Annihilation of Ext, part 2

This blog post will be used in a later one. Please skip ahead to the next one.

Let R be a regular ring. Let R —> S = R/J be a quotient. Assume we have f_1, …, f_c in J and derivations D_1, …, D_c : R —> R as well as an element z’ in R such that z’ J is contained in (f_1, …, f_c) + J^2. Let B be the Koszul algebra on f_1, …, f_c over R and let

z = det(D_i(f_j))

be as in the previous blog post.

We can extend B —> S to a Tate resolution. Thus we may assume we have

R —> B —> A —-> S

where A is gotten from B by adjoining variables and extending the differential. In particular A —> S is a quasi-isomorphism and A is free over R and over B as a graded module and B —> A is the inclusion of a direct summand (as a graded B-module).

Lemma: There exists an n >= 1 such that (zz’)^n annihilates Cone(B ⊗ S —> A ⊗ S) in D(S). (Tensor products over R.)

Proof: After inverting zz’ the immersion is regular by Tag 0GEE. This uses that R is regular! For J = (f_1, …, f_c) and f_1, …, f_c a regular sequence the map is a quasi-isomorphism as both sides compute Tor^R_*(S, S). EndProof

Remark: For a while I tried to see if n = 1 is sufficient. I haven’t yet found a counter example. I think stuff in the literature may say that this is true if S is CM.

Lemma: Let M’ in D(S). Let M be a dg B-module which is graded free and which comes with a quasi-isomorphism M —> M’ of dg B-modules. The cone of the map M ⊗_B S —> M’ is annihilated by (zz’)^n

Proof. By standard things the module M ⊗_B S is quasi-isomorphic to M’ ⊗_B A. Then we can replace B by B’ = B ⊗_R S and A by A’ = A ⊗_R S. Thus we have to show that the cone on M’ ⊗_{B’} A’ —> M’ is annihilated by (zz’)^n. By the lemma above we know that the cone C’ of the map B’ —> A’ is annihilated by (zz’)^n. Now we have the short exact sequence of B’-modules

0 —> M’ —> M’ ⊗_{B’} A’ —> M’ ⊗_{B’} C’ —> 0

whose first arrow is a splitting to the arrow M’ ⊗_{B’} A’ —> M’. Anyway, this shows that the cone we are looking at is isomorphic to a shift of M’ ⊗_{B’} C’ which proves what we want. EndProof

Denote i_* : D(S) —> D(R) the pushforward and denote i* : D(R) —> D(S) the pullback. Let M be an object of D(S). We have a counit map

e : i*i_*M —> M

Lemma: For any S-module M’ there is a map s : M’ —> i*i_*M’ whose composition with e is equal to multiplication by z^{2n + 1} (z’)^{2n}.

Proof. Denote a_* : D(S) —> D(B), a* : D(B) —> D(S), b_* : D(B) —> D(R), b* : D(R) —> D(B) the usual functors. By the previous lemma we see that there is a map a*M —> M’ whose cone is annihilated by (zz’)^n. Then it suffices to prove the lemma for a*M but with the multiplier being z (in this argument we get the 2n powers of z and z’ in the statement). So we want to construct a map

a*M —> i*i_*a*M = a*b*b_*a_*a*M

For this we can first use the unit for a to get a*b*b_*M —> a*b*b_*a_*a*M. Then we can use the construction of the previous post to get a*M —> a*b*b_*M and this is where z comes in! EndProof

Annihilating Ext, part 1

This blog post will be used in a later one. Please skip ahead to the next one.

Let f_1, …, f_c be elements of a ring R. We view the Koszul algebra B = K(R, f_1, …, f_c) as a differential graded R-algebra sitting in cohomological degrees -c, …, 0. So we have R = B^0 and B^{-1} is free over R with a basis x_1, …, x_c such that d(x_i) = f_i.

Of course, if f_1, …, f_c is a regular sequence, then B —> R/(f_1, …, f_c) is a quasi-isomorphism. But we are interested in the general case too.

Denote i_* : D(B) —> D(R) the pushforward and denote i* : D(R) —> D(B) the pullback. Let M be an object of D(B). We have a counit map

e : i*i_*M —> M

We want to find an element z in R such that for every M in D(B) there is a map s : M —> i*i_*M whose composition with e is multiplication by z on M. In other words, we want to split e up to multiplication by z.

