## Natural coherent sheaves on algebraic varieties

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Let X be a smooth projective variety of dimension d over an algebraically closed field k. Can you please list some examples of natural locally free or coherent sheaves that you can always construct on X, regardless whether X has some particular structure or is of general type?

I can only name the two obvious examples: the structural sheave and the sheaf of differentials (and obvious combinations of them, of course, which do not count).

I can believe that these are the only ones for curves, as the genus is the only discrete invariant available.

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The canonical sheaf and the tangent sheaf count as "combinations"? – Martin Brandenburg Mar 7 2011 at 15:37

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@Martin: by "combination" vic means something like "any other sheaf derived from these". The canonical is the determinant and the tangent is the dual of the sheaf of differentials. – Sándor Kovács Mar 7 2011 at 16:31

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You can take sheaves of differential operators $Diff_N$ of finite order $\le N$. It is related to the tangent sheaf $T_X$, but I don't see an obvious functor $T_X\mapsto Diff_N$. But I agree there aren't too many obvious examples. – Donu Arapura Mar 7 2011 at 16:31

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If you're working over $\mathbb C$ you should be able to construct lots of singular metrics on the line bundles of $X$. Every one of those gives a coherent multiplier ideal sheaf on $X$. (Of course this won't work if you admit the existence of algebraically closed fields different from $\mathbb C$.) – Gunnar Magnusson Mar 7 2011 at 16:53

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@Sándor: Ah yes, you only get ideal sheafs for pseudo-effective bundles, thanks for that (we can put a singular metric on any line bundle, but their curvature currents need to satisfy a positivity condition to define ideal sheaves - see tinyurl.com/493y6q3 starting with sections 4, 5 and 6). @unknowngoogle: No. Or at least very seldom. I've heard some talk of less singular metrics than others, but nothing widespread. I don't know of any Hermite-Einstein style criterions for singular metrics that would give a canonical choice either. – Gunnar Magnusson Mar 7 2011 at 20:09
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## 9 Answers

What about the locally free sheaf which is the direct sum of all n-torsion invertible sheaves, for some fixed integer n? Of course on many smooth, projective varieties this is zero. But it is a "canonically defined" locally free sheaf which is sometimes nonzero. And I see no obvious way to build this from the structure sheaf or the cotangent sheaf.

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-   Good point! – Sándor Kovács Mar 7 2011 at 22:27
Hi Jason. That's a good point, but your sheaf is only canonically defined up to isomorphism, right? It seems that the OP has to decide what exactly their question is... – James Borger Mar 8 2011 at 5:45
Thanks Jason, this is also interesting. James: there is no "correct answer" to my question, as every one can take his own interpretation. As for me is concerned, I already learned a lot reading all your answers, but I'll be happy to hear more, of course!  – vic Mar 9 2011 at 4:57

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One formal approach to a problem like this is motivated by the "formal geometry" of Gelfand-Fuchs (which is the subject of other MO questions). We can define a "natural bundle" on a class of spaces to be one associated to a representation of a structure group of the given geometry. This is a common approach in differential geometry, where there's a lot known about natural differential operators etc. But in any case in the current context we can define a natural bundle on smooth n-dimensional varieties as one associated to a representation of the group of changes of coordinates on a formal n-dimensional disc. Every smooth variety has a canonical principal bundle for this group (with fiber over $x\in X$ the variety of isomorphisms between the completion of $X$ at $x$ and the formal disc), and so given a representation of the group we get a vector bundle on any variety of the given dimension.

All of the bundles discussed in the answers above are of this form (including jet bundles, sheaves of differential operators etc). If you take this as a definition of a natural bundle it's easy to prove the conjecture that they're all extensions of powers of bundles of forms: the group of changes of coordinates is an extension of $GL_n$ (the structure group of the tangent bundle) by a pro-unipotent group (changes of coordinate with derivative the identity). Hence all representations have filtrations with associated graded bundles associated to the (frame bundle of the) tangent bundle in the usual sense (ie all the Schur functors of the tangent bundle, like forms).

