Properties of sets of Subspaces with Constant Intersection Dimension

TL;DR

This paper investigates the properties and bounds of sets of subspaces with constant intersection dimensions, providing new theoretical limits and examples for their structure in finite vector spaces.

Contribution

It establishes new upper bounds for the sum of the dimensions of the span and intersection of subspace sets with constant intersection dimension.

Findings

Derived upper bounds for im S + im I in various cases

Provided spectrum results for specific parameter ranges

Constructed examples reaching large intervals of im S + im I

Abstract

A $(k,k-t)$-SCID (set of Subspaces with Constant Intersection Dimension) is a set of $k$-dimensional vector spaces that have pairwise intersections of dimension $k-t$. Let $\mathcal{C}=\{\pi_1,\ldots,\pi_n\}$ be a $(k,k-t)$-SCID. Define $S:=\langle \pi_1, \ldots, \pi_n \rangle$ and $I:=\langle \pi_i \cap \pi_j \mid 1 \leq i < j \leq n \rangle$. We establish several upper bounds for $\dim S + \dim I$ in different situations. We give a spectrum result for the case $(n-1)(k-t)\leq k$ and for the case $n\leq\frac{q^{t(n-\eta)}-1}{q^t-1}$, giving examples of $(k,k-t)$-SCIDs reaching a large interval of values for $\dim S + \dim I$.

Tables

Table 1. Table 1: Summary of the best bounds found for dim S + dim I dimension 𝑆 dimension 𝐼 \dim S+\dim I , for different values of n 𝑛 n , k 𝑘 k and t 𝑡 t .

Condition	Upper bound for $dimS+dimI$	Sharpness?
$\left(k-t\right)\left(n-1\right)\le k$	$nk$	yes
$k\ge 2t$, $n\ge 3$	$2k+2\left(n-2\right)t-\left(n-3\right)$	unknown
$\left(k-t\right)\left(n-1\right)>k$	$nk$	no

Equations

$$S$$

$$I$$

$$\dim S+\dim I\le nk.$$

$$\dim I=\dim I^{\prime }+\dim A-\dim (A\cap I^{\prime }).$$

$$\dim I\le \dim I^{\prime }+k-t+\delta .$$

$$\dim S=\dim S^{\prime }+\dim {\pi }_n-\dim ({\pi }_n\cap S^{\prime }).$$

$$\dim S\le \dim S^{\prime }+k-(k-t+\delta )=\dim S^{\prime }+t-\delta .$$

$$\dim S+\dim I$$

$$\le (n-1)k+k$$

$$\le nk,$$

$$\dim S+\dim I=nk$$

$$\dim ⟨U_i,V_{lj}\mid i\in \{1,\dots ,n\}\text{ and }1\le l<j\le n⟩$$

$$=n(k-(n-1)(k-t))+\frac{n(n-1)}{2}(k-t).$$

$$\sum _{i=1}^n\dim U_i+\sum _{1\le i<j\le n}\dim V_{ij}=n(k-(n-1)(k-t))+\frac{n(n-1)}{2}(k-t).$$

$$\dim I=\sum _{1\le l<j\le n}\dim V_{lj}=\frac{n(n-1)}{2}(k-t).$$

$$\dim S=n(k-(n-1)(k-t))+\frac{n(n-1)}{2}(k-t).$$

$$\dim I=\dim I^{\prime }+k-t+\delta ,$$

$$\dim S=nk-\dim I=nk-\frac{n(n-1)}{2}(k-t),$$

$$\dim I$$

$$\dim S$$

$$\dim S+\dim I\le (n-1)k+2t.$$

$$\dim S^{\prime }+\dim I^{\prime }\le (n-1)k.$$

$$\dim I=\dim I^{\prime }+\dim A-\dim (A\cap I^{\prime }).$$

$$\dim I\le \dim I^{\prime }+k-t+\delta -(k-2t)=\dim I^{\prime }+t+\delta .$$

$$\dim S\le \dim S^{\prime }+\dim {\pi }_n-\dim A\le \dim S^{\prime }+t-\delta .$$

$$\dim S+\dim I$$

$$\le (n-1)k+2t.$$

$$2k+2(n-2)t-(n-3)$$

$$<2k+(n-3)k+2t$$

$$=(n-1)k+2t,$$

$$\dim S+\dim I\le 2k+2(n-2)t-(n-3).$$

$$\dim S^{\prime }+\dim I^{\prime }\le 2k+2(n-3)t-(n-4).$$

$$\dim S\le \dim S^{\prime }+\dim {\pi }_n-\dim A=\dim S^{\prime }+t-\delta .$$

$$\dim {\pi }_n$$

Full text

1 Introduction

Let $\mathcal{V}$ be a vector space over a finite field $\mathbb{F}_{q}$. Let $k$ and $t$ be integers such that $t\leq k$. A set $\mathcal{C}$ of $k$-dimensional subspaces of $\mathcal{V}$ that have pairwise intersections of dimension $k-t$ is called a $(k,k-t)$-SCID. The acronym SCID stands for a set of Subspaces with Constant Intersection Dimension.

SCIDs are introduced in [3], where a similar definition is given in terms of projective spaces instead of vector spaces. For our purposes however, the ambient space will always be a vector space $\mathcal{V}$ over a finite field $\mathbb{F}_{q}$ of order $q$. The dimension of this vector space $\mathcal{V}$ does not need to be predefined.

In the domain of coding theory, SCIDs are better known as equidistant codes. These codes are relevant in a random network coding setting, where information is sent through a network with varying topology. This network is depicted as a directed multigraph where the information has to be transmitted from the sources to the sinks through some intermediary nodes. Within network coding, these intermediary nodes apply coding to the received inputs, instead of simply routing them. It was shown in [1] that the maximal information rate of a network with one source can be achieved by applying this technique. When the order of the ground field is large enough, it is sufficient to only apply linear network coding, where the nodes transmit linear combinations of the input they receive [8]. Random network coding is when the nodes output random linear combinations of the input, instead of using a predefined scheme. The concept and benefits of random network coding in a multi-source setting are explored in [5], and later on approached mathematically through subspace codes in [6].

