# Morphism $H^0(X,K_X^\ast/\mathcal O_X^\ast)\to H^1(X,\mathcal O_X^\ast)$ from exact sheaf sequence: why does it have this form?

We start with a holomorphic complex manifold $X$. We have the sheaf of invertible holomorphic functions (i.e. everywhere nonzero holomorphic functions) $\mathcal O_X^\ast$, where the operation is the product. We have $K_X^\ast$, the sheaf of invertible meromorphic functions (i.e. meromorphic functions that do not vanish identically on any open $U\subseteq X$), with the product again. Remember that a meromorphic function on an open $U\subseteq X$ is a map $x\mapsto f_x$ where $f_x$ is in the field of fractions of the ring $\mathcal O_{X,x}$ (germs of holomorphic functions at $x$) such that for any $x\in U$ there is an open $U'\subseteq U$ and holomorphic functions $g,h\in\mathcal O_X(U)$ such that for any $y\in U'$ we have $f_y=\frac{g_y}{h_y}$. With all that, we clearly have an exact sequence of sheaves:

$$0\to\mathcal O_X^\ast\xrightarrow{i}K_X^\ast\xrightarrow{p}\frac{K_X^\ast}{\mathcal O_X^\ast}\to0,$$

where I called $i$ the inclusion and $p$ the projection onto the cokernel of $i$. This induces a long exact sequence in sheaf cohomology:

$$0\to H^0(X,\mathcal O_X^\ast)\to H^0(X,K_X^\ast)\to H^0\left(X,\frac{K_X^\ast}{\mathcal O_X^\ast}\right)\xrightarrow{\delta}H^1(X,\mathcal O_X^\ast) \to H^1(X,K_X^\ast)\to\dots.$$

The map I am interested in is the one I labeled $\delta$. The codomain of $\delta$ can be identified with the Picard group $\operatorname{Pic}(X)$ of line bundles with the operation given by the tensor product. Hence, an image of something via $\delta$ will be given by a collection of open sets $U_\alpha\subseteq X$ which form an open cover, and a family of transition functions $f_{\alpha\beta}:U_{\alpha\beta}\to\mathbb C$, which will have to be holomorphic and satisfy the cocycle condition $f_{\alpha\beta}f_{\beta\gamma}=f_{\alpha\gamma}$ for any $\alpha,\beta,\gamma$ in the set of indices of the open cover. Since all we need are the intersections and the functions, I will specify a line bundle by $\{(U_\alpha\cap U_\beta,f_{\alpha\beta})\}$.

The domain of $\delta$ is the group of Cartier divisors, or global sections of that quotient sheaf, which, besides being identifiable with Weil divisors, can be specified by a collection $\{(U_\alpha,f_\alpha)\}$ where $\{U_\alpha\}$ is an open cover and $\frac{f_\alpha}{f_\beta}\in\mathcal O_X^\ast(U_\alpha\cap U_\beta)$ for all $\alpha,\beta$.

With all of that, my professor's notes tell me that:

$$\delta(\{(U_\alpha,f_\alpha)\})=\left\{\left(U_\alpha\cap U_\beta,\frac{f_\beta}{f_\alpha}\right)\right\}.$$

While this is certainly a morphism with the same domain and codomain as $\delta$, I can't seem to be able to work out that form.

Here is what I tried. I went back to the proof of the existence of the long exact sequence in cohomology, and I drew the following diagram:

$$\require{AMScd}$$

$$\begin{CD} 0 @>>> \mathcal O_X^\ast(X) @>i>> K_X^\ast(X) @>p>> \frac{K_X^\ast}{\mathcal O_X^\ast}(X) @>>> 0 \\ @. @Vd_O^{-1}VV @Vd_K^{-1}VV @Vd_{KO}^{-1}VV \\ 0 @>>> ds\,\mathcal O_X^\ast(X) @>i_0>> ds\,K_X^\ast(X) @>p_0>> ds\,\frac{K_X^\ast}{\mathcal O_X^\ast}(X) @>>> 0 \\ @. @Vd_O^0VV @Vd_K^0VV @Vd_{KO}^0VV \\ 0 @>>> ds\operatorname{coker}d_O^{-1}(X) @>i_1>> ds\operatorname{coker}d_K^{-1}(X) @>p_1>> ds\operatorname{coker}d_{KO}^{-1}(X) @>>> 0 \\ @. @Vd_O^1VV @Vd_K^1VV @Vd_{KO}^1VV \\ @. \vdots @. \vdots @. \vdots \end{CD}$$

