I have a question about an argument used in the proof on Lemma 7.4 (Chap. III) from Hartshorne's "Algebraic Geometry":
Rmk:We work with the setting from 7.3: So $X $ is projective, therefore closed subscheme of $P:= \mathbb{P}^N_k$ of codimension $r$.
In the proof – using the observation that the $\mathscr{J}^i=\mathscr{Hom}_P(O_V,\mathscr{I}^i) $ are injective – we conclude that the complex $\mathscr{J}^{\bullet}$ splits to $\mathscr{J}^{\bullet}=\mathscr{J}^{\bullet}_1 \oplus \mathscr{J}^{\bullet}_2$ where
–$\mathscr{J}^{\bullet}_1$ is in degrees $0 \le i \le r$ (non zero) and is exact (so $h^i(\mathscr{J}^{\bullet}_1)=0$ for $0 \le i \le r$
–$\mathscr{J}^{\bullet}_2$ is in degrees $i \ge r$ (non zero)
Then we can conclude (unsure why) that $$\omega^{\circ}_X = ker(d^r:\mathscr{J}^{r}_2 \to \mathscr{J}^{r+1}_2)$$
I think that this is because by definition $\omega^{\circ}_X = \mathscr{Ext}^r _P(O_X, \omega_P)$. Then following the argumentation above we get $\mathscr{Ext}^r _P(O_X, \omega_P)=h^i(Hom_X(O_X,\mathscr{J}^{\bullet}) = h^i(\mathscr{J}^{\bullet}) $ since $Hom_X(O_X,\mathscr{J}^{\bullet}) \cong \mathscr{J}^{\bullet}$. Is this a correct argument?
The QUESTION is that I don't understand why this implies
$$Hom_X(\mathcal{F},\omega^{\circ}_X)= Ext^r _P(\mathcal{F}, \omega_P)$$?
My ideas:
By above & definition of cohomology we know $$Ext^r _P(\mathcal{F}, \omega_P)=h^r(Hom_X(\mathcal{F}, \mathscr{J}^{\bullet}))= \frac{ker(\bar{d}^r: Hom_X(\mathcal{F}, \mathscr{J}^{r}_1 \oplus \mathscr{J}^{r}_2) \to Hom_X(\mathcal{F}, \mathscr{J}^{r+1}_2))}{im(\bar{d}^{r-1}: Hom_X(\mathcal{F}, \mathscr{J}^{r-1}_1) \to Hom_X(\mathcal{F}, \mathscr{J}^{r}_1 \oplus \mathscr{J}^{r}_2))}$$
But why does the last expression coinside with $Hom_X(\mathcal{F},ker(d^r:\mathscr{J}^{r}_2 \to \mathscr{J}^{r+1}_2))$?
Best Answer
First, you should know that the way Hartshorne sets up the decomposition $J=J_1\oplus J_2$, there are no nonzero maps from any term of $J_1$ to any term of $J_2$ or vice-versa. So the kernels and cokernels of this sequence are just the direct sums of the kernels and cokernels from the complexes $J_1$ and $J_2$, and the homology of $J$ is just the direct sum of homologies from $J_1$ and $J_2$. Since $J_1$ is exact, that means it contributes nothing to the homology at any step, and since $J_2^i$ is zero for $i<r$, that means the homology at the $r^{th}$ term is just $\ker(J_2^{r}\to J_2^{r+1})$, which explains why this is isomorphic to $\omega_X^\circ$.
Next, you should observe that $J_1$ being a split exact sequence of injective objects means that it's isomorphic to a direct sum of complexes of the form $0\to A \stackrel{id}{\to} A \to 0$ up to a shift in degree with $A$ injective. So $\operatorname{Hom}(F,J_1)$ can be decomposed as a direct sum of complexes of the form $0\to\operatorname{Hom}(F,A)\stackrel{id}{\to} \operatorname{Hom}(F,A)\to 0$ up to shifts in degree, and this means it has trivial homology. So we get that the complexes $\operatorname{Hom}^\bullet(F,J)=\operatorname{Hom}^\bullet(F,J_1\oplus J_2) = \operatorname{Hom}^\bullet(F,J_1)\oplus \operatorname{Hom}^\bullet(F,J_2)$ and $\operatorname{Hom}^\bullet(F,J_2)$ have equal homology. But the first computes $\operatorname{Ext}^\bullet_P(F,\omega_P)$ while the second (after a grading shift by $r$) computes $\operatorname{Ext}^\bullet_X(F,\omega_X^\circ)$ since the shifted $J_2$ is an injective resolution of $\omega_X^\circ$ by the work we did in the first paragraph. Then $\operatorname{Ext}^0_X(F,\omega_X^\circ)=\operatorname{Hom}_X(F,\omega_X^\circ)$ and we're done.