Is this one infinite dimensional manifold? First of all, just to give context to the question: I've been reading some articles in Physics, and those articles imply without proof that one space is one infinite dimensional manifold. One of those articles is this one and the space I'm talking about is the "space of shapes" the article talks about.
The set in question is the set $Q$ of parametrizations of smooth surfaces in $\mathbb{R}^3$, that is, the injective smooth functions $S: U\subset \mathbb{R}^2\to \mathbb{R}^3$.
Now, I don't really want to accept on faith that this is one infinite dimensional manifold, but althought I've studied analysis on manifolds it was just on finite dimension, and I really don't even know where to start to tackle this. I don't even really know which natural topology could be given to this set.
I also heard that there are several types of infinite dimensional manifold, so I don't know of which type we are treating here.
So, is it possible to give this set a natural structure of infinite dimensional manifold? If yes, how can this be done?
 A: First, a clarification. 
Let $S\subset\mathbb{R}^3$ be a two-dimensional compact submanifold. The "space of shapes" corresponding to $S$, according to the article you linked to, is the set $\mathscr{S}$ of $\mathbb{R}^3$-valued $C^1$ functions on $S$ which are injective and non-degenerate (in the meaning that the differential has full rank). In notation, $\mathscr{S}$ consists of $f:S\to \mathbb{R}^3$, injective, of class $C^1$, and $\mathrm{d}f$ has rank 2. 
To see its manifold structure, first observe that the space $C^1(S:\mathbb{R}^3)$ of continuously differentiable $\mathbb{R}^3$-valued functions on $S$ is naturally a Banach space. 
The fact that $\mathscr{S}$ is a Banach manifold follows from the fact that it is an open subset of $C^1(S:\mathbb{R}^3)$: observe simply that given $f\in \mathscr{S}$, for any $g\in C^1(S:\mathbb{R}^3)$ you can find some $\epsilon >0$ (here we used the compactness of $S$) such that for all $\delta < \epsilon$, the function $f \pm \delta g \in \mathscr{S}$. 

In the case where $S$ is non-compact one has to be a bit more careful; the space $C^1(S:\mathbb{R}^3)$ is no longer Banach, but is Frechet. But considering compactly supported perturbations you again see, more-or-less analogously to above, that the corresponding $\mathscr{S}$ is an open subset and hence a Frechet manifold. 
