We embed the rotation group, $SO(3)$ into the Lorentz group, $O(1,3)$ : $SO(3) \hookrightarrow O(1,3)$ and then determine the six generators of Lorentz group: $J_x, J_y, J_z, K_x, K_y, K_z$ from the rotation and boost matrices.

From the number of the generators we realize that $O(1,3)$ is a six parameter matrix Lie group.

But are there any other way to know the number of parameters of the Lorentz group in the first place?


3 Answers 3


From special relativity we know that a Lorentz transformation: \begin{equation} x'^\mu = \Lambda^\mu {}_\nu x^\nu \end{equation} preserves the distance: \begin{equation} g^{\mu \nu} \Delta x_\mu \Delta x_\nu = g^{\mu \nu} \Delta x_\mu' \Delta x_\nu' \end{equation} The above two equations imply: \begin{equation} g^{\mu \nu} = g^{\rho \sigma}\Lambda_\rho {}^\mu \Lambda_\sigma {}^\nu \end{equation} Now, let us consider an infinitesimal transformation: \begin{equation} \Lambda_\nu {}^\mu = \delta_\nu{}^\mu + \omega_\nu{}^\mu + O(\omega^2) \end{equation} such that we can write: \begin{equation} \begin{aligned} g^{\mu \nu} & = g^{\rho \sigma}\Lambda_\rho {}^\mu \Lambda_\sigma {}^\nu \\& = g^{\rho \sigma} \left( \delta_\rho{}^\mu + \omega_\rho{}^\mu + \cdots \right)\left( \delta_\sigma{}^\nu + \omega_\sigma{}^\nu + \cdots \right) \\& = g^{\mu \nu} + g^{\mu \sigma} \omega_\sigma{}^\nu + g^{\rho \nu} \omega_\rho{}^\mu + O(\omega^2) \\& = g^{\mu \nu} + \omega^{\mu\nu} + \omega^{\nu \mu} + O(\omega^2) \end{aligned} \end{equation} and so: \begin{equation} \omega^{\mu\nu} = - \omega^{\nu \mu} \end{equation} Thus, the matrix $\omega$ is a $4 \times 4$ antisymmetric matrix, which corresponds to $6$ independent parameters (i.e. the $3$ parameters corresponding to boosts and the $3$ parameters corresponding to rotations).

  • 5
    $\begingroup$ I.e. 3 rotation angles and 3 parameters for the 3-velocity of a boost. $\endgroup$
    – innisfree
    Feb 19, 2014 at 18:28
  • $\begingroup$ @innisfree thanks! I've edited my original message just to avoid potential confusion. $\endgroup$
    – Hunter
    Feb 19, 2014 at 18:37
  • $\begingroup$ Why the downvote? Is there something wrong with the physics/maths in the original message? $\endgroup$
    – Hunter
    Feb 19, 2014 at 19:08
  • 1
    $\begingroup$ Downvotes happen randomly on this site as it undergoes quantum interactions with the outside World. I can't see anything wrong with your answer, which to me is clearly the most fundemental and best of all the three here, answering the OP's needs directly. Moreover, if there is something wrong, I'd for one like to know about it, which is why I think downvoting without leaving a comment is pointless to everyone's learning. However, in this case, I'd wager a large amount that there is nothing in the slightest wrong here. $\endgroup$ Feb 21, 2014 at 1:03
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    $\begingroup$ @WetSavannaAnimalakaRodVance thanks for your message. I agree that the most annoying part of the downvote is that I don't understand the reason behind it and so I wondered for a while if I had made a mistake in my original message. $\endgroup$
    – Hunter
    Feb 21, 2014 at 7:12

It's the same way you know there are three parameters in $SO(3)$. The equation $\Lambda^T \eta \, \Lambda = \eta$ has $(n^2+n)/2$ independent scalar equations. To see this, write the equation in component form: $\Lambda^{\mu\nu} \Lambda_\mu{}^\rho = \eta^{\nu\rho}$. Now we see there are $n^2$ scalar equations equations, but because $\eta$ is symmetric and the left hand side is symmetric in $\nu$ and $\rho$ as well, the equations related by switching $\nu$ and $\rho$ are the same. Thus we have established that there are $(n^2+n)/2$ independent scalar equations.

Since $\Lambda$ has $n^2$ components, we get $n^2-(n^2+n)/2 = n(n-1)/2$ degrees of freedom. In $3D$ this comes out to $3\cdot 2 /2=3$, and in $4D$ this comes out to $4 \cdot 3/2=6$.

