Timeline for Singular points of algebraic varieties and parametrization by Puiseux series
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Sep 8, 2014 at 18:05 | comment | added | Felipe Voloch | @IgorKhavkine The normalization provides a way of constructing a non-singular curve mapping birationally to the given curve. This is in Hartshorne I.6 or Fulton 7.5. The fact that on a non-singular curve, every coordinate function is a power series in one of them is the implicit function theorem or exercise 7.18 in Fulton. | |
| Sep 8, 2014 at 17:34 | comment | added | Igor Khavkine | @FelipeVoloch, as I mentioned, I don't have a thorough background in algebraic geometry, so even some elementary things may not be obvious to me. Thanks for mentioning the "normalization" of a curve. I'll have to look it up. If you would like to add something about that to your answer, I'd be grateful as well. | |
| Sep 8, 2014 at 16:01 | comment | added | Felipe Voloch | @IgorKhavkine The passage from Taylor to Puiseux is the same as in my answer. To get Taylor, just use that the normalization of a curve is smooth. | |
| Sep 8, 2014 at 15:46 | comment | added | Igor Khavkine | @FelipeVoloch, in that case, could please give a reference where Puiseux parametrization of hyper-"space curves" is discussed? I've only ever seen the plane-curve case discussed in textbooks. I would really appreciate that. Thanks! | |
| Sep 8, 2014 at 15:08 | history | edited | Felipe Voloch | CC BY-SA 3.0 |
Removed "plane curve" in response to mistake pointed out in comments.
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| Sep 8, 2014 at 15:06 | comment | added | Felipe Voloch | @IgorKhavkine As I explained in my responses to Vivek, the argument can be modified to work. You cut by a linear subspace such that the intersection is a curve and the argument works. | |
| Sep 8, 2014 at 14:57 | comment | added | Igor Khavkine | @FelipeVoloch, just for the record, I wanted to reiterate that your solution does not work in the general case, which is what I'm interested in, and only does when "cutting by a 2-plane" is possible. In any case, thanks for you contribution! | |
| Sep 8, 2014 at 11:04 | comment | added | Felipe Voloch | @IgorKhavkine The expression of $1/r$ in terms of invariants of $V$ I didn't address and I agree it's harder. | |
| Sep 8, 2014 at 5:36 | comment | added | Igor Khavkine | @FelipeVoloch, I think I have to agree with Vivek. I don't think it can't be that simple. In particular, Milnor in his Singular points of complex hypersurfaces spends some non-trivial amount of time proving the so-called "curve selection lemma", which concludes that there exists at least one such curve $\gamma$ whose endpoint $q$ is a generic point of $V$ (but doesn't cover arbitrary fixed $q$, unfortunately). So I don't think it simply comes down to "slice by a 2-plane". The matter of expressing the Piuseux exponent $1/r$ in terms of invariants of $V$ also seems somewhat non-trivial to me. | |
| Sep 8, 2014 at 4:41 | comment | added | Vivek Shende | yeah, I just wanted to point out that you can't literally cut to a plane curve, since I at least find plane curves are easier to think about. | |
| Sep 8, 2014 at 4:00 | comment | added | XL _At_Here_There | @FelipeVoloch,since I may read the post and your answer for more than one time,I will not delete my comment and ask you to keep the comment on my comment so as to find the post and your answer easily. | |
| Sep 8, 2014 at 3:58 | comment | added | Felipe Voloch | @VivekShende for your example, cut the variety with the hyperplane $x_1=x_4,x_2=x_5,x_3=x_6$ | |
| Sep 8, 2014 at 3:55 | comment | added | Vivek Shende | Suppose $x,y \ne 0, 1$. Then by nondegeneracy of the Vandermonde determinant, any remaining point $(w, w^2, w^3, z, z^2, z^3)$ must have $(w, z) \in \{0,1,x\} \times \{0,1,y\}$. If $x = 0, 1$, then in fact $(w,w^2,w^3)$ has to be a scalar multiple of $(1,1,1)$; this only happens $w = 0, 1$. Thus in any case your plane hits $V$ in at most 9 points and in particular contains no curves. | |
| Sep 8, 2014 at 3:55 | comment | added | Felipe Voloch | @XL_at_China Thanks, fixed. Vivek, right, slice with a linear space of appropriate dimension so that the result is a curve (not necessarily plane) containing the two points. The same argument works. | |
| Sep 8, 2014 at 3:52 | history | edited | Felipe Voloch | CC BY-SA 3.0 |
added 6 characters in body
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| Sep 8, 2014 at 3:51 | comment | added | Vivek Shende | Why should such a plane exist? Strictly speaking, taking $V$ to be a non-planar curve is already a counterexample. If that's cheating, consider the product of two rational normal cubics, i.e. the variety which is the image in $\mathbb{C}^6$ of $\mathbb{C}^2$ by the map $(s,t) \mapsto (t, t^2, t^3, s, s^2, s^3)$. Say I take any plane through the points $p = (0,0,0,0,0,0)$ and $q = (1,1,1,1,1,1)$. Three points determine a plane; let $(x, x^2, x^3, y, y^2, y^3)$ be a third one. | |
| Sep 8, 2014 at 3:44 | comment | added | XL _At_Here_There | "I don't that's a big deal",any typos? | |
| Sep 8, 2014 at 0:13 | history | answered | Felipe Voloch | CC BY-SA 3.0 |