Timeline for Lifting of quadrics containing a curve
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 1 at 16:50 | comment | added | Will Sawin | @user267839 This is essentially equivalent to normality - if a meromorphic function satisfies a monic polynomial equation then it has nonnegative valuation, i.e. it has effective divisor, so if the ring fails to be integrally closed you get a counterexample to this property. | |
| Jul 1 at 16:24 | comment | added | user267839 | So essentially, that's exactly then the case if for every line bundle $O_X(D)$ associated with a divisor $D$ the global sections can be characterized precisely as those meromorphic functions $f$ on $X$, satisfying divisorial ineq $(f)+D \ge 0$ (ie depend only on what happens purely in codim $1$; the bigger codimensional stuff "not matter"; in normal case as you wrote by Hartogs argument) | |
| Jul 1 at 15:11 | comment | added | Will Sawin | @user267839 This always works if $X$ is normal. The point is that $t/s$ is a meromorphic section of $\mathcal L^e$ whose divisor is $F-E$ which is effective, hence this section has no poles at any codimension $1$ points, and thus extends to the whole space by normality. | |
| Jul 1 at 15:04 | comment | added | user267839 | In situation above we had the rather spefic situation $E \sim 2D$ or equivalently $E-D \sim D$ and this was sufficient to find a $r$ with $\operatorname{div}(r) \sim E-D$ which does the job. But what about more general case? The weakest assumptions I can think of is if the induced $f:X \to \operatorname{Proj}(\oplus_k H^0(X, \mathcal{L}^k))$ (...assuming welldefined ie $L$ base pt free) is surjctive and ideal $(s)$ (as homog ideal) is reduced. Then we can use the basic relation "above" in the Proj $V(t) \supset V(s)$ iff $t \in \sqrt{(s)}=(s)$, so $s \vert t$. Does this argumentation work? | |
| Jul 1 at 15:03 | comment | added | user267839 | level of ring elements." is literally true? (of course, only the "=>" direction is interesting) In other words, when for $X$ integral scheme and $\mathcal{L}(D)$ line bdl corresp to an eff divisor $D \ge 0$, for $E=\operatorname{div}(s), F=\operatorname{div}(t)$ with $s \in H^0(X, \mathcal{L}^d), t \in H^0(X, \mathcal{L}^{d +e})$ from $F \ge E$ (as Weil divs) we can deduce $ s \vert t$ inside ring $\oplus_k H^0(C, \mathcal{L}^k)$? | |
| Jul 1 at 15:03 | comment | added | user267839 | I would like to dig up again your statement on equivalence of divisibility of Weil divisors and ring elements for curves (or more generally projectve schemes) As you remarked in subsequent comment that's not correct in full generality, but in specific situation above, as "accidently" $\operatorname{div}(q|C)$ lin equiv to twice the hyperplane class. So it appears here to be more or less "accidently". But what I'm wondering about under which weakest as possible assumptions such statement/"slogan" "Dividing on the level of Weil divisors is equivalent to dividing on | |
| Mar 17 at 2:10 | comment | added | Will Sawin | @user267839 All I can see is that the same argument gives liftability statements for higher-degree forms, e.g. $k$-normality gives liftability of degree $k+1$ forms on the hyperplane. | |
| Mar 17 at 0:48 | comment | added | user267839 | Yes, right, I see. A side note: So summarizingly already the $1$-normality grants this lifting property described in the question. If we impose the much stronger "projective normality", or say "normality up to level $n$, ie pullback maps are surjective for all $k \le n$, intuitively we should expect even stronger - in what direction ever - liftability properties for the sections. Can the statement above about the liftability property of sections be strengthened (in certain appropriate way) if we impose these stronger assumptions on "normality"? | |
| Mar 17 at 0:23 | vote | accept | user267839 | ||
| Mar 17 at 0:19 | comment | added | Will Sawin | @user267839 Or, I see the problem. It's of course not true for every effective Weil divisor of degree $D$. But the point is that $\operatorname{div}(q|C)$ is a divisor linearly equivalent to twice the hyperplane class and $\operatorname{div}(a\mid_C)$ is a divisor linearly equivalent to the hyperplane class so their difference is linearly equivalent to the hyperplane class. | |
| Mar 17 at 0:18 | comment | added | Will Sawin | @user267839 Dividing on the level of Weil divisors is equivalent to dividing on the level of ring elements. There exists a section $\overline{r}$ of a given line bundle with divisor $R$ if and only if $R$ is effective and linearly equivalent to the divisor of that line bundle. Effectiveness is assumed and linear equivalence is obvious. | |
| Mar 17 at 0:18 | comment | added | user267839 | In other words the problem I see in the argument so far is that it's not clear why every effective Weil divisor in $C$ of degree $d$ comes from a global section in $H^0(C, \mathcal O_C(1))$. Doesn't the "degree of obstruction" for this not exactly lie in the genus of the curve? Do you see how circumvent this problem (if it is one)? | |
| Mar 17 at 0:09 | comment | added | user267839 | associated divisor $\text{div}(q \mid_C)-\text{div}(a \mid_C):=R$ is effective of degree $d$ (...as you remarked $q|_C$ vanishes on $H \cap C$ and has degree $2d$), why $R$ should come from a section, ie why there should exist a $\overline{r}\in H^0(C, \mathcal O_C(1))$ with $R=\text{div}(r)$? | |
| Mar 17 at 0:06 | comment | added | user267839 | There is one point I not completely understand: When you write "that $a \mid_C \in H^0(C, \mathcal O_C(1))$ divides $q\mid_C \in H^0(C, \mathcal O_C(2))$ ", you mean to "dividing" on level of associated Weil divisors to $a \mid_C, c \mid_C$ in the sense that for effective Weil divs $D, D' \subset C$ we say $D$ divides $D'$ iff $D'-D := R \ge 0$ is effective, and not on level of ring elements inside $\bigoplus_k H^0(C, \mathcal O_C(k))$ right? The problem I see so far in your argument, is because, even if the | |
| Mar 16 at 23:39 | vote | accept | user267839 | ||
| Mar 16 at 23:56 | |||||
| Mar 16 at 20:12 | history | answered | Will Sawin | CC BY-SA 4.0 |