A subspace code is a code that has vector subspaces as codewords. Note that this is different from the classical case, where the codewords are vectors. The distance between two codewords $U$ and $V$ from a subspace code is called the subspace distance and is defined as $d(U,V)=\dim U+\dim V-2\dim(U\cap V)$. When all codewords have the same dimension, the code is called a constant dimension code. When in addition all pairwise intersections have the same dimension, then the distance between any two codewords is constant. In this case we say we have an equidistant code, see e.g. [4]. It should now be clear why these codes correspond to SCIDs.

Note that in the definition of a SCID, it is required that all pairwise intersections have the same dimension. It is not necessary that they all coincide. If this is the case, then the SCID is called a $(k,k-t)$-sunflower. The $(k-t)$-space that all the elements of the sunflower have in common, is often called the center of the sunflower. Sunflowers have been investigated before, see for instance [2].

Another special case of SCIDs occurs when $k=t$, i.e. when every two distinct elements intersect trivially. A $(k,0)$-SCID is called a partial $k$-spread. Note that a partial $k$-spread is also a $(k,0)$-sunflower with trivial center. Partial spreads are studied thoroughly within the domain of finite geometry, see for example [9] and [10].

From now on, we will assume that a SCID contains at least two elements. The following lemma follows directly from the definitions:

Lemma 1.1**.**

A subset of a $(k,k-t)$-SCID (resp. $(k,k-t)$-sunflower), containing at least two elements, is again a $(k,k-t)$-SCID (resp. $(k,k-t)$-sunflower).

Let $\mathcal{C}=\{\pi_{1},\ldots,\pi_{n}\}$ be a $(k,k-t)$-SCID with $n$ elements. Define the following two spaces:

[TABLE]

Intuitively, when the space $S$ has large dimension, the elements of $\mathcal{C}$ are further apart, causing the dimension of $I$ to be smaller and vice versa. This raises the question: is it possible to give an upper bound on $\dim{S}+\dim{I}$?

The article is structured as follows: In Section 2, we establish several upper bounds for different situations. In Section 3, we give a spectrum result for the case $(n-1)(k-t)\leq k$ and for the case $n\leq\frac{q^{t(n-\eta)}-1}{q^{t}-1}$, giving examples of $(k,k-t)$-SCIDs reaching a large interval of values for $\dim S+\dim I$.

2 Upper bounds on $\dim S+\dim I$

In this section, we justify the intuition from the previous section by giving upper bounds on the sum $\dim S+\dim I$. When an upper bound is established on this sum, it is clear that for a large dimension of $S$, the dimension of $I$ must be small, and vice versa. We give several upper bounds and compare them for different values of $n\geq 2$, $k$ and $t$. At the end of this section, a summary of the best bounds is given in Table 1.

Theorem 2.1 gives a bound that is valid for all values of $n\geq 2$, $k$ and $t$.

Theorem 2.1**.**

Let $\mathcal{C}=\{\pi_{1},\ldots,\pi_{n}\}$ be a $(k,k-t)$-SCID, $n\geq 2$. Define $S:=\langle\pi_{1},\ldots,\pi_{n}\rangle$ and $I:=\langle\pi_{i}\cap\pi_{j}\mid 1\leq i<j\leq n\rangle$. Then:

[TABLE]

Proof.

The proof is by induction on the number of spaces $n$. For the induction base, assume $n=2$. Then $\mathcal{C}=\{\pi_{1},\pi_{2}\}$. Hence, $S=\langle\pi_{1},\pi_{2}\rangle$ has dimension $k+t$ and $I=\pi_{1}\cap\pi_{2}$ has dimension $k-t$. In this case, $\dim S+\dim I=2k$, agreeing with the theorem.

Now assume the theorem is true for $n-1$. Define $\mathcal{C^{\prime}}:=\{\pi_{1},\ldots,\pi_{n-1}\}$, then $\mathcal{C^{\prime}}$ is a $(k,k-t)$-SCID by Lemma 1.1. Hence, if we define $S^{\prime}:=\langle\pi_{1},\ldots,\pi_{n-1}\rangle$ and $I^{\prime}:=\langle\pi_{i}\cap\pi_{j}\mid 1\leq i<j\leq n-1\rangle$, then $\dim S^{\prime}+\dim I^{\prime}\leq(n-1)k$, by the induction hypothesis.

Define $A:=\langle\pi_{1}\cap\pi_{n},\pi_{2}\cap\pi_{n},\ldots,\pi_{n-1}\cap\pi_{n}\rangle$, so $A$ is the space spanned by all intersections of $\pi_{1},\ldots,\pi_{n-1}$ with the space $\pi_{n}$. Then $k-t\leq\dim A\leq k$, since $\pi_{1}\cap\pi_{n}\subseteq A\subseteq\pi_{n}$. Note that $I=\langle I^{\prime},A\rangle$, such that:

[TABLE]

Let $0\leq\delta\leq t$ be such that $\dim A=k-t+\delta$, then:

[TABLE]

On the other hand, from $S=\langle S^{\prime},\pi_{n}\rangle$, it follows:

[TABLE]

But $A\subseteq\pi_{n}\cap S^{\prime}$, hence $\dim(\pi_{n}\cap S^{\prime})\geq\dim A=k-t+\delta$. Together with $\dim\pi_{n}=k$, this results in the following inequality:

[TABLE]

Combining (1) and (2) with the induction hypothesis, we find:

[TABLE]

which concludes the proof. ∎

The natural question that arises now is whether this bound is sharp. In Theorem 2.2, a construction of a SCID reaching this upper bound is given, under the assumption that $(n-1)(k-t)\leq k$. Hence, under this assumption, the bound given in Theorem 2.1 is sharp.

Theorem 2.2**.**

Let $\mathcal{C}=\{\pi_{1},\ldots,\pi_{n}\}$ be a $(k,k-t)$-SCID. Define $S:=\langle\pi_{1},\ldots,\pi_{n}\rangle$ and $I:=\langle\pi_{i}\cap\pi_{j}\mid 1\leq i<j\leq n\rangle$. Let $(n-1)(k-t)\leq k$. Then

[TABLE]

if and only if there exist $(k-t)$-spaces $V_{ij}$, for each $1\leq i<j\leq n$, and $(k-(n-1)(k-t))$-spaces $U_{i}$, for each $1\leq i\leq n$, such that the following conditions hold:

For each $1\leq i<j\leq n$, $V_{ij}=\pi_{i}\cap\pi_{j}$. 2. 2.

