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Let $X$ be a smooth manifold. What is the appropriate topology on $C^\infty(X)$ such that a linear functional $\lambda$ on $C^\infty(X)$ is continuous iff it can be represented as a limit of the form $$\lambda f = \lim_{\ell\to\infty} \sum_{k=1}^{n_\ell}\lambda_{k\ell}f(x_{k\ell})$$ with finite $n_\ell$, suitable numbers $\lambda_{k\ell}$, and suitable points $x_{k\ell}\in X$?

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  • $\begingroup$ As far as I understand, the usual topology of the locally convex projective limit $C^\infty(X)=\underset{U}{\lim} C^\infty(U)$, $U\cong {\mathbb R}^n$ does not satify you? $\endgroup$ – Sergei Akbarov Jul 3 '18 at 13:15
  • $\begingroup$ @SergeiAkbarov: Maybe it would satisfy me if the topology were expressed in a less abstract way and if I'd see the argument why my statement is then true. $\endgroup$ – Arnold Neumaier Jul 3 '18 at 13:40
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Arnold, if $l$ is not obliged to belong to $\mathbb N$, i.e. the index set for $l$ can be arbitrary (for $l\in\mathbb N$ this seems to be also true, but this requires a verification), then the usual (the weakest locally convex) topology of $C^\infty(M)$ generated by the projections to the spaces $C^\infty(U)$, where $U$ is diffeomorphic to ${\mathbb R}^n$ (or to an open subset in ${\mathbb R}^n$, the topology on such $C^\infty(U)$ is described in the Kirillov-Gvishiani book), satisfies your requirement, since the linear combinations of delta-functionals are dense in $(C^\infty(M))'$ (this in its turn follows from the fact that delta-functionals separate elements of $C^\infty(M)$).

In this topology a net of functions $f_\nu$ tends to a function $f$ in $C^\infty(M)$ if and only if for each subset $U\subseteq M$, diffeomorphic to $\mathbb R^n$, the restrictions $f_\nu\big|_U$ tend to $f\big|_U$ uniformly on each compact set $K\subseteq U$ by each partial derivative $\partial^m$ generated by the local chart for $U$: $$ \forall m\in{\mathbb N}^n\qquad \sup_{x\in K}\bigg|\partial^m\Big(f_\nu\big|_U-f\big|_U\Big)(x)\bigg|\underset{\nu\to\infty}{\longrightarrow}0 $$ (I assume everywhere that $\mathbb N$ contains zero).

There is another "abstract way" to describe this topology that I find "more visual", I told about it here:

  1. For each function $f\in {\mathcal C}^\infty(M)$ let us define its support as the closure of the set of the points where $f$ does not vanish: $$ \text{supp}f=\overline{\{x\in M:\ f(x)\ne 0\}}. $$ An equivalent definition: $\text{supp}f$ is the set of the points in $M$ where $f$ has non-zero germs: $$ \text{supp}f=\{x\in M:\ f\not\equiv 0\ (\text{mod}\ x)\}. $$

  2. Let us define differential operators (see e.g. S.Helgason's book) on $M$ as linear mappings $D:{\mathcal C}^\infty(M)\to {\mathcal C}^\infty(M)$ which do not extend the support of functions: $$ \text{supp}Df\subseteq \text{supp}f,\quad f\in{\mathcal C}^\infty(M). $$ Equivalently, $D$ is local, i.e. the value of $Df$ in a point $x\in M$ depends only on the germ of $f$ in $x$: $$ \forall f,g\in{\mathcal C}^\infty(M)\quad \forall x\in M\qquad f\equiv g\ (\text{mod}\ x)\quad\Longrightarrow\quad Df(x)=Dg(x). $$

  3. Then we say that a sequence of functions $f_n$ converges to a function $f$ in ${\mathcal C}^\infty(M)$ $$ f_n\overset{{\mathcal C}^\infty(M)}{\underset{n\to\infty}{\longrightarrow}}f $$ if and only if for each differential operator $D:{\mathcal C}^\infty(M)\to {\mathcal C}^\infty(M)$ the sequence of functions $Df_n$ converges to $Df$ in the space ${\mathcal C}(M)$ of continuous functions with the usual topology of uniform convergence on compact sets in $M$: $$ Df_n\overset{{\mathcal C}(M)}{\underset{n\to\infty}{\longrightarrow}}Df $$

The definition of the differential operator in this construction belongs (as far as I know) to Jaak Peetre. We can replace it by the one that Jet Nestruev uses for abstract rings, and the topology on $C^\infty(M)$ will be the same. Similarly, we can replace $D$ by arbitrary compositions of vector fields on $M$ $$ D=X_1\circ...\circ X_p, \quad p\in {\mathbb N}, $$ where vector fields are defined as derivations of the ring $C^\infty(M)$, i.e. linear operators $X:C^\infty(M)\to C^\infty(M)$ with the Leibnitz property $$ X(f\cdot g)=X(f)\cdot g+f\cdot X(g). $$

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  • $\begingroup$ I like the final definition under 3. How to see from that definition that the linear combinations of delta-functionals are dense in (C∞(X))′ without first proving equivalence to the first definition? $\endgroup$ – Arnold Neumaier Jul 3 '18 at 14:35
  • $\begingroup$ Arnold, I doubt that this can be seen without proving the equivalence. In my head the explanation is the following. First, we prove that $C^\infty(M)$ is a Fréchet space for $\sigma$-compact $M$. Second, we prove that for arbitrary $M$ the space $C^\infty(M)$ is isomorphic to the direct product of a family of spaces $C^\infty(M_i)$ with $\sigma$-compact $M_i$. Together this means that $C^\infty(M)$ is always stereotype. After that we conclude that the bipolar theorem (see Schaefer IV, 1.5) holds for the pair $\Big(C^\infty(M),C^\infty(M)^\star\Big)$. $\endgroup$ – Sergei Akbarov Jul 3 '18 at 15:20
  • $\begingroup$ As a corollary, if a subspace $P\subseteq C^\infty(M)^\star$ has zero polar in $C^\infty(M)$, then its bipolar $P^{\circ\circ}$, which is the $C^\infty(M)$-weak closure of $P$, and thus the closure of $P$ in $C^\infty(M)^\star$, coincides with $C^\infty(M)^\star$. In other words, $P$ must be dense in $C^\infty(M)^\star$. Finally, we take $P$ as the linear span of delta-functions, and we obtain what we need. $\endgroup$ – Sergei Akbarov Jul 3 '18 at 15:32
  • $\begingroup$ The only reference that comes to my mind is Kirillov-Gvishiani springer.com/gb/book/9780387906386 But they consider only the case when $M=U$ is an open subset in ${\mathbb R}^n$. However, from their Theorem 26 in Chapter III (where they prove that $C^\infty(U)$ which they denote by ${\mathcal E}(U)$ is a Fréchet space) it follows what you need, you should just follow the scheme that I described. $\endgroup$ – Sergei Akbarov Jul 3 '18 at 17:54
  • $\begingroup$ I understand. I think there must be a textbook where all the details are explained accurately. We should ask people to give a reference. $\endgroup$ – Sergei Akbarov Jul 3 '18 at 22:30

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