Notation. Given two sets of rank sequences, say $A = \operatorname {\mathrm {RS}}({\mathbf {F}})$ and $B = \operatorname {\mathrm {RS}}({\mathbf {G}})$ , we will write $A+B$ to refer to the set of sequences that may be expressed as a sum of a sequence in A and a sequence in B. Similarly, we will write $nA$ to refer to the set $A+A+\cdots + A,$ where the sum has n terms.

2.1 The BGG correspondence

The key tool for our results is the Bernstein–Gel’fand–Gel’fand correspondence [Reference Bernšteĭn, Gel’fand and Gel’fand3], which allows us to translate questions about linear free complexes of modules over a symmetric algebra into questions about modules over an exterior algebra. We will cherry-pick what we need of this rich subject; for further detail, see [Reference Eisenbud9, §7B] and [Reference Eisenbud, Fløystad and Schreyer10].

Let $\mathbb {k}$ be a field, let V be a $\mathbb {k}$ -vector space with basis $x_1, \ldots , x_n$ , and let W be the dual vector space of V with basis $e_1, \ldots , e_n$ . Let $E = \mathbb {k}\langle e_1, \ldots , e_n\rangle $ denote the exterior algebra on W. Let S denote the symmetric algebra $\operatorname {\mathrm {Sym}}(V)$ , and identify S with the polynomial ring $\mathbb {k}[x_1, \ldots , x_n]$ . We will assume that the $x_i$ are graded in degree $1$ , and the $e_i$ are graded in degree $-1$ . Unless otherwise stated, all tensor products are assumed to be over the ground field $\mathbb {k}$ .

We define a pair of functors $\mathbb {L}$ and $\mathbb {R}$ as follows:

$$ \begin{align*} \mathbb{L}: \{\text{Graded } E\text{-modules}\} &\to \{\text{Linear complexes of free } S\text{-modules}\}\\ N &\longmapsto (\ldots\to S\otimes N_d\xrightarrow[]{\partial_d} S\otimes N_{d-1}\to\dots) \end{align*} $$

with differential $\partial _d$ defined by linearly extending

$$\begin{align*}1\otimes f\mapsto \sum_{i=1}^n x_i\otimes fe_i \end{align*}$$

and

$$ \begin{align*} \mathbb{R}: \{\text{Graded } S\text{-modules}\} &\to \{\text{Linear complexes of free } E\text{-modules}\}\\ M &\longmapsto (\ldots\to E\otimes M_d\xrightarrow[]{\partial_d} E\otimes M_{d+1}\to\dots) \end{align*} $$

with differential $\partial _d$ defined by linearly extending

$$\begin{align*}1\otimes g\mapsto \sum_{i=1}^n e_i\otimes gx_i. \end{align*}$$

Example 2.3. Consider the module $N = \langle e_1, e_2e_3 \rangle E$ , where $E = \mathbb {k}\langle e_1, e_2, e_3, e_4\rangle $ . We will use the following $\mathbb {k}$ -bases for the graded pieces of N:

$$\begin{align*}\begin{array}{ll} \operatorname{degree} \ -1 &: e_1. \\ \operatorname{degree} \ -2 &: e_1e_2, \ e_1e_3, \ e_1e_4, \ e_2e_3. \\ \operatorname{degree} \ -3 &: e_1e_2e_3, \ e_1e_2e_4, \ e_1e_3e_4, \ e_2e_3e_4. \\ \operatorname{degree} \ -4 &: e_1e_2e_3e_4. \end{array} \end{align*}$$

Tracing through the definition of $\mathbb {L}$ we can see, for example, that

$$\begin{align*}{\partial}_{-2}(1 \otimes e_1e_2) = \sum\limits_{i=1}^4 x_i \otimes e_1e_2e_i = x_3 \otimes e_1e_2e_3 + x_4 \otimes e_1e_2e_4.\end{align*}$$

The entirety of $\mathbb {L}(N)$ is the complex:

$$\begin{align*}0 \rightarrow S \otimes N_{-1} \xrightarrow{ \begin{bmatrix} x_2 \\ x_3 \\ x_4 \\ 0 \end{bmatrix} } S \otimes N_{-2} \xrightarrow{ \begin{bmatrix} x_3 & -x_2 & 0 & x_1 \\ x_4 & 0 & -x_2 & 0 \\ 0 & x_4 & -x_3 & 0 \\ 0 & 0 & 0 & x_4 \end{bmatrix} } S \otimes N_{-3} \xrightarrow{ \begin{bmatrix} x_4 & -x_3 & x_2 & -x_1 \end{bmatrix} } S \otimes N_{-4} \rightarrow 0. \end{align*}$$

The functors $\mathbb {L}$ and $\mathbb {R}$ as we have defined them may be extended to arbitrary complexes via a totalization procedure, thus yielding functors on the corresponding bounded derived categories. The BGG correspondence states that if we consider $\mathbb {L}$ and $\mathbb {R}$ as functors on the bounded derived categories then they are adjoint, implying that the derived categories of bounded linear complexes of finitely generated graded E-modules and S-modules are equivalent. This equivalence tells us that $\mathbb {L}$ and $\mathbb {R}$ are exact. What is more, the functor $\mathbb {L}$ gives a bijection on objects under which

1. 1. Any linear complex ${\mathbf {F}}$ of S-modules may be expressed as $\mathbb {L}(N)$ for some E-module N [Reference Eisenbud, Fløystad and Schreyer10].

2. 2. Subcomplexes of ${\mathbf {F}}$ correspond to E-submodules of N.

We also make use of the following relationship between $\mathbb {L}$ and $\mathbb {R}$ .

