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maybe this question makes no sense and I just cannot accept the fact that dual the line bundle is different from the respective line bundle itself. Since it looks like that manifolds are more intuitive than algebraic varieties, let's consider smooth complex compact manifolds.

Anyway, picking the most ordinary non-trivial line bundle over a complex manifold , namely the tautological line bundle, it's know that $0 = H^0(\mathbb{P}^n, \mathscr{O}(-k)) \not\cong H^0(\mathbb{P^n}, \mathscr{O}(k)) =$"homogeneous polynomials of degree $k$", because if both have non-trivial global sections then both must be trivial (because the manifold is compact). Another way of seeing this is by computing the cocyles of the respective bundles.

Is there an intuitive way of seeing why the above one and the dual of the above one are so different (by drawing or seeing where the glueing fails when trying to create a global section)?

Where does it fail when I try creating an isomorphism between some a line bundle and it's dual by picking fiberwise isomorphisms (as vector spaces of complex dimension $1$)?

If we just consider the smooth structure (without the holomorphic one),what happens to the global sections of both bundles in the above example? Where the fiberwise isomorphism fails to be a vector bundle (of rank $2$) isomorphism?

Thanks in advance.

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    $\begingroup$ The fiberwise isomorphism fails because there's no canonical way to identify $V$ with $V^*$. $\endgroup$
    – user27126
    Commented Apr 8, 2014 at 3:45
  • $\begingroup$ @Sanchez Actually there is Riesz isomorphism for spaces Hilbert spaces, so it must fail in smoothness or even continuity (don't know how to see this though). $\endgroup$
    – user40276
    Commented Apr 8, 2014 at 3:48
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    $\begingroup$ A vector bundle in the smooth category has a Riemannian metric, and that allows you to write down explicitly an isomorphism to the dual bundle. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Apr 8, 2014 at 4:48
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    $\begingroup$ @user40276 Are you familiar with the understanding of line bundles in terms of divisors? If you are, then if you think about $\mathcal{O}(n)$ as $nH_0$ (where $H_0$ is the hyperplane $V(x_0)$) and $\mathcal{O}(-n)$ as $-nH_0$, then the global sections of $\mathcal{O}(n)$ are rational functions which must only have one possible pole, at $H_0$, and there the pole can be no worse than of order $n$. Global sections of $\mathcal{O}(-n)$ says that the global sections must have NO poles, and a zero of order at least $n$ at $H_0$. But, you can't have no poles if you have zeros, if you're non-zero. $\endgroup$
    – Alex Youcis
    Commented Apr 8, 2014 at 4:54
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    $\begingroup$ @user40276 So, morally if you think about bundles as being specifications of functions with "pole/zero" data, then it's clear why $\mathcal{O}(-k)$ and $\mathcal{O}(k)$ are different. The fact that the specification is $\text{div}(f)\geqslant -D$, for elements of $f\in K(X)$, says that if $-D$ is effective, then $D$ itself is specifying untenable global data. But, if $D$ is effective, there is no a priori unrealistic conditions being imposed on global sections. This also gives you an intuition about what is the difference between complex and real bundles. On real manifolds there aren't $\endgroup$
    – Alex Youcis
    Commented Apr 8, 2014 at 4:56

1 Answer 1

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Since I find some comments to the question a bit ambiguous, let me emphasize:

For $k\gt0$ the line bundles $\mathcal O(k)$ and $\mathcal O(-k)$ are not isomorphic in the $C^0$-category and thus a fortiori not isomorphic in the $C^\infty$-category.

In other words, the non-isomorphism you are asking about has nothing to do with algebraic geometry and can already been read on the underlying topological manifold and underlying topological bundles.
The most convincing proof is through the use of the first Chern classes:$$ c_1(\mathcal O(k))=k\neq c_1(\mathcal O(-k))=-k\in H^2(\mathbb P^n(\mathbb C),\mathbb C)=\mathbb C$$ Beware that if a real line bundle has a riemannian metric, then it is isomorphic to its dual but if a complex line bundle $E$ has a hermitian metric (which is the case for $E=\mathcal O(k)$) you can only say that its dual $E^\ast$ is isomorphic to its conjugate bundle $\overline E$, but not to $E$ itself.

Bibliography
By far the best resource for these results is Milnor-Stasheff's justly celebrated Characteristic Classes, especially §§ 13, 14.

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    $\begingroup$ Georges, while your answer is certainly 100% correct content wise, I disagree that the original question has absolute nothing to do with algebraic geometry. While it is certainly true that the Chern classes tell the bundles apart, it is hard to understand intuitively what the difference in Chern classes (beyond "they have different orientation twisting") tells us. It's a kind of technical fact which is, as hard, if not harder, to reconcile than the global sections argument. I do think that understanding why the bundles are specifying different pole data, and so shouldn't be the same, is more $\endgroup$
    – Alex Youcis
    Commented Apr 8, 2014 at 7:31
  • $\begingroup$ intuitive, despite using ideas (algebraic/complex geometry) which are, technically, unneeded. Regardless, this is a good answer, clarifying some of the ambiguous comments (including mine!) about the differences between the real and complex cases. +1! $\endgroup$
    – Alex Youcis
    Commented Apr 8, 2014 at 7:32
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    $\begingroup$ Dear @Alex, yes, you are right: the statement that this non-isomorphism has nothing to do with algebraic geometry is a bit of an overstatement. I just wanted to emphasize that the non-isomorphism already exists at the topological level and is not due to some strange definition/convention invented by weird algebraic geometers :-) Anyway, thanks for your attention and your interesting comment. $\endgroup$
    – Georges Elencwajg
    Commented Apr 8, 2014 at 7:52
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    $\begingroup$ They are a wily bunch. Thank heavens we have the non-algebraic geometers like you to safeguard us from their devissage and their pro-etale sites. :) $\endgroup$
    – Alex Youcis
    Commented Apr 8, 2014 at 7:54
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    $\begingroup$ Dear @Alex: yes, be very afraid of their wily schemes... $\endgroup$
    – Georges Elencwajg
    Commented Apr 8, 2014 at 9:12

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