For each $1\leq i\leq n$, $\pi_{i}=\langle U_{i},\pi_{i}\cap\pi_{j}\mid i\neq j\rangle$. 3. 3.

The dimension of the span of all the spaces above is maximal, i.e.,

[TABLE]

Proof.

First note that to find the dimension of the space in the third condition, we can just sum up the dimensions of the spaces $U_{i}$ and $V_{lj}$. This gives us the formula in the third condition:

[TABLE]

For the first part of the proof, assume that all the enlisted conditions hold for $\mathcal{C}$. Then we want to prove that $\dim S+\dim I=nk$. The third condition implies that the span of all spaces $V_{ij}$, with $1\leq i<j\leq n$, must be maximal. So to find the dimension of $I$, we just need to sum up the dimensions of the spaces $V_{ij}$:

[TABLE]

On the other hand, the first two conditions imply that $S=\langle\pi_{1},\ldots,\pi_{n}\rangle=\langle U_{i},V_{lj}\mid i\in\{1,\ldots,n\}\text{ and }1\leq l<j\leq n\rangle$. The third condition immediately implies:

[TABLE]

Combining (3) and (4), we get that $\dim S+\dim I=nk$. This completes the first part of the proof.

The remainder of the proof is again by induction on the size $n$ of $\mathcal{C}$. For the induction base, assume $n=2$. Then it is not hard to see that the enlisted conditions must hold.

Now assume that the theorem is true for any $(k,k-t)$-SCID with size $n-1$ and that we have a $(k,k-t)$-SCID $\mathcal{C}=\{\pi_{1},\ldots,\pi_{n}\}$ of size $n$ such that $\dim S+\dim I=nk$. Now define $\mathcal{C^{\prime}}:=\{\pi_{1},\ldots,\pi_{n-1}\}$, $S^{\prime}:=\langle\pi_{1},\ldots,\pi_{n-1}\rangle$ and $I^{\prime}:=\langle\pi_{i}\cap\pi_{j}\mid 1\leq i<j\leq n-1\rangle$. Then, by Lemma 1.1, $\mathcal{C^{\prime}}$ is a $(k,k-t)$-SCID.

Note that we only can have $\dim S+\dim I$ maximal, if equality holds in (1) and (2) in every induction step of the proof of Theorem 2.1. Adding up these two equalities, we find that $\dim S+\dim I=\dim S^{\prime}+\dim I^{\prime}+k$, implying that also $\dim S^{\prime}+\dim I^{\prime}=(n-1)k$ must be maximal. Applying the induction hypothesis on $\mathcal{C^{\prime}}$ now gives us the following:

There exist $(k-t)$-spaces $V_{ij}$, for each $1\leq i<j\leq n-1$, such that $V_{ij}=\pi_{i}\cap\pi_{j}$. 2. 2.

There exist $(k-(n-2)(k-t))$-spaces $U^{\prime}_{i}$, for each $1\leq i\leq n-1$, such that $\pi_{i}=\langle U^{\prime}_{i},\pi_{i}\cap\pi_{j}\mid i\neq j\rangle$. 3. 3.

The span of the spaces above has maximal dimension.

Define $A:=\langle\pi_{1}\cap\pi_{n},\pi_{2}\cap\pi_{n},\ldots,\pi_{n-1}\cap\pi_{n}\rangle$, and $\delta\geq 0$ such that $\dim A=k-t+\delta$. As remarked before, equality must hold in (1) in the proof of Theorem 2.1:

[TABLE]

hence, $\dim(A\cap I^{\prime})=0$. Now define $V_{in}:=\pi_{i}\cap\pi_{n}$, for all $1\leq i<n$. Then $\dim(A\cap I^{\prime})=0$ implies that $V_{in}\subseteq U^{\prime}_{i}$, so the third property implies that the span $\langle V_{ij}\mid 1\leq i<j\leq n\rangle$ has maximal dimension. Analogously as in the first part of the proof, we now find that $\dim I=\frac{n(n-1)}{2}(k-t)$.

Now choose $(k-(n-1)(k-t))$-spaces $U_{i}\subseteq U^{\prime}_{i}$, for $1\leq i<n$, such that $U^{\prime}_{i}=\langle U_{i},V_{in}\rangle$. Also choose a $(k-(n-1)(k-t))$-space $U_{n}$ such that $\pi_{n}=\langle V_{in},U_{n}\mid 1\leq i<n\rangle$. For these choices of $V_{ij}$, with $1\leq i<j\leq n$, and $U_{i}$, with $1\leq i\leq n$, the first two conditions are fulfilled.

Note that $S=\langle U_{i},V_{lj}\mid i\in\{1,\ldots,n\}\text{ and }1\leq l<j\leq n\rangle$, while for the dimension of $S$ we have:

[TABLE]

which is exactly the sum of the dimensions of the spaces $V_{ij}$, with $1\leq i<j\leq n$ and the spaces $U_{i}$, with $1\leq i\leq n$. This is precisely what the third condition states. ∎

Note that the condition $(n-1)(k-t)\leq k$ is necessary for the construction in Theorem 2.2 to work. Moreover, it follows from the proof that if this condition doesn’t hold, then there exists no $(k,k-t)$-SCID with $\dim S+\dim I=nk$. This means that the bound given in Theorem 2.1 is sharp if and only if the condition $(n-1)(k-t)\leq k$ holds.

The objective now is to gain more insight in what happens if the condition $(n-1)(k-t)\leq k$ doesn’t hold. For this purpose, it is useful to consider the case where $n=3$.

In the case $2(k-t)>k$, the bound from Theorem 2.1, although valid, cannot be sharp. Lemma 2.3 provides an upper bound that is also valid for all values of $k$ and $t$, but which is in fact an improvement in the case $2(k-t)>k$.

