# How do I know if some objects represent the Lorentz Group?

1. May 1, 2015

### Spinnor

I'm thinking of an object or objects. How do I show that the objects form a representation of the Lorentz group in 1+1 D spacetime?

Thanks for any help!

2. May 1, 2015

### fzero

First let me detour into the relationship between a group and it's representations so we that we have some concepts straight.

For a general group, we would start with some definition of the group that allows us to write down some "canonical" presentation of the group elements so that we see the product rules. For a Lie group, we can also study the elements that are infinitesimally close to the the identity element, so $g \sim 1 + t$, then the set of elements that we use to express the quantities $t$ are known as the generators of the group. The product rules for the group elements imply certain commutation relations for the generators that are used to define what we call the algebra associated to the group. A representation usually means that we find a collection of matrices that satisfy the same commutation relations.

For example, for $SO(3)$, we could have started with the global rotations in 3d as the group elements, then we could have the differential angular momentum operators as our generators. A representation would be the 3x3 orthogonal matrices, but there are an infinite number of other representations that we could write down using larger matrices. In the original presentation of the group we had generators that acted as linear differential operators on functions, but what we've narrowly called representations here are linear operators that act on a vector space.

To get back to your original question about the 1+1d Lorentz group, you should start by asking precisely what this group is and find a representation connected to your starting definition. You will then consider the object(s) in question and look for a way to map them to your canonical representation of the group. The algebra must be the same and you might also have to consider global aspects, such as whether the parameters are defined on a closed or open interval.

3. May 2, 2015

### Spinnor

Thank you Fzero. Looks like I may have asked the wrong question. It now seems that the objects I'm thinking of are hopefully isomorphic to spinors and not the representation of the group. I'm trying to find objects that act like spinors in 1+1D spacetime when looked at in different rest frames. So I guess I want to know the relationship between spinors in 1+1D spacetime and the Lorentz group and its representation(s).

Thanks again for your time and patience with my shifting question.

4. May 2, 2015

### Spinnor

From a Google search "spinor representation of the lorentz group" it looks like spinors are the representations? I'm confused.

5. May 2, 2015

### fzero

Spinors are a particular class of representations of the Lorentz group. A simple example of how they work can be seen for $SO(3)$. There, as I remarked above, we find 3 generators leading to a 3-dimensional Lie algebra. The corresponding 3-dimensional representation is a set of real antisymmetric matrices that act on vectors in $\mathbb{R}^3$. From this vector representation, we can construct various tensor representations.

However, if we were to to look at $SU(2)$, we again find 3 generators of a Lie algebra that is isomorphic to the Lie algebra of $SO(3)$. But now we have a 2-dimensional representation consisting of 2x2 Hermitian matrices acting on $\mathbb{C}$. This is the fundamental spinor representation of $SO(3)$. Tensor combinations of the spinor representation lead to new spinor representations, as well as representations that we can identify with the representations that we could construct from the vector representation.

To further understand the connection between the spinors and $SO(3)$ requires knowing some global properties of the group, which are discussed here. The crux of the connection is that $SU(2)$ is topologically the universal cover of $SO(3)$. A vector in $\mathbf{R}^3$ when rotated through a given angle by $2\pi$ comes back to itself. It turns out that a spinor is a strange object that actually picks up a minus sign when we rotate through $2\pi$. There is a bit too much to explain here, but the details can probably be found in a decent QFT book like Weinberg or Nakahara's geometry text.

As for the 1+1 dim Lorentz group, the detailed discussion above turn out to be a bit of overkill. In 1+1 dim we just have a single boost generator. The spinor representation is 1-dimensional acting on $\mathbf{R}$. There are no rotations to act on the spinor.

I'm not quite sure where to go with this. Perhaps you want to explain more details about what you are doing in 1+1 dim or you want to do something more obviously nontrivial in higher dimensions.

6. May 2, 2015

### Spinnor

Sunday I will try and regroup and figure out where I'm trying to get to and maybe with help I can pointed in the right direction.

7. May 3, 2015

### strangerep

Dunno if this will help, but, to clarify some terminology,...

A "representation" in this context means a mapping from the abstract group elements to concrete matrices acting on a vector space. The vector space is then said to "carry the representation". In your case, the space of spinors "carries" a representation of the Lorentz group in terms of 2x2 matrices acting on that space.

8. May 3, 2015

### Spinnor

From http://arxiv.org/pdf/0912.2560.pdf we are given,

Λ = e(iθµνMµν) , which in the 1+1D spacetime simplifies to
Λ = e(iθM)?

The same Λ then acts on both 2 dimensional vectors in 2 dimensional Minkowski spacetime, but also the space of 2 component complex spinors?

