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Perhaps the following algebraic formulation could work::Let $X$ and $Y$ be two compact topological space and $A=C(X)$ and $B=C(Y)$ be the $C^{*}$ algebra of complex valued continuoes functions on $X$ and $Y$ resp.Choose (and fix) two dense subalgebra $A'$ of $A$ and $B'$ of $B$. every continuous function on $f:X \rightarrow Y$ define a natural morphism $f^{*}$from $B$ to $A$ we can say that $f$ is $A'-B'$ differentiable if $f^{*}(B) \subset A$. This could be considered as a natural generalization of standard differentability for maps between manifold.
@Igor perhaps this is an alternative proof for parallelizability of $TS^{2}$:According to your statements it is sufficient to prove $TS^{2} \oplus TS^{2} $ is a trivial bundle on $S^{2}. But this a consequence of the fact "the complexification of tangent bundle of spheres is trivial", this is said in Sawans paper "f.g projective module and vector bundles" But what is a proof for this statement of Swan's paper?
There is no any obstruction for compactification to obtain a complex manifold again? There is no any obstruction for holomorphic extension? I would appreciate if you explain more clear. Where you used the polynomial ness of $\sigma$. Could you please give me a reference for some background?
According to the original formulation of my question we search for complex manifold M (Not orbifold)So do you think that the answer of my question is "No"? I mean that is it possible to prove that there is no a complex manifold with those properies?
Thank you.puting $a_{n}=1$ we simplify the question as follows: we identify all monic polynomial of degree n with $\mathbb{C}^{n}$. We search for covering space $M \rightarrow \mathbb{C}^{n}$. It seems that I donot underestand something. Does your covering map cover wholle C^n? could you please more explain?
