Showing that the isomorphism of the general linear group of a vector space with the group of invertible matrices is smooth

Question

This is Example 7.3(e) from John Lee's Introduction to Smooth Manifolds.

If $V$ is any real or complex vector space, $GL(V)$ denotes the set of invertible linear maps from $V$ to itself. It is a group under composition. If $V$ has finite dimension $n$, any basis for $V$ determines an isomorphism of $GL(V)$ with $GL(n,\mathbb{R})$ or $GL(n,\mathbb{C})$ so $GL(V)$ is a Lie group. The transition map between any two such isomorphisms is given by a map of the form $A\mapsto BAB^{-1}$ (where $B$ is the transition matrix between the two bases), which is smooth. Thus the smooth manifold structure on $GL(V)$ is independent of the choice of basis.

From here, I think the isomorphism of $GL(V)$ with $GL(n,\mathbb R)$ is $f: T \mapsto [T]_\beta$ where $[T]_\beta$ is the matrix representation of the invertible linear map in $V$ w.r.t the basis $\beta$. However, how can I show that this map is smooth? I can't think of a chart in the domain of $GL(V)$ containing $T$, $(U,\phi)$ and a chart $(V,\psi)$ containing $[T]_\beta$ such that $\psi \circ f \circ \phi^{-1}$ is smooth. I would greatly appreciate if anyone can help me with the construction of charts here.

Answer 1

I suggest to use an alternative approach based on the following three simple facts:

1. Each linear map $f : \mathbb R^m \to \mathbb R^m$ is smooth.

2. Each finite-dimensional real vector space $W$ has a canonical smooth structure. In fact, if $\dim W = m$, a smooth atlas $\mathcal A_W$ is obtained by taking all linear isomorphisms $W \to \mathbb R^m$, and this atlas generates the canonical smooth structure.

3. Each linear isomorphism $\phi : W \to W'$ between finite-dimensional real vector spaces is a diffeomorphism. To see this, simply use $\mathcal A_W$ and $\mathcal A_{W'}$ to represent $\phi$ and $\phi^{-1}$ in terms of charts.

In particular, the vector space $\operatorname{End}_\mathbb R(V)$ of all linear endomorphisms on $V$ has a canonical smooth structure.

For each isomorphism $\phi : V \to \mathbb R^n$ we define $$\phi_\# : \operatorname{End}_\mathbb R(V) \to \operatorname{End}_\mathbb R(\mathbb R^n), \phi_\#(f) = \phi \circ f \circ \phi^{-1}.$$ This is an isomorphism of vector spaces.

The matrix representation of linear maps $g : \mathbb R^n \to \mathbb R^n$ with respect to the standard basis of $\mathbb R^n$ gives a canonical isomorphism of vector spaces $$\mu : \operatorname{End}_\mathbb R(\mathbb R^n) \to \operatorname{M}(n, \mathbb R) .$$ Hence $$\phi_* = \mu \circ \phi_\# : \operatorname{End}_\mathbb R(V) \to \operatorname{M}(n, \mathbb R)$$ is an isomorphism of real vector spaces and therefore a diffeomorphism.

Usually one reads phrases like

we can identify $\operatorname{M}(n, \mathbb R)$ with $\mathbb R^{n^2}$.

This is true, but there are many such identifications. A nice one is writing the $n$ row vectors of a matrix $A = (a_{ij})$ one after the other which produces the isomorphism $$\rho : \operatorname{M}(n, \mathbb R) \to \mathbb R^{n^2}, \rho(A) = (a_{11},\ldots,a_{1n},a_{21},\ldots,a_{2n},\ldots, a_{n1},\ldots,a_{nn}) .$$ This is a diffeomorphism of smooth manifolds. Of course we can take any other isomorphism $\rho' : \operatorname{M}(n, \mathbb R) \to \mathbb R^{n^2}$, but the above one has the benefit that it makes immediately clear that the determinant $\det : \operatorname{M}(n, \mathbb R) \to \mathbb R$ is continuous (even smooth). Just recall that $\det A = \sum_{\sigma \in \mathfrak S_n} \operatorname{sign}(\sigma) a_{1\sigma(1)} \ldots a_{n\sigma(n)}$.

The continuity of $\det$ shows that $\operatorname{GL}(n, \mathbb R)$ is open in $\operatorname{M}(n, \mathbb R)$, thus it inherits a smooth structure from $\operatorname{M}(n, \mathbb R)$. We conclude that $\operatorname{GL}(V) = \phi_* ^{-1}(\operatorname{GL}(n, \mathbb R))$ is open in $\operatorname{End}_\mathbb R(V)$, thus it also inherits a smooth structure from $\operatorname{End}_\mathbb R(V)$.

All we need here is that $\operatorname{GL}(V)$ is open in $\operatorname{End}_\mathbb R(V)$. However, the use of a base is implicit in the above construction (using $ \phi$).

Our construction also shows that $\phi_*$ restricts to a diffeomorphism $\operatorname{GL}(V) \to \operatorname{GL}(n, \mathbb R)$.

The situation for a complex vector space $V$ is similar. For the moment let us ignore smooth structures. As above we get an an isomorphism of complex vector spaces

$$\phi_* : \operatorname{End}_\mathbb C(V) \to \operatorname{M}(n, \mathbb C)$$ which depends on the choice of an isomorphism $\phi : V \to \mathbb C^n$.

This is pure linear algebra, no additional structures are involved so far.

Now we need the following two simple facts:

1. Each complex vector space $V$ can be regarded as a real vector space. If $\dim_\mathbb C V = n$, then $\dim_\mathbb R V = 2n$. Let us write $[V]$ when we regard $V$ as a real vector space.

2. Each $\mathbb C$-linear map $f : V \to W$ between complex vector spaces $V, W$ is $\mathbb R$-linear. More precisely, $f : [V] \to [W]$ is $\mathbb R$-linear.
   In particular, if $f$ is an isomorphism between the complex vector spaces $V, W$, then it is also an isomorphism between the real vector spaces $[V],[W]$.

Both $\operatorname{End}_\mathbb C(V)$ and $\operatorname{M}(n, \mathbb C)$ are complex vector spaces. Thus, regarding them as real vector spaces, they receive canonical smooth structures. This makes $\phi_* : \operatorname{End}_\mathbb C(V) \to \operatorname{M}(n, \mathbb C)$ a diffeomorphism.

Using the determinant $\det : \operatorname{M}(n, \mathbb C) \to \mathbb C$ we see that $\operatorname{GL}(n,\mathbb C)$ is open in $M(n,\mathbb C)$ and conclude that $\operatorname{GL}(V)$ is open in $\operatorname{End}_\mathbb C(V)$.