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Suppose $Y$ is a projective variety over a field $k$. Fix an embedding $\iota: Y \hookrightarrow \mathbb{P}^n_k$ for some $n$, and consider the local cohomology sheaves $\mathcal{H}^j_Y(\omega_{\mathbb{P}^n})$ of the canonical sheaf on $\mathbb{P}^n$ supported on $Y$. I believe that the sheaf cohomology groups

$H^i(\mathbb{P}^n, \mathcal{H}^{n+1-j}_Y(\omega_{\mathbb{P}^n}))$

should be independent of $n$ and of the embedding, for all $i > 0$. (By considering embeddings $Y = \mathbb{P}^0 \hookrightarrow \mathbb{P}^0$ and $Y = \mathbb{P}^0 \hookrightarrow \mathbb{P}^1$, is is not hard to see that this claim fails for $i = 0$.)

As in Hartshorne's paper On the de Rham cohomology of algebraic varieties, one strategy for proving such a statement is to consider two embeddings $\iota: Y \hookrightarrow \mathbb{P}^n$ and $\iota': Y \hookrightarrow \mathbb{P}^m$, together with the diagonal embedding $\Delta: Y \hookrightarrow X = \mathbb{P}^n \times \mathbb{P}^m$. If $\pi: X \rightarrow \mathbb{P}^n$ is the projection, we have $\iota = \pi \circ \Delta$. By symmetry, it would be enough to show that, given such data, we must have

$H^i(\mathbb{P}^n, \mathcal{H}^{n+1-j}_Y(\omega_{\mathbb{P}^n})) \simeq H^i(X, \mathcal{H}^{n+m+1-j}_Y(\omega_X))$ for all $i > 0$.

The sheaves $\mathcal{H}^{n+m+1-j}_Y(\omega_X)$ (indeed, any quasi-coherent $\mathcal{O}_X$-modules supported on $\Delta(Y)$) are $\pi_*$-acyclic: given any $P \in \mathbb{P}^n$, $\pi^{-1}(\{P\}) \cap \Delta(Y)$ contains at most one point, so we can use, e.g., the theorem on formal functions to show the higher direct images have zero stalks. Therefore we have

$H^i(X, \mathcal{H}^{n+m+1-j}_Y(\omega_X)) \simeq H^i(\mathbb{P}^n, \pi_*\mathcal{H}^{n+m+1-j}_Y(\omega_X))$ for all $i \geq 0$.

To compute the local cohomology sheaves of $\omega_X$ and $\omega_{\mathbb{P}^n}$, we can use their Cousin resolutions (since the sheaves are line bundles, these are injective resolutions). In fact, if $E^{\bullet}(\omega_X)$ (resp. $E^{\bullet}(\omega_{\mathbb{P}^n})$) are these resolutions, we have a quasi-isomorphism $\pi_*E^{\bullet}(\omega_X)[m] \xrightarrow{\sim} E^{\bullet}(\omega_{\mathbb{P}^n})$ of complexes of sheaves on $\mathbb{P}^n$. The reason for this is that the map $\pi$ can be viewed as a "projective space" $\mathbb{P}^m_{\mathbb{P}^n} \rightarrow \mathbb{P}^n$, and so the corresponding trace map is a derived category isomorphism (Proposition III.4.3 of Residues and duality) This trace map takes the form $\mathbf{R}\pi_*\pi^!E^{\bullet}(\omega_{\mathbb{P}^n}) \rightarrow E^{\bullet}(\omega_{\mathbb{P}^n})$, and that the left-hand side coincides with $\pi_*E^{\bullet}(\omega_X)[m]$ is essentially worked out on pp. 31-32 of Hartshorne's de Rham paper.

It follows that we can use $\pi_*E^{\bullet}(\omega_X)[m]$ as an injective resolution of $\omega_{\mathbb{P}^n}$. Then we get

$\mathcal{H}^{n+1-j}_Y(\omega_{\mathbb{P}^n}) = h^{n+m+1-j}(\underline{\Gamma}_Y(\pi_*E^{\bullet}(\omega_X)))$, where $\underline{\Gamma}_Y$ is the sheafified "sections with support on $Y$" functor.

On the other hand, since $E^{\bullet}(\omega_X)$ is an injective resolution of $\omega_X$, we can compute $\pi_*\mathcal{H}^{n+m+1-j}_Y(\omega_X)$ as $h^{n+m+1-j}(\pi_*\underline{\Gamma}_YE^{\bullet}(\omega_X))$. Here I've used the fact that $\pi_*$ and cohomology objects commute when applied to a complex of quasi-coherent sheaves on $X$ supported on $\Delta(Y)$, again by the acyclicity argument.

