Minimal free resolutions of sub-permanents

# On minimal free resolutions of sub-permanents and other ideals arising in complexity theory

## Abstract.

We compute the linear strand of the minimal free resolution of the ideal generated by sub-permanents of an generic matrix and of the ideal generated by square-free monomials of degree . The latter calculation gives the full minimal free resolution by [1]. Our motivation is to lay groundwork for the use of commutative algebra in algebraic complexity theory. We also compute several Hilbert functions relevant for complexity theory.

###### Key words and phrases:
Computational Complexity, Free Resolution, Determinant, Permanent
###### 2001 Mathematics Subject Classification:
68Q17, 13D02, 14L30, 20B30
Landsberg supported by NSF DMS-1405348
Schenck supported by NSF DMS-1312071
Weyman supported by NSF DMS-1400740

## 1. Introduction

We study homological properties of two families of ideals over polynomial rings: the ideals generated by square-free monomials of degree in variables and the ideals generated by sub-permanents of an generic matrix. Recall that the permanent of an matrix is the polynomial

 permm(Y)=∑σ∈Smy1,σ(1)y2,σ(2)⋯ym,σ(m),

where denotes the symmetric group on elements.

We obtain our results via larger ideals (resp. ). The ideal is generated by all monomials of degree in variables. The ideal is generated by permanents of matrices produced from where repetition of rows and columns is allowed. Invariantly, is the ideal generated by and is the ideal generated . The main result in each case says that the linear strand of resolution of (resp. ) is the subcomplex of the linear strand of the resolution of (resp. of ) consisting of elements of regular weights (cf. §2).

Our motivation comes from complexity theory. We seek to find differences between the homological behavior of ideals generated by minors (i.e., subdeterminants) of the generic matrix and the ideals generated by subpermanents. The ideal generated by square-free monomials arises as the -th Jacobian ideal of the monomial .

### Acknowledgments

We thank the anonymous referee for numerous useful suggestions, including simplifications of several proofs. Efremenko, Landsberg and Weyman thank the Simons Institute for the Theory of Computing, UC Berkeley, for providing a wonderful environment during the fall 2014 program Algorithms and Complexity in Algebraic Geometry to work on this article.

## 2. Preliminaries

### 2.1. Representation Theory

For proofs of the statements here, see, e.g., [3] or [9, Ch. 2]. We work exclusively over the complex numbers , although our results hold for an arbitrary field of characteristic . If is a –vector space of dimension , a choice of basis determines a maximal torus of diagonal matrices and a labeling of weights for the torus by -tuples . A weight is dominant if . Irreducible representations of are in one-to-one correspondence with dominant weights . Let denote the irreducible representation associated to . Write for the size of . A weight is regular if each is equal to or . The regular weights will play an important role in stating our results.

When is a dominant weight with , we say that is a partition of , and we write . When dealing with partitions we often omit the trailing zeros. Associated to a partition is its Young diagram which consists of left–justified rows of boxes, with boxes in the –th row: for example, the Young diagram associated to is

 \Yvcentermath1\yng(5,2,1).

The transpose of a partition is obtained by transposing the corresponding Young diagram. For the example above, .

Given finite dimensional –vector spaces , the Cauchy formulas describe the decomposition of the symmetric and exterior powers of into a sum of irreducible –representations, see, e.g., [9, Cor. 2.3.3]:

 (1) Symn(F⊗G)=⨁λ⊢nSλF⊗SλG, n⋀(F⊗G)=⨁λ⊢nSλF⊗Sλ′G.

Let denote the ideal generated by .

### 2.2. GLm and Sm representations

Let equipped with its standard basis. The symmetric group is then contained in as the permutation matrices. Consider the irreducible representation where is a partition of . Inside of we have the -submodule spanned by the elements of regular weight . This submodule is denoted , the Specht module corresponding to . The representations are the distinct irreducible representations of (see, e.g., [5]). Write

 (SλCm)reg=[λ].

Recall that for finite groups , and an -module , is the induced -module. For and ,

 (SλCn)reg≡IndSnSm×Sn−m([λ]⊗[n−m]).

Introduce the notation , and if is a partition of , write for the module that is as an -module and trivial as an -module.

### 2.3. Finite free resolutions

Let be the ring of polynomials in variables equipped with its grading by degree. Let denote its -th graded component. Let denote the maximal ideal . Let be a graded -module. A complex of free graded -modules

 F:0→FNdN→FN−1→⋯→F1d1→F0

is a minimal free resolution of if the only homology of is , and are maps of degree such that for all .

