Finite Fields and Their Applications 24 (2013) 15–28

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Finite Fields and Their Applications

www.elsevier.com/locate/ffa

The Splitting Subspace Conjecture

Eric Chen a,∗, Dennis Tseng b

a Princeton University, Princeton, NJ 08544, USAb Massachusetts Institute of Technology, Cambridge, MA 02139, USA

a r t i c l e i n f o a b s t r a c t

Article history:Received 25 August 2012Revised 25 April 2013Accepted 15 May 2013Available online 1 June 2013Communicated by L. Storme

MSC:05A9911T7115A03

Keywords:Finite fieldsSplitting subspacesFinite dimensional vector spacesq-Chu–Vandermonde identityEnumeration

We answer a question of Niederreiter concerning the enumerationof a class of subspaces of finite dimensional vector spaces overfinite fields by proving a conjecture of Ghorpade and Ram.

© 2013 Elsevier Inc. All rights reserved.

1. Introduction

We positively resolve the Splitting Subspace Conjecture, stemming from a question of Niederreiter(1995) [5, p. 11] and stated by Ghorpade and Ram [3,4]. This conjecture was inspired by earlier workfrom Zeng, Han and He [6] and Ghorpade, Hasan and Kumari [2]. We first define the notion of aσ -splitting subspace.

* Corresponding author. Fax: +1 678 752 7664.E-mail addresses: ecchen@princeton.edu (E. Chen), DennisCTseng@gmail.com (D. Tseng).

1071-5797/$ – see front matter © 2013 Elsevier Inc. All rights reserved.http://dx.doi.org/10.1016/j.ffa.2013.05.006

16 E. Chen, D. Tseng / Finite Fields and Their Applications 24 (2013) 15–28

Definition 1.1. In the vector space Fqmn over the finite field Fq , given a σ ∈ Fqmn such that Fqmn =Fq(σ ), an (m-dimensional) subspace W of Fqmn is a σ -splitting subspace if

W ⊕ σ W ⊕ · · · ⊕ σ n−1W = Fqmn .

For example, {1, σ n, σ 2n, . . . , σ (m−1)n} spans a σ -splitting subspace. If n = 1, then Fqmn is the onlyσ -splitting subspace; if m = 1, then each 1-dimensional subspace of Fqmn is σ -splitting.

Conjecture (Ghorpade–Ram). The number of σ -splitting subspaces is

qmn − 1

qm − 1qm(m−1)(n−1).

This follows as Corollary 3.4 from our main result, Theorem 3.3. The next two sections are devotedto proving this theorem. We will first construct a recursion that gives the cardinality of certain setsof flags of subspaces

W1 ⊇ W2 ⊇ · · · ⊇ Wk

with agreeable properties relative to σ . In particular, the recursion will give the cardinality of the setof such flags with dim(W1) = m and W j = ⊕ j−1

i=0 σ i W1 for 1 � j � k = n; the corresponding W1 arethen exactly the σ -splitting subspaces. Solving this recursion, we obtain our general result and finallydiscuss some special cases.

2. Recursion

In this section, we fix notation and define the sets of flags of subspaces

[(a1,1,a1,2), (a2,1,a2,2), . . . , (ar,1,ar,2)

]to be counted. We then construct the recursion in Lemma 2.7 and demonstrate a bijection betweena special class of these flags and the σ -splitting subspaces, thus connecting our recursion to theoriginal problem.

For the remainder of this article, unless otherwise noted, consider more generally the vector spaceFqN (= F

Nq ) over the finite field Fq , given σ ∈ FqN such that FqN = Fq(σ ). Also recall the definitions

for the q-analogues [n]q := 1 + q + · · · + qn−1, [n]!q := [1]q[2]q · · · [n]q and[n

k

]q := [n]!q

[n−k]!q[k]!q .

We begin by recalling a well-known, key property of the linear transformation v �→ σ v .

Proposition 2.1. Multiplication by a primitive element σ generating Fq(σ ) = FqN preserves no subspacesexcept {0} and FqN .

Proof. If W is such a subspace with nonzero w ∈ W , then w∑N−1

i=0 aiσi ∈ W for all choices of ai ∈ Fq ,

so W = FqN . �To motivate the recursion below, consider the case n = 2 in the Splitting Subspace Conjecture, in

which σ -splitting subspaces are the m-dimensional spaces W such that W +σ W = Fq2m . Equivalently,

these are the m-dimensional spaces W for which W ∩ σ−1W = {0}. More generally, we might try tocount the set S(k,k′) of k-dimensional spaces W such that dim(W ∩ σ−1W ) is a fixed number k′between 0 and k − 1. It turns out that at one endpoint, S(k,k − 1) is easy to count and we cancharacterize the structure of the subspaces in S(k,k − 1) (see Section 4).