Lemma 2.3**.**

Let $\mathcal{C}=\{\pi_{1},\pi_{2},\pi_{3}\}$ be a $(k,k-t)$-SCID. Define $S:=\langle\pi_{1},\pi_{2},\pi_{3}\rangle$ and $I:=\langle\pi_{1}\cap\pi_{2},\pi_{1}\cap\pi_{3},\pi_{2}\cap\pi_{3}\rangle$. Then $\dim S+\dim I\leq 2(k+t)$.

Proof.

Define $\epsilon\geq 0$, such that $\dim(\pi_{1}\cap\pi_{2}\cap\pi_{3})=k-t-\epsilon$.

Then:

[TABLE]

such that $\dim S+\dim I\leq k+t+t-\epsilon+k-t+2\epsilon=2k+t+\epsilon$. Note however that $\epsilon\leq k-(k-t)=t$, such that $\dim S+\dim I\leq 2k+2t$. ∎

Note that the equality $\dim S+\dim I=2(k+t)$ only occurs when $\epsilon=t$. In that case, we have $\pi_{1}=\langle\pi_{1}\cap\pi_{2},\pi_{1}\cap\pi_{3}\rangle$, $\pi_{2}=\langle\pi_{1}\cap\pi_{2},\pi_{2}\cap\pi_{3}\rangle$ and $\pi_{3}=\langle\pi_{2}\cap\pi_{3},\pi_{1}\cap\pi_{3}\rangle$.

Inspired by this lemma, we prove a new bound on $\dim S+\dim I$ that is valid for all values of $n\geq 3$, $k$ and $t$.

Theorem 2.4**.**

Let $\mathcal{C}=\{\pi_{1},\ldots,\pi_{n}\}$ be a $(k,k-t)$-SCID, with size $n\geq 3$. Define $S:=\langle\pi_{1},\ldots,\pi_{n}\rangle$ and $I:=\langle\pi_{i}\cap\pi_{j}\mid 1\leq i<j\leq n\rangle$. Then:

[TABLE]

Proof.

Define $S^{\prime}:=\langle\pi_{1},\ldots,\pi_{n-1}\rangle$ and $I^{\prime}:=\langle\pi_{i}\cap\pi_{j}\mid 1\leq i<j\leq n-1\rangle$. Then, by Theorem 2.1, we know that:

[TABLE]

Now define $A:=\langle\pi_{1}\cap\pi_{n},\ldots,\pi_{n-1}\cap\pi_{n}\rangle$ and let $\delta\geq 0$ be such that $\dim A=k-t+\delta$.

Note that $I=\langle I^{\prime},A\rangle$, such that:

[TABLE]

Define $B:=\langle\pi_{1}\cap\pi_{2},\pi_{1}\cap\pi_{3},\ldots,\pi_{1}\cap\pi_{n-1}\rangle$, the space spanned by all the intersections with $\pi_{1}$, except for $\pi_{1}\cap\pi_{n}$. Then $\pi_{1}\cap\pi_{2}\subseteq B$, such that $\dim B\geq k-t$. Moreover we have $\langle\pi_{1}\cap\pi_{n},B\rangle\subseteq\pi_{1}$, such that:

$\dim\pi_{1}\geq\dim(\pi_{1}\cap\pi_{n})+\dim B-\dim(\pi_{1}\cap\pi_{n}\cap B)$

$\Downarrow$

$k\geq k-t+k-t-\dim(\pi_{1}\cap\pi_{n}\cap B)$

$\Downarrow$

$\dim(\pi_{1}\cap\pi_{n}\cap B)\geq k-2t.$

Since $\pi_{1}\cap\pi_{n}\cap B\subseteq A\cap I^{\prime}$, it follows that $\dim(A\cap I^{\prime})\geq k-2t$. Combining this with (6) we get:

[TABLE]

On the other hand, $S=\langle S^{\prime},\pi_{n}\rangle$ and $A\subseteq S^{\prime}\cap\pi_{n}$, such that:

[TABLE]

Combining (7) and (8) with (5), we find:

[TABLE]

∎

Comparing this new bound to the bound given in Theorem 2.1, we can now distinguish three cases:

• $k<2t$

In this case, $nk<(n-1)k+2t$, so the bound from Theorem 2.1 is the best bound we have. By Theorem 2.2, this bound is sharp if and only if the inequality $(n-1)(k-t)\leq k$ holds.

• $k=2t$

Now we have $nk=(n-1)k+2t$, such that the new bound is the same as the bound given in Theorem 2.1. Note that in this case $(n-1)(k-t)\leq k\Leftrightarrow n\leq 3$, such that the bound is only sharp for $n\leq 3$. For $n>3$, we will show in Theorem 2.5 that $\dim S+\dim I\leq nk-(n-3)$. We don’t know whether this bound is sharp.

• $k>2t$

Then $(n-1)k+2t<nk$, such that the new bound is an improvement compared to Theorem 2.1. In Theorem 2.5, we will show that $\dim S+\dim I\leq 2k+2(n-2)t-(n-3)$. Note that, given $k>2t$, we have for $n\geq 3$:

[TABLE]

such that Theorem 2.5 indeed gives us a better bound for $n\geq 3$.

Theorem 2.5**.**

For $n\geq 3$, let $\mathcal{C}=\{\pi_{1},\ldots,\pi_{n}\}$ be a $(k,k-t)$-SCID, with $k\geq 2t$. Define $S:=\langle\pi_{1},\ldots,\pi_{n}\rangle$ and $I:=\langle\pi_{i}\cap\pi_{j}\mid 1\leq i<j\leq n\rangle$. Then:

[TABLE]

Proof.

The proof is by induction on $n$.

For the induction base, consider $n=3$. If $k>2t$, it follows from Lemma 2.3 that $\dim I+\dim S\leq 2k+2t$, agreeing with the theorem. If $k=2t$, then we have by Theorem 2.1 that $\dim S+\dim I\leq 3k=2k+2t$.

