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May 23, 2015 at 7:10 comment added Qifeng LI From the answer above, it can be seen that when $n$ is large enough comparing to $d$, then the counterexample appears. Then my new question is that: let $X$ be as in the first question, then whether can we show that $d(d+1)\leq 2n-2$? Or equivalently, is the variety parameterizing the lines on $X$ passing through a fixed general point covered by lines ? If this is true, then we can say that the degrees of this kind of special hypersurfaces $X$ are small enough.
Mar 23, 2015 at 15:05 vote accept Qifeng LI
Mar 14, 2015 at 18:47 history edited Jason Starr CC BY-SA 3.0
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Mar 14, 2015 at 16:33 history edited Jason Starr CC BY-SA 3.0
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Mar 14, 2015 at 16:00 history edited Jason Starr CC BY-SA 3.0
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Mar 12, 2015 at 6:09 comment added Jason Starr That quotient is a stack, not a geometric invariant theory quotient. Also, the point of my answer, and of the article I linked, is that it is easier to prove the conclusion than to understand the morphism $\zeta$. This is just an incidence correspondence computation: form the incidence correspondence of pairs of a hypersurface in $\mathbb{P}^n$ together with a linear $\mathbb{P}^4$ section. If you do not (yet) know how to make incidence correspondence arguments, then you have more fundamental questions to answer than the one above.
Mar 12, 2015 at 4:17 comment added Qifeng LI Thanks for your answer! But there are two questions about your answer. Firstly, I doubt that hypersurfaces in P^{4} that have multiplicity>=2 are semistable elements or not (if not, then the surjectivity of \zeta has no help to my question). Secondly, we can only get that \zeta is dominate instead of that it is surjective. Hence, your answer can not convince me.
Mar 11, 2015 at 17:51 history edited Jason Starr CC BY-SA 3.0
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S Mar 11, 2015 at 16:19 history answered Jason Starr CC BY-SA 3.0
S Mar 11, 2015 at 16:19 history made wiki Post Made Community Wiki by Jason Starr