MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

When dealing with moduli spaces of, say connections or metrics, I am using the notions of Frechet spaces/manifolds/groups. I have become familiar with Banach manifolds (I think), but Frechet manifolds less so. In my studies, we want to apply the Inverse and Implicit Function Theorems, which work on Banach spaces, but apparently fail for Frechet spaces (and as a result we pass to Sobolev completions).

Why does it fail (intuitively), or why should I expect it to fail? What is the hinge that allows it to work for Banach but not Frechet? Do I have to do a lot more to get an analogous result in the land of Frechet? This should also help me see the differences between a Banach and Frechet manifold, because right now it seems subtle.

share|cite|improve this question
I don't know if this helps, but one side effect of this is that submersions of Fréchet manifolds are defined not in terms of surjective maps of tangent spaces, but in terms of surjective - in fact split surjective - maps of charts. I image the fact not all onto maps of Fréchet spaces are (continuously!) split has a key part in this. – David Roberts Jun 29 '13 at 3:48
@DavidRoberts Not all onto maps of Banach spaces are continuously split either. The quotient map $\ell^\infty \to \ell^\infty/c_0$ is not split, for example. – Loop Space Jun 29 '13 at 20:56
Ok strike that last comment then. (I've revealed my almost total ignorance of functional analysis, I hang my head in shame) – David Roberts Jun 29 '13 at 23:11
up vote 23 down vote accepted

The usual proof of the inverse function theorem in the setting of Banach spaces uses the Banach fixed point theorem. We cannot make sense of the Banach fixed point theorem in a Fréchet space, since a Fréchet space is merely metrizable: there is no preferred choice of metric. I do not know exactly how crucial the Banach fixed point theorem actually is for proving the inverse function theorem in Banach spaces. Hopefully someone can come along and explain whether this is really the "hinge" that makes the inverse function theorem work.

Since you say IFT "apparently" fails for Fréchet spaces, I'm guessing you haven't seen a counterexample. Let $C^\infty[-1,1]$ denote the Fréchet space of smooth real-valued functions on $[-1,1]$. Consider the smooth map $$P: C^\infty[-1,1] \longrightarrow C^\infty[-1,1],$$ $$f \mapsto f - xff'.$$ Now $$DP(f)g = g - xgf' - xfg',$$ so $DP(0) = I$. Since $P(0) = 0$, if the inverse function theorem holds then $P$ should be invertible in some neighborhood of $f = 0$, with $P^{-1}(0) = 0$. To see that this is not true, consider the family of functions $$g_n(x) = \frac{1}{n} + \frac{x^n}{n!}.$$ Now $g_n \longrightarrow 0$ in $C^\infty[-1,1]$, but we claim that $g_n$ is not in the image of $P$ for any $n$. To see this, write the power series expansion $$f(x) = a_0 + a_1x + a_2x^2 + a_3 x^3 + \cdots.$$ Then we have $$Pf(x) = a_0 + (1 - a_0)a_1 x + (a_2 - a_1^2 - 2a_0a_2)x^2 + (a_3 - 3a_1a_2 - 3a_0a_3)x^3 + \cdots.$$ Suppose $Pf(x) = g_n(x)$. Then $a_0 = \tfrac{1}{n}$, and an inductive argument shows that $a_1 = a_2 = \cdots = a_{n-1} = 0$, giving $$Pf(x) = \frac{1}{n} + (1 - na_0)a_nx^n + \cdots.$$ The order $n$ term is then zero, contradicting the fact that $Pf(x) = g_n(x)$. Hence $g_n$ does not lie in the image of $P$ for any $n$. Since $P(0) = 0$, $DP(0) = I$ is invertible, and $g_n \longrightarrow 0$, this provides a counterexample to the inverse function theorem.

The generalization of the inverse function theorem to Fréchet spaces is the Nash-Moser theorem. You need some much better conditions on the map $f$ you want to invert in this case: First, $f$ must be invertible in a neighborhood, not just at a single point. Second, the Fréchet spaces involved and $f$ must satisfy the technical condition of "tameness." Finally, the inverses of $f$ in the neighborhood of interest must also be tame. There are counterexamples to show that each of these extra conditions is necessary.

