User chrislazda - MathOverflow

Answer by ChrisLazda for Why is a proper, affine morphism finite?

2013-03-27T17:00:48Z

Have a look at Ravi Vakil's notes on Algebraic geometry, 18.1.8, 18.9.A (and possibly tracing through the results used in these sections) http://math.stanford.edu/~vakil/216blog/. The point is that $f:X\rightarrow Y$ is affine, then $X\cong \underline{\mathrm{Spec}}(f_*\mathcal{O}_X)$, and if $f$ is proper, then $f_*\mathcal{O}_X$ is coherent. Hence $X$ is the relative spectrum of a coherent $\mathcal{O}_Y$-algebra, and hence finite.

Overconvergent/infinitesimal site, base change and six operations

2013-03-26T02:06:41Z

This question is about 6 operations formalism for 'crystalline' cohomology theories - more specifically the infinitesimal cohomology of smooth $\mathbb{C}$-varieties, and the overconvergent cohomology of varieties in positive characteristic. I'm not really sure of a good reference for the former, for the latter I'm talking about le Stum's site theoretic approach to rigid cohomology, as developed in his paper "The Overconvergent Site".

In particular, I'm interested in whether or not one might hope for a 6 operations formalism for the cohomology of the overconvergent site. One thing that suggests to me that this is not the 'right' set-up to get 6 operations is that I think I can convince myself that for any Cartesian square of $k$-varieties ($k$ a perfect field of characteristic $p>0$) $$ \begin{matrix} X' & \overset{g'}\rightarrow &X \\ f'\downarrow &&\downarrow f \\ Y' & \overset{g}{\rightarrow} &Y\end{matrix}$$ the induced base change morphism $$ g^*_{\mathrm{An}^\dagger} \mathbb{R}f_{\mathrm{An}^\dagger*}E \rightarrow \mathbb{R}f'_{\mathrm{An}^\dagger*}g'_{\mathrm{An}^\dagger}^*E$$ is an isomorphism. This seems 'wrong' to me - one should only expect such a base change for a proper morphism $f$. My question is whether or not this should be taken seriously as a reason why 6 operations with good 'topological' properties won't exist in this context, or is there something I've missed?

I'd also be interested to know whether or not 6 operations has been worked out for the infinitesimal site of $\mathbb{C}$-varieties (or varieties over any alg. closed char $0$ field). In the introduction to another paper of his, "Constructible $\nabla$-modules on curves", le Stum says that constructible sheaves have a definition in terms of the infinitesimal site (due to Deligne, but unpublished), but what about complexes and six operations? Is there anything known in this direction?

Also, just for reference, the argument that we have base change for any Cartesian square is basically just combining the paragraph before 1.4.2 of http://perso.univ-rennes1.fr/bernard.le-stum/Publications_files/OverconvergentSite.pdf with 7.28.1 of http://stacks.math.columbia.edu/tag/04IT. One can also see it in terms of realisations.

EDIT: It may be worth making clear that I am familiar with both Caro's theory of 6 operations for overholonomic $F\text{-}\mathcal{D}^\dagger$-modules, and the theory of algebraic $\mathcal{D}$-modules in char $0$. I am more interested in whether or not, for example in char $0$, one can do all this purely in terms of the infinitesimal site.

Answer by ChrisLazda for How to see the geometry and arithmetic of tannakian fundamental groups?

2013-03-07T16:46:22Z

To answer your second question, for any nilpotent neutral Tannakian category $\mathcal{C}$, (i.e. one in which every object is an iterated extension of the unit object $\underline{1}$), with fibre functor $\omega$ and assocaited pro-unipotent group scheme $G=G(\mathcal{C},\omega)$, there is an isomorphism

$\mathrm{Lie}(G^\mathrm{ab})^*\cong \mathrm{Ext}^1_\mathcal{C}(\underline{1},\underline{1})$

So in our case, taking $\mathcal{C}$ to be the category of unipotent lisse $\mathbb{Q}_\ell$-sheaves on some variety over an algebraically closed field of characteristic $\neq\ell$, we can recover the first étale cohomology as the dual of the abelianisation of the pro-unipotent étale fundamental group, exactly as one might expect from the Hurewiz theorem.

Pullbacks of intermediate/middle extensions and Gabber's purity theorem

2013-03-06T13:49:06Z

I am currently trying to understand intermediate extensions of perverse sheaves, specifically the proof of Gabber's purity theorem, which states that the intermediate extension of a pure perverse sheaf is pure.

As part of the proof, in §5 of "Faisceuax pervers" by Bernstein, Beilinson & Deligne, I've come across the following claim, which is not elaborated on in the paper and which I am struggling to understand.

