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Proper orthochronous Loentz transform are Lorentz transforms that satisfy the conditions (sign convention of Minkowskian metric $+---$) $$\det \Lambda=+1, \qquad \Lambda^0{}_0 \geq +1.$$ How to prove they form a subgroup of Lorentz group? All books I read only give this result, but no derivation.

Why is this condition $\det \Lambda=+1, \ \Lambda^0{}_0 \geq +1$ enough for a Lorentz transform to be proper orthochronous?

The temporal component of a transformed vector is $$x'^0=\Lambda^0{}_0 x^0+\Lambda^0{}_1 x^1+\Lambda^0{}_2 x^2+\Lambda^0{}_3 x^3,$$ the positivity of $\Lambda^0{}_0$ alone does not seem at first glance sufficient for the preservation of the sign of temporal component.

And how to prove that all Lorentz transform satisfying such simple conditions can be generated from $J_i,\ K_i$?

For those who think that closure and invertibility are obvious, keep in mind that $$\left(\bar{\Lambda}\Lambda \right)^0{}_0\neq \bar{\Lambda}^0{}_0\Lambda^0{}_0,$$ but instead $$\left(\bar{\Lambda}\Lambda \right)^0{}_0= \bar{\Lambda}^0{}_0\Lambda^0{}_0+\bar{\Lambda}^0{}_1\Lambda^1{}_0+\bar{\Lambda}^0{}_2\Lambda^2{}_0+\bar{\Lambda}^0{}_3\Lambda^3{}_0.$$

And I'm looking for a rigorous proof, not physical "intuition".

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Sorry, Karsus, after having read everything that is here, I would still insist that the closure is completely obvious, and this question deserved a minus one. The proper orthochronous group is exactly the subgroup of the Lorentz group that is connected to the identity (and connectedness can't spoil the closure because one may simply merge two paths). Alternatively, the sum of the four terms in your last equation is clearly greater or equal than one by a triangle identity if the two Lambda matrices are Lorentz and orthochronous. –  Luboš Motl Sep 14 '12 at 7:16
@LubošMotl: It's physically obvious, but he wants a rigorous proof that makes it equally mathematically obvious. It's something I didn't like, having to do ridiculous algebra on such non-invariant things like positive time components, for something so physically clear. +1, because I think one should be nicer and twenty years ago, I would have likely asked this too. –  Ron Maimon Sep 14 '12 at 7:18
@LubošMotl: Well, maybe I should have asked about $\mathrm{SO}^{+}(1,3)$ instead of proper orthochronous Lorentz group to avoid so many "physical" argument when I want a mathematical one. And what triangle identity should I use? If it is that obvious, could you bother posting it up for the stupid people like me? –  C.R. Sep 14 '12 at 8:22
You all seem to believe that proper orthochronous Lorentz transform is in itself proof of group structure. So I reformulate the question: why is Lorentz transforms satisfying this condition proper orthochronous. –  C.R. Sep 14 '12 at 9:04
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4 Answers

up vote 11 down vote accepted

Let the Minkowski metric $\eta_{\mu\nu}$ in $d+1$ space-time dimensions be

$$\tag{1}\eta_{\mu\nu}~=~{\rm diag}(1, -1, \ldots,-1).$$

Let the Lie group of Lorentz transformations be denoted as $O(1,d;\mathbb{R})=O(d,1;\mathbb{R})$. A Lorentz matrix $\Lambda$ satisfies (in matrix notation)

$$\tag{2} \Lambda^t \eta \Lambda~=~ \eta. $$

Here the superscript "$t$" denotes matrix transposition. Note that the eq.(2) does not depend on whether we use east-coast or west-coast convention for the metric $\eta_{\mu\nu}$.

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

$$\tag{3} \Lambda ~=~ \left[\begin{array}{cc}a & b^t \cr c &R \end{array} \right],$$

where $a=\Lambda^0{}_0$ is a real number; $b$ and $c$ are real $d\times 1$ column vectors; and $R$ is a real $d\times d$ matrix.

Now define the set of orthochronous Lorentz transformations as

$$\tag{4} O^{+}(1,d;\mathbb{R})~:=~\{\Lambda\in O(1,d;\mathbb{R}) | \Lambda^0{}_0 > 0 \}.$$

The proof that this is a subgroup can be deduced from the following string of exercises.

Exercise 1: Prove that

$$\tag{5} |c|^2~:= ~c^t c~ = ~a^2 -1.$$

Exercise 2: Deduce that

$$\tag{6} |a|~\geq~ 1.$$

Exercise 3: Use eq.(2) to prove that

$$\tag{7} \Lambda \eta^{-1} \Lambda^t~=~ \eta^{-1}. $$

Exercise 4: Prove that

$$\tag{8} |b|^2~:= ~b^t b~ = ~a^2 -1.$$

Next let us consider a product

$$\tag{9} \Lambda_3~:=~\Lambda_1\Lambda_2$$

of two Lorentz matrices $\Lambda_1$ and $\Lambda_2$.

Exercise 5: Show that

$$\tag{10} b_1\cdot c_2~:=~b_1^t c_2~=~a_3-a_1a_2.$$

Exercise 6: Prove the double inequality

$$\tag{11} -\sqrt{a_1^2-1}\sqrt{a_2^2-1} ~\leq~ a_3-a_1a_2~\leq~ \sqrt{a_1^2-1}\sqrt{a_2^2-1},$$

which may compactly be written as $| a_3-a_1a_2|~\leq~\sqrt{a_1^2-1}\sqrt{a_2^2-1}$.

