In what sense is a matrix Lie group a type of manifold?

Question

I was trained as a physicist, rather than a mathematician. So, I apologize if my question is naive. I believe I have the answer to my question, but I want to make sure that my understanding is correct.

I have studied a little bit of the second edition of Brian Hall's book, "Lie Groups, Lie Algebras, and Representations". On p. 4 of his book, Hall defines a matrix Lie group as a group whose elements are invertible square matrices. (A matrix Lie group has additional properties, which I will not repeat here.) On p. 25, Hall defines a Lie group as a type of manifold. (A Lie group of course has additional properties, but I will not repeat these.)

Hall notes that all matrix Lie groups are Lie groups, although the converse is not true. Since all matrix Lie groups are Lie groups, and since Lie groups are manifolds, matrix Lie groups must in some sense be manifolds.

As a physicist, I intuitively understand a manifold to be an object which looks locally like $\mathbb{R}^n$. (Any neighborhood of a point on a manifold can be mapped to $\mathbb{R}^n$ by a one-to-one continuous map.) I understand that mathematicians have a more precise definition for a manifold as a type of topological space, but I do not know topology.

Naively, one might think that matrix Lie groups could not possibly be manifolds. Examples of manifolds are Euclidean spaces and hypersurfaces embedded in Euclidean spaces (although not all manifolds have embeddings into higher dimensional Euclidean spaces). Intuitively, one would think that a point in a Euclidean space or a point on a hypersurface in no sense is a matrix. Since the elements of matrix Lie groups are matrices, one intuitively would think that the elements of matrix Lie groups are different than the points in a manifold. Thus, one would come to the erroneous conclusion that matrix Lie groups cannot be manifolds.

As I stated, I believe I know the resolution to this issue, but I want to make sure my understanding is correct. I believe the resolution is as follows.

For a matrix Lie group whose elements are $n \times n$ real matrices, one could simply identify each entry of each matrix with a coordinate of $\mathbb{R}^{n^2}$. This identification would map the elements of the matrix Lie group to $\mathbb{R}^{n^2}$. The mapping would not in general cover all of $\mathbb{R}^{n^2}$. Rather, the mapping would cover some hypersurface in $\mathbb{R}^{n^2}$.

For a matrix Lie group whose elements are $n \times n$ complex matrices, one could perform a similar identification. One could identify each real part of each matrix with a coordinate in $\mathbb{R}^{2n^2}$. One then could identify each imaginary part of each matrix with the remaining coordinates. This would be a mapping of the elements of the group to $\mathbb{R}^{2n^2}$. The mapping would cover some hypersurface in $\mathbb{R}^{2n^2}$.

Thus, for a matrix Lie group with either real or complex matrix entries, the identification I described would map the elements of the group onto a hypersurface in $\mathbb{R}^N$, where $N$ is some integer. Such a hypersurface would be a manifold embedded in $\mathbb{R}^N$. The dimension of the manifold would be the number of real parameters characterizing the matrix Lie group.

There obviously is a one-to-one correspondence between each element of a matrix Lie group and each point of the corresponding manifold. Due to this one-to-one correspondence, I believe we are justified in calling a matrix Lie group a manifold.

I believe Hall's book outlines the correspondence between a matrix Lie group and a manifold, as I described it. However, the book is difficult for me to read, since I am not trained as a mathematician. I also have to return the book to the library soon. Before I return the book, I want to make sure I correctly understand how a matrix Lie group is a manifold.

I know that the correspondence I described is not sufficient to show that a matrix Lie group is a Lie group, since being a manifold is not the only property of a Lie group. I simply want to make sure I am correctly understanding how matrix Lie groups are manifolds.

Thank you for your help.

Answer 1

Using Hall’s definition it is not trivial at all, that matrix Lie groups are actually Lie groups, which are by definition manifolds.

The fact that matrix Lie groups are Lie groups follows from Cartan’s closed-subgroup theorem, which states that if G is a Lie group and H is a subgroup of G which is closed subset of G then H Lie subgroup of G, i.e. H has the structure of a manifold such that the natural map $\iota:H\rightarrow G$ $\iota(h)=h$ is a smooth embedding and a group homomorphism. As a corollary one gets that H itself is a manifold.

In your case $G=GL(n,\mathbb{C}) $ and the definition of matrix Lie groups are precisely subgroups of G which are closed subsets of G. To see that $GL(n,\mathbb{C})$ is a manifold first notice that the space of $n\times n$ matrices with complex entry is just the space $\mathbb{C}^{n^2}$, just put the rows next to each other to get a vector. Now look at the map $\text{det}:\mathbb{C}^{n^2}\rightarrow \mathbb{C}$, this is smooth since it is a polynomial, and $GL(n,\mathbb{C})=\{ x\in \mathbb{C}^{n^2} \ | \ \text{det}(x)\neq 0\} $, i.e. it is an open set of $\mathbb{C}^{n^2}$, thus every point has a neighborhood which is locally looks like $\mathbb{R}^{2n^2}$ (just use the identity map). Now we also have to check that the multiplication is smooth but again it is just the polynomials of the coordinates so it is smooth.

Answer 2

I have not read a general full proof of why matrix groups are Lie groups, but it really helps to look at an instructive example.

Let $G = SO(3)$. This is defined as the group of all real $3$ by $3$ matrices $g$ such that $g^T g = 1$ (with the right-hand side denoting the $3$ by $3$ identity matrix). If we denote by $v_1$, $v_2$ and $v_3$ the three columns of $g$, the equation $g^T g = 1$ consists of the following constraints:

1. $v_i$ has unit Euclidean norm, for $i = 1, 2, 3$.
2. $v_i$ is orthogonal to $v_j$ (i.e. their Euclidean inner/dot product vanishes) for $1 \leq i < j \leq 3$.

So in total, 1. gives 3 constraints and 2. also gives another 3 constraints. Thus $g^T g = 1$ consists of $6$ real constraints. Thus one would expect the dimension of $G$ to be $3^2 - 6 = 3$.

Actually, there are theorems in mathematics which, when applied to this case, would also tell you that $G$ is a smooth $3$-dimensional manifold. There are 2 cousin theorems which could be applied here: the inverse function theorem and the implicit function theorem. In this case, it is easier to apply the implicit function theorem, or a special case, which is known as the constant rank theorem.

Moreover, if like me you like to see explicit maps, then it is actually not too hard to show that $SO(3)$ is diffeomorphic to $\mathbb{R}P^3$ (real projective $3$-space), as a smooth manifold. Well, I say it is not too hard, but I guess I mean, if I give you some hints, then it is not too difficult to find such a map explicitly.

My favorite way to see this is to first consider $SU(2)$, which one could easily show that it is diffeomorphic to $S^3$, as a smooth $3$-manifold. Then one could show that the quotient $SU(2)/\{\pm 1\}$ is diffeomorphic to $G = SO(3)$. But $\mathbb{R}P^3$ can be described as the quotient of $S^3$ by the antipodal map, so that $G$ is indeed diffeomorphic to real projective $3$-space $\mathbb{R}P^3$.

I like to think of smooth manifolds as curved "spaces" without singularities, like spheres, tori ("surface" of a donut), etc. The technical definition can take a while to master. You can think that a smooth $n$-manifold is obtained by gluing little open subsets of Euclidean $n$-dimensional space together in a way that is smooth and doesn't have singularities.

If you are interested in mathematics and physics, you may appreciate a book like Penrose's "Road to Reality". I thought it was a fun read, but it is a big book.