Edit: it's interesting to note that if we take representations of this group which extend to the Lie algebra of all derivations of formal power series in n variables (ie we have an action of $\partial_x$'s on the module as well) we get "all" natural flat bundles on smooth varieties --- such as the sheaves of all jets or all differential operators.

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Beautiful answer!  +1 – Qfwfq Mar 7 2011 at 17:49
It seems to me that his viewpoint assumes that the construction of the sheaf is local'', e.g. the restriction to an open analytic subset $U$ only depends on $U$. – ABayer Mar 7 2011 at 17:52
Absolutely - it's a universal local (or even formal) construction, based on the fact that we know what smooth varieties look like formally. It would be interesting to see if there are any natural bundles defined in a nonlocal way, eg using an integral kernel not supported on the diagonal. If you're willing to be only quasicoherent you could use things like the structure sheaf of $X\times X\setminus \Delta$ eg -- take your favorite local construction $F$ and assign $x\mapsto H^*(X\setminus x,F)$.. – David Ben-Zvi Mar 7 2011 at 18:17
.. or if we want coherent sheaf, the following seem natural constructions: take again $F$ to be a natural bundle in the local sense (eg forms or jets), fix an integer and consider $x \mapsto H^*(X,F(nx))$ -- more formally, $\pi_{2,*}(\pi_1^*(F)(n\Delta))$ where the $pi_i:X\times X\to X$ are the projection. These seem like  "nonlocal natural coherent sheaves".. – David Ben-Zvi Mar 7 2011 at 20:03
I assume you mean s.th. like $p_{2, *}(p_1^*(F) \otimes I_{n\Delta}$ where $I_{n\Delta}$ is the ideal sheaf of the $n$-th infinitesimal neighborhood. But I am not sure a purist would count these as new'' sheaves, e.g. for $n = 1$ we would get the kernel of  $H^0(F) \otimes \mathcal O_X \to F$. I imagine vic might count these as a combination of $F$ and $O_X$. – ABayer Mar 7 2011 at 22:07

I'll put out the following possibly bold, possibly totally stupid conjectures. In any case, the statements below are not intended as mathematically rigorous statements, but may have some truth in them. Cheers!

Conjecture 1 Any naturally defined coherent sheaf on all smooth projective varieties is related to the sheaf of differentials via tensor operations and up to torsion.

A more concrete version is

Conjecture 2 Let $\mathscr L_X$ be a naturally defined line bundle on all smooth projective varieties $X$. Then some tensor power (possibly negative or zero) of $\mathscr L_X$ agrees with some tensor power of the canonical bundle.

Remark Jason pointed out that there are naturally defined torsion line bundles and Arend's idea of defining sheaves as push-forwards would produce sheaves supported on proper closed subvarieties. This is the main motivation for the "up to torsion" part of the first conjecture and for taking powers of the natural line bundle and the canonical bundle. Also notice that David Ben-Zvi's construction produces sheaves that satisfy these conjectures and the extensions in David Speyer's answer produce sheaves whose determinants are powers of the canonical sheaf.

To get to Conjecture 1 from Conjecture 2 one could argue the following way. Since the claim is "up to torsion" we can mod out by the torsion and assume that our sheaf is torsion-free. Now since we are on a smooth variety this implies that it is locally free in codimension $2$ and actually, again by the "up to torsion" principle, we may assume that it is reflexive, that is, take the reflexive hull, or in other words, the push-forward of the restriction to the open set where it is a locally free sheaf. In other words, we may perform all tensor operations as if we had locally free sheaves and in particular, the (reflexive hull of the) determinant will be a line bundle. In other words, up to torsion, we obtained a natural line bundle. If that is either the structure sheaf or the canonical bundle, then we're in business.