E. Chen, D. Tseng / Finite Fields and Their Applications 24 (2013) 15–28 17

We might next try to count the set S(k,k − 2) of k-dimensional subspaces W such that dim(W ∩σ−1W ) = k − 2. Let P (k,k − 2) be the set of pairs (W1, W2) of subspaces such that dim(W1) = k,dim(W2) = k − 2, and W2 + σ W2 ⊂ W1. Then on one hand (with the previous three conditionsimplicit),

P (k,k − 2)

= {(W1, W2): dim

(W1 ∩ σ−1W1

) = k − 2}︸ ︷︷ ︸

A

∪{(W1, W2): dim

(W1 ∩ σ−1W1

) = k − 1}︸ ︷︷ ︸

B

.

On the other hand (choosing W2 in the pairs first),

P (k,k − 2)

= {(W1, W2): dim(W2 + σ W2) = k − 1

}︸ ︷︷ ︸C

∪{(W2 + σ W2, W2): dim(W2 + σ W2) = k

}︸ ︷︷ ︸D

.

For A, choosing W1 uniquely determines W2, so |A| = |S(k,k − 2)|, and similarly |D| =|S(k − 2,k − 4)|. For B , there are

[k−1k−2

]q

choices of W2 ⊂ W1 ∩ σ−1W1 once we choose W1,

so |B| = [k−1k−2

]q|S(k,k − 1)|. For, C , after we choose W2, there are

[2m−k+11

]q ways to choose a

k-dimensional subspace of F2mq that contains W2 + σ W2, so |C | = [2m−k+1

1

]q|S(k − 2,k − 3)|. Sub-

stituting in the above equations, we therefore obtain

∣∣S(k,k − 2)∣∣ +

[k − 1

k − 2

]q

∣∣S(k,k − 1)∣∣ =

[2m − k + 1

1

]q

∣∣S(k − 2,k − 3)∣∣ + ∣∣S(k − 2,k − 4)

∣∣.So we have reduced our problem of computing |S(k,k − 2)| to computing |S(k − 2,k − 4)|, which isof lower dimension.

It is similarly possible to deduce a recursion for the case n = 2 that computes the number ofσ -splitting subspaces through double counting. There is no more conceptual difficulty in going fromthe case n = 2 to general n, but we will need to manufacture sets suitable for developing a recursionthrough double counting and establish concise notation that avoids confusion between the differentsets in the recursion.

We proceed by defining notation to describe the objects to be counted by the general recursion.

Definition 2.2. Let A1, A2, . . . , Ak be sets of subspaces of FqN . Define [A1, A2, . . . , Ak] to be the set ofall k-tuples (W1, W2, . . . , Wk) such that

W i ∈ Ai for 1 � i � k,

W i ⊇ W i+1 + σ W i+1 for 1 � i � k − 1.

If Ai is the set of all subspaces of FqN with dimension di , then Ai is denoted within the bracketsas di . For example, [3, A2] denotes all tuples (W1, W2) such that dim(W1) = 3, W2 ∈ A2 and W1 ⊇W2 + σ W2.

Definition 2.3. For nonnegative integers a, b with N > a > b or a = b = 0,

(a,b) := {W ⊆ FqN : dim(W ) = a and dim

(W ∩ σ−1W

) = b}.

18 E. Chen, D. Tseng / Finite Fields and Their Applications 24 (2013) 15–28

For example, (1,0) is the set of all 1-dimensional subspaces and (2,1) is the set of all 2-dimen-sional subspaces W such that dim(W ∩ σ−1W ) = 1.

Observe that we have now defined the sets of flags [(a1,1,a1,2), (a2,1,a2,2), . . . , (ar,1,ar,2)] to becounted by the recursion. In order to construct the recursion, however, we require one additionaldefinition.

Definition 2.4. Given sets [A1,1, A1,2], [A2,1, A2,2], . . . , [Ar,1, Ar,2] as defined above, let

⟨[A1,1, A1,2], [A2,1, A2,2], . . . , [Ar,1, Ar,2]⟩

denote the set of 2r-tuples of subspaces (W1,1, W1,2, W2,1, W2,2, . . . Wr,1, Wr,2) such that

(W i,1, W i,2) ∈ [Ai,1, Ai,2] for 1 � i � r,

W i,2 ⊇ W i+1,1 for 1 � i � r − 1.

For example, 〈[3,2], [2,1]〉 is the set of all 4-tuples of subspaces (W1, W2, W3, W4) such that

dim(W1) = 3, dim(W2) = 2, dim(W3) = 2, dim(W4) = 1,

W1 ⊇ W2 + σ W2, W3 ⊇ W4 + σ W4,

W2 ⊇ W3.

Note that W2 = W3 in this case. We use the following proposition extensively in constructing therecursion.

Proposition 2.5. For nonnegative integers N > a > b or a = b = 0,

[a,b] =max(a−1,0)⋃

i=b

[(a, i),b

](1)

=max(b−1,0)⋃

j=0

[a, (b, j)

]. (2)

Proof. It follows from Proposition 2.1 and the definitions of [ , ], ( , ). �We next define an ordering on the tuples labelling the sets of subspaces

[(a1,1,a1,2), (a2,1,a2,2), . . . , (ar,1,ar,2)

].

