User matt - MathOverflow

Lifting to Characteristic 0 not over W

2011-09-26T01:50:25Z

I thought of this several months ago and forgot about it. Now I rethought of it again and I just can't find it anywhere in the literature, so I'll ask here.

Is it known whether or not there exists a (smooth, proper, ...) variety over a field $k$ (perfect? alg. closed?) of positive characteristic that lifts to characteristic $0$ over some ramified extension of $W(k)$ and also lifts to $W_2(k)$, but does not lift over $W(k)$ itself?

In other words, if it lifts to characteristic $0$ (in a way related to $W(k)$) and it lifts to $W_2(k)$ must it lift via $W(k)$ itself?

I have looked at examples that lift to $W_2(k)$ but not to $W_3(k)$ and hence not over $W(k)$, but they don't seem to lift to char $0$ at all. I've also looked examples that lift over a ramified extension but not $W$, but these can easily be shown to not lift to $W_2$ as well.

Answer by Matt for Fiction books about mathematicians?

2012-07-08T21:29:08Z

The Broken God by David Zindell is a sci-fi novel about a universe in which the top ruling class is called "The Order of Mystic Mathematicians and Other Seekers of the Ineffable Flame". They have to go through extreme training in advanced math because they use topology to navigate through the universe using something called the Vild. To my knowledge it is the only work of fiction that uses the term "topology" in its mathematical sense in the first few pages.

$\ell$-adic Weil cohomology theory

2012-01-06T18:22:23Z

I have a reference or counterexample request. Suppose $k$ is a field and $\ell\neq char(k)$. There are several common references that show that $H^i_{et}(-, \mathbb{Q}_\ell )$ is a Weil cohomology theory when $k$ is separably closed.

I've spent several hours skimming through Milne's Etale Cohomology, the 1994 Motives volume, SGA articles, online searches, etc and I can't seem to determine whether or not $\ell$-adic cohomology forms a Weil cohomology theory when you don't assume you are in some "geometric" situation by making assumptions on the field.

Is there a reference that proves this is still a Weil cohomology theory or is it just false in this case? Thanks.

(This might be in SGA somewhere, but my skimming of French is rather slow and any specific related statement I find tends to throw in being over an algebraically closed field.)

Edit/Update: Everyone is commenting on the non-finitely generatedness, so I'll be more specific. That isn't really the interesting thing to me. Do you still have some sort of cycle class map that behaves nicely (functorially)? For instance, that paper Timo listed seems to imply that as long as finiteness is satisfied when you plug in a particular variety, everything else seems to be fine, but I haven't had time to seriously look at it yet.

What is the difference between Grothendieck groups K_0(X) vs K^0(X) on schemes?

2010-04-21T21:34:51Z

More specifically, I was wondering if there are well-known conditions to put on $X$ in order to make $K_0(X)\simeq K^0(X)$. Wikipedia says they are the same if $X$ is smooth. It seems to me that you get a nice map from the coherent sheaves side to the vector bundle side (the hard direction in my opinion) if you impose some condition like "projective over a Noetherian ring". Is this enough? In other words, is the idea to impose enough conditions to be able to resolve a coherent sheaf, $M$, by two locally free ones $0\to \mathcal{F}\to\mathcal{G}\to M\to 0$?

Answer by Matt for Frobenius splitting and derived Cartier isomorphism

2011-05-31T02:51:23Z

Keeping the notation of the question $X$ is a smooth variety over an algebraically closed field $k$ of characteristic $p>\dim X$. Just following along from Decomposition of the de Rham Complex by V Srinivas, we see that the obstruction to lifting the pair $(X,F)$ to a pair $(X^{(2)}, F^{(2)})$ where $X^{(2)}$ is a lift of the variety to $W_2(k)$ and $F^{(2)}$ is a lift of Frobenius consistent with all diagrams is exactly the class $\zeta\in \mathrm{Ext}^1(\Omega_{X/k}^1, B_X^1)$ that corresponds to the sequence $0\to B_X^1\to Z_X^1\to \Omega_{X/k}^1\to 0$.

If we look at the sequence $0\to \mathcal{O}_X\to F_* \mathcal{O}_X\to B_X^1\to 0$ and take $\mathrm{Hom}(\Omega^1, -)$ we get a connecting homomorphism in the long exact sequence $\mathrm{Ext}^1(\Omega^1, B^1)\stackrel{\delta}{\to} \mathrm{Ext}^2(\Omega^1, \mathcal{O}_X)$. It is well known that the obstruction to lifting lies in $\mathrm{Ext}^2(\Omega^1, \mathcal{O}_X)\simeq H^2(X, \mathcal{T}_X)$, but what is not well-known is that the obstruction class in this case is exactly the image of $\zeta$ under $\delta$. So $\delta$ acts as sort of a forgetful map for obstruction to lifting the pair to obstruction for lifting the variety without lifting Frobenius.

