Isomorphisms (and non-isomorphisms) of holomorphic degree $1$ line bundles on $\mathbb{CP}^1$ and elliptic curves

I have two highly-coupled questions concerning holomorphic line bundles, and so I will go ahead and ask them together. The first concerns line bundles on $\mathbb{CP}^1$ and the other concerns line bundles on a complex torus (elliptic curve).

On a Riemann surface $X$, I can always define a so-called "point bundle", to use the terminology of Gunning's book on vector bundles (Princeton notes series). I can always pick a point $p$ and define a holomorphic line bundle $N_p$ of degree $1$, whose nonzero holomorphic sections vanish to order $1$ at $p$ and are nonvanishing elsewhere.

1. On $X=\mathbb{CP}^1$, there is only one such line bundle up to isomorphism, and we call this $\mathcal O(1)$, the hyperplane bundle. I can prove this in the standard way, using the long sequence in cohomology associated to the exponential sequence, i.e. $H^1(\mathcal O^*)\cong H^2(\mathbb Z)\cong\mathbb Z$ on the projective line. (The first isomorphism is the degree map and the second comes from the fact that the underlying topological space is compact and real $2$-dimensional.)

But is there a way to see more directly that if $N_p$ and $N_q$ are two point bundles on $\mathbb{CP}^1$, then they must be isomorphic, even when $p$ and $q$ are distinct? I thought about rotating the underlying $2$-sphere along the equatorial circle connecting $p$ and $q$, but does this rotation of the sphere lift to an isomorphism of bundles (and is it holomorphic)? Is there a better way to see the isomorphism?

2. On an elliptic curve $X$, the isomorphism classes of holomorphic line bundles of a fixed degree are parametrised by another elliptic curve, the Jacobian. If we take the Jacobian of degree $1$ line bundles, and we take a point $p\in X$, then we get a map from $X$ to the Jacobian by sending $p$ to the class $[N_p]$ belonging to the corresponding point bundle. This map is an isomorphism. Why? In other words, if $p$ and $q$ are distinct points in $X$, then why are $N_p$ and $N_q$ non-isomorphic as holomorphic line bundles? (Again, I don't want to use the cohomology of the exponential sequence. It gives me that the Jacobian for a fixed degree is a torus of complex dimension $1$, the quotient of a copy of $\mathbb C$ by a lattice $\mathbb Z^2$, and that's great. But I want to see directly why the two line bundles cannot admit an isomorphism between them.)

• I do realize these are probably trivial questions for the experts. I don't wish to waste anyone's time. However, I would appreciate as direct an answer as possible (rather than a "go work it out" or go "look it up" answer --- I have been both working it out and looking it up, and now I need help). Thank you. – MathsByTheSea Dec 30 '14 at 22:25

Now for an elliptic curve, suppose you had an isomorphism between these bundles. Then $N_p \otimes N_q^{-1}$ would admit a nonzero section. This is a function with at most one pole of order 1 and such a function does not exist on a genus 1 curve (say by the Hurwitz formula). Same for any other higher genus curve.
• Hunter --- thanks for this. Why am I allowed to use a meromorphic function for the isomorphism between $N_p$ and $N_q$ on $\mathbb{CP}^1$? – MathsByTheSea Dec 30 '14 at 22:52
• My understanding is that two isomorphic line bundles' transition functions differ by $\delta h$, where $h=(h_0,h_1)$ is a $0$-cochain for the sheaf of non-vanishing holomorphic functions adapted a standard $2$-set covering of the sphere. When we apply the boundary operator $\delta$, we get $(\delta h)_{0,1}=h_1h_0^{-1}$. This is holomorphic on the intersection, but globally meromorphic. Is that what is happening here? (I am taking the open set $U_0$ to be centred at $p$ and the other to be centred at $q$, and the intersection to contain neither of these points.) – MathsByTheSea Dec 30 '14 at 23:33
• Yes, when I said $(h_0,h_1)$ is a cochain for the sheaf of non-vanishing holomorphic functions, I thought that implied they are holomorphic on all of $U_0$ and $U_1$, respectively. – MathsByTheSea Dec 30 '14 at 23:47
• I think I've got it, anyway: if $U_0$ is centred at $p$ and excludes $q$, then I take a holomorphic function $h_0$ on an open set $V_0$ that properly contains both $U_0$ and $q$, such that $h_0$ vanishes at $q$ and is holomorphic non-vanishing on $U_0$. Then we do something similar regarding $h_1$ (it vanishes at $p$ but is holomorphic non-vanishing on $U_1$). The isomorphism in the direction $N_p\to N_q$ is $h_0h_1^{-1}$. – MathsByTheSea Dec 31 '14 at 0:09