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In the introduction to 'A convenient setting for Global Analysis', Michor & Kriegl make this claim: "The study of Banach manifolds per se is not very interesting, since they turnout to be open subsets of the modeling space for many modeling spaces."

But finite-dimensional manifolds are found to be interesting even though they can be embedded in some Euclidean space (of larger dimension). (Actually this seems to me, to make the above claim intuitively plausible, so that claim should be no more than we should expect).

But they do go on to say that "Banach manifolds are not suitable for many questions of Global Analysis, as ... a Banach Lie group acting effectively on a finite dimensional smooth manifold it must be finite dimensional itself.", which does seem a rather strong limitation.

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    $\begingroup$ There is a big difference between the finite-dimentional case and the other: in the latter, manifolds are open subsets of the modeling space, while in the former manifolds rarely are open sets of the modeling space (they embed in a euclidean space of a generally much larger dimension, and even more rarely as open sets thereof) $\endgroup$ Commented Mar 9, 2012 at 4:02
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    $\begingroup$ Usually the infinite dimensional objects one encounters in geometry are Fréchet manifolds (e.g. diffeomorphism groups or spaces of maps with the smooth topology) and these are really the objects of interest. Banach manifolds are a useful tool for studying infinite dimensions and actually proving anything because there you have the implicit function theorem. In particular if you have an elliptic problem, e.g. holomorphic curves, it's much easier to think of the solution space inside a Banach manifold of maps and then use elliptic regularity it prove it's really inside the smooth locus. $\endgroup$ Commented Mar 9, 2012 at 7:09
  • $\begingroup$ @Evans: you mean infinite-dimensional vector spaces? $\endgroup$ Commented Dec 4, 2012 at 21:06
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    $\begingroup$ A finite-dimensional Banach manifold is called a Finsler manifold and is of interest. $\endgroup$
    – Deane Yang
    Commented Jul 1, 2014 at 14:46
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    $\begingroup$ The unit sphere in a Banach space is a Banach manifold. Even if one can realize it as an open subset of the Banach space, it seems more reasonable to think of the sphere as a Banach manifold. Note that Uhlenbeck decisively uses this point of view in her proof that, for example, Laplace eigenvalues are generically simple. $\endgroup$ Commented Jul 15, 2017 at 15:20

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In his remarkable thesis Douady proved that, given a compact complex analytic space $X$, the set $H(X)$ of analytic subspaces of $X$ has itself a natural structure of analytic space .
If $X=\mathbb P^n(\mathbb C)$ for example, then $H(X)$ is the Hilbert scheme $ Hilb(\mathbb P^n(\mathbb C))$.
However the problem is much more difficult for non algebraic $X$.
Douady solved it by massive use of Banach analytic manifolds, the most important of them being the grassmannian of complemented closed subspaces of a Banach space.

The thesis starts with the candid statement of its aim: "Le but de ce travail est de munir son auteur du grade de docteur-ès-sciences mathématiques et l'ensemble H(X) des sous espaces analytiques compacts de X d'une structure d'espace analytique", that is to endow its author with the title of doctor in mathematics and the the set H(X) of compact analytic subspaces of X with the structure of analytic space.

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I am of two minds on this topic. It is much easier to work on Banach manifolds because the implicit function theorem on such spaces has a simple formulation. On the other hand, as the examples of gauge theory or the theory of pseudo-holomorphic curves show, in these contexts one works not with one Banach manifold, but with several, determined by stronger and stronger Sobolev norms. One important part of the game is to conclude that objects with a priori weaker Sobolev regularity are in fact smooth. This feels very much like we are implicitly working on a Frechet manifold.

One draw back of Banach spaces is that they do not have many smooth functions on them, and the notion of real analycity on such spaces is problematic. Let's take the example of Seiberg-Witten equations. These are quadratic equations in its variables, so intuitively they ought to be real analytic, though I do not know how to formulate this rigorously in a Sobolev context.

Why do I care about real analycity? In the real analytic context one can formulate an intersection theory involving not necessarily smooth objects. For example, the point $0\in\mathbb{R}$ is a solution of the quadratic equation $x^2=0$. It is a degenerate zero, and from the point of view of intersection theory it has multiplicity $0$. My hope is that this real analytic point of view would allow one to deal with mildly degenerate solutions of the the Seiberg-Witten equations, and assign multiplicities to such solutions.

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  • $\begingroup$ Multiplicity 0? Not 2? $\endgroup$ Commented Mar 11, 2012 at 14:23
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    $\begingroup$ In the real case the multiplicity is zero. To see this look at the equation $x^2-\varepsilon=0$. For $\varepsilon>0$ it has two solutions, one counter with multiplicity $1$ and the other with multiplicity $-1$ and thus the intersection number of the graph of $x^2-\varepsilon$ with the $x$-axis is zero. If $\varepsilon <0$, then the equation has no solution, and the above intersection number is also zero. $\endgroup$ Commented Mar 11, 2012 at 16:53
  • $\begingroup$ Got it. $\ \ \ $ $\endgroup$ Commented Mar 12, 2012 at 1:29

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