Theorem 2.6. (Reciprocity Theorem) [Reference Eisenbud, Fløystad and Schreyer10, Th. 3.7] Let M be a graded S-module, and let N be a graded E-module. Then

$$\begin{align*}N\to \mathbb{R}(M) \end{align*}$$

is an injective resolution if and only if

$$\begin{align*}\mathbb{L}(N)\to M \end{align*}$$

is a free resolution.

2.2 Tate resolutions

The following construction, when considered in tandem with the BGG correspondence, will play a key role in Section 3.2.

Definition 2.7. For any module N over any ring, we can combine a projective resolution ${\mathbf {P}}$ of N and an injective resolution ${\mathbf {I}}$ of N in the following way

to produce a Tate resolution.

More detail about general Tate resolutions can be found in [Reference Eisenbud, Fløystad and Schreyer10], but we are most interested in Tate resolutions of modules over E, where injective and projective modules are both free. In this case, we can take ${\mathbf {P}}$ to be a minimal free resolution of N and ${\mathbf {I}}$ to be the dual of the minimal free resolution of the dual of N to create a unique doubly infinite exact complex of free modules where the image of $P_0$ is isomorphic to N. We will call this doubly infinite complex the Tate resolution ${\mathbf {T}}(N)$ .

Example 2.8. If $N = E/\langle e_1, \ldots , e_n \rangle \cong \mathbb {k}$ , then the Cartan resolution $({\mathbf {C}},{\partial })$ is a projective resolution of N (cf. [Reference Eisenbud9, Cor. 7.10]). The dual of ${\mathbf {C}}$ is an injective resolution of $\mathbb {k}$ (which is its own dual), so stitching the two together yields the Tate resolution of $\mathbb {k}$ . Below is a snippet of ${\mathbf {T}}(\mathbb {k})$ in the $n=3$ case:

where $ {\partial }_0 = \begin {bmatrix} e_1 e_2 e_3 \end {bmatrix}$ , ${\partial }_1 = \begin {bmatrix} e_1 & e_2 & e_3 \end {bmatrix}$ , and ${\partial }_2 = \begin {bmatrix} e_1 & e_2 & 0 & e_3 & 0 & 0 \\ 0 & e_1 & e_2 & 0 & e_3 & 0 \\ 0 & 0 & 0 & e_1 & e_2 & e_3 \end {bmatrix}. $

In general, the differential ${\partial }_s$ in the Cartan resolution can be computed by indexing the columns of ${\partial }_s$ with the degree-s monomials in the $x_i$ ’s and the rows by the degree $s-1$ monomials in the $x_i$ ’s. Then, if column i is indexed by a monomial m and row j is indexed by a monomial $m'$ , the $(i,j)$ th entry of ${\partial }_s$ is $e_k$ if $m/m' = x_k$ if $m' \mid m$ and $0$ if $m \nmid m'$ . In Example 2.8, the indexing monomials for the entries of the ${\partial }_s$ are listed in graded reverse lexicographic order with $x_1> x_2 > x_3$ .

Remark 2.9. Because E is free over $\mathbb {k}\langle e_1, \ldots , e_m \rangle $ for $m \leq n$ , extending scalars from $\mathbb {k}\langle e_1, \ldots , e_m \rangle $ to E is faithfully flat. This means the Tate resolution ${\mathbf {T}}(E/\langle e_1, \ldots , e_m \rangle )$ has the same structure as the Tate resolution ${\mathbf {T}}(\mathbb {k}\langle e_1, \ldots , e_m\rangle /\langle e_1, \ldots , e_m \rangle )$ as a complex of $\mathbb {k}\langle e_1, \ldots , e_m\rangle $ -modules. That is, the complex of $\mathbb {k}\langle e_1, \ldots , e_m\rangle $ -modules ${\mathbf {T}}(\mathbb {k})$ and the complex of E-modules ${\mathbf {T}}(\mathbb {k}\langle e_{m+1}, \ldots , e_n \rangle )$ have modules of the same rank and twists, and differentials with the same entries, regardless of the ambient ring. For example, the Tate resolution ${\mathbf {T}}(\mathbb {k}\langle e_4\rangle )$ over $\mathbb {k} \langle e_1, \ldots , e_4\rangle $ will “look” the same as the one shown in Example 2.8, with all E’s replaced by $\mathbb {k}\langle e_1, \ldots , e_4 \rangle $ .

3.1 The Koszul complex

Definition 3.1. The Koszul complex ${\mathbf {K}}(x_1, \ldots , x_m)$ is the graded exact complex

$$\begin{align*}{\mathbf{K}}(x_1, \ldots, x_m): 0 \rightarrow S(-m) \xrightarrow{{\partial}_m} \cdots \xrightarrow{{\partial}_3} S(-2)^{\binom{m}{2}} \xrightarrow{{\partial}_2} S(-1)^m \xrightarrow{{\partial}_1} S^1 \rightarrow 0, \end{align*}$$

where we index basis elements of by size d subsets of m. For $T~=~\{i_1, \ldots , i_d\}$ , the differential ${\partial }_d$ acts on $e_T$ by ${\partial }_d(e_T) = \sum _{j = 1}^{d} (-1)^{j} x_{i_j} e_{T \backslash i_j}$ .

Example 3.2. The Koszul complex ${\mathbf {K}}(x_1,x_2,x_3)$ is given by

$$\begin{align*}{\mathbf{K}}(x_1, x_2, x_3): 0 \rightarrow S(-3) \xrightarrow{{\partial}_3} S(-2)^3 \xrightarrow{{\partial}_2} S(-1)^3 \xrightarrow{{\partial}_1} S \rightarrow 0, \end{align*}$$

where

Looking at the above example, we can immediately identify some subcomplexes of the Koszul complex. If $m \leq n$ , the complex ${\mathbf {K}}(x_1, \ldots , x_m)$ is a subcomplex of ${\mathbf {K}}(x_1, \ldots , x_n)$ – one can see the Koszul complex ${\mathbf {K}}(x_1,x_2)$ boxed in red in Example 3.2. One can also simply truncate ${\mathbf {K}}(x_1, x_2, x_3)$ after two modules and omit the $S(-3)$ module at the end, or even omit $S(-3)$ and some summands of the $S(-2)^3$ in the next spot. This observation yields certain sequences that we can be sure must occur as rank sequences of subcomplexes of ${\mathbf {K}}(x_1, \ldots , x_m)$ but to obtain a more complete classification we can peer through the BGG lens and, in particular, use the following key fact.

