The multi-slice rank method and polynomial bounds for orthogonal systems in $\mathbb{F}_{q}^{n}$

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by  Eric Naslund

THE MULTI-SLICE RANK METHOD AND POLYNOMIAL BOUNDS FOR
ORTHOGONAL SYSTEMS IN Fn
q
ERIC NASLUND
Abstract. Let Fn be thesubset of self-orthogonal vectors in Fn, which has size =
qn−1+O(√qn).SIn⊂thiqspaperweusetherecentlydeveloped slice ranqk method[11] to|sSh|ow
7
that any set E of size
1
0 ⊂S E >(n+1)(q−1)(k−1)
| |
2 contains a k-tuple of distinct mutually orthogonal vectors, where n k and q = pr with
n p k. The key innovation is a more general version of the slice ra≥nk2of a tensor, which
≥ (cid:0) (cid:1)
a we call the multi-slice rank. We use a combinatorial argument to create an appropriate
J indicator tensor for the orthogonality of k-vectors, however for k 4 this tensor will have
≥
6 large slice rank, and so to obtain our results we need to use the less restrictive multi-slice
1 rank. Additionally, we use this method to generalize a recent result of Ge and Shangguan
[7], and provethat any set A Fn of size
] ⊂ q
O n+(q 1)(k 1)
A > − −
C | | (q 1)(k 1) !
− −
. containsa k-right-corner, that is distinct vectorsx ,...,x ,x wherex x ,...,x
h 1 k k+1 1 k+1 k
at xk+1 are mutually orthogonal, as long as n≥ k+21 and q=pr with p>k.− −
m (cid:0) (cid:1)
[

1 1. Introduction
v
5 Problem 1. What is the size of the largest subset E ⊂ Fnq which contains no distinct k-tuple
7 of mutually orthogonal vectors?
4
4 In2008Vinh[13]generalized aresultof IosevichandSenger [8], proving thatforn ≥ 2k−1,
0 if E ⊂ Fn satisfies
. q
01 |E| ≫ qn2+k−1,
7 then E contains
:1 (1+o(1)) |E|kq−(k2)
v k!
i mutually orthogonal k-tuples in E. Here q = pr is a prime power, and Fn denotes the vector
X q
space of dimension n over the finite field with q elements. In Iosevich and Senger’s [8, Lemma
r
a 3.2] paper, they constructed a set E ⊂ Fn of size
q
|E| ≥ cqn2+1
for some c > 0 such that no two of it’s vectors are orthogonal. In light of Vinh’s bound, when
k = 2 this construction is tight up to a constant, and so it seems that problem 1 is resolved
in this case. However, critical to this lower bound construction is the fact that none of the
vectors in this set are self-orthogonal, that is no x ∈ E satisfies n x2 = 0. Let S denote
i=1 i
the set of all self-orthogonal vectors in Fn, also referred to as the set of isotropic vectors (see
q
P
[1]). We note that
n
1
|S|= qn−1+ χ(x2)
q  
χX∈ Fˆq xX∈Fq
 