Definition 2.6. First, define an ordering on the ordered pairs of the form (a,b) such that (a1,b1) �(a2,b2) if a1 > a2 or a1 = a2 and b1 < b2. Next, define an ordering on tuples of the form[(a1,1,a1,2), (a2,1,a2,2), . . . , (ar,1,ar,2)] such that the order is lexicographic in terms of the orderedpairs (ai,1,ai,2) from left to right. Finally, define an ordering on the same tuples for s � 0 such that

[(a1,1,a1,2), (a2,1,a2,2), . . . , (ar+s,1,ar+s,2)

] � [(a1,1,a1,2), (a2,1,a2,2), . . . , (ar,1,ar,2)

].

For example, (3,1) � (3,2) � (2,0) and [(6,5), (4,2)] � [(6,5), (4,3)] � [(5,2), (2,0)].

E. Chen, D. Tseng / Finite Fields and Their Applications 24 (2013) 15–28 19

The recursion in Lemma 2.7 will give the cardinality of sets of flags of subspaces labelled as[(a1,1,a1,2), (a2,1,a2,2), . . . , (ar,1,ar,2)] in terms of the cardinality of sets labelled by tuples beforethem in the ordering. The base case is [(0,0)], containing one element.

Lemma 2.7. Suppose that

N > a1,1 > a1,2 � a2,1 > a2,2 � · · · � ar,1 > ar,2 � 0 = ar+1,1 = ar+1,2 = · · · = ar+s,1 = ar+s,2,

and after setting a0,1 = a0,2 = N, that

ai−1,1 � 2ai,1 − ai,2 for 1 � i � r.

(Otherwise, [(a1,1,a1,2), (a2,1,a2,2), . . . , (ar,1,ar,2)] is empty.)Additionally, let

C = {( j1, . . . , jr): max(ai+1,2,2ai,2 − ai,1) � ji � max(ai,2 − 1,0), 1 � i � r

},

D = {(k1, . . . ,kr): ai,2 � ki � ai,1 − 1, 1 � i � r

}.

Then

∣∣[(a1,1,a1,2), (a2,1,a2,2), . . . , (ar+s,1,ar+s,2)]∣∣

= ∣∣[(a1,1,a1,2), (a2,1,a2,2), . . . , (ar,1,ar,2)]∣∣

=∑

( j1,..., jr)∈C

∣∣[(a1,2, j1), (a2,2, j2), . . . , (ar,2, jr)]∣∣ r∏

i=1

[ai−1,2 − (2ai,2 − ji)

ai,1 − (2ai,2 − ji)

]q

−∑

(k1,...,kr)∈D\(a1,2,...,ar,2)

∣∣[(a1,1,k1), (a2,1,k2), . . . , (ar,1,kr)]∣∣ r∏

i=1

[ki − ai+1,1

ai,2 − ai+1,1

]q.

We give an example before proving the theorem. Let r = 2; we compute |[(3,1), (1,0)]| by count-ing |〈[3,1], [1,0]〉| in two different ways. Applying (1) from Proposition 2.5 gives

∣∣⟨[3,1], [1,0]⟩∣∣ = ∣∣⟨[(3,2),1],[(1,0),0

]⟩∣∣ + ∣∣⟨[(3,1),1],[(1,0),0

]⟩∣∣.Above, if (W1, W2, W3, W4) ∈ 〈[(3,2),1], [(1,0),0]〉 then W2 = W3 and W4 = {0}. So we have

∣∣⟨[(3,2),1],[(1,0),0

]⟩∣∣ = ∣∣[(3,2), (1,0)]∣∣.

Likewise for (W1, W2, W3, W4) ∈ 〈[(3,1),1], [(1,0),0]〉 then W2 = W3 and W4 = {0}. So we have

∣∣⟨[(3,1),1],[(1,0),0

]⟩∣∣ = ∣∣[(3,1), (1,0)]∣∣,

and

∣∣⟨[3,1], [1,0]⟩∣∣ = ∣∣[(3,2), (1,0)]∣∣ + ∣∣[(3,1), (1,0)

]∣∣.

20 E. Chen, D. Tseng / Finite Fields and Their Applications 24 (2013) 15–28

Next, applying (2) from Proposition 2.5 gives

∣∣⟨[3,1], [1,0]⟩∣∣ = ∣∣⟨[3, (1,0)],[1, (0,0)

]⟩∣∣.If (W1, W2, W3, W4) ∈ 〈[3, (1,0)], [1, (0,0)]〉, then W3 = W2, W4 = {0} and thus W1 is a

3-dimensional subspace containing the 2-dimensional space W2 + σ W2. So

∣∣⟨[3, (1,0)],[1, (0,0)

]⟩∣∣ = ∣∣[(1,0), (0,0)]∣∣[N − 2

1

]q.

We then have, after rearranging, that

∣∣[(3,1), (1,0)]∣∣ = ∣∣[(1,0), (0,0)

]∣∣[N − 2

1

]q− ∣∣[(3,2), (1,0)

]∣∣.Note that [(3,2), (1,0)], [(1,0), (0,0)] come before [(3,1), (1,0)] in the ordering on tuples.

