Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $f\colon X\to Y$ be a smooth map between smooth manifolds. Then the pull-back operation $f^*\colon C^\infty(Y)\to C^\infty(X)$ is a linear continuous operator when $C^\infty$ is equipped with the usual topology of of uniform convergence on compact subsets of all partial derivatives.

It tuns out that sometimes $f^*$ can be extended to larger spaces of generalized functions. More precisely, fix a closed $\mathbb{R}_{>0}$-invariant subset $\Lambda\subset T^*Y\backslash 0$ of the cotangent bundle with removed zero section. Let $C^{-\infty}_\Lambda(Y)$ denote the space of generalized functions with the wave front set contained in $\Lambda$. This space $C^{-\infty}_\Lambda(Y)$ is equipped with some standard locally convex linear topology (see e.g. Ch. 6 in the book "Geometric Asymptotics" by Guillemin and Sternberg).

Let us assume that the map $f$ is transversal to $\Lambda$ in the sense that for any $x\in X$ if $(f(x),\eta)\in \Lambda$ then $(df_x^*)(\eta)\ne 0$. Then one defines a ("natural") linear map

$$f^*\colon C^{-\infty}_\Lambda(Y)\to C^{-\infty}(X)\,\,\,\,\,\,\,\,\,\, (1)$$

extending the usual pull-back on smooth functions (see the above mentioned book. Another very good reference is Hormander's ”The analysis of linear partial differential operators, I”; see especially Theorem 8.2.4.)

The point is that in the above literature the map (1) is proven to be sequentially continuous, namely it maps convergent sequences to convergent ones.

QUESTION. Is the pull-back map (1) topologically continuous?

The difference between usual topological continuity (e.g. continuity in the usual sense of maps of topological spaces) and sequential continuity seems to me to be quite subtle.

Edit: Definition of topology on $C^{-\infty}_\Lambda(X)$. Covering $X$ by open charts and using partition of unity we may assume that $X= \mathbb{R}^n$. For any $N\in\mathbb{N}, \phi \in C^\infty_c(\mathbb{R}^n)$, and any closed $\mathbb{R}_{>0}$-invariant subset $V\subset \mathbb{R}^n$ such that $$\Lambda\cap (supp(\phi)\times V)=\emptyset$$ let us define the semi-norm on $C^{-\infty}_\Lambda(\mathbb{R}^n)$ by $$||u||_{\phi,N,V}=sup_{\xi\in V}|\xi|^N|\widehat{\phi u}(\xi)|,$$ where $\hat F$ denotes the Fourier transform of the function $F$. Then one equips $C^{-\infty}_\Lambda(X)$ with the weakest locally convex topology which is stronger than the weak topology on $C^{-\infty}(X)$ and such that all semi-norms $\{||\cdot||_{\phi,N,V}\}$ are continuous.

share|improve this question
Have you looked in Chap. 1 of Duistermaat's "Fourier Integral Operators"? There he proves that the pullback is continuous in the appropriate topologies. –  Liviu Nicolaescu Oct 2 '13 at 14:45
If both spaces are locally convex Hausdorff and $C^{-\infty}_\Lambda(Y)$ is bornological, then continuity follows from sequential one. This is e.g. in Exercise 3.7.7 on page 226 in Horváth's book Topological Vector Spaces and Distributions. –  TaQ Oct 2 '13 at 20:11
@Liviu Nicolaescu: Many thanks for the reference. I just had a look at Duistermaat's book. The author claims explicitly that the pull-back is topologically continuous (Proposition 1.3.3)! However when I tried to follow the argument, I realized that the proof is very concise, and the author just claims the continuity without proving it. Again, the sequential continuity easily follows from his argument, but not the topological one. In fact, he uses topology which is defined slightly differetly than Hormander's one, but I strongly suspect that the two topologies are equivalent. –  MKO Oct 3 '13 at 10:17
@TaQ: I was told that $C_\Lambda^{-\infty}(Y)$ is not bornological unfortunately. –  MKO Oct 3 '13 at 10:19
@ semyon alesker. Could you please edit your question so that the definition of the space $C^{-\infty}_\Lambda(Y)$ becomes explicit. Quite often PDE authors define the distribution spaces to be $E_\sigma'$ when $E$ is the corresponding "test" function space. Instead, Horváth defines them to be $E_\beta'$, and these spaces are bornological. In Horváth's book, there are general result concerning continuity of the transpose ${}^{\rm t}u:F'\to E'$ of a linear map $u:E\to F$ when the dual spaces are given different kinds of topologies. I cannot say more about your problem (cont.) –  TaQ Oct 4 '13 at 9:02

2 Answers 2

up vote 1 down vote accepted

Recently I found a counterexample to a somewhat stronger version of the question I originally asked, but which I actually needed. Namely for a smooth map $f\colon X\to Y$ which is transversal to $\Lambda\subset T^*Y\backslash 0$ (see e.g. Hormander's book for the definitions) the map $f^*\colon C^{-\infty}_{\Lambda}(Y)\to C^{-\infty}_{f^*\Lambda}(X)$ may not be topologically continuous (notice the subscript $f^*\Lambda$ in the target).

