Calculating the decomposition of a vector bundle over rational curve 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?
 A: 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 $N = O(-1)^2$.
A: 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)

       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

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. 
A: 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.
