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 $S^{\mathbb{N}}$.