Now assume that the theorem is true for $(k,k-t)$-SCIDs with $n-1$ elements. Then it is in particular true for $\mathcal{C^{\prime}}:=\{\pi_{1},\ldots,\pi_{n-1}\}$. Define $S^{\prime}:=\langle\pi_{1},\ldots,\pi_{n-1}\rangle$ and $I^{\prime}:=\langle\pi_{i}\cap\pi_{j}\mid 1\leq i<j\leq n-1\rangle$. Then the induction hypothesis implies:

[TABLE]

Define $A:=\langle\pi_{1}\cap\pi_{n},\ldots,\pi_{n-1}\cap\pi_{n}\rangle$ to be the space spanned by the intersections with the space $\pi_{n}$. Then $\dim A=k-t+\delta$, for a certain value $\delta\geq 0$. Note that $S=\langle S^{\prime},\pi_{n}\rangle$ and that $A\subseteq\pi_{n}\cap S^{\prime}$, thus we have:

[TABLE]

We can now repeat the same argument as in the previous proof, to find that $\dim(A\cap I^{\prime})\geq k-2t$. We distinguish between two cases:

• •

Case 1: $\dim(A\cap I^{\prime})=k-2t$.

For any $i,j$, with $1\leq i<j\leq n-1$, we have $\langle\pi_{i}\cap\pi_{n},\pi_{j}\cap\pi_{n}\rangle\subseteq\pi_{n}$, implying:

[TABLE]

We find $\dim(\pi_{i}\cap\pi_{j}\cap\pi_{n})\geq k-2t$. Since $\dim(A\cap I^{\prime})=k-2t$ and $\pi_{i}\cap\pi_{j}\cap\pi_{n}\subseteq A\cap I^{\prime}$, we have that $\pi_{i}\cap\pi_{j}\cap\pi_{n}=A\cap I^{\prime}$. Since this argument is independent from the choices of $i$ and $j$, it follows that all intersections $\pi_{i}\cap\pi_{j}\cap\pi_{n}$ must coincide, for $1\leq i<j\leq n-1$. Hence, the intersections $\pi_{i}\cap\pi_{n}$ form a $(k-t,k-2t)$-sunflower inside $\pi_{n}$.

Now let $X$ be a $t$-dimensional space skew to $\pi_{n}$, such that $\pi_{1}\cap\pi_{2}=\langle\pi_{1}\cap\pi_{2}\cap\pi_{n},X\rangle$. Then, $\pi_{1}=\langle\pi_{1}\cap\pi_{n},X\rangle\subseteq\langle\pi_{n},X\rangle$ and $\pi_{n}\subseteq\langle\pi_{n},X\rangle$. For any $i$, with $1<i<n$, we have $\pi_{i}\cap\pi_{1}\subseteq\pi_{i}$, $\pi_{i}\cap\pi_{n}\subseteq\pi_{i}$ and

[TABLE]

This implies that $\pi_{i}=\langle\pi_{i}\cap\pi_{1},\pi_{i}\cap\pi_{n}\rangle\subseteq\langle\pi_{1},\pi_{n}\rangle\subseteq\langle\pi_{n},X\rangle$. Hence, $S\subseteq\langle\pi_{n},X\rangle$ and $I\subseteq\langle\pi_{n},X\rangle$.

We have that $\dim\langle\pi_{n},X\rangle=k+t$, which implies that $\dim I\leq k+t$ and $\dim S\leq k+t$, such that $\dim S+\dim I\leq 2(k+t)$. This bound is lower than the one stated in the theorem.

• •

Case 2: $\dim(A\cap I^{\prime})>k-2t$.

We have that $I=\langle I^{\prime},A\rangle$, such that

[TABLE]

Combining (10) and (• ‣ 2) with (9) we get:

[TABLE]

∎

We conclude this section with a summary of the bounds for different values of $n$, $k$ and $t$, given by Table 1.

3 Spectrum results

In this section we construct examples of SCIDs for several values of $\dim S+\dim I$. In the second subsection, we’ll assume that the condition $(n-1)(k-t)\leq k$ holds, and adapt the construction from Theorem 2.2 to construct new SCIDs with $\dim S+\dim I=nk-\epsilon$. Then later in the third subsection, we will drop the assumption $(n-1)(k-t)\leq k$ and use field reduction to construct sunflowers, which are particular examples of SCIDs, for even smaller values of $\dim S+\dim I$. The concept of field reduction is explained in the first subsection. We finish this section with a summary of the spectrum results we obtained.

3.1 Field reduction

Field reduction is a powerful tool in finite geometry. The method is described for projective spaces in [11] and for projective spaces and polar spaces in [7]. For our purposes however, we will consider field reduction in vector spaces.

The idea relies on the fact that $\mathbb{F}_{q^{t}}$ is a $t$-dimensional vector space over $\mathbb{F}_{q}$. Hence, all 1-dimensional subspaces of a vector space $V(n,q^{t})$ correspond to $t$-dimensional subspaces of the vector space $V(nt,q)$. Moreover, the set of all 1-dimensional spaces of $V(n,q^{t})$ corresponds to a $t$-spread in $V(nt,q)$. This spread is often called a Desarguesian spread.

Note that a $d$-dimensional subspace in $V(n,q^{t})$ corresponds to a $dt$-dimensional subspace in $V(nt,q)$. Hence, we have the following lemma:

Lemma 3.1**.**

Using field reduction, a set of 1-dimensional subspaces in $V(n,q^{t})$ spanning a $d$-dimensional space corresponds to a partial $t$-spread in $V(nt,q)$ spanning a $dt$-dimensional space.

3.2 Spectrum result on SCIDs

In the proof of Theorem 2.2, we constructed a SCID such that both $\dim I$ and $\dim S$ were maximal. This way the sum $\dim S+\dim I$ was maximal as well. In this section we want to construct SCIDs with smaller values for this sum. In order to do this, we adapt the construction from Theorem 2.2 in such a way that $\dim I$ decreases, while $\dim S$ stays as large as possible.

The idea behind the proof of Theorem 3.2 is that instead of having all pairwise intersections span a space of maximal dimension, we now demand that there is some overlap between three spaces of the SCID. This causes $\dim I$ to decrease. On the other hand, we want $\dim S$ to be as large as possible under this requirement. So apart from the overlap, we need all spaces of the SCID to span maximal dimension.

