# Question(s) on Lie Groups

1. Oct 1, 2011

### teddd

Hi everybody!
Ok, so from a few days i've begun a group theory class, and i have to say i love the subject.

In particular i happend to like Lie groups, but there are things that are not cristal clear to me, hope you'll help to figure'em out!

First of all, Lie groups are continuous group, so their set of elements form a topological group.
Now, the definition of the Lie group (at least the one given to me) says:

Take any neighbourhood of the set V of the group element which contains the unit element.
Every elements g of the group G in the neighbour V can be identified with some parameters $(\alpha_1,\alpha_2...\alpha_n)$ ,called canonical coordinates, in a 1-1 corrispondence,
and if g1,g2 and g=g1g2 lie in V then $g=g(\alpha'_1,\alpha'_2...\alpha'_n)$ where the $\alpha'_k$ are analytic function of the parameters of g1g2.

This is an n-dimensional Lie group.

Now, the questions!!!

1)
Is it right to say that since we're restricting to a certain neighbour (the one which contains the unit element) we're actually ignoring (or puttin aside, choose the term you prefer) the global topology of the group? I mean, if we only work in a fixed neighbour we are always working in a topological space of an n-dimensional ball!

2)
Take the group of the rotation in two dimension.
IN the realization which employs the unimodular complex number, i.e. $\mathbb{T}=\{z\epsilon\mathbb{C},zz^*=1\}$ the group elements have the form $g(\theta)=e^{i\theta}$.
Restricting to [0,2∏] the parameter you need to identify an element is just the angle theta, so that this Lie group is a 1-dimensional group (and thus will have just one generator).
But now I ask: i do not have to restrict to a neighbour of the unit element to make that work, i just can take all of the set [0,2∏] becaouse within that the identification parameters-group IS 1-1, so effectively i can take V=[0,2∏], and so I can see the global topological structure (which is a circle minus a point that is, if we start from 0, ∏----wait a minute is that the reason?? so that i'm working on a line and not a circle becaouse i cannot close the latter (hope i've explained...))
How is that??

3)This one comes from an explicit example.
Take SU(2), the group of all the matrix with determinant=+1, which certainly IS a Lie group.
An element g of such a group can be written as
$$g=\left( \begin{array}{cc} v^*&-u^*\\ u&v \end{array} \right)$$
with $uu^*+vv^*=1$.
Now my teacher's note, after writing that matrix, says that this is clearly a 3-dimensional group.
How do you see that?? At first sight i'm pushed to say that a generic element g is determined only by u and v!
Then he (the teacher) writes down the generators, which are 3 (proportional to the Pauli marices, you know this better than me) making clear that this is a 3 dimensional LIe group.
But what are then the 3 parameters to identify one element of the group, and why u and v alone are not enough?
There is a strange precisation too: it's said that the 3 euler angles are NOT the canonical coordinates. But why? what special features do the canonical coordinates have to have?

4)
Finally, the last question!
It regrds the exponential representation.
Now, why on earth do i have to take only those element close to the unit one??
Take the unimodualr complex group in question 2.
Now, if the element of such group is called $\phi$, the exponential map is just $T(\phi)=e^{i\phi}$, and that is valid in all the set of elements on which there is a 1-1 corrispondence between $\phi$ and the representation chosen!

I thing that, for now, that's all!!

2. Oct 2, 2011

### zhentil

In (3), u and v are complex numbers, so you have four independent real variables and one real constraint.

In (4), the exponential map is a map from the Lie Algebra (the tangent space at the identity) into the group G. Not every element of every Lie group is in the image of such a map. If G is not connected, there's no way to hit every element.

(1) says that every point has a neighborhood which "looks" like a neighborhood of the identity, both in terms of topology and in terms of the group structure. This does not (per se) have anything to do with the global topology, but does place restrictions on which topological manifolds admit Lie structures.

(2) is kind of a mess. The circle is a Lie group. Verifying the definition of a Lie group is not supposed to tell you that it's a circle. What you've shown is that the exponential map is not injective.

3. Oct 2, 2011

### Jamma

Zhentil has answered your question(s) well, but I'd add that in 2), you can't take the whole group for the local neighbourhood since it won't be a homeomorphism onto it's image, you are effectively mapping the circle onto the interval, which won't be a homemorphism. You will need two at least two charts.

Just think of it as being "group structure + manifold structure" such that the group structure is compatible with the continuous structure (this is how I like to summarise most mathematical definitions!).