Let $k$ be a finite field, and suppose that I have an open immersion $j:U\rightarrow X$ of $k$-varieties, and a 'projection' $f:X\rightarrow \mathbb{A}^1_k$. Fix a perverse $\overline{\mathbb{Q}}_\ell$ sheaf $\mathcal{F}$ on $U$. Then the claim is that for almost all closed points $v\in \mathbb{A}^1_k$, taking intermediate extensions of $\mathcal{F}$ commutes with pulling back to the fibre over $v$. In other words, for almost all $v$, looking at the commutative diagram

$f^{-1}(v)\cap U \overset{i}{\rightarrow} U$

$\begin{matrix} &&\downarrow j && \downarrow j \end{matrix}$

$\begin{matrix}f^{-1}(v)&\overset{i}{\rightarrow} &X\end{matrix}$

then $i^*\mathcal{F}[-1]$ and $i^*(j_{!*}\mathcal{F})[-1]$ are both perverse, and $j_{!*}(i^*\mathcal{F}[-1])=i^*(j_{!*}\mathcal{F})[-1]$ .

Why is this true?

I also have another closely related question, which comes up in trying to understand Delinge's proof of Weil II in terms of perverse sheaves. Suppose that $Y$ is smooth and connected, and that I have a lisse $\overline{\mathbb{Q}}_\ell$-sheaf $\mathcal{F}$ on some relative curve $X\rightarrow Y$, which admits a good compactification $j:X\hookrightarrow \overline{X}$ into a smooth and proper curve $\overline{X}$ over $Y$ whose complement is finite étale over $Y$. Then does taking intermediate extensions of $\mathcal{F}$ commute with pulling back to closed points of $y$? In other words, if I have a closed point $y\in Y$ then should I expect to have something like $(j_{!*}\mathcal{F}[-\dim X])_y \cong j_{!+}(\mathcal{F}_y[-\dim X_y])$ as perverse sheaves on $\overline{X}_y$? Does this basically follow from the first question, or at least from its method of proof, by repeatedly cutting $Y$ with divisors?

Any help with either of these two questions would be greatly appreciated!

Answer by ChrisLazda for Essential geometric morphisms on the étale site.

2013-03-05T08:27:55Z

As Jonathan said, the answer is always, and the left adjoint $f_!$ has a simple description as the "extension by $\emptyset$" functor. Think of an open immersion $j:U\rightarrow X$ of topological spaces, then we have the usual extension by $\emptyset$ functor by sheafififying the presheaf

$j_!\mathcal{F}(W)=\mathcal{F}(W)$ if $W\subset U$

$j_!\mathcal{F}(W)=\emptyset$ otherwise.

It's an easy calculation to show that this is left adjoint to $j^*$.

Translating this to the étale topology, if $f:X\rightarrow Y$ is an étale morphism of schemes, we have the extension by $\emptyset$ functor by sheafififying the presheaf

$f_!\mathcal{F}(W\overset{f}{\rightarrow} Y)=\coprod_{g\in\mathrm{Hom}_Y(W,X)}\mathcal{F}(W\overset{g}{\rightarrow} X)$

and again it's fairly easy to verify that $f_!$ does what we want it to.

It's worth noting that to get an adjoint for abelian sheaves, we need to replace this coproduct by the coproduct in the category of abelian sheaves, i.e. direct sum. Thus $f_!\mathcal{F}$ depends on whether we are considering $\mathcal{F}$ as a sheaf of sets or of abelian groups.

It's possibly also worth noting that we don't get a geometric morphism $f^*:\mathrm{Sh}(Y)\leftrightarrows \mathrm{Sh}(X):f_!$ because $f_!$ doesn't preserve limits. Despite this, the version of $f_!$ for abelian sheaves is actually exact!

Model category structures on dga's in a ringed topos

2013-01-25T18:07:10Z

In the introduction to his paper "Towards a non-abelian $p$-adic Hodge theory", Olsson says that for any ringed topos $(\mathcal{T},\mathcal{O})$ with $\mathcal{O}$ a sheaf of $\mathbb{Q}$-algebras, the category $\mathrm{dga}_{\mathcal{O}}$ of $\mathcal{O}$-dga's has a model category structure, where the weak equivalences are quasi-isomorphisms, and the fibrations are surjections with level wise injective kernel (injective as objects in the category of $\mathcal{O}$-modules).

Now, it seems to me the proof of this fact goes something along the following lines. Since the category of $\mathcal{O}$-modules has enough injectives, then the category of positively graded chain complexes has a similarly defined model category structure. One takes generating sets cofibrations and acyclic cofibrations in this model category of chain complexes, and one applies the 'free algebra functor' from complexes to dga's and the small object argument to get generating sets of cofibrations and acyclic cofibrations in the category of dga's.

My question is the following.

Do we really need to restrict to $\mathbb{Q}$-algebras here? Or will this argument work for any ringed topos $(\mathcal{T},\mathcal{O})$? For example, will the above definitions of weak equivalences and fibrations define a model category structure on the category of sheaves of $\mathbb{Z}/\ell^n$-modules in the étale topos of some scheme?

