{\beta+\gamma}{2}. \end{eqnarray*}

Combining it with L(A) we obtain group homomorphism from SL(2,R) to SO(2,1). Explicitly

(5)   \begin{equation*}L(A')=\begin{bmatrix} \frac{\alpha ^2-\beta ^2-\gamma ^2+\delta ^2}{2} & \alpha \beta -\gamma \delta & \frac{-\alpha ^2-\beta ^2+\gamma ^2+\delta ^2}{2} \\ \alpha \gamma -\beta \delta & \beta \gamma +\alpha \delta & -\alpha \gamma -\beta \delta \\ \frac{-\alpha ^2+\beta ^2-\gamma ^2+\delta ^2}{2} & -\alpha \beta -\gamma \delta & \frac{\alpha ^2+\beta ^2+\gamma ^2+\delta ^2}{2} \end{bmatrix}.\end{equation*}

So far, so good, but now there comes real magic!

Real magic

We do not have to travel around the world, through SU(1,1) and hyperboloids. We will derive the last formula directly using a method that is similar to the one we used when deriving the map from quaternions to rotation matrices in Putting a spin on mistakes.

The Lie algebra sl(2,R) of the Lie group SL(2,R) consists of real 2\times 2 matrices of zero trace – we call them “generators”. We already met three particular generators X_1,X_2,X_3, for instance in SL(2,R) generators and vector fields on the half-plane, and I will skip this time primes as they have been denoted previously

(6)   \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*}

Every element of sl(2,R) is a real linear combination of these three. So, X_1,X_2,X_3 can be considered as a basis of sl(2,R). For instance in SL(2,R) generators and vector fields on the half-plane we constructed a particular generator Y_+ and Y_- defined as

(7)   \begin{eqnarray*}Y_+&=&\frac{1}{2}(X_1+X_3)=\begin{bmatrix}0&0\\1&0\end{bmatrix},\\ Y_-&=&\frac{1}{2}(X_1-X_3)=\begin{bmatrix}0&1\\0&0\end{bmatrix}. \end{eqnarray*}

The Lie algebra sl(2,R) is a three-dimensional real vector space. But it is more than just a vector space. The group SL(2,R) acts on this space by what is called “the adjoint representation”. This is true for the Lie algebra of any Lie group. Here we have a particular case. Namely, if X is in sl(2,R), that is if X has trace zero, and if A is in SL(2,R), that is the determinant of A is one, then AXA^{-1} is also of zero trace (we do not need the property of determinant one for this). The map X\mapsto AXA^{-1} is a linear map. Thus we have action, let us call it \mathcal{L}, of SL(2,R) on sl(2,R):

(8)   \begin{equation*} \mathcal{L}(A)X\mapsto AXA^{-1}.\end{equation*}

Remark: I will now be skipping primes that I was using to distinguish matrices from SL(2,R) from matrices from SU(1,1).

Evidently (from associativity of matrix multiplication) we have

    \[\mathcal{L}(A_1A_2)=\mathcal{L}(A_1)\mathcal{L}(A_2).\]

Usually instead of \mathcal{L}(A) one writes \mathrm{Ad}(A) and uses the term “adjoint representation”. In short: the group acts on its Lie algebra by similarity transformations. Similarity transformation of a generator is another generators. Even more, by expanding exponential into power series we can easily find that

(9)   \begin{equation*}e^{t AXA^{-1}}=Ae^{tX}A^{-1}.\end{equation*}

So sl(2,R) is a three dimensional real vector space and SL(2,R) acts on it by linear transformations.

But that is not all. In sl(2,R) we can define a very nice scalar product (X,Y) as follows

(10)   \begin{equation*}(X,Y)=\frac{1}{2}\mathrm{tr} (XY),\end{equation*}

where \mathrm{tr} (XY) is the trace of the product of matrices X and Y.

Why is this scalar product “nice”? What is so nice about it? It is nice, because with this scalar product the transformations \mathcal{L}(A) are all isometries – they preserve this scalar product:

(11)   \begin{equation*}(\mathcal{L}(A)X,\mathcal{L}(A)Y)=(X,Y).\end{equation*}

The derivation of this last property follows from the definitions and from the property that similarity transformations do not change the trace.

