Let $k$ be an odd number, and consider the Veronese embedding of degree $2$ $$ \mathbb{P}^{2^lk-1}\rightarrow \mathbb{P}^{N-1},$$ where $N = \binom{2^lk+1}{2}$. Now, we compose this embedding with projection to a generic projective space of dimension $2^{2l-1}k^2-1$. Let $V$ be the image through of this composition. Let $\omega_1$ be the canonical bundle of $V$ and let $\omega_2$ be the canonical bundle of $\mathbb{P}^{2^{2l-1}k^2-1}$. Then, $$\omega_1\simeq\mathcal{E}xt^r(i_{*}\mathcal{O}_V,\omega_2),$$ where $r$ is the codimension of $V$. In this situation, there exists unique line bundles $L_1,L_2$ such that $L_i^2\simeq\omega_i$. Is there an analogous formula as with the canonical bundles relating $L_1$ with an $r$-th derived functor involving $L_2$?
1 Answer
I'm assuming that you chose the numbers so the general projection of the Veronese embedding would still be an embedding, otherwise the claim about the existence of $L_1$ would need more explanation. In other words, let's assume that $V\simeq \mathbb P^{2^lk-1}$.
What you are writing is actually not entirely correct. The left hand side should be $i_*\omega_1$. Of course, you could follow the usual convention that the $i_*$ is suppressed, but you didn't follow that on the right hand side... Anyway, that's not a big deal.
Now, a trivial answer, which I suspect is not what you expect is that yes there is a similar formula, namely $$i_*L_1^2\simeq\mathcal{E}xt^r(i_{*}\mathcal{O}_V,L^2_2).\tag{$\#$}$$ Perhaps you would want $L_2$ in place of $\omega_2$? You can do that, but you will not get $L_1$. The way Ext groups and sheaves work, you can "flip" line bundles from one side to the other, so you can rewrite this as $$i_*L_1^2\simeq\mathcal{E}xt^r(i_{*}\mathcal{O}_V,L_2)\otimes L_2.\tag{$\star$}$$ Now, you can try to figure out how $L_1$ and $L_2$ are related.
By the projection formula $i_*(L_1^2\otimes i^*L_2^{-1})\simeq i_*L_1^2\otimes L_2^{-1}$ so by $(\star)$ we get that $$i_*(L_1^2\otimes i^*L_2^{-1})\simeq \mathcal{E}xt^r(i_{*}\mathcal{O}_V,L_2).$$ Now, you can write down $L_2$ in terms of $\mathscr O(1)$ of the ambient projective space and then its restriction is just multiplied by the degree of $V$, so you can write $i^*L_2$ in terms of the $\mathscr O(1)$ of $V$ and do the same for $L_1$. It seems to me that this will end up as some power of $L_1$, so eventually you get something like $$i_*L_1^q\simeq \mathcal{E}xt^r(i_{*}\mathcal{O}_V,L_2),$$ where this $q$ is computed from your particular choices of dimensions.
Similarly, you can get a formula where $L_1$ takes the place of $\omega_1$. For that, you need that there is an $L_3$ on $V$ such that $L_1=L_3^2$. I believe in your situation that holds if $l>1$. Actually the point is that you need an $L_4$ on the ambient projective space so that $L_1=i^*L_4$. If that holds, then $(\#)$ tells you that $$i_*L_1\otimes L_4\simeq i_*L_1^2\simeq\mathcal{E}xt^r(i_{*}\mathcal{O}_V,L^2_2),$$ so $$i_*L_1\simeq\mathcal{E}xt^r(i_{*}\mathcal{O}_V,L^2_2\otimes L_4^{-1}).$$ You can write both $L_2$ and $L_4$ as powers of $\mathscr O(1)$ to simplify this.
The same arguments actually can be used for a more general situation: Consider an embedding $\mathbb P^m\simeq V\overset i\hookrightarrow W\simeq \mathbb P^M$. Assume that there exist (unique) line bundles $\mathscr L_1$ on $V$ and $\mathscr L_2$ on $W$ such that $\omega_V\simeq \mathscr L_1^{\otimes r_1}$ and $\omega_W\simeq \mathscr L_2^{\otimes r_2}$. Then you can find a similar connection between $\mathscr L_1$ and $\mathscr L_2$.
Also, the isomorphism you are referring to is actually a special case of Grothendieck duality and can be written (in a bit more complicated way) as $$ i_*\omega_V^{\bullet}\simeq Ri_*R\mathscr Hom_V(\mathscr O_V, \omega_V^\bullet)\simeq R\mathscr Hom(Ri_*\mathscr O_V,\omega_W^\bullet), $$ where $\omega_V^\bullet\simeq \omega_V[m]$ and $\omega_W^\bullet\simeq \omega_W[M]$ are the dualizing complexes of $V$ and $W$ respectively. As $V$ is non-singular (you really only need that it is Gorenstein), the left hand side is only supported in degree $-m$, so it follows that so is the right hand side, and the degree $-m$ cohomology sheaf of that complex is exactly that Ext sheaf you wrote down. The advantage of this formulation is that this also includes the fact that the other Ext sheaves are zero and explains why it is the codimension that comes up. It also applies in similar situations where $V$ and $W$ are allowed to be other than projective spaces and perhaps even reducible and not equidimensional.
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$\begingroup$ Thanks a lot for your answer. Unfortunately, this si not what I need. I am interested in replacing $\omega_1$ by $L_1$ in the formula $$i^{*}\omega_1 \simeq \mathcal{E}xt^r(I^{*}\mathcal{O}_V,\omega_2).$$ Can something similar be done in this case? $\endgroup$ Commented Feb 15, 2023 at 9:53
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$\begingroup$ @JavierSendra, see the update. $\endgroup$ Commented Feb 15, 2023 at 18:21