  • $\begingroup$ Can you please elaborate the answer a bit more? Especially this portion: "The equation $\Lambda^T \eta \, \Lambda = \eta$ has $(n^2+n)/2$ independent scalar equations since the product is guaranteed to be symmetric." $\endgroup$
    – rainman
    Feb 19, 2014 at 18:51
  • $\begingroup$ I elaborated on that part. And thanks for the edits. $\endgroup$ Feb 19, 2014 at 19:11

You've got two very good answers from Hunter and NowIGetToLearnWhatAHeadIs. However, it's probably useful to know that this beast $O(1,3)$ is isomorphic or locally isomorphic (i.e. has the same Lie algebra) to a surprising number of other interesting groups, which each give you a slightly different way to think about it. First note that its identity connected component $SO^+(1,3)$ of orthochronous, proper Lorentz transformations (those that keep the orientation of space and time the same, also called the "restricted" Lorentz group) of course determines the Lie algebra.

  1. $SO^+(1,3)\cong {\rm Aut}(\hat{\mathbb{C}}) \cong PSL(2,\mathbb{C})$ is isomorphic to the Möbius group of all Möbius transformations, in turn isomorphic to the group of all conformal transformations of the unit sphere. So it is defined by $z\mapsto \frac{a\,z+b}{c\,z+d}$ with $a,\,b,\,c,\,d\in\mathbb{C}$ and $a\,d-b\,c=1$. So there are three independent complex parameters, i.e. six independent real parameters;

  2. The double cover of $PSL(2,\mathbb{C})$, namely $SL(2,\mathbb{C})$ (still locally isomorphic to $SO^+(1,3)$) is the group of all $2\times 2$ matrices of the form:

$$\exp\left(\frac{1}{2}\left[\left(\eta^1 + i\theta \gamma^1\right) \sigma_1 + \left(\eta^2 + i\theta \gamma^2\right) \sigma_2 + \left(\eta^3 + i\theta \gamma^3\right) \sigma_3\right]\right)$$

where $\sigma_j$ are the Pauli spin matrices, $\theta$ is the angle of rotation, $\gamma^1,\,\gamma^2,\,\gamma^3$ are the direction cosines of the rotation axis and $\eta^1,\,\eta^2,\,\eta^3$ the components of the rapidities of the Lorentz transformation. So it's just like the general matrix $\exp\left(\frac{\theta}{2}\left(\gamma^1 \sigma_1 + \gamma^2 \sigma_2 + \gamma^3 \sigma_3\right)\right)$ in $SU(2)$ but with three complex parameters, rather than three real ones ($\theta \gamma^j$) for $SU(2)$. So again we see six real parameters.

  • $\begingroup$ I am confused. Isn't the Mobius group you describe in your first bullet point $SL(2, \mathbb{C})$ itself. (This is what I have read from CFT books). What is the difference between $PSL$ and $SL$? $\endgroup$
    – user7757
    Jun 17, 2014 at 13:48
  • $\begingroup$ @ramanujan_dirac No. There are two matrices in $SL(2,\,\mathbb{C})$ which stand for the same Möbius transformation $\frac{a\,z+b}{c\,z+d}$, namely both matrices $\pm\left(\begin{array}{cc}a&b\\c&d\end{array}\right)$: multiply top and bottom of the Möbius transformation by -1 and you still have the same function: do the same for the matrices and their determinant is still 1. $PSL(2,\,\mathbb{C})$ is the projective special linear group of $2\times 2$ complex matrices and $SL(2,\,\mathbb{C})$ is its double cover. $SL(2,\,\mathbb{C})$ acts on the Riemann sphere by spinor maps when the ... $\endgroup$ Jun 17, 2014 at 14:11
  • $\begingroup$ ... Riemann sphere's co-ordinates are represented by $X = x \sigma_1 + y \sigma_2 + z \sigma_3$ where $x^2+y^2+z^2=1$ and the transformation is $X\mapsto \gamma\,X\,\gamma^\dagger$ where $\gamma\in SL(2,\,\mathbb{C})$. Check out en.wikipedia.org/wiki/… and en.wikipedia.org/wiki/M%C3%B6bius_transformation $\endgroup$ Jun 17, 2014 at 14:14
  • $\begingroup$ The matrix with the minus sign will be have determinant -1 right? So it won't fall in $SL(2, \mathbb{C})$ $\endgroup$
    – user7757
    Jun 17, 2014 at 14:16
  • $\begingroup$ @ramanujan_dirac No, if you multiply all elements by -1, the determinant is still +1 $\endgroup$ Jun 17, 2014 at 14:17

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