Orthochronous Lorentz transform are Lorentz transforms that satisfy the conditions (sign convention of Minkowskian metric $+---$) $$ \Lambda^0{}_0 \geq +1.$$ 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 $ \Lambda^0{}_0 \geq +1$ enough for a Lorentz transform to be 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".


5 Answers 5


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

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

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)

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

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

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

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

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

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

Exercise 1: Prove that

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

Exercise 2: Deduce that

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

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

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

Exercise 4: Prove that

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

Next let us consider a product

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

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

Exercise 5: Show that

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

Exercise 6: Prove the double inequality

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

which may compactly be written as $$| a_3-a_1a_2|~\leq~\sqrt{a_1^2-1}\sqrt{a_2^2-1}.\tag{12}$$

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

$$ a_1\neq 0 ~\text{and}~ a_2\neq 0~\text{have same signs} \quad\Rightarrow\quad a_3>0. \tag{13}$$ $$ a_1 \neq 0~\text{and}~ a_2\neq 0~\text{have opposite signs} \quad\Rightarrow\quad a_3<0. \tag{14}$$

Exercise 8: Use eq. (13) to prove that $O^{+}(1,d;\mathbb{R})$ is stabile/closed under the multiplication map.

Exercise 9: Use eq. (14) 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.$^{\dagger}$


  1. S. Weinberg, Quantum Theory of Fields, Vol. 1, 1995; p. 57-58.

$^{\dagger}$A mathematician would probably say that eqs. (13) and (14) show that the map

$$O(1,d;\mathbb{R})\quad \stackrel{\Phi}{\longrightarrow}\quad \{\pm 1\}~\cong~\mathbb{Z}_2\tag{15}$$

given by

$$\Phi(\Lambda)~:=~{\rm sgn}(\Lambda^0{}_0)\tag{16}$$

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

$$ {\rm ker}(\Phi)~:=~\Phi^{-1}(1)~=~O^{+}(1,d;\mathbb{R}) \tag{17}$$

is always a normal subgroup.

For a generalization to indefinite orthogonal groups $O(p,q;\mathbb{R})$, see this Phys.SE post.

  • 1
    $\begingroup$ Comment: There is a generalization to indefinite orthogonal groups $O(p,q;\mathbb{R})\ni \Lambda=\begin{pmatrix} a & b \cr c& d\end{pmatrix}$ with $O^+(p,q;\mathbb{R}):=\{\Lambda\in O(p,q;\mathbb{R})\mid \det(a)>0\}$. Then $|\det(a)|\geq 1$, $|\det(d)|\geq 1$, and $\det(\Lambda)~=~{\rm sgn}\det(a)~{\rm sgn}\det(d)$. $\endgroup$
    – Qmechanic
    Commented Jun 21, 2019 at 10:30
  • $\begingroup$ Wait, how do you do the Cauchy-Schwarz step (Ex. 6) for the general case of the indefinite Orthogonal group? $\endgroup$ Commented Jul 26, 2019 at 12:37

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).

  • $\begingroup$ does this prove also that $O_+$ leaves the pseudosphere invariant? $\endgroup$
    – jj_p
    Commented Jun 18, 2013 at 7:08
  • $\begingroup$ That's the definition of the Lorentz group, I don't know what there is to prove. $\endgroup$
    – Ron Maimon
    Commented Jun 18, 2013 at 14:11
  • $\begingroup$ I meant upper component: does this prove that $O_+$ leaves the upper hyperboloid invariant? $\endgroup$
    – jj_p
    Commented Jun 19, 2013 at 10:22
  • $\begingroup$ @Nicolo': Yes . $\endgroup$
    – Ron Maimon
    Commented Jun 19, 2013 at 14:49
  • 1
    $\begingroup$ This doesn't generalise to the case of $O^+(m,n)$ where $n>1$, because the hyperboloid is then connected. $\endgroup$ Commented Jul 30, 2019 at 7:59

An easy way to do this is to prove $\Lambda \in \rm O(3,1)$ preserves the sign of $v^0$ for a timelike vector $v \in \mathbb R^4$ iff $\Lambda \in \rm O^+(3,1) = \{\Lambda \in \mathrm O(3,1) : \Lambda^0{}_0 > 0\}$. Once you've proved this, it falls out immediately that $\mathrm O^+(3,1)$ is a subgroup, since if $\Lambda$ and $\Lambda'$ preserve the direction of time, so too must $\Lambda^{-1}$ and $\Lambda \Lambda'$. That $\mathrm O^+(3,1)$ is a normal subgroup is also simple to prove.

The first step is to prove $$ (\Lambda^0{}_0)^2 = 1 + \Lambda^0{}_i \Lambda_0{}^i $$ for any $\Lambda \in \mathrm O(3,1)$, where Latin indices sum over $1,2,3$. Then it is easy to show using the Cauchy-Schwarz inequality that for any timelike $v \in \mathbb R^4$ with $v^0 > 0$ we have $$ \Lambda^0{}_\mu v^\mu > 0 $$ where Greek indices sum over $0,1,2,3$. Now let $\Lambda' \in \mathrm{O}^-(3,1)$, and $v$ be timelike with $v^0 < 0$. It follows $-v$ is timelike with $v^0 > 0$ and $-\Lambda' \in \mathrm{O}^+(3,1)$, thus $$ \begin{align*} \Lambda^0{}_{\mu}(-1)v'^\mu &> 0 \\ \implies \Lambda^0{}_{\mu}v'^\mu &< 0 \\ (-1)\Lambda'^0{}_{\mu} v^\mu &> 0 \\ \implies \Lambda'^0{}_{\mu} v^\mu &< 0 \\ (-1)\Lambda'^{0}{}_{\mu}(-1)v'^\mu &> 0 \\ \implies \Lambda'^0{}_{\mu}v'^\mu &> 0 \end{align*} $$ Hence $\Lambda \in \rm O(3,1)$ preserves the sign of $v^0$ for a timelike vector $v \in \mathbb R^4$ iff $\Lambda \in \rm O^+(3,1) $.


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.


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.

  • 1
    $\begingroup$ Closure is definitely not easy, and that is what I'm asking. $\endgroup$
    – Siyuan Ren
    Commented Sep 14, 2012 at 5:14
  • $\begingroup$ I want elementary linear algebra approach. No involvement of generators. $\endgroup$
    – Siyuan Ren
    Commented Sep 14, 2012 at 5:50
  • $\begingroup$ @Misha: You say "no physical transformation may be imagined which does not form a group.", and yet there are things in physics which are not associative: theoreticalatlas.wordpress.com/2011/07/20/… so presumably you don't consider the examples at that link to count as physical transformations, not even velocity-addition which is closed but not associative. $\endgroup$
    – vtt
    Commented Sep 14, 2012 at 18:56
  • 1
    $\begingroup$ @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. $\endgroup$
    – Misha
    Commented Sep 15, 2012 at 5:48
  • $\begingroup$ @Karsus Ren sorry, I have no time to write the whole stuff. Are you satisfied with Qmechanic answer? He did an excellent job writing this in details. $\endgroup$
    – Misha
    Commented Sep 15, 2012 at 7:09

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