Timeline for Irreducible components of the Hilbert scheme

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9 events

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Jun 24, 2012 at 12:10	comment	added	Jason Starr		@Arend: The OP says "quartics", not "plane quartics". Indeed if you interpret the question with "plane quartics", it is absurd.
Jun 23, 2012 at 14:16	vote	accept	Naga Venkata
Jun 23, 2012 at 9:10	comment	added	Arend Bayer		I don't understand the map on Jason's open subset either. I assume $Hilb_{P_2}$ is the Hilbert scheme of plane quartics. If $C \subset X$ is planar, that does not mean that $2C \subset X$ is also. You need to adjust $P_2$ to be the Hilbert polynomial of the first infinitesimal neighborhood of $C \subset X$. You can compute this by determining the degree of the normal bundle of $C$ (via adjunction), but the answer will depend on the degree $d$ of $X$.
Jun 23, 2012 at 5:34	answer	added	Dan Petersen		timeline score: 1
Jun 22, 2012 at 17:36	comment	added	Jason Starr		@Will -- Yes, it is obvious. It is 9 linear conditions on quartic surfaces to contain a given smooth plane conic, yet there is only an 8-parameter family of plane conics in $\mathbb{P}^3$.
Jun 22, 2012 at 17:33	comment	added	Will Sawin		Is it obvious that not every quartic surface contains a plane conic?
Jun 22, 2012 at 17:18	comment	added	Jason Starr		I disagree that you have such a morphism. You have a rational transformation which is regular on the open subset parameterizing pairs $(C,X)$ such that $C$ is a Cartier divisor in $X$. Also, when $d$ equals $4$, clearly the image of $\text{Hilb}_{P_2,Q}$ equals all of $\text{Hilb}_Q$ (consider hyperplane sections), whereas the image of $\text{Hilb}_{P_1,Q}$ will be a proper closed subset.
Jun 22, 2012 at 16:25	history	edited	Naga Venkata	CC BY-SA 3.0	added 191 characters in body; added 1 characters in body
Jun 22, 2012 at 16:04	history	asked	Naga Venkata	CC BY-SA 3.0