# Lie algebras of the Galilean group: Problem three chapter 2 of Wienberg vol. I

1. Oct 21, 2007

### Jim Kata

1. The problem statement, all variables and given/known data

Derive the commutation relations for the generators of the Galilean group directly from group multiplication law (without using our results for the Lorentz group). Include the most general set of central charges that cannot be eliminated by redefinition of the group generators.

The general Galilean transformation is given by
$${\mathbf{\bar x}} = {\mathbf{Gx}} + {\mathbf{a}}$$
where $${\mathbf{a}} \in \mathbb{R}^4$$ and
$${\mathbf{G}} = \left( {\begin{array}{*{20}c} 1 & 0 & 0 & 0 \\ {v_x } & {r_{xx} } & {r_{xy} } & {r_{xz} } \\ {v_y } & {r_{yx} } & {r_{yy} } & {r_{yz} } \\ {v_z } & {r_{zx} } & {r_{zy} } & {r_{zz} } \\ \end{array} } \right)$$ where $$\vec v \in \mathbb{R}^3$$ is our velocity of our coordinate system and

$$\left( {\begin{array}{*{20}c} 1 & 0 & 0 & 0 \\ 0 & {r_{xx} } & {r_{xy} } & {r_{xz} } \\ 0 & {r_{yx} } & {r_{yy} } & {r_{yz} } \\ 0 & {r_{zx} } & {r_{zy} } & {r_{zz} } \\ \end{array} } \right) \in SO(3,\mathbb{R})$$

Taking the Unitary representation of this coordinate transformation $${\mathbf{U}}\left( {{\mathbf{G}},{\mathbf{a}}} \right)$$ we have a multiplication rule of

$${\mathbf{U}}\left( {{\mathbf{\bar G}},{\mathbf{\bar a}}} \right){\mathbf{U}}({\mathbf{G}},{\mathbf{a}}) = \exp \left( {i\phi } \right){\mathbf{U}}\left( {{\mathbf{\bar GG}},{\mathbf{\bar Ga}} + {\mathbf{\bar a}}} \right)$$ (equation 1) , in our projective space.

$${\mathbf{U}}\left( {{\mathbf{G}},{\mathbf{a}}} \right) = 1 + i\frac{1} {2}\omega _{ij} {\mathbf{J}}^{ij} + i\delta v_i {\mathbf{K}}^i + i\varepsilon _i {\mathbf{P}}^i - i\varepsilon _0 {\mathbf{H}} + \cdots$$ (equation 2)

infinitesimally

$$\bar G_\mu ^\alpha G_\beta ^\mu = (\delta _\mu ^\alpha + \bar \omega _\mu ^\alpha )(\delta _\beta ^\mu + \omega _\beta ^\mu )$$
and
$$\bar G_\beta ^\alpha a^\beta + \bar a^\alpha = \left( {\delta _\beta ^\alpha + \omega _\beta ^\alpha } \right)\varepsilon ^\beta + \bar \varepsilon ^\alpha$$

using this we get

$$\begin{gathered} {\mathbf{U}}({\mathbf{\bar GG}},{\mathbf{\bar Ga}} + {\mathbf{\bar a}}) = 1 + i\frac{1} {2}\left( {\omega _{ij} + \bar \omega _{ij} + \bar \omega _{ir} \delta ^{rs} \omega _{sj} } \right){\mathbf{J}}^{ij} + i\left( {\delta v_i + \delta \bar v_i + \bar \omega _{ir} \delta ^{rs} \delta v_s } \right){\mathbf{K}}^i + \hfill \\ i\left( {\varepsilon _i + \bar \varepsilon _i + \delta \bar v_i \varepsilon _0 + \bar \omega _{ir} \delta ^{rs} \varepsilon _s } \right){\mathbf{P}}^i - i\left( {\varepsilon _0 + \bar \varepsilon _0 } \right){\mathbf{H}} + \cdots \hfill \\ \end{gathered}$$ (equation 3)

$$\phi = f^{ij} \bar \varepsilon _i \varepsilon _j + f^{ij,k} \bar \omega _{ij} \varepsilon _k + f^{ij,rs} \bar \omega _{ij} \omega _{rs} + \cdots$$ (equation 4)

