Relation between line bundles over $\mathbb{P}^1$ in geometry and topology For an algebraic geometry assignment I recently had to prove that the isomorphism classes of line bundles over $\mathbb{P}_k^n$ for any field $k$ are the powers of the tautological line bundle, or in other words the Picard group is given by Pic($\mathbb{P}_k^n) \cong \mathbb{Z}$.  I was trying to consider simple examples to reconcile this with my intuition about line bundles from algebraic topology.
Taking $k = \mathbb{R}$ and $n = 1$, we have Pic($\mathbb{RP}^1) \cong \mathbb{Z}$.  Naively, I might have thought that since $\mathbb{RP}^1 \cong S^1$, there would be only the two isomorphism classes of (topological) line bundles over the circle.  I was wondering: what are the relations between line bundles in algebraic geometry and topology in this case, and why the discrepancy?  
I figure maybe the real numbers are not a very well behaved field in algebraic geometry. If $k = \mathbb{C}$ and we again take $n = 1$, do the classifications of bundles match up on $\mathbb{CP}^1 \cong S^2$ in the geometric and topological contexts?  
 A: You need to be careful what kind of $\mathrm{Pic}$ you are talking about!
The point is that if $X/\mathbb{R}$ is a finite type $\mathbb{C}$-scheme then $X(\mathbb{R})$ (resp. $X(\mathbb{C})$) is a real (resp. complex) manifold (the former being only 'locally a manifold' depending on what sort of axioms of a point-set topological nature you impose). This then allows you to define maps
$$\mathrm{Pic}_\mathrm{alg}(X)\to \mathrm{Pic}_\mathrm{smooth}(X(\mathbb{R}))\to\mathrm{Pic}_{\mathrm{cont.}}(X(\mathbb{R}))\qquad (1)$$
and maps
$$\mathrm{Pic}_\mathrm{alg}(X_\mathbb{C})\to \mathrm{Pic}_{\mathrm{hol.}}(X(\mathbb{C}))\to\mathrm{Pic}_{\mathrm{cont.}}(X(\mathbb{C}))\qquad (2)$$
But, in general, these maps needn't all be isomorphisms!
For example if $M$ is a smooth real manifold then there is actually an isomorphism
$$\mathrm{Pic}_\mathrm{smooth}(M)\xrightarrow{\approx}\mathrm{Pic}_{\mathrm{cont.}}(M)\cong H^1_\mathrm{sing}(M,\mathbb{Z}/2\mathbb{Z})$$
The latter isomorphism comes from the fact that the classifying space of continuous real line bundles is $\mathbb{RP}^\infty$ which is a $K(\mathbb{Z}/2\mathbb{Z},1)$. The former isomorphism can be thought about in two ways:


*

*The fact that we have smooth approximation for maps $M\to \mathbb{RP}^\infty$

*The fact that smooth real line bundles should be classified by $\mathcal{O}_M^\times$. There is then a SES
$$0\to \mathcal{O}_M\to \mathcal{O}_M^\times\to \underline{\mathbb{Z}/2\mathbb{Z}}\to 0$$
of sheaves and the fact that $H^1(M,\mathcal{O}_M)=H^2(M,\mathcal{O}_M)=0$ because the sheaves $\mathcal{O}_M$ are so-called 'fine' (and so are acyclic).


