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I expect that this is a very easy question, but somehow I can't get it. What is the dimension of the moduli space of complete intersections of degree 2 and 4 in $\mathbb{P}^5$? The answer should be 89.

I apologize again that if the question is too easy.

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  • $\begingroup$ Where do you get 89 from? I seem to get 91. $\endgroup$
    – user64687
    May 27 '14 at 12:51
  • $\begingroup$ This paper: arxiv.org/abs/hep-th/0506196 page 12 How are you getting 91? Seems pretty close :) $\endgroup$
    – user148177
    May 29 '14 at 3:16
  • $\begingroup$ Dear user148177, I am not inclined to doubt those authors. So let me think a bit more about whether I messed up somewhere. $\endgroup$
    – user64687
    May 29 '14 at 7:41
  • $\begingroup$ May I ask what approach you took to this problem? $\endgroup$
    – user148177
    Jun 1 '14 at 11:40
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Here's how I arrived at my answer.

First, choose a quadric in $\mathbf P^5$. The parameter space for such choices has dimension ${5+2 \choose 2} -1$ = 20.

Next, choose a quartic in $\mathbf P^5$, modulo the already chosen quadric. (Two quartics forms which differ by a multiple of the chosen quadric will give the same complete intersection.) The parameter space for such choices has dimension $ \left( {5+4 \choose 4} - {5+2\choose 2} \right) -1$ = 126-21-1=104.

OK, so the parameter space of complete intersections has dimension 104+20=124. But now we want the moduli space, so we divide out by the action of $PGL(6)$. That has dimension $6^2-1=35$. So in the end we get a space of dimension $124-35=89$.

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  • $\begingroup$ Looks great, thanks! $\endgroup$
    – user148177
    Jun 2 '14 at 12:39
  • $\begingroup$ Dear @user148177, you're welcome! I am glad I was able to find those extra (-1)'s... $\endgroup$
    – user64687
    Jun 2 '14 at 13:03

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