For an overview of the Nash-Moser theorem, see Hamilton's The inverse function theorem of Nash and Moser.

share|cite|improve this answer
Could you say how this example change for $C^1[-1,1]$? That is a Banach space after all. – Igor Khavkine Jun 29 '13 at 10:23
Then the target space is $C^0$, and the differential is no longer invertible (it is inclusion). – BS. Jun 29 '13 at 11:12
@ Henry T. Horton: To me, your argument only could show that the image under $P$ of the subset of analytic functions in $C^\infty([{-1},1])$ does not contain a neighbourhood of zero. – TaQ Jun 29 '13 at 14:11

The simplest counterexample: $C^\infty(\mathbb R)$ with compact $C^\infty$-topology. $F(f)= e^f$ has $DF(f)g = e^f.g$ which is invertible everywhere. But the image of $F$ is not open, it consists of all positive functions, whereas any nonempty open set of $C^\infty$ functions is controlled only on a compact subset of $\mathbb R$ and thus contains functions which are negative far off.

The intuition for an IFT to hold is: You have to generate enough Banach space situation around your equation by using special circumstances like a priori estimates for elliptic PDE's and smoothing operators that you can use Banach's fixed point theorem.

share|cite|improve this answer
It's worth pointing out that the same thing can be said about $C^0(\mathbb{R})$ with the compact-open topology. The Fréchet-ness of this space comes from the non-compactness of the domain $\mathbb{R}$ rather than from considering infinite differentiability. – Igor Khavkine Jun 29 '13 at 10:25
To my knowledge, the example of Peter Michor is essentially already in a classical article of James Eels, possibly "A setting for global analysis", Bulletin AMS 72 (5): 751–807. 1966. – TaQ Jun 29 '13 at 14:21

One point that seems important to me is the very definition of differential. To write a first-order Taylor formula $$\varphi(v+h) = \varphi(v)+D\varphi(v)\cdot h + o(h)$$ you have to define what $o(h)$ means, that is to be able to measure vectors uniformly over the space. In a Fréchet space, you cannot do that or rather, you can do that in too many different ways, so you define $D\varphi(v)$ by restricting to one-dimensional ones. But clearly for many differential calculus theorems, including local inversion, you would need uniformity in the $o(h)$.

share|cite|improve this answer
Originally I thought I understood this, but can you briefly clarify "you can measure vectors uniformly over the space in too many different ways"? – Chris Gerig Jun 30 '13 at 3:08
Assume your topology is derived from some metric $d$. Then $d^{1/2}$ is still a metric, with the same balls as before, in particular it induces the same topology. But now the quantitative meaning of $o(h)$ is not the same as before. This kind of manipulation of the metric is possible because we did not ask for positive homogeneity, contrary to Banach spaces. – Benoît Kloeckner Jun 30 '13 at 10:25

A natural (and perhaps surprising at first sight) source of counter-examples to IFT are the Fréchet-Lie groups of diffeomorphisms of smooth closed manifolds, and the exponential map from vector fields to diffeomorphisms (both smooth).

Even in the simplest case of the circle, the image of the exponential is not a neighbourhood of identity (whereas its differential at $0$ is -- by definition -- the identity).

The reason for this is "dynamical" : if a member $\phi_t=\exp(tX)$ of a one parameter subgroup of $\mathrm{Diff}(S^1)$ has no fixed point, it is conjugated to a rotation by a diffeomorphism sending the -- necessarily -- nowhere vanishing $X$ to a "constant" vector field.

If such a $\phi_t$ has one $n$-periodic point, then all points are $n$-periodic. But any neighbourhood of the identity contains (for large enough $n$) a fixed point free diffeomorphism whith isolated $n$-periodic points.

share|cite|improve this answer
Sorry to be obtuse, but I don't quite follow the contradiction that is implicit in your last two paragraphs. If $\phi$ is $n$-periodic and "close" to the identity, why can't $\phi=\exp(tX)$ for some $X$ and $t$? – Igor Khavkine Jun 29 '13 at 12:53
I have edited the last paragraph to (hopefully) clarify it. – BS. Jun 29 '13 at 13:18
I see now, thanks! The issue is with the n-periodic points being isolated. For any diffeomorphism of the form $\exp(tX)$, if there is no fixed point, then either all points are $n$-periodic for some $n>0$ or none are. While your last paragraph indicates that there exist diffeomorphisms with no fixed points but with isolated $n$-periodic points, and they exist in any neighborhood of the identity. – Igor Khavkine Jun 29 '13 at 15:18

Chris Gerig: "... we want to apply the Inverse and Implicit Function Theorems, which work on Banach spaces, ... Why does it fail (intuitively), or why should I expect it to fail? What is the hinge that allows it to work for Banach but not Frechet? Do I have to do a lot more to get an analogous result in the land of Frechet?"