χ 6= χ
0
Date: January 12 2017.
1
THE MULTI-SLICE RANK METHOD 2
where χ ranges over the characters of F , and so
q
|S| = qn−1+O qn2
(cid:16) (cid:17)
since the Gauss sum x∈Fqχ(x2) has magnitude q21. Our main theorem is:
Theorem 1. Let S ⊂PFn be the set of all x ∈ Fn satisfying n x2 = 0. Let k > 1 be given,
q q i=1 i
and suppose that n ≥ k and q = pr satisfies p ≥ k. If E ⊂ S has size
2 P
(cid:0) (cid:1) n+(q−1)(k−1)
|E| > ,
(q−1)(k−1)
(cid:18) (cid:19)
then E contains a k-tuple of k distinct mutually orthogonal vectors.
The right hand side above is a polynomial in the dimension and so the result holds for very
sparse subsets of S. In particular, the right hand side is bounded above by
(n+1)(q−1)(k−1),
which is at most
O (1+ǫ)n
ǫ,k,q
for any ǫ > 0, where the subscript denotes that the implicit constant could depend on ǫ,k
and q. To prove theorem 1, we use the slice rank method which was introduced by Tao [11]
following the breakthrough of Croot, Lev and Pach [3]. The slice rank method for 3-tensors is
asymmetrized version of Croot, Lev and Pach’s argument, and in his blog post, Tao usedit to
reprove Ellenberg and Gijswijt’s bounds for arithmetic progression-free subsets of Fn [4]. This
p
method has seen applications to a variety of problems including the Erdő-Szemerédi sunflower
problem [10, 6], and the group theoretic approach to fast matrix multiplication [2], and we
refer the reader to [2, Section 4] for an in depth discussion of the slice rank as well as its
connection to geometric invariant theory. The slice rank method allows us to prove theorem 1
for k = 2 and k = 3, however to handle k ≥ 4 we need additional flexibility, and we introduce
a generalization of the slice rank which we call the multi-slice rank. Using a combinatorial
argument coming from the group action of S on the set of k-tuples of vectors in Fn, we create
k q
a tensor I satisfying
k
1 x ,...,x are distinct and mutually orthogonal
1 k
I (x ,...,x ) = (−1)k(k−1)! if x = ··· = x ,
k 1 k  1 k
0 otherwise
so that Ik is diagonal when restricted to E×···×E, where E is any set that does not contain
a k-tuple of distinct mutually orthogonal vectors. For k ≥ 4, I will have very large slice
k
rank, and so the slice rank method fails to obtain non-trivial bounds. However I will have
k
polynomially small multi-slice rank, and so the multi-slice rank method allows us to prove
theorem 1 for k ≥ 4.
Recently, Ge and Shangguan [7] used the slice rank method to improve upon a result of
Bennett [1], showing that any set A⊂ Fn of size
q
n+q
|A| > +3
q−1
(cid:18) (cid:19)
contains a right-angle, that is distinct vectors x,y,z ∈ A satisfying
hy−x,z−xi = 0.
Usingthemulti-slice rankmethod wegeneralize this result to k-right-corners. Wesay thatthe
vectors x ,...,x ,x form a k-dimensional right corner if the k vectors x −x ,...,x −
1 k k+1 1 k+1 k
x form a mutually orthogonal k-tuple. In other words, x ,...,x must meet x at
k+1 1 k k+1
mutually right angles.
THE MULTI-SLICE RANK METHOD 3
Theorem 2. Let k be given, q = pr with p > k, and n ≥ k+1 . If A⊂ Fn has size
2 p
n+(q−1)(k−1)(cid:0) (cid:1)
|A| > ,
(q−1)(k−1)
(cid:18) (cid:19)
then A contains a k-dimensional right corner.
The proof of theorem 2relies heavily on theflexibility of the multi-slice rank fork ≥ 3. Just
as before, when we have 4 or more variables involved, the constructed indicator tensor will
have polynomial small multi-slice rank yet exponentially large slice rank. In [9], we provide
another application of the multi-slice rank method, and prove exponential upper bounds for
the Erdős-Ginzburg-Ziv [5] constant of (Z/pZ)n.
In subsection 2.1, we review the slice rank, as defined in [11] and [2, Section 4.1], and in
subsection 2.2 we introduce the multi-slice rank and prove the critical lemma, which states
that the multi-slice rank of a diagonal tensor equals the number of non-zero diagonal entries.
In 3.1 we introduce the indicator function H (x ,...,x ), which satisfies
k 1 k
1 if x ,...,x are distinct
1 k
H (x ,...,x ) = (−1)k(k−1)! if x = ··· = x .
k 1 k  1 k
0 otherwise
This function allows us to modify our tensor so that it picks up only k-tuples of distinct