So sl(2,R) is a three dimensional real vector space with a scalar product. But in sl(2,R) we have our basis X_i,\,(i=1,2,3). It is easy to calculate scalar products of the basis vectors. We get the following matrix for the matrix G with entries

(12)   \begin{equation*}G_{ij}=(X_i,X_j):\end{equation*}

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

Thus sl(2,R) has all the properties of the Minkowski space with two space and one time dimensions. The generator X_3 has “time direction”, while X_1,X_2 are “space directions”.

Once we have a basis there, we can calculate the components of the transformations \mathcal{L}(A) in this basis:

(14)   \begin{equation*}\mathcal{L}(A)X_i=\sum_{j=1}^3 \mathcal{L}(A)_{ji}X_j,\,(i=1,2,3).\end{equation*}

I used Mathematica to calculate \mathcal{L}(A)_{ji} for A=\left[\begin{smallmatrix}\alpha&\beta\\ \gamma&\delta\end{smallmatrix}\right]. Here is the result:

(15)   \begin{equation*} \mathcal{L}(A)=\begin{bmatrix} \frac{\alpha ^2-\beta ^2-\gamma ^2+\delta ^2}{2} & \alpha \beta -\gamma \delta & \frac{-\alpha ^2-\beta ^2+\gamma ^2+\delta ^2}{2} \\ \alpha \gamma -\beta \delta & \beta \gamma +\alpha \delta & -\alpha \gamma -\beta \delta \\ \frac{-\alpha ^2+\beta ^2-\gamma ^2+\delta ^2}{2} & -\alpha \beta -\gamma \delta & \frac{\alpha ^2+\beta ^2+\gamma ^2+\delta ^2}{2} \end{bmatrix}.\end{equation*}

Here I admit that in this last formula I did copy and paste from Eq. (5) above. Because indeed that is what happened in the calculation – the result came identical. And that is the real magic. We do not need external space-time and hyperboloids. Everything is already contained in the group itself and in its Lie algebra!

Geodesics on the upper half-plane – parametrization

Posted on by arkajad

The last note ended with the following problem:

Thus: geodesics are circles. Or better: straight lines are circles! In fact: half-circles, because their centers are on the x-axis, and our arena is only upper half-plane.

Except that we have missed some solutions. In Conformally Euclidean geometry of the upper half-plane there was a sentence:

In the next note we will start calculating “straight lines”, or “geodesics” of our geometry. Some of them are almost evident candidates: vertical lines, perpendicular to the real line. But what about the other ones?

Well, we have these other ones, but how the vertical lines fit our reasoning above?

If \gamma(t)=(x(t),y(t)) is a vertical line, then x(t)=x_0, therefore \dot{x}=0. In Geodesics on the upper half-plane – Part 2 circles we arrived at equations

(1)   \begin{equation*} (\dot{\gamma}(t),K_1(\gamma(t)))=\frac{\dot{x}}{y^2(t)}=\mathrm{const},\end{equation*}

(2)   \begin{equation*} (\dot{\gamma}(t),K_2(\gamma(t)))=\frac{\dot{x}(t)x(t)+\dot{y}(t)y(t)}{y^2(t)}=\mathrm{const},\end{equation*}

and took the ratio of the second to the first. But for vertical lines taking the ratio is not allowed. The first equation is satisfied with the constant on the right hand side equal to zero. The second equation reduces to

(3)   \begin{equation*}(\dot{\gamma}(t),K_2(\gamma(t)))=\frac{\dot{y}(t)}{y(t)}=\mathrm{const}.\end{equation*}

If we now recall Eq. (2) from Conformally Euclidean geometry of the upper half-plane :

(4)   \begin{equation*}ds=\sqrt{\frac{dx^2+dy^2}{y^2}}=\sqrt{\frac{\dot{x}^2+\dot{y}^2}{y^2}}dt,\end{equation*}

we see that ds and dt on the vertical line must be proportional:

(5)   \begin{equation*}s=ct+s_0.\end{equation*}

Whenever this last equation holds, one says that t is an “affine parameter”: it is proportional to the arc length, possibly translated. In fact that is part of the definition of the geodesic that enters the “Noether’s theorem” that we are using. Usually we choose the proportionality constant equal to one.