Define the graded Betti numbers of by

 Fi=⊕j≥0S(−i−j)βi,j

where denotes a copy of with generator in degree .

If is an ideal generated in degree , then . The linear strand of is a subcomplex

 Flin:0→FlinNdN→FlinN−1→⋯→Flin1d1→Flin0

where . The graded Betti numbers have the interpretation in terms of functors

 βi,j=dimCTorSi(S/S+,M)i+j

where the subscript denotes the homogeneous component. This means that can be calculated as

 (2) TorSi(S/S+,M)i+j=Hi(C(x1,…,xN;M))i+j

where is the Koszul complex defined by and the differential by

 d(ej1∧⋯∧eji⊗m)=i∑u=1(−1)u+1ej1∧⋯∧^eju∧⋯∧eji⊗xjum.

## 3. The linear strands of the minimal free resolutions of the ideals Ipermn,k

### 3.1. The resolution of I1×k(n,n)

The ideal is the ideal generated by , where . The minimal free resolution of this ideal is known (see, e.g., [7], where it is denoted by ). The linear components of this resolution are generated by

 (3) F––linj=⨁a+b=jS(k+b,1a)E.

So the -th linear term is .

Since the resolution is -equivariant, each module in the complex has a double weight decomposition induced by the restricted action of pairs of diagonal matrices.

### 3.2. The main result

We work over .

Define a sub-complex of the complex given by (3) by setting to be the subspace of spanned by the basis elements of regular content. Note that is indeed a sub-complex of . Let , denote the span of the first basis vectors.

###### Theorem 3.1.

When , the complex is the linear part of the minimal free resolution of the ideal . Moreover, .

As an -module,

 (4) H––linj=IndSn×Sn~Sκ+j×~Sκ+j(⨁a+b=j˜[κ+b,1a]Eκ+j⊗˜[κ+a,1b]Fκ+j).
###### Proof.

Consider the complex giving the linear strand of the ideal . The term consists of the generators of , the space .

Inside is the ideal generated by the sub-permanents which consists of the subspace of regular weights. Note that the set of regular vectors in any (where we assume in order for the set of such vectors to be nonempty) spans a -submodule.

The linear strand of the -the term in the minimal free resolution of the ideal is also a -submodule of . We claim this sub-module is generated by the span of the regular vectors. In what follows denotes the sub-permanent formed from rows and columns .

We work by induction, the case was discussed above. Assume the result has been proven up to homological degree and consider the homological degree and homogeneous degree . The generators of the -th module in the linear strand of the resolution of have to be contained in linear part of , so all its weights are either regular, or such that one of the row indices is , and/or one of the column indices is , and all other are zero or . Call such a weight sub-regular. It remains to show that no linear syzygy with a sub-regular weight can appear. To do this we show that no sub-regular weight vector in maps to zero in .

First consider the case where both the and weights are sub-regular, then (because the space is a -module), the weight must appear in the syzygy. The only way for this to appear is to have a term divisible by . But, since is not a zero-divisor in , such a term cannot map to zero because our syzygy is a syzygy of degree zero multiplied by . But by minimality no such syzygy exists.

Finally consider the case where there is a vector of weight appearing. Here it is more convenient to look at the calculation of the free resolution using the Koszul complex. Such a syzygy would give a Koszul cycle with summands of the form

 (5) z=∑tatea1,t,b1,t∧⋯∧eaj,t,bj,t⊗p(It;Jt)

where are subpermanents formed from distinct rows and columns and . The total weight is and the Koszul differential is zero. Consider this differential. The coefficients of all the basis elements of have to be zero. There are three kinds of basis elements: the indices can contain number twice, once or not contain at all. Consider the basis elements not containing , say the element . The only elements that can appear on the right hand side of the tensor product in are the elements , for .

###### Lemma 3.2.

Let . The elements , for are linearly independent in .

###### Proof.

After re-labeling, the lemma amounts to showing the polynomials are linearly independent, where is the permanent of the matrix obtained by removing the -th column of

 ⎛⎜ ⎜ ⎜ ⎜⎝x1,1x1,2…x1,k+1x2,1x2,2…x2,k+1…………xk,1xk,2…xk,k+1⎞⎟ ⎟ ⎟ ⎟⎠.