Basically, it does not fail for Fréchet spaces. The basic fixed point argument indeed can be transplanted to the setting of mappings between Fréchet spaces as is done here. However, the thus obtained inverse or implicit function theorems seem to be of no use as is indicated here. The basic reason for things going wrong in this kind of try of a generalization is that a corresponding differentiability compatible with the fixed point argument becomes so strong that there are no practical maps satisfying such a strong differentiability requirement. Behind the phrase "Banach space situation" of Peter Michor lies the essential thing for getting useful inverse and implicit function theorems guaranteeing suitable regularity properties of the (local) inverse or implicit function. The need for creating the so called Nash−Moser theorems lies in the phenomenon of "loss of derivatives", this term being introduced by Jürgen Moser, if I remember correctly. Basically, one has here a setting where in a vector space $X$ one has a finite decreasing sequence $B_0,B_1,\ldots B_k$ of "disks" such that the corresponding generated subspaces $E_{\,i}=X_{B_i}$ become Banach when equipped with the norms making these disks the (closed) unit balls. The "loss of derivatives" means that one has a function $f$ mapping $E_{\,i+1}\to E_{\,i}$. In the fixed point iteration one then "loses derivatives" when from $x_n\in E_{\,i}$ one goes to $x_{\,n+1}\in E_{\,i-i_0}$ and to compensate this one uses the smoothing operators which takes one back to the "higher" space. In a practical situation one may have for example Sobolev spaces $E_{\,i}=H^{s_i}(\Omega)$ with $s_i<s_{i+1}$ and a nonlinear partial differential operator mapping $E_{\,i+1}\to E_{\,i}$.

share|cite|improve this answer

In a Banach setting the proof of the Implicit Function Theorem (IFT) is based on Newton iteration. Each iteration step requires the solution of a linearized equation. When applying this, for example, to a partial differential equation $F(u)=f$, problems arise, even if the base manifold is compact, if the linearizations $DF(u)$ (assumed invertible) are not elliptic. The reason is "loss of derivatives" in the iterative corrections $DF(u)^{-1}F(u)$. Application by $F$ decreases (Sobolev or Hölder) differentiability by the order of $F$, and this loss is regained by $DF(u)^{-1}$ only if $DF(u)$ is elliptic. If $DF(u)$ is not elliptic, then the techniques of Nash and Moser are used to obtain an IFT. Here the iteration scheme is modified by mollification: In each iteration step $u$ is replaced by $S_\theta u$ for a suitable mollification parameter $\theta$, $S_\theta u\to u$ in each Hölder space as $\theta\to\infty$. It is essential that norm estimates such as $$\|S_\theta u\|_\alpha\lesssim\theta^{\alpha-\beta}\|u\|_\beta,\quad\alpha\geq\beta,$$ allow control of higher order derivatives of the mollifications of $u$ by lower order derivatives of $u$. (It would be of no help to assume $u\in C^\infty$ from the beginning because this gives no control of higher order derivatives.) The method of Nash was simplified and applied by Hörmander; see chapter 3 of the book of Alinhac and Gerard (GSM 82) for an exposition and references. Hamilton abstracted the estimates as tameness properties and gave a generalization of the IFT to Frechet spaces.

So, my intuition about IFT in a Frechet setting comes from the special, but significant, situation which I just described: The non-ellipticity of the derivative/linearization of the differential operator and the resulting loss in derivatives during iteration are the cause of trouble. A remedy if is approximation by suitable mollications.

share|cite|improve this answer

As Henry T. Horton already pointed out, there exists a Fréchet version of the inverse function theorem. Therefore, the easiest way to get an intuition about what can go wrong with the classical Banach-inverse function theorem in the Fréchet context is to consider the stronger requirements of the Nash-Moser-Theorem:

  • The spaces under consideration have to be 'tame': This is a technical requirement saying that your spaces are not to far away from Banach spaces. (Actually, one can proof the Nash-Moser theorem relatively easy for the space of exponentially falling sequences in some Banach space and then one tries to extend these results to a bigger class of Fréchet spaces - these are the tame ones.) As far as I know, there exists result about weakening this condition, see e.g. Ekeland, An inverse function theorem in Fréchet spaces.

  • The function has to be invertible in a whole neighborhood and not only in one point: The reason for this requirement is that the invertible operators are no longer open in the space of all linear operators.

  • You need some 'tame' estimates on the function: This can be more or less compared to the $C^\infty$- regularization after you solved a problem in the Sobolev-spaces and thus corresponds to the a-priori estimates made there.

Furthermore, if you are interested in moduli spaces in the language of Fréchet manifolds you should have a look at

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.