elements. In subsections 3.2 and 3.3, we prove theorems 1 and 2, respectively, using the
multi-slice rank and the indicator H .
k
2. The Slice Rank and Multi-slice rank
2.1. The Slice Rank. We begin by recalling the tensor rank and the definition of the rank
of a two variable function. Let X,Y be finite sets, and suppose that F is a field. The rank of
a function
F : X ×Y → F
is defined to be the smallest k such that
k
F(x,y) = f (x)g (y)
i i
i=1
X
for some functions f ,g . For finite sets X ,...,X , a function
i i 1 n
h :X ×···×X → F,
1 n
h6= 0, of the form
h(x ,...,x ) = f (x )f (x )···f (x )
1 n 1 1 2 2 n n
is called a rank 1 function, and the tensor rank of
F :X ×···×X → F
1 n
is defined to be the minimal k such that
k
F = g
i
i=1
X
where the g are rank 1 functions. Following the breakthrough of Croot, Lev and Pach [3],
i
Tao [11] introduced the notion of the slice rank of a tensor:
Definition 1. Let X ,...,X be finite sets. We say that the function
1 n
h :X ×···×X → F,
1 n
h6= 0, is a 1-slice if
h(x ,...,x ) = f(x )g(x ,...,x ,x ,...,x )
1 n i 1 i−1 i+1 n
THE MULTI-SLICE RANK METHOD 4
for some 1 ≤ i ≤n. The slice rank, or 1-slice rank, of
F :X ×···×X → F
1 n
is the smallest k such that
k
F = g
i
i=1
X
where the g are 1-slices.
i
The following lemma was proven by Tao, and used to great effect:
Lemma 1. Let X be a finite set, and let Xk denote the k-fold Cartesian product of X with
itself. Suppose that
F :Xk → F
is a diagonal tensor, that is
F(x ,...,x ) = c δ (x )···δ (x )
1 n a a 1 a n
a∈A
X
for some A ⊂ X, c 6= 0, where
a
1 x = a
δ (x) = .
a
(0 otherwise
Then
slice-rank(F) = |A|.
Proof. We refer the reader to [11, Lemma 1] or [2, Lemma 4.7]. (cid:3)
2.2. The Multi-Slice Rank. In what follows, we propose a new more general definition of
the slice rank. Given variables x ,...,x and a set S ⊂ {1,...,n}, S = {s ,...,s }, we use
1 n 1 k
−→
the notation x to denote
S
x ,...,x ,
s1 sk
and so for a function g of k variables, we have that
−→
g(x )= g(x ,...,x ).
S s1 sk
Definition 2. Let X ,...,X be finite sets. We say that the function
1 n
h :X ×···×X → F,
1 n
h6= 0, is a b-slice for 1≤ b ≤ n if
2
−→ −→
h(x ,...,x ) = f(x )g x
1 n S {1,...,n}\S
for some f,g and S ⊂ {1,2,...,n} of size |S| = b. M(cid:0)ore genera(cid:1)lly, we say that F is a slice if
it is a b-slice for some 1≤ b ≤ n.
2
For example, if X is a finite set, then the function h :X7 → F given by
h(x ,x ,x ,x ,x ,x ,x )= f(x ,x ,x )g(x ,x ,x ,x )
1 2 3 4 5 6 7 2 5 7 1 3 4 6
is a slice, and in particular it is a 3-slice with S = {2,5,7}. This leads us to the definition of
the multi-slice rank:
Definition 3. Let X ,...,X be finite sets. The multi-slice rank of a function
1 n
F :X ×···×X → F
1 n
is the minimal k such that
k
F = c g
i i
i=1
X
where the g are slices.
i
THE MULTI-SLICE RANK METHOD 5
It is immediate that any tensor F with slice-rank(F) = k will satisfy multi-slice-rank(F) ≤
k.
Remark 1. For a finite set X and a field F, consider the function
F :X ×X ×X ×X → F
given by
1 x =y and z = w
F(x,y,z,w) = ,
(0 otherwise
that is
F(x,y,z,w) = δ(x,y)δ(z,w)
where δ(x,y) denotes the function that is 1 when x = y, and 0 otherwise. Then F will have
multi-slice-rank(F) = 1 and slice-rank(F) = |X| [12].
Our key observation is that the multi-slice rank of a diagonal tensor is full.
Lemma 2. Let X be a finite set, and let Xk denote the k-fold Cartesian product of X with
itself. Suppose that
F :Xk → F
is a diagonal tensor, that is
F(x ,...,x ) = c δ (x )···δ (x )
1 n a a 1 a n
a∈A
X
for some A ⊂ X where c 6= 0, and
a
1 x = a
δ (x) = .
a
(0 otherwise
Then
multi-slice-rank(F) = |A|.
Proof. It’s evident that the multi-slice rank is at most |A|, and so our goal is to prove the
lower bound. The proof proceeds by induction on the number of variables. When n = 2, this
is the usual notion of rank, and so the result follows. Suppose that we can write
k
−→ −→
F(x ,...,x )= c f (x )g x
1 n i i Si i {1,...,n}\Si
i=1
X (cid:0) (cid:1)
where c 6= 0 and k < |A|, and suppose without loss of generality that |S | ≤ n for each i. If
i i 2
there is no i such that |S | = 1, then choose an arbitrary variable, say x , and average over
i 1
than coordinate. Then we have that
k
F(x ,...,x ) = c δ (x )···δ (x ) = c f˜(−x−−−→)g˜ −→x ,
1 n a a 2 a n i i Si\{1} i {2,...,n}\Si
xX1∈X aX∈A Xi=1 (cid:0) (cid:1)
for functions f˜,g˜ given by averaging f,g over x . This contradicts the inductive hypothesis
1
since c δ (x )···δ (x ) will have multi-slice rank equal to |A| > k. Now, suppose that
a∈A a a 2 a n
there exists some S such that |S | = 1. Then S = {j} for some j ∈ {1,...,n}. Let U be the
i i i
P
set of indices for which we have S = {j} = S for each u ∈ U, so that in particular i ∈ U.
u i
Define
V = h: X → F : f (x )h(x )= 0 for all u ∈ U .
u j j
 