The BGG correspondence thus tells us that subcomplexes of the Koszul complex are in correspondence with submodules of the exterior algebra E twisted by $(-n)$ , so our question about the possible rank sequences of subcomplexes of ${\mathbf {K}}$ is transformed into a question about the possible Hilbert functions of submodules of E itself (after the appropriate twist). This perspective immediately reveals that subcomplexes of ${\mathbf {K}}$ are less restricted than one might guess from the $n=3$ case. Indeed, Example 2.3 shows that we can obtain a subcomplex of ${\mathbf {K}}(x_1, \ldots , x_4)$ whose rank sequence is $(1,4,4,1,0)$ , which is not the rank sequence of a smaller Koszul complex and furthermore cannot be obtained by truncating free summands from the tail of K.

This observation also underscores the complexity of the structural question of classifying all subcomplexes in the case of the Koszul complex. Such a task would be equivalent to classifying all ideals in E. Though the feasibility of such classification is yet unknown, it is worth noting that the parallel question of classifying ideals in S is impossible by Vakil’s Murphy’s Law [Reference Vakil19].

By work of Aramova–Herzog–Hibi [Reference Aramova, Herzog and Hibi1, Th. 4.1], possible Hilbert sequences of submodules of the exterior algebra are exactly those corresponding to f-vectors of simplicial complexes as described by the Kruskal–Katona theorem. We can use these results to characterize the possible rank sequences for a subcomplex of the Koszul complex with the following notation.

Definition 3.4. If a is a positive integer, then, for every positive integer i, a has a unique Macaulay expansion

$$\begin{align*}a = \binom{a_i}{i} + \binom{a_{i-1}}{i-1}+\cdots +\binom{a_j}{j}, \end{align*}$$

where $a_i>a_{i-1}>\cdots >a_j\geq j\geq 1$ . Define

$$\begin{align*}a^{(i)}:= \binom{a_i}{i+1} + \binom{a_{i-1}}{i}+\cdots +\binom{a_j}{j+1}. \end{align*}$$

Theorem 3.5. A non-negative integer sequence $(r_0, r_1, \ldots , r_n)$ is in $\operatorname {\mathrm {RS}}({\mathbf {K}}(x_1, \ldots , x_n))$ if and only if it is the all zeros sequence or if $r_0=1$ and r satisfies

$$\begin{align*}0\leq r_{i+1}\leq r_i^{(i)} \text{ for } 1 \leq i \leq n-1. \end{align*}$$

3.2 The Eagon–Northcott complex

The Eagon–Northcott complex [Reference Eagon and Northcott7] plays the same role for determinantal ideals that a Koszul complex plays for a sequence of ring elements. We provide a brief presentation here that describes the complex for S-modules; more details can be found in [Reference Eisenbud9, Appendix A2H]. Throughout, we will choose bases for our free modules so that we can represent these maps as matrices.

Definition 3.7. Let $F = S^f$ and $G=S^g$ , with $g \leq f$ , and $\alpha : F \rightarrow G$ a map represented by a $g \times f$ matrix A with respect to bases $\{e_1, \ldots , e_f\}$ of F and $\{\varepsilon _1, \ldots , \varepsilon _g\}$ of G. Then the Eagon–Northcott complex of the map $\alpha $ is the complex

$$\begin{align*}{\mathbf{EN}}(\alpha): 0 \rightarrow EN_{f-g+1} \xrightarrow{d_{f-g+1}} EN_{f-g} \xrightarrow{d_{f-g}} \cdots \xrightarrow{d_3} EN_2 \xrightarrow{d_2} EN_1 \xrightarrow{\Lambda^g \alpha} \Lambda^g G, \end{align*}$$

where $EN_{k+1} = (\operatorname {\mathrm {Sym}}_k G)^* \otimes \Lambda ^{g+k} F$ and $d_{k+1}: (\operatorname {\mathrm {Sym}}_k G)^* \otimes \Lambda ^{g+k} F \rightarrow (\operatorname {\mathrm {Sym}}_{k-1} G)^* \otimes \Lambda ^{g+k-1} F$ is the map

$$ \begin{align*} (\varepsilon_{1}^{p_1} \dots \varepsilon_{g}^{p_g})^* &\otimes e_{s_1} \wedge \cdots \wedge e_{s_{g+k}} \mapsto \\ &\quad \sum\limits_{i=1}^{g+k} (-1)^{i-1} \left[ \sum\limits_{j=1}^{g} A_{j,s_i} (\varepsilon_1^{p_1} \dots \varepsilon_j^{p_j-1} \dots \varepsilon_g^{p_g})^* \right] \otimes e_{s_1} \wedge \dots \wedge \widehat{e_{s_i}} \wedge \cdots \wedge e_{s_{g+k}} \end{align*} $$

for $k \geq 1$ , where $p_1 + \dots + p_g = k$ and we adopt the convention that $\varepsilon _j^p = 0$ if $p < 0$ .

Note that, if we represent $\alpha $ by the matrix A, then $A_{j,s} = \varepsilon _j^*(\alpha (e_s))$ , and that using a different basis to express A will give an isomorphic complex.