The reasoning I can offer for Conjecture 2 is the following: If there is a natural line bundle, then we can ask whether it is ample (or its inverse is) and for those varieties that it is we obtain a natural embedding (after taking some power). Once we have this we can look at the corresponding Hilbert schemes and try to construct moduli spaces. For those varieties on which this mysterious line bundle is not ample we can still define a corresponding Kodaira dimension and study Iitaka fibrations and eventually work toward a corresponding classification theory. I don't think any of this has happened except for the version using the canonical sheaf. I believe that suggests that there are no other non-trivial natural line bundles.

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Is there though any deeper reason as to why people study classification theory in terms of the canonical bundle or is it simply because the canonical bundle is one of the few canonically constructible bundles on a variety? – Frank Mar 7 2011 at 17:18

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I think the ultimate reason is that that's the only (or at least most obvious) line bundle that exists on every variety. Of course, it helps that it also plays a major role in duality, but one might argue that that is a consequence of being canonical. In other words, if the dualizing sheaf would be a totally different (line) bundle, then we'd have two to play around with (and in some sense, neither would be the canonical choice). – Sándor Kovács Mar 7 2011 at 17:53
-1? Could you explain? – Sándor Kovács Mar 7 2011 at 17:54
Thanks; duality does seems to be a good reason to expect the canonical bundle to be special but still these could all be consequences of making this initial choice. PS: It was not me that downvoted! – Frank Mar 7 2011 at 18:03
@Frank: I did not think it was (the downvote). I noticed accidentally; it happened while I was writing the response to your comment. It's not a big deal, but it would be nice to know the reason. Cheers! – Sándor Kovács Mar 7 2011 at 18:34
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In characteristic $p > 0$ one can take the Frobenius (iterated) pushforward of the structure sheaf and various sheaves of differentials. These are always locally free by an old result of Kunz. For toric varieties (and related varieties) you can explicitly compute these Frobenius pushforward sheaves, see a paper by Thomsen.

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Thanks, but this is again not what I am looking for: these sheaves arise from the two ones I mention in my question. Besides, I would like the sheaves to really be available in any characteristic, including characteristic zero. – vic Mar 7 2011 at 15:57
You are right, those sheaves are constructed from the sheaves you mentioned. I guess I felt that they were distinct enough to warrant special mention. – Karl Schwede Mar 8 2011 at 2:00

This is really a comment that is too long for the comment box, in an attempt to more clearly define the problem.

I assume that "obvious combinations" includes the result of applying any Schur functor to $T^* X$ and $T_* X$, and any direct sum of such.

Does it include nontrivial extensions of such? Because there are some. For example: Let $\mathcal{I}$ be the ideal sheaf of the diagonal on $X \times X$. Then we have a short exact sequence on $X \times X$: `$$0 \to \mathcal{I}/\mathcal{I}^2 \to \mathcal{O}/\mathcal{I}^2 \to \mathcal{O}/\mathcal{I} \to 0.$$` Pushing this to the first copy of $X$, we have a short exact sequence: `$$0 \to \Omega^1(X) \to A \to \mathcal{O}_X \to 0$$` where $A$ is an extension which is usually nontrivial. (In particular, it is nontrivial if any of the Chern classes of $T^* X$ are nontrivial.) You can play similar games with higher tensor powers of $\mathcal{I}$ and get other canonical nontrivial extensions between tensor powers of $T^* X$.

Let me suggest breaking your question up into several parts, of successively greater optimism. I will leave the word "canonical" undefined for now. For the reasons Karl mentions, all of these are guesses are in characteristic zero.

Guess 1: The only canonical classes in $K_0(X)$ are integer combinations of the Schur functors of $T_* X$ and $T^* X$.

Guess 2: The only canonical classes in $K_0(X)$ which are classes of vector bundles are nonnegative integer combinations of the Schur functors of $T_* X$ and $T^* X$.

Guess 3: Every canonical method of assigning a vector bundle to an algebraic variety has an filtration whose successive quotients are the Schur functors of $T_* X$ and $T^* X$.

