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Oct 2, 2012 at 17:14 comment added Sasha Of course knowing the generators is not enough, but if you know all the syzygies it may help. For example, in case of $Gr(2,5)$ the resolution has form $O(-5) \to O(-3)^5 \to O(-2)^5 \to O$. From this you see immediately that $\det N^* = O(-5)$ which does the job. Of course we are lucky here that the resolution is so simple. Alternatively, we could argue that there is an exact sequence $0 \to \Lambda^2N^* \to O(-3)^5 \to O(-2)^5 \to N^* \to 0$, which also allows to compute the degree of $N^*$.
Oct 2, 2012 at 13:53 comment added Damian Rössler I agree but that is because it is easily computable in that case. How do you propose to compute the normal bundle of a general smooth submanifold, if you know some generators of its homogenous ideal ?
Oct 2, 2012 at 13:51 comment added Sasha @Damian: Even for complete intersections you compute the canonical class by computing the degree of the normal bundle first.
Oct 2, 2012 at 10:20 comment added Damian Rössler I don't think that the normal bundle is any easier to compute than the canonical one...
Oct 2, 2012 at 10:00 history answered Sasha CC BY-SA 3.0