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The following definitions are taken from Marsden et al. Manifolds, Tensor Analysis, and Applications.

Definition 1: Let $S$ be a set. A chart on $S$ is a bijection $\varphi$ from a subset $U$ of $S$ to an open subset of a Banach space. We sometimes denote $\varphi$ by $(U,\varphi)$, to indicate the domain $U$ of $\varphi$. A $C^k$ atlas on $S$ is a family of charts $\mathcal A = \{ (U_i, \varphi_i) : i \in I \}$ such that:

(i) $S = \bigcup \{ U_i : i \in I \}$

(ii) Any two charts in $\mathcal A$ are compatible in the sense that the overlap maps between members of $\mathcal A$ are $C^k$ diffeomorphisms: for two charts $(U_i, \varphi_i)$ and $(U_j, \varphi_j)$ with $U_i \cap U_j \ne \emptyset$, we form the overlap map: $\varphi_{ji} = \varphi_j \circ \varphi_i^{-1}{|\varphi_i(U_i \cap U_j)}$, where $\varphi_i^{-1}{|\varphi_i(U_i \cap U_j)}$ means the restriction of $\varphi_i^{-1}$ to the set $\varphi_i(U_i \cap U_j)$. We require that $\varphi_i(U_i \cap U_j)$ is open and that $\varphi_{ij}$ be a $C^k$ diffeomorphism.

Definition 2: Two $C^k$ atlases $\mathcal A_1$ and $\mathcal A_2$ are equivalent if $\mathcal A_1 \cup \mathcal A_2$ is a $C^k$ atlas. A $C^k$ differentiable structure $\mathcal D$ on $S$ is an equivalence class of atlases on $S$. The union of the atlases in $\mathcal D$, $\mathcal A_D = \bigcup \{ \mathcal A : \mathcal A \in \mathcal D \}$ is the maximal atlas of $\mathcal D$, and a chart $(U,\varphi) \in \mathcal A_{\mathcal D}$ is an admisssible local chart. [...] A differentiable manifold $M$ is a pair $(S, \mathcal D)$ [...]

They then define what a submanifold is:

A submanifold of a manifold $M$ is a subset $B \subseteq M$ with the property that for each $b \in B$ there is an admissible chart $(U,\varphi)$ in $M$ with $b \in U$ which has the submanifold property, namely, $$ \varphi : U \to E \times F, \mbox{ and } \varphi(U \cap B) = \varphi(U) \cap (E \times \{ 0 \}). $$

By the definition of an atlas it is implied that all charts map into the same Banach space (but I guess this must be mentioned for $U_i \cap U_j = \emptyset$ in the definition; but maybe I have understood something wrongly). So, if a manifold admits a submanifold, then we can decompose this Banach space as $E \times F$ (or as a direct sum, which would be isomorphic, i.e. splits)?

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    $\begingroup$ Your interpretation seems correct - the prototypical example for a $k$-dimensional submanifold of an $n$-dimensional manifold ($k \le n < \infty$) is $\mathbb R^n = \mathbb R^k \times \mathbb R^{n-k}$. Thus it should be true that $E \times F$ is the Banach space upon which the ambient manifold is modelled; i.e. a submanifold should be modelled on a split subspace of the Banach space. $\endgroup$ – Anthony Carapetis Mar 21 '16 at 0:00
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First, one can always decompose a well-behaved Banach space (of dimension greater than $1$) into a product simply by choosing a basis. So "admitting a submanifold" does not tell us much about non-exotic examples. (I'm using wishy-washy language because I haven't thought about whether one can have an uncountable-dimensional Banach space which might not admit any countable basis at all, and what it would mean to have a manifold modeled on such a vector space.)

Second, if you're not already familiar with finite-dimensional manifolds, I would suggest replacing every instance of "Banach space" in these definitions with "$\Bbb{R}^n$" and thinking about that. This fixes dimension and disallows infinite-dimensional manifolds. It's a useful way to get intuition for why people have made these definitions before jumping to the more general definitions. (As a rule, more general definitions are created to encompass specific examples which arose after the original definition was created, but if you're not familiar with those examples, then the more general version of the concept can seem pretty impenetrable when you first encounter it.)

Third, your intuition may be leading you toward the notion of a tubular neighborhood. If you have a connected submanifold, then there is neighborhood of that submanifold which (locally) decomposes as a product of manifolds. (Exercise: Why did I emphasize "connected"? What happens if we don't insist on connectedness?)

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    $\begingroup$ Tubular neighborhoods only locally decompose as products (i.e. they're only locally trivial). Only for submanifolds with trivial normal bundle do they globally decompose as a product. $\endgroup$ – Pedro Mar 21 '16 at 0:52
  • $\begingroup$ @Pedro Yep yep, thanks for clarifying that. Edited into my answer. $\endgroup$ – Neal Mar 21 '16 at 10:14

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