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Orthochronous Lorentz transformations are time-preserving and $SL(2,\mathbb{R})$

+ 3 like - 0 dislike
62 views

Let's consider the psuedosphere/hyperboloid in $\mathbb{R}^{1,2}$ given by

$$x^2+y^2-z^2=-R^2.$$

We know that the Lorentz group

$$O(1,2)=\{ A \in Mat(3,\mathbb{R}): A^tGA=G \},$$

where $G=diag(-1,-1,1)$ leaves the pseudosphere invariant. Now we are interested in the following facts:

  1. How can we show that the orthochronous Lorentz group $O_+(1,2)=\{ A: a_{33}>0 \}$ is subgroup and, more important, maps upper cone to upper cone?

  2. What is the relation between groups $O_+(1,2)$ and $SL(2,\mathbb{R})$?

This post imported from StackExchange Physics at 2015-01-30 14:30 (UTC), posted by SE-user jj_p
asked Jun 17, 2013 in Mathematics by jj_p (150 points) [ no revision ]
retagged Feb 1, 2015 by dimension10
To help you start: for 1.) you will first of all need to show that if $g,h \in O_+$, then $g^{-1}$ and $g \cdot h$ are also in $O_+.$ This requires some algebra.

This post imported from StackExchange Physics at 2015-01-30 14:30 (UTC), posted by SE-user Vibert
OK, in the 'duplicate' it is explained first part of 1, i.e. why it is group. Is it also obvious that it maps upper cone to upper cone from that or from the topological answer therein?

This post imported from StackExchange Physics at 2015-01-30 14:30 (UTC), posted by SE-user jj_p
No, that's a separate problem. But you can adapt the same method as indicated in the duplicate to obtain the result.

This post imported from StackExchange Physics at 2015-01-30 14:30 (UTC), posted by SE-user Vibert
could you give me a hint in that direction?

This post imported from StackExchange Physics at 2015-01-30 14:30 (UTC), posted by SE-user jj_p

2 Answers

+ 4 like - 0 dislike

I) The proof that the orthochronous Lorentz group $O^{+}(1,d; \mathbb{R})$ form a group (which is closed/stabile under multiplication and inversion) is given in this Phys.SE post.

II) Next we would like to prove the following.

A Lorentz transformation takes a timelike vector $\tilde{x}=(x^0,x)$ with $|x| < |x^0|$ to a timelike vector $\tilde{x}^{\prime}=(x^{\prime 0},x^{\prime})$ with $|x^{\prime}| < |x^{\prime 0}|$.

Proof. This follows from the fact that a Lorentz transformation preserves the Minkowski norm. Thus to prove that an orthochronous Lorentz transformation

$$\tag{1} \Lambda ~=~ \begin{bmatrix}a & b^t \cr c &R \end{bmatrix}~\in~O(1,d; \mathbb{R}) $$

(which by definition has $a=\Lambda^0{}_0>0$) takes a future timelike vector $\tilde{x}=(x^0,x)$ with

$$\tag{2} |x| ~<~ x^0$$

to a future timelike vector $\tilde{x^{\prime}}=(x^{\prime 0},x^{\prime})$ with $|x^{\prime}| < x^{\prime 0}$, it is enough to prove that

$$\tag{3} 0~<~x^{\prime 0} ~=~ a x^0 + b\cdot x .$$

But the inequality (3) follows from the following inequality

$$\tag{4} -2\frac{b}{a} \cdot \frac{x}{x^0}~\leq~ \left(\frac{b}{a}\right)^2 +\left(\frac{x}{x^0}\right)^2 ~<~ \frac{a^2-1}{a^2} + 1 ~<~2. $$

Here we used the fact that $ b\cdot b =a^2-1$ and the inequality (2). End of proof.

III) Thus there only remains OP's last question:

What is the relation between $SL(2,\mathbb{R})$ and $SO^{+}(1,2;\mathbb{R})$?