While it is possible that a Frobenius split variety has non-zero obstruction class $\zeta$ (no example comes to mind right now) and hence the pair doesn't lift, this splitting assumption actually gives us lots of information when coupled with the above information.

We see that $0\to \mathcal{O}_X\to F_* \mathcal{O}_X\to B_X^1\to 0$ splitting gives us $\mathrm{Ext}^1(\Omega^1, F_*\mathcal{O}_X)\twoheadrightarrow \mathrm{Ext}^1(\Omega^1, B^1)\stackrel{\delta}{\to} \mathrm{Ext}^2(\Omega^1, \mathcal{O}_X)$, so in fact $\delta=0$. This means that it doesn't matter whether or not we can lift the pair, all we had to know was that the obstruction to lifting $X$ was the image of $\zeta$ under $\delta$ which is $0$.

Thus any smooth Frobenius split variety lifts to $W_2(k)$ and since we assumed $p>\dim X$ we also get that the Hodge-de Rham spectral sequence degenerates at $E_1$ by work of Deligne and Illusie.

Formal Schemes Mittag Leffler

2011-04-25T16:51:51Z

Here is a question that I'm just copying from Math Stack Exchange that I asked awhile ago. It has just been sitting there unanswered, and although I haven't really thought about it since I posted it, I'm still very interested in a nice example if it exists.

Suppose $(\mathfrak{X}, \mathcal{O}_\mathfrak{X})$ is a Noetherian formal scheme and let $\mathcal{I}$ be an ideal of definition. Then we have a system of schemes $X_n=(|\mathfrak{X}|, \mathcal{O}_\mathfrak{X}/\mathcal{I}^n)$.

If the inverse system $\Gamma (X_n, \mathcal{O}_{X_n})\to \Gamma (X_{n-1},\mathcal{O}_{X_{n-1}})$ satisfies the Mittag-Leffler condition (the images eventually stabilize), then we get some particularly nice properties such as $Pic(\mathfrak{X})=\lim Pic(X_n)$.

More generally, we don't have to be worried about converting between thinking about coherent sheaves on the formal scheme and thinking about them as compatible systems of coherent sheaves on actual schemes.

My question:

Is there a known example of a formal scheme for which that system of global sections does not satisfy the Mittag-Leffler condition?

One thing to note is that it can't be affine (the maps are all surjective) or projective (finite dimensionality forces the images to stabilize).

A subquestion is whether or not there is a general reason to believe such an example exists. People I talk to usually say things along the lines of: you definitely have to be careful here because in principle this could happen. But no one seems to have ever thought up an example.

Lastly (still related...I think), is there a known example where you can't think of coherent (or maybe invertible) sheaves as systems because the two aren't the same?

Closed vs Rational Points on Schemes

2010-02-02T22:51:59Z

Background: When Ueno builds the fully faithful functor from Var/k to Sch/k he mentions that the variety $V$ can be identified with the rational points of $t(V)$ over $k$. I know how to prove this on affine everything and will work out the general case at some future time.

The question that this got me thinking about was if $X$ is a $k$-scheme where $k$ is algebraically closed, then are the $k$-rational points of $X$ just the closed points? This is probably extremely well known, but I can't find it explicitly stated nor can I find a counterexample.

For $k$ not algebraically closed, I can come up with examples where this is not true. So in general is there some relation between the closed points and rational points on schemes (everything over $k$)?

This would give a bit more insight into what this functor does. It takes the variety and makes all the points into closed points of a scheme, then adds the generic points necessary to actually make it a legitimate scheme. General tangential thoughts on this are welcome as well.

Answer by Matt for Examples of common false beliefs in mathematics.

2010-05-04T23:29:51Z

The standard projection map in a first course in topology is open. How could it not be closed? I always forget the standard homework exercise in which people first try to use this non-fact.

Answer by Matt for introductory book on spectral sequences

2010-04-24T07:09:44Z

I may as well shamelessly plug my blog posts http://hilbertthm90.wordpress.com/2010/01/18/exact-couples/ .

They probably don't offer any insights not already in the other responses, and it is a blog, so, well, the typesetting isn't great. On the other hand, it is a blog, so it might be less formal.

Answer by Matt for What should be taught in a 1st course on Riemann Surfaces?