 xXj∈X 
This vector space has dimension at least |X|−|U|. Let v ∈ V have maximal support, and set
 
Σ = {x ∈ X : v(x) 6= 0}. Then |Σ|≥ dimV ≥ |X|−|U|, since otherwise there exists nonzero
r ∈ V vanishing on Σ, and the function v+r would have a larger support than v. Multiplying
THE MULTI-SLICE RANK METHOD 6
both sides of our expression by v(x ) and summing over x reduces our tensor dimension by
j j
1. Indeed
v(x )F(x ,...,x )= c δ (x )···δ (x )δ (x )···δ (x ) v(x )δ (x ) ,
j 1 n a a 1 a j−1 a j+1 a n j a j
 
xXj∈X aX∈A xXj∈X
 
and since the sum v(x )δ (x ) will be non-zero for at least |X|−|U| values of a ∈ X,
xj∈X j a j
the multi-slice rank of the above must be at least |A|−|U| by the inductive hypothesis. Since
P
−→
v(x )f (x )= 0
j i Si
xXj∈X
for each i ∈ U, it follows that
k
−→ −→
v(x ) c f (x )g x
j i i Si i {1,...,n}\Si
xXj∈X Xi=1 (cid:0) (cid:1)
will be a sum of at most k−|U| slices, and hence it has multi-slice rank at most k−|U|. This
implies that |A|−|U| < k−|U|, which is a contradiction, and the lemma is proven. (cid:3)
We conclude this section with the following lemma:
Lemma 3. Let X be a finite set, and suppose that F : Xk → F has tensor rank T and
G: Xk → F has multi-slice rank R, then
multi-slice-rank(G·T)≤ RT.
Proof. This follows immediately from writing F as a sum of T slices and G as a sum of R rank
1 functions, since the product results in at most RT slices. (cid:3)
3. The Main Theorems
Let S denote the set of self-orthogonal vector. Our goal is to find a function
I : Sk → F ,
k q
of low multi-slice rank satisfying
c x ,...,x are mutually orthogonal and distinct
1 1 k
(3.1) I (x ,...,x )= c x = ··· = x
k 1 k  2 1 k
0 otherwise
where c2 6= 0. If E ⊂ S is a set without any distinct mutually orthogonal k-tuples, then when
we restrict I to Ek, I will be a diagonal tensor taking the value c on the diagonal, and so
k k 2
by lemma 2, I must have multi-slice rank at least |E|. It then follows that any set E ⊂ S of
k
size
E > multi-slice-rank(I ),
k
must contain a k-tuple of k distinct mutually orthogonal vectors.
Define
F : Sk → F
k q
by
(3.2) F (x ,...,x ) = 1−hx ,x iq−1 .
k 1 k j l
j<l≤k
Y (cid:0) (cid:1)
For example,
n q−1 n q−1 n q−1
F (x,y,z) = 1− x y 1− y z 1− z x ,
3 i i i i i i
   
