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This is a result physics books tell all the time, that the branch of proper Lorentz transformations with positive first entry forms the identity component of Lorentz group.

In mathematical language, for all $\Lambda\in\mathrm{SO}(1,3)$ and $(\Lambda)_{00} \geq 0$ (indices: 0,1,2,3), we have $\Lambda\in\mathrm{SO}^{+}(1,3)$. The converse is also true. But why? $(\Lambda)_{00} \geq 0$ is a very weak condition, and I have tried for hours but cannot even prove such $\Lambda$'s is closed under multiplication. Could anyone give me a hint?

In addition, is this result generalizable to arbitrary $\mathrm{SO}(1,n)$?

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2 Answers 2

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It is perhaps easiest to think in terms of frames.

Let $\{e_0, e_1, \ldots, e_n\}$ be a basis (we treat them as explicit column vectors in the standard coordinates of the vector space) of $\mathbb{R}^{1+n}$ such that under the Minkowski metric you have $$ m(e_0, e_0) = -1\qquad m(e_i,e_j) = \delta_{ij}, i,j \geq 1, \qquad m(e_0,e_j) = 0$$ (so that it forms an orthonormal basis of the Minkowski metric). You have that the matrix $$ \Lambda = \begin{pmatrix} e_0 & e_1 &\ldots & e_n\end{pmatrix}$$ formed with columns the basis vectors is an element of $O(1,n)$. Similarly, from the definition of $O(1,n)$ (that is preserves the Minkowski metric) you see that every element must be of the form above: that its individual column vectors form an orthonormal basis of Minkowski space.

Since the determinant is a continuous function, we have that the subsets of $O(1,n)$ for which $\det$ are $\pm1$ must be disconnected. Define $SO(1,n)$ to be the component with the positive determinant.

Now, since $m(e_0,e_0) = -1$, we expand $$ m(e_0,e_0) = - \left[(e_0)_0\right]^2 + \sum_{i = 1}^n \left[(e_0)_i\right]^2 \geq - \left[(e_0)_0\right]^2 $$ where $(e_0)_\mu$ are the components of the column vector $e_0$. This implies that $\left[(e_0)_0\right]^2 = (\Lambda_{00})^2 \geq 1$. Hence $\Lambda_{00}$ is either greater or equal to positive 1, or less than or equal to negative 1. Since for a connected component, $\Lambda_{00}$ is a continuous function, we must have that the sets $O(1,n) \cap \{\Lambda_{00} > 0\}$ and $O(1,3)\cap \{\Lambda_{00} < 0\}$ are disconnected. We call the component where $\Lambda_{00} > 0$ by $O^+(1,3)$.

We claim that $O^+(1,n)$ forms a subgroup. We argue geometrically. Let $\Lambda$ be as above.

Claim 1. Let $u,v,w\in \mathbb{R}^{1+n}$ be such that $m(u,u), m(v,v), m(w,w)$ are all negative (so they are all time-like). Then $m(u,v)\cdot m(v,w)$ and $m(u,w)$ have the opposite signs.

Proof: complete $v$ to an orthonormal basis $\{v_0 = v, v_1, v_2, \ldots, v_n\}$. We have that $u = - m(v,u) v_0 + \sum m(v_i,u) v_i$ and similarly $w$. The claim follows immediately.

Claim 2. If $v,w$ are time-like vectors such that $v-w$ is space-like. Then $m(v,w) < 0$.

Proof: Observe that $$ m(v,w) + m(v, v-w) + m(w,v) + m(w,w-v) = m(v,v) + m(w,w) < 0 $$ and that the left hand side rearranges to $$ 2 m(v,w) + m(v-w,v-w) \geq 2 m(v-w) $$

Claim 3. If $\Lambda \in O^+(1,n)$, and $v$ is a time-like vector such that the $0$ component of $v$ is positive, then so is $\Lambda\cdot v$. (That is, $\Lambda$ preserves time orientation.)

