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Consider the rational curve (conic) given by image of the map

$$ u([z,w])=[z^2,-z^2,w^2,-w^2,zw] \in \mathbb{P}^4 $$

which lies in quintic 3-fold $X: x_1^5+\cdots+x_5^5- x_1\cdots x_5=0$.

By Grothendick theorem and the fact that $X$ is Calabi-Yau, we know that $u^*N_C^X= O(a) \oplus O(b)$, for some $a+b=-2$, where $N_C^X$ is the normal bundle of $C=imgae(u)$ in $X$.

How should I calculate $a,b$ for this (or any other) explicitly given map?

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    $\begingroup$ You should be careful - it is not necessarily the case that the inclusion of \OO(2) into u^*T_X splits. Also, your first calculation of u^*TP^4 seems fine... the \OO(10) term will not appear. $\endgroup$
    – mdeland
    Jan 19, 2011 at 17:39
  • $\begingroup$ @ Andrei: You are right about 1/5 factor,I'll edit my question $\endgroup$
    – Mohammad Farajzadeh-Tehrani
    Jan 19, 2011 at 18:59
  • $\begingroup$ @ mdeland: Why not ? u^*TX= TC+ u^* N_C where N_C is the normal bundle and TC is the tangent bundle to this smooth curve. because map is degree 2, and normal bundle of quintic is of degree 5 this gives a sub-bundle of degree 10 in pullback. $\endgroup$
    – Mohammad Farajzadeh-Tehrani
    Jan 19, 2011 at 19:03
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    $\begingroup$ Mohammed, the normal bundle is a quotient of the tangent bundle. This doesn't mean that $\mathcal{O}(10)$ will appear in the decomposition of $u^*T\mathbb{P}^4$, just that it has $\mathcal{O}(10)$ as a quotient. The existence of the short exact sequence $\mathcal{O} \to \mathcal{O}(1)^{\oplus 2} \to \mathcal{O}(2)$ should convince that having a line bundle as a quotient of a vector bundle doesn't mean it will appear in the decomposition of the same vector bundle. (This comment is really the dual of mdeland's comment.) $\endgroup$
    – Arend Bayer
    Jan 19, 2011 at 19:04
  • $\begingroup$ Sorry you are right. stupid mistake.I'll edit my question. So then How can I find a and b satisfying $N_C^X=O(a)\oplus O(b)$? $\endgroup$
    – Mohammad Farajzadeh-Tehrani
    Jan 19, 2011 at 22:53

3 Answers 3

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Let $(z,w) \mapsto (f_1(z,w),\dots,f_5(z,w)$, $\deg f_i = s$, be a map $P^1 \to P^4$ and $g(x_1,\dots,x_5)$, $\deg g = d$, be an equation of a hypersurface containing the image. Then the normal bundle is the middle cohomology of the following complex $$ O(1)^2 \to O(s)^5 \to O(ds) $$ where the first map is given by the matrix $(\partial f_i/\partial z,\partial f_i/\partial w)$, and the second map is given by the matrix $(\partial g/\partial x_i)(f_1,\dots,f_5)$. In the case of a conic in a quintic the complex takes form $$ O(1)^2 \to O(2)^5 \to O(10), $$ so the normal bundle should be $O(-1+a) \oplus O(-1-a)$. Then it is clear that $a = \dim H^0(N) = \dim H^1(N)$, you should use the above complex (with explicit differentials) to compute $H^0(N)$. Passing to the cohomology of the above complex you obtain $$ k^4 \to k^{15} \to k^{11}. $$ Then $H^1(N)$ is the cokernel of the second map. So, you just have to compute the rank of this map. My computation shows that in your specific case the map is surjective, hence $a = 0$. So, the final answer is $N = O(-1)^2$.

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  • $\begingroup$ Could you elaborate on where that complex came from? And this is not a short exact sequence, how can you 'pass to cohomology'? $\endgroup$
    – J.C. Ottem
    Jan 20, 2011 at 9:43
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    $\begingroup$ By passing to cohomology I mean the hypercohomology spectral sequence. Since $N$ is the only cohomology of the complex, it converges to $H^\bullet(N)$. The complex is a combination of well known exact sequences $0 \to O \to O(1)^2 \to T_{P^1} \to 0$, $0 \to O \to O(s)^5 \to u^*T_{P^4} \to 0$, $0 \to T_{P^1} \to u^*T_{P^4} \to N_{P^1/P^4} \to 0$ and $0 \to N_{P^1/X} \to N_{P^1/P^4} \to u^*N_{X/P^4} \to 0$ and an isomorphism $N_{X/P^4} \cong O(d)$. $\endgroup$
    – Sasha
    Jan 20, 2011 at 9:55
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    $\begingroup$ To combine the sequences you should observe that the map $T_{P^1} \to u^*T_{P^4}$ lifts to $O(1)^2 \to O(s)^5$, and the lifting is given by my first map, so the cokernel of the first map is $N_{P^1/P^4}$. Then you should also observe that the composition $O(s)^5 \to u^*T_{P^4} \to u^*N_{X/P^4}$ is given by my second map. $\endgroup$
    – Sasha
    Jan 20, 2011 at 9:56
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Since you know the explicit equation of the conic, you can compute everything by using Macaulay2.

