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Algebraic isomorphism classes of vector bundles of rank $r$ on $\mathbb P^1_\mathbb C$ are in bijective correspondence with $r$-tuples of integers $a=(a_1,\dots,a_r)$ such that $a_1\geq a_2\geq \dots a_r$. Two distinct such "classes" $a,b$ define the same topological vector bundle if and only if the corresponding vector bundles have the same degree (i.e. $\sum a_i=\sum b_i$).

This fact says that the algebraic classification produces "more" isomorphism classes than the topological one (in other words, if $F\cong_{alg}E$, then $F\cong_{top}E$).

But in A. Langer's notes on moduli of sheaves I read the phrase:

"Later, vector bundles on $\mathbb P^n_\mathbb C$ were classified for $n \leq 6$, but it is not known which of these vector bundles can be realized as algebraic vector bundles."

This suggests that there should be an inclusion

$$\{\textrm{Algebraic isomorphism classes}\}\subset\{\textrm{Topological isomorphism classes}\},$$ which is contradicting the fact that the algebraic classification is "finer".

Question. What am I misunderstanding? Which one is actually finer, and in which sense, between the two classifications?

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    $\begingroup$ Of course algebraic isomorphism is finer; I read the quote from Langer as saying simply that there are topological isomorphism classes of vector bundles on $\mathbf{P}^n$ which do not contain any algebraic bundles. In other words map from the set of alg. iso. classes to top. iso. classes is not surjective. $\endgroup$
    – user64687
    Oct 8, 2013 at 21:39
  • $\begingroup$ Thank you very much, I see your point. $\endgroup$
    – Brenin
    Oct 9, 2013 at 22:19

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As suggested by @Asal Beag Dubh (thanks again!), the point is that in the case of vector bundles on $\mathbb P^1$ there is a surjective map

$$\{\textrm{Algebraic iso classes}\}\twoheadrightarrow \{\textrm{Topological iso classes}\},$$ as the right hand side "equals" $\{\textrm{Algebraic iso classes}\}/$degree.

General situation (on $\mathbb P^n$, say): we know that there can be several holomorphic structures on the same topological bundle; in other words, $[E]_{alg}\subseteq [E]_{top}$, but no one knows whether $[E]_{alg}$ is empty for a fixed topological bundle $E$. In other words, the map

$$\{\textrm{Algebraic iso classes}\}\to \{\textrm{Topological iso classes}\}$$ might not be surjective, i.e. there might exists a topological bundle $E$ such that no algebraic vector bundle is isomorphic to $E$.

Still, I would be glad to see:

  • an example of this behavior on a variety that is not a projective space;
  • an example of a variety $X$ such that the topological and the algebraic classification of vector bundles on $X$ coincide, in the sense that none is finer than the other.
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