Example 3.8. For this example, let $S= \mathbb {k}[x_1, x_2, x_3]$ . Consider the example $\alpha :~S^4~\rightarrow ~S^2$ represented by the matrix $A = \begin {bmatrix} x_1 & x_2 & x_3 & 0 \\ 0 & x_1 & x_2 & x_3 \\ \end {bmatrix}.$

Making the appropriate identifications for each module, we can see that ${\mathbf {EN}}(\alpha )$ is the resulting graded complex

$$\begin{align*}{\mathbf{EN}}(\alpha): 0 \rightarrow S(-4)^3 \xrightarrow{d_3} S(-3)^8 \xrightarrow{d_2} S(-2)^6 \xrightarrow{d_1} S. \end{align*}$$

Note that the ideal of maximal minors in Example 3.8 is the ideal $\langle x_1, x_2, x_3 \rangle ^2$ . In general, the Eagon–Northcott complex minimally resolves any power of the maximal ideal ${\mathfrak {m}}^d$ by constructing the complex for the $d \times (n+d-1)$ matrix (e.g., per [Reference Eisenbud8, Exer. A2.17d])

$$\begin{align*}M^{n,d} = \begin{bmatrix} x_1 & x_2 & \ldots & x_n & 0 & \ldots & \ldots & 0 \\ 0 & x_1 & x_2 & \ldots & x_n & 0 & \ldots & 0 \\ \vdots & \ddots & \ddots & \ddots & & \ddots & \ddots & \vdots \\ 0 & \ldots & 0 & x_1 & x_2 & \ldots & x_n & 0 \\ 0 & \ldots & \ldots & 0 & x_1 & x_2 & \ldots & x_n \\ \end{bmatrix}. \end{align*}$$

With this notation, the matrix A in Example 3.8 is $M^{3,2}$ .

Our eventual goal is to understand, up to rank sequence, the possible subcomplexes of the Eagon–Northcott complex in a similar way as we did with the Koszul complex. We will see that these sequences are much more difficult to classify completely, but that we can still find restrictions that narrow down the possibilities. To do this, we will once again leverage the BGG correspondence in order to understand subcomplexes of Eagon–Northcott complexes. In order to do this, we must restrict ourselves to the linear maps–and therefore the degree d strand–of the Eagon–Northcott complex. Note, however, that with the exception of the first map, the Eagon–Northcott complex is always linear, and any subcomplex of the degree d strand will extend to a subcomplex of the entire Eagon–Northcott complex.

We define the complex $L_{n,d}$ to be the resolution of the ideal $\langle x_1, \ldots , x_n \rangle ^d$ as an S-module as presented by ${\mathbf {EN}}(M^{n,d})$ and note that $L_{n,d}$ is linear. Indeed, for $d\geq 2$ , $L_{n,d}$ is the degree–d strand of ${\mathbf {EN}}(M^{n,d})$ shifted by one homological degree, while for $d=1$ the entire complex is linear, so $L_{n,1}$ is not the entire linear strand since it is missing the first map. This complex corresponds to a specific E-module $N_{n,d}$ under the BGG correspondence, that is, $L_{n,d} =\mathbb {L}(N_{n,d})[-n+1]$ . Therefore, classifying subcomplexes of $L_{n,d}$ corresponds to understanding the E-submodules of $N_{n,d}(-n+1)$ .

In this way, our search for possible rank sequences of subcomplexes of $L_{n,d}$ is translated to a search for possible Hilbert functions of submodules of $N_{n,d}$ (again with appropriate twist). We denote this set of the possible Hilbert functions of submodules of $N_{n,d}$ by $\operatorname {HF}(N_{n,d})$ and prove this set satisfies certain constraints. Some persnickety bookkeeping is necessary proceeding.

Theorem 3.13. The set of possible Hilbert functions of submodules of $N_{n,d}$ is restricted by the following containment:

$$\begin{align*}\operatorname{HF}(N_{n,d}) \subseteq \operatorname{HF}(N_{n,d-1}) + \operatorname{HF}(\overline{N_{n-1,d}}). \end{align*}$$

Proof. First, we prove that

$$\begin{align*}0 \rightarrow N_{n,d-1} \rightarrow N_{n,d} \rightarrow \overline{N_{n-1,d}} \rightarrow 0 \end{align*}$$

is a short exact sequence. Note that ${\mathfrak {m}}^d$ is the S-module $S_{\geq d}$ , from which we get the exact sequence of S-modules

$$\begin{align*}0 \rightarrow S_{\geq d-1}(-1) \rightarrow S_{\geq d} \rightarrow S^{\prime}_{\geq d} \rightarrow 0, \end{align*}$$

where $S' = \mathbb {k}[x_1, \ldots , x_{n-1}]$ is an S-module in the usual way: $x_n \cdot f = 0$ for any $f \in S'$ .

Because $\mathbb {L}$ preserves exactness, this means that

$$\begin{align*}0 \rightarrow \mathbb{R}(S_{\geq d-1}(-1)) \rightarrow \mathbb{R}(S_{\geq d}) \rightarrow \mathbb{R}(S^{\prime}_{\geq d}) \rightarrow 0 \end{align*}$$

is a short exact complex of linear complexes of E-modules. In particular, we know what the kernel of each complex is: exactly the corresponding module N. Therefore,

$$\begin{align*}0 \rightarrow N_{n,d-1} \rightarrow N_{n,d} \rightarrow \overline{N_{n-1,d}} \rightarrow 0 \end{align*}$$

is indeed a short exact sequence of E-modules.

