# Are Banach Manifolds intrinsically interesting?

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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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) –  Mariano Suárez-Alvarez Mar 9 '12 at 4:02
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. –  Jonny Evans Mar 9 '12 at 7:09
@suarez-alvarez: got you, thanks. –  Mozibur Ullah Dec 4 '12 at 21:05
@Evans: you mean infinite-dimensional vector spaces? –  Mozibur Ullah Dec 4 '12 at 21:06

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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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.
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. –  Liviu Nicolaescu Mar 11 '12 at 16:53
Got it. $\ \ \$ –  Allen Knutson Mar 12 '12 at 1:29