! ! !
i=1 i=1 i=1
X X X
   
THE MULTI-SLICE RANK METHOD 7
is a polynomial of degree 3(q−1) in n variables. This function satisfies
1 x ,...,x are mutually orthogonal
1 k
F (x ,...,x )= .
1 k
(0 otherwise
We note that x ,...,x are not required to be unique for F(x ,...,x ) to equal 1, and in
1 k 1 k
particular if x = ··· = x , then F (x ,...,x ) = 1 since hx,xi = 0 for every x ∈ S. Ideally,
1 k k 1 k
we would like to take I = F , however this only works when k = 2, and does not hold due to
k k
the repetitions for k ≥ 3. For example, when k = 4, if x,y ∈ Fn satisfyhx,yi= 0, then
q
F (x,x,y,y) = 1,
4
and we need to try to modify F to make it zero in this case.
k
3.1. Constructing an indicator function for distinct k-tuples. Let X be a finite set, F
a field, and let Xk = X ×···×X denote the Cartesian product of X with itself k times. For
every σ ∈ S , define
k
f :X ×···×X → F
σ
to be the function that is 1 if (x ,...,x ) is a fixed point of σ, and 0 otherwise.
1 k
Lemma 4. We have the identity
1 if x ,...,x are distinct
1 k
sgn(σ)f (x ,...,x ) = ,
σ 1 k
(0 otherwise
σX∈Sk
where sgn(σ) is the sign of the permutation.
Proof. By definition,
sgn(σ)f (x ,...,x ) = sgn(σ)
σ 1 k
σX∈Sk σ∈SXtab(−→x)
−→ −→
where Stab(x) ⊂ S is the stabilizer of x. Since the stabilizer is a product of symmetric
k
−→
groups, this will be non-zero precisely when Stab(x) is trivial, and hence x ,...,x must be
1 k
distinct. This vector is then fixed only by the identity element, and so the sum equals 1. (cid:3)
Lemma 5. Let C ⊂ S , C = Cl(1 2 ··· k), be the conjugacy class of the k-cycles in S , and
1 k 1 k
define
H (x ,...,x )= sgn(σ)f (x ,...,x ).
k 1 k σ 1 k
σX∈ Sk
σ ∈/ C
1
Then
1 if x ,...,x are distinct
1 k
H (x ,...,x )= (−1)k(k−1)! if x = ··· = x .
k 1 k  1 k
0 otherwise
Proof. This follows from lemma 4and the fact that the conjugacy class of the k-cycle in Sk
has (k−1)! elements each with sign equal to (−1)k−1. (cid:3)
This function can be used to zero-out those tuples of vectors with repetitions, and so when
X = Fn, we define
q
I (x ,...,x ) := F (x ,...,x )H (x ,...,x ),
k 1 k k 1 k k 1 k
and this will satisfy precisely the desired properties of equation (3.1) with c = 1 and c =
1 2
(−1)k(k−1)!. Since we want c 6= 0, this imposes the constraint p ≥ k.
2
When k = 2,
H (x ,x ) = 1,
2 1 2
when k = 3
H (x ,x ,x ) = 1−δ(x ,x )−δ(x ,x )−δ(x ,x )
3 1 2 3 1 2 2 3 3 2
THE MULTI-SLICE RANK METHOD 8
and when k = 4
H (x ,x ,x ,x )=1−δ(x ,x )−δ (x ,x )−δ(x ,x )−δ(x ,x )−δ(x ,x )−δ(x ,x )
4 1 2 3 4 1 2 2 2 3 3 4 4 1 1 3 2 4
+2δ(x ,x ,x )+2δ(x ,x ,x )+2δ(x ,x ,x )+2δ(x ,x ,x )
1 2 3 2 3 4 3 4 1 4 1 2
+δ(x ,x )δ(x ,x )+δ(x ,x )δ(x ,x )+δ(x ,x )δ(x ,x )
1 2 3 4 1 3 2 4 1 4 2 3
where
1 x = ··· = x
1 k
δ(x ,...,x ) = .
1 k
(0 otherwise
As a function on X3, slice-rank(δ(x ,x )) = 1, but on X4 the function
1 2
δ(x ,x )δ(x ,x )
1 2 3 4
hasslicerankequalto|X|. Thisfunction does however havemulti-slice rank equalto 1. When
writing H (x ,...,x ) as a linear combination of δ functions, it will always have multi-slice
k 1 k
rank at most k!, since it is a sum of k! slices. However, starting from k = 4 there will be
functions in the sum whose slice rank is maximal, and this is why the multi-slice rank is
needed to prove theorems 1 and 2 for k ≥ 4 and k ≥ 3 respectively.
3.2. Proof of Theorem 1.
Proof. Let F be defined as in (3.2), and let H :Sk → F be defined as in lemma 5. Define
k k q
I (x ,...,x )= F (x ,...,x )H (x ,...,x ).
k 1 k k 1 k k 1 k
This function satisfies
1 x ,...,x are mutually orthogonal and distinct
1 k
I (x ,...,x ) = (−1)k(k−1)! when x = x =··· = x ,
k 1 k  1 2 k
0 otherwise
and so we assume that q = pr where p ≥ k, since otherwise the above function would be zero

