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It is often stated that rotations in the 3 spatial dimensions are examples of Lorentz transformations.

But Lorentz transformations form a group named the Lorentz Group, $O(1,3)$ which is a group a $4 \times 4$ matrices, $\Lambda$ having the following property:

$$ \Lambda^T g \Lambda = g$$

where $g$ is the metric tensor.

Now rotations matrices for the 3 spatial dimensions are $3 \times 3$ matrices and form $SO(3)$. How can they be in the $O(1,3)$ ?

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Think that the Lorentz group is made of spatial rotations, space-time rotations (boosts) and space/time reversal. – jinawee Feb 17 '14 at 19:11
up vote 7 down vote accepted

One can embed the $3\times3$ rotation matrices

$$R~\in~ SO(3)~:=~\{R\in{\rm Mat}_{3\times 3}(\mathbb{R}) \mid R^tR~=~{\bf 1}_{3\times 3}~\wedge~ \det(R)=1 \}$$

into the $4\times4$ Lorentz matrices

$$\Lambda~\in~ O(1,3)~:=~\{\Lambda\in{\rm Mat}_{4\times 4}(\mathbb{R}) \mid\Lambda^t\eta \Lambda~=~\eta \}$$


$$SO(3)~\ni~R~\stackrel{\Phi}{\mapsto}~ \Lambda ~=~ \left[\begin{array}{cc}1 & 0 \cr 0 &R \end{array} \right]~\in~ O(1,3).$$

It is not hard to see that this embedding $\Phi:SO(3)\to O(1,3)$ is an injective group homomorphism

$$\Phi(R_1 R_2)=\Phi(R_1)\Phi(R_2), \qquad R_1,R_2~\in~SO(3).$$

The pertinent group operations are for both groups just matrix multiplication.

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Just a comment. Using this embedding one can prove a technically important theorem. Given $\Lambda \in O(1,3)$ there exist $R_1,R_2 \in SO(3)$ such that $\Lambda = \Phi(R_1)\Lambda_3 \Phi(R_2)$ for a special Lorentz transformation along $z$. $R_1,R_1, \Lambda_3$ are uniquely determined by $\Lambda$. – Valter Moretti Feb 18 '14 at 15:16
I have a question. What is the purpose of the embedding? Phased differently, we have $O(1,3)$ and why do we need to embed the $SO(3)$ in it? – omephy Feb 18 '14 at 15:57
Perhaps the following comment is helpful: Often in Physics we are interested in the symmetries of the theory (which we investigate). A relativistic theory should be invariant under Lorentz transformations. In particular it should be invariant under spatial rotations. – Qmechanic Feb 18 '14 at 17:10
@Qmechanic: Thanks. I was searching for the answer of this question in books, but haven't found anything. Good insight! – omephy Feb 18 '14 at 17:18

Yes, this is a result rigorously stated as: There's a proper subgroup of $O(1,3)$ isomorphic to $SO(3)$. It's made up of the set of Lorentz transformations of the form:

$$\left(\begin{array}{cc} 1 & 0\\ 0 & R(3) \end{array}\right)$$

where $R(3)\in SO(3)$,

together with the internal operation of matrix multiplication.

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