Proof of Lemma 2.7. The proof of the lemma is a generalization of the above process. The first equalityis clear. The size of [(a1,1,a1,2), (a2,1,a2,2), . . . , (ar,1,ar,2)] is computed by applying Proposition 2.5

∣∣⟨[a1,1,a1,2], [a2,1,a2,2], . . . , [ar,1,ar,2]⟩∣∣

=∑

(k1,...,kr)∈D

∣∣⟨[(a1,1,k1),a1,2],[(a2,1,k2),a2,2

], . . . ,

[(ar,1,kr),ar,2

]⟩∣∣

=∑

(k1,...,kr)∈D

∣∣[(a1,1,k1), (a2,1,k2), . . . , (ar,1,kr)]∣∣ r∏

i=1

[ki − ai+1,1

ai,2 − ai+1,1

]q. (R)

Expanding in the other way, we get

∣∣⟨[a1,1,a1,2], [a2,1,a2,2], . . . , [ar,1,ar,2]⟩∣∣

=∑

( j1,..., jr)∈C

∣∣⟨[a1,1, (a1,2, j1)],[a2,1, (a2,2, j2)

], . . . ,

[ar,1, (ar,2, jr)

]⟩∣∣

=∑

( j1,..., jr)∈C

∣∣[(a1,2, j1), (a2,2, j2), . . . , (ar,2, jr)]∣∣ r∏

i=1

[ai−1,2 − (2ai,2 − ji)

ai,1 − (2ai,2 − ji)

]q. (L)

Then 0 = (L) − (R), and adding |[(a1,1,a1,2), (a2,1,a2,2), . . . , (ar,1,ar,2)]| to both sides of the equalityproduces the stated result of the lemma. �

Finally, we relate the flags in [(a1,1,a1,2), (a2,1,a2,2), . . . , (ar,1,ar,2)] to σ -splitting subspaces.

Proposition 2.8. Let m and n be positive integers. Then

[((n − 1)m, (n − 2)m

),((n − 2)m, (n − 3)m

), . . . , (2m,m), (m,0)

](3)

={(

n−2⊕σ i W ,

n−3⊕σ i W , . . . , W ⊕ σ W , W

): dim

(n−1⊕

σ i W

)= mn

}.

i=0 i=0 i=0

E. Chen, D. Tseng / Finite Fields and Their Applications 24 (2013) 15–28 21

In particular, when N = mn so that FqN = Fqmn ,

∣∣[((n − 1)m, (n − 2)m),((n − 2)m, (n − 3)m

), . . . , (2m,m), (m,0)

]∣∣is the number of σ -splitting subspaces.

Proof. If W is an m-dimensional subspace with dim(⊕n−1

i=0 σ i W ) = mn, then

(n−2⊕i=0

σ i W ,

n−3⊕i=0

σ i W , . . . , W ⊕ σ W , W

)

∈ [((n − 1)m, (n − 2)m

),((n − 2)m, (n − 3)m

), . . . , (2m,m), (m,0)

].

For the other inclusion, suppose that

(Wn−1, . . . , W1) ∈ [((n − 1)m, (n − 2)m

),((n − 2)m, (n − 3)m

), . . . , (2m,m), (m,0)

].

Then for 1 � k � n − 2,

dim(Wk+1) = (k + 1)m = 2km − (k − 1)m = dim(Wk + σ Wk).

So Wk+1 = Wk + σ Wk for 1 � k � n − 2. Also, W2 = W1 ⊕ σ W1 as W1 ∩ σ W1 = {0}.Suppose that Wk = ⊕k−1

i=0 σ i W1. Then, since dim(Wk+1) = dim(Wk + σ Wk) = (k + 1)m, we obtain

Wk+1 = Wk + σ Wk = W1 + σ W1 + · · · + σ k W1 =k⊕

i=0

σ i W1.

When k = n−1, we have that Wn−1 +σ Wn−1 = ⊕n−1i=0 σ i W1, since Wn−1 +σ Wn−1 is mn-dimensional,

so we have proved the first claim.If N = mn, there is a bijection between the set of flags (3) and the set of σ -splitting subspaces, so

the second claim follows. In particular, the subspaces corresponding to (m,0) in the flags describedby (3) are then exactly the σ -splitting subspaces. �

We conclude that to answer the Splitting Subspace Conjecture, it suffices to enumerate the el-ements of (3) when N = mn. As neither the base case |[(0,0)]| nor Lemma 2.7 depends on the σchosen, we also have the following corollary.

Corollary 2.9. The number of σ -splitting subspaces in Fqmn over Fq is independent of choice of primitiveelement σ .

Remark. In fact, the converse of Proposition 2.1 is true as well; the linear endomorphisms of FqN thatpreserve no subspaces other than {0} and all of FqN are exactly those which act as multiplication bya primitive element σ that generates the extension Fq(σ ) = FqN .

Note that any linear operator T together with the vector space FNq can be viewed as a finitely

generated Fq[x] module M , where x acts as T . Since Fq[x] is a principal ideal domain, we can use

the primary decomposition of M to find M ∼= ⊕ki=1 Fq[x]/(pi(x)ri ), where pi is a polynomial for each

i and ri is a positive integer.

