# Definition of chart for Lie groups

1. Feb 22, 2016

### mnb96

Hello,

I'm reading a book on Lie group theory, and before giving the definition of a Lie group G, the author defines the concept of chart as a pair (U(g), f) where:

i) U(g) is a neighborhood of g∈G
ii) f : U(g)→f(U(g))⊂ℝn is an invertible map such that f(U(g)) is an open subset of ℝn.

My question is: how can we even define a neighborhood of an element of the group G if we haven't assigned yet a topology to it? Should we perhaps assume that the topology on G is the one induced by the inverse map f-1?

Thanks.

2. Feb 22, 2016

### Staff: Mentor

How did he define neighborhood? This is already a topological term, e.g. if G is assumed a manifold or a metric space.

3. Feb 23, 2016

### mnb96

I am pretty sure he did not define anywhere "neighborhood".

What I was trying to ask is best explained with an example: let's consider the group G=SO(2) of 2x2 rotation matrices. We know that G is a group but we don't know yet if it is a Lie group. Now, we should find charts for G.

The problem is that, in order to define a chart we need to know what is a neighborhood of an arbitrary element g∈G. In other words, given a rotation matrix we need a criterion to know what are the matrices that are "close" to it, but this is impossible unless we assume a topology on G. Thus, I deduced that we can never prove that a group is a Lie group unless we assume/assign a topology to it. Am I right?

If I am right, the only way I can see to assign a topology to SO(2) is to induce it from the by the parametrization from the space of coordinates to SO(2).

Last edited: Feb 23, 2016
4. Feb 23, 2016

### micromass

You are right, he needs to assign a topology on the group first. In the case of $SO_2$, there is a canonical topology with straightforward neighborhoods. But you cannot do anything without this topology.
There is however a way to characterize manifolds and Lie groups without accepting an a priori topology, but then the concept of a neighborhood should not be used.

5. Feb 23, 2016

### mnb96

I think the problem here is that, forgetting about Lie groups, I am still confused about the definition of differentiable manifold.
It seems to me that for every different book that I open, I find a "different" (though probably equivalent) definition of differentiable manifold.

So far, the only definition that I personally found logically sound is the one given in Do Carmo's book "Differential Geometry of Curves and Surfaces". Interestingly, I noticed that the author does not even mention "atlases" and "charts" at all. He just uses "parametrizations", which are defined through one-to-one maps $\mathbf{x}_\alpha : U_\alpha \subset \mathbb{R}^n \rightarrow G$ of open sets Uα into a set G.
Since his definition invokes the concept of "open sets" in ℝn, he only needs to assume an Euclidean topology in the coordinate space: no need to invoke any a-priori topology in G.

Instead, what really causes me troubles are those definitions of manifold that are based on "charts" and "atlases". The reason is that, charts (if I understood correctly) are defined only for topological spaces, and they are one-to-one maps $\varphi_\alpha : V_\alpha \subset G \rightarrow U_\alpha \subset \mathbb{R}^n$ from open sets of G to open sets of the Euclidean coordinate space. Thus, definitions of manifolds that rely on the concept of chart (contrary to Do Carmo's definition) apparently assume a topology in G.

- Which of the two definition should I rely on?
- Many books give "chart"-based definitions of a Lie group. What is the topology that they assume?

Last edited: Feb 23, 2016
6. Feb 23, 2016

### lavinia

A parameterization is the inverse map of a chart. Defining the topology in terms of one is the same as defining it in terms of the other.

For charts for instance, one assumes that the space has some topology and then one describes the topology by saying what the open sets are. For a topological manifold they are generated by neighborhoods that are homeomorphic to open sets in Euclidean space. Every point has such a neighborhood.

If one chooses a homeomorphism from the neighborhood into Euclidean space this is called a coordinate chart. If one goes the other direction, by choosing the homeomorphism from an open set in Euclidean space onto the neighborhood, this is called a parameterization. Either way you get a space with the topology of a manifold.

