Dtyjlui

This post should be self contained - it is from a concern of page 13 of this notes.

Let $q ge n$, $V_n(Bbb R^q)$ be the subspace of $n$ -linearly independent vectors in $(Bbb R^q)^n$.

We define $G_n(Bbb R^q)$ to be the set of $n$ planes in $Bbb R^q$.

It is given the quotient topology of the map
$$ q:V_n(Bbb R^q) rightarrow G_n(Bbb R^q)$$
sending $n$ vectors to the space spanend by it.

Now fix $X in G_n(Bbb R^q)$, and let $U subseteq G_n(Bbb R^q)$, be the open set such that for $Y in U$, the projection map $pi:Bbb R^q rightarrow X$, induces an isomoprhism $pi|_Y:Y rightarrow X$.

I want to prove the following map is continuous:

Fix a basis ${x_i }$ of $X$
$$D:q^{-1}(U) rightarrow q^{-1}(U)$$
such that $(y_1, ldots, y_n)$ is mapped to $(y'_1 ,ldots, y'_n)$ where $pi(y'_i) = x_i$ and $q(y'_1,ldots, y'_n) = q(y_1,ldots, y_n)$.

It is claimed that (from the notes) this is continuous. But I do not see an easy expression of this map.

I think the author was not well-advised to omit the proof.

We begin by recalling some facts from linear algebra:

Let $V$ be a vector space over a field $K$. Each matrix $A = (a_{ij}) in M_n(K)$ induces a linear map $A : V^n to V^n, A(v_1,dots,v_n) = (sum_{j=1}^n a_{1j}v_j,dots,sum_{j=1}^n a_{nj}v_j)$. Note that in this definition $dim(V)$ is arbitrary. If $mathbf{v} = (v_1,dots,v_n) in V^n$ forms a basis of $V$ (which requires $dim(V) = n$), then each $mathbf{w} = (w_1,dots,w_n) in V^n$ admits a unique $A(mathbf{v},mathbf{w}) in M_n(K)$ such that $A(mathbf{v},mathbf{w})(mathbf{v}) = mathbf{w}$. We have $A(mathbf{v},mathbf{w}) in GL_n(K)$ if and only if $(w_1,dots,w_n)$ forms a basis of $V$. In that case $A(mathbf{v},mathbf{w})$ realizes the change of basis from $mathbf{v}$ to $mathbf{w}$.

Each linear map $f : V to W$ induces a linear map
$$f^n : V^n to W^n, f^n(v_1,dots,v_n) = (f(v_1),dots,f(v_n)) .$$
It is readily verified that
$$f^n(A(mathbf{v})) = A(f^n(mathbf{v})) .$$

We now come to the proof of continuity. Given a fixed basis $mathbf{x} = (x_1,dots,x_n)$ of $X$, the orthogonal projection $pi : mathbb{R}^q to X$ can be written as $pi(y) = sum_{j=1}^n xi_j(y) x_j$ with linear maps $xi_j :mathbb{R}^q to mathbb{R}$. This gives us a linear map
$$phi : (mathbb{R}^q)^n to M_n(mathbb{R}), phi(y_1,dots,y_n)_{ij} = xi_j(y_i) .$$
For each $mathbf{y} = (y_1,dots,y_n) in q^{-1}(U)$ the span $q(mathbf{y})$ is mapped by $pi$ isomorphically onto $X$. Hence $pi^n(mathbf{y}) = (pi(y_1),dots,pi(y_n)) = (sum_{j=1}^n xi_j(y_1) x_j,dots,sum_{j=1}^n xi_j(y_n) x_j)$ is a basis of $X$. Thus the matrix $phi(mathbf{y})$ realizes the change of basis from $mathbf{x}$ to $pi^n(mathbf{y})$, i.e. we have $phi(mathbf{y})(mathbf{x}) = pi^n(mathbf{y})$. We conclude
$$phi(q^{-1}(U)) subset GL_n(mathbb{R}) .$$
Since linear maps between finite-dimensional vector spaces (endowed with any norm) are continuous and inverting matrices in $ GL_n(mathbb{R})$ is continuous, we see that
$$psi : q^{-1}(U) to GL_n(mathbb{R}), psi(mathbf{y}) = phi(mathbf{y})^{-1}$$
is continuous. The matrix $psi(mathbf{y})$ realizes the change of basis from $pi^n(mathbf{y})$ to $mathbf{x}$, i.e. we have $psi(mathbf{y})(pi^n(mathbf{y})) = mathbf{x}$.

For $mathbf{y} in q^{-1}(U)$ define
$$D(mathbf{y}) = psi(mathbf{y})(mathbf{y})$$
where we regard $psi(mathbf{y})$ as a linear map $q(mathbf{y})^n to q(mathbf{y})^n$. Since $mathbf{y}$ is a basis of $q(mathbf{y})$, also $D(mathbf{y})$ is a basis of $q(mathbf{y}) in U$. Hence $D(mathbf{y}) in q^{-1}(U)$, i.e. we have defined a function
$$D : q^{-1}(U) to q^{-1}(U) .$$

We have
$$pi^n(D(mathbf{y})) = pi^n(psi(mathbf{y})(mathbf{y})) = psi(mathbf{y})(pi^n(mathbf{y})) = mathbf{x}$$
which shows that our $D$ is the same as the author's.

To see that $D$ is continuous note that the coordinate functions $D_i : q^{-1}(U) to mathbb{R}^q$ are given by $D_i(mathbf{y}) = sum_{j=1}^n psi(mathbf{y})_{ij}y_j = sum_{j=1}^n psi(mathbf{y})_{ij}p_j(mathbf{y})$ with (continuous!) coordinate projections $p_j : (mathbb{R}^q)^n to mathbb{R}^q$.