# Non-isomorphism of topological line bundles on a Riemann surface, from first principles only

Although this question is in the same vein as my previous query,

it is nonetheless a different question.

Consider a compact, connected, smooth Riemann surface $X$.

Consider two line bundles now: $M=X\times\mathbb C$, the trivial line bundle, and a line bundle $N$ defined in the following way. Pick a point $x\in X$ and an open neighbourhood $U$ of $x$. Equip $U$ with a (continuous) coordinate function $z$ that vanishes only at $x$. The cover $\{U,V\}$, where $V=X\backslash\{x\}$, and a transition function $g_{UV}:(y,v)\mapsto(y,z(y)v)$, where $y$ is in $U\cap V$, are enough data to define $N$.

There are several ways to see that $M$ and $N$ are non-isomorphic as topological line bundles. For one, $M$ admits a global non-vanishing section while $N$ does not. In terms of characteristic classes, the first Chern classes $M$ and $N$ are different. Since $X$ is a Riemann surface, $H^2(X;\mathbb Z)$ is integral, and so we can rephrase this by saying that the degree of $M$ is $0$ and the degree of $N$ is $1$.

But how can I see directly, from first principles by writing down bundle maps, that $M$ and $N$ cannot be isomorphic? These are fairly simple bundles after all.

Idea:

Let's first refine the cover from $M$ so that it is the same as that of $N$. In other words, $M$ is the line bundle given by the cover $\{U,V\}$ and the transition $t_{UV}=(\mbox{id},1)$.

If $h:M\to N$ is an invertible bundle map, then it acts on $t_{UV}$ as $h_Vt_{UV}h_U^{-1}$, where $h_U$ is a map from $U\times\mathbb C$ to itself (acting as identity on $U$, and linearly and non-vanishing on $\mathbb C$), restricted to $U\cap V$. Likewise for $h_V$.

If I take $h_U$ to be $(y,v)\mapsto(y,z(y)v)$ and $h_V$ to be $(\mbox{id},1)$, then these transform $t_{UV}$ into the $g_{UV}$ originally specified for $N$.

The problem must be with the vanishing of $z$ at $y=x$. However, is it possible to use a function $Z$ that agrees with $z$ on $U\backslash\{x\}$ but which doesn't vanish at $x$, without assuming that they are analytic?

• There is no way you could possibly see this because "compact connected topological space" doesn't guarantee that $H^2(X, \mathbb{Z})$ is nonzero. This is the group of isomorphism classes of complex line bundles on $X$, so if it's zero, then every line bundle on $X$ is trivial. If you're interested in compact Riemann surfaces (and compactness is essential as well, and your construction does not use it) you should stick to that case. – Qiaochu Yuan Jan 2 '15 at 21:22
• Thanks, Qiaochu. I've edited the question so that there is no mention now of arbitrary topological spaces. Only Riemann surfaces are mentioned now. Do you have any thoughts on the final part of the question? – MathsByTheSea Jan 2 '15 at 21:27
• Is there something about unique extensions of continuous functions to closures of their domains that I can use here? – MathsByTheSea Jan 3 '15 at 1:30