22 E. Chen, D. Tseng / Finite Fields and Their Applications 24 (2013) 15–28

If T preserves no proper subspaces of FqN , then k = 1. Also, r1 = 1 or else p1(T )M is a propersubmodule of M . Therefore, we have M is equal to Fq[x]/(p1(x)), where p1 is an irreducible polyno-mial. This is exactly what it means for x(= T ) to act as the primitive element of the field extensionFq(σ ) = FqN = F

Nq with minimal polynomial p1(x).

Remark. More generally, given an arbitrary invertible linear operator T on Fqmn over Fq , we mightconsider how many “T -splitting” subspaces exist; that is, the number of m-dimensional subspaces Wsuch that

W ⊕ T W ⊕ · · · ⊕ T n−1W = Fqmn .

We may then redefine ( , ), [ , ], 〈 , 〉 by replacing the expressions W + σ W with W + T W andW ∩ σ−1W with W ∩ T −1W .

Recall from Proposition 2.1 and Lemma 2.7 that when T (v) = σ v , the nonzero numbers|[(a1,1,a1,2), (a2,1,a2,2), . . . , (ar,1,ar,2)]| can be computed from the base case |[0,0]| = 1. But if T isany invertible linear operator, there may exist nonempty sets of the form [(a1,a1), (a2,a2), . . . , (ar,ar)]where ar �= 0. In fact, such sets cannot be computed recursively. For example

∣∣⟨[4,4], [2,2]⟩∣∣ = ∣∣⟨[(4,4),4],[(2,2),2

]⟩∣∣ = ∣∣[(4,4), (2,2)]∣∣

= ∣∣⟨[4, (4,4)],[2, (2,2)

]⟩∣∣ = ∣∣[(4,4), (2,2)]∣∣.

We may still apply Lemma 2.7 in the case of general T , however, with the cardinalities of the sets offlags in [(a1,a1), (a2,a2), . . . , (ar,ar)] as additional base cases.

Remark. Motivated by Section 4 of [4], one might try to derive pointed analogues of the recursion.However, at least the naive approach yields nothing interesting. In particular, given a fixed vector v ,it is not difficult to show that the number of tuples in [(a1,1,a1,2), (a2,1,a2,2), . . . , (ar,1,ar,2)] whose�th subspace contains v is equal to

qa�,1 − 1

qN − 1

∣∣[(a1,1,a1,2), (a2,1,a2,2), . . . , (ar,1,ar,2)]∣∣.

We could then use the same double counting argument used in the proof of Theorem 2.7 in orderto derive a pointed analogue. It is not immediately obvious, but after doing the computation we justobtained the same equation (R) = (L) as in the proof of Theorem 2.7, except scaled by a factor ofqa�,1 −1

qN −1.

3. Solution to the recursion

We proceed in this section to solve the recursion above by verifying the explicit formula givenin Theorem 3.3. Recall the definitions given in Section 2 for [k]q , [k]!q , and

[nk

]q . Further recall

that (a;q)m := ∏ai=0 (1 − aqi). The next two lemmas are special cases of the following q-Chu–

Vandermonde identity for N a nonnegative integer [1, p. 354]:

2φ1

(a,q−N ; c;q,

cqN

a

)= (c/a;q)N

(c;q)N,

where 2φ1 denotes the basic hypergeometric series

2φ1

(a,q−N ; c;q,

cqN

a

):=

N∑ (q−N ;q)m(a;q)m

(q;q)m(c;q)m

(cqN

a

)m

.

m=0

E. Chen, D. Tseng / Finite Fields and Their Applications 24 (2013) 15–28 23

Lemma 3.1. If C � B − 1 � D − 1 � A − 1 are nonnegative integers, then

B−1∑s=C

[A − B − 1

B − s − 1

]q

[B

s

]q

[s

C

]q

[A − (2B − s)

D − (2B − s)

]qq(B−s)(B−s−1)

= [B]q

[D − C]q

[B − 1

C

]q

[A − B − 1

D − B − 1

]q

[D − C

B − C

]q.

Lemma 3.2. If C � D � B − 1 � A − 1 are nonnegative integers, then

B−1∑s=D

[A − B − 1

B − s − 1

]q

[B

s

]q

[s

C

]q

[s − C

D − C

]qq(B−s)(B−s−1)

= [B]q

[A − D]q

[B − 1

C

]q

[B − C − 1

D − C

]q

[A − D

B − D

]q.

Both lemmas may be shown by taking s �→ B − 1 − s to obtain two identites equivalent to theoriginals, which are then rewritten in standard form so that the q-Chu–Vandermonde identity may beapplied.

We now give our main theorem.

Theorem 3.3. Suppose that

N > a1,1 > a1,2 � a2,1 > a2,2 � · · · � ar,1 > ar,2 � 0,

a0,1 = a0,2 = N, ar+1,1 = ar+1,2 = 0.