Plugging equations 2, 3, and 4 into equation 1 and working out the commutation relations and applying all the possible Jacobi identities I get:

$$\begin{gathered} =[{\mathbf{K}}^i ,{\mathbf{H}}] = i(C^i {\mathbf{1}} + {\mathbf{P}}^i ) \hfill \\ [{\mathbf{K}}^i ,{\mathbf{K}}^j ] = [{\mathbf{J}}^{ij} ,{\mathbf{H}}] = [{\mathbf{P}}^i ,{\mathbf{H}}] = [{\mathbf{P}}^i ,{\mathbf{P}}^j ] = 0 \hfill \\ [{\mathbf{J}}^{ij} ,{\mathbf{K}}^k ] = i\left( {\delta ^{ik} (L^j {\mathbf{1}} + {\mathbf{K}}^j ) - \delta ^{jk} (L^i {\mathbf{1}} + {\mathbf{K}}^i )} \right) \hfill \\ [{\mathbf{J}}^{ij} ,{\mathbf{P}}^k ] = i\left( {\delta ^{ik} (C^j {\mathbf{1}} + {\mathbf{P}}^j ) - \delta ^{jk} (C^j {\mathbf{1}} + {\mathbf{P}}^j )} \right) \hfill \\ [{\mathbf{J}}^{ij} ,{\mathbf{J}}^{rs} ] = i\left( {\delta ^{ir} (D^{js} {\mathbf{1}} + {\mathbf{J}}^{js} ) + \delta ^{is} (D^{rj} {\mathbf{1}} + {\mathbf{J}}^{rj} ) - \delta ^{js} (D^{ri} {\mathbf{1}} + {\mathbf{J}}^{ri} ) - \delta ^{jr} (D^{is} {\mathbf{1}} + {\mathbf{J}}^{is} )} \right) \hfill \\ [{\mathbf{K}}^i ,{\mathbf{P}}^j ] = 0 \hfill \\ \end{gathered}$$

where the stuff next to the $${\mathbf{1}}$$'s are my central charges. Now this is fine and dandy and matches what's on wikipedia for Galilean transformations http://en.wikipedia.org/wiki/Galilean_transformation, but have a problem with commutation relation $$[{\mathbf{K}}^i ,{\mathbf{P}}^j ] = 0$$ it's lacking the central extension $${\mathbf{M}}$$, the mass of the system. Now, from what I got, I can redefine my variables to eliminate my central charges. Namely:

$$\begin{gathered} {\mathbf{\tilde P}}^i \equiv C^i {\mathbf{1}} + {\mathbf{P}}^i \hfill \\ {\mathbf{\tilde K}} \equiv L^i {\mathbf{1}} + {\mathbf{K}}^i \hfill \\ {\mathbf{\tilde J}}^{ij} \equiv D^{ij} {\mathbf{1}} + {\mathbf{J}}^{ij} \hfill \\ \end{gathered}$$

giving
$$\begin{gathered} = [{\mathbf{\tilde K}}_i ,{\mathbf{\tilde K}}_j ] = [{\mathbf{\tilde J}}_i ,{\mathbf{H}}] = [{\mathbf{\tilde P}}_i ,{\mathbf{H}}] = [{\mathbf{\tilde P}}_i ,{\mathbf{\tilde P}}_j ] = 0 \hfill \\ [{\mathbf{\tilde J}}_i ,{\mathbf{\tilde J}}_j ] = i\varepsilon _{ijk} {\mathbf{\tilde J}}_k \hfill \\ [{\mathbf{\tilde J}}_i ,{\mathbf{\tilde K}}_j ] = i\varepsilon _{ijk} {\mathbf{\tilde K}}_k \hfill \\ [{\mathbf{\tilde J}}_i ,{\mathbf{\tilde P}}_j ] = i\varepsilon _{ijk} {\mathbf{\tilde P}}_k \hfill \\ [{\mathbf{\tilde K}}_i ,{\mathbf{\tilde P}}_j ] = 0 \hfill \\ \end{gathered}$$
where

$${\mathbf{\vec J}} = \{ {\mathbf{J}}^{23} ,{\mathbf{J}}^{31} ,{\mathbf{J}}^{12} \}$$

My question is where does $${\mathbf{M}}$$ arise in the commutation relation $$[{\mathbf{\tilde K}}_i ,{\mathbf{\tilde P}}_j ]$$
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 24, 2007

### Jim Kata

Sorry to be a nag, but HELP!!!! I'm drunk tangent time

According to representation theory there are two ways in which a projective space can occur.