So my map $\delta$ will be snaking its way from (almost-)top-right (not the zero) to bottom-(almost-far-)left. I am starting with $\{(U_\alpha,f_\alpha)\}$. First thing is to go down with $d_{KO}^{-1}$. So I'm viewing my $\{(U_\alpha,f_\alpha)\}$ as a discontinuous section of the quotient sheaf, i.e. a map associating to each $x\in X$ the germ of my thing at $x$. Said germ should be an equivalence class of germs of meromorphic functions modulo holomorphic germs. So I thought this would give me the map $x\mapsto(f_\alpha)_x\cdot\mathcal O_{X,x}^\ast=[(f_\alpha)_x]_{\operatorname{mod}\mathcal O_{X,x}^\ast}$. Then I have to pull this back to $ds\,K_X^\ast(X)$, so I thought I could choose $x\mapsto(f_\alpha)_x$ where $\alpha$ is the minimum index such that $x\in U_\alpha$ (I am assuming the indexes are well-ordered since, by the Well-Ordering Principle, any set can be well-ordered). Then I have to go down again. I will do so in two steps. The first step is projecting that map to the cokernel of $d_K^{-1}$. I originally thought I'd just take the equivalence class of the map above modulo $K_X^\ast(X)$, but now I remember that the quotient sheaf is a sheafification, so the global sections of this cokernel may not be equivalence classes of discontinuous global sections of $K_X^\ast$. But this is getting decidedly too incomprehensible.

So how do I get to that form up there?

Update

The comment exchange with @KennyWong below (1, 2 and 3) convinced me that, with a little caveat on the covers, one can use diagrams such as:

$$\begin{CD} 0 @>>> \mathcal O_X^\ast(X) @>i>> K_X^\ast(X) @>p>> \frac{K_X^\ast}{\mathcal O_X^\ast}(X) @>>> 0 \\ @. @VVV @VVV @VVV \\ 0 @>>> C^0(\mathcal U,\mathcal O_X^\ast) @>i_0>> C^0(\mathcal U,K_X^\ast) @>p_0>> C^0\left(\mathcal U,\frac{K_X^\ast}{\mathcal O_X^\ast}\right) @>>> 0 \\ @. @V\delta_O^0VV @V\delta_K^0VV @V\delta_{KO}^0VV \\ 0 @>>> C^1(\mathcal U,\mathcal O_X^\ast) @>i_0>> C^1(\mathcal U,K_X^\ast) @>p_0>> C^1\left(\mathcal U,\frac{K_X^\ast}{\mathcal O_X^\ast}\right) @>>> 0 \\ @. @V\delta_O^1VV @V\delta_K^1VV @V\delta_{KO}^1VV \\ @. \vdots @. \vdots @. \vdots \end{CD}$$

to obtain an exact sequence exactly like the one I mentioned above, but with Čech cohomology groups. In that sequence, it is clear that $\delta$ has the given form. The question now remains open if there is a not-too-messy way to incarnate the $\delta$ from sheaf cohomology, and perhaps prove that, if we compose it with the isomorphism $H^1\cong\check H^1$ coming from a weak version of Leray's theorem, we get the given form.

Update 2

I had an email exchange with my professor, and says that, even in the cover-independent Čech cohomology, the sequence is not exact in general, and that the above argument only works for $\check H^0\to\check H^1$. It appears that he is at least partly right: the morphisms may always be defined, but there are cases where the sequence fails to be exact at $\check H^2$.

