Bases for spaces of smooth functions

Question

Let $S$ denote the space of rapidly decreasing sequences, which means sequences $a=(a_k)_{k=1}^\infty$ such that the numbers $p_d(a)=\sup\{k^d|a_k| : 1\leq k<\infty\}$ are finite for all $d\in\mathbb{N}$. We give this space the topology generated by the family of seminorms $p_d$.

Now let $M$ be a compact smooth closed manifold, and consider the space $C^\infty(M)$. For any differential operator $L:C^\infty(M)\to C^\infty(M)$ (of any nonnegative order) we have a seminorm $p_L(f)=\|Lf\|_\infty$, and we give $C^\infty(M)$ the topology determined by this family of seminorms.

By a basis for $C^\infty(M)$ I mean a sequence of functions $f_k$ such that the rule $a\mapsto\sum_ka_kf_k$ gives an isomorphism $S\to C^\infty(M)$ of topological vector spaces.

I think it is known that $C^\infty(M)$ always has a basis.

• Is this true, and if so, what is a good reference? I think I have seen it in the literature, but I cannot find it right now.
• Is there a reasonably effective criterion to check whether a given sequence is a basis?
• Suppose that $M$ is a subspace of $\mathbb{R}^m$ defined by polynomial equations (and is still compact and smooth). Is there an effective way to find a basis consisting of polynomial functions? In particular, can I just use a Gröbner basis with respect to degree-lexicographic order?

Here is a little background, partly taken from some notes of Dietmar Vogt:

http://www2.math.uni-wuppertal.de/~vogt/vorlesungen/fs.pdf

When $M=S^1$ we can just take $f_{2k+1}(\cos(\theta),\sin(\theta))=\cos(k\theta)$ and $f_{2k}(\cos(\theta),\sin(\theta))=\sin(k\theta)$. This gives an isomorphism $S\to C^\infty(S^1)$, and of course we can precompose this with any of the many automorphisms of $S$, so $C^\infty(S^1)$ has many different bases. If $(f_j)$ is a basis for $C^\infty(M)$ and $(g_k)$ is a basis for $C^\infty(N)$ then the functions $h_{jk}(x,y)=f_j(x)g_k(y)$, enumerated in a suitable order, will give a basis for $C^\infty(M\times N)$.

If $V$ is any nuclear Frechet space, then a theorem of Komura and Komura shows that $V$ is isomorphic to a subspace of $S^{\mathbb{N}}$. I do not understand all the issues here, but it seems like there is not too much difference between $S$, $S^{\mathbb{N}}$ and subspaces of $S^{\mathbb{N}}$. It is certainly known that $C^\infty(M)$ is always a Frechet space. Vogt's notes show that when $U$ is a nonempty open subset of $\mathbb{R}^m$, the space $C^\infty(U)$ is isomorphic to $S^{\mathbb{N}}$. If we choose $U$ to be a tubular neighbourhood of an embedded copy of $M$, then $M$ will be a retract of $U$ and so $C^\infty(M)$ will be isomorphic to a summand in $C^\infty(U)$, and thus to a summand in $S^{\mathbb{N}}$.

Answer 1

The paper

• MR0688001 Reviewed Vogt, Dietmar Sequence space representations of spaces of test functions and distributions. Functional analysis, holomorphy, and approximation theory (Rio de Janeiro, 1979), pp. 405–443, Lecture Notes in Pure and Appl. Math., 83, Dekker, New York, 1983. (Reviewer: M. Valdivia)

shows that for a compact manifold $M$, the Frechet space $C^\infty(M)$ is always linearly isomorphic to the space $\mathcal s$ of rapidly deceasing sequences.

Added later:

If you consider the $L^2$ orthonormal basis $\phi_k$ for the Laplacian of a Riemannian metric on $M$ as suggested by Liviu Nicolaescu, then any $f\in C^\infty(M)$ is of the form $f=\sum_k f_k\phi_k$, and $f_k\in \ell^2$ initially. But $1+\Delta$ (geometric $\Delta$ here) is an isomorphism between the Sobolev spaces $H^k(M)$ and $H^{k-2}(M)$ for each $k$. By Weyl's formula the eigenvalues $\lambda_k$ of $\Delta$ satisfy $\lambda_k \sim C k^{2/\dim(M)}$ for $k\to \infty$; see page 155 of the book of Chavel, `Eigenvalues in Riemannian geometry'. Since $\bigcap_k H^k(M)=C^\infty(M)$ on a compact manifold, we see that $$(1+\Delta)^m f = \sum_k f_k(1+ \lambda_k)^{m}\phi_k$$ with coefficients again in $\ell^2$, for each $m$. Thus the coefficients $f_k(1+Ck^{2/\dim(M)})^m\in \ell^2$ for each $m$, and the $f_k$ are rapidly decreasing. Moreover, any rapidly decreasing sequence of coefficients gives a function in $C^\infty(M)$. This proves again that $C^\infty(M)\cong \mathcal s$, even with the basis of eigenfunctions for any Laplacian.