when x = ··· = x . It then follows from lemma 2 that any set E ⊂ S satisfying
1 k
|E| > multi-slice-rank(I )
k
must contain a k-tuple of distinct mutually orthogonal vectors. In what follows, we bound the
multi-slice rank of I . For any t, we may write
k
n q−1
F (x ,...,x ) = 1− x x ,
t 1 j ji li
 
!
j<l≤t i=1
Y X
 
and expand this product as a polynomial. The total number of terms in the expansion of this
polynomial will be at most the total number of terms in the polynomial
(1+z1+···+zn)(q−1)t(t−21).
t(t−1)
Letting β = (q−1), it is a classical combinatorics result that the above has at most
2
n+β 1
β
≤ (n+β)
β β!
(cid:18) (cid:19)
terms. Since this was a bound on the total number of terms, this bounds the tensor rank of
F , and by lemma 3 combined with the fact that multi-slice-rank(H ) ≤ k!, we immediately
t k
retrieve the bound
n+β
multi-slice-rank(I ) ≤ k! .
k
β
(cid:18) (cid:19)
By being slightly more careful, we can obtain significantly better results. For σ ∈S let f be
k σ
defined as in subsection 3.1, and consider
f (x ,...,x )F (x ,...,x ).
σ 1 k k 1 k
THE MULTI-SLICE RANK METHOD 9
Let t be the number of cycles in the permutation decomposition of σ, and let r denote the
number of fixed points. Then f will be a product of t − r multi-variable delta functions.
σ
Suppose that f (x ,...,x ) = 1. Then there will be exactly t distinct elements among the x ,
σ 1 k i
and so we have the equality
−→
f (x ,...,x )F (x ,...,x )= f (x ,...,x )F (x )
σ 1 k k 1 k σ 1 k t S
where S ⊂ {1,2,...,k} is a subset of indices such that no two elements of S are in the same
cycle of the permutation σ. By slicing in between the cycles, it follows that the multi-slice
rank of this tensor will be at most the slice-rank of the t-dimensional tensor F (z ,...,z ).
t 1 t
Thus
−→
multi-slice-rank(f (x ,...,x )F (x )) ≤ multi-slice-rank(F ),
σ 1 k t S t
where the right hand side is viewed as a t-dimensional tensor. This implies the bound
k
k
(3.3) multi-slice-rank(I )≤ multi-slice-rank(F )
k i
i
i=2(cid:20) (cid:21)
X
where k denotes the unsigned Stirling numbers of the first kind, which count the number
i
of permhutiations of length k made of exactly i disjoint cycles, and multi-slice-rank(Fi) refers
to the multi-slice rank of F when viewed as an i-dimensional tensor. Returning to (3.2), the
i
product form for F ,
t
n q−1
F (z ,...,z ) = 1− z z ,
t 1 t ji li
 
