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Let $f:X \rightarrow \mathbb{P}^1_k$ be a hyperelliptic curve, where k is a field of characteristic not 2. $\mathbb{P}^1_k$ is the union of $U = Spec k[t]$ and $V = Spec k[s]$ where $s= 1/t$. This gives us $f^{-1}(V) \cup f^{-1}(U) = X$. On pg. 292 of Liu's book, Liu does the following:

  1. He shows that $U' = Spec k[t,y]/(y^2-P(t)$ where $P(t)$ has no square factors, and that $V' = Spec k[s,Z]$ where $P_1(s) = P(1/s)s^{2r}$, where $r = [(deg(P(s)+1)/2]$ and $P_1(s)$ is what we get by restricting (this is just the standard affine opens of the hyperelliptic curve.

  2. He shows that $\Omega^1_{U'/U} = k[t]/(P(t))$ and $\Omega^1_{V'/V} = k[t]/(P_1(t))$. All is clear so far, but here is my problem. He then writes:

"As $\mathbb{P}^1_k = U \cup \{s=0\}$, we have $$H^0(X,\Omega^1_{X/\mathbb{P}^1_k}) = k[t]/(P(t)) \oplus k[s]_m /P_1(s)$$ $m = sk[s]$."

Now, my question is: Why does this hold on global sections? I thought the standard approach was just to calculate on the two affine opens and check that they coincide on intersections?

Edit: Maybe I should use the adjunction formula? But I can't see how really.

Tedar
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  • What is the restriction map from $\Omega^1_{V'/V}$ to $U' \cap V'$? When the degree of $P$ is odd $\Omega^1_{U'/U}$ and $\Omega^1_{V'/V}$ have different dimensions over k (d versus d+1), so when you glue $U'$ and $V'$ to construct a global relative differential, this is where that extra dimension comes from - the single point over $s=0$. – Tomi Tyrrell Aug 26 '13 at 18:57
  • @Thom Tyrell: I am not sure I understand. I do agree with that the dimensions are different when the case is odd - the restriction maps are not given in the text. But why does $X = U \cup {s=0}$say anything? – Tedar Aug 26 '13 at 19:02
  • If you work out the restriction maps it might make the isomorphism more clear. In the odd case, with the decomposition $X = U' \cup {(s,z) = (0,0)}$, we can give global sections by gluing sections from the open cover of U' plus a basis around $(0,0)$. A global section is then a relative differential on $U'$, and a collection of differentials on open nbhds of $(0,0)$ - i.e. a germ. Perhaps the "as" is misleading, but there should be a slick way to see the direct sum decomposition. – Tomi Tyrrell Aug 26 '13 at 19:25
  • @ThomTyrrell: OK, but in the odd case - why should the colleciton of differentials on open nbhds of (0,0) glue nicely with the ones on U? Is this something one could see with restriction maps? – Tedar Aug 26 '13 at 19:28
  • With the restriction maps you can explicitly take a section on U' and a germ at $(0,0)$, and produce a family of sections on neighborhoods of $(0,0)$. The map from $\Omega^1_{V'/V}$ to the stalk at $(0,0)$ sends a polynomial to its constant term. So, we can extend germs (constants) to open nbhds as the constant terms of polynomials in $s$ with the higher order terms coming from the section on $U'$ restricted to $U' \cap V'$. – Tomi Tyrrell Aug 26 '13 at 19:46
  • @ThomTyrell: After some calculations - are the restrictions simply the identity? I.e they are the same, up to a unit on the intersection. – Tedar Aug 26 '13 at 20:09
  • Sorry - I meant localization with respect to t and s. – Tedar Aug 26 '13 at 20:16
  • The ds and dZ play a role in the restriction/transition maps. – Tomi Tyrrell Aug 26 '13 at 20:35
  • @Thom Tyrell: If you have the time, I would appreciate an answer on how the restrictions work out. I use the description as $k[t]/(P(t))$ and $k(s)/(P_1(s))$ . – Tedar Aug 26 '13 at 20:38

1 Answers1

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As $\Omega^1_{X/\mathbb P^1_k}$ is a skyscraper sheaf, you don't need to glue, but just take direct sum.

Cantlog
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