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I work on complex numbers. Let Q \subset \PP^4 be a quartic threefold that contains a double line l. I want to compute the normal sheaf of l in Q, and I seem to find only two possible cases, i.e.

O_l(-1) \oplus O_l and O_l(-2)\oplus O_l(1)

can someone tell me which is the right one? There should be some paper by Iskovskikh about that but I am not able to find a copy of that.

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1 Answer 1

Let $X$ be your threefold (I assume it is smooth) and $\ell \subset X$ a line. We have the Gauss map $\ell \rightarrow \mathbb{P}^2$ which associates ot a point x the tangent $T_x X$. Here the target $\mathbb{P}^2$ is the space of hyperplanes containing $\ell$.

Since the map is given by derivatives of an equation of $X$, it has degree $3$, so it is either 1:1 on a cubic or 3:1 on a line. In the first case the cubic spans the whole $\mathbb{P}^2$, so the intersection of all tangent hyperplanes along $\ell$ is $\ell$ itself. In the second case there is a plane $P \supset \ell$ which is everywhere tangent along $\ell$.

I believe if the double line is contained in $X$ we are in the second case. Now these two cases can be used to distinguish the normal of $\ell$ in $X$ as follows. First we can compute by adjunction $c_1(N_{\ell, X}) = -1$. Since a vector bundle on a line splits, we must have $N_{\ell, X} = \mathcal{O}(e_1) \oplus \mathcal{O}(e_2)$, with $e_1 + e_2 = -1$. Since this is a subbundle of $N_{\ell, \mathbb{P}^4} = \bigoplus \mathcal{O}(1)$, each $e_i \leq 1$, leaving the two cases you mentioned, namely $(e_1, e_2) = (0, -1)$ or $(1, -2)$.

Now it comes the part where I actually did not do the computations, but it should be the same as the case of a cubic fourfold, where I have worked everything out. Namely you can distinguish the two cases according to the number of sections of $N_{\ell, X}$. You have to write explicitly a generic section of $N_{\ell, \mathbb{P}^4}$; these form a space of dimension $6$.

Then you impose that such a section is actually tangent to $X$; this lowers the dimension by something which depends on the derivatives of $X$. If you do the computation, you will find that this dimension is different according to the two cases I have described above. I believe it turns out that $h^0(\ell, N_{\ell, X} = 1$ in the first case and $2$ in the second (which should be the one where the line is double).

A final remark: to perform the above computation it may be easier to observe that the normal exact sequence for the inclusion of the line into $X$ splits and work with sections of $T_X |_\ell$.

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Thank you for your answer! Let us make clear a few points: when you say "I assume $X$ smooth" you mean smooth out of $l$ I suppose. Then what do you mean by "degree" of a map? I agree that the restriction of the gauss map is given by cubics but if it is 1:1 then the degree is 1. –  IMeasy Feb 19 '10 at 15:19
    
No, I'm actually using that it is smooth along l. Indeed I am using the Gauss map to make the computations. This is not in contrast with the assumption that it contains a double line, if this is what you have in mind (for instance a plane contains a double line). By degree here I am meaning the degree of the divisor giving the map, which is the product of the topological degree of the map and the degree of the image. In this case saying it is three just means the map is given by polynomials of degree 3. Sorry for not being precise about this. –  Andrea Ferretti Feb 19 '10 at 15:48
    
yeah, very good. I just missed that point. I just supposed it was singular... this simplifies everything and I agree with the rest of your writing. Thank you, that was of great help! –  IMeasy Feb 19 '10 at 17:06
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