Naturally our treatment will have some overlap with Trimok's correct answer. We use the sign convention $(+,-,-,-)$ for the Minkowski metric $\eta_{\mu\nu}$.

IV) First let us identify the Minkowski space $M(1,3;\mathbb{R})$ with the space of Hermitian $2\times2$ matrices $u(2)$. In detail, there is a bijective isometry from the Minkowski space $(M(1,3;\mathbb{R}),||\cdot||^2)$ to the space of Hermitian $2\times2$ matrices $(u(2),\det(\cdot))$, $$\mathbb{R}^4~=~M(1,3;\mathbb{R}) ~\cong ~ u(2) ~:=~\{\sigma\in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid \sigma^{\dagger}=\sigma \} ~=~ {\rm span}_{\mathbb{R}} \{\sigma_{\mu} \mid \mu=0,1,2,3\}, $$ $$ M(1,3;\mathbb{R})\ni\tilde{x}~=~(x^0,x^1,x^2,x^3) \quad\mapsto \quad\sigma~=~x^{\mu}\sigma_{\mu}~\in~ u(2), $$ $$\tag{5} ||\tilde{x}||^2 ~=~x^{\mu} \eta_{\mu\nu}x^{\nu} ~=~\det(\sigma), \qquad \sigma_{0}~:=~{\bf 1}_{2 \times 2},$$

see also this Phys.SE post.

V) There is a group action $\rho: SL(2,\mathbb{C})\times u(2) \to u(2)$ given by

$$\tag{6} g\quad \mapsto\quad\rho(g)\sigma~:= ~g\sigma g^{\dagger}, \qquad g\in SL(2,\mathbb{C}),\qquad\sigma\in u(2). $$

A straightforward calculation shows that the two groups $SL(2,\mathbb{R}) \equiv Sp(2,\mathbb{R})$ and

$$SU(1,1)~=~\left\{\left. \begin{bmatrix} a & b \\ b^{*} & a^{*} \end{bmatrix}\right| a,b\in \mathbb{C}, |a|^2-|b|^2=1\right\}$$ $$ \tag{7} ~=~\left\{\left. \begin{bmatrix} f\sqrt{|b|^2+1} & b \\ b^{*} & f^{*}\sqrt{|b|^2+1} \end{bmatrix} \right| f,b\in \mathbb{C}, |f|=1\right\}~\cong~S^1\times \mathbb{C}$$

are the stabilizer subgroups (also called the isotropy subgroups) of the $x^2$-coordinate and the $x^3$-coordinate, respectively. Since there is no spatially preferred direction, the two subgroups are isomorphic. (The explicit isomorphism is given in Ref. 1.) The two subgroups are path connected but not simply connected. In detail, the fundamental group is

$$\tag{8} \pi_1(SL(2,\mathbb{R}),*)~=~\pi_1(SU(1,1),*)~=~\pi_1(S^1\times \mathbb{C},*)$$ $$~=~\pi_1(S^1,*)\oplus \pi_1(\mathbb{C},*)~=~\mathbb{Z}.$$

VI) We now restrict attention to the $1+2$ dimensional case. Let us identity the Minkowski space $M(1,2;\mathbb{R})~\subseteq~ M(1,3;\mathbb{R})$ as the hyperplane $x^2=0$. The corresponding hyperplane in $u(2)$ is the set

$$\tag{9} s(2)~:=~\{ \sigma \in {\rm Mat}_{2\times 2}(\mathbb{R}) \mid \sigma^t =\sigma \} $$

of real symmetric $2\times2 $ matrices.

