Posted by: Steven Sam | February 24, 2009

## Boij–Söderberg theory I: preliminaries

I wanted to say something about Boij–Söderberg theory. But before I can even explain what that means, I have to give some definitions in homological commutative algebra. I might as well also use this as an excuse to talk about some results in that area.

Fix a field $K$ and let $A = K[x_1, \dots, x_n]$ be a polynomial ring in $n$ variables. Let $M$ be a module over $A$. We say that a chain complex

$\cdots \to F_i \to F_{i-1} \to \cdots \to F_1 \to F_0 \to M \to 0$

is a free resolution of $M$ if each $F_i$ is a free module, and if the complex is exact in each degree, i.e., the image of each map is equal to the kernel of the one following it. It is easy to see that every module has a free resolution: pick some set of generators for $M$, and let $F_0$ be the free module generated by them, so that we have a surjection $F_0 \to M$. Now pick a set of generators for the kernel of this map, and let $F_1$ be the free module generated by them, so that we have a map $F_1 \to F_0 \to M$ which is exact at $F_0$, and repeat as necessary (possibly forever).

But we won’t really be interested in this unless $M$ is graded. The polynomial ring has a natural grading $A = \bigoplus_{i \ge 0} A_i$ where $A_i$ is the $K$-vector space generated by degree $i$ monomials. By grading, I just mean that $A_i \cdot A_j \subseteq A_{i+j}$ for all $i$ and $j$. So a (positively) graded module will be one with a decomposition $M = \bigoplus_{i \ge 0} M_i$ such that $A_i \cdot M_j \subseteq M_{i+j}$ for all $i$ and $j$. Elements in $M_i$ for some $i$ will be called homogeneous. The right notion of a homomorphism of graded modules $M \to N$ will be degree 0 maps: those that send $M_i$ to $N_i$ for all $i$. We could also talk about degree $d$ maps, but instead we will use the notation $M(d)$ to denote a grading shift: $M(d)_i = M_{d+i}$. So what we might call a degree $d$ map is just a degree 0 map $M(-d) \to N$. So now that our modules are graded, we will be interested in graded free resolutions. In general, these will look like

$(*) \quad \cdots \to \bigoplus_d A(-d)^{\beta_{i,d}} \to \cdots \to \bigoplus_d A(-d)^{\beta_{0,d}} \to M \to 0$

(we will see later why I use $-d$ instead of $d$, where the $\beta_{i,j}$ will denote the number of generators in each graded piece (most of these numbers will be 0 of course). The existence of graded free resolutions works as before: instead of just picking random generators each time, we restrict our attention to homogeneous ones.

Okay, so how about some examples. The first module to resolve is $K$ considered as the quotient module $A / (x_1, \dots, x_n)$, so that it has a trivial action, and all of its elements are degree 0. Define $F_k$ to be the free module on generators $e(I)$ where $I$ ranges over all $k$-subsets of $\{1, \dots, n\}$, and each generator has degree $k+1$, i.e., $F_k = A(-k-1)^{\binom{n}{k}}$. Define a map $d \colon F_k \to F_{k-1}$ on the free generators by $\displaystyle d(e(i_1, \dots, i_k)) = \sum_{j=1}^k (-1)^{j-1} x_{i_j} e(i_1, \dots, \hat{\imath}_j, \dots, i_k)$ where $x_{i_j}$ corresponds to the variable in $A$, and $\hat{\imath}_j$ means remove it from the list. It’s an exercise to check that $0 \to F_n \to F_{n-1} \to \cdots \to F_1 \to F_0 \to K \to 0$ is a graded free resolution of $K$. This is the Koszul complex associated to the sequence $(x_1, \dots, x_n)$ and was first introduced in the context of Lie algebra cohomology (in this case, our Lie algebra would be a trivial Lie algebra of dimension $n$). The point is that the Koszul complex is exact precisely because $(x_1, \dots, x_n)$ forms a regular sequence for $A$, i.e., $x_1$ is not a zerodivisor, $x_i$ is not a zerodivisor in the ring $A/(x_1, \dots, x_{i-1})$ for $i=2,\dots,n$, and $A/(x_1,\dots,x_n) \ne 0$. The connection between the Koszul complex and regular sequences goes further, but we’ll talk about that another time maybe.