For the manifold to be differentiable though, more is needed. One needs to be able to do calculus and in particular, one needs to take derivatives in different coordinate systems and compare them using the Chain Rule. For instance, the Chain Rules allows you to compare derivatives in the plane using either Cartesian coordinates or polar coordinates.

What this means is that the transition functions between two coordinates systems must themselves be differentiable. So one must require that a parameterization followed by a coordinate map be differentiable. Without this condition, the space is still a manifold but it is not a differentiable manifold. It is a topological manifold.

Given a topological manifold it may be possible to choose coordinate charts so that changes of coordinates are differentiable. In this case, the topological manifold has been given the structure of a differentiable manifold. There may be more than one way to do this. Each choice of smoothly overlapping charts is called an atlas.

A Lie Group has even more structure. It is a differentiable manifold and also a group. And in addition, the group operations are differentiable.

Last edited: Feb 23, 2016
7. Feb 23, 2016

### mnb96

Hi Lavinia.

Thanks. Your answer was quite clear and complete. There is just one sentence that I wish you could elaborate a bit more:

Well, let's say I would like to choose a parametrization. Then I will have a homeomorphism from an open set in Euclidean space onto some neighborhood of G.

Question 1) What is the topology of G? Is it the topology induced by the parametrization?

Question 2) If I would like to choose charts instead of parametrizations then I will have to assume a-priori some (arbitrary?) topology in G, am I right? What guarantees me that I will exactly choose the same topology as the one induced by the parametrizations?

8. Feb 23, 2016

### lavinia

No it already has a topology and there are many different parameterizations. The parameterization is just a homeomorphism from an open set in one topological space onto an open set in another topological space.

On the other hand you could build the topology out of open sets described by parameterizations. You would need to cover the entire manifold with them.

Because both are homeomorphisms and are just inverses off each other. Both say that a set is homeomorphic to an open set in Euclidean space.

Last edited: Feb 23, 2016
9. Feb 23, 2016

### mnb96

1) If we have a Lie group G, say SO(2), what is its topology? You say that G already has a topology but I still fail to see any topology in it, as in my mind G is just a collection of an infinite amount of 2x2 matrices and I can't see any criterion to define neighborhoods of a given matrix.

2) Is it so that in order to define a manifold, we need a topology both for the space in which the manifold "lives" and for the space of coordinates? If so, then the definition in Do Carmo's book is incomplete, because he only mentions the Euclidean topology of the coordinate space, am I right?

10. Feb 23, 2016

### lavinia

Matrices can be thought of as subsets of Euclidean space. For instance, 2x2 matrices are subsets of $R^4$. They inherit a topology from Euclidean space as subsets. Matrix multiplication is a polynomial in each coordinate and so is differentiable.

For SO(2), the rotation matrices, you can parameterize them with a single variable, $θ$ which is the angle of rotation. A typical element of SO(2) the is parameterized as $X_{11} = cos(θ)$ , $X_{12} = sin(θ)$ , $X_{21} = -sin(θ)$ , $X_{22} = cos(θ)$

Note that $θ$ maps an open interval, $(0,2π)$ ,diffeomorphically onto SO(2) minus the identity matrix. If you try to cover all of SO(2) you will not have a diffeomorphism of an open set. You need a second parameterization to cover all of it.

The unit circle in the complex plane is a Lie group given by complex multiplication. Its topology is inherited as a subset of the plane. One can parameterize it by $θ \rightarrow e^{iθ}$ and again one needs to miss one point to have a diffeomorphism. The map that takes $e^{iθ}$ to the matrix $X_{11} = cos(θ)$ , $X_{12} = sin(θ)$ , $X_{21} = -sin(θ)$ , $X_{22} = cos(θ)$ is a Lie group isomorphism. That is: it is a group isomorphism that is also a diffeomorphism.

So SO(2) is the same Lie group as the unit circle in the complex plane.