Theorem 3.2**.**

If $(n-1)(k-t)\leq k$ and $0\leq\epsilon\leq k-t$, then there exists a $(k,k-t)$-SCID $\{\pi_{1},\ldots,\pi_{n}\}$, such that

[TABLE]

with $S:=\langle\pi_{1},\ldots,\pi_{n}\rangle$ and $I:=\langle\pi_{i}\cap\pi_{j}\mid i\neq j\rangle$.

Proof.

If $\epsilon=0$, then a construction is given by Theorem 2.2.

For $\epsilon>0$, consider a $(k,k-t)$-SCID $\mathcal{C}=\{\pi_{1},\ldots,\pi_{n}\}$ constructed as in the proof of Theorem 2.2. We will slightly adapt this SCID in order to construct a new SCID $\mathcal{C^{\prime}}=\{\pi^{\prime}_{1},\ldots,\pi^{\prime}_{n}\}$. In this new SCID, there will be an $\epsilon$-dimensional overlap between the three spaces $\pi^{\prime}_{1}$, $\pi^{\prime}_{2}$ and $\pi^{\prime}_{n}$.

Consider an $\epsilon$-dimensional subspace $E\subseteq\pi_{1}\cap\pi_{n}$ and two $(k-t-\epsilon)$-dimensional subspaces $D_{1}\subseteq\pi_{1}\cap\pi_{2}$ and $D_{2}\subseteq\pi_{2}\cap\pi_{n}$. From the conditions in Theorem 2.2, it follows that $\langle E,D_{1},D_{2}\rangle$ has maximal dimension. We’ll choose our spaces from $\mathcal{C^{\prime}}$ such that the space $E$ is the overlap between $\pi^{\prime}_{1}$, $\pi^{\prime}_{2}$ and $\pi^{\prime}_{n}$, and such that $\pi^{\prime}_{1}\cap\pi^{\prime}_{2}=\langle E,D_{1}\rangle$ and $\pi^{\prime}_{2}\cap\pi^{\prime}_{n}=\langle E,D_{2}\rangle$.

By shifting the intersections like this, we loose $\epsilon$ dimensions from each of the three spaces $\pi_{1}$, $\pi_{2}$ and $\pi_{n}$. To compensate for this loss, choose $\epsilon$-dimensional spaces $P_{1}$, $P_{2}$ and $P_{n}$, such that these spaces span maximal dimension together with the elements of $\mathcal{C}$, i.e., $\dim\langle P_{1},P_{2},P_{n},\pi_{i}\mid 1\leq i\leq n\rangle$ is maximal.

Remember that by Theorem 2.2, there exist $(k-(n-1)(k-t))$-dimensional spaces $U_{1},\ldots,U_{n}$, such that for each $1\leq i\leq n$, the space $U_{i}$ is skew to $\langle\pi_{1},\ldots,\pi_{i-1},\pi_{i+1},\ldots,\pi_{n}\rangle$ and such that $\pi_{i}=\langle U_{i},\pi_{i}\cap\pi_{j}\mid i\neq j\rangle$. Now we have all components to define the spaces of $\mathcal{C}^{\prime}$:

[TABLE]

Now we first want to show that this defines in fact a $(k,k-t)$-SCID. For the dimensions of the spaces $\pi^{\prime}_{1},\ldots,\pi^{\prime}_{n}$, note that we can just sum up the dimensions of the spaces in their definitions above. This follows directly from the third condition in Theorem 2.2. We now find:

[TABLE]

Moreover, for the intersections we have that $\pi^{\prime}_{i}\cap\pi^{\prime}_{j}=\pi_{i}\cap\pi_{j}$, for $1\leq i<j\leq n$, except for $\pi^{\prime}_{1}\cap\pi^{\prime}_{2}$ and $\pi^{\prime}_{2}\cap\pi^{\prime}_{n}$. For these intersections we have that $\pi^{\prime}_{1}\cap\pi^{\prime}_{2}=\langle D_{1},E\rangle$ and $\pi^{\prime}_{2}\cap\pi^{\prime}_{n}=\langle D_{2},E\rangle$. Note that $E\subseteq\pi_{1}\cap\pi_{n}$. Hence all pairwise intersections of the spaces $\pi^{\prime}_{1},\ldots,\pi^{\prime}_{n}$ have dimension $k-t$. This implies that $\mathcal{C^{\prime}}=\{\pi^{\prime}_{1},\ldots,\pi^{\prime}_{n}\}$ is indeed a $(k,k-t)$-SCID.

Now define $S:=\langle\pi^{\prime}_{1},\ldots,\pi^{\prime}_{n}\rangle$ and $I:=\langle\pi^{\prime}_{i}\cap\pi^{\prime}_{j}\mid 1\leq i<j\leq n\rangle$. Note that, since $E\subseteq\pi^{\prime}_{1}\cap\pi^{\prime}_{n}$:

[TABLE]

where $(i,j)\in\{1,\ldots,n\}^{2}\setminus\{(1,2),(2,n)\}$. By the definitions above and the third condition of Theorem 2.2, we can sum up the dimensions of these spaces to find the dimension of $I$:

[TABLE]

On the other hand, note that:

[TABLE]

where $1\leq l\leq n$ and $(i,j)\in\{1,\ldots,n\}^{2}\setminus\{(1,2),(2,n)\}$. Note again that $E\subseteq\pi^{\prime}_{1}\cap\pi^{\prime}_{n}$. Again by the third condition of Theorem 2.2 and by the way we defined these spaces, we can find the dimension of $S$ by summing the dimensions:

[TABLE]

Combining (3.2) and (3.2), we get:

[TABLE]

We can conclude that $\mathcal{C^{\prime}}$ is a $(k,k-t)$-SCID fulfilling the condition of the theorem. ∎

The idea of this proof can be generalized to construct SCIDs with lower values of the sum $\dim S+\dim I$. In the first part of the proof of Theorem 3.3, we let intersections coincide instead of just having some overlap. This again causes $\dim I$ to decrease. Meanwhile we keep $\dim S$ as large as possible by choosing spaces that span maximal dimension.

In the second part of the proof, we adapt the construction from the first part by using a similar technique to Theorem 3.2.

