Update

OK, I’ve worked through your comments on the Stacks project once more. This is good because my semester starts here tomorrow and I won’t have nearly as much time to do these things during the semester. However, I’ll try to be a bit more attentive to your comments as they come by, because I did find some to be quite a bit more interesting than I initially thought they would be. Also, maybe we can have an ongoing discussion this way too if you like. Anyway, please keep them coming!

Cohomology and motives

This summer I am trying to write a little bit about Weil cohomology theories for the Stacks project. My motivation is that I want to add the theorem that over a field of characteristic zero de Rham cohomology is one. And in turn, I became motivated to do this, because of the discussions I had with our graduate students in my course Ask me anything. Because after all: what is a “good cohomology theory” for algebraic varieties?

While working on this, I found that I had never properly understood the reason for all the conditions imposed by Kleiman in his paper. I still haven’t understood, I think. In this post I will discuss a basic question I had as I worked through this material and the answers I came up with so far.

Let k be an algebraically closed field; this will be the base field for our smooth projective varieties X, Y, Z. Let F be a field of characteristic 0; this will be the coefficient field for our cohomologies. Feel free to assume F is algebraically closed too. A Weil cohomology theory H^* assigns a graded commutative F-algebra H^*(X) to X in a contravariant manner. It comes with additional structure, namely trace maps and cycle classes, and it has to satisfy a bunch of axioms:

(A) Poincare duality: if dim(X) = d the trace map H^{2d}(X)(d) —> F is an isomorphism, H^i nonzero only for 0 ≤ i ≤ 2d, H^i is finite dimensional and the pairing H^i(X) x H^{2d – i}(X)(d) —> F defined using cup product and the trace map is nondegenerate,

(B) Kunneth formula: H^*(X x Y) = ⨁ H^i(X) ⊗ H^j(Y) and this is compatible with trace maps.

(C) Cycle classes are compatible with pullbacks, pushforwards, intersection products, and trace of the cycle class of a point on Spec(k) is 1.

Remark 1: I’ve written the axioms slightly differently from Kleiman in his paper “Algebraic cycles and the Weil conjecture”. In particular I am keeping track of Tate twists and I am adding the axiom that trace is compatible with Kunneth (which you can deduce from the other axioms — but will become important later).

Remark 2: In some references compatibility of cycle classes and pushforwards is omitted. But without this compatibility I immediately get stuck, for example I don’t know how to compute the class of the diagonal without this axiom. If you know a proof of this compatibility from the other axioms, please drop me an email.

Let M_k be the symmetric monoidal category of Chow motives over k (using rational equivalence). Let h(X) be the motive of X and recall that h(-) is contravariant. Let us write 1(1) for the Tate motive, i.e., the inverse of the invertible object h^2(P^1) of M_k.

A key consequence of the axioms (A, (B), (C) is that H^*(X) = G(h(X)) for some symmetric monoidal functor G from M_k to the symmetric monoidal category of graded F-vector spaces such that G(1(1)) is nonzero only in degree -2.

In fact, let’s start with a G as above and consider the contravariant functor H^*(X) = G(h(X)) from smooth projective varieties to graded F-vector spaces. What can we say about H^*? It turns out that you get cup products, trace maps, cycles classes and all the axioms (A), (B), (C) as formulated above except for possibly “the trace map … isomorphism” and “H^i nonzero only for 0 ≤ i ≤ 2d”.

Question: Can we make an example of a G where the corresponding H^*(X) does not have the “correct” betti numbers?

For example, we can ask whether there could be a G such that H^*(X) is nonzero in negative degrees for some X? Or we could ask whether we can find a G such that H^0(X) has dimension > 1. I think the answer is “yes” when k is the complex numbers and conjecturally “no” when the ground field k is the algebraic closure of a finite field.

Remark 3: When k is the algebraic closure of a finite field, a paper of Katz and Messing proves the betti numbers are the same for any Weil cohomology theory satisfying more axioms (namely some Lefschetz type thing). As far as I can tell, this result doesn’t apply to answer my question above, since I’m even weakening the assumptions of a Weil cohomology theory.

