# What is Little Group ?

1. Feb 20, 2012

### yicong2011

What is "Little Group"?

In my Quantum Field Theory class, I too often meet with the term "Little Group".
Unfortunately, I cannot find a good description of Little Group until now. I just know it is a subgroup of Lorentz Group.

Can anyone have any brief description of this concept? Or any good reference on it?

Thanks a lot.

2. Feb 21, 2012

### DrDu

Re: What is "Little Group"?

No, it is not a subgroup of the Lorentz group but rather of the Poincare group although it is also used in other context such as in crystallography.
It is the group which leaves a given k vector invariant. Hence different k vectors have different little groups.
See, e.g. Gordon Hamermesh, Group theory
or Eugene Wigners book on group theory as he introduced the concept in relativistic QM.
A more modern introduction is Sternberg, Group theory

3. Feb 21, 2012

### Bill_K

Re: What is "Little Group"?

No, in its usual application in quantum field theory, the "little group" is a subgroup of the Lorentz group. In general if you have a group G which acts on a space X, and an element x in X, the little group of x is the subgroup of G that leaves x invariant.

For example the Lorentz group acts on the space of 4-vectors. If x is taken to be a timelike vector, the little group is the SO(3) subgroup of the Lorentz group in the 3-space orthogonal to x. If x is spacelike or null, the little group will be SO(2,1) or E(2) respectively.

4. Feb 21, 2012

### negru

Re: What is "Little Group"?

Or if you're talking about massless particles, you can write their momentum as the product of two spinors \lambda \tilde{\lambda}

Then if you multiply lambda by t, and tilde lambda by 1/t, you leave the momentum unchanged.

5. Feb 21, 2012

### chrispb

Re: What is "Little Group"?

More group-theoretically, a little group is the group that leaves some particular state invariant. Poincare transformations act on good old quantum mechanical states; the little group of the state of one massive particle in its rest frame is therefore the SO(3) of rotations around it.

6. Feb 21, 2012

### DrDu

Re: What is "Little Group"?

Are you sure that E2 is a sub-group of the Lorentz group?
Anyhow, what I really wanted to say was that the little groups are used to construct representations of the Poincare group and not of the Lorentz group.

7. Feb 21, 2012

### George Jones

Staff Emeritus
Re: What is "Little Group"?

Yes, in this context the (double cover) of E2 is a subgroup of the (cover of the) homogeneous Lorentz group.
Yes.

8. Feb 22, 2012

### DrDu

Re: What is "Little Group"?

Interesting! I am not much into the theory of the Lorentz Group. Could you give me an outline of how to see this? Where do the translations in the plane come from?

9. Feb 22, 2012

### tiny-tim

10. Feb 22, 2012

### naima

Re: What is "Little Group"?

You can also find in "//arxiv.org/abs/hep-th/0211208" [Broken] details on E2 and T2

Last edited by a moderator: May 5, 2017
11. Feb 22, 2012

### DrDu

Re: What is "Little Group"?

Thank you, that's an interesting link. But while it is clear that E2 can be obtained from O(3) by an Inonu Wigner transformation, this leads out of O(3) and maybe also out of the homogeneous Lorentz group?

Last edited by a moderator: May 5, 2017
12. Feb 22, 2012

### George Jones

Staff Emeritus
Re: What is "Little Group"?

Give me a bit of time, and I'll post the explicit construction.

13. Feb 22, 2012

### George Jones

Staff Emeritus
Re: What is "Little Group"?

The universal cover of the restricted Lorentz group is $SL\left( 2,\mathbb{C}\right)$. If a 4-vector is written as $X=x^{0}+x^{i}\sigma _{i}$, then the action of $A \in SL\left( 2,\mathbb{C}\right)$ on $X$ is $A X A^\dagger$. The little group of the lightlike 4-vector $X=1+\sigma_3$ is the subgroup of $SL\left( 2,\mathbb{C}\right)$ that consists of matrices of the form
$$\begin{pmatrix} e^{i\theta} & b\\ 0 & e^{-i\theta} \end{pmatrix},$$
where $\theta$ is an arbitrary real number and $b$ is an arbitrary complex number.