The counterexample is quite simple. Let $f\colon \mathbb{R}^2\to \mathbb{R}$ be the projection to the first coordinate. Let $\Lambda=\mathbb{R}\times (\mathbb{R}\backslash 0)$. Then $C^{-\infty}_{\Lambda}(\mathbb{R})=C^{-\infty}(\mathbb{R})$ is the usual Schwartz space with the usual weak topology. Furthermore $f^*\Lambda=\{(x,y;\xi,0)\}$. Then the map $f^*\colon C^{-\infty}_\Lambda(\mathbb{R})\to C^{-\infty}_{f^*\Lambda}(\mathbb{R}^2)$ is not topologically continuous.

Strictly speaking I do not yet have a counterexample to my original question, but this one already suffices my purposes.

share|improve this answer
@ semyon alesker: I do not understand your counterexample to your modified question. In your (original) question the set $\Lambda$ is required to be closed in $T^*Y$ minus the zero section. In your counterexample to your modified question you obviously should have $f_*:(x,y;\xi,\eta)\mapsto(x,\xi)$ under which the inverse image of $\mathbb R\times(\mathbb R\backslash 0)$ is $\Lambda_1=\{\,(x,y;\xi,\eta):x,y,\xi,\eta\in\mathbb R\text{ and }\xi\not=0\,\}$ which is not closed (w.r.t. the relative topology) in the complement of the zero section $\{\,(x,y;0,0):x,y,\in\mathbb R\,\}$ . –  TaQ Oct 12 '13 at 15:25
So how did you get the $f^*\Lambda=\{(x,y;\xi,0)\}$ ? –  TaQ Oct 12 '13 at 15:27
The problem was that I did not define $f^*\Lambda$. By definition, $f^*\Lambda:=\{(x,\xi)\in T^*X\backslash 0|\, \exists \eta\in \Lambda\cap T^*_{f(x)}Y s.t. \xi=(df_x)^*\eta\}$, where $(df_x)^*$ denotes the dual of the differential of $f$ at the point $x$. –  MKO Oct 12 '13 at 16:05

I deleted my previous answer since "googling" the book "Geometric Asymptotics" by Guillemin and Sternberg revealed that my provisions there were not in accordance with the presentation in the book. Chapter VI begins on page 305, and from page 306 it becomes plausible (but not clear) that by "$C^{-\infty}$" the authors refer to the dual of $C_0^{\,\infty}$ and not to that of $C^{\,\infty}$ which I assumed earlier. Unfortunately, I couldn't see page 333, but Proposition 3.7 on page 335 claims that $f^*v$ for $f:X\to Y$ can be resonably defined as a certain limit of a sequence when $v$ is a distribution in $C_Z^{-\infty}(Y)$ . Here $Z$ is the $\Lambda$ of OP. I didn't find any claim of this extended pull-back operation being even sequentially continuous $C_Z^{-\infty}(Y)\to C^{-\infty}(X)$ .

However, having thought more of the problem, it has began to seem to me that continuity might be able to be proven in the case where $X$ and $Y$ have the same dimension and $f$ is "almost" a diffeomorphism. When the dimensions are not equal, it seems that one only can prove that for $f^*:C^{\,\infty}(Y)\to C^{\,\infty}(X)$ its restriction to every bounded set in $C^{-\infty}(Y)$ is continuous when the spaces $C^{\,\infty}$ are equipped with the initial topologies by the embeddings $C^{\,\infty}(Y)\to C_Z^{-\infty}(Y)$ and $C^{\,\infty}(X)\to C^{-\infty}(X)$ . This allows the extended pull-back to be reasonably defined, and possibly also sequential continuity (but not continuity) of $f^*:C_Z^{-\infty}(Y)\to C^{-\infty}(X)$ to be established.

share|improve this answer
I think that if $f\colon X\to Y$ is a diffeomorphism then $f^*\colon C^{-\infty}_{\Lambda}(Y)\to C^{-\infty}_{f^*\Lambda}(X)$ is topologically continuous. In particular $f^*\colon C^{-\infty}_{\Lambda}(Y)\to C^{-\infty}(X)$. –  MKO Oct 12 '13 at 6:18
By "almost a diffeomorphism" I referred to the situation where $f:X\to Y$ is such that $X$ can be covered by finitely many open sets $U_0,U_1,\ldots\,U_k$ such that $f\,|\,U_i$ is a diffeomorphism $U_i\to f[U_i]$ for $i=1,\ldots\,k$ and the "singularities" of $f$ are contained in a compact set $K\subseteq U_0$ . –  TaQ Oct 12 '13 at 9:59

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.