Then

∣∣[(a1,1,a1,2), . . . , (ar,1,ar,2)]∣∣ =

[N1

]q[a1,1

1

]q

∏r−1i=0

[ ai,1−ai+1,1−1ai+1,1−ai+1,2−1

]q

[ai+1,1ai+1,2

]q

[ai+1,2ai+2,1

]q∏r−1

i=1

[ ai,1−1ai+1,1−1

]q

qE , (4)

where

E =r∑

i=1

(ai,1 − ai,2)(ai,1 − ai,2 − 1).

Corollary 3.4 (Splitting Subspace Conjecture). We have, when N � mn, the equality

∣∣[((n − 1)m, (n − 2)m), . . . , (2m,m), (m,0)

]∣∣ =[N

1

]q[m

1

]q

[N − mn + m − 1

m − 1

]qqm(m−1)(n−1).

In particular, when N = mn,

∣∣[((n − 1)m, (n − 2)m), . . . , (2m,m), (m,0)

]∣∣ =[mn

1

]q[m

1

]q

qm(m−1)(n−1).

24 E. Chen, D. Tseng / Finite Fields and Their Applications 24 (2013) 15–28

Proof of Theorem 3.3. We verify that (4) satisfies the recursion in Lemma 2.7.Recall that

L =∑

( j1,..., jr)∈C

∣∣[(a1,2, j1), (a2,2, j2), . . . , (ar,2, jr)]∣∣ r∏

i=1

[ai−1,2 − (2ai,2 − ji)

ai,1 − (2ai,2 − ji)

]q,

R =∑

(k1,...,kr)∈D

∣∣[(a1,1,k1), (a2,1,k2), . . . , (ar,1,kr)]∣∣ r∏

i=1

[ki − ai+1,1

ai,2 − ai+1,1

]q.

We first check equality when ar,2 �= 0 so that the expressions obtained for (L) and (R) using (4) donot contain negative q-binomials.

Substituting (4) and applying Lemma 3.1 to the resulting independent sums in (L) with

A = ai−1,2, B = ai,2, C = ai+1,2, D = ai,1, s = ji

gives

L =[N

1

]q[a1,2

1

]q

∏ri=1

∑ji

[ai−1,2−ai,2−1ai,2− ji−1

]q

[ai,2ji

]q

[ jiai+1,2

]q

[ai−1,2−(2ai,2− ji)

ai,1−(2ai,2− ji)

]qq(ai,2− ji)(ai,2− ji−1)

∏r−1i=1

[ ai,2−1ai+1,2−1

]q

=[N

1

]q[a1,2

1

]q

r∏i=1

[ai,2]q

[ai,1 − ai+1,2]q

[ai,2 − 1

ai+1,2

]q

[ai−1,2 − ai,2 − 1

ai,1 − ai,2 − 1

]q

[ai,1 − ai+1,2

ai,2 − ai+1,2

]q

r−1∏i=1

[ai,2 − 1

ai+1,2 − 1

]−1

q.

Substituting (4) and applying Lemma 3.2 to the resulting independent sums in (R) with

A = ai−1,1, B = ai,1, C = ai+1,1, D = ai,2, s = ki

gives

R =[N

1

]q[a1,1

1

]q

∏ri=1

∑ki

[ai−1,1−ai,1−1ai,1−ki−1

]q

[ai,1ki

]q

[ kiai+1,1

]q

[ ki−ai+1,1ai,2−ai+1,1

]qq(ai,1−ki)(ai,1−ki−1)

∏r−1i=1

[ ai,2−1ai+1,2−1

]q

=[N

1

]q[a1,1

1

]q

r∏i=1

[ai,1]q

[ai−1,1 − ai,2]q

[ai,1 − 1

ai+1,1

]q

[ai,1 − ai+1,1 − 1

ai,2 − ai+1,1

]q

[ai−1,1 − ai,2

ai,1 − ai,2

]q

r−1∏i=1

[ai,1 − 1

ai+1,1 − 1

]−1

q.

After tedious simplification we obtain

L = R = [N]q[N − a1,2 − 1]!q[N − a1,1]!q[ar,2]!q

r∏i=1

[ai,1 − ai+1,2 − 1]!q[ai,1 − ai,2 − 1]!q[ai,1 − ai,2]!q

r∏i=2

1

[ai−1,2 − ai,1]!q . (5)

Finally, we deal with the case ar,2 = 0, when the expression obtained by directly applying (4) to(L) may contain negative q-binomials ((R) is unaffected). Suppose r > 1. By definition, we know that

∣∣⟨[a1,1,a1,2], [a2,1,a2,2], . . . , [ar,1,0]⟩∣∣ = ∣∣⟨[a1,1,a1,2], [a2,1,a2,2], . . . , [ar−1,1,ar−1,2]⟩∣∣[ar−1,2

ar,1

].

q

E. Chen, D. Tseng / Finite Fields and Their Applications 24 (2013) 15–28 25

This means that

L =∑

( j1,..., jr)∈C

∣∣[(a1,2, j1), (a2,2, j2), . . . , (ar,2, jr)]∣∣ r∏

i=1

[ai−1,2 − (2ai,2 − ji)

ai,1 − (2ai,2 − ji)

]q

= ∣∣⟨[a1,1,a1,2], . . . , [ar−1,1,ar−1,2]⟩∣∣[ar−1,2

ar,1

]q.

