$\mathbb{R}^{m\times n}$ being isomorphic to $\mathbb{R}^{mn}$

differential-topologygroup-theorymanifoldsmatricessmooth-manifolds

I am reading Tu's book titled "Introduction to Manifolds". In Chapter 2, paragraph 5.3, example 5.15 states that

…Since $\mathbb{R}^{m\times n}$ is isomorphic to $\mathbb{R}^{m n}$, we give it the topology of $\mathbb{R}^{m n}$…

I have a few questions about this statement:

  1. If we have to say that $\mathbb{R}^{m\times n}$ is isomorphic to $\mathbb{R}^{m n}$, that means that these two spaces are different? How are they different? What is the first space and what is the second?

  2. we give the topology of $\mathbb{R}^{m n}$ to what? To $\mathbb{R}^{m\times n}$? And if so, why are we allowed to give the topology of some space to another, given that they are isomorphic?

  3. How is all that useful for what is next, i.e. defining the general linear group of $n\times n$ matrices with real entries?

Any help will be appreciated…

Best Answer

Q1:

Both $\mathbb{R}^{m\times n}$ and $\mathbb{R}^{m n}$ are vector spaces over the ground field $\mathbb R$ with dimensio $m \cdot n$. However, the elements of $\mathbb{R}^{m\times n}$ are $(m \times n)$-matrices $A = (a_{ij})$ and the elements of $\mathbb{R}^{m n}$ are tupels $(x_1,\ldots, x_{mn})$. The two vector spaces are isomorphic, but not identical. An isomorphism is given by writing down the rows of a matrix one after the other, that is $$\phi : \mathbb{R}^{m\times n} \to \mathbb{R}^{m n}, \phi((a_{ij}) = (a_{11},\ldots,a_{1n},a_{21}, \ldots, a_{2n}, \ldots, a_{m1},\ldots,a_{mn}) .$$

Note that there are many other isomorphisms. For example we can write down the columns of a matrix one after the other.

Euclidean spaces $\mathbb R^k$ have a standard topology induced by the Euclidean norm $\lVert (x_1,\ldots,x_k) \rVert = \sqrt{\sum_{i=1}^k x_i^2}$.

We can now give $\mathbb{R}^{m\times n}$ the unique topology making $\phi$ a homeomorphism. But doesn't this topology depend on the choice of a linear isomorphism $\phi$ between $\mathbb{R}^{m\times n}$ and $\mathbb{R}^{m n}$? It is an easy exercise to show that this is not the case: For all isomorphisms $\phi : \mathbb{R}^{m\times n} \to \mathbb{R}^{m n}$ we get the same topology. By the way, this topology is induced on the Euclidean matrix norm $\lVert (a_{ij}) \rVert = \sqrt{\sum_{i,j} a_{ij}^2}$.

Q2:

This has been already answered. Generally speaking, if we have a bijection $\beta : S \to X$ from a set $S$ to a topological space $X$, we can give $S$ the unique topology making $\beta$ a homeomorphism. Explicitly, if $\mathcal T_X$ is the topology of $X$, then we give $S$ the topology $\mathcal T_S = \beta^{-1}(\mathcal T_X) = \{ \beta^{-1}(U) \mid U \in \mathcal T_X \}$.

Q3:

The general linear group $\operatorname{GL}(n,\mathbb R)$ is a certain subset of $\mathbb{R}^{n\times n}$. Since we endowed $\mathbb{R}^{n\times n}$ with a topology, $\operatorname{GL}(n,\mathbb R)$ becomes a topological space (with the subspace topology inherited from $\mathbb{R}^{n\times n}$). In fact it is an open subspace of $\mathbb{R}^{n\times n}$ and therefore a manifold.

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