Now suppose that $N \subseteq N_{n,d}$ is a submodule with Hilbert function $h(N)$ . The image of N in $\overline {N_{n-1,d}}$ is also a submodule, which we will denote $N'$ . Let $N"$ be the kernel of the induced map $N\to N'$ . It is a submodule of $N_{n,d-1}$ , so we have a short exact sequence

$$\begin{align*}0\to N"\to N \to N' \to 0. \end{align*}$$

Because Hilbert functions sum over short exact sequences, we have $h(N) = h(N') + h(N")$ , so the Hilbert function of $N\subseteq N_{n,d}$ is realized as a sum of Hilbert functions of submodules of $\overline {N_{n-1,d}}$ and $N_{n,d-1}$ . Thus, we see that

$$\begin{align*}\operatorname{HF}(N_{n,d})\subset \operatorname{HF}(\overline{N_{n-1,d}}) + \operatorname{HF}(N_{n,d-1}).\\[-40pt] \end{align*}$$

Note that the containment in Theorem 3.13 is not an equality. We have shown that any Hilbert function of a submodule of $N_{n,d}$ may be realized as a sum of Hilbert functions of submodules of $\overline {N_{n-1,d}}$ and $N_{n,d-1}$ , but there may be submodules of $\overline {N_{n-1,d}}$ and $N_{n, d-1}$ the sum of whose Hilbert functions is not the Hilbert function of a submodule of $N_{n,d}$ . Indeed, the following example shows that the containment is strict in even a very small case.

Example 3.14. For this example, we will use $E = \mathbb {k}\langle e_1, e_2 \rangle $ and consider the E-module $N_{2,2}$ . From above, we have

$$\begin{align*}\operatorname{HF}(N_{2,2}) \subseteq \operatorname{HF}(\overline{N_{1,2}}) + \operatorname{HF}(N_{2,1}). \end{align*}$$

Recall that $N_{2,2} = \operatorname {\mathrm {coker}}\begin {bmatrix} e_1\\e_2\end {bmatrix}$ , $\overline {N_{1,2}} = \operatorname {\mathrm {coker}}\begin {bmatrix} e_1\end {bmatrix}$ , and $N_{2,1} = \operatorname {\mathrm {coker}}\begin {bmatrix} e_1e_2\end {bmatrix}$ . The module $\overline {N_{1,2}}$ has Hilbert function $(1,1,0)$ , while the submodule $0 \subset N_{2,1}$ has Hilbert function $(0,0,0)$ , so we get the sum $(1,1,0)+(0,0,0)$ as a potential Hilbert function for a submodule of $N_{2,2}$ . However, there is no submodule of $N_{2,2}$ with Hilbert function $(1,1,0)$ by the following argument.

The module $N_{2,2}$ has two generators in degree $0$ , which we will call $\alpha $ and $\beta $ . In degree $-1,$ we have $e_1\alpha , e_1\beta , $ and $e_2\alpha $ , with $e_1\alpha = -e_2\beta $ . Suppose we have a submodule $N\subset N_{2,1}$ with one generator in degree $0$ . If we denote this degree $0$ generator of N by $\zeta $ , then we have that $\zeta = a\alpha + b\beta $ for some $a,b\in \mathbb {k}$ . This gives us $2$ elements in degree $-1$ : $e_1\zeta = ae_1\alpha + be_2\beta $ and $e_2\zeta = ae_2\alpha +be_2\beta = (ae_2-be_1)\alpha $ . We can see that these are linearly independent since the only relation in N is the relation $e_1\alpha = -e_2\beta $ , so the Hilbert function of N cannot be $(1,1,0)$ .

Our goal now will be to restate Theorem 3.13 as a statement about $\operatorname {\mathrm {RS}}(L_{n,d})$ in terms of complexes for which we have a complete characterization of possible rank sequences, namely Koszul complexes. We first introduce some helpful notation: for a nonnegative integer m, we will write $E_{(m)} = E/\langle e_{m+1}, \ldots , e_n\rangle $ . Note that $E_{(m)} \cong \mathbb {k}$ when $m=0$ . Furthermore, we will write $I_i$ to refer to the ideal generated by the single degree $-i$ monomial $e_1e_2\dots e_i$ and we will make the convention that $I_0$ is the unit ideal. We will write ${\mathbf {K}}_{(m)}$ to mean the Koszul complex on m variables for $m\leq n$ .

Lemma 3.16. For $1\leq j\leq n$ , we have the equality of sets

$$\begin{align*}\operatorname{HF}(\overline{N_{1,d}}) = \operatorname{HF}(E_{(n-1)}). \end{align*}$$

Lemma 3.17. For $n,d\geq 2$ , the Hilbert functions of submodules of $N_{n,d}$ are restricted by the following containment:

$$\begin{align*}\operatorname{HF}(N_{n,d}) \subseteq \sum_{i=1}^n\binom{n+d-2-i}{n-i}\operatorname{HF}(\overline{N_{i,1}}). \end{align*}$$

Proof. Theorem 3.13 gives the containment

$$\begin{align*}\operatorname{HF}(N_{n,d}) \subset \operatorname{HF}(\overline{N_{n-1, d}}) + \operatorname{HF}(N_{n,d-1}). \end{align*}$$

We can then iterate until we have $\operatorname {HF}(N_{n,d})$ expressed completely in terms of $\operatorname {HF} (\overline {N_{1,j}})$ and $\operatorname {HF} (\overline {N_{i,1}})$ for $2\leq i,j\leq n$ . Ultimately, we reduce to

$$\begin{align*}\operatorname{HF}(N_{n,d})\subset \sum_{i=2}^n \alpha_{i,1}\operatorname{HF}(\overline{N_{i,1}}) + \sum_{j=2}^d \alpha_{1,j}\operatorname{HF}(\overline{N_{1,j}}), \end{align*}$$

where $\alpha _{i,j}$ counts the number of times that $N_{i,j}$ appears in the sum. This quantity $\alpha _{i,j}$ is the number of times that $(i,j)$ appears as the result of repeatedly subtracting $(1,0)$ and $(0,1)$ from $(n,d)$ , with the caveat that, since $(1,i+1)$ is a base case, we never reach $(1,i)$ by subtracting $(0,1)$ from $(1,i+1)$ , and similarly for $(1,j)$ . One can thus interpret $\alpha _{i,1}$ as the number of integer lattice paths from $(i,2)$ to $(n,d)$ and $\alpha _{1,j}$ as the number of integer lattice paths from $(2,j)$ to $(n,d)$ . The number of such lattice paths from $(i,j)$ to $(n,d)$ is given by $\binom {n-i+d-j}{n-i}$ . This gives

$$\begin{align*}\operatorname{HF}(N_{n,d})\subset \sum_{i=2}^n \binom{n+d-2-i}{n-i}\operatorname{HF}(\overline{N_{i,1}}) + \sum_{j=2}^d \binom{n+d-2-j}{n-2}\operatorname{HF}(\overline{N_{1,j}}). \end{align*}$$