Since ar−1,2 � ar,1 > 0, we may apply our previous result to obtain

∣∣⟨[a1,1,a1,2], . . . , [ar−1,1,ar−1,2]⟩∣∣[ar−1,2

ar,1

]q

=[

ar−1,2

ar,1

]q

[N]q[N − a1,2 − 1]!q[N − a1,1]!q[ar−1,2]!q

r−1∏i=1

[ai,1 − ai+1,2 − 1]!q[ai,1 − ai,2 − 1]!q[ai,1 − ai,2]!q

r−1∏i=2

1

[ai−1,2 − ai,1]!q .

We wish to show that this is equal to

[N]q[N − a1,2 − 1]!q[N − a1,1]!q[ar,2]!q

r∏i=1

[ai,1 − ai+1,2 − 1]!q[ai,1 − ai,2 − 1]!q[ai,1 − ai,2]!q

r∏i=2

1

[ai−1,2 − ai,1]!q

when ar,2 = 0. Take the quotient to find

[N]q[N−a1,2−1]!q[N−a1,1]!q[ar,2]!q

∏ri=1

[ai,1−ai+1,2−1]!q[ai,1−ai,2−1]!q[ai,1−ai,2]!q

∏ri=2

1[ai−1,2−ai,1]!q[ar−1,2

ar,1

]q

[N]q[N−a1,2−1]!q[N−a1,1]!q[ar−1,2]!q

∏r−1i=1

[ai,1−ai+1,2−1]!q[ai,1−ai,2−1]!q[ai,1−ai,2]!q

∏r−1i=2

1[ai−1,2−ai,1]!q

=[ar−1,2]!q [ar,1−1]!q

[ar,1−1]!q[ar,1]!q1

[ar−1,2−ar,1]!q[ar−1,2ar,1

]q

= 1,

as desired. Therefore, when ar,2 = 0, the equality

L =∑

( j1,..., jr)∈C

∣∣[(a1,2, j1), (a2,2, j2), . . . , (ar,2, jr)]∣∣ r∏

i=1

[ai−1,2 − (2ai,2 − ji)

ai,1 − (2ai,2 − ji)

]q

= [N]q[N − a1,2 − 1]!q[N − a1,1]!q[ar,2]!q

r∏i=1

[ai,1 − ai+1,2 − 1]!q[ai,1 − ai,2 − 1]!q[ai,1 − ai,2]!q

r∏i=2

1

[ai−1,2 − ai,1]!q= R

still holds.Finally, suppose r = 1. Then

L = ∣∣[(0,0)]∣∣[ N

a1,1

]q=

[N

a1,1

]q.

If we plug in 〈[a1,1,0]〉 into (5), then we get[ N

a1,1

]q, as desired. �

26 E. Chen, D. Tseng / Finite Fields and Their Applications 24 (2013) 15–28

Corollary 3.5. The numbers

∣∣⟨[a1,1,a1,2], [a2,1,a2,2], . . . , [ar,1,ar,2]⟩∣∣

are given by

L = R = [N]q[N − a1,2 − 1]!q[N − a1,1]!q[ar,2]!q

r∏i=1

[ai,1 − ai+1,2 − 1]!q[ai,1 − ai,2 − 1]!q[ai,1 − ai,2]!q

r∏i=2

1

[ai−1,2 − ai,1]!q .

4. Special case: (k,k − 1)

Recall from Definition 2.3 that in the vector space FqN over Fq and for k < N that

(k,k − 1) := {W ⊆ FqN : dim(W ) = k and dim

(W ∩ σ−1W

) = k − 1}.

Note that when r = 1 and a1,1 = k, a1,2 = k − 1, with k � N − 1, the formula (4) gives

∣∣(k,k − 1)∣∣ =

[N

1

]q,

a number independent of k.

Proposition 4.1. There is a bijection between sets of the form (k1,k1 −1) and (k2,k2 −1) when k1,k2 � N −1.

Proof. It suffices to show that there exists a bijection between (k,k − 1) and (k − 1,k − 2) for 2 � k �N − 1. Define

φ : (k − 1,k − 2) → (k,k − 1),

W �→ W + σ W .

The map φ is well defined:

dim(W + σ W ) = k,

dim((W + σ W ) ∩ (

σ−1W + W)) = k − 1.

The second equality follows from the fact that (W + σ W ) ∩ (σ−1W + W ) contains W and has di-mension strictly less than k by Proposition 2.1.

Next, φ is injective: if W1, W2 ∈ (k − 1,k − 2) and W1 +σ W1 = W2 +σ W2 = W ′ ∈ (k,k − 1), thenW ′ ∩ σ−1W ′ = W1 = W2.

Finally, φ is surjective: if W ′ ∈ (k,k − 1), then W ′ ∩ σ−1W ′ ∈ (k − 1,k − 2) since (W ′ ∩ σ−1W ′) +σ(W ′ ∩ σ−1W ′) ⊆ W ′; in fact (W ′ ∩ σ−1W ′) + σ(W ′ ∩ σ−1W ′) = W ′ . �Corollary 4.2. Every subspace W ∈ (k,k − 1) can be expressed uniquely as V +σ V + · · ·+σ k−1 V , where Vis a 1-dimensional subspace.