$$\begin{pmatrix} e^{i\theta} & b\\ 0 & e^{-i\theta} \end{pmatrix} \rightarrow \begin{pmatrix} e^{i2\theta} & b\\ 0 & 1 \end{pmatrix}$$
is a two-to-one homomorphism between groups of matrices. Write $b = u+iv$ for real $u$ and $v$, and $\left( x , y \right) \in \mathbb{R}^2$ as the column
$$\begin{pmatrix} x+iy\\ 1 \end{pmatrix}.$$

$$\begin{pmatrix} e^{i2\theta} & u+iv)\\ 0 & 1 \end{pmatrix} \begin{pmatrix} x+iy\\ 1 \end{pmatrix} = \begin{pmatrix} x\cos2\theta - y\sin2\theta + u + i\left( x\sin\theta + y\cos2\theta +v \right)\\ 1 \end{pmatrix}$$

Last edited: Feb 22, 2012
14. Feb 23, 2012

### DrDu

Re: What is "Little Group"?

Thank you George, very interesting.
So setting theta=0 yields the pure translations by b. The elements A are of the same form as the X, $X=x^{0}+x^{i}\sigma _{i}$. If I remember correctly, imaginary parts of the x^i correspond to boosts. So a pure translation is of the form $X=x^{0}+b/2\sigma _{x}+ib/2 \sigma_y$. So the translation is a 50/50 mixture of rotation and boost.

15. Feb 23, 2012

### strangerep

Re: What is "Little Group"?

The explicit generators of the E2 little group are
$$J_3 ~,~~~~ A ~:=~ J_2 + K_1 ~,~~~~ B ~:= -J_1 + K_2$$
so the commutation relations are
$$[J_3, A] = iB ~,~~~~ [J_3,B] = -iA ~,~~~~ [A,B] = 0 ~.$$

16. Feb 26, 2012

### Hans de Vries

Re: What is "Little Group"?

Just to elaborate a bit:

The A and B, acting on a momentum in the z-direction, represent centrifugal
accelerations K compensated by counter rotations J. The total effect of A or
B is therefor zero. The general case is.
$$A\cos \phi ~+~B\sin\phi$$
Where $\phi$ is the angle which determines the direction in the x-y plane of the
centrifugal acceleration. Small nitpick about his signs: In a right-handed
coordinate system they should be:
$$J_3 ~,~~~~ A ~:=~ J_2 - K_1 ~,~~~~ B ~:= -J_1 - K_2$$
Note that if you reverse the direction of the momentum, (k,0,0-k) instead
of (k,0,0,k), that the signs of the generators also change. In this case you
get indeed.
$$J_3 ~,~~~~ A ~:=~ J_2 + K_1 ~,~~~~ B ~:= -J_1 + K_2$$
One should expect this because under spatial inversion K behaves like a
vector and J like a pseudo vector.

Regards, Hans

Last edited: Feb 26, 2012
17. Feb 27, 2012

### yicong2011

Re: What is "Little Group"?

Ah, Thanks a lot, guys. Thanks for glorious detail explanation.

18. May 13, 2012

### sfn17

Re: What is "Little Group"?

I'm preparing a little sketch on the irreps of the Poincaré group according to Wigner's classification. Unfortunately, this subject seems to be not too popular in the textbooks and in review articles. No author whom I found describes clearly and correctly the structure of the little group of the photon. The most part of them resort just to rescaling the angle (i.e. $\phi\to\phi/2$), some give wrong statements on the semidirect structure.

I spent some time to puzzle everything together. Now, it's quite easy to explain. I try to make some comments. I am much indebted to the comments of George Jones.

According to Simms, I will denote the little group of the photon by Δ.

That indicates, that Δ is also a sort of covering group. At least, its center will be $\left\{ I_2,-I_2\right\}$, $I_2$ being the identity element of $SL\left( 2,\mathbb{C}\right)$.

This is the group Δ.

To get rid of using a chart of $U(1)$ (and thereby to avoid any problems with rescaling), I prefer the equivalent definition of the group Δ as follows:

The little group Δ of the lightlike 4-vector $X=1+\sigma_3$ is the subgroup of $SL\left( 2,\mathbb{C}\right)$ that consists of matrices of the form
$$\begin{pmatrix} u & u^{-1}b\\ 0 & u^{-1} \end{pmatrix},$$
where $u,b$ are arbitrary complex numbers but with $\left|u\right|=1$.

The introduction of an additional factor $u^{-1}$ in the entry (1,2) simplifies the exhibition of the semidirect structure.