Of course we can use Eq. (3) in order to determine the parameter t. Choosing the constant equal to 1, we have

(6)   \begin{equation*}\frac{dy}{y}=dt,\end{equation*}

therefore

(7)   \begin{equation*}\log y=t+t_0.\end{equation*}

To my surprise Bjab has discovered this all by himself – see the discussion under the previous post

Therefore, when discussing geodesics, we will assume that they are always parametrized by their arc length or, in other words, that the tangent vector \dot{\gamma}(t) is of unit length. In our case that is equivalent to

(8)   \begin{equation*}(\dot{\gamma}(s),\dot{\gamma}(s))_{\gamma(s)}=\frac{\dot{x}(s)^2+\dot{y}(s)^2}{y(s)^2}=1.\end{equation*}

Remark: There are many important examples of pseudo-Riemannian metrics, that are not positive definite. In such a case “length square” along a geodesic line is normalized to +1 or -1 or 0, so that there are three kinds of geodesics. In physics this happens for space-time metrics with Minkowski signature, and in multi-dimensional Kaluza-Klein theories. We will discuss another such case in the following posts.

We can now use these insights also in the case of circular geodesics. Before we have taken the ratio of two “conservation law” equations (1,2). Now, that we know we have a circle, we can write circle equation

(9)   \begin{eqnarray*}x&=&r\cos (\phi)+x_0,\\ y&=&r\sin( \phi),\label{eq:yr}\end{eqnarray*}

where \phi is some function of the arc length parameter s, and substitute into Eqs. (1,2).

We have

(10)   \begin{eqnarray*}\dot{x}&=&-r\sin (\phi)\, \dot{\phi}\\ \dot{y}&=&r\cos (\phi)\,\dot{\phi},\label{eq:dy}\end{eqnarray*}

therefore Eq. (8) reduces to

(11)   \begin{equation*} \frac{\dot{\phi}}{\sin\phi}=\pm 1.\end{equation*}

It is now a straightforward exercise to verify that with Eqs. (9–11) equations (1,2) are satisfied automatically.

Now that we know geodesics on the upper half-plane, we can draw them for pleasure. There is a famous pattern known as Dedekind’s tesselation

Dedekind tessellation

I was able to reproduce a part of it, namely the part described on p. 3 in the paper on SL(2,Z) by Keith Conrad

SL(2,Z) tessellation

But I would like to be able to reproduce the pretty image from website of Jerzy Kocik from Southern Illinois University

Dedekind tessellation Click on the image to view it full size

And I do not know yet how to do it.

Update: After several hours I managed to produce this:

My poor version – very primitive

Geodesics on the upper half-plane – Part 2 circles

Posted on by arkajad

In the last note Geodesics on the upper half-plane – Part 1 Killing vectors we verified that SL(2,R) transformations are isometries of the upper half-plane endowed with the Riemannian metric defined by the line element

(1)   \begin{equation*}ds^2=\frac{dx^2+dy^2}{y^2}.\end{equation*}

The corresponding scalar product between tangent vectors being

(2)   \begin{equation*}(\xi,\eta)_{z}=\frac{\xi_x\eta_x+\xi_y\eta_y}{y^2}.\end{equation*}

The idea is to use a “classical result” from differential geometry (a version of Noether’s theorem) that we have quoted from the online book by Sean Carroll:

Before applying this theorem to our case let us first see how it works for the standard plane Euclidean geometry. Translations are certainly isometries there. Streamlines of the vector fields generating horizontal and vertical translations form a square grid.

That straight lines intersect these grid lines under a constant angle is evident. In fact, that could serve as an alternative definition of a straight line.

So far so good.

But rotations are also isometries of the Euclidean geometry.

Looking at the image that shows the intersections of a straight lines with streamlines of the vector field generating rotation – nothing is evident. The angle of intersection is certainly not constant. We have to look at the theorem more closely.

The theorem states that the scalar product between Killing vector field (generating isometry) and the tangent vector to the geodesic is constant.

