Is $C[0,1]$ a manifold?

Question

I know that $C[0,1]$, as a topological space induced by the metric $d(f,g)=\sup_x |f(x)-g(x)|$, is Hausdorff, second countable, and has cardinality same as $\mathbb R$. But is it a manifold?

By manifold, I mean a topological space that is Hausdorff and second countable. The chart map from an open neighbourhood of a point in $C[0,1]$ to a open $n$ dimensional euclidean space. Is $C[0,1]$ a manifold? What is the dimension?

Answer 1

An infinite-dimensional normed space $X$, such as $C[0,1]$ is not a topological manifold over any $\mathbb{R}^n$, because there is no homeomorphism between an open subset $U\subset X$ and an open subset $V\subset \mathbb{R}^n$. To see why, take a compact subset of $V$ with nonempty interior; its image in $U$ would have to be also be a compact set with nonempty interior. But there are no such sets in $X$, because closed balls are not compact. The latter follows from Riesz' lemma.

For completeness, I also quote a comment by Thomas:

it's a Banach manifold (modeled over the Banach space $C[0,1]$). You can do differential geometry also in the infinite dimensional case, though you have to be careful with some statements which are considered trivial in finite dimensions (like the statement that every linear functional is continous or that every finite dimensional vector space is isomorphic to it's dual). -- Thomas