The infinite-dimensional separable Hilbert space $H$ has the unusual property that it is diffeomorphic to its unit sphere $S^{\infty}$. Therefore, $H$ admits the round metric as a complete and bounded Riemannian structure, which only applies to compact Riemannian manifolds in finite-dimensional case.

According to corollary 4 in "Infinite-dimensional Manifolds are Open Subsets of Hilbert Space" by David W. Henderson (see this download link), every connected infinite-dimensional separable Hilbert manifold is homeomorphic to some $\lvert K \rvert \times H$, where $K$ is a connected countable locally finite simplicial complex. Since "homotopic" equals "diffeomorphic" for Hilbert manifolds (corollary 2 and 3 in the same article), the product Riemannian structure is complete and bounded if $\lvert K \rvert$ is homotopic to a closed smooth manifold. There are also examples where $K$ has to be infinite-dimensional, such as the infinite complex projective space $\mathbb{CP}^{\infty}=S^{\infty}/S^1$ with its Fubini-Study metric.

Is it true that every infinite-dimensional, separable Hilbert manifold admits a complete and bounded Riemannian structure? In particular, does the infinite torus $\mathbb{T}^{\infty}=H/\oplus_{i=1}^{+\infty} \mathbb{Z}$ (the group acts by translation) admit one?