Is the dimension of a Hilbert space as a manifold different from the dimension as a vector space?

Question

A Hilbert space $H$ has a natural topology $\tau$ induced from its inner product. If $(H,\tau)$ constitutes a manifold, is the manifold dimension of $H$ different from its vector space dimension? In what cases does this hold and in what cases does this not hold?

Answer 1

No. First, if $H$ has the structure of a smooth or even topological manifold, then $H$ has finite vector space dimension (manifolds are locally compact but infinite dimensional Hilbert spaces are not). However, every finite dimensional Hilbert space (and in fact every finite dimensional real topological vector space) is homeomorphic to $\mathbb{R}^n$ where $n$ is the vector space dimension. Thus, $H$ is a topological manifold iff it is finite dimensional in which case its manifold dimension must be the same as its vector space dimension. The same reasoning applies for complex Hilbert spaces and manifolds.