Say that

 (6) k+1∑s=1bsx1,sPs=0

for some scalars . We need to show that all are zero. By symmetry it suffices to show that . So set for . Using the Laplace expansion of permanents along the first row, and writing for the permanent obtained by removing row and columns we can rewrite (6) as

 b1x1,1(x1,2P1,2+x1,3P1,3)+b2x1,2(x1,1P1,2+x1,3P1,3)+b3(x1,1P1,3+x1,2P2,3)=0

which gives for , so . ∎

Lemma 3.2 implies that in all the summands in of (5) with , both appearances of the index have to occur among . Now consider the coefficient of the basis element where occurs among , say, in . We obtain a linear combination of the elements for . But these elements are trivially linearly independent (all monomials occurring in them are different) so all coefficients are zero.

The rest of Theorem 3.1 follows because if is a partition of then the weight subspace of , considered as an -module, is (see, e.g., [5]), and the space of regular vectors in is . In the formula for the dimension of , the factor is explained by inducing. The dimensions of hook Specht modules are binomial coefficients, So we need to prove that

 ∑a+b=j(k+j−1a)(k+j−1b)=(2(k+j−1)j)

This has a combinatorial explanation. Given balls, white and black, both sides of the equation calculate number of choices of of them: the left side partitions into how many white (a) and how many black (b) are chosen. ∎

###### Remark 3.3.

For small and , computer computations show no additional first syzygies on the sub-permanents of a generic matrix (besides the linear syzygies) in degree less than the degree of the Koszul relations . For example, for and , there are cubic generators for the ideal and minimal first syzygies of degree six. There can be at most Koszul syzygies, so there must be additional non-Koszul first syzygies.

## 4. The minimal free resolutions of the ideals Isqf;n,k

### 4.1. The resolutions of I1×k(1,n)

Next we consider the case , . In this case and the ideal is just the ideal generated by all monomials of degree . The resolution of this ideal is well-known, see, e.g., [7] or [2].

The whole resolution is linear and -equivariant, and its -th term is

 (7) Fj=S(k,1j)F⊗S(−k−j).

### 4.2. The resolution of Isqf;n,k

We work over .

Define a subcomplex of the complex given by (7) by setting to be the subspace of spanned by the basis elements of regular weight. Note that

 Hlin=⊕jH––linj⊗S(−k−j)

is indeed a subcomplex of .

###### Theorem 4.1.

The complex is the linear part of the minimal free resolution of the ideal . We have the -module decomposition

 (8) H––linj=IndSn~Sκ+j˜[κ,1j],

which has dimension .

###### Proof.

We want to show that is the linear strand of the resolution of . We proceed by induction on . The case is clear because the generators of are precisely the generators of with regular weights. Assume we proved the result for and consider the -th module in the resolution. As with the subpermanent case, it is enough to consider the elements of a subregular weight as linear relations between elements of regular weight are either of regular or subregular weight.

Consider the syzygies in homological dimension and in homogeneous degree in term of cycles in the Koszul complex . These will be cycles of the form

 z=∑tatea1,t∧…∧eaj,t⊗xu1,txu2,t…xuk,t

where the total weight is and all monomials are of regular weights, and are scalars. In each summand with we are forced to have one among and one among . So we can assume that in each summand with we have and . We have . But, looking at the coefficient of with respect to the basis vector we see that its coefficient is just which forces to be zero. The dimension formula follows as in Theorem 3.1. ∎

###### Remark 4.2.

The easiest way to see that the resolution of the ideal is linear and to see the ranks of the modules is to observe that for the matrix

 M(A,X)=⎛⎜⎝a1,1x1…a1,nxn………ak,1x1…ak,nxn⎞⎟⎠

where

 A=⎛⎜⎝a1,1…a1,n………ak,1…ak,n⎞⎟⎠

is a matrix of scalars with all maximal minors non-zero, the ideal of maximal minors of the matrix is just , so the resolution in question is an Eagon-Northcott complex ([2] or [9, 6.1.6]). After this paper was submitted, a characteristic free description of the resolution of , with explicit differentials appeared in [4].

For , let denote the ideal generated by the partial derivatives of of order .

### 5.1. Size two subpermanents

###### Theorem 5.1.

Let denote the degree component of the ideal generated by the size two sub-permanents of an matrix, so . Then

 dimC[xi,j]/Ipermn,2t= (n2+t−1t)−[(nt)2+n2+(t−1)((n22)−(n2)2)+2(t−12)((n2)2+n(n3)) +2nt−1∑j=3(t−1j)(nj+1)],

Where recall that for , in which case the formula is minus the value of the Hilbert polynomial at .

