3

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?

JSCB
  • 13,698
  • 15
  • 67
  • 125
  • 1
    It is a complete normed space, so I guess that it is trivially a manifold. – Siminore Nov 30 '15 at 15:55
  • 2
    What definition do you take for a manifold? – mathcounterexamples.net Nov 30 '15 at 15:56
  • 3
    manifold over what space? Points have no neighborhoods that are finite-dimensional. You could think of if as a manifold, but not over $\mathbb R^n$ for any finite $n$. – Mirko Nov 30 '15 at 16:06
  • 1
    As pointed out by @Siminore, 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 dimenstional vector space is isomorphic to it's dual). – Thomas Nov 30 '15 at 16:32
  • @Thomas Sure. However I imagine that you take for atlas the one having a single chart equal to $(C[0,1],Id)$ which doesn't provide a manifold having much different properties than $C[0,1]$ itself! – mathcounterexamples.net Nov 30 '15 at 16:38
  • By manifold, I mean a topological manifold that is hausdorff and second countable.The chart map from an open neighbourhood of a points in $C[0,1]$ to a open n dimensional euclidean space. Is $C[0,1]$ a manifold? What is the dimension? – JSCB Dec 01 '15 at 00:59
  • I strongly believe that it is NOT a topological manifold, but how to prove this? – JSCB Dec 01 '15 at 01:01

1 Answers1

2

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

Community
  • 1