Theorem 3.3**.**

If $(n-1)(k-t)\leq k$, $0\leq\epsilon\leq k-t$ and $2\leq\eta\leq n-1$, then there exists a $(k,k-t)$-SCID $\{\pi_{1},\ldots,\pi_{n}\}$ such that:

[TABLE]

with $S:=\langle\pi_{1},\ldots,\pi_{n}\rangle$ and $I:=\langle\pi_{i}\cap\pi_{j}\mid i\neq j\rangle$.

Proof.

For $\eta=2$, this is exactly Theorem 3.2. So assume $\eta>2$. We distinguish between two cases, based on the value of $\epsilon$.

Case 1: $\epsilon=0$

Let $\mathcal{C}$ be a $(k,k-t)$-SCID as constructed in Theorem 2.2. Just like in the previous proof, we will make adaptations to $\mathcal{C}$ in order to obtain a new $(k,k-t)$-SCID $\mathcal{C}^{\prime}=\{\pi^{\prime}_{1},\ldots,\pi^{\prime}_{n}\}$ fulfilling the conditions of the theorem. For this new SCID, there will be a $(k-t)$-dimensional subspace that all spaces $\pi^{\prime}_{1},\ldots,\pi^{\prime}_{\eta-1}$ and $\pi^{\prime}_{n}$ have in common.

Now define $D:=\pi_{1}\cap\pi_{n}$, this will be the common subspace. But if we shift the elements of the SCID in such a way that all spaces $\pi^{\prime}_{1},\ldots,\pi^{\prime}_{\eta-1}$ and $\pi^{\prime}_{n}$ have $D$ in common, then we loose $(k-t)(\eta-2)$ dimensions in each of these spaces. To compensate this, choose $(k-t)(\eta-2)$-dimensional spaces $P_{1},\ldots,P_{\eta-1}$ and $P_{n}$, such that these spaces span maximal dimension with the elements of $\mathcal{C}$, i.e., $\dim\langle P_{1},\ldots,P_{\eta-1},P_{n},\pi_{i}\mid 1\leq i\leq n\rangle$ is maximal.

Note that, by Theorem 2.2, there exist $(k-(n-1)(k-t))$-dimensional spaces $U_{1},\ldots,U_{n}$, such that for each $1\leq i\leq n$, the space $U_{i}$ is skew to $\langle\pi_{1},\ldots,\pi_{i-1},\pi_{i+1},\ldots,\pi_{n}\rangle$ and such that $\pi_{i}=\langle U_{i},\pi_{i}\cap\pi_{j}\mid i\neq j\rangle$. Now we have all components to define the spaces of $\mathcal{C^{\prime}}$:

[TABLE]

We now want to show that this indeed defines a $(k,k-t)$-SCID. To find the dimensions of the spaces $\pi^{\prime}_{1},\ldots,\pi^{\prime}_{n}$, note that we can simply sum up the dimensions of the spaces in their definitions above. This follows from the third condition in Theorem 2.2. We now find:

[TABLE]

For the pairwise intersections, we have $\pi^{\prime}_{i}\cap\pi^{\prime}_{j}=\pi_{i}\cap\pi_{j}$, if $1\leq i\leq n$ and $\eta\leq j<n$. All other pairwise intersections are equal to the space $D$. We can conclude that all the pairwise intersections of the spaces $\{\pi^{\prime}_{1},\ldots,\pi^{\prime}_{n}\}$ have dimension $k-t$, implying $\mathcal{C^{\prime}}:=\{\pi^{\prime}_{1},\ldots,\pi^{\prime}_{n}\}$ is a $(k,k-t)$-SCID.

Now define $S:=\langle\pi^{\prime}_{1},\ldots,\pi^{\prime}_{n}\rangle$ and $I:=\langle\pi^{\prime}_{i}\cap\pi^{\prime}_{j}\mid 1\leq i<j\leq n\rangle$. Note that

[TABLE]

where we can add the dimensions of the intersections $\pi^{\prime}_{i}\cap\pi^{\prime}_{j}$ and $D$, to find $\dim I$. Note that the number of intersections $\pi^{\prime}_{i}\cap\pi^{\prime}_{j}$ occurring in this expression is

[TABLE]

From this follows the dimension of $I$:

[TABLE]

where the first value $k-t$ comes from $\dim D$. On the other hand, for $S$ we have:

[TABLE]

By construction, we can find the dimension of $S$ by summing up all dimensions of the spaces occurring in the expression above. Hence, we find:

[TABLE]

Note that $2\eta(n-\eta)+(n-\eta)(n-\eta-1)=n(n-1)-\eta(\eta-1)$. Combining this with (14) and (3.2), we find:

[TABLE]

Hence, $\mathcal{C^{\prime}}$ is a $(k,k-t)$-SCID fulfilling the condition of the theorem.

Case 2: $\epsilon>0$

Let $\mathcal{C^{\prime}}=\{\pi^{\prime}_{1},\ldots,\pi^{\prime}_{n}\}$ be a $(k,k-t)$-SCID as constructed in the previous case for $\epsilon=0$. Let the spaces $D$, $P_{i}$ and $U_{j}$, for $i\in\{1,\ldots,\eta-1,n\}$ and $j\in\{1,\ldots,n\}$, be as defined in the first case. We will again adapt $\mathcal{C^{\prime}}$ to construct a $(k,k-t)$-SCID $\mathcal{C^{\prime\prime}}=\{\pi^{\prime\prime}_{1},\ldots,\pi^{\prime\prime}_{n}\}$ meeting the desired conditions. For this, we generalize the technique used in the proof of Theorem 3.2.

Let $E\subseteq D=\pi^{\prime}_{1}\cap\pi^{\prime}_{n}$ be an $\epsilon$-dimensional subspace, this will be the overlap between the spaces $\pi^{\prime\prime}_{1},\ldots,\pi^{\prime\prime}_{\eta}$ and $\pi^{\prime\prime}_{n}$. Next, consider $(k-t-\epsilon)$-dimensional subspaces $D_{1}\subseteq\pi^{\prime}_{1}\cap\pi^{\prime}_{\eta},\ldots,D_{\eta-1}\subseteq\pi^{\prime}_{\eta-1}\cap\pi^{\prime}_{\eta}$ and $D_{n}\subseteq\pi^{\prime}_{\eta}\cap\pi^{\prime}_{n}$. We will choose the elements of $\mathcal{C^{\prime\prime}}$ in such a way that $\pi^{\prime\prime}_{i}\cap\pi^{\prime\prime}_{\eta}=\langle D_{i},E\rangle$, for $i\in\{1,\ldots,\eta-1,n\}$.