Construction of a weird G when both k and F are the field of complex numbers. Namely, let (H’)^* be the functor which sends X to (H’)^*(X) = ⨁ H^{p, q}(X) viewed as a bigraded vector space over F. This is a Weil cohomology theory: by Hodge theory it is just the same as sending X to its usual cohomology with complex coefficients. By the discussion above we get G’ such that for every motive M we have a bigrading G'(M) = ⨁ (G’)^{p, q}(M). (One could say that the target category is the category of graded complex Hodge structures, suitably defined as a symmetric monoidal category.) Of course as a graded F-vector space we have that the degree of the summand (G’)^{p, q}(X) is p + q, in other words, we have

(G’)n(M) = ⨁ n = p + q (G’)p, q(M)

Now I am going to define a weird G by setting G(M) equal to the graded F-vector space with degree n part equal to

Gn(M) = ⨁ n = 3p – q (G’)p, q(M)

Observe that G(1(1)) is still in degree -2! (A technical point is that because I chose 3 and -1 odd the functor G is still compatible with the commutativity constraints which you have to check in order to see that G is a symmetric monoidal functor.) Now, if E is an elliptic curve then H^*(E) = G(h(E)) has betti numbers dim H^{-1}(E) = 1, dim H^0(E) = 1, dim H^1(E) = 0, dim H^2(E) = 1, dim H^3(E) = 1 and everything else zero. Cool!

On the other hand, let p be a prime number and suppose that k is the algebraic closure of the field with p elements. Then I think conjecturally all objects in M_k ⊗ F are semi-simple and all simple objects are “1 dimensional”. Moreover, I think that those simple guys are classified by Weil p-numbers α in F up to multiplication by roots of unity (these α are algebraic numbers, but aren’t necessarily algebraic integers, and have a weight which is an integer which can be negative). Say M(α) is the motive corresponding to α. Now the point is that given an α of weight zero, its nth root for any n > 1 is also a Weil p-number of weight zero. Thus the 1-dimensional F-vector space G(M(α)) had better be in degree 0. And this then fixes the degrees for all motives because the motive corresponding to α = p^{-1} is 1(1) which we are requiring to be in degree -2.

Enjoy!

Workshop announcement

Next summer (2020) sometime in late July or early August Wei Ho, Pieter Belmans, and I are going to have another Stacks project workshop. If you google around you’ll find the website for the previous iteration of the workshop; this one will be structured similarly. But we’ve only just started working on the organization and everything is still in flux. Feel free to email me with suggestions for things we could do, or mentors we could invite, etc. Thanks!

Ask me anything

This semester (Spring 2019) I taught our first year, second semester algebraic geometry course for graduate students as a series of ask me anything sessions. This blog post is a report on how it worked and whether I would advise you to do the same when you teach your course.

Some background info: The enrollment was 10 students (including 4 undergraduates) but the course was attended by a few additional graduate students. Here is a link to the course webpage.

Why did I do this? First of all, I have taught algebraic geometry for graduate students many, many times and I have never taught it the same way twice. When I teach these courses, I am trying to “level up” our graduate students: I am trying to show that there is a well defined language of algebraic geometry with very sound foundations, that you can enjoy building more complicated things on top of this, that you can read about fantastic new methods/results in papers and books, and that you can do all of these things yourself actually. Maybe I can put it simply this way: I want to get them to the point where it would be fun for me (and hopefully others) to have them as graduate students. So what is more fun than a student who comes up with good questions? My second motivation was a more personal one, namely I wanted to have a bit more direct contact than usual with our first year graduate students.

How did it work? Well, I simply went to the lecture room and asked if there were any questions. After an awkward silence, usually lasting not more than one or two minutes, somebody would ask a question. Most of the time I would answer. Then another question, and so on. Often the answers were longish because of the backtracking necessary to give sufficient background in order for the answer to make sense.

Initially, the students were reading along and asked questions which allowed me to more or less develop basics of scheme theory. I tried to stimulate questions from people who preferred to think about only classical varieties, as I think this is a perfectly sound way to do algebraic geometry. Roughly in the middle of the course, in one of the lectures I limited myself to 3 minutes per answer. This lecture I think worked rather well, because I ended up getting more different questions from more different people in the room. Later I introduced another ingredient, namely, 20 questions for schemes! The students surprisingly quickly guessed I was thinking about the degree 5 Fermat surface over the field with 7 elements. I repeated this the next time, where the student, again surprisingly quickly, found that I was thinking of the spectrum of the dual numbers over the complex numbers.

An ingredient of the course which I quite enjoyed was that I was able to talk about the de Rham complex and its cohomology, both in characteristic p and in characteristic 0. Namely, the natural question arose to what extend we can find the “usual cohomology” if you are an algebraic geometer who isn’t willing to use inequalities between real numbers (such as myself). I was then also able to tie this in with coherent duality and the relationship between dualizing sheaves and sheaves of differentials.

