# Space of Germs of Holomorphic Function

A bit of a general question, but here goes. Morally, what is the space of germs of a holomorphic function?

I know that a germ is simply an equivalence class of function elements, where we regard two function elements as equivalent at a point if they agree on some open neighbourhood of that point. Moreover I know the definition that the space of germs is simply the union of these equivalence classes for all functions $f$ and points $x$.

This is all a bit abstract at the moment though. I can't see how germs are useful, or how I might calculate them for a concrete function. Has anyone got any nice examples of calculations of the space of germs? And could someone explain the overarching idea behind them in Riemann Surfaces?

Many thanks.

## 1 Answer

Consider the set of all germs of holomorphic functions at a specific point $x$. If you know of stalks, then this is just $\mathcal{O}_x$. You can think of this set as a collection of all "possible functions" which are holomorphic at that point.

To be precise, you can think of it as the set of all power series which converge in a little neighborhood of $x$. So $\mathcal{O}_x$ is isomorphic to the ring $\mathbb{C}\{z-x\}$ of all convergent power series in $z-x$.

Of course, two functions define the same germ if their power series about $x$ are the same. This gives a method to calculate the germs of a function $f$ at points $x$.

Now the union of all germs of all functions is useful for instance because it allows for the construction of a maximal analytic continuation of a given holomorphic function:

For a Riemann Surface $X$, a point $x\in X$ and a function germ $f$ at $x$, you get a Riemann Surface $Y$ together with an unbranched holomorphic map $Y\rightarrow X$ and a holomorphic function $F$ on all $Y$, such that the germ of $F$ is in some natural way the same as that of $f$.

Now the space of germs allows for the construction of a "maximal $Y$". You can think of it as the largest possible domain of definition for $f$. Because there may be different continuations of a representative of $f$ in different neighborhoods, you have to consider them all and combine them to get your larger space $Y$.

The actual construction and proofs can be read, for instance, in O. Forster's "Lectures on Riemann Surfaces".

• Thanks - I hadn't thought in terms of power series, so that's really useful! I'll look up the construction that you suggested. – Edward Hughes May 29 '12 at 15:57
• I've been looking at the specific example $f(z)=\sqrt{1+\sqrt{z}}$ and trying to find its germs at $z=1$. I'm not sure how to find the relevant power series here though. Could you possibly let me know what your reasoning would be in this case to point me in the right direction? Very many thanks! – Edward Hughes May 29 '12 at 20:07
• You could try to calculate some derivatives of $f$, come up with an idea for the $n$-th derivative and prove that by induction. Or you could use existing power series for $\sqrt{z}$ and $\sqrt{1+z}$ and plug them into each other. This is both rather ugly and would take a fair amount of work, I suppose. But it's definitely doable. Don't expect a large radius of convergence though, because $\sqrt{z}$ is not holomorphic on $\mathbb{R}^-_0$. – Gregor Botero May 30 '12 at 10:31
• Of course you could also pose this as a new question, maybe someone has a more elegant solution ready. I currently don't. – Gregor Botero May 30 '12 at 10:32
• I may indeed post it as a new question. I tried the approach you suggested in your comment but it was really messy. I agree that it should work, but surely there's a more elegant way...! – Edward Hughes May 30 '12 at 10:35