Proof: Let $s_0$ be the column vector $(1,0,0\ldots)$. By assumption on the $0$ component, we have $m(s_0,v) < 0$. By the isometry property, $\Lambda\cdot v$ is time-like. By definition, $\Lambda \cdot v - |m(s_0,v)| e_0$ is space-like. $e_0$ is time-like. By Claim 2 we have that $m(e_0, \Lambda\cdot v) < 0$. By assumption we have $m(e_0,s_0) < 0$. So by Claim 1 we must also have $m(\Lambda \cdot v, s_0) < 0$, as desired.

Claim 4. $O^+(1,n)$ is a subgroup.

Proof: the identity clearly belongs to it. It suffices to show that matrix multiplication is closed. Let $\Lambda,\Lambda'$ be elements of $O^+(1,n)$. The $00$ component of the product $\Lambda\Lambda'$ is obtained by $s_0^T\Lambda\Lambda' s_0$. That is to say, for $\Lambda\Lambda'$ to have positive $00$ component it suffices that $m(\Lambda\Lambda's_0,s_0) < 0$. But this follows from Claim 3.


The above implies that $O^+(1,n)$ and $SO(1,n)$ are subgroups of $O(1,n)$. Hence their intersection $SO^+(1,n) = O^+(1,n) \cap SO(1,n)$ is a subgroup of $O(1,n)$. Furthermore, by definition, the connected component of the identity of $O(1,n)$ entirely contained in $SO^+(1,n)$. Thus it remains to show that $SO^+(1,n)$ is (path) connected. One can in principle show this by considering the exponential map applied to the corresponding Lie algebra, but a more intuitive way is as follows:

using the above characterisation of $\Lambda$ as a frame, it suffices to find a continuous family of frames connecting the standard basis $\{s_0, s_1, \ldots, s_n\}$ to the basis $\{e_0, e_1, \ldots, e_n\}$. A path can be sketched as follows: first to an $SO(n)$ rotation on $\{e_0, \ldots, e_n\}$ such that $e_0$ is in the span of $\{s_0, s_1\}$. This is path connected. Then a hyperbolic translation (Lorentz boost) can bring $e_0$ back to $s_0$; this step requires that $\Lambda\in O^+(1,3)$. Now finish with another $SO(n)$ rotation aligning the remainder of the axes. This uses the fact that $SO(n)$ is connected, and that $\Lambda\in SO^+(1,3)$, so that once $e_0$ is "brought back" to $s_0$, the remainder of the axis are of the same orientation as $\{s_1, \ldots, s_n\}$.

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The action of $SO(1,3)$ leaves invariant the quadratic form $t^2-x^2-y^2-z^2$. It is easily seen that the orbits of $SO(1,3)$ are precisely the sets of vectors $(t,x,y,z)$ for which this form takes a fixed value $q\in\mathbf R$. If $q=0$ this is a cone, otherwise it is a hyperboloid. But there is a qualitative difference between the case $q<0$, where the hyperboloid is connected, and the case $q>0$ where the hyperboloid has $2$ connected components, one on each side of the hyperplane defined by $t=0$. Since the action of $SO(1,3)$ is continuous, it follows that $SO(1,3)$ can be separated into two open subsets according to whether it sends a particular vector $v_0=(1,0,0,0)$ (where the form is positive) to a vector in one or the other of the connected components of its orbit, i.e., a vector with $t>0$ or with $t<0$ (in fact with $t\geq 1$ or with $t\leq-1$). Since the stabilizer subgroup of $v_0$ is isomorphic to $SO(3)$, and therefore connected, it follows that these two open subsets of $SO(1,3)$ are in fact the connected components of $SO(1,3)$.

And as a consequence there is nothing special about the choice of $v_0$: doing the same for any other vector where the quadratic form is positive (and whose orbit therefore has two components) would lead to the same separation of $SO(1,3)$ into two parts.

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