For each of the lectures, I wrote, from memory, an account of the mathematics discussed during each meeting. You can find these on the webpage of the course linked above. As you can see, we covered an acceptable range of topics during the lectures. Also, I gave problem sets based on the topics discussed previously. (Since you do not know what the students are going to ask, you cannot give problems foreshadowing the upcoming material.) Finally, I decided to give an oral exam to the graduate students and assigned final papers to the undergraduates.

What worked well?

  1. The students were amazing: they asked good questions whose answers they really desired to know.
  2. You don’t need to prepare the lectures!
  3. You get more buy in from your students.
  4. You can cover more diverse topics.
  5. Different answers can be at different levels (some of the students may already know more theory and you can occasionally discuss material which is more advanced).
  6. You can react to the interests of the students in real time.
  7. This structure gives you an opportunity to skip worn out paths and reorder the material drastically.
  8. Students will help you prove things as you are working through them on the board.
  9. I found that I was doing more examples this way. For example, we discussed the blowing up of the cone over a conic along a line through the vertex to show that blowups mayn’t be what you think they are from nice pictures in books.
  10. You can skip over annoying verification without the students noticing. You may think this is a drawback, but more and more I think this is a useful and necessary evil of teaching algebraic geometry.

What didn’t work so well?

  1. You don’t always remember the slickest proof of every result you talk about.
  2. The order of the arguments isn’t always the right one.
  3. Students may never be sure that they have completely grokked all of the material up to a certain point. The material is not presented in a linear fashion. They’ll have to do a lot of work themselves.

The last point is perhaps the most serious drawback. However, IMHO it is impossible to teach algebraic geometry and truly cover all the details needed to get a sound theory. It is necessary for those who intend to work in the field or intend to use AG in a serious manner to sit down and work/read through a good deal of the material by themselves.

Would I recommend teaching a graduate course in Algebraic Geometry this way? Actually, no. First of all, if you are teaching algebraic geometry for the first time, I recommend choosing a good text to work with and sticking fairly close to it; or choose a topic, for example linear algebraic groups and representations, choose a good book for it, and aim the development of your algebraic geometry towards the topic (maybe the book already discusses some algebraic geometry in the first chapter). On the other hand, if you have taught algebraic geometry already multiple times, then you know why it is a difficult thing to do. You probably have already found a best possible method of teaching the course for you, and there is nothing I can say to change your mind!

Also, personally, the next time I will yet again use a completely different method!

Log de Rham cohomology

Let p be a prime number. Let k be a perfect field of characteristic p. Let U be a smooth variety over k. Choose a compactification U ⊂ X over k such that X is smooth over k and such that the divisor D = X – U is a strict normal crossings divisor D = D_1 ∪ … ∪ D_n. Then we can define the log de Rham complex Ω_X^*(log D) and try to define

H^*_{dR, log}(U) = H^*_{dR}(X, Ω_X^*(log D)

I would like to know is whether there is a published/online proof of the independence of the choice of the compactification provided one has a sufficiently strong form of resolution of singularities (RS). I did the calculation myself on a napkin (see explanation below), but it’d be great if somebody can point to a more honest writeup. Of course I searched the web for a while… also I think one of my students told me this calculation works and somebody else (maybe Illusie himself?) told me a student of Illusie worked on a better version of this a while ago? Does this ring a bell?

What do I mean by RS? Well, I mean that we can go from any compactification to any other one by a sequence of good blowing ups and good blowing downs. A good blowing up of an X as above is one which has an irreducible smooth center Z contained in the boundary (of course) such that for any I ⊂ {1, …, n} the intersection Z ∩ ⋂_{i in I} D_i is either all of Z or empty or a smooth closed subscheme of Z of codimension equal to the number of elements of I. (Aside: I think that embedded resolution of singularities will imply RS.)

Assuming RS the independence claimed above boils down to a local calculation. Think of affine r + s space A^{r + s} as the spectrum of k[x_1,…,x_r,y_1,…,y_s] with divisor D given by x_1…x_r = 0. A good blowing up looks etale locally like the blowing up of A^{r + s} in the ideal generated by x_1,…,x_{r’}, y_1, …, y_{s’} for some 1 ≤ r’ ≤ r and 0 ≤ s’ ≤ s. This blowing up is clearly equal to the blowing up of A^{r’ + s’} times the other factors. By a suitable Kunneth argument for logarithmic complexes this reduces us to the case r = r’ > 0 and s = s’. OK, so denote b : W → A^{r + s} this blowing up with exceptional divisor E isomorphic to P^{r + s – 1}. What I did was compute the cokernels of the maps

b^*Ω^i_{A^{r + 1}}(log D) → Ω^i_W(log b^{-1}D)

for all i. My napkin calculation for i = 1 showed the cokernel to be equal to Q = O_E(-1)^s. For notational convenience set S = O_E^r. Then for i = 2 my calculation gave a cokernel with a filtration having 3 graded pieces, namely

S ⊗ Q, ∧^2(Q)(1), ∧^2(Q).