VII) There is a group action $\rho: SL(2,\mathbb{R})\times s(2) \to s(2)$ given by

$$\tag{10} g\quad \mapsto\quad\rho(g)\sigma~:= ~g\sigma g^t, \qquad g\in SL(2,\mathbb{R}),\qquad\sigma\in s(2), $$

which is length preserving, i.e. $g$ is a pseudo-orthogonal (or Lorentz) transformation. In other words, there is a Lie group homomorphism

$$\tag{11} \rho: SL(2,\mathbb{R}) \quad\to\quad O(s(2),\mathbb{R})~\cong~ O(1,2;\mathbb{R}).$$

Since $\rho$ is a continuous map from a path connected set $SL(2,\mathbb{R})$, the image $\rho(SL(2,\mathbb{R}))$ is also path connected. We conclude that Lie group homomorphism

$$\tag{12} \rho: SL(2,\mathbb{R}) \quad\to\quad SO^{+}(s(2),\mathbb{R})~\cong~ SO^{+}(1,2;\mathbb{R})$$

maps into the restricted Lorentz group $SO^{+}(1,2;\mathbb{R})$. [Here we have used the easily established fact that the Lorentz group $O(1,2;\mathbb{R})$ has at least four connected components because $\Lambda^0{}_{0}\neq 0$ and $\det(\Lambda)\neq 0$. We do not assume the fact that there is precisely four connected components.] It is trivial to check that the kernel

$$\tag{13} {\rm ker}(\rho)~=~\rho^{-1}({\bf 1}_{s(2)})~=~\{\pm {\bf 1}_{2 \times 2}\}~\cong~\mathbb{Z}_{2}.$$

Let

$$\tag{14} \tilde{\rho}: SL(2,\mathbb{R})/\mathbb{Z}_2 \quad\to\quad SO^{+}(1,2;\mathbb{R})$$

denote the corresponding injective Lie group homomorphism. Thus if we could prove that $\rho$ is surjective/onto, i.e. that the image ${\rm Im}(\rho)\equiv\rho(SL(2,\mathbb{R}))$ is precisely the restricted Lorentz group, cf. Section X below, we would have proved that

$SL(2,\mathbb{R})$ is the double cover of the restricted Lorentz group $SO^{+}(1,2;\mathbb{R})$.

Note that $SL(2,\mathbb{R})$ is not a universal cover, since we just saw in Section V that

$$\tag{15}\pi_1(SL(2,\mathbb{R}),*)~=~\mathbb{Z}.$$

The universal covering group $\overline{SL(2,\mathbf{R})}$ is an example of a finite-dimensional Lie group that is not a matrix group.

VIII) One may show that the exponential map $\exp: sl(2,\mathbb{R}) \longrightarrow SL(2,\mathbb{R})$ is not surjective

$$\tag{16}{\rm Im}(\exp) ~=~\left\{M\in SL(2,\mathbb{R}) \mid {\rm Tr}(M)> -2\right\} ~\cup~ \left\{-{\bf 1}_{2 \times 2}\right\} ~\subsetneq~ SL(2,\mathbb{R}).$$

It is a small miracle that plus/minus the exponential map $\pm \exp: sl(2,\mathbb{R}) \longrightarrow SL(2,\mathbb{R})$ is indeed surjective, which enough for our purposes, cf. the $\mathbb{Z}_{2}$-kernel (13).

IX) Next let us consider the following Lemma for arbitrary spatial dimensions $d$.

Lemma. Any restricted Lorentz transformation is a product of a pure rotation and a pure boost.

Proof. Let us decompose a Lorentz matrix $\Lambda$ into 4 blocks

$$\tag{17} \Lambda ~=~ \begin{bmatrix}a & b^t \cr c &R \end{bmatrix} ,$$

where $a=\Lambda^0{}_0\neq 0$ is a real number; $b$ and $c$ are real $d\times 1$ column vectors; and $R$ is a real $d\times d$ matrix. First argue from $\Lambda^t \eta \Lambda=\eta$, or equivalently from $\Lambda \eta^{-1} \Lambda^t=\eta^{-1}$, that

$$\tag{18} a^2~=~b^tb+1,\qquad c~=~\frac{Rb}{a}, \qquad b~=~\frac{R^tc}{a}. $$

Next argue that

$$\tag{19} B(b)~:=~ \begin{bmatrix} a & b^t \\ b & {\bf 1}_{d\times d}+\frac{bb^t}{a+1} \end{bmatrix}, \qquad a~:=~\sqrt{b^tb+1}~\geq~1, $$

is a Lorentz matrix with an inverse matrix

$$\tag{20} B(-b)~=~\begin{bmatrix} a & -b^t \\ -b & {\bf 1}_{d\times d}+\frac{bb^t}{a+1} \end{bmatrix}, \qquad B(b)B(-b)~=~{\bf 1}. $$

Such matrices correspond to pure (finite) boosts. Use this to prove the Lemma. Hint: The matrix $\Lambda B(-b)$ is on block diagonal form. End of proof.