In this case is Λ the "concrete" matrix? If so what are the abstract elements in this case?

Historically, which representation came first?

9. May 3, 2015

### fzero

OK, this will help be a bit more concrete. The $\Lambda$ are elements of the Lorentz group, while $M$ is the generator of the Lorentz algebra. A specific choice of $M$ as a matrix (along with the vector space it acts on) is what is called a representation of the Lorentz algebra (it is also a representation of the group). The 2-dimensional representation of $M$ as a 2x2 real matrix acting on a real vector space would be referred to as the fundamental representation for this Lorentz group.

Now $\Lambda$ is how the Lorentz group acts on spacetime vectors:

$$x^\mu \rightarrow {\Lambda^\mu}_\nu x^\nu.~~~(*)$$

The action on spinors will not in general be the same. To describe how the Lorentz group acts on other types of objects, let's be a little abstract and discuss a general case where we have a function $\Phi_a(x^\mu)$ defined on spacetime. The index $a$ labels the definite representation that the function transforms in. So if $\Phi_a$ was just a scalar-valued function (a real or complex number), then it would belong to the trivial representation. But $\Phi_a$ can also be a spinor, vector or other tensor-valued function. The Lorentz transformation rule for $\Phi_a$ when the coordinates are transformed by (*) is

$$\Phi_a(x) \rightarrow L_{ab}(\Lambda) \Phi_b (\Lambda^{-1}x).$$

Here $L_{ab}(\Lambda)$ is an appropriate matrix in the representation for $\Phi$ with parameters determined by $\Lambda$. In the argument of the function, we have $\Lambda^{-1}$ appearing because this is an active transformation. The new value of $\Phi$ at the point $x$, which we could call $\Phi'(x)$, must be compared to the value of $\Phi$ at the point $\Lambda^{-1}x$ that was mapped to the point $x$.

To complete the expression, I will note that we can reexpress

$$\Phi_b (\Lambda^{-1}x) = e^{i\theta_{\mu\nu} J^{\mu\nu}} \Phi_b(x),$$

where $J_{\mu\nu}$ is a differential operator

$$J_{\mu\nu} = i (x_\mu \partial_\nu - x_\nu \partial_\mu).$$

You can compare these to the differential operators for angular momentum in quantum mechanics and their familiar action on the spherical harmonics. In fact our description of $\Phi_a(x)$ is completely analogous to spherical harmonics, which are functions of the points on the sphere and have indices labeling what representation of $SO(3)$ they belong to.

Now if we put this all together, we find that

$$\Phi_a(x) \rightarrow L_{ab}(\Lambda) e^{i\theta_{\mu\nu} J^{\mu\nu}} \Phi_b(x).$$

If we are considering a spinor as a type of function on spacetime, then we need to determine what $L_{ab}$ is. However, as I mentioned in a previous post, the fundamental spinor representation in 1+1 space time is actually just one real dimensional. The reason for this is analogous to the way in 3+1 a Dirac spinor (4 dimensional) can be projected onto 2 Weyl spinors (2 dimensional). The notes that you linked discuss that. In 1+1, what would have been a 2 dimensional Dirac spinor projects to a pair of real Weyl spinors (which are technically called Majorana-Weyl).

The upshot is that, for the basic spinor in 1+1, the $L_{ab}$ is just the identity. Under a LT, the only transformation on the spinor comes from the $e^{i\theta_{\mu\nu} J^{\mu\nu}}$ part having to do with the coordinates themselves.

10. May 3, 2015

### Spinnor

Thank you for that detailed reply!

In 1+1 dimensional spacetime its seems that the spinor is not doing very much? We need it for solutions to the Dirac equation in 1+1 but can you say what it "tells" us, or what information it conveys that is already not contained in the phase, exp(-i[Et-px])? Is it just a "bookkeeping" device?

That may be a pretty vague question.

11. May 3, 2015

### fzero

At the level of transformations under the LG, etc. I agree that the spinor (or other tensor field) doesn't look any different from the scalar, since we only have the boost and there is no "room" in 1+1 to do anything. Where you see a difference is in the dynamics, since the field equation for a (massless) scalar is

$$\partial^\mu \partial_\nu \phi = 0,$$

while the spinor satisfies some version of the Dirac equation

$$\gamma^\mu \partial_\mu \psi =0.$$

You may complain that I wrote this for a 2-dim Dirac fermion and I've been telling you all along that the spinor is really 1d, but if I were careful to work out the appropriate projection, then we would have some specific linear combination of $\partial_{0,1}$ appearing in the EOM. The main point is that it is 1st order vs 2nd order for the scalar.