4. Oct 3, 2011

### Deveno

just to clarify a bit: by considering neighborhoods of {e}, you are not "ignoring the global toplogy". you're saying that the neighborhoods of {e} are just like the neighbohoods of {g}, since the map x--->gx is is a homeomorphism (with inverse y-->g^-1y).

so a group structure introduces a certain homogeneity into the topology on a manifold. for example, on S^1, every neighborhood of any point on the circle is just a neighborhood of (1,0) "rotated".

5. Oct 3, 2011

### Jamma

Well, he has the right idea about "ignoring the global topology" to be honest. It's just how for manifolds you check that the neighbourhoods are homeomorphic to Euclidean space but you don't care about the global topology- how can you tell, for example, that your Lie group is not the real line instead of S^1 from just the information that all neighbourhoods are homeomorphic to the real line?

At least, I think this is what he meant. But yes, the homogeneity induced by the homemorphisms x--->gx is very important.

6. Oct 3, 2011

### Ben Niehoff

Yes and no. It is easy to show that a neighborhood of the identity is homeomorphic to R^n, because group members close to the identity have a simpler structure. It is required, as part of the definition of a "manifold", that every point have a neighborhood homeomorphic to R^n. So, in order to show that a Lie group is a manifold, we observe that this holds in a neighborhood of the identity. To show that it holds elsewhere, we use the group action $h \mapsto gh$ to map the neighborhood of the identity to anywhere we like. Since the group action is free and transitive, the group axioms then imply that every point of the group have a neighborhood homeomorphic to R^n.

If you want to calculate global topological invariants such as homotopy groups, etc., you can do that. But that will require more work.

Be careful with your notation. You should have $\theta \in [0, 2\pi)$, not $[0, 2\pi]$.

Because the circle is topologically nontrivial, it cannot be covered by a single coordinate patch. You need at least two patches, say $\theta \in [0, 2\pi)$ and $\varphi \in [-\pi, \pi)$, with transition functions on the overlaps. All the global topological information is contained in the transition functions; the transition functions tell you how to glue the two half-open line segments together such that they close into a circle.

u and v are complex numbers, so that makes 4 parameters. There is one determinant condition, which reduces it to 3 independent parameters.

Yes, you can take the exponential of any matrix. But the point of the exponential map is that it relates a neighborhood of the identity (an infinitesimal piece of the group) to a whole neighborhood of the identity (a finite piece of the group). The exponential map tells you how to travel away from the identity, essentially, for a finite distance. This is important, because it allows us to use Lie algebras (a much simpler structure) to study the properties of Lie groups.

7. Oct 4, 2011

### teddd

Well, you have all be extremely kind answer me, i'm very grateful!!

And i'm getting the problem stright.

1) That's clear now! i did not think of making the mapping $h:G\rightarrow G$ that picks an element $g$ of the group and maps into another element $k$ of the same group!
I guess that, for example, such a mapping could be of the kind: $k=a+g$, where k and g are group elements identified on the manifold with their coordinates and a a fixed parameter
(isn't this like a change of coordinates?)

2)
This phrase is beautiful, makes all clear.

3) This was a stupid question.

4) This stills gives me some doubt.
My question is now: how much "far" can I go?
In the definition it's required to "stay close" to the unit element, but how is that intended?
(Now i know that i can move around the manifold with the mapping onto the group itself, but i guess this is not the way!)

Thank you again friends, and excuse me if I heve been slow to reply, I had something important to take care of!

8. Oct 4, 2011

### Ben Niehoff

"How far" depends on the details of the group. In SU(2), the exponential map can reach the entire group, because SU(2) is connected. In O(3), the exponential map can only reach half the group, because O(3) has two disconnected pieces: one having determinant 1, and one having determinant -1. As far as I remember, the exponential map always gives you the "identity component", which is the piece of the group manifold connected to the identity element.

To get the rest of the group, you have to append extra factors corresponding to the group elements that map between the various disconnected pieces of the group. For example, every element of O(3) can be written

$$g = R \exp X$$
where X is in the Lie algebra of O(3), and R is either the identity or a reflection.

9. Oct 7, 2011

### teddd

Ok, that makes sense.

But, as a last thing, can you explain why
in question 3) ?

I'm still wondering what proprieties must have a set of coordinates to be canonical coordinates

Thanks a lot fellas!