• Rather than using sheaf cohomology, have you considered using Cech cohomology on the open cover $\{ U_\alpha \}$? – Kenny Wong Mar 25 '17 at 22:34
• So in Cech cohomology, the connecting map $\delta$ acts on $(U_\alpha, [f_\alpha]) \in H^0(K^\star / O^\star)$ by (i) pulling back via $p$ (gives you $(U_\alpha, f_\alpha) \in \check H^0(K^\star)$), (ii) acting by the Cech boundary map (gives you $(U_{\alpha\beta}, f_\alpha/f_\beta) \in \check H^1(K^\star)$), (iii) pulling back via $i$ (gives you $(U_{\alpha\beta}, f_\alpha/f_\beta) \in \check H^1(O^\star)$). – Kenny Wong Mar 25 '17 at 22:38
• @KennyWong Trouble is, the theorem giving the exact sequence is about sheaf cohomology, not Čech cohomology, meaning that either I should use sheaf cohomology, or the professor should at least mention one can construct a long exact sequence in Čech cohomology also. I do know that $H^1$ and $\check H^1$ are isomorphic in the case of $\mathcal O_X^\ast$ (indeed, proving $\operatorname{Pic}(X)\cong H^1(X,\mathcal O_X^\ast)$ is proved by first showing the isomorphism with $\check H^1$ and then using a weak version of Leray's theorem to prove those two are isomorphic. (cont.) – MickG Mar 25 '17 at 22:43
• (cont.) PS I assume $U_{\alpha\beta}$ means $U_\alpha\cap U_\beta$. Also, it seems that the professor swapped the indexes in the cocycle. – MickG Mar 25 '17 at 22:43
• Actually the snake lemma DOES apply for Cech cohomology! There is one caveat though: you can't using Cech cohomology on a fixed open cover; you have to take the direct limit in which the open covers become "increasingly refined". – Kenny Wong Mar 25 '17 at 22:44

## 1 Answer

We can define that morphism on the cover-dependent version of Čech cohomology, and then pass it to the colimits. That is, for every cover we have a morphism, and we can induce a colimit-group morphism by saying take $c\in\check H^0(X,\frac{K_X^\ast}{\mathcal O_X^\ast})$, take a representative $\tilde c\in\check H^0(\mathcal U,\frac{K_X^\ast}{\mathcal O_X^\ast})$ for some cover $\mathcal U$, apply your $\delta$ and then pply the projection $\check H^1(\mathcal U,\mathcal O_X^\ast)\to\check H^1(X,\mathcal O_X^\ast)$. Now let me introduce you to my first-ever 3D tikz-cd diagram: Pardon the cut right side but it was just too big for a page width. Disclaimer: I was not too zealous with checking the names of the various maps, so there might be some inconsistencies.

That mess of a diagram has the following interesting property:

Every possible square in it is commutative.

This means that, if I consider the $\delta$ going diagonally from $\check H^0(\mathcal U,K_X^\ast/\mathcal O_X^\ast)$ to $\check H^1(\mathcal U,\mathcal O_X^\ast)$, and try to view it as $H^0\to H^1$by snaking our ways from lines to columns (or is it vice versa?), we are assured it is exactly the map from the long exact sequence in cohomology. Precisaly, the path leading from $\check H^0$ to $H^1$ via $\delta$ is (back in dim 3)-down-(back in dim 3)-right-up-, whereas if we go via $H^0$ the path is right-up-(back in dim 3)-right-(back in dim 3) -- hopefully that explains the paths. And by commutativity I'm willing to bet these two paths are identical. This means that, if we apply $\delta$ and then the morphism from Čech to sheaf cohomology, or apply the morphism first and then the sheaf-cohomology-long-exact-sequence morphism, we get the same thing. Composing with the inverse of the isomorphism $H^0\to\check H^0$, we see that the sheaf-cohomology-long-exact-sequence morphism is equivalent to going back to Čech, applying $\delta$, and then going to sheaf cohomology. QED.