# Relation between Lie Algebras and Gauge Groups

1. Jul 13, 2010

### YangMills

Alright, I understand that there are redundant degrees of freedom in the Lagrangian, and because transformations between these possible "gauges" can be parametrized by a continuous variable, we can form a Lie Group.

What I am not so firm upon is how Lie Algebras, specifically, the Lie Algebra of the group generators, relates to this. I understand that a Lie Algebra is basically a vector space over a field with a binary operation, but I don't see how this can be used to analyze the corresponding Lie group.

Essentially, how do we go from gauge group to gauge field?

2. Jul 13, 2010

### Haelfix

You want to look up the Baker-Campbell-Hausdorff (BCH) formula for the link between Lie algebras and their corresponding groups.

3. Jul 14, 2010

### tom.stoer

First I would like to mention that Lie groups appear as global symmetries as well as local gauge symmetries.

Look at ordinary rotations in R³; this symmetry group is called SO(3), which means the group elements are special orthogonal matrices, i.e. with det = 1. One can parameterize these rotations as follows

$$R(\theta_a) = e^{i\theta^aT^a}$$

$$\theta_a$$ are three angles parameterizing the rotations
$$R(\theta_a)$$ is a 3*3 rotation matrix (don't confuse this with the Euler angles; they are related, but not the same)
$$T^a$$ are 3*3 matrices, the so-called generators of the group SO(3); at the same time they are the basis of the so(3) algebra (you know what an algebra is, I guess)

One can define the matrix exponential via a Taylor series; the first two ternms read

$$R(\theta_a) = 1 + i\theta^aT^a$$

So the Lie algebra so(3) is something like the tangent space of Lie group at identity. This is possible for all Lie groups, not only for SO(3). Because a lot of calculations in the context of Lie groups and a lot of investigations regarding their properties can be done using the generators = using the Lie algebra the latter one is very important. Roughly speaking the local properties of the Lie group are fixed via the Lie algebra, whereas global properties are beyond the scope of the Lie algebra.

Last but not least: what are local gauge symmetries? SO(3) is the global symmetry group of an ordinary sphere S² embedded into R³. One can rotate the sphere as a whole and it will look exactly the same, regardless which rotations one uses.

Now think about the following possibility: for each point in 3-space (or for each point on the sphere S²) define an own set of rotation angles, which means the angles $$\theta_a$$ depend on the coordinates $$x_i$$ in 3-space

$$R(\theta_a) \to R(\theta_a(x_i))$$

Of course this is no longer a symmetry of the sphere S² because it gets distorted. Two points on the the sphere (two vectors from the center to the two points) define a certain angle between them before the rotations; after the rotation the angle between these two vectors may have changed, so this new operation is not a symmetry operation any more.

In gauge theory one introduces a different object which is invariant under the local gauge symmetry; it's not simply the angle between two vectors. But note that the Lie algebra is still the same as the generators remain constant:

$$R(\theta_a(x_i)) = e^{i\theta^a(x_i)T^a}$$

Last edited: Jul 14, 2010
4. Jul 14, 2010

### YangMills

Unfortunately I don't see how you reach that. I though tangent spaces were generalizations of the "tangent" (e.g. line, plane) to higher dimensions. So why are you adding operators?

5. Jul 14, 2010

### tom.stoer

I take the exponential and write down the first two terms of its (defining) Taylor series. Then I use the definition of an algebra: a vector space A with the additional property that you can define a multiplication * of elements x,y such that x*y lies in A.

As a basis of this vector space I use the (normalized) generators $$T^a$$ of the Lie algebra. I can add them (because they are matrices), I can multiply them (because they are matrices), and I can look at the properties of $$\theta^a T^a$$; it is nothing else but a vector in A.

Because I used the Taylor expansion for $$\theta^a T^a$$ at $$\theta^a = 0$$ which corresponds to $$R = 1$$ I can say that this is something like an n-dim vector space.

6. Jul 15, 2010

### tom.stoer

That's because of the following: Each generator $$T^a$$defines a one-parameter subgroup $$e^{isT^a} \sim 1 + isT^a$$

Tangent vectors at the identity (i.e. for s=0) in the a-direction are defined via

$$\frac{d}{ds}\left(e^{isT^a}\right)_{s=0} \sim iT^a$$

Last edited: Jul 15, 2010
7. Jul 15, 2010

### YangMills

Ah, I get it now. Basically, we expand R about R(0) = 1 in the Taylor fashion, giving us a series of matrices (generators), which constitute a basis for a field. If we started with a gauge group, we now have a gauge field.

Thank you very much.

Last edited: Jul 15, 2010
8. Jul 15, 2010

### tom.stoer

One last remark: the mathematicians are able to prove this w/o using the exponential. But I think for a quick explanation it's much simpler.

9. Jul 21, 2010

### YangMills

Sorry that it's taken so long for me to get back. One last question:

Can you direct me to that more advanced explanation? It sounds interesting.

10. Jul 22, 2010

### tom.stoer

I suggest to study some scripts (or books) regarding group theory in physics. Here are some links; the web site is in German, but many scripts are English; check the "Höhere Mathematik für Physiker" section

http://physik-skripte.de/#3