# Elements of the zero-th Čech cohomology group versus global holomorphic sections

Something that is confusing (well, to me) has come up in the course of asking other questions.

Let $\pi:V\to X$ be a holomorphic vector bundle of rank $r$ on a complex manifold $X$, such that $V$ is defined by an open cover $\{U_\alpha\}$ and holomorphic transition functions $g_{\alpha\beta}:U_\alpha\cap U_\beta\to\mbox{GL}(r,\mathbb C)$.

If $s:X\to V$ is a global holomorphic section of $V$, then the local data $s_\alpha:U_\alpha\to\mathbb C^r$ of $s$ must glue together as $s_\beta=g_{\alpha\beta}s_\alpha$ over the overlaps $U_\alpha\cap U_\beta$.

Global holomorphic sections are also said to be elements of $H^0(X;\mathcal O(V))$, where $H^i$ is Čech cohomology and $\mathcal O(V)$ is the sheaf of holomorphic sections of $V$.

I'm going to assume that the cover $\{U_\alpha\}$ is a "good" cover, for Čech purposes.

Literally, $H^0(X;\mathcal O(V))=\mbox{ker}[C^0(\mathcal O(V))\stackrel{\delta}{\longrightarrow}C^1(\mathcal O(V))]$, where $\delta$ is the Čech coboundary operator, and $C^p(\mathcal O(V))$ is the vector space of $0$-cochains for $\mathcal O(V)$ with respect to $\{U_\alpha\}$.

Hence, a section $s$ of $V$ is in $H^0(X;\mathcal O(V))$ if and only if $(\delta s)_{\alpha\beta}=s_\beta-s_\alpha=0$, which says that $s_\alpha$ and $s_\beta$ must agree on the nose over $U_\alpha\cap U_\beta$.

This is different than saying that $s_\beta=g_{\alpha\beta}s_\alpha$ on the overlap.

Having $s_\alpha=s_\beta$ seems to be a very strong restriction on a section. The existence of such a section suggests that $g_{\alpha\beta}=I_r$ on each overlap, and hence $V$ would be trivial.

But for instance: the holomorphic line bundle $\mathcal O(1)$ (i.e. hyperplane bundle) on $\mathbb{CP}^1$ is non-trivial, and at the same time is supposed to have $H^0(\mathbb{CP}^1;\mathcal O(1))\cong\mathbb{C}^2$.

• I think your formula for $\delta$ is wrong. It should be the thing that makes everything match up. – Qiaochu Yuan Jan 3 '15 at 1:20
• Hmmm...not so sure about that. See the responses from Hoot and Ted below. – MathsByTheSea Jan 3 '15 at 3:27
• No, the coboundary is just fine. See my answer below. – Ted Shifrin Jan 3 '15 at 3:29
• I interpreted $s_{\alpha}$ in the OP's expression for $\delta$ to mean the same thing that it meant earlier. Of course if you change the interpretation then everything is fine. – Qiaochu Yuan Jan 3 '15 at 5:12
• I meant degree nonnegative. Also, Mariano, I appreciate your vantage point, but please don't say the "rest is just technicalities". I am a physicist by training who is trying to become a better mathematician, and a better geometer in particular. The technicalities are everything to me right now, and they are precisely what I am here for. :-) – MathsByTheSea Jan 3 '15 at 18:38

Taking the Čech point of view requires a lot of vigilance; I think it really gets confusing when you start thinking about divisors and what it means for a section of $\mathcal{O}(D)$ to vanish somewhere. Let's ignore trivializations for the moment.

I hope it's okay to work with sheaves of sections instead of bundles; I think the notation is going to get worse if I try to do it the other way. To that end I'm going to write $U_{\alpha\beta}$ for $U_\alpha \cap U_\beta$. I've got a locally free sheaf $\mathscr{F}$ and to calculate $H^0$ I'm given sections $s_\alpha \in H^0(U_\alpha, \mathscr{F})$ and I need to check whether $s_\alpha|_{U_{\alpha\beta}} - s_\beta|_{U_{\alpha\beta}} = 0$.

If I choose trivializations $\tau_\alpha\colon \mathscr{F}|_{U_\alpha} \to \mathscr{O}_{U_\alpha}^{\oplus r}$ then I can write $s_\alpha = \tau_\alpha^{-1}(v_{\alpha})$ and the condition is $$0 = \tau_\alpha^{-1}(v_\alpha)|_{U_{\alpha\beta}} - \tau_\beta^{-1}(t_\beta)|_{U_{\alpha\beta}} = \tau_\alpha^{-1}(v_\alpha|_{U_{\alpha\beta}}) - \tau_\beta^{-1}(v_\beta|_{U_{\alpha\beta}}).$$ If your aim is to compare tuples of functions on $X$ then you should probably pick some trivialization over $U_{\alpha\beta}$ and stick with it. If I choose $\tau_\alpha$ for this and hit the equation with it then the requirement is the one you had before: $$v_\alpha|_{U_{\alpha\beta}} - (\tau_\alpha|_{U_{\alpha\beta}} \circ \tau_\beta|_{U_{\alpha\beta}}^{-1})(v_\beta|_{U_{\alpha\beta}}) = v_\alpha|_{U_{\alpha\beta}} - g_{\beta\alpha}(v_\beta|_{U_{\alpha\beta}}).$$ Anyway, there's nothing particularly insightful in the above but I think the way to get past the doubts is to just write it all out and then carry on using transition functions as carelessly as life will allow. For the calculation of $H^0(\mathcal{O}_{\mathbb{P}^1}(1))$ what ends up happening is that you get functions $f(z)$ and $g(1/z)$ on an overlap and it may seem that the only way to for these to agree is for them to both be the same constant; but we need to multiply $g(1/z)$ by the transition function $z$ in order to be able to compare it with $f(z)$! One might have had this trouble way back when learning the definition of a manifold.

• Nice --- thanks. I appreciate the vigilance required now, and I also appreciate the vigilance of your response! – MathsByTheSea Jan 3 '15 at 1:28
• Just to be absolutely sure: when you write $s_\alpha=\tau_\alpha^{-1}(v_\alpha)$, do you mean $t_\alpha$ instead of $v_\alpha$, just to square things up with subsequent notation? – MathsByTheSea Jan 3 '15 at 21:21
• Yes, I slipped up. One sec. – Hoot Jan 3 '15 at 21:43

No, you're confusing notations. For the transition function view, $s_\alpha\colon U_\alpha\to \Bbb C^r$ is a representation of the section in the trivialization. On the other hand, when you come to the Čech viewpoint, $s_\alpha$ refers to an actual section of $V$ on $U_\alpha$.

• So, in other words, there is no issue with how the boundary operator $\delta$ is defined in my question? Rather, the problem is that the notion of $s_\alpha$ is inconsistent across the two "halves" of the question? So the representations are related by the transition functions, while the bona fide sections of $V$ on the $U_\alpha$ must agree on the nose on overlaps? I think Hoot's answer agrees with what you are saying, in any case. – MathsByTheSea Jan 3 '15 at 3:26
• Your coboundary is fine. I just wanted to give it to you as succinctly as possible! – Ted Shifrin Jan 3 '15 at 3:29
• That's great, I appreciate it. Thanks for the help! :-) – MathsByTheSea Jan 3 '15 at 3:33