Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

How can I determine the generators of the Poincare Group, $P(1,3)$ explicitly? Here $P(1,3)$ means a matrix Lie group.

share|improve this question
    
add comment

1 Answer 1

up vote 11 down vote accepted

The general idea.

Let's restrict the discussion to matrix Lie Groups for simplicity. Determining the generators of a given Lie group $G$ simply means (by definition) determining a basis for its Lie algebra $\mathfrak g$. Here's a standard method for finding such a basis:

  1. Recall that the Lie algebra $\mathfrak g$ of a matrix Lie group $G$ is defined as the set of all matrices $X$ for which $e^{s X}$ is an element of $G$ for all real numbers $s$.

  2. Use the properties of elements of $G$ to constrain the elements $X$; the resulting admissible elements $X$ are precisely the elements of the Lie algebra $\mathfrak g$ which is a vector space of matrices.

  3. Determine a basis for this resulting vector space.

An example: $\mathrm{SO}(2)$.

Elements of the rotation group in two dimensions are precisely those $2\times 2$ real matrices $R$ for which \begin{align} R^tR = I, \qquad \det R = 1. \end{align} Now, suppose that $X$ is an element of its Lie algebra $\mathfrak{so}(2)$, then $e^{sX}$ is an element of $\mathrm{SO}(2)$ which means that \begin{align} e^{s X} (e^{sX})^t = I, \qquad \det (e^{sX}) = 1 \end{align} for all $s\in\mathbb R$. Now we use the facts that $(e^M)^t = e^{M^t}$ and $\det e^M = e^{\mathrm{tr}M}$ to assert that \begin{align} e^{s(X + X^t)} = I, \qquad e^{s(\mathrm{tr} X)} = 1 \end{align} which implies that \begin{align} X = -X^t, \qquad \mathrm{tr} X = 0. \end{align} Thus, the Lie algebra $\mathfrak{so}(2)$ is given by the set of all real, antisymmetric, traceless $2\times 2$ matrices. This is a one-dimensional vector space of matrices whose general element can be written as a scalar multiple of \begin{align} J = \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix} \end{align} which is therefore the generator we were looking for.

The Poincare group.

The same procedure can we used to determine an explicit matrix form for the generators of the Poincare group. Since, in that case, the Poincare group can be written as a semi-direct product of the Lorentz group $\mathrm{SO}(3,1)$ with the four-dimensional translation group $\mathbb R^4$, one often finds the generators of the Lorentz and translation groups individually for convenience.

When finding the generators of the Lorentz group, you simply need to use the analogous properties to $RR^t = I$ and $\det R = 1$ that define its elements. For example, in this case, the defining property is that the elements of $\mathrm{SO}(3,1)$ preserve the Minkowski scalar product; \begin{align} \Lambda^\mu_{\phantom\mu\alpha} \Lambda^\nu_{\phantom\nu\beta}\eta^{\alpha\beta} = \eta^{\mu\nu} \tag{$\star$} \end{align} where $\eta^{\mu\nu} = \mathrm{diag}(-1,+1,+1,+1)$. This can be written in matrix form, and then the same procedure used above for $\mathrm{SO}(2)$ can be used to find the Lie algebra and generators.

How it's done in practice by physicists.

In order to determine the algebra as outlined above using the invariance condition $(\star)$, physicists will often write a Lorentz group element as \begin{align} \Lambda^\mu_{\phantom\mu\nu} = \delta^\mu_\nu + \omega^\mu_{\phantom\mu\nu} \end{align} where $\omega = (\omega^\mu_{\phantom\mu\nu})$ is an "infinitesimal" matrix. Note that this is simply the same thing as writing out the matrix exponential to first order in the parameter $s$ in the procedure above, and identifying the first order term as the Lie algebra element. Then, one can plug this expression into the invariance condition $(\star)$ and determined what properties $\omega$ needs to obey by comparing the terms that are first order in $\omega$. This is equivalent to the procedure outlined above for $\mathrm{SO}(2)$, but it's often more computationally convenient.

If you go through this computation for the Lorentz group, you should find that \begin{align} \omega_{\mu\nu} = -\omega_{\nu\mu} \end{align} where $\omega_{\mu\nu} = \eta_{\mu\alpha}\omega^\alpha_{\phantom\alpha\nu}$.

share|improve this answer
    
Just curious, How about directly beginning from the generators, $$ J^{\mu\nu} = i(x^\mu\partial^\nu - x^\nu\partial^\mu) $$ –  user35952 Feb 11 at 6:32
    
Doesn't Poincare-group also include translations along with Lorentz transformations ? –  user35952 Feb 11 at 6:33
1  
@user35952 Are you making a suggestion with the first comment, or is it a question perhaps? I don't quite understand. As for translations, yes it does include translations. I mention this in the section entitled "The Poincare group." –  joshphysics Feb 11 at 6:36
1  
@user35952 Well my understanding was that the OP wanted to know how to determine the generators in the first place. You've given, from the start, the expressions for particular representations of the Lorentz generators, so there's nothing left to do. Perhaps you're asking how one could obtain those differential representations of the generators in the first place? –  joshphysics Feb 11 at 8:44
1  
@Ome Well for the Lorentz generators, it's pretty straightforward to obtain matrix representations of the rotation generators $J_i$ and boost generators $K_i$ in "usual coordinates." Have a look action section II.3 of Zee's QFT in a nutshell. I don't have any reference suggestions for light cone coordinates though, although in principle you should just be able to obtain them from the "usual coordinate" expressions by a similarity transformation. –  joshphysics Feb 11 at 9:17
show 6 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.