Showing tha $\mathbb{R}^{a \times b}_{\mathrm{O}}$ is a submanifold of $\mathbb{R}^{a \times b}_{\mathrm{LI}}$. Let $\mathbb{R}^{a \times b}_{\mathrm{LI}}$ denote the set of $a \times b$ matrices with linearly independent columns. Let $\mathbb{R}^{a \times b}_{\mathrm{O}}$ denote the set with orthonormal columns. Suppose we wish to show that $\mathbb{R}^{a \times b}_{\mathrm{O}}$ is a submanifold of $\mathbb{R}^{a \times b}_{\mathrm{LI}}$, one way would be to use the Gram-schmidt function $$\Gamma : \mathbb{R}^{a \times b}_{\mathrm{LI}} \rightarrow \mathbb{R}^{a \times b}_{\mathrm{O}}$$ and argue that it's image has to be a submanifold.

Question. Is there an easy way of doing this? If not, what would be a good way of showing that $\mathbb{R}^{a \times b}_{\mathrm{O}}$ is a submanifold?

 A: Idea: consider the matrices as vector tuples $A = (A_i)$. Define a function with coordinates $f_{ij}$ by
$$f_{i,j}(A) = \langle A_i,A_j\rangle.$$
Your submanifold is defined by the equations
$$f_{i,j}(A) = \delta_{i,j}.$$
Now, use the regular value theorem.
A: It's better to mimic the argument that $O_n(\mathbb{R})$ is a submanifold of $\operatorname{GL}_n(\mathbb{R})$ which corresponds to the case $a = b$. Namely, let $S = \{ C \in M_{b}(\mathbb{R}) \, | \, C = C^T \}$ denote the vector space of $b \times b$ symmetric matrices and consider the map $T \colon M_{a \times b}(\mathbb{R}) \rightarrow S$ given by $T(A) = A^TA$. Then $\mathbb{R}_{O}^{a \times b} = T^{-1}(I_b)$ so it is enough to show that $I_b$ is a regular value of the map $T$. Let $A \in \mathbb{R}_{O}^{a \times b}$ and let $C \in S$. We need to show that the equation
$$ dT|_{A}(B) = \frac{d}{dt}(A + tB)^T(A + tB)|_{t = 0} = B^T A + A^T B = C $$
has a solution. By choosing $B = \frac{AC}{2}$ we have
$$ B^T A + A^T B = \frac{1}{2} (C^T A^T A + A^T A C) = \frac{1}{2} (C^T + C) = C$$
and so $dT|_{A} \colon M_{a \times b}(\mathbb{R}) \rightarrow S$ is onto and $\mathbb{R}^{a \times b}_{O}$ is an embedded submanifold of $M_{a \times b}(\mathbb{R})$ of dimension
$$ \dim M_{a \times b}(\mathbb{R}) - \dim S = ab - \frac{b(b+1)}{2} = \frac{b(2a - b - 1)}{2}.$$
Since $\mathbb{R}^{a \times b}_{\operatorname{LI}}$ is an open subset of $M_{a \times b}(\mathbb{R})$ and $\mathbb{R}^{a \times b}_{O} \subset \mathbb{R}^{a \times b}_{\operatorname{LI}}$ we see that $\mathbb{R}^{a \times b}_{O}$ is also an embedded submanifold of $\mathbb{R}^{a \times b}_{\operatorname{LI}}$ of the same dimension.