First, the Hilbert polynomial:

###### Theorem 5.2.

For the ideal of permanents of an matrix, the Hilbert polynomial of is

 (9) n∑i=0fi(t−1i),

where is the entry in the vector

 [n2,(n22)−(n2)2,2(n2)2+2n(n3),2n(n4),2n(n5),…,2n(nn)].
###### Proof.

[6, Thm. 3.2] gives a Gröbner basis for , the radical of , and by [6, Thm. 3.3], has finite length, so vanishes in high degree. The Hilbert polynomial only measures dimension asymptotically, so

By [6], for any diagonal term order, the Gröbner basis for is given by quadrics of the form

 xijxkl+xkjxil with i

and five sets of cubic monomials

 (10) xi1j1xi1j2xi2j3i1>i2j1i2j1j2xi1j1xi2j2xi3j2i1j2xi1j1xi2j2xi3j3i1j2>j3.

The key observation is that all the cubic monomials are square-free, as are the initial terms of the quadrics. Thus the initial ideal of is a square-free monomial ideal and corresponds to the Stanley-Reisner ideal of a simplicial complex . By [8, Lemma 5.2.5], the Hilbert polynomial is as in Equation (9), where is the number of -dimensional faces of . As the vertex set of corresponds to all lattice points with , it is immediate that .

Since is a non-face if , , no edge connects a southwest lattice point to a northeast lattice point. Hence, the edges of consist of all pairs with and , of which there are .

Next, consider the triangles of . Equation (10) says there are no triangles in of the types in Figure 1. Also, there are no triangles which contain an edge connecting vertices at positions and with ,. Thus, the only triangles in are right triangles, but with hypotenuse sloping from northwest to southeast. For a lattice point at position there are exactly right triangles having as their unique north-most vertex. In the rightmost column , there are no such triangles, in the next to last column there are such triangles. Continuing this way yields a total count of

 (n−1)(n2)+(n−2)(n2)+⋯2(n2)+(n2)=(n2)2

such right triangles, and taking into account the right triangles for which is the unique south-most vertex doubles this number.

However, this count neglects thin triangles–those which have all vertices in the same row or column. Since the number of thin triangles is , the final count for the triangles of is

 2(n2)2+2n(n3).

For tetrahedra, the conditions of Equation (10) imply that there can only be thin tetrahedra, and an easy count gives such. The same holds for higher dimensional simplices, and concludes the proof. ∎

###### Corollary 5.3.

For the ideal of permanents of an matrix, the Hilbert function of is, when ,

 (11) HF(Sym(V)/Ipermn,2,t)=(nt)2+HP(Sym(V)/Ipermn,2,t),

and it equals the Hilbert polynomial for .

###### Proof.

The Hilbert function of in degree is by [6, Thm. 3.3]. The result follows by combining Equation 9 with the short exact sequence

 0⟶√Ipermn,2/Ipermn,2⟶Sym(V)/Ipermn,2⟶Sym(V)/√Ipermn,2⟶0,

and additivity of the Hilbert function. ∎

For the purposes of comparing with other ideals, we rephrase this as:

###### Corollary 5.4.

. For :

 dimIpermn,2t= (n2+t−1t)−[(nt)2+n2+(t−1)((n22)−(n2)2)+2(t−12)((n2)2+n(n3)) +2nt−1∑j=3(t−1j)(nj+1)].

### 5.2. Hilbert functions for ideals of square-free monomials

Although these can be deduced from our resolutions, we present the Hilbert functions and polynomials for the ideals generated by square-free monomials.

###### Proposition 5.5.

The Hilbert function of in degree is

 (12) dimI(x1⋯xn),κκ+t=n−κ∑j=0(nκ−j)(κ+t−1κ+j−1)
###### Proof.

The ideal in degree has a basis of the distinct monomials of degree containing at least distinct indices. When we divide such a basis vector by the denominator will have degree at most . For each , the space of possible numerators with a denominator of degree that is fixed, has dimension , and there are possible denominators. Summing over gives the result. ∎

For the Hilbert function of the coordinate ring, we have the following expression:

###### Proposition 5.6.

The Hilbert function of in degree is

 (13) dim(Sym(Cn)/Ix1⋯xn,κ)t=n−κ−2∑j=0(nj+1)(t−1j),

if , and if .

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