Example: suppose that f_1, …, f_r is a regular sequence and that the map R —> R/(f_1, …, f_r) has a section. Then R —> B has a section too and we get what we want with z = 1.

Let M be a (right) dg module over B which is free as a graded module. Any dg module over B is quasi-isomorphic to one of these, so there is no loss of generality. The map e is the map

multiplication : M ⊗ B —> M

where the tensor product is over R. We are going to construct a map s : M —> M ⊗ B using derivations. Before we continue, observe that we do have a map of dg B-modules

ξ : M —> (M ⊗ B)[-c]

Namely, we can send m in M to

(sign)m ⊗ 1 (x_1 ⊗ 1 – 1 ⊗ x_1) … (x_c ⊗ 1 – 1 ⊗ x_c)

where the sign is (-1) to the power cdeg(m). The reason this works is that x_i ⊗ 1 – 1 ⊗ x_i is killed by the differential. There are some sign rules for the multiplication on B ⊗ B: we have (x_i ⊗ 1)(1 ⊗ x_j) = x_i ⊗ x_j and we have (1 ⊗ x_j)(x_i ⊗ 1) = – x_i ⊗ x_j.

Let D : R —> R be a derivation. Then we can extend D to a degree zero map D’ : B —> B which satisfies the Leibniz rule by setting D'(x_i) = 0. Of course D’ does not commute with d in general.

Suppose that M is a (right) dg module over B which is free as a graded module. (Any dg module over B is quasi-isomorphic to one of these.) Then we can similarly find a degree zero map D’ : M —> M which satisfies the Leibniz rule over D’.

In both cases consider the map θ : B —> B and θ : M —> M defined by the formula &theta = D’ o d – d o D’. Then θ has degree 1, defines a map B —> B[1] and M —> M[1] of complexes, θ : B —> B is a derivation, and θ : M —> M satisfies the Leibniz rule θ(mb) = θ(m)b + (sign) m θ(b) where the sign is (-1) to the power the degree of m.

A simple calculation shows that θ(x_i) = D(f_i).

Next, suppose we have c derivations D_1, …, D_c. Then we get c maps θ_1, …, θ_c : M —> M[1]. Then we can consider the composition

M — ξ –> (M ⊗ B)[-c] — θ_1 … θ_c ⊗ 1 –> M ⊗ B

Unless I made a calculation error (which is very possible) the composition of these maps with the map e : M ⊗ B —> M is equal to multiplication by

z = det(D_i(f_j))

Thus we conclude what we want with z as above.

The conclusion of this is a precise version of something we all already know: if we have a closed embedding i of codimension c and we have c tangent fields spanning the normal bundle, then we can split the counit map i*i_*M —> M using those tangent fields.

Elkik’s Theorem 7

In the title I am referring to a result in the paper “Solutions d’équations à coefficients dans un anneau hensélien”.

I’ve completely forgotten how to make mathematical symbols appear in blog posts. So this blog post will use ugly workarounds to get the mathematics across. Sorry!

Let A be a Noetherian ring. Let I be an ideal of A. Let A —> B be a ring map such that B is I-adically complete and such that B/IB is of finite type over A/I. Then we say A —> B is rig-smooth if the “completed” naive cotangent complex NL_{B/A}^\wedge 0AJL has the following property: over the complement of V(IB) in Spec(B) it only has cohomology in degree 0 and this cohomology is a finite locally free module there. (I intend to add this definition to the Stacks project soon.)

Elkik’s Theorem 7: If I is a principal ideal and A —> B is rig-smooth, then B is the I-adic completion of a finite type A-algebra.

Question: Does this theorem also hold if I is not principal?

This question was asked by Temkin in this paper in Remark 3.1.3 part (iii).

I do not know any application of this if it were true. Moreover, I am having a very hard time even coming up with a possible scenario in which you could apply this and not one of the many closely related things that I do know how to prove. But this is of course just one of those things one goes through as a mathematician: once a question has been asked it is hard to stop thinking about it.

Anyway, I would be grateful for any ideas of comments you might have about the question. Please either email me or leave a comment on this post. Thanks!