Since $N_{1,j} = \mathbb {k} \langle e_1 \rangle / \langle e_1 \rangle = \mathbb {k}$ , we see that $\overline {N_{1,j}} = \mathbb {k} \otimes \mathbb {k} \langle e_2, \ldots , e_n \rangle \cong E_(n-1)$ , so we can write the sum

$$\begin{align*}\sum_{j=2}^d \binom{n+d-2-j}{n-2}\operatorname{HF}(\overline{N_{1,j}}) = \left( \sum_{j=2}^d \binom{n+d-2-j}{n-2}\right)\operatorname{HF}(E_{(n-1)}). \end{align*}$$

We can reindex and convert via the hockey stick identity to see that

$$\begin{align*}\sum_{j=2}^d \binom{n+d-2-j}{n-2} = \sum_{k=n-2}^{n+d-4}\binom{k}{n-2} = \binom{n+d-3}{n-1} \end{align*}$$

so we have

$$ \begin{align*} \operatorname{HF}(N_{n,d})&\subset \binom{n+d-3}{n-1}\operatorname{HF}(E_{(n-1)}) + \sum_{i=2}^n \binom{n+d-2-i}{n-i}\operatorname{HF}(\overline{N_{i,1}})\\ &= \sum_{i=1}^n \binom{n+d-2-i}{n-i}\operatorname{HF}(\overline{N_{i,1}}), \end{align*} $$

where the incorporation of the first term into the sum uses the fact that $E_{(n-1)} \cong \overline {N_{1,1}}.$

Lemma 3.18. For $1\leq i\leq n$ , we have the containment of sets

$$\begin{align*}\operatorname{HF}(\overline{N_{i,1}}) \subseteq \sum_{j=0}^{i-1} \operatorname{HF}(E_{(n-j-1)}(j)). \end{align*}$$

Proof. When $d = 1$ , the module $N_{i,1}$ is easily computable as $N_{i,1} = \operatorname {\mathrm {coker}} \begin {bmatrix} e_1 \dots e_i \end {bmatrix} = \mathbb {k} \langle e_1, \ldots , e_i \rangle / I_i$ , and so $\overline {N_{i,1}} = (E_{(i)}/I_i)\otimes \mathbb {k} \langle e_{i+1}, \ldots , e_n \rangle = E/I_i$ .

Now, we can reduce using the short exact sequences of $E/I_i$ -modules

$$\begin{align*}0\to \langle e_i\rangle E/I_i \to E/I_i\to E/(\langle e_i\rangle + I_i)\to 0. \end{align*}$$

But $\langle e_i\rangle E/I_i \cong E_{(n-1)}/I_{i-1}(1)$ and that $E/(\langle e_i\rangle + I_i) \cong E_{(n-1)}$ , so by a similar argument as we have used previously in the proof of Theorem 3.13, we may now write

$$\begin{align*}\operatorname{HF}(E/I_m)\subseteq \operatorname{HF}(E_{(n-1)}/I_{m-1}(1)) + \operatorname{HF}(E_{(n-1)}). \end{align*}$$

Now, we can split $\operatorname {HF}(E_{(n-1)}/I_{m-1}(1))$ and proceed inductively to get

$$\begin{align*}\operatorname{HF}(E/I_i) \subseteq \sum_{j=0}^{i-1}\operatorname{HF}(E_{(n-j-1)}(j)).\\[-50pt] \end{align*}$$

Theorem 3.19. For $n,d\geq 2$ , the rank sequence of any subcomplex of $L_{n,d}$ can be written as a positive integral sum of rank sequences of Koszul subcomplexes on fewer than n variables. In particular,

$$\begin{align*}\operatorname{\mathrm{RS}}(L_{n,d}) \subseteq \sum_{j=0}^{n-1}\binom{n-j+d-2}{d-1}\operatorname{\mathrm{RS}}({\mathbf{K}}_{(n-j-1)}). \end{align*}$$

Proof. Combining Lemmas 3.17 and 3.18, we can see that

$$\begin{align*}\operatorname{HF}(N_{n,d}) \subseteq \sum\limits_{i=1}^{n} \sum\limits_{j=0}^{i-1} \binom{n+d-2-i}{d-2} \operatorname{HF}(E_{(n-j-1)}(j)). \end{align*}$$

Twisting each side of this equality by $(-n+1)$ gives

$$\begin{align*}\operatorname{HF}(N_{n,d}(-n+1)) \subseteq \sum\limits_{i=1}^{n} \sum\limits_{j=0}^{i-1} \binom{n+d-2-i}{d-2} \operatorname{HF}(E_{(n-j-1)}(-n+j+1)), \end{align*}$$

which, fed through the functor $\mathbb {L}$ , yields a containment of sets of rank sequences:

$$\begin{align*}\operatorname{\mathrm{RS}}(L_{n,d}) \subseteq \sum\limits_{i=1}^{n} \sum\limits_{j=0}^{i-1} \binom{n+d-2-i}{d-2} \operatorname{\mathrm{RS}}({\mathbf{K}}_{(n-j-1)}). \end{align*}$$

