Timeline for Linear system on singular plane curve
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| May 18, 2020 at 15:54 | history | edited | user267839 | CC BY-SA 4.0 |
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| May 15, 2020 at 22:09 | history | edited | user267839 | CC BY-SA 4.0 |
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| Apr 23, 2020 at 15:20 | comment | added | Zach Teitler | Let us continue this discussion in chat. | |
| Apr 23, 2020 at 15:10 | comment | added | user267839 | I don't understand which meaning your expression $R = L - \sum m_i(m_i-1) p_i$ has. Where does it live? $R$ is a Weil divisor in $C$ that it should live in Abelian group $\operatorname{Div}(C)$, or not? On the other hand $L$ lives in $\operatorname{Div}(\mathbb{P}^2)$. So I not see how to interpret your $R = L - \sum m_i(m_i-1) p_i$ | |
| Apr 23, 2020 at 15:10 | comment | added | user267839 | then we get $L \cap C= \sum _{p \in C \ : p \in \operatorname{Supp}(L \cap C)}a_p p= \sum_i a_{p_i}p_i + \sum_{p \in \operatorname{Supp} \: p \not \in \{p_1,...,p_n\}}a_p p$ which is first of all a Weil divisor in $C$. Now the text suggests that $R$ is $\sum_{p \in \operatorname{Supp} \: p \not \in \{p_1,...,p_n\}}a_p p$ since it describes $L \cap C$ as consisting of $p_i$ with $a_{p_i} \ge m_i(m_i-1)$ by the argument you have explained before and the residual $R$. | |
| Apr 23, 2020 at 15:09 | comment | added | user267839 | @ZachTeitler: I'm a bit confused now: $L \in |L|$ is a divisor in $\mathbb{P}^2$. That is $L = \sum_{l=1}^k l_k D_k$ where $l_k \in \mathbb{Z}$ and $D_k$ certain prime divisors of $\mathcal{P}^2$. The intersection $L \cap C$ seems to be a bad notation for intersection product of divisors $L,C$ and not the set thereoretical intersection. We assume $L$ and $C$ haven't common irred component, | |
| Apr 23, 2020 at 14:14 | comment | added | Zach Teitler | Subtracting a fixed point doesn’t change the dimension of a linear system. | |
| Apr 23, 2020 at 13:30 | comment | added | user267839 | What we need is the following: we take arbitrary (Weil) divisor $L \in |L|$ and a.d. $R \in |R|$ and want to compare the dimensions of $k$-spaces $\Gamma(\mathbb{P}^2, \mathcal{L}(L)$ and $\Gamma(C, \mathcal{L}(R))$. I not see from your argument why these should have same dimensions? | |
| Apr 23, 2020 at 13:24 | comment | added | user267839 | @ZachTeitler: And why does it imply $\dim |R| = \dim |L|$? | |
| Apr 23, 2020 at 13:21 | history | edited | user267839 | CC BY-SA 4.0 |
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| Apr 23, 2020 at 3:49 | comment | added | Zach Teitler | It should be that for $L \in |L|$ (dubious notation) the corresponding $R$ is $R = L - \sum m_i(m_i-1) p_i$, not $R = L - \sum \operatorname{mult}_{p_i}(L \cap C) p_i$. That is, if $L \cap C$ has multiplicity $> m_i(m_i-1)$ at a point $p_i$, then $R$ is obtained by still subtracting $m_i(m_i-1) p_i$, not the higher multiple. I hope that this sheds some light on why $\dim |R| = \dim |L|$. | |
| Apr 22, 2020 at 23:20 | comment | added | user267839 | any idea why $\dim \vert R \vert= \dim \vert L \vert$? | |
| Apr 22, 2020 at 22:58 | comment | added | user267839 | @ZachTeitler: I see, it's of course $3$ here. The reason that leaded me to this wrong suspicion in question 1 on equality instead of inequality $\ge m_i(m_i − 1)$ was the caclulation of the degree $\operatorname{deg} \vert R \vert = d(d-1) + \sum_i m_i(m_i-1)$. It looks like application of Bezout’s theorem. And I think the point is that it's only true that for general member $L \in \vert L \vert$ multiplicity of $L∩C$ in $p_i$ satisfies only the equality $=m_i(m_i−1)$? I think that causes my confusion... | |
| Apr 22, 2020 at 22:35 | comment | added | Zach Teitler | Regarding the first question, here is an example that might help understand why the statement is an inequality rather than an equation: the curve $C$ defined by $f = y^2 - x^2 + x^3$ has multiplicity $m=2$ at the origin and the curve (line) $L$ defined by $g = y-x$ has multiplicity $m-1=1$ at the origin. So the statement says that the multiplicity of $C \cap L$ at the origin is $\geq m(m-1) = 2$. What do you think is actually the multiplicity? | |
| Apr 22, 2020 at 20:50 | comment | added | Evgeny Shinder | Regarding the last question: if you want a polynomial $f(x,y)$ and all its derivatives up to $m$ to vanish at a point $p$, it gives $\binom{m+2}{2}$ independent linear conditions on the coefficients. Indeed, if we assume $p = (0,0)$, then the condition is that monomials of $f$ of degrees up to $m$ vanish, and there are $1 + 2 + \dots + (m+1) = \binom{m+2}{2}$ of these. | |
| Apr 22, 2020 at 20:18 | history | asked | user267839 | CC BY-SA 4.0 |