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 $\mathcal O_Y$?

  • $\begingroup$ I still couldn't figure out 1) and 2). I understand that taking dual commutes with exterior powers, but why only for the highest exterior power, viz Cor 31 of therisingsea.org/notes/TensorExteriorSymmetric.pdf. However, I posted an answer for the last question (the easy one) below. $\endgroup$ – Rodrigo Jun 17 '13 at 17:07
  • $\begingroup$ For 2) it looks like he is using tensor hom adjunction, but that doesn't preserve isomorphisms, so I'm confused. Maybe somehow in this case it does? $\endgroup$ – Seth Feb 2 '15 at 16:26
  • $\begingroup$ Ah, he is using the fact that invertible sheaves form a group with tensor product and multiplying by the inverse on both sides, which is the dual. $\endgroup$ – Seth Feb 6 '15 at 5:01

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.).

  • $\begingroup$ Perhaps you should explicitly mention that exterior powers are preserved by base change (and hence localisation). $\endgroup$ – Zhen Lin Jun 11 '13 at 21:53
  • $\begingroup$ This is already needed for the definition of exterior powers in the ad-hoc sense of Hartshorne, therefore I didn't mention it. $\endgroup$ – Martin Brandenburg Jun 12 '13 at 8:26
  • $\begingroup$ Thank you Martin. I am figuring out the isomorphism right now, but why is it important that exterior powers are preserved by base change? $\endgroup$ – Rodrigo Jun 12 '13 at 20:51
  • $\begingroup$ @ZhenLin, can you elaborate a little bit on what you mean by base change? I can't find any necessary extra assumptions for the isomorphism that Martin wrote down. In fact, not even that this natural paring only works for the n-th exterior power, where n=r is the rank of the module. Indeed, Prop 30 & Cor 31 of these notes therisingsea.org/notes/TensorExteriorSymmetric.pdf suggest that one always has the isomorphisms for all n≥1. $\endgroup$ – Rodrigo Jun 15 '13 at 21:29
  • $\begingroup$ @MartinBrandenburg would you say that I have explained my notation (Y, X, I, etc.) well enough so far? $\endgroup$ – Rodrigo Jul 1 '13 at 13:26

To get the isomorphism $\mathcal {I/I^2} \cong \mathcal L^{-1} \otimes \mathcal O_Y$ note that $\mathcal O_Y = \mathcal O_X/\mathcal I$. So the result follows from the same reasoning as Hartshorne 8.9.1 $\mathcal O_{\Delta X}$-module structure on $\mathcal I$ := the kernel of the diagonal morphism.


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