1. Algebraically- you can not eliminate the central charge infinitesimally close to the identity.

(I think my previous post has shown that I can eliminate the central charges algebraically)

2. Topologically- due to the fact that your space is not simply connected there is a phase term arising from the fundamental group.

That just leaves the topological case.

Question?
The Galilean group is not simply connected. What is the universal cover of the Galilean group? My thoughts are that the universal cover is responsible for this <b>M</b> thing. Someone help!

3. Oct 25, 2007

### meopemuk

Hi Jim,

perhaps you may find useful my derivation for the Poincare group in subsection 3.2.2 of http://www.arxiv.org/abs/physics/0504062. It is very detailed and fills 10 pages. I guess it should be relatively easy to see which steps should be modified in the case of the Galilei group. I haven't done the Galilei group myself.

Eugene.

4. Oct 25, 2007

### Jim Kata

Ok, cool. I guess this vindicates me in a way. I get the same results as the guy in your reference, but what's with this central extension M. I'll take a closer look at your reference, but at first glance it looks like he got the same result as me.

5. Oct 25, 2007

### meopemuk

Are you sure that after applying Jacobi identities you get

$$[K_i , P_j] =0$$?

In the case of Poincare algebra, the central charge is not eliminated from this commutator (see. eq. (3.49) in the non-relativistic limit)

$$[K_i , P_j] = E_{ij}$$

My guess is that the same should be true in the Galilei algebra, and that by adding constants to generators you should be able to reduce the right hand side to

$$E_{ij} = M \delta_{ij}$$

as required.

Eugene.

6. Oct 25, 2007

### Jim Kata

Well, look at equation 1.23 in your reference and look equation <img
src=http://upload.wikimedia.org/math/7/3/9/739fb4bddf8c97bf5ed6302314b1de6d.png> in the wiki resource, and look at the fact that there is no $$\delta v_i \varepsilon _j$$ crossover term in my post. No, I'm pretty sure the M comes from the fact that the Galilean group is not simply connected. There is a big difference between the Poincare group and Galilean group. Namely, that you can eliminate all central charges from the Poincare group, but not the Galilean group. Truthfully, I don't know what I mean by that. I think I mean that the covering group of the Galilean group has another generator namely M.

7. Oct 25, 2007

### Jim Kata

8. Oct 25, 2007

### meopemuk

I still urge you to take a closer look at subsection 3.2.2. The Galilei algebra case can be obtained by simply letting $c \to \infty$ in this derivation. Then instead of eq. (3.49) you would get

$$[\tilde{K}_i, \tilde{P}_j] = E_{ij}^{(8)}$$

The rest of derivation is basically the same up to the last equation on page 97, which now reads

$$[K_i, P_j] = -i \hbar \delta_{ij} E^{(8)}$$

Now, in contrast to the Poincare algebra case, the right hand side does not contain the Hamiltonian $\tilde{H}$, so the central charge cannot be eliminated by adding a constant to $\tilde{H}$. For this reason the central charge $E^{(8)}$ remains in the Lie algebra of the Galilei group. In a more common notation

$$[K_i, P_j] = -i \hbar \delta_{ij} M$$

The remaining central charge simply means that projective representations of the Galilei group are not reducible to linear unitary representations. This has nothing to do with the global properties of the group (i.e. that it is not simply connected). Commutation relations in the Lie algebra are determined by local properties near the group identity.

Eugene.

9. Oct 25, 2007

### Jim Kata

wow, I think that's it. Thanks!

10. Oct 25, 2007

### Jim Kata

Ok, I know your right, but this is what's tripping me up. Stupid, I know, but I'm dumb

$$[K_i ,P_j ] = iM\delta _{ij}$$ The left side of this equation is skew symmetric and the right side of this equation appears to be symmetric. How is this possible?

11. Oct 25, 2007

### meopemuk

The left side is skew symmetric only in the trivial sense $[K_i, P_j] = - [K_j, P_i]= 0$ if $i \neq j$. So, there is no contradiction.

Eugene.

12. Oct 25, 2007

gotcha ;-)