Hilbert manifold

Remarks

Example: let $H$ be a real Hilbert space and $S = \set{x \in H : \norm{x} = 1}$ its unit sphere. For $p \in S$, the map $x \mapsto x - \ip{x}{p} p$ restricted to $\set{x \in S : \ip{x}{p} > 0}$ is a chart onto the open unit ball of the closed hyperplane $p^\perp$, with inverse $y \mapsto y + \sqrt{1 - \norm{y}^2}\, p$. Both maps are smooth, $p^\perp$ is a Hilbert space, and these charts cover $S$, so $S$ is a smooth Hilbert manifold.

Non-example: the Banach space $\ell^1$ with its identity chart is a Banach manifold, but no compatible chart takes values in a Hilbert space. A compatible chart $\varphi$ into a Hilbert space $H$ would have a derivative at a point that is a linear isomorphism of Banach spaces from $\ell^1$ onto $H$. But $\ell^1$ is not isomorphic to a Hilbert space.

Every finite-dimensional smooth manifold is a Hilbert manifold modelled on $\R^n$.