Guess 4, somewhat vague: All of the extensions of vector bundles occurring in guess 3 are built from constructions on nilpotent thickenings of the diagonal in $X \times X$.

I don't know whether any of these guesses are true, but they might help focus the discussion.

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The bundle $A$ is known as the first jets. Similarly, you can take the first jets of any bundle as well as the higher order jets. – Sasha Mar 7 2011 at 16:26
What about some "natural" sheaves given by specifying some property of the stalks of the structure sheaf, such as $\mathcal{I}_{\mathrm{Sing}(X)}$ ? – Qfwfq Mar 7 2011 at 16:47
Ops, sorry, $X$ is assumed to be smooth... – Qfwfq Mar 7 2011 at 16:48
Thanks David, this is very interesting, I'll read more about these non-trivial extensions. Do you recommend any paper? I'd like to understand better these criteria for non-triviality. – vic Mar 9 2011 at 4:54

Given a map $f \colon X \to Y$, there are of course various ways to construct sheaves on $X$ from $f$ - e.g. relative differentials, or any other sheaf obtained from the cotangent complex of $f$, etc.

So any canonically constructed map gives canonically constructed sheaves. An example would be the Albanese map. For varieties of general type, one could also construct sheaves associated to the map to the canonical model (but of course this needs more care, as this is only a rational map).

One could also construct sheaves on $X$ from natural morphisms to $X$, but I don't right away see a way to get a coherent sheaf in that manner. (A cheap way to get a quasi-coherent sheaf would, for example, be to take the push-forward of the structure sheaf of the universal curve of the union of all Kontsevich spaces of stable maps from curves of a fixed genus $g$.)

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The jet bundle $J_k (X)$ has been mentioned several times. It is NOT a ''combination'' of the tangent bundle, because it cannot be recovered from the tangent bundle (not by bundle techniques, at least). To fix the notation, a point in $J_k (X)$ over $x$ is a germ of a rational function on $X$, and two are equivalent if they coincide up to order $k$ at $x$. There are epimorphisms $J_k (X) \to J_{k-1}$ with kernel $Sym^k (T^{\ast} X)$, and of course $J_1 (X) = T^{\ast}(X)$. Dually, we can consider jet bundles of germs of curves into $X$. These jet bundles give some more credibility to the four guesses that David made.

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In characteristic $p>0$, you can not only take push-forwards by the Frobenius (see Karl Schwede's answer), but also pull-backs.

Of course pulling the structure sheaf back gives the structure sheaf, but you can pull-back higher rank bundles, e.g. $\Omega_X^i$. The bundles $F^{s*} \Omega^i_X$ have been studied in the paper

Brückmann, P., Müller, D. "Birational invariants in the case of prime characteristic". Manuscripta Math. 95 (1998), no. 4, 413–429. (MR1618186)

As the title suggests, they produce birational invariants generalizing the plurigenera.

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Do $\mathcal{D}$-modules, such as the sheaf of rings of differential operators $\mathcal{D}_X$ itself, count as combinations?

And the jet bundles $\mathrm{Jet}^k_X$ ?

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Then, if $Z:=\mathrm{Sing}(X)$, the coherent sheaf of ideals $\mathcal{I}_Z$ is coherent (though not locally free in general), and $Z$ is somewhat "naturally" attached to $X$. – Qfwfq Mar 7 2011 at 16:46

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Ops, sorry, $X$ was supposed to be smooth. – Qfwfq Mar 7 2011 at 16:48
btw, $\mathcal{D}_X$ is only quasi-coherent.. (isn't it?) – Qfwfq Mar 7 2011 at 16:57
$\mathcal{D}_X$ is coherent for holomorphic operators, if I recall correctly, so I guess by some sort of GAGA stuff it's coherent in the algebraic case as well? – Ketil Tveiten Mar 22 2011 at 14:55
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