For i = 3 we get graded pieces

∧^2(S) ⊗ Q, S ⊗ ∧^2(Q)(1), S ⊗ ∧^2(Q), ∧^3(Q)(2), ∧^3(Q)(1), ∧^3(Q).

And so on. If correct (caveat emptor), these cokernels have zero cohomology in all degrees. (Note that Q has rank s which is ≤ dim(E) because r > 0.) Hence the displayed arrow defines an isomorphism on cohomology and we get the desired isomorphism on logarithmic de Rham cohomology because Rb_* b^* = id on locally free coherent modules.

Looking forward to your comments!

Affineness results

For whatever reason I really enjoy results that tell us certain schemes are affine. Here is a list of a number of results of this nature in the Stacks project (but only those which deal with schemes — there are analogues of most of these results when we look at algebraic spaces and algebraic stacks):

Tag 02O0 A scheme whose underlying space is finite discrete is affine.

Tag 01PV The nonvanishing locus of a section of a line bundle on an affine scheme is affine.

Tag 0C3A Let Y be a locally closed subscheme of an affine scheme X and assume there is an affine open U of X such that Y ∩ U is affine and such that Y ∖ U is closed in X. Then Y is affine.

Tag 04DE If X → Y is a homeomorphism onto a closed subset of the affine scheme Y then X is affine.

Tag 01XF Vanshing of higher cohomology for quasi-coherent modules implies affine.

Tag 0EBE If X is quasi-affine and H^i(X, O_X) = 0 for i > 0 then X is affine.

Tag 0EBR Suppose you have a reflexive rank 1 module L over a local ring A and a section s of L such that s^n is contained in mA L[n]. Then the locus where s doesn’t vanish is affine. This generalizes the case of invertible modules mentioned above.

Tag 05YU If X → Y is surjective and integral (for example finite) and X is affine, then Y is affine.

Tag 09NL If a scheme X is the union of finitely many affine closed subschemes, then X is affine.

Tag 0A28 If X is a curve and not proper, then X is affine.

Tag 0F3R If f : X → Y is a morphism of affine schemes which has a positive weighting w, then the set V of points y of Y such that the total weight over y is maximal is an affine open of Y. For example, if f is etale, then V is the maximal open of Y over which f is finite etale. Other cases where one has a weighting are discussed in Lemmas Tag 0F3D and Tag 0F3E

Tag 0EB7 The complement of a 1 dimensional closed subset of the spectrum of a 2 dimensional normal excellent Noetherian local ring is affine.

Tag 0ECD Let f : X → Y be a finite type morphism of excellent affine schemes over a field with X normal and Y regular. Then the locus V in X where f is etale is affine. (This should be true without assuming Y to be over a field.) This result is a strengthening of purity of ramification locus which itself is a result of Gabber you can find in section Tag 0EA1.

I hope you enjoy this kind of result as well! If you know addtional results of this nature, please leave a comment or send me an email. Thanks!

Coh proper henselian schemes

In this blog post we (partially?) answer one of the questions posed in this mathoverflow post. Namely, let A be a complete discrete valuation ring with uniformizer t and fraction field K. Let X = P^1_A be the projective line over A. Let (X^h, O^h) be the henselian scheme you get by henselizing along t = 0.

Lemma. If the characteristic of K is zero, then H^1(X^h, O^h) is not zero.

Proof. Recall that the underlying topological space of X^h is the projective line over the residue field of A. Consider the standard open covering of this projective line. Then the first Cech cohomology for O^h with respect to this covering is the cokernel of the map A[x]^h x A[1/x]^h → A[x, 1/x]^h. Here the henselizations are taken with respect to the ideal generated by t. Since for H^1 Cech cohomology always injects into cohomology, it suffices to show that this cokernel is nonzero.

