# How can a group of matrices form a manifold?

So for example, $GL(n,\mathbb{R})$ group. It is said that this group can be considered as manifold - but I do not get how this is possible. How does one then assign a neighborhood of a matrix, and talk of compactness? (of course manifold can be disconnected, but.)

• The $n\times n$-matrices can be thought of as points in $\mathbb{R}^{n^2}$. The invertible ones form an open subset (disconnected by the zero set of the determinant function). – Jyrki Lahtonen Jun 14 '13 at 13:33
• If you identify a $n\times n$ matrix with a $n^2$-vector, then your group can be thought as a vector space. Thus you have a natural topology and you can talk about neighborhoods – Abramo Jun 14 '13 at 13:34

An $n \times m$ matrix is by definition a function $A: \{1, \ldots, m\} \times \{1, \ldots, n\} \to \mathbb{R}$. Most of the time we write $A_{ij}$ for its value at $(i,j)$ instead of $A(i,j)$. If you think about the definition of a vector, this is equivalent to looking at points of $\mathbb{R}^{mn}$. So for $n \times n$ square matrices, they are just points of $\mathbb{R}^{n^2}$.

On $\mathbb{R}^k$ we have a bunch of norms. It's a basic theorem of functional analysis that these are all equivalent and define equivalent topologies (this is essentially what the point of equivalence is). Perhaps the simplest is just the $\ell_1$-norm: the sum of the absolute values of the entries of the matrix.

So now we can talk about distance and other similar notions like compactness. For example, we have the function $\det: \mathbb{R}^{n^2} \to \mathbb{R}$. By its definition, it is a polynomial in the entries of the matrix. Hence it is a very nice smooth function. The group $\text{GL}(n,\mathbb{R})$ is

$$\text{GL}(n, \mathbb{R}) = \{ A \in \mathbb{R}^{n^2} \mid \det(A) \neq 0 \}$$

and so is the preimage of an open set under a continuous function. Therefore it is an open set.

The special linear group consists of the matrices with determinant $\pm 1$, and so similarly forms a closed subgroup.

For a compact example, consider the orthogonal matrices. An orthogonal matrix, if you think about it, is one that satisfies $n^2$ linear equations in its entries. So they'll form a closed set (a subgroup, even). Moreover, we can see that they are actually bounded (the column sums cannot exceed $1$), and so by the Heine-Borel theorem AKA the characterisation of compactness in $\mathbb{R}^k$ we know that they are a compact subgroup.

A nice application of this topological nonsense is the Cayley-Hamilton theorem (any matrix is a root of its characteristic polynomial). Working over $\mathbb{C}$, one can show that the diagonalizable matrices are in fact dense in $\mathbb{C}^{n^2}$. The Cayley-Hamilton theorem is true for these, and hence extends to all complex matrices. Moreover, one can show by some algebraic nonsense that this implies it's true for all commutative rings (with unit).

• nice answer. Do you happen to have a reference on the Cayley Hamilton argument. I'm interested in the topologically nonsensical aspect. I have plenty of books with the algebra nonsense. – James S. Cook Jun 14 '13 at 14:30
• this particular writeup of the density part is succinct: math.stackexchange.com/a/397915/52216 i seem to recall that there's a terry tao or gowers version somewhere, but i can't seem to find it – citedcorpse Jun 14 '13 at 14:35

I give this answer to expand on the comments of Jyrki Lahtonen and Abramo.

Let $M = gl(n, \mathbb{R})$ which is the set of $n \times n$ matrices over $\mathbb{R}$. A global coordinate chart $x_{ij}$ is given by the component mapping: $$x_{ij}(A) = e_i^TAe_j$$ where $e_i$ serves as the standard basis of $\mathbb{R}^n$. It is easy to show this is a bijection from $M$ to $\mathbb{R}^{n^2}$ and it gives $M$ the structure of an $n^2$-dimensional manifold.

The mapping $det: M \rightarrow \mathbb{R}$ has $GL(n, \mathbb{R})$ as the complement of the fiber over $0$ of this map; $GL(n, \mathbb{R}) = M-det^{-1} \{ 0 \}$. As such it is simply a restriction of $M$ to a subset. I leave it to you to ascertain if $GL(n, \mathbb{R})$ so constructed is open or closed. I'll give you a hint: the determinant map is formed by polynomials of coordinate maps on $M$ hence is clearly continuous and $\{0 \}$ is a closed set.