# Stiefel Manifold is a Manifold

We consider the real Stiefel manifold $$V_k(\mathbb{R}^n) := \{(v_1, v_2, ..., v_k) \in S^{n-1} \times ... \times S^{n-1} \vert = \delta_{ij} \}$$.

I want to know how to show that $$V_k(\mathbb{R}^n)$$ is a topological manifold without using differential geometric methods.

Indeed, interpreting $$V_k(\mathbb{R}^n)$$ as $$\lbrace X\in\mathbb{R}^{n\times k}\big\vert X^{\text{T}}X=I_k \}$$ one can show that it is a submanifold of $$\mathbb{R}^{n\times k}$$. But using this argument one has to show that the derivative of the map $$X^TX-I_k$$ has full rank so we make here unfortunately usage of differential geometry. But I'm looking for pure topological arguments.

Hausdorff and second countableness is ok. The problem is to find for an arbitrary $$x \in V_k(\mathbb{R}^n)$$ a local neighbourhood with $$U_x \cong \mathbb{R}^m$$ with $$m= kn +1/2(k+1)k$$.

• Just follow Example 4.54 in Allen Hatcher's AT book (trivialization as fiber bundles)? – user10354138 Nov 11 '18 at 4:21