The key point is that the product in Δ resembles much to that in the Euclidean group E(2), but not fully. Namely we have
$$\begin{pmatrix} u_1 & u_1^{-1}b_1\\ 0 & u_1^{-1} \end{pmatrix}\begin{pmatrix} u_2 & u_2^{-1}b_2\\ 0 & u_2^{-1} \end{pmatrix} = \begin{pmatrix} u_1u_2 & u_2^{-1}u_1b_2+u_1^{-1}b_1\\ 0 & u_1^{-1}u_2^{-1} \end{pmatrix} =\begin{pmatrix} u_1u_2 & (u_1u_2)^{-1}(u_1^2b_2+b_1)\\ 0 & (u_1u_2)^{-1} \end{pmatrix}.$$
The difference lies in the $u_1^2$.

Let's describe the special Euclidean group SE(2) by real $3\times3$-matrices: SE(2) consists of the elements
$$\begin{pmatrix} R & b \\ 0 & 1 \end{pmatrix}$$
with $R\in SO(2)$, $b\inℝ^2$. Note: Reflections should be excluded since they cannot be represented by any $e^{i\theta}$.

Ordinary matrix multiplication reproduces the structure as semidirect product of rotations and translations in two dimensions correctly:
$$\begin{pmatrix} R_1 & b_1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} R_2 & b_2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} R_1R_2 & R_1b_2+b_1 \\ 0 & 1 \end{pmatrix}.$$
Here, the rotation $R_1$ does not appear as $R_1^2$.
To have then some sort of isomorphism to Δ, in the entry (1,2) should be a $u_1$ instead of $u_1^2$. It seems that the elements $u$ and $-u$ have to be identified.

This identification is achieved by the following homomorphism

that means that I have got the homomorphism $\phi$:
$$\phi:\quad \begin{pmatrix} u & b \\ 0 & u^{-1} \end{pmatrix} \mapsto \begin{pmatrix} u^2 & b \\ 0 & 1 \end{pmatrix}.$$
(That I now suppress the additional factor $u^{-1}$ at entry (1,2) is of course of no significance.)

Note: I don't make use of any angle $\theta$ nor of some rescaling. Defining the group multiplication with elements written in the form $g(\theta,b)$ suffers from tackling the addition of two angles to a value exceeding $2\pi$ or even $4\pi$. Somewhere in the literature, the mere extension of the interval $\left[0,2\pi\right]$ to $\left[0,4\pi\right]$ together with the replacement $\theta$ by $\theta/2$ seems to imitate the descent to a group covered by Δ.

The kernel of $\phi$ is obviously $\left\{I_2,-I_2\right\}$.
So, in taking the cosets of this kernel we get an isomorphism $\iota$
$$\iota:\quad \left\{\begin{pmatrix} u & b \\ 0 & u^{-1} \end{pmatrix}, \begin{pmatrix} -u & -b \\ 0 & -u^{-1} \end{pmatrix} \right\} \mapsto \begin{pmatrix} u^2 & b \\ 0 & 1 \end{pmatrix}.$$
The group formed by the elements $\begin{pmatrix}u^2 & b \\ 0 & 1\end{pmatrix}$ with $\left|u\right|=1$ is of course the same as the group formed by the elements $\begin{pmatrix}v & b \\ 0 & 1\end{pmatrix}$ with $\left|v\right|=1$.

To prove that this group is isomorphic to the proper Euclidean group $SE(2)$, one needs now the identification $v=e^{i\theta}$ with $\theta\in[0,2\pi[$. Then one can proceed in the manner as demonstrated by George Jones. No factor 2 is needed.

I conclude that the group Δ is a twofold covering group of the Euclidean group $SE(2)$. The covering homomorphism (Δ onto an isomorphic copy of $SE(2)$) is given by $\phi$ defined above.

I do not state that the little group Δ is the universal covering group of $SE(2)$ since I don't know whether Δ is simply connected. So, I refrain from writing $\widetilde{SE(2)}$ instead of Δ.

Last edited: May 13, 2012
19. May 13, 2012

Re: What is "Little Group"?

Shouldn't the middle term of the above be
$$\begin{pmatrix} u_1u_2 & u_2^{-1}u_1b_2+u_1^{-1}\underline{u_2^{-1}}b_1\\ 0 & u_1^{-1}u_2^{-1} \end{pmatrix} ?$$
(I think the final one is correct.)

20. May 13, 2012

### sfn17

Re: What is "Little Group"?

Oh yes, you're right! Thank you very much!
I was lost in the forest of TeX...

21. May 13, 2012