Calculating the decomposition of a vector bundle over rational curve - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T20:59:41Z http://mathoverflow.net/feeds/question/52535 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/52535/calculating-the-decomposition-of-a-vector-bundle-over-rational-curve Calculating the decomposition of a vector bundle over rational curve Mohammad F.Tehrani 2011-01-19T17:23:11Z 2011-03-05T20:41:37Z <p>Consider the rational curve (conic) given by image of the map</p> <p>$$u([z,w])=[z^2,-z^2,w^2,-w^2,zw] \in \mathbb{P}^4$$</p> <p>which lies in quintic 3-fold $X: x_1^5+\cdots+x_5^5- x_1\cdots x_5=0$.</p> <p>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 $X$.</p> <p>How should I calculate $a,b$ for this (or any other) explicitly given map?</p> http://mathoverflow.net/questions/52535/calculating-the-decomposition-of-a-vector-bundle-over-rational-curve/52597#52597 Answer by Sasha for Calculating the decomposition of a vector bundle over rational curve Sasha 2011-01-20T09:32:33Z 2011-01-20T09:32:33Z <p>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 $N = O(-1)^2$.</p> http://mathoverflow.net/questions/52535/calculating-the-decomposition-of-a-vector-bundle-over-rational-curve/52598#52598 Answer by Sándor Kovács for Calculating the decomposition of a vector bundle over rational curve Sándor Kovács 2011-01-20T09:43:21Z 2011-01-20T17:33:41Z <p>Sheldon Katz has computed this in general in <a href="http://www.numdam.org/item?id=CM_1986__60_2_151_0" rel="nofollow">On the finiteness of rational curves on quintic threefolds</a>. 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.</p> http://mathoverflow.net/questions/52535/calculating-the-decomposition-of-a-vector-bundle-over-rational-curve/52619#52619 Answer by Francesco Polizzi for Calculating the decomposition of a vector bundle over rational curve Francesco Polizzi 2011-01-20T13:53:30Z 2011-01-20T15:48:41Z <p>Since you know the explicit equation of the conic, you can compute everything by using Macaulay2.</p> <p>The following script should be clear:</p> <pre><code>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 &lt;--- 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) 4 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 </code></pre> <p>The output reads</p> <p>$h^0(X, N_{C|X}^{*})=4, \quad h^1(X, N_{C|X}^{*})=0, \quad h^0(X, N_{C|X})=0, \quad h^1(X, N_{C|X})=0$,</p> <p>hence <code>$N_{C|X}=\mathcal{O}_{\mathbb{P}^1}(-1) \oplus \mathcal{O}_{\mathbb{P}^1}(-1)$</code>, according to Sasha's and Sandor's answers. </p>