2010-04-15T06:34:47Z

Although it is sort of indirectly related, it might be nice to talk about some introductory abelian variety things (as in the first few pages of Mumford's Abelian Varieties). The motivation would come from proving the equivalence of the definition of genus as the dimension of the Jacobian variety of the curve. When I took a "curves" class, I would have liked to see this rather than thinking the course was "self-contained".

Do not be afraid to show glimpses of huge areas of math that were motivated by the study of curves, even if you don't have time to do more than just mention it. I would have been far more excited and motivated to learn some of these things if I had seen it as motivated by curves, rather than the other way around (studying abelian varieties as interesting in their own right and only later learning a motivation).

Comment by Matt

2013-06-19T16:38:08Z

I assume there is a typo? Should this be: not every $F: \widehat{C}\to Sets$ is of the form $\widehat{G}$ for some $G: C\to Sets$?

Comment by Matt

2013-06-16T04:26:02Z

Oops. Of course. In the one-dimensional case the set of torsors is identified with the Brauer group of the field under the standard connecting homomorphism, which in this case is $C_1$. Sorry. The other part of the argument is the same. I was trying to remember why genus $0$ curves over finite fields always have points.

Comment by Matt

2013-06-16T02:31:02Z

Maybe I'm being really dumb, but why does no rational points imply bad reduction? I know the standard argument for genus $1$, but it involves Lang's theorem.

Comment by Matt

2013-06-13T00:11:17Z

It is not upper semicontinuous because the fiber dimension at infinity is $0$, but on any neighborhood of $0$ all other fibers are $1$-dimensional. The dimension function is actually lower semicontinuous here.

Comment by Matt

2013-06-11T16:05:36Z

As stated, this is obviously not a question.

Comment by Matt

2013-06-07T01:53:07Z

I know I should keep to comments that only relate to the question, but my recommendation is to go to the best place you get in and not just apply to places that have what you think you want to do. Most people (especially if you only have a small exposure) end up changing their minds after the first year of grad school. You'll want good people in all sorts of fields if this happens.

Comment by Matt

2013-05-02T16:27:37Z

Thank you!! That was exactly the point I was missing.

Comment by Matt

2013-05-01T18:04:02Z

Is there some obvious Galois cohomology way to see this? Since twists are parametrized by $H^1(G_{\overline{K}/K}, Aut(X))$ doesn't this say that this group is trivial?

Comment by Matt

2013-04-06T19:40:00Z

Ah, thanks! I've clearly been indoctrinated by only thinking about liftability of smooth varieties.

Comment by Matt

2013-04-06T16:15:04Z

I'm confused Angelo. The question says $X$ is affine and the obstruction to such a lifting lies in $H^2$ which vanishes, so doesn't such a lift always exist?

Comment by Matt

2013-03-25T02:21:28Z

Even if it is well-known, isn't it still fairly standard to include these types of things as corollaries? It shows that your work has applications to things that people care about and gives alternate proofs of known results. Both these things seem worth it.

Comment by Matt

2013-03-20T16:04:51Z

I actually have no idea what your question is from this. Saying what your question is about is quite a bit different than asking a question directly.

Comment by Matt

2013-03-15T23:11:48Z

Alright, then I think you mean that $X'$ should be over $\mathbb{Z}_p$ otherwise the dimensions won't make any sense. Every smooth projective curve always lifts to characteristic $0$ because the obstructions to deforming both the curve and an ample line bundle lie in $H^2$ which vanishes since it is a curve. So by Grothendieck's Existence Theorem the formal lift algebraizes. Surfaces are more delicate. There are known results like every K3 surface lifts to characteristic $0$, but there are also known surfaces that do not lift.

Comment by Matt

2013-03-15T15:52:42Z

This might be standard, but a variety over $\mathbb{Q}_p$ with "special fiber ..." just means a scheme $\mathfrak{X}/\mathbb{Z}_p$ whose generic fiber is $X'$ and special fiber is $X$ right? There are lots of examples of when this can't happen. Sometimes they formally lift, but aren't projective. Sometimes the deformation theory is highly obstructed. I'm not sure there can ever be some general conditions on $X$ that would guarantee a lift. Are you interested in a more specific type of variety or at least of a particular dimension.

Comment by Matt

2013-03-15T01:43:04Z

If the map is $f:X\to S$, then you could define $Pic_{X/S}(T)=H^0(T, R^1f_{T*}\mathbb{G}_m)$. The notation is really that we're defining the functor of points of something. When doing that it is pretty standard to say the $T$-points are blah up to equivalence where equivalence means blah. So that notation is just meant to say the points are elements of $Pic(X\times T)$ where two line bundles are equivalent if they differ by the pullback of something from $T$. The quotient is by an equivalence relation and not necessarily a subgroup. Thanks Ryan, that was the obvious thing to try.