Also note that we may conjugate a pure boost matrix with a pure rotation matrix to obtain a pure boost matrix in a preferred direction. The Lorentz algebra is

$$\tag{21} so(1,d;\mathbb{R})~=~\left. \left\{ \begin{bmatrix} 0 & b^t \\ b & r \end{bmatrix} \right| r^t ~=~ -r \right\}. $$

The exponential map is surjective on the set of pure boost:

$$\tag{22} \exp \begin{bmatrix} 0 & b^t \\ b & {\bf 0}_{d\times d} \end{bmatrix} ~=~ B\left(\frac{\sinh|b|}{|b|}b\right), \qquad |b|~:=~\sqrt{b^tb}~\geq~0.$$

Moreover, one may prove that the exponential map $\exp: so(d,\mathbb{R})\to SO(d,\mathbb{R})$ for pure rotations is surjective. For $d=2$ this is trivial.

[Below we only consider the case $d=2$.]

X) Finally, we are able to prove the following Lemma

Lemma. The group homomorphism $\rho: SL(2,\mathbb{R})\to SO^{+}(1,2;\mathbb{R})$ is surjective.

Proof. Note that boosts along the $x^3$-axis corresponds to

$$\tag{23} g(\beta)~:=~\begin{bmatrix}\exp\left(\frac{\beta}{2}\right) & 0 \cr 0 &\exp\left(-\frac{\beta}{2}\right) \end{bmatrix}\in SL(2,\mathbb{R}),$$

while rotations corresponds to

$$\tag{24} g(\theta)~:=~\begin{bmatrix}\cos\frac{\theta}{2} & \sin\frac{\theta}{2} \cr -\sin\frac{\theta}{2} &\cos\frac{\theta}{2} \end{bmatrix}\in SL(2,\mathbb{R}).$$

Given an arbitrary restricted Lorentz matrix $\Lambda\in SO^+(1,2;\mathbb{R})$, we saw in Section IX that it can be decomposed as (rotation)(boosts along the $x^3$-axis)(rotation'). Hence it can be hit by the $\tilde{\rho}$ group homomorphism

$$\tag{25} \Lambda ~=~ \tilde{\rho}\left(\tilde{g}(\theta)\tilde{g}(\beta)\tilde{g}(\theta^{\prime})\right).$$

End of proof.

XI) We have the following commutative diagram

$$\tag{26} \begin{array}{rlcrl} &&\tilde{\rho}_{\ast}&& \cr &sl(2,\mathbb{R}) & \longrightarrow &so(1,2;\mathbb{R}) \cr \pm\exp &\downarrow &\circlearrowright&\downarrow&\exp\cr &SL(2,\mathbb{R})/\mathbb{Z}_{2} & \longrightarrow &SO^+(1,2;\mathbb{R})\cr &&\tilde{\rho}&&\end{array} $$

All horizontal arrows are bijections. In particular, the above shows the following theorem.

Theorem. The exponential map $\exp: so(1,2;\mathbb{R}) \to SO^+(1,2;\mathbb{R})$ is surjective.

References:

  1. V. Bargmann, Irreducible Unitary Representations of the Lorentz Group, Ann. Math. 48 (1947) 568-640. The pdf file is available here. We mostly use results from p. 589-591.
This post imported from StackExchange Physics at 2015-01-30 14:30 (UTC), posted by SE-user Qmechanic
answered Jun 22, 2013 by Qmechanic (2,790 points) [ no revision ]
+ 1 like - 0 dislike

For 1) @Vibert gives you the indications.