So finally, we are comparing the sheaves $h^{n+m+1-j}(\pi_*\underline{\Gamma}_YE^{\bullet}(\omega_X))$ and $h^{n+m+1-j}(\underline{\Gamma}_Y(\pi_*E^{\bullet}(\omega_X)))$ on $\mathbb{P}^n$. The functors $\underline{\Gamma}_Y$ and $\pi_*$ do not commute, and in fact the global sections of these sheaves need not be the same. Is there a way to see that their higher sheaf cohomologies on $\mathbb{P}^n$ are the same? In general, is there a homological-algebraic gadget to handle the situation of complexes of sheaves which are not quasi-isomorphic, but whose cohomology objects should have identical higher sheaf cohomology?

Edited to add (5/4) If $k$ has characteristic $p > 0$, or if $Y$ is smooth and $k$ has characteristic zero, I believe the independence holds. This is because the $k$-dimension of $H^i(\mathbb{P}^n, \mathcal{H}^{n+1-j}_Y(\omega_{\mathbb{P}^n}))$ is the Lyubeznik number $\lambda_{i+1,j}(Y)$ of $Y$, known already to be independent of the embedding in these cases. The embedding-independence of Lyubeznik numbers in characteristic zero for singular $Y$ is still open, and the above question comes from an attempt to attack this in a characteristic-free way (of course, it says nothing about $\lambda_{0,j}(Y)$ or $\lambda_{1,j}(Y)$, which are sure to be harder).

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    $\begingroup$ I have some doubts about this. For a smooth projective variety $Y$ of dimension $d$ and a closed immersion $\iota$, there is only one nonzero "local cohomology" sheaf, namely $\mathcal{H}^{n-d}_Y(\omega_{\mathbb{P}^n})$, and this equals the colimit of the dualizing sheaves of all infinitesimal thickenings of $Y$ in $\mathbb{P}^n$. Because of this, the higher cohomology has to do with the higher cohomology of symmetric powers of the normal bundle of $\iota$. I believe this might depend on the embedding. $\endgroup$ – Jason Starr May 4 '17 at 13:25
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    $\begingroup$ Continuing that comment: $H^{d+i}(\mathbb{P}^n,\mathcal{H}_Y^{n-d}(\omega_{\mathbb{P}^n}))$ is independent of the embedding for $i\geq 0$ (the group is zero if $i>0$, and it is canonically isomorphic to $k$ for $i=0$). So any examples would need to have $d\geq 2$. The examples that worry me have $d\geq 3$. If there is a pair of embeddings both of which have $h^1$ vanish for $\oplus_{d\geq 1}\text{Sym}^d(N_{Y/\mathbb{P}^n})\otimes \omega_Y$, yet precisely one has nonzero $h^2$, then $H^2(\mathbb{P}^n,\mathcal{H}^{n-d}_Y(\omega_{\mathbb{P}^n}))$ is different for the two embeddings. $\endgroup$ – Jason Starr May 4 '17 at 13:39
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    $\begingroup$ For $d$ equal to $5$, for a smooth projective $5$-fold with vanishing Dolbeault cohomology groups $H^{p,q}=H^q(Y,\Omega^p_{Y/k})$ with $(p,q) = (4,2),(4,3),(5,2),$ and $(5,3)$, for two embeddings by very ample invertible sheaves $\mathcal{L}_i$, $i=1,2$ with $h^1$ equal to $0$, but with $h^2(Y,\mathcal{L}_1)\neq 0$ yet $h^2(Y,\mathcal{L}_2)=0$, then for the associated first order infinitesimal neighborhoods, $Y_1$, resp. $Y_2$, $H^2(Y,\omega_Y)\to H^2(Y_1,\omega_{Y_1})$ is not surjective, yet $H^2(Y,\omega_Y)\to H^2(Y_2,\omega_{Y_2})$ is surjective. That suggests dependence on the embedding. $\endgroup$ – Jason Starr May 4 '17 at 14:58
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    $\begingroup$ @Jason Starr: thanks for your comments. I will make sure to think about your specific example. I think a counterexample $Y$ can't be smooth, though: see the edit above. $\endgroup$ – Nick Switala May 4 '17 at 20:23
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It is false! As mentioned in the edit, a positive answer to this question would imply the Lyubeznik numbers $\lambda_{i,j}$ of a projective scheme are independent of the defining projective embedding for $i \geq 2$. The recent preprint

https://arxiv.org/abs/1803.07448

by T. Reichelt, M. Saito, and U. Walther, gives counterexamples (even equidimensional ones) in characteristic zero.

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