I can't see where the argument breaks down, but I may not have understood it well enough.

Reference for rigid analytic GAGA

2013-02-08T12:56:01Z

I'm looking for a reference for the following result.

Theorem. Let $K$ be a complete, non-archimedean field, and let $X/K$ be a projective scheme, with analytification $X^\mathrm{an}$. Then the analytification functor from coherent $\mathcal{O}_X$-modules to coherent ${\mathcal{O}}_{X^\mathrm{an}}$ -modules is an equivalence of categories.

While I've seen this sort of statement in a lot of introductory notes on rigid analytic geometry (most attributing it to Keihl), none of them seem to give a published reference. Any help would be much appreciated.

Answer by ChrisLazda for regular singularities

2013-01-31T15:39:37Z

We can see what $\ker(\nabla^\mathrm{an})$ will be by the following reasoning (which can be made a bit more precise):

$\nabla(f)=0\Leftrightarrow \frac{df}{f}=-\alpha\frac{dt}{t}\Leftrightarrow \log(f)=-\alpha\log(t)+c\Leftrightarrow f=c_0t^{-\alpha}$. So local sections of $\ker(\nabla^\mathrm{an})$ will then just be multiples of some fixed branch of the function $t\mapsto t^{-\alpha}$.

Global sections of this will exist iff $\alpha\in\mathbb{Z}$, and so in this case we can see that $H^0(\mathbb{G}_m^\mathrm{an},(\mathcal{O}^\mathrm{an},\nabla^\mathrm{an}))$ will be zero if $\alpha\notin\mathbb{Z} $ and $\mathbb{C}\cdot t^{-\alpha}$ otherwise.

On the other hand, algebraic de Rham cohomology will just be the cohomology of the complex of $k[t,t^{-1}]$-modules

$0\rightarrow k[t,t^{-1}]\overset{f\mapsto f'+\frac{f\alpha}{t}}{\rightarrow} k[t,t^{-1}]\rightarrow0$

Thus elements of $H^0(\mathbb{G}_{m,k},(\mathcal{O},\nabla))$ will be (multiples of) global algebraic branches of $t\mapsto t^{-\alpha}$. Again, if $\alpha\in\mathbb{Z}$, then this $H^0$ is just $k\cdot t^{-\alpha}$ and if $\alpha\notin\mathbb{Z}$ it will be zero.

I'm not sure how to see 'directly' that the two $H^1$'s will be the same though.

Tensor product of $\mathcal{D}$-modules and constructible sheaves

2012-11-30T13:56:29Z

The Riemann-Hilbert correspondence, as proved by Kashiwara and Mebkhout, says that for X a smooth algebraic variety over $\mathbb{C}$ there is an equivalence of triangulated categories

$D^b_c(X,\mathbb{C})\cong D^b_\mathrm{rh}(\mathcal{D}_X)$

between the bounded derived category of complexes of $\mathbb{C}$-modules on $X$ with constructible cohomology sheaves, and the bounded derived category of complexes of coherent $\mathcal{D}_X$-modules with regular holonomic cohomology sheaves.

Moreover, this equivalence respects the 6 operations $f^* , \mathbf{R}f_* , f^!, \mathbf{R}f_!, \boxtimes, \mathbb{D} $ of usual and extraordinary direct and inverse images, exterior tensor product and duality.

$\mathbf{Question:}$ Does the Riemann-Hilbert correspondence also preserve the interior tensor product?

On the constructible side, the interior tensor product is $\Delta_X^*(-\boxtimes-) $ where $\Delta_X$ is the diagonal immersion, but on the holonomic side the interior tensor product is $\Delta^!_X(-\boxtimes-)[d_X]$. So its seems to a novice like me that we're getting different operations. Or is there some comparison between $f^!$ and $f^*$ for a closed immersion $f$ that I've missed?

Modules with connection over $p$-adic laurent series rings

2012-09-19T14:39:03Z

If $X$ is a smooth rigid analytic space over a $p$-adic field $K$ (of characteristic zero), then every coherent $\mathcal{O}_X$-module with integrable connection is locally free. In his paper "Finiteness theorems for the cohomology of an overconvergent isocrystal on a curve" Crew gives a proof of this result in the particular case of a 1-dimensional annulus over $K$ as follows.

Let $I$ be a closed interval of $[0,\infty)$, and $A_I$ the ring of Laurent series in 1 variable over $K$ which are convergent for $\vert x \vert\in I$. Let $M$ be a finitely generated $A_I$-module with connection, we wish to prove that $M$ is a projective $A_I$-module. Since $A_I$ is a PID it suffices to show that $M$ is torsion free. Let $f$ generate the annihilator of $M_{\mathrm{tors}}$, one can easily show using the axioms of a connection that $f'\in (f)$. (Here $f'$ means differentiation with respect to the variable).