From this we see that we can refine $(1)$ to 
$$\mathrm{Pic}_\mathrm{alg}(X)\to \mathrm{Pic}_\mathrm{smooth}(X(\mathbb{R}))\xrightarrow{\approx}\mathrm{Pic}_{\mathrm{cont.}}(X(\mathbb{R}))$$
but this former map needn't be an isomorphism. As you pointed out, if $X=\mathbb{P}^1_\mathbb{R}$ then this fomer map is 
$$\mathrm{Pic}_\mathrm{alg.}(\mathbb{P}^1_\mathbb{R})\cong \mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}= \mathrm{Pic}_{\mathrm{smooth}}(\mathbb{RP}^1)$$
where explicitly the map takes $\mathcal{O}(n)$ to the trivial bundle if $n$ is even and the Mobius bundle if $n$ is odd! The point being that while the algebraic structures on $\mathcal{O}$ and $\mathcal{O}(2n)$ (as well as $\mathcal{O}(1)$ and $\mathcal{O}(2n+1)$) are not algebraically equivalent, they are smoothly equivalently. Try it for yourself with overlap maps $x$ and $x^3$!
In the $\mathbb{C}$-case if you assume that $X$ is, in addition, proper then you actually get an isomorphism
$$\mathrm{Pic}_{\mathrm{alg.}}(X)\xrightarrow{\approx}\mathrm{Pic}_{\mathrm{hol.}}(X(\mathbb{C}))$$
this follows from Serre's GAGA results. Moreover, we can describe the map $\mathrm{Pic}_{\mathrm{hol.}}(X(\mathbb{C}))\to \mathrm{Pic}_{\mathrm{cont.}}(X(\mathbb{C}))$ quite nicely. Namely, the complex continuous line bundles on $X(\mathbb{C})$ can be described as $H^2_\mathrm{sing}(X,\mathbb{C})$. Again, the reason for this is that the classifying space of complex line bundles is $\mathbb{CP}^\infty$ which is a $K(\mathbb{Z},2)$. The holomorphic line bundles on $X(\mathbb{C})$ are classified by $H^1(X(\mathbb{C}),\mathcal{O}_{X(\mathbb{C})}^\times)$. The connection between the two is given by the exponential sequence
$$0\to \underline{\mathbb{Z}}\to \mathcal{O}_{X(\mathbb{C})}\to\mathcal{O}_{X(\mathbb{C})}^\times\to 0$$
where the second map is $f\mapsto \exp(2\pi i f)$. Taking the long exact sequence in cohomology we get (part of the ) long exact exponential sequence
$$H^1_\mathrm{sing}(X,\mathbb{Z})\to H^1(X(\mathbb{C}),\mathcal{O}_{X(\mathbb{C})})\to H^1(X(\mathbb{C}),\mathcal{O}_X^\times)\to H^2_\mathrm{sing}(X(\mathbb{C}),\mathbb{Z})\to H^2(X(\mathbb{C}),\mathcal{O}_{X(\mathbb{C})})$$
And, in fact, the natural diagram
$$\begin{matrix}\mathrm{Pic}_{\mathrm{hol.}}(X(\mathbb{C})) & \to & \mathrm{Pic}_{\mathrm{cont.}}(X(\mathbb{C})\\ \downarrow & & \downarrow\\ H^1(X(\mathbb{C}),\mathcal{O}_{X(\mathbb{C})}^\times) & \to & H^2_\mathrm{sing}(X(\mathbb{C}),\mathbb{Z})\end{matrix}$$
commutes with the vertical maps being isomorphisms. This map is called the Chern class of an algebraic/holomorphic line bundle.
So, if $X$ is a smooth projective (geometrically) connected curve of genus $g$ we see that the map 
$$\mathrm{Pic}_{\mathrm{alg.}}(X(\mathbb{C}))\to \mathrm{Pic}_{\mathrm{cont.}}(X(\mathbb{C}))$$
is surjective with kernel a quotient of a vector space of dimension $g$. So, if $X=\mathbb{P}^1_\mathbb{R}$ the kernel is trivial and we get the desired isomorphism
$$\mathrm{Pic}_{\mathrm{alg.}}(\mathbb{P}^1_\mathbb{C})\cong \mathrm{Pic}_{\mathrm{cont.}}(X(\mathbb{C}))\cong H^2(\mathbb{CP}^1,\mathbb{Z})\cong \mathbb{Z}$$
In fact, what's generally true is that if $X/\mathbb{R}$ is a smooth projective (geometrically) connected curve then the (portion of )the long exact exponental sequence can be written
$$\begin{matrix}H^1_\mathrm{sing}(X(\mathbb{C}),\mathbb{Z}) & \to & H^1(X(\mathbb{C}),\mathcal{O}_{X(\mathbb{C})}) & \to & \mathrm{Pic}(X_\mathbb{C}) & \to & H^2(X(\mathbb{C}),\mathbb{Z}) & \to & H^2(X(\mathbb{C}),\mathcal{O}_{X(\mathbb{C})})\\ \downarrow & & \downarrow & & & &\downarrow & & \downarrow\\ \mathbb{Z}^{2g} & &\mathbb{C}^g & & &  & \mathbb{Z} & & 0\end{matrix}$$
And, if you take it on faith (this is the beginning of Hodge theory!) that $\mathbb{Z}^{2g}$ is embedded into $\mathbb{C}^g$ as a lattice, then we see that we get a short exact sequence
$$0\to \mathbb{C}^g/\mathbb{Z}^{2g}\to \mathrm{Pic}(X_\mathbb{C})\to \mathbb{Z}\to 0$$
so that $\mathrm{Pic}(X_\mathbb{C})$ looks like a disconnected complex Lie group with component group $\mathbb{Z}$ and identity component an abelian variety (i.e. a compact complex Lie group). This identity component is called the Jacobian of $X_\mathbb{C}$ and is denoted $\mathrm{Jac}(X_\mathbb{C})$. The map from 
$$\mathrm{Pic}(X_\mathbb{C})\to \mathbb{Z}=\pi_0(\mathrm{Pic}(X_\mathbb{C}))$$
is just the degree map. Of course, since $\mathbb{Z}$ is projective and discrete this sequence non-canonically splits to give you that $\mathrm{Pic}(X_\mathbb{C})\cong \mathrm{Jac}(X_\mathbb{C})\times\mathbb{Z}$.
For example, if you take $X=E$ an elliptic curve, then it turns out that $\mathrm{Jac}(E_\mathbb{C})\cong E_\mathbb{C}$!
All of this then starts the fascinating journey into the Albanese variety/Picard scheme of a smooth proper scheme!