Here’s another example. Let $n = 2$, and write the variables as $x,y$. Let $M = K[x,y] / (y^2, xy, x^4)$. Then one can write down a minimal resolution

$0 \to A(-3) \oplus A(-5) \to A(-2)^2 \oplus A(-4) \to A \to M \to 0$

where the $A$ corresponds to the fact that $M$ can be generated by one element, and the terms $A(-2)^2 \oplus A(-4)$ are the minimal relations $y^2 = xy = x^4 = 0$ which have degrees 2,2,4, respectively. But this map is not injective because these relations satisfy second-order relations. Write $e_1 = y^2, e_2 = xy, e_3 = x^4$. Then we have three relations $xe_1 = ye_2$, $x^3e_2 = ye_3$, and $x^4e_1 = y^2e_3$. Of course, the third relation is a consequence of the first two, and the first two relations are of degrees 3 and 5, respectively, which accounts for the term $A(-3) \oplus A(-5)$. Finally, there are no third-order relations (check!), so our resolution stops after two steps.

Another concept which will be of importance is that of a minimal resolution. First write $\mathfrak{m} = (x_1, \dots, x_n)$. A resolution is minimal if the image of each map $d_i \colon F_i \to F_{i-1}$ lies inside of $\mathfrak{m} \cdot F_{i-1}$. It seems like a funny definition, but this is a clean way of saying that at step of creating the free resolution, we always chose a minimal set of generators, or more precisely, that a basis of $F_i$ is mapped to a minimal set of generators for $\mathrm{coker}(d_i)$. To see that this is equivalent, consider the map $F_i / \mathfrak{m} F_i \to \mathrm{coker}(d_i) / \mathfrak{m} \mathrm{coker}(d_i)$, which is surjective by the fact that $F_\bullet$ is a resolution. The codomain is a vector space over $K = A/\mathfrak{m}$, and by degree considerations, a basis for this vector space lifts to minimal set of generators for $\mathrm{coker}(d_i)$. The fact that this map is also injective is equivalent to the requirement that $d(F_i) \subseteq \mathfrak{m} F_i$.

One reason that the notion of a minimal resolution is a good one is that it is unique up to isomorphism. The $i$th module $F_i$ is the $i$th syzygy module of $M$. Furthermore, minimal resolutions are always finite, in fact, $F_i = 0$ for all $i>n$. This is known as the Hilbert syzygy theorem, and can be proven using the Koszul complex above, and the symmetry of the Tor functor. First, we know that $\mathrm{Tor}_i^A(K,M) = 0$ for all $i>n$ and all modules $M$ by the fact that the Koszul complex has length $n$. So if $F_\bullet$ is a minimal resolution of $M$, then we can tensor it with $K$ and compute homology to calculate $\mathrm{Tor}_i^A(M,K)$. But by minimality, the differentials in $F_\bullet \otimes_A K$ are all zero, and hence we must have $F_i \otimes_A K = 0$ for $i>n$, which implies that $F_i = 0$ for $i>n$. Motivated by the uniqueness and finiteness of minimal resolutions, we can associate some invariants to graded modules based on their minimal resolutions. The first is the Betti table $\beta(M)$ of $M$ which consists of the numbers $\beta_{i,j}$ from (*). Because of the minimality condition (i.e., minimal degrees of generators have to increase by at least 1 at each step), the convention is to let the $i$th column and $(i-j)$th row of $\beta(M)$ be $\beta_{i,j}$.

One thing which may be annoying (I guess depending on how much you care) is that for a general module $M$, its syzygy modules will be generated by elements of different degrees. So we’ll say that a module is pure if each syzygy module is generated by elements all of the same degree. For example, the Koszul complex shows that $K$ is a pure module. Now here’s a surprising fact: we can always write $\beta(M) = r_1\beta(M_1) + \cdots + r_N\beta(M_N)$ where the $r_i$ are positive rational numbers, and the $M_i$ are pure modules. The addition and multiplication by rational numbers means pointwise addition and pointwise scalar multiplication. This was conjectured by Mats Boij and Jonas Söderberg in 2006 because it implies the multiplicity conjecture (which I’ll explain next time). The proof of their conjecture is due to David Eisenbud and Frank-Olaf Schreyer and came about in 2008. The proof is quite remarkable (and establishes some even more surprising results) and I’ll explain the important parts of it in my next post (or two). The proof was originally for Cohen–Macaulay modules (which I’ll define next time, and explain why this condition is relevant), but it turns out that it isn’t necessary if you’re willing to let the pure Betti tables of the $M_i$ have different numbers of columns.

This series is continued here.

-Steven