BTW: Groups can always be given the discrete topology in which case they are automatically Lie groups. From this point of view any finite group is a Lie group. But one could also give SO(2) the discrete topology. As such it would be a different Lie group than the unit circle in the complex plane. As a manifold a discrete Lie group is zero dimensional.

The topology is determined by the collection of all coordinate charts. Each separately is homeomorphic to an open set in Euclidean space and the entire topology is determined by all of their finite intersections and arbitrary unions.

The manifold does not necessarily live in another space. Its topology can be determined from the coordinate charts alone. Often though, manifolds are described as subsets of other spaces. But they can also be described from their coordinate charts without an ambient space. For instance, on the sphere one can draw polar coordinates from the north and south poles. These two charts and the way that they overlap determine the topology of the sphere.

Last edited: Feb 23, 2016
11. Feb 23, 2016

### mnb96

Thanks Lavinia for the detailed answer.

The examples that you reported seem to confirm what I was trying to say: in fact, in order to define a topology on the Lie group, you have always parametrized its elements.

I think the rest of your post clarifies my doubts.

12. Feb 23, 2016

### lavinia

Yes but you could go the other way as well. E.G. for SO(2) one has the chart that takes the inverse cosine of the first matrix element.
OK Consider the map that takes the logarithm of a point on the unit circle in the complex plane then divides by $i$. This gives a coordinate chart for each choice of a branch for the logarithm. For $SO(2)$ you could take the inverse cosine of the first coordinate.

The topology of SO(2) is the subspace topology inherited from $R^4$ but this defines it extrinsically as a subspace of another space.

13. Feb 23, 2016

### zinq

It's certainly not necessary to have parametrizations to define the topology on a Lie group.

For another example consider the rotations of 3-space R3. Which are the maps from R3 to itself that take the origin to itself, preserve all distances, and preserve the orientation of 3-space. These form the Lie group known as SO(3).

Then one rotation R is near another one S if R moves all points so they end up near where S moves them.

Or more formally, a sequence of rotations {R1, R2, R3, ...} approaches rotation S in the limit if for each point x of R3, the points R1(x), R2(x), R3(x), ..., Rn(x),... approach the point S(x) as n approaches ∞.

Here we made use of the known topology of R3, but we never needed any parametrizations of the rotation group SO(3).

14. Feb 24, 2016

### mnb96

Thanks zinq. Interesting observation.

By the way, I still haven't fully understood if the topology of a Lie group is something that we can infer in a canonical way from the sole elements and the group operation, or if it is something that we need to define arbitrarily (e.g. see Lavinia's example of SO(2) with discrete topology).

15. Feb 24, 2016

### zinq

I have recently seen the example of the Lie groups Rn, for n ≥ 1. As abstract groups, (i.e., forgetting about their topology) these are all isomorphic. (Technically, this can be seem from the fact that, as vector spaces over the rationals, ℚ, all the Euclidean spaces of dimension ≥ 1 have the same dimension: the continuum, which implies they are isomorphic as abstract groups.)

But they are not isomorphic as Lie groups because their underlying manifolds are not even homeomorphic, no less diffeomorphic.

* * *​

Conversely, it is also true that a given smooth manifold can have non-isomorphic Lie group structures on it. Probably the lowest-dimensional example is, on the one hand, the Lie group R2, and on the other hand the group defined as

G = L[a, b] | (a,b) ∈ R2}​

where L[a, b] may be thought of as the function

L[a, b]: RR

via

L[a, b](x) = ea x + b​

and the group operation is defined by composition of functions:

L[a, b] o L[c, d] = L[a+c, ead+b].​

Since R2 is commutative and G is not, that shows they cannot be isomorphic as abstract groups, and hence not as Lie groups, either.

(((
A more sophisticated example is, on the one hand SO(4) and on the other hand the product group SO(3) × SU(2), both of which have as underlying manifolds the cartesian product P3 x S3.

Since SO(4) can be shown not to be a product group, these cannot be isomorphic as Lie groups.
)))