A general straight line \gamma(t)=(x(t),y(t)) on the plane has equation:

(3)   \begin{eqnarray*} x(t)&=&at+x_0,\\ y(t)&=&bt+y_0. \end{eqnarray*}

The tangent vector (\dot{x},\dot{y}) has components (a,b)

Rotation is defined by

(4)   \begin{eqnarray*} x(\phi)&=&\cos(\phi)x-\sin(\phi)y,\\ y(\phi)&=&\sin(\phi)x+\cos(\phi)y. \end{eqnarray*}

The corresponding vector field X(x,y) is obtained by taking derivatives with respect to \phi at \phi=0:

(5)   \begin{equation*} X(x,y)=(-y,x).\end{equation*}

It is streamlines of this vector field that we have drawn in the figure above.

Let us now calculate the scalar product of X and the vector tangent to our straight line at a point (x(t).y(t)) on our line:

(6)   \begin{equation*} (\dot{\gamma}(t),X(\gamma(t))=a(-y(t))+b(x(t))=a(-bt-y_0)+b(at+x_0)=-ay_0+bx_0=\mathrm{const}.\end{equation*}

Somewhat miraculously the non-constant (t-dependent) terms have cancelled out, and the scalar product is indeed constant. Not so evident, but nevertheless true. Perhaps there is some simple argument predicting this result, but I do not know it. Either use Noether’s theorem, or use direct calculation as above.

Let us now move from the Euclidean geometry of the plane to the hyperbolic geometry of upper half-plane. In SL(2,R) generators and vector fields on the half-plane we have focused our attention on two Killing vector fields (now we know that they are “Killing”), we will call them K_1,K_2 here:

(7)   \begin{eqnarray*} K_1(x,y)&=&(1,0),\\ K_2(x,y)&=&(x,y). \end{eqnarray*}

The first one corresponds to horizontal translation, the second one to uniform dilation. Their streamlines are

Suppose now \gamma(t)=(x(t),y(t)) is a geodesic, with tangent vector \dot{\gamma}(t)=(\dot{x}(t),\dot{y}(t)). Its scalar products with K_1 and K_2 should be constant. But now we remember that in calculating the scalar product we must use Eq. (2), so there will be denominator:

(8)   \begin{equation*} (\dot{\gamma}(t),K_1(\gamma(t)))=\frac{\dot{x}}{y^2(t)}=\mathrm{const},\end{equation*}

(9)   \begin{equation*} (\dot{\gamma}(t),K_2(\gamma(t)))=\frac{\dot{x}(t)x(t)+\dot{y}(t)y(t)}{y^2(t)}=\mathrm{const}.\end{equation*}

Their ratio should be therefore constant:

(10)   \begin{equation*}\frac{\dot{x}(t)x(t)+\dot{y}(t)y(t)}{\dot{x}(t)}=\mathrm{const},\end{equation*}

or

(11)   \begin{equation*}\dot{x}(t)x(t)+\dot{y}(t)y(t)=\mathrm{const }\, \dot{x}(t).\end{equation*}

This looks rather simple. In fact it should remind us about the equation of a circle. A circle with center on some point x_0 on x-axis and radius r has equation:

(12)   \begin{equation*}(x-x_0)^2+y^2=r^2.\end{equation*}

Assuming x=x(t),y=y(t) and differentiating with respect to t we get

(13)   \begin{equation*} 2\dot{x}(t)(x(t)-x_0)+2\dot{y}(t)y(t)=0,\end{equation*}

or

(14)   \begin{equation*} \dot{x}(t)x(t)+\dot{y}(t)y(t)=x_0\,\dot{x}(t).\end{equation*}

This is exactly Eq. (11)

Thus: geodesics are circles. Or better: straight lines are circles! In fact: half-circles, because their centers are on the x-axis, and our arena is only upper half-plane.

Except that we have missed some solutions. In Conformally Euclidean geometry of the upper half-plane there was the following sentence:

In the next note we will start calculating “straight lines”, or “geodesics” of our geometry. Some of them are almost evident candidates: vertical lines, perpendicular to the real line. But what about the other ones?

Well, we have these other ones, but how the vertical lines fit our reasoning above?

This is a homework. It is somewhat tricky though ….