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.