!
j<l≤t i=1
Y X
 
we may bound the multi-slice rank by counting the number of polynomials in z that can
1
occur. Expanding the above product, it will be a linear combination of terms of the form
(ze11···zet1)···(ze1n···zetn)
11 t1 1n tn
where e ≤ q−1 for all i,j and
ij
n
e ≤ (q−1)(t−1)
ij
j=1
X
for every i. It then follows that
n
multi-slice-rank(F ) ≤ # ~v ∈ {0,1,...,q−1}n : v ≤ (q−1)(t−1) .
t i
( )
i=1
X
Let
n
C =# ~v ∈ {0,1,...,q−1}n : v ≤(q−1)(t−1) .
t i
( )
i=1
X
This can be thought of as putting (q−1)(t−1) balls into n+1 buckets, where the (n+1)st
bucket represents the constant term. By the balls and buckets theorem,
n+(q−1)(t−1)
C ≤ ,
t
(q−1)(t−1)
(cid:18) (cid:19)
however this significantly over counts the number of such vectors. Each vector v satisfying
n
(q−1)(t−2) < v ≤ (q−1)(t−1)
i
i=1
X
will be counted exactly once. Each vector v satisfying
n
(q−1)(t−3) < v ≤ (q−1)(t−2)
i
i=1
X
THE MULTI-SLICE RANK METHOD 10
will be counted exactly n+1 = n+1 times, since we may put an extra (q−1) balls into any
1
of the n+1 buckets and arrive at the same vector. Each vector v satisfying
(cid:0) (cid:1)
n
(q−1)(t−r−2) < v ≤ (q−1)(t−r−1)
i
i=1
X
will be counted exactly n+r times, since we have r extra (q−1)’s to put into n+1 buckets.
r
Thus
(cid:0) (cid:1)
n+(q−1)(k−1) n+1 n+0 n+2 n+1
= C + − C + − C +··· ,
k k−1 k−2
(q−1)(k−1) 1 0 2 1
(cid:18) (cid:19) (cid:18)(cid:18) (cid:19) (cid:18) (cid:19)(cid:19) (cid:18)(cid:18) (cid:19) (cid:18) (cid:19)(cid:19)
that is
n+(q−1)(k−1) n+1 n+2
(3.4) = C +nC + C + C +···
k k−1 k−2 k−3
(q−1)(k−1) 2 3
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
k
n−1+j
(3.5) = C .
k−j
j
j=0(cid:18) (cid:19)
X
We remark that from this identity, one could use generating series to work out C exactly as
k
a sum of binomial coefficients, and hence get an exact expression for
k
k
C .
t
t
i=t (cid:20) (cid:21)
X
However thisexpressionhasanunpleasant form, andsotohavethetheorem inasimplerform,
we proceed by proving the following lemma:
Lemma 6. Let k denote the unsigned Stirling numbers of the first kind and let n = k .
i 2
Then h i
(cid:0) (cid:1)
k n
≤
k−i i
(cid:20) (cid:21) (cid:18) (cid:19)
for all 1 ≤ i≤ k.
Proof. Let σ ,...,σ denote the k transpositions in S . Every element of S with k − i
1 n 2 k k
cycles is the product of exactly i distinct transpositions. Hence
(cid:0) (cid:1)
k n
≤ ,
k−i i
(cid:20) (cid:21) (cid:18) (cid:19)
and the lemma is proven. (cid:3)
By equation (3.4) and the above lemma, for n ≥ k it follows that
2
(cid:0) (cid:1)k
n+(q−1)(k−1) k
≥ C ,
t
(q−1)(k−1) t
(cid:18) (cid:19) t=2(cid:20) (cid:21)
X
and so
n+(q−1)(k−1)
multi-slice-rank(I ) ≤ ,
k
(q−1)(k−1)
(cid:18) (cid:19)
and the proof is complete. (cid:3)