Let me list the things I think I can prove (caveat emptor):

  1. It is true if I is principal but more generally when the complement of V(IB) in Spec(B) is affine
  2. It is true if B is rig-etale, see 0AKA
  3. It is true if A is an excellent ring
  4. There exists an ideal J = (b_1, …, b_m) in B such that V(J) = V(IB) and such that the I-adic completion of the affine blow-up algebras B[J/b_i] are completions of finite type A-algebras

The final statement says that locally on the rigid space associated to the formal scheme Spf(B) we have the algebraizability (spelling?) that we want. The second statement is what we use in the Stacks project to reprove a strong version of Artin’s algebraization of dilatations. I am hoping to upgrade the relevant chapter of the Stacks project and add a discussion of Artin’s theorem on contractions.

Thanks for reading!

Worked through all your comments

Thanks to all those who helped out. As usual, many of the comments were more interesting than they appeared at first sight. The most helpful are comments which (1) address the mathematics directly, (2) point out precisely what the problem is, (3) if possible give a definite suggestion as to how to fix or improve things, (4) are left on the page of the smallest unit that is being commented on (so if commenting on a lemma go to the webpage that only displays that lemma). You can leave mathematics in the comment (using latex) and when this is done I can steal your latex code when I make my edits which helps! Of course, the editor for leaving comments isn’t perfect, but it mostly works.

Thanks again!

Another thing: if people ask questions about the mathematics, I don’t always answer those questions. But you, dear reader, can answer those questions too. I welcome that!

Purity (Part 3)

This is a continuation of the post Purity (part 2). Thanks to an email from János Kollár we now know that the answer to the question is no for relative dimension ≥ 2 as I will explain in this post. All mistakes in this post are mine (of course).

Take a large integer n. The minimal versal deformation space (Y, y) of the rank 2 locally free module O + O(n) on P^1 has dimension n – 1 and is smooth. Let X’ be the projectivization of the universal deformation over Y x P^1. Then X’ → Y is a smooth projective family of Hirzebruch families. The fibre Σ = X’_y is the Hirzebruch surface Σ → P^1 which has a section σ with self-intersection -n. Recall that the Picard group of Σ is generated by σ and a fibre F. Consider the invertible module L on X’ whose restriction to Σ is D = -K + (n – 2)F. A computation shows that

  1. D σ = 0
  2. H^1(O_Σ(D)) = H^2(O_Σ(D)) = 0
  3. H^0(O_Σ(D)) has dimension 3n + 3 and gives an embedding of the contraction of σ in Σ into P^{3n + 2}

However, every other fibre of X’ → Y is a Hirzebruch surface whose directrix has self-intersection > -n. Hence -K + (n – 2)F will not contract the directrix on any other fibre. We conclude that L defines a factorization X’ → X → Y where we are constracting σ on Σ to a point x in X and nothing else. Thus f : X → Y is a morphism of varieties, Y is smooth, X is a normal variety, and f is smooth at all points except at x. Thus we see

codim Sing(f) = n – 1 + 2 = n + 1

Since the question was whether codim Sing(f) ≤ 1 + 2 we see the answer is very much no in the case of relative dimension 2.

For relative dimension d ≥ 2 we take the morphism X x A^{d – 2} → Y which has a singular locus still of codimension n + 1 and hence this shows the answer is no as soon as n > d.

As far as I know the question remains unanswered for relative dimension 1 (besides the one subcase of relative discussed in the previous post on this topic). Please let me know if you have an idea or an example.

PS: As János points out the morphism f constructed above is even flat and the fibres of f have rational singularities. Thus it seems unlikely there is a class of singularities strictly bigger than the lci ones (see previous post) for which purity holds.

PPS: Another question is whether examples like this can help us find examples of other purity statements gone wrong.

Talk by Will Sawin TODAY

I would like to advertise this talk in the Stanford AG Seminar.

Will Sawin – The Shafarevich Conjecture for Hypersurfaces in Abelian Varieties

Time: Mar 27, 2020 07:00 PM Eastern Time (US and Canada)

To join the zoom meeting please email me and I will send you the link.

Abstract: Faltings proved the statement, previously conjectured by Shafarevich, that there are finitely many abelian varieties of dimension n, defined over the rational numbers (or another fixed number field), with good reduction outside a fixed finite set of primes, up to isomorphism. In joint work with Brian Lawrence, we prove an analogous finiteness statement for hypersurfaces in a fixed abelian variety with good reduction outside a finite set of primes. I will give a broad, intuitive introduction to some of the ideas in the proof, which combines tools from different areas in algebraic geometry, arithmetic geometry, and number theory, and builds heavily on recent work of Lawrence and Venkatesh.

