Timeline for quadrics containing the tangential variety of a curve
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Apr 12, 2023 at 16:57 | comment | added | JackYo | $\mathbb{G}(1,\mathbb{P}^3)$ and therefore in $Q^4$ by Plücker identification. So it contains any tangent line of $C$. But I would like to know if one can show this claim only with algebraic methods. Do you see how? | |
| Apr 12, 2023 at 16:51 | comment | added | JackYo | Now we can use the geometric fact that a line $L \subset \mathbb{P}^5$ is contained in Grassmannian $\mathbb{G}(1,\mathbb{P}^3) \subset \mathbb{P}^5$ iff there exist a point $p \in \mathbb{P}^3$, and a hyperplane $H \subset \mathbb{P}^3$ such that $L:= \{\mathcal{l} \in \mathbb{G}(1,\mathbb{P}^3) \vert p \in \mathcal{l} \subset H \}$. The tangent at $[v(t) \wedge v'(t)]$ is given as the set $S:=\{\mathcal{l} \in \mathbb{G}(1,\mathbb{P}^3) \vert v(t) \in \mathcal{l} \subset \langle v(t), v'(t), v''(t) \rangle \} $. And by the fact above it is contained in Grassmannian | |
| Apr 12, 2023 at 16:48 | comment | added | JackYo | we can locally parametrize $C' \subset \mathbb{P}^3$ via $t \mapsto v(t):=(v_1(t),v_2 (t),v_3(t)) \in \mathbb{A}^3 \subset \mathbb{P}^3 $. Then the Gauss map $G: C' \to \mathbb{G}(1,\mathbb{P}^3), c \mapsto T_c C'$ is explicitely given locally as $ v(t) \mapsto [v(t) \wedge v'(t)]$. The tangent at $[v(t) \wedge v'(t)]$ is given as the set $T_{[v(t) \wedge v'(t)]}C:=\{\mathcal{l} \in \mathbb{G}(1,\mathbb{P}^3) \vert v(t) \in \mathcal{l} \subset \langle v(t), v'(t), v''(t) \rangle \} $ | |
| Apr 12, 2023 at 16:47 | comment | added | JackYo | @DmitriPanov: Could you give some hints how to show that for the curve $C \subset \mathbb{G}(1,\mathbb{P}^3)$ in your construction the tangent variety $TC$ is contained in $Q^4$ ($\cong \mathbb{G}(1,\mathbb{P}^3)$ under Pluecker)? Is is possible to show it pure algebraically? Here an analytic approach which came into my mind using local parametrization of $C$ (and I'm curious if it's possible to avoid analytic tools): | |
| Jul 13, 2011 at 13:20 | vote | accept | Jie Wang | ||
| Jul 13, 2011 at 13:19 | vote | accept | Jie Wang | ||
| Jul 13, 2011 at 13:20 | |||||
| Jul 13, 2011 at 13:19 | vote | accept | Jie Wang | ||
| Jul 13, 2011 at 13:19 | |||||
| Jul 11, 2011 at 19:28 | comment | added | Jie Wang | Yes. By $g^r_d$ I mean a linear system of (projective) dimesion $r$ and degree $d$. This linear system maps $C$ into $\mathbb{P}^r$ (assuming it is very ample). So by general $g^r_d$ I mean exactly what you said: you could deform it in $\mathbb{P}^r$ so that $TC$ is not contained in a quadric. Thanks | |
| Jul 11, 2011 at 14:49 | comment | added | Dmitri Panov | What is $g^r_d$? Do I understand correctly that you want to know if each non-degenerate curve $C$ in $\mathbb P^n$ can be deformed so that $TC$ is not contained in any quadric? If this is indeed the question, I have to think more about it. You might consider again to make this assumption explicit in your question (or ask a follow up question). | |
| Jul 11, 2011 at 13:39 | comment | added | Jie Wang | Thanks. That example will work.I apprciate it. What about if I modify my question by supposing $C$ is a general curve embedded in $mathbb{P}^r$ by a general $g^r_d$? | |
| Jul 10, 2011 at 21:10 | history | edited | Dmitri Panov | CC BY-SA 3.0 |
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| Jul 10, 2011 at 18:35 | comment | added | Dmitri Panov | The answer is completely rewritten. | |
| Jul 10, 2011 at 18:33 | history | undeleted | Dmitri Panov | ||
| Jul 10, 2011 at 18:33 | history | edited | Dmitri Panov | CC BY-SA 3.0 |
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| Jul 9, 2011 at 21:53 | history | deleted | Dmitri Panov | ||
| Jul 9, 2011 at 18:07 | comment | added | Jie Wang | Thanks. But I think my curve is nondegenerate, which means it is not contained in any hyperplane. | |
| Jul 9, 2011 at 18:01 | history | answered | Dmitri Panov | CC BY-SA 3.0 |