Proof. This follows from the argument in Proposition 4.1 when k1 = k and k2 = 1. If we follow thebijections (1,0) → (2,1) → ·· · → (k,k − 1) in the proof of Proposition 4.1 above, we find that V ∈(1,0) maps exactly to V + σ V + · · · + σ k−1 V . �

E. Chen, D. Tseng / Finite Fields and Their Applications 24 (2013) 15–28 27

5. A q = 1 analogue

We might ask what (4) counts when q = 1; indeed the situation translates from enumeratingsubspaces of vector spaces to enumerating subsets of sets.

Instead of subspaces of FqN , we consider subsets of {1, . . . , N}. Rather than multiplying by theelement σ , we let σ = (12 · · · N) cyclically permute the elements of {1, . . . , N} so that σ preserves noproper subset, in analogy with Proposition 2.1; in fact, any permutation of {1, . . . , N} preserving noproper subset is cyclic. When N = mn, the number of m-element subsets W of {1, . . . , N} such that⋃n−1

i=0 σ i W = {1, . . . , N} is clearly n; this is true in a more general setting. We retain the [ , ], ( , ), 〈 , 〉notation with the definitions restated in the context of subsets of {1, . . . , N} below.

Definition 5.1. Let A1, A2, . . . , Ak be sets of subsets of {1, . . . , N}. Define [A1, A2, . . . , Ak] be the set ofall k-tuples (W1, W2, . . . , Wk) such that

W i ∈ Ai for 1 � i � k,

W i ⊇ W i+1 ∪ σ W i+1 for 1 � i � k − 1.

If Ai is the set of all subsets of {1, . . . , N} with cardinality di , then Ai is denoted within thebrackets as di .

Definition 5.2. For nonnegative integers a, b with N > a > b or a = b = 0,

(a,b) := {W ⊆ {1, . . . , N}: |W | = a,

∣∣W ∩ σ−1W∣∣ = b

}.

Definition 5.3. Given sets [A1,1, A1,2], [A2,1, A2,2], . . . , [Ar,1, Ar,2] as defined above, let

⟨[A1,1, A1,2], [A2,1, A2,2], . . . , [Ar,1, Ar,2]⟩

denote the set of 2r-tuples of subsets (W1,1, W1,2, W2,1, W2,2, . . . Wr,1, Wr,2) such that

(W i,1, W i,2) ∈ [Ai,1, Ai,2] for 1 � i � r,

W i,2 ⊇ W i+1,1 for 1 � i � r − 1.

Lemma 2.7 and our formulas in Theorem 3.3 and Corollaries 3.4 and 3.5 are still valid here bysetting q = 1; the q-binomials counting ways to extend subspaces become binomial terms countingways to enlarge subsets. However, we can directly count some special cases.

An appropriate adaptation of Proposition 2.8 gives

[((n − 1)m, (n − 2)m

),((n − 2)m, (n − 3)m

), . . . , (2m,m), (m,0)

]=

{(n−2⋃i=0

σ i W ,

n−3⋃i=0

σ i W , . . . , W ∪ σ W , W

):

∣∣∣∣∣n−1⋃i=0

σ i W

∣∣∣∣∣ = mn and |W | = m

}.

Counting in two different ways the number of ordered pairs (W ,k) where W satisfies the con-ditions in the set above and k is an element of W , one by fixing k first and the other by fixing Wfirst, yields |[((n−1)m, (n−2)m), . . . , (m,0)]| = N

m

(N−mn+m−1m−1

), which is the same as the formula from

Corollary 3.4 when we set q = 1.We may also count the elements of (m,k) directly by counting the number of ordered pairs (W ,a)

where W ∈ (m,k) and a /∈ W in two ways, one by fixing W first and the other by fixing a first. Thisyields N

N−m

(N−mm−k

)(m−1k

) = Nm

(N−m−1m−k−1

)(mk

). For q a power of a prime, (4) yields

28 E. Chen, D. Tseng / Finite Fields and Their Applications 24 (2013) 15–28

∣∣(m,k)∣∣ =

[N1

]q[m

1

]q

[N − m − 1

m − k − 1

]q

[m

k

]qq(m−k)(m−k−1),

which is the q-analogue of the expression obtained above.

Acknowledgments

This research was conducted at the 2012 summer REU (Research Experience for Undergraduates)program at the University of Minnesota, Twin Cities, and was supported by NSF grants DMS-1001933and DMS-1148634. We would like to thank Gregg Musiker, Pavlo Pylyavskyy, Vic Reiner, and Den-nis Stanton, who directed the program, for their support, and express particular gratitude to DennisStanton both for introducing us to this problem and for his indispensable guidance throughout theresearch process. We would further like to thank Alex Miller for his assistance in editing this reportas well as Sudhir Ghorpade and Samrith Ram for their helpful comments.

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