For 2) The group $0(1,2)$ - with signatures (+ - -) has 4 disjoint components which can be characterized by :

$$M_1 = Diag (1, 1, 1)$$ $$M_2 = Diag (1, -1, -1)$$ $$M_3 = Diag (-1, 1, 1)$$ $$M_3 = Diag (- 1, -1, -1)$$

$S0(1,2)$ corresponds to matrix of determinant 1, so $S0(1,2)$ has 2 disjoint components ($M_1, M_2$)

$0^+(1,2)$ - which conserve the sign of the 1st coordinate - has 2 disjoint components ($M_1, M_2$)

$S0^+(2,1)$ - has 1 component ($M_1$)

$SL(2,\mathbb{R})$ is connected (so only 1 component), but it is not simply connected.

So, it is not possible to have a isomorphism between $SL(2,\mathbb{R})$ and $0^+(1,2)$ because the number of disjoint components is different.

We could think about an isomorphism between $SL(2,\mathbb{R})$ and $S0^+(1,2)$, but in fact the isomorphism is between $SL(2,\mathbb{R})$ and $Spin^+(1,2)$, while there is an isomorphism between $PSL(2,\mathbb{R})$ and $S0^+(1,2)$, see Wikipedia

Note that $SL(2,\mathbb{R})$, $SU(1,1)$, and $Sp(2,\mathbb{R})$ are isomorphic, see this question.

This post imported from StackExchange Physics at 2015-01-30 14:30 (UTC), posted by SE-user Trimok
answered Jun 17, 2013 by Trimok (950 points) [ no revision ]
Most voted comments show all comments
So the right thing to look for is an isomorphism between $PSL(2,\mathbb{R})$ and $SO(1,2):$ can you point me to a proof of that?

This post imported from StackExchange Physics at 2015-01-30 14:30 (UTC), posted by SE-user jj_p
The isomorphism is between $PSL(2,\mathbb{R})$ and $S0^+(1,2)$

This post imported from StackExchange Physics at 2015-01-30 14:30 (UTC), posted by SE-user Trimok
You could find a demonstration page 16 of this ref. In this ref, note that $S0_o(1,2)$ is the same thing that $S0^+(1,2)$, and that the $Z_2$ quotient is there because in the formula (32), the result is the same if you choose $-g$ instead of $g$

This post imported from StackExchange Physics at 2015-01-30 14:30 (UTC), posted by SE-user Trimok
Right. This makes perfet sense, and it is the analogous of what is done in 4 dimensions. Still I'm not sure how to prove that $O_+$ leaves the upper cone invariant.

This post imported from StackExchange Physics at 2015-01-30 14:30 (UTC), posted by SE-user jj_p
Right, in 4 dimensions, the isomorphism is between $PSL(2,\mathbb{C})$ and $S0_o(1,3)$.

This post imported from StackExchange Physics at 2015-01-30 14:30 (UTC), posted by SE-user Trimok
Most recent comments show all comments
Each of the 2 components corresponding to $M_1$ and $M_2$ (in my answer) keep the orientation for the first coordinate (it is clear when looking at the matrices). The full $0^+(1,2) = SO(1,2)$, is a subgroup of $0(1,2)$. Being the the union of these 2 disjoint components which preserve the orientation of the 1st coordinate, obviously the full $0^+(1,2)$ preserve also the orientation of the 1st coordinate.

This post imported from StackExchange Physics at 2015-01-30 14:30 (UTC), posted by SE-user Trimok
For instance, $M_1$ and $M_2$ are special elements of $0^+(1,2)$. And you can check that $M_1 M_1 = M_1$, $M_1 M_2 = M_2$, $M_2 M_1 = M_2$, $M_2 M_2 = M_1$

This post imported from StackExchange Physics at 2015-01-30 14:30 (UTC), posted by SE-user Trimok

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