We can switch the order of the double sum, reindex, and apply the hockey stick identity once again to conclude the proof:

$$\begin{align*} \sum\limits_{i=1}^n \sum\limits_{j=0}^{i-1} \binom{n+d-2-i}{d-2} \operatorname{\mathrm{RS}}({\mathbf{K}}_{(n-j-1)}) &\subseteq \sum\limits_{j=0}^{n-1} \sum\limits_{i=j+1}^n \binom{n+d-2-i}{d-2} \operatorname{\mathrm{RS}}({\mathbf{K}}_{(n-j-1)}) \\ &= \sum\limits_{j=0}^{n-1} \left( \sum\limits_{i=0}^{n-j-1} \binom{d-2+i}{d-2} \right) \operatorname{\mathrm{RS}}({\mathbf{K}}_{(n-j-1)}) \\ &= \sum\limits_{j=0}^{n-1} \binom{n-j+d-2}{d-1} \operatorname{\mathrm{RS}}({\mathbf{K}}_{(n-j-1)}). \\[-50pt] \end{align*}$$

Example 3.20. Let $n=4, d=3$ . The complex $L_{4,3}$ resolves the ideal of maximal minors of the matrix

$$\begin{align*}\begin{bmatrix} x_1 & x_2 & x_3 & x_4 & 0 & 0\\ 0 & x_1 & x_2 & x_3 & x_4 & 0\\ 0 & 0 & x_1 & x_2 & x_3 & x_4 \end{bmatrix} \end{align*}$$

and has the form

$$\begin{align*}0 \to S^{10} \to S^{36} \to S^{45} \to S^{20} \to 0. \end{align*}$$

We can use Theorem 3.19 to rule out some integer sequences as possible rank sequences for subcomplexes of $L_{4,3}$ . The containment in Theorem 3.19 states that

$$ \begin{align*} \operatorname{\mathrm{RS}}(L_{4,3}) &\subseteq\sum_{j=0}^3\binom{5-j}{2}\operatorname{\mathrm{RS}}({\mathbf{K}}_{(3-j)})\\ &= 10\operatorname{\mathrm{RS}}({\mathbf{K}}_{(3)}) + 6\operatorname{\mathrm{RS}}({\mathbf{K}}_{(2)}) + 3\operatorname{\mathrm{RS}}({\mathbf{K}}_{(1)}) + \operatorname{\mathrm{RS}}({\mathbf{K}}_{(0)}), \end{align*} $$

that is, any rank sequence of a subcomplex must be expressible as a sum of 10 rank sequences of subcomplexes of the Koszul complex on 3 variables, 6 rank sequences of subcomplexes of the Koszul complex on 2 variables, 3 rank sequences of subcomplexes of the Koszul complex on 1 variable, and 1 rank sequence of a subcomplex of the Koszul complex on 0 variables.

If we consider the Koszul complex on three variables, Theorem 3.5 tells us that any rank sequence $(r_0, r_1, r_2, r_3)$ of a subcomplex of ${\mathbf {K}}_{(3)}$ with $r_3 = 1$ must have $r_2 = 3$ . This means that for a rank sequence $(r_0, r_1, r_2, r_3)$ of a subcomplex of $L_{4,3}$ , we must have $r_2\geq 3r_3$ . For instance, we may say for certain that the sequence $(10,16,20,8)$ is not a possible rank sequence of a subcomplex of $L_{4,3}$ , since $20<3\cdot 8$ . In this way, we are able to use our complete characterization of rank sequence of Koszul subcomplexes to eliminate certain potential rank sequences from consideration in the Eagon–Northcott case.

4.1 The general Koszul complex

The Koszul complex can be defined more generally to give a minimal free resolution of a complete intersection. For $f_1, \ldots , f_m \in S$ , a regular sequence of homogeneous elements, we replace the differential in definition 3.1 by ${\partial }_d(e_T) = \sum _{j = 1}^{d} (-1)^{j} f_{i_j} e_{T - i_j}$ , adjusting the twists accordingly.

Theorem 4.1. Let $f_1, \ldots , f_m$ be a regular sequence of homogeneous polynomials in S. An integer sequence $r = (r_0, \ldots , r_m)$ is in $\operatorname {\mathrm {RS}}({\mathbf {K}}(f_1, \ldots , f_m))$ if and only if it satisfies

$$\begin{align*}0 \leq r_{i+1} \leq r_i^{(i)} \text{ for } 1 \leq i \leq m-1. \end{align*}$$

Proof. We will show that $\operatorname {\mathrm {RS}}({\mathbf {K}}(f_1, \ldots , f_m)) = \operatorname {\mathrm {RS}}({\mathbf {K}}(x_1, \ldots , x_m))$ , then apply Theorem 3.5.

Let ${\mathbf {K}}$ be the Koszul complex on the variables $x_1, \ldots , x_m$ . For ${\mathbf {K}}'$ a general Koszul complex ${\mathbf {K}}(f_1, \ldots , f_m)$ over the ring $S' = \mathbb {k}[y_1, \ldots , y_n]$ , there is a map ${\mathbf {K}} \to {\mathbf {K}}'$ induced by the map $S\to S'$ sending $x_i$ to $f_i$ .

If ${\mathbf {F}}$ is a subcomplex of ${\mathbf {K}}$ , then the image of ${\mathbf {F}}$ under this map is a subcomplex of ${\mathbf {K}}$ with the same rank sequence. So any possible rank sequence of a subcomplex of ${\mathbf {K}}$ must also be possible for a subcomplex of ${\mathbf {K}}'$ .

