How to prove that orthochronous Lorentz transformations $O^+(1,3)$ form a group?

Question

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 $J_i,\ K_i$?

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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".

Answer 1

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)

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

Here, the superscript "$t$" denotes matrix transposition. Note that the eq. $\eqref{eq:2}$ 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. $\eqref{eq:2}$ 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

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

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 $\eqref{eq:11}$ that

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

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

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

References:

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

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$^{\dagger}$A mathematician would probably say that eqs. $\eqref{eq:13}$ and $\eqref{eq:14}$ 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.

Answer 2

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).

Answer 3

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) $.

Answer 4

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.

Answer 5

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.