Posted by: Steven Sam | April 2, 2009

## Boij–Söderberg theory III: proofs

It’s been a while since the last post, but I want to give an indication of how the Eisenbud and Schreyer proved the Boij–Söderberg conjectures. This fell into roughly 3 steps. There are some constructions that are involved which I will not mention, but may mention in a future post. In particular, I will take it as a given that for any given degree sequence d, there exists a Cohen–Macaulay module M whose Betti diagram is pure with degree sequence d. Recall the setup from the previous post: we picked degree sequences $\overline{d}$ and $\underline{d}$ to restrict our attention to a finite-dimensional space of Betti tables.

Step 1: Identify the exterior facets of the cone spanned by the pure diagrams.

This step was done by Boij and Söderberg in their original paper. Recall that we put a partial ordering on the set of degree sequences given by pointwise comparison and that the cone spanned by the pure diagrams offers a geometric realization of this poset. In particular, the facets are given by chains of degree sequences which are obtained uniquely by removing a single element $\pi(d_i)$ from a maximal chain $\pi = (\pi(d_1) < \cdots \pi(d_q))$. The uniqueness part means that there do not exist two different maximal chains such that one can remove a single element from each to get the same resulting chain. Let i denote the index of the degree sequence that was removed. They classified the exterior facets into three types:

Case 1: We remove the maximal or minimal element.

Case 2: The degree sequences $\pi(d_{i-1})$ and $\pi(d_{i+1})$ differ in exactly one entry.

Case 3: The degree sequences $\pi(d_{i-1})$ and $\pi(d_{i+1})$ differ in exactly two entries.

In both cases 1 and 2, there is a unique coordinate x of the Betti table that is nonzero for $\pi(d_i)$ but zero for the other degree sequences in the chain. In this case, x gives a linear functional which is positive on the set of all Betti tables and is zero on the facet in question. The nontrivial case of course is case 3. The first thing Eisenbud and Schreyer show in this case is that there is a hyperplane which contains this facet and those degree sequences which are greater than or equal to $\pi(d_{i+1})$, and furthermore that this hyperplane is unique. Similarly, one can obtain a hyperplane which contains all degree sequences less than or equal to $\pi(d_{i-1})$. The construction of the linear functionals is inductive and we won’t repeat it here. The point is that given this uniqueness, one can hope to reconstruct these linear functionals in a different way which illucidates that they are nonnegative on the set of Betti tables. We sketch this in the next section.

Step 2: Define a bilinear pairing between Betti tables and cohomology tables.

This may seem to stray from our goal, but in fact, this is the crucial observation to showing that the case 3 linear functionals in Step 1 are in fact positive on all Betti tables. The point will be that it is easy to show positivity for this pairing, and then one need only pick the right cohomology tables to get the desired facet-defining equations. Here of course I will sketch only the constructions and give some ideas, since there are far too many details to discuss.

First, let $\beta$ be a graded Betti table, and let $\gamma$ be a graded cohomology table (i.e., the degrees of some cochain complex). Eisenbud and Schreyer introduce a bilinear functional $\langle \beta, \gamma \rangle = \sum_{\{ i,j,k \mid j \le i\} } (-1)^{i-j} \beta_{i,k} \gamma_{j,-k}$. In the case that $\beta$ comes from a free resolution F and $\gamma$ comes from a cochain complex E, we can write this sum as $\sum_j \chi(F_{\ge j} \otimes \mathrm{H}^j(E))$. Here $F_{\ge j}$ denotes the truncation of F to degrees at least j, and $\chi$  denotes the usual Euler characteristic. The next step is to examine the double complex $F \otimes E$ where we think of F as having cohomological indices, i.e., $F^i = F_{-i}$. We assume that E is bounded and consists of free modules since we only need this case. Since F is a resolution, the total cohomology of this complex is zero in negative cohomological degrees (seen by using the spectral sequence that starts by computing cohomology of the rows $F_\bullet \otimes E^j$). But we’re really interested in some truncation of this double complex, so we go to the next page of the spectral sequence starting in the other direction (along columns) and then remove the things with positive cohomological index (they don’t appear in our bilinear form). Then in fact our bilinear form can be identified with the Euler characteristic of the total cohomology of this truncation. But we know that the total cohomology vanishes in negative degrees, and we’ve removed the positive part, so in fact our bilinear form is just the dimension of the degree 0 part of the total cohomology of the truncated double complex, and hence nonnegative. Doing some more analysis, one can determine when this total cohomology vanishes, but we’ll skip that.

Unfortunately, this isn’t quite good enough. For one thing, we have not used the fact that F should be a minimal free resolution. To get around this, Eisenbud and Schreyer introduce a second bilinear form which is some kind of augmentation of the one above. Then using nonnegativity of the original bilinear form, they show that the augmented one is nonnegative when F is a minimal free resolution, and this is form is the correct thing to proceed with.

Step 3: Find the appropriate cochain complexes.

What is needed are cochain complexes which have the right vanishing conditions (which I didn’t explain) subject to the above bilinear form. Briefly I can say where these come from. The first way to get interesting cochain complexes is via linear monads for vector bundles on (n-1)-dimensional projective space (here n is the number of variables of our polynomial ring). What this means precisely is if $\mathcal{E}$ is a vector bundle on $\mathbf{P}^{n-1}$ such that its dual bundle $\mathcal{E}^*$ is a-regular (a-regular means that $\mathrm{H}^i \mathcal{E^*}(a-j)) = 0$ for i>0), then there exists a linear complex E such that $E^k = A(a+k)^{b_k}$ (here A is our polynomial ring, and the $b_k$ are just some integers) whose cohomology is close to the cohomology of $\mathcal{E}$. To be more precise, $\mathrm{H}^i(E) = \sum_d \mathrm{H}^i \mathcal{E}(d)$ for i < n-1, and $\mathrm{H}^{n-1}(E) = \sum_{d > -a-n} \mathrm{H}^{n-1}\mathcal{E}(d)$. Without going too much deeper, I will just end by saying then that the vector bundles whose linear monads give the desired linear functionals are those with supernatural cohomology. To be more precise, this means that 1) each twist of the vector bundle has at most one nonzero cohomology group, and 2) the roots of its Hilbert polynomial are all integers. Eisenbud and Schreyer show that given any prescribed set of integers, there indeed exists a supernatural vector bundle whose roots are precisely those integers (this is in some sense, dual to the problem of construction pure CM modules with a given degree sequence). To get the ones that give our linear functionals, the roots are chosen according to the chain of degree sequences that the facet is defined by. As an unexpected bonus, one can use the bilinear functionals to show that every cohomology table of a vector bundle on $\mathbf{P}^{n-1}$ is a positive linear combination of the supernatural cohomology tables.

And this is about as much as I want to say. Sorry if this post seemed much more dense than the others, but hopefully this gives some kind of picture of what goes behind this particular theorem. The interersted reader should of course consult Eisenbud and Schreyer’s paper.

I may say some more about the existence of supernatural vector bundles and pure CM modules in a later post. They exist over any arbitrary field, but in characteristic 0, one can get $\mathbf{GL}(n)$-equivariant sheaves and modules using the Borel–Weil–Bott theorem, and I’ll try to say something about the equivariant constructions later.

-Steven