To see that the possible rank sequences for subcomplexes of ${\mathbf {K}}'$ are exactly those that are possible for subcomplexes of ${\mathbf {K}}$ , we need to check that given a subcomplex ${\mathbf {F}}'$ of ${\mathbf {K}}'$ , the differentials of ${\mathbf {F}}'$ are described by matrices over the subalgebra $R = \mathbb {k}[f_1, \ldots , f_m] \subseteq S'$ .

For each i, we have

where the vertical maps are given by matrices over $\mathbb {k}$ . So the differential ${\partial }$ is a matrix over R if and only if ${\partial }'$ is. But entries of ${\partial }'$ are linear in the $f_i$ , so they are defined as matrices over R.

Now given a subcomplex ${\mathbf {F}}'$ of ${\mathbf {K}}'$ , we need only replace each $F^{\prime }_i$ by a free S-module of the same rank and each $f_i$ in the differential by $x_i$ to obtain a subcomplex ${\mathbf {F}}$ of ${\mathbf {K}}$ with the same rank sequence.

4.2 More general Eagon–Northcott complexes

Just as the Koszul complex can be generalized to give a minimal free resolution of a complete intersection, the Eagon–Northcott complex can be generalized to give a minimal free resolution of certain Cohen–Macaulay algebras of the form $S/I$ , where I has the maximum possible codimension. We can relate the behavior of subcomplexes of the Eagon–Northcott complex resolving ${\mathfrak {m}}^d$ to the behavior of subcomplexes of a general Eagon–Northcott complex as follows. First, we consider a motivating example.

Example 4.2. Let $Y = [y_{i,j}]$ be a $p \times q$ generic matrix with $p \leq q$ . Then there is a containment of sets

$$\begin{align*}\operatorname{\mathrm{RS}}({\mathbf{EN}}(Y)) \subset \operatorname{\mathrm{RS}}({\mathbf{EN}}(M^{q-p+1,p})).\end{align*}$$

Let $n = pq$ , so our matrix is a map $S^q\to S^p$ for $S = \mathbb {k}[y_{i,j}] \cong \mathbb {k}[x_1, \ldots , x_n]$ . This specializes to the matrix

$$\begin{align*}M^{q-p+1,p} = \begin{bmatrix} x_1 & x_2 & \cdots &x_{q-p+1} & 0 &\cdots & 0\\ 0 & x_1 & \cdots & x_{q-p} & x_{q-p+1} & \cdots & 0\\ \vdots & & \ddots & & \ddots & \ddots & \vdots\\ 0 & \cdots & 0 & x_1 & \cdots & x_{q-p} & x_{q-p+1} \end{bmatrix} \end{align*}$$

under a map that we will call $\varphi $ . The maximal minors of this matrix define the ideal $(x_1, \ldots , x_{q-p+1})^p$ .

The map $\varphi $ gives us a map of complexes ${\mathbf {EN}}(Y) \to {\mathbf {EN}}(M^{q-p+1,p})$ . What is more, under $\varphi $ any subcomplex of ${\mathbf {EN}}(Y)$ gives a subcomplex of ${\mathbf {EN}}(M^{q-p+1,p})$ .

This gives us a containment

(4.1) $$ \begin{align} \operatorname{\mathrm{RS}}({\mathbf{EN}}(Y)) \subset \operatorname{\mathrm{RS}}({\mathbf{EN}}(M^{q-p+1,p})). \end{align} $$

Note that the generic nature of Y had no bearing on the argument in Example 4.2, so a more general statement relating general Eagon–Northcott complexes to the complex ${\mathbf {EN}}(M^{n,d})$ holds by the same reasoning.

Theorem 4.3. Let Z be a $p \times q$ matrix whose maximal minors define an ideal I whose codimension is $q-p+1$ and where $S/I$ is Cohen–Macaulay (equivalently, I has grade $q-p+1$ ). Then there is a containment of sets

$$\begin{align*}\operatorname{\mathrm{RS}}({\mathbf{EN}}(Z)) \subset \operatorname{\mathrm{RS}}({\mathbf{EN}}(M^{q-p+1,p})).\end{align*}$$

While this theorem relates rank sequences of subcomplexes of the entire complexes ${\mathbf {EN}}(Z)$ and ${\mathbf {EN}}(M^{q-p+1,p})$ rather than their degree d strands, Remark 3.9 tells us that our restrictions on $\operatorname {\mathrm {RS}}(L_{q-p+1,p})$ , together with the above theorem, still give us valuable information about $\operatorname {\mathrm {RS}}({\mathbf {EN}}(Z))$ . It should be noted, however, that the result in Theorem 4.3 is a strict containment, as demonstrated in the following example.

Example 4.4. Let $S = \mathbb {k}[x,y,z,w]$ . Consider the Eagon–Northcott complex on the matrix $\begin {bmatrix} 0 & y & z\\ y & z & 0 \end {bmatrix}$ , which resolves the square of the maximal ideal in the subalgebra $\mathbb {k}[y,z]\subset S$ . This is a specialization of the Eagon–Northcott complex on $\begin {bmatrix} x & y & z\\ y & z & w \end {bmatrix}$ obtained via the map $\varphi : S\to S$ defined by $x,w\mapsto 0$ and $y,z\mapsto y,z$ , so we have the following map of complexes:

Consider the following subcomplex of ${\mathbf {F}}$ :

which has $\operatorname {\mathrm {rs}}({\mathbf {G}}) = (1,2,1)$ . The subcomplex ${\mathbf {G}}$ is realized as the image of

under $\varphi ^*$ . However, one can check that ${\mathbf {G}}'$ is not a subcomplex of ${\mathbf {F}}'$ . Moreover, a straightforward linear algebra computation confirms that there is no subcomplex of ${\mathbf {F}}'$ with rank sequence $(1,2,1)$ , so $\operatorname {\mathrm {RS}}({\mathbf {F}})\subsetneq \operatorname {\mathrm {RS}}({\mathbf {G}})$ .