Minimal Cech coverings

Please don’t read this unless you want to be distracted in these difficult times and you enjoy thinking about elementary problems in etale cohomology. I am going to pose a challenge computing etale cohomology by Cech coverings.

The setting. We have a quasi-projective scheme X over an algebraically closed field k, a prime number ℓ different from the characteristic of k and a constructible sheaf F of Z/ℓ Z vector spaces. Next we have a cohomology class ξ in the etale cohomology group H^i(X, F). By a result of Artin, the cohomology of F over X is the same as the Cech cohomology with respect to etale coverings, but (of course) you have to take the colimit of all etale coverings. Thus we know there exists an etale covering cU = {U_i —> X} and a degree i Cech cohomology class ξ’ for cU and F which maps to ξ in H^i(X, F).

The problem I want to think about is this: is there a notion of minimal etale covers cU such that you can represent ξ by a Cech cohomology class for that covering.

Let’s do an example. Take X = P^1 the projective line and take F the constant sheaf Z/ ℓ Z. Then there is a nonzero class ξ in H^2(X, F). Then we can try and minimize (one at a time) the following numerical invariants of Cech coverings \cU = {U_i —> X}_{i = 1, …, n} such that ξ can be represented as a Cech cohomology class on them:

  1. the sum of the number of connected components of U_i,
  2. the sum of the geometric genera of the components of the U_i,
  3. the sum of all the betti numbers of the U_i, or
  4. the sum of the degrees of the morphisms U_i —> X.

I will edit this and put other suggestions here in the future

Rmk: Observe that if cU = {U_i —> X}_{i =1, …, n} is an etale covering, then so is {U —> X} where U is the disjoint union of the U_i and these covering have identical Cech complexes. Hence whatever invariant we use, it shouldn’t distinguish between these coverings, for example we shouldn’t use “n” as an invariant.

Rmk: for etale cohomology we have to use “regular” Cech cohomology, not alternating Cech cohomology.

Rmk: It might be better to work with etale hypercoverings and define numerical invariants for those.

Back to the example, here is a covering that works. Let t be a coordinate on P^1. Let U_0 = G_m with coordinate x mapping to t = x^ℓ in P^1. Let U_1 = G_m with coordinate y mapping to t = y^ℓ + 1. Let U_2 = G_m with coordinate z mapping to t = z^ℓ / (z^ℓ – 1). The (double) fibre product of U_0 and U_1 over P^1 is connected and similarly for U_0 and U_2 and for U_1 and U_2. However, the (triple) fibre product of U_0 and U_1 and U_2 over P^1 is disconnected because a computation shows that (x^{-1}yz)^ℓ = 1 and hence x^{-1}yz has to be an ℓth root of 1. This suggests that we can write down an interesting Cech cohomology class in degree 2 for this etale Cech covering and indeed this is true. To actually prove that ξ can be represented by a Cech cohomology class for this covering, you can proceed as follows: first compute the obstruction to representing ξ by a Cech cohomology class for the open covering P^1 = (P^1 – {0}) cup (P^1 – {∞}). This is an element of H^1(P^1 – {0, ∞}, F). Then show this obstruction dies when passing to the covering {U_0 –> P^1, U_1 —> P^1, U_2 —> P^1} which refines the two part open covering because the H^1 element dies on U_i \times_{P^1} U_j for all i, j. Anyway, my solution gives the invariants as in list above:

  1. For this invariant I get 3 because U_0, U_1, U_2 are connected
  2. For this invariant I get 0 because U_0, U_1, U_2 are rational
  3. For this invariant I get 6 because U_0, U_1, U_2 are G_m
  4. For this invariant I get 3ℓ because the maps U_i —> P^1 have degree ℓ

Challenge: Can you improve for any of these invariants on the numbers I get? I’ve tried to improve on the first invariant without success so far.

Artwork for a T-shirt

OK, we are still planning to have the Stacks project workshop this summer. I am slowly trying to work on having a new t-shirt or sweat shirt. I asked my youngest and middle son to make some artwork. Here is what they came up with for the front and back of the shirt.

I’ve ordered 3 shirts with these pictures on it for me and my sons; I will post a picture here when we get them. But a question I have is what text (if any) to put with these pictures. What I came up with was

  • Front: Climbing mathematics…
  • Back: You can keep going

This doesn’t really work very well. Instead what one of my sons came up with was

  • On the stack
  • No looking back

and I think it’s better. What do you think? Do you have a suggestion?