In mathematics, a Banach manifold is a manifold modeled on Banach spaces. Thus it is a topological space in which each point has a neighbourhood homeomorphic to an open set in a Banach space (a more involved and formal definition is given below). Banach manifolds are one possibility of extending manifolds to infinite dimensions.

A further generalisation is to Fréchet manifolds, replacing Banach spaces by Fréchet spaces. On the other hand, a Hilbert manifold is a special case of a Banach manifold in which the manifold is locally modeled on Hilbert spaces.

## Definition

Let $X$ be a set. An atlas of class $C^{r},$ $r\geq 0,$ on $X$ is a collection of pairs (called charts) $\left(U_{i},\varphi _{i}\right),$ $i\in I,$ such that

1. each $U_{i)$ is a subset of $X$ and the union of the $U_{i)$ is the whole of $X$ ;
2. each $\varphi _{i)$ is a bijection from $U_{i)$ onto an open subset $\varphi _{i}\left(U_{i}\right)$ of some Banach space $E_{i},$ and for any indices $i{\text{ and ))j,$ $\varphi _{i}\left(U_{i}\cap U_{j}\right)$ is open in $E_{i};$ 3. the crossover map
$\varphi _{j}\circ \varphi _{i}^{-1}:\varphi _{i}\left(U_{i}\cap U_{j}\right)\to \varphi _{j}\left(U_{i}\cap U_{j}\right)$ is an $r$ -times continuously differentiable function for every $i,j\in I;$ that is, the $r$ th Fréchet derivative
$\mathrm {d} ^{r}\left(\varphi _{j}\circ \varphi _{i}^{-1}\right):\varphi _{i}\left(U_{i}\cap U_{j}\right)\to \mathrm {Lin} \left(E_{i}^{r};E_{j}\right)$ exists and is a continuous function with respect to the $E_{i)$ -norm topology on subsets of $E_{i)$ and the operator norm topology on $\operatorname {Lin} \left(E_{i}^{r};E_{j}\right).$ One can then show that there is a unique topology on $X$ such that each $U_{i)$ is open and each $\varphi _{i)$ is a homeomorphism. Very often, this topological space is assumed to be a Hausdorff space, but this is not necessary from the point of view of the formal definition.

If all the Banach spaces $E_{i)$ are equal to the same space $E,$ the atlas is called an $E$ -atlas. However, it is not a priori necessary that the Banach spaces $E_{i)$ be the same space, or even isomorphic as topological vector spaces. However, if two charts $\left(U_{i},\varphi _{i}\right)$ and $\left(U_{j},\varphi _{j}\right)$ are such that $U_{i)$ and $U_{j)$ have a non-empty intersection, a quick examination of the derivative of the crossover map

$\varphi _{j}\circ \varphi _{i}^{-1}:\varphi _{i}\left(U_{i}\cap U_{j}\right)\to \varphi _{j}\left(U_{i}\cap U_{j}\right)$ shows that $E_{i)$ and $E_{j)$ must indeed be isomorphic as topological vector spaces. Furthermore, the set of points $x\in X$ for which there is a chart $\left(U_{i},\varphi _{i}\right)$ with $x$ in $U_{i)$ and $E_{i)$ isomorphic to a given Banach space $E$ is both open and closed. Hence, one can without loss of generality assume that, on each connected component of $X,$ the atlas is an $E$ -atlas for some fixed $E.$ A new chart $(U,\varphi )$ is called compatible with a given atlas $\left\{\left(U_{i},\varphi _{i}\right):i\in I\right\)$ if the crossover map

$\varphi _{i}\circ \varphi ^{-1}:\varphi \left(U\cap U_{i}\right)\to \varphi _{i}\left(U\cap U_{i}\right)$ is an $r$ -times continuously differentiable function for every $i\in I.$ Two atlases are called compatible if every chart in one is compatible with the other atlas. Compatibility defines an equivalence relation on the class of all possible atlases on $X.$ A $C^{r)$ -manifold structure on $X$ is then defined to be a choice of equivalence class of atlases on $X$ of class $C^{r}.$ If all the Banach spaces $E_{i)$ are isomorphic as topological vector spaces (which is guaranteed to be the case if $X$ is connected), then an equivalent atlas can be found for which they are all equal to some Banach space $E.$ $X$ is then called an $E$ -manifold, or one says that $X$ is modeled on $E.$ ## Examples

• If $(X,\|\,\cdot \,\|)$ is a Banach space, then $X$ is a Banach manifold with an atlas containing a single, globally-defined chart (the identity map).
• Similarly, if $U$ is an open subset of some Banach space then $U$ is a Banach manifold. (See the classification theorem below.)

## Classification up to homeomorphism

It is by no means true that a finite-dimensional manifold of dimension $n$ is globally homeomorphic to $\mathbb {R} ^{n},$ or even an open subset of $\mathbb {R} ^{n}.$ However, in an infinite-dimensional setting, it is possible to classify "well-behaved" Banach manifolds up to homeomorphism quite nicely. A 1969 theorem of David Henderson states that every infinite-dimensional, separable, metric Banach manifold $X$ can be embedded as an open subset of the infinite-dimensional, separable Hilbert space, $H$ (up to linear isomorphism, there is only one such space, usually identified with $\ell ^{2)$ ). In fact, Henderson's result is stronger: the same conclusion holds for any metric manifold modeled on a separable infinite-dimensional Fréchet space.

The embedding homeomorphism can be used as a global chart for $X.$ Thus, in the infinite-dimensional, separable, metric case, the "only" Banach manifolds are the open subsets of Hilbert space.