Smooth curves in a Frechet space

Question

Is the space $C^{\infty}([0,1];C^{\infty}(S^1))$ equal with the space $C^{\infty}([0,1]\times S^1)$ ? I am interested in characterizing the smooth curves in the space $C^{\infty}(S^1)$ where $S^1$ is the circle.

If my first question has a negative answer is the space $C^{\infty}([0,1];C^{\infty}(S^1))$ a tame space in the sense of R.S. Hamilton ? Two possible gradings on it are:

$$\|u\|_n=\sup_{i=\overline{0,n}}\sup_{t\in[0,1]}\|u^{(i)}(t,x)\|_{C^{n}(S^1)}$$ and: $$\|u\|_n=\sup_{i=\overline{0,n}}\sup_{t\in[0,1]}\|u^{(i)}(t,x)\|_{C^{n-i}(S^1)}$$ the derivative $u^{(i)}(t,x)$ being taken after $t.$ I can not figure out if with one of these gradings is a tame space or not.

Answer 1

Yes, the spaces are isomorphic as Fréchet spaces. This is often called the exponential law and holds for every compact manifolds $M$ and $N$, $$C^\infty(M \times N) = C^\infty(M, C^\infty(N))$$ and as a corollary you obtain that both space are tame in the sense of Hamilton.

A proof of this exponential map can be found in the book "The Convenient Setting of Global Analysis" by Andreas Kriegl and Peter W. Michor. Or see Glöckner's lecture notes around Proposition 4.5.4 for further comments about the exponential law.