I'm thinking about a problem in Ravi Vakil's algebraic geometry notes http://math.stanford.edu/~vakil/216blog/ (Exercise 3.2B) and I'm having trouble with the second part of the exercise as my understanding of complex analysis is quite rudimentary.

If to every open set $U\subset \mathbb{C}$ we associate the ring of holomorphic functions which admit a holomorphic square root, this defines a presheaf on $\mathbb{C}$ (I can see this).

Apparently however this is not a sheaf -- supposedly it fails to satisfy the gluing property.

To prove this, we need to take an open cover $\{U_i\}$ of some open $U\subseteq \mathbb{C}$, a holomorphic function $f_i:U_i\to \mathbb{C}$ for each $i$ which has a holomorphic square root (i.e. there exists a holomorphic function $g_i$ for each $i$ such that $f_i(z)=g_i(z)^2$ for all $z\in U_i$), which all agree nicely on the intersections of the various open sets, but such that it is impossible to find a holomorphic function $f:U\to \mathbb{C}$ with a holomorphic function which restricts to $f_i$ on each $U_i$.

Any help would be great -- my problem is in showing that such a function cannot exist.

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    $\begingroup$ As a start: Take small enough patches of a non-simply connected region. Look at the Wikipedia pages on analytic continuation and the monodromy theorem to see what is going on. $\endgroup$ – t.b. May 11 '11 at 9:09
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    $\begingroup$ To be a bit more specific: Take an annulus around the unit circle. Start with $f(z) = z$ in a small neighborhood of $1$ and continue the standard square root analytically along the unit circle, once going clockwise towards $-1$ and once counterclockwise. Can these square roots "match up" at $-1$? $\endgroup$ – t.b. May 11 '11 at 9:17
  • $\begingroup$ As a last remark: The last comment is of course not quite enough to show what you want, but it should give the idea. $\endgroup$ – t.b. May 11 '11 at 9:29
  • $\begingroup$ But how do we know there are NO functions that work? Aren't you just giving a particular example? $\endgroup$ – Clinton Boys May 11 '11 at 9:30
  • $\begingroup$ How did you see the existence of a holomorphic square root is equivalent to the existence of a holomorphic logarithm? $\endgroup$ – Clinton Boys May 11 '11 at 9:46

Since I messed up in the comments, I'm giving an argument as an answer.

First, I'm claiming that $f(z) = z$ doesn't have a square root in the annular region $U = \{ 1 - \varepsilon \lt |z| \lt 1 + \varepsilon\}$. Recall that for a holomorphic function $g$ without zeroes on the unit circle we have $\int_{|z| = 1} \frac{g'(z)}{g(z)} \,dz \in 2 \pi i \mathbb{Z}$. Now we know that $\int_{|z| = 1} \frac{f'(z)}{f(z)}\,dz = 2\pi i$. On the other hand if $f = g^2$, we would have $\int_{|z| = 1} \frac{f'(z)}{f(z)}\,dz = 2\int_{|z|=1} \frac{g'(z)}{g(z)}\,dz \in 4\pi i \mathbb{Z}$, a contradiction.

Now clearly, $f(z) = z$ can be patched together from functions that admit square roots, so the presheaf in question can't be a sheaf.

  • $\begingroup$ As I understand it can be patched together from functions that admit square roots in the following way: Take an open cover of the annulus by open balls - clearly the constant function has a square root on these balls (holomorphic because....), and the functions obviously agree on the intersection, and the constant function defined on the whole of $U$ restricts down nicely to all these balls. But the function defined on all the big set can't have a holomorphic square root (by your argument). How do we know, though, that there isn't some other function which does, which still restricts the same? $\endgroup$ – Clinton Boys May 11 '11 at 11:08
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    $\begingroup$ @Clinton: I'm not looking at the constant function. I'm looking at the function $f : z \mapsto z$. This function has a square root in any region that "doesn't wind around zero", for instance by cutting up the annulus along a ray from zero to infinity and amking use of logarithms. In fact, there are exactly two square roots differing by a factor $-1$. A function is uniquely determined by its restrictions to open sets. The sheaf condition is about the converse. $\endgroup$ – t.b. May 11 '11 at 11:30

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