Crew now argues as follows: "Since $A_I$ is a PID and contains the rational numbers, $(f)$ must be the unit ideal, and hence $M_{\mathrm{tors}}=0$".

Why does this implication hold? That is, why does the fact that $A_I$ is a PID and a $\mathbb{Q}$-algebra, together with the fact that $f'\in (f)$, imply that $(f)=A_I$?

For example, would this implication still hold if I worked instead over the ring $K\otimes_{\mathcal{O}_K} \mathcal{O}_K[[t]]$ of power series with bounded coefficients?

Answer by ChrisLazda for A question about the Tannakian etale fundamental group of a curve

2012-08-24T15:22:40Z

Al ulrich says, this is not true in general. If $X$ is non-compact (i.e. affine), then the Lie algebra of U is isomorphic to the free Lie algebra on $H^1_\mathrm{et}(X,\mathbb{Q}_p)^\vee$. Hence in the case of $\mathbb{P}^1\setminus {0,1,\infty }$, where the first etale cohomology is $\mathbb{Q}_p(-1)^{\oplus 2}$, it follows that $U^n/U^{n+1}$ is isomorphic to $\mathbb{Q}_p(n)^{r_n}$ where $r_n$ is the dimension of the $n$th graded part of the free Lie algebra on a 2-dimensional vector space, but for $X$ with a higher genus compactification, what you claim will fail even for $U^1/U^2\cong H^1_\mathrm{et}(X,\mathbb{Q}_p)^\vee$.

Concerning higher dimensional varieties, it might be worth looking at Chapter 4 of Hadian-Jazi's thesis, http://hss.ulb.uni-bonn.de/2010/2217/2217.htm, which looks at extending Kim's method to higher dimensions.

Comment by ChrisLazda

2013-05-07T07:48:36Z

Yes, this is how you show that categories of sheaves/$\mathcal{O}_X$-modules have enough injectives, see for example the first few sections of Chapter III of Hartshorne.

Comment by ChrisLazda

2013-04-08T09:42:20Z

I can't think of a reference, but I have a vague recollection that this is how you do it: since the Cech complex of a sheaf is functorial, if we have a complex of sheaves $\mathcal{F}^{\cdot}$ we can take the Cech complex of each term to get a double complex $C^{\cdot,\cdot}$. Now just take the associated simple complex, $\mathrm{Tot}(C^{\cdot,\cdot})$ - in suitably nice situations, this computes hypercohomology of $\mathcal{F}^{\cdot}$.

Comment by ChrisLazda

2013-03-26T10:47:04Z

At least for F-isocrystals, 6 operations has now been worked out by Caro - he has a good theory of overholonomic F-D modules which are stable under all operations and contain the category of overconvergent F-isocrystals. On quasi-projective varieties he has proved stability of holonomicity (with F-strucure). I am specifically curious as to whether one might hope to make 6 operations work within le Stum's framework, because this base change business seems to suggest not (to me anyway).

Comment by ChrisLazda

2013-03-18T16:29:44Z

At least for varieties over a field of char 0 that is.

Comment by ChrisLazda

2013-03-18T16:28:54Z

A good place to look might be Hartshorne's paper on de Rham cohomology of algebraic varieties. archive.numdam.org/ARCHIVE/PMIHES/… In particular, Section 1.7 defines the object you talk about, and much of the paper is devoted to the study of its cohomology.

Comment by ChrisLazda

2013-03-06T13:40:24Z

Thanks for the detailed answer, but I'm not sure I'm convinced. The link in David's comment suggests that in char p, dga's just don't form a model category. I can't see where the argument goes wrong though..

Comment by ChrisLazda

2013-03-06T13:38:20Z

Okay, so it seems that the answer is quite an emphatic no. However, I like the idea of using simplicial dga's instead, thanks for suggesting it!

Comment by ChrisLazda

2013-02-08T22:45:05Z

Great, thanks. It was Köpf's paper that eventually came up after digging through Conrad's papers.

Comment by ChrisLazda

2013-02-08T14:28:48Z

His notes on rigid geometry didn't seem to have a reference. Rooting around some of his papers has done the job though, thanks!

Comment by ChrisLazda

2012-12-25T23:46:37Z

The pro-algebraic hull is the inverse limit over all finite dimensional representations of G of the Zariski closure of the image of G in GL_n. It's maybe worth pointing out that pro-algebraic hulls can be real beasts. For example, if G=Z, the integers, and k=C, the complexes, then the C-points of pro-algebraic hull is isomorphic to the direct sum of the additive group C and the group of all endomorphisms of the multiplicative group C^*.

Comment by ChrisLazda

2012-09-20T11:58:01Z

Thank-you very much!