Let $V$ be a finite dimensional real inner product space and $U$ a real inner product space of countable dimension. Why is the space of linear isometries from $V$ to $U$ paracompact?

## 1 Answer

Certainly if $U$ is a finite-dimensional inner product space, then $\text{Isom}(V, U)$ is paracompact (being a closed subspace of $\text{Lin}(V, U) \cong U^n$).

If $U$ has countably infinite dimension, then $U$ is a sequential colimit of finite-dimensional inner product subspaces $U_n$ and inclusions between them, and

$$\text{Isom}(V, U) \cong \text{colim}_n \text{Isom}(V, U_n)$$

since each isometry $V \to U$ must factor (for some $n$) as $V \to U_n \hookrightarrow U$ for a uniquely determined isometry $V \to U_n$. So $\text{Isom}(V, U)$ equipped with the colimit topology is a sequential colimit along closed embeddings of paracompact Hausdorff spaces. But such a colimit is again paracompact Hausdorff (see Proposition 4.2 here).