The following script should be clear:

i1 : k=ZZ/32003;

i2 : ringP1=k[x, y];

i3 : ringP4=k[z1, z2, z3, z4, z5];

i4 : I= ideal(z1^5+z2^5+z3^5+z4^5+z5^5-z1*z2*z3*z4*z5);

o4 : Ideal of ringP4

i5 : ringQuintic=ringP4/I;

i6 : conicMap=map(ringP1, ringQuintic, {x^2, -x^2, y^2, -y^2, x*y});

o6 : RingMap ringP1 <--- ringQuintic

i7 : conic=image conicMap;

i8 : IC=ideal conic;

o8 : Ideal of ringQuintic

i9 : ConormalModuleConic = IC/IC^2;

i10 : ConormalSheafConic= sheaf ConormalModuleConic;

i11 : NormalSheafConic= dual sheaf ConormalModuleConic;

i12 : HH^0(ConormalSheafConic)

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o12 = k

o12 : k-module, free

i13 : HH^1(ConormalSheafConic)

o13 = 0

o13 : k-module

i14 : HH^0(NormalSheafConic)

o14 = 0

o14 : k-module

i15 : HH^1(NormalSheafConic)

o15 = 0

o15 : k-module

The output reads

$h^0(X, N_{C|X}^{*})=4, \quad h^1(X, N_{C|X}^{*})=0, \quad h^0(X, N_{C|X})=0, \quad h^1(X, N_{C|X})=0$,

hence $N_{C|X}=\mathcal{O}_{\mathbb{P}^1}(-1) \oplus \mathcal{O}_{\mathbb{P}^1}(-1)$, according to Sasha's and Sandor's answers.

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  • $\begingroup$ Its enough to know $h^1(N) = 0$. $\endgroup$
    – Sasha
    Jan 20, 2011 at 14:19
  • $\begingroup$ Yes I know, but I wanted to be sure that the script was correct... $\endgroup$
    – Francesco Polizzi
    Jan 20, 2011 at 14:23
  • $\begingroup$ Cubic should be Quintic! $\endgroup$
    – Sasha
    Jan 20, 2011 at 15:40
  • $\begingroup$ interesting!... $\endgroup$
    – Mohammad Farajzadeh-Tehrani
    Jan 20, 2011 at 16:24
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    $\begingroup$ Macauley2 is a standard computer-algebra system for commutative algebra and algebraic geometry. You can download it for free at math.uiuc.edu/Macaulay2, and a good manual is "Computations in algebraic geometry with Macaulay 2", edited by David Eisenbud, Daniel R. Grayson, Michael E. Stillman, and Bernd Sturmfels. The script I wrote is very elementary, and if you read it you can guess by yourself what it is doing. If you install the software on your pc and you try to run the script, you will see that it works very naturally $\endgroup$
    – Francesco Polizzi
    Jan 20, 2011 at 16:58
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Sheldon Katz has computed this in general in On the finiteness of rational curves on quintic threefolds. See Appendix B on pp. 158-159. He gives a list of the possibilities based on the equations. It is rather simple to check the condition. According to my computation the answer in the above case is $\mathcal O_{\mathbb P^1}(-1)\oplus \mathcal O_{\mathbb P^1}(-1)$, but it is easy to produce explicit examples with other normal bundles.

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  • $\begingroup$ Maybe you mean $\mathcal{O}_{P^1}(-1) \oplus \mathcal{O}_{P^1}(-1)$? $\endgroup$
    – Francesco Polizzi
    Jan 20, 2011 at 13:56
  • $\begingroup$ Yes, $C\simeq \mathbb P^1$, and less $\#$ of characters. :) But I will correct it. $\endgroup$
    – Sándor Kovács
    Jan 20, 2011 at 17:33
  • $\begingroup$ I did not want to be pedantic :-) But since $C$ is $\mathbb{P}^1$ embedded as a conic, one could read $\mathcal{O}_C(-1)$ as a line bundle of degree $-2$, whereas in this way there is no ambiguity $\endgroup$
    – Francesco Polizzi
    Jan 20, 2011 at 17:41
  • $\begingroup$ Sure. I understood that. I just meant to explain the reason I wrote it that way. But you're absolutely right. $\endgroup$
    – Sándor Kovács
    Jan 20, 2011 at 18:50

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