The last thing I'll say that is that, in some sense, the algebraic bundles on $X/\mathbb{R}$ smooth projective are much more closely related to the holomorphic bundles on $X(\mathbb{C})$ than the continuous bundles on $X(\mathbb{R})$! In fact, there's the so-called 'Picard-Brauer sequence' which contains the terms
$$0\to \mathrm{Pic}(X)\to \mathrm{Pic}(X_\mathbb{C})^{\mathrm{Gal}(\mathbb{C}/\mathbb{R})}\to \mathrm{Br}(\mathbb{R})(\cong\mathbb{Z}/2\mathbb{Z})$$
From this we see that the algebraic bundles on $X$ embed into the algebraic bundles on $X_\mathbb{C}$ (which is equal to the holomorphic bundles on $X(\mathbb{C})$) and that up to a $\mathbb{Z}/2\mathbb{Z}$-term they exactly hit the Galois invariants of the algebraic bundles on $X_\mathbb{C}$. 
In the case of $X=\mathbb{P}^1_\mathbb{R}$ this sequence is not very interesting. It looks like 
$$0\to \mathbb{Z}\to\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$$
where the map $\mathbb{Z}\to\mathbb{Z}$ is an isomorphism and the map $\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}$ is trivial.
But, if instead of $\mathbb{P}^1_\mathbb{R}$ you took it's only non-trivial twist $X:=V(x^2+y^2+z^2)\subseteq\mathbb{P}^2_\mathbb{R}$ then your sequence actually looks like 
$$0\to 2\mathbb{Z}\to \mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}\to 0$$
The point being that if you take any degree $2$-point on $X$ then $\mathrm{Pic}(X)\cong \{\mathcal{O}(np):n\in\mathbb{Z}\}$. But, since $p$ is a point of degree $2$ when you base change to $\mathbb{C}$ you get that $p$ splits into two points--$q_0$ and its Galois conjugate $\sigma(q_0)$ so that $\mathcal{O}(p)$ maps to $\mathcal{O}(q_0)\otimes \mathcal{O}(\sigma(q_0))\cong \mathcal{O}(2)$. 
A: The difference between real and complex projective space is what exactly you mean by a 'line bundle.' In algebraic geometry, we work over algebraically closed fields, and there is a rough analogy between what happens over any $\bar{k}$ and what happens over $\mathbb{C}$. However, the real numbers are not algebraically closed, and smooth bundles over smooth manifolds are going to be different. This is why you are seeing a difference between what happens for the circle and the sphere - there are no holomorphic bundles at all over the circle because the circle is not a complex manifold in the first place.
Life isn't completely bad though. If you have a complex line bundle, you can show that it is determined by its first Chern class, but in fact it is determined not just up to holomorphic isomorphism, but in fact up to $C^\infty$ isomorphism. This is done explicitly in Griffiths and Harris somewhere in the chapter on divisors and line bundles.
In general, construction of a bundle is related to the combinatorics of a cover of your space. A very important related concept is that of Cech cohomology, which agrees with singular cohomology for manifolds. The Chern class can be realized as a Cech cohomology class, and so for each such choice of a class, you can cook up a bundle. Here's a few words on what happens specifically in the case of $2$-sphere.
You can define a bundle by just prescribing the isomorphisms on the intersections that become the overlap maps of your bundle. Since the $2-$sphere has a cover by two disks intersecting in a circle, the description of your bundle is the same as choosing a map from this circle into the group of units in the complex numbers, as these are the only isomorphisms of a complex line bundle. Thus complex line bundles on the $2-$sphere are deterimned by homotopy classes of maps $S^1 \to S^1$ which is just $\pi_1(S^1)$, and this is $\mathbb{Z}$, agreeing with what you got in your algebraic geometry result.
As correctly pointed out in the comments, the machinery of aglebraic geometry does work just fine over non-algebraically closed fields. However, many results are formulated and proved only for algebraically closed fields, and false otherwise. Thus, notions like line bundles, the Picard group, and anything else even remotely scheme-theoretic can be defined, but care must be taken to know which results depend on algebraic closure and which do not.
