Is the Zariski topology equipped with Eisenstein's metric an analytic submanifold?

36 Views Asked by At

Using $M=(C(\mathbb{R}),T_z)$ with the norm $(x,y) \to \log(\partial_x+\partial_y)$, we can easily define a derivative using distributions. I was wondering: Does this make $M$ an analytic manifold sitting inside $\mathbb{R}^\mathbb{R}$?

Edit: The derivative I refer to maps an $f \in C(\mathbb{R})$ to the spectrum of it's Eisenstein matrix: $f \to Sp(M(f))$.

I guess my question is: Does this space have a countable basis with respect to that derivative? This would then make it analytic, correct?