Exercise 7: Deduce from the double inequality (11) that

$$\tag{12} a_1\neq 0 ~\text{and}~ a_2\neq 0~\text{have same signs} \quad\Rightarrow\quad a_3>0. $$ $$\tag{13} a_1 \neq 0~\text{and}~ a_2\neq 0~\text{have opposite signs} \quad\Rightarrow\quad a_3<0. $$

Exercise 8: Use eq.(12) to prove that $O^{+}(1,d;\mathbb{R})$ is stabile/closed under the multiplication map.

Exercise 9: Use eq.(13) to prove that $O^{+}(1,d;\mathbb{R})$ is stabile/closed under the inversion map.

The Exercises 1-9 show that the set $O^{+}(1,d;\mathbb{R})$ of orthochronous Lorentz transformations form a subgroup.$^1$

$^1$A mathematician would probably say that eqs. (12) and (13) show that the map

$$O(1,d;\mathbb{R})\quad \stackrel{\Phi}{\longrightarrow}\quad \{\pm 1\}~\cong~\mathbb{Z}_2$$

given by

$$\Phi(\Lambda)~:=~{\rm sgn}(\Lambda^0{}_0)$$

is a group homomorphism between the Lorentz group $O(1,d;\mathbb{R})$ and the cyclic group $\mathbb{Z}_2$, and a kernel

$$ {\rm ker}(\Phi)~:=~\Phi^{-1}(1)~=~O^{+}(1,d;\mathbb{R}) $$

is always a normal subgroup.

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Excellent answer. One point: the $a_{1,2}\neq 0$ condition in eqns 12 and 13 is redundant, given eqn 6 (though I'm guessing you know that). –  James Sep 14 '12 at 19:33
Right, the conditions $a_{1,2}\neq 0$ in eqs (12) and (13) are only imposed for semantic reasons when discussing same/opposite signs. –  Qmechanic Sep 14 '12 at 19:51
Thank you! Exactly what I need. –  C.R. Sep 15 '12 at 7:56
Summary of this answer: Cauchy inequality. –  C.R. Sep 15 '12 at 14:08
@KarsusRen: This answer is correct, but my answer is better: this stuff is exactly what should not be necessary, and isn't if you use topology. –  Ron Maimon Sep 16 '12 at 21:12
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Your problem bugged me too a long time ago, so I know what you are asking about. The "proper" part is easy from the fact that determinants multiply under matrix multiplication, so restricting to unit determinant is simple. The positive sign of the time component is proved topologically.

The Lorentz group moves the unit time vector somewhere on the hyperboloid:

$$ t^2 - x^2 = 1 $$

In however many dimensions. This is a disconnected space, there are two components--- the ones with t>0 and t<0. To prove disconnected, you can see that there are no real solutions to the equation with $-1<t<1$, and the intermediate value theorem requires that any path connecting the top hyperboloid with the bottom pass through the middle.

This means that any transformation where the image of the unit time vector reverses the sign of time is disconnected from the identity. If you look at the component of the Lorentz group connected to the identity, it must not reverse the sign of the time vector, and the property of being continuously connected to the identity is preserved under multiplication and inverses, by an easy argument (connect to the identity and take pointwise product/inverse).

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does this prove also that $O_+$ leaves the pseudosphere invariant? –  jj_p Jun 18 '13 at 7:08
That's the definition of the Lorentz group, I don't know what there is to prove. –  Ron Maimon Jun 18 '13 at 14:11
I meant upper component: does this prove that $O_+$ leaves the upper hyperboloid invariant? –  jj_p Jun 19 '13 at 10:22
@Nicolo': Yes . –  Ron Maimon Jun 19 '13 at 14:49
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Misha's answer is correct and complete.

However, let me give you the physical argument that explains why you do not find the proof in any book. The proper orthochronous transformations are spatial rotations and pure Lorentz transformations (or boosts). And it is clear from a physical point of view that these transformations verify the group laws: closure, existence of inverse (opposite angle or velocity) and identity.

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All group axioms are satisfied and obvious. Closure is easy to prove, Associativity is easy to prove, Identity is obvious and Inverse is obvious.

In principle, no physical transformation may be imagined which does not form a group. The group definition is mostly inspired by the idea of movements from physics: rotations, shifts, Lorentz transform, etc.

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Closure is definitely not easy, and that is what I'm asking. –  C.R. Sep 14 '12 at 5:14
Extend your question and I will extend my answer. I see two ways 1) All elements may be represented as a product of basis boost and rotations. The closure is obvious for them. 2) Do some linear algebra. Which way you prefer? –  Misha Sep 14 '12 at 5:42
I want elementary linear algebra approach. No involvement of generators. –  C.R. Sep 14 '12 at 5:50
@KarsusRen: The only slightly non-obvious thing (in terms of rigorous proofs) is the topological bit for positivity. –  Ron Maimon Sep 14 '12 at 6:15
@vtt Magmas, loops and others may be constructive to use sometimes, somewhere for something which is not physical transform/coordinate change/etc. You have a wrong idea of what associativity means. –  Misha Sep 15 '12 at 5:48
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