On page 182 Hartshorne argues that $\omega_X \otimes \mathcal O_Y = \omega_Y \wedge^r (\mathcal I / \mathcal I^2)$, where Y is a nonsingular subvariety of codimension r in the nonsingular variety X over k, and $\omega$ is the canonical sheaf. Then, he writes: "Since formation of the highest exterior power commutes with taking the dual sheaf, we find $\omega_Y \cong \omega_X \otimes \wedge^r \mathcal N_{Y/X} $". Here $\mathcal N$ is the normal sheaf. There are two things that I would like to ask at this point:
1) Why does only the highest exterior power commute with the dual sheaf?
2) Assuming anyway that they commute, after taking the dual sheaf, which is exact in this case (cf. Is the dualizing functor $\mathcal{Hom}( \cdot, \mathcal{O}_{X})$ exact?), we would be now taking the highest exterior power of the SES written just before the proposition, i.e. $$0\to \mathcal T_Y \to \mathcal T_X \otimes \mathcal O_Y \to \mathcal N_{Y/X} \to 0$$which should give us now $\bigwedge \mathcal T_X \otimes \mathcal O_Y \cong \bigwedge \mathcal T_Y \otimes \mathcal N_{Y/X}$. How does one now get the isomorphisms $\bigwedge \mathcal T_X \otimes \mathcal O_Y \cong \omega_Y$ and $\bigwedge \mathcal T_Y \cong \omega_X$?
Edit: now I realize that maybe Hartshorne is not looking at the dual SES, but possibly he is going from the isomorphism $\omega_X \otimes \mathcal O_Y = \omega_Y \otimes \wedge^r (\mathcal I / \mathcal I^2)$ to the result $\omega_Y \cong \omega_X \otimes \wedge^r \mathcal N_{X/Y}$ by taking the dual on the congruence rather than on the exact sequence itself. How does this work though?
Finally, just to finish the proof, I would like to ask how one gets the isomorphism $\mathcal {I/I^2} \cong \mathcal L^{-1} \otimes \mathcal O_Y.$ In other words, how is factoring by $\mathcal {I^2}$ equivalent to tensoring by $\mathcal O_Y$?
Best Answer
1) comes down to a statement from multilinear algebra: if $V$ is a free $k$-module of rank $n$, there is a natural pairing $\wedge^n(V) \otimes \wedge^n(V^*) \to k$ given by mapping $v_1 \wedge \dotsc \wedge v_n \otimes \phi_1 \wedge \dotsc \wedge \phi_n$ to the determinant of the matrix $(\phi_i(v_j))$. It is perfect, i.e. induces a natural isomorphism $\wedge^n(V)^* \cong \wedge^n(V^*)$. You can check this directly using bases.
For the rest of the questions, you should explaind your notation ($Y,X,I$, etc.).