What does this mean? Well, by Artin approximation the ring A[x, 1/x]^h is the set of algebraic elements of the t-adic completion of A[x, 1/x] as defined in the previous blog post. A similar statement holds for A[x]^h and A[1/x]^h. See Tag 0A1W for an explanation. Thus we see immediately that the result of the previous blog post exactly gives the nonvanishing of the cokernel. QED.

What is mildly interesting is that this counter example doesn’t work if the characteristic of K is p > 0. Moreover, in this blog post we proved that one does have theorem B for henselian affine schemes in characteristic p. It could still be true that there is some good theory of henselian schemes and quasi-coherent modules on them in positive characteristic. Let me know if you have one. Thanks!

Algebraic Laurent series

In this post I discuss a funny observation about algebraic Laurent series. In a future post I will explain why it is also an interesting fact. Thanks to Will Sawin and Raymond Cheng for helping me figure this out! As usual, all mistakes are mine.

Let A be a complete discrete valuation ring with uniformizer t and fraction field K. Let x be a variable. Consider an element f in the t-adic completion of A[x, 1/x]. Then we can write

f = sum an xn

where the sum is over all integers n (postive and negative) and where an is in A and tends to zero t-adically as the absolute value of n goes to infinity. We say f is algebraic if there exists a relation

sum Pi fi = 0

where the sum is finite and the Pi are rational functions of x not all zero. Of course by clearing denominators we may assume Pi is in A[x] and not all of them equal to zero. Finally, given f we can of course write

f = fplus + fminus

where fplus is the sum of an xn for n ≥ 0 and fminus is the sum of an xn for n < 0.

Question. If f is algebraic, must fplus and fminus be algebraic?

It turns out that if K has positive characteristic, then the answer is yes and if K has characteristic zero, then the answer isn’t yes in general. Let’s get to work.

Char p. If f is algebraic, then we can find a relation of the form

sum Pi fpi = 0

for some polynomials Pi not all zero. Then looking at powers of x we observe that

sum Pi fpluspi

is a polynomial in x! This of course means that fplus is algebraic. QED.

Char 0. This is more difficult. We claim that the square root f of (1 + tx)(1 + t/x) which starts as 1 + 1/2t(x + 1/x) + … is a counter example. To see this we consider the differential operator

L = 2x(1 + tx)(t + x)D – t(x^2 – 1)I

where D = d/dx and I is the identity operator. An easy computation shows that L(f) = 0. Looking at powers of x the reader easily shows that

L(fplus) = a + bx

for some nonzero a, b in A. In fact, a computation shows that we have

a = t * sum_{i = 0, 1, 2, …} binomial(1/2, i)^2 * t^(2*i)

and

b = sum_{i = 1, 2, 3, …} -t^(2*i) * binomial(2*i – 2, i – 1) * 4 * 3 * (3 – 8*i + 4*i^2)^(-1) * 2^(-i*4) * binomial(2*i, i)

(It isn’t necessary to verify this for what we’re going to say next.)

Lemma. If fplus is algebraic, then fplus is the sum of a rational function and a K-multiple of f.

Proof. Observe that the operator L’ = (a + bx)D – bI annihilates a + bx. Hence we see that fplus is a solution of L’L. Choose an embedding of K into the complex numbers C. On some affine open U of P^1_C we see that L defines a local system V of rank 1 with finite monodromy; in fact the monodromy group is of order 2 because L is of order 1 with 4 regular singular points whose residues are each a half integer (you can also see this without computation by considering how we chose f in the first place). On the other hand, L’ defines a local system V’ of rank 1 with trivial monodromy (as it has a solution which is a rational function). Finally, L’L defines an extension W of V’ by V, i.e., we have a short exact sequence 0 → V → W → V’ → 0. The monodromy of W is either finite and then the extension is split, or the monodromy is infinite and the extension is nonsplit. Of course fplus defines a horizontal section of W over some small disc. Since L(fplus) is nonzero, we conclude that fplus isn’t in V. If fplus is algebraic (before embedding into C then of course also after embedding K into C), then we conclude that the monodromy of W is finite and the extension is split over all of U. This means there is a rational function h over C with L’L(h) = 0. By a change of fields argument (omitted), this implies we can find a rational function h over K with L’L(h) = 0. Then any solution, and in particular fplus, is a complex linear combination of h and f. QED

OK, so now we know that we have fplus = P/Q + c f for some constant c and some P, Q in A[x] with Q not zero. Multiplying through by Q we obtain Q * fplus = P + c * Q * f. Looking at very negative powers of x we conclude that c = 0 but looking at very positive powers of x we conclude that c = 1. This contradiction finishes the proof.