1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.