Shifting the spaces like this again causes a loss of dimensions. Note that we loose $(\eta-1)\epsilon$ dimensions from $\pi^{\prime}_{\eta}$. To compensate, choose an $(\eta-1)\epsilon$-dimensional space $Y$, such that $Y$ spans maximal dimension together with the spaces of $\mathcal{C}$, i.e., $\dim\langle Y,\pi_{i}\mid 1\leq i\leq n\rangle$ is maximal.

From each of the first $\eta-1$ spaces and the $n$th space we loose $\epsilon$ dimensions. To compensate, choose $\epsilon$-dimensional spaces $X_{1},\ldots,X_{\eta-1}$ and $X_{n}$, such that these spaces span maximal dimension together with $Y$ and the spaces of $\mathcal{C^{\prime}}$. I.e. $\dim\langle X_{1},\ldots,X_{\eta-1},X_{n},Y,\pi^{\prime}_{i}\mid 1\leq i\leq n\rangle$ is maximal.

Now we have everything we need to define $\mathcal{C^{\prime\prime}}=\{\pi^{\prime\prime}_{1},\ldots,\pi^{\prime\prime}_{n}\}$:

[TABLE]

By the way we defined the spaces in the definitions above, we can just sum up the dimensions of the subspaces to find the dimensions of the spaces $\pi^{\prime\prime}_{i}$:

[TABLE]

For the pairwise intersections we have $\pi^{\prime\prime}_{i}\cap\pi^{\prime\prime}_{j}=\pi^{\prime}_{i}\cap\pi^{\prime}_{j}$, as long as $i$ and $j$ are different from $\eta$. For $\eta<i<n$, we have $\pi^{\prime\prime}_{\eta}\cap\pi^{\prime\prime}_{i}=\pi^{\prime}_{\eta}\cap\pi^{\prime}_{i}$. For $1\leq i<\eta$, $\pi^{\prime\prime}_{i}\cap\pi^{\prime\prime}_{\eta}=\langle E,D_{i}\rangle$ and similarly we have $\pi^{\prime\prime}_{\eta}\cap\pi^{\prime\prime}_{n}=\langle E,D_{n}\rangle$. We can conclude that all the pairwise intersections of the spaces $\{\pi^{\prime\prime}_{1},\ldots,\pi^{\prime\prime}_{n}\}$ have dimension $k-t$, implying $\mathcal{C^{\prime\prime}}:=\{\pi^{\prime\prime}_{1},\ldots,\pi^{\prime\prime}_{n}\}$ is indeed a $(k,k-t)$-SCID.

Now define $S:=\langle\pi^{\prime\prime}_{1},\ldots,\pi^{\prime\prime}_{n}\rangle$ and $I:=\langle\pi^{\prime\prime}_{i}\cap\pi^{\prime\prime}_{j}\mid 1\leq i<j\leq n\rangle$. Note that, since $E\subseteq D$:

[TABLE]

where $1\leq i\leq n$, $\eta<j<n$ and $i\neq j$. Note that the number of different intersections $\pi^{\prime\prime}_{i}\cap\pi^{\prime\prime}_{j}$ occurring in the expression above is:

[TABLE]

By construction, we can again sum the dimensions of the spaces occurring in the expression above to find the dimension of $I$:

[TABLE]

On the other hand, note for $S$ that:

[TABLE]

where $1\leq i\leq n$, $\eta<j<n$ and $i\neq j$. Again, the construction allows us to calculate the sum of the dimensions of all the spaces in the expression above to find the dimension of $S$. We now have:

[TABLE]

Combining (3.2) and (3.2) and noting that

[TABLE]

we find:

[TABLE]

This shows that $\mathcal{C^{\prime\prime}}$ meets the desired conditions, finishing the proof. ∎

As long as the condition $(n-1)(k-t)\leq k$ holds, we have established examples of $(k,k-t)$-SCIDs with $\dim S+\dim I=N$, for any integer $N\in[nk-(n-2)(k-t),nk]$.

3.3 Spectrum result on sunflowers

What we actually did in the constructions of the previous section, was making $\dim I$ smaller while keeping $\dim S$ as large as possible. This method eventually gives rise to a maximal $(k,k-t)$-sunflower, for the case $\dim S+\dim I=nk-(n-2)(k-t)=2k+(n-2)t$.

Note that for any sunflower we have $\dim I=k-t$, which is the smallest possible dimension for $I$. To construct SCIDs with $\dim S+\dim I\leq 2k+(n-2)t$, it is not possible to further reduce $\dim I$. But since we’re dealing with a maximal sunflower, we can reduce $\dim S$. In that case we still have a sunflower, which is an example of a SCID.

From now on, we drop the condition $(n-1)(k-t)\leq k$. The essence of this section lies in field reduction and the following lemma:

Lemma 3.4**.**

The existence of a $(k,k-t)$-sunflower spanning dimension $d$ is equivalent to the existence of a partial $t$-spread in a $(d-k+t)$-dimensional space, spanning that $(d-k+t)$-dimensional space.

Proof.

Let $S$ be the space spanned by the elements of the sunflower and let $C$ be its center. Then $\dim S=d$, $\dim C=k-t$, and all elements of the sunflower have dimension $k$.

Now consider the quotient space $S/C$. The elements of the sunflower all contain the center $C$, so in the quotient space they have dimension $k-(k-t)=t$. Since all elements of the sunflower have precisely $C$ as their pairwise intersections, their quotient equivalents must intersect trivially. So they must form a partial $t$-spread in $S/C$.

Moreover, since the elements of the sunflower span the space $S$, their quotient equivalents must span $S/C$, which has dimension $d-(k-t)=d-k+t$. ∎

Remember that we are working in a vector space $\mathcal{V}$ over the field $\mathbb{F}_{q}$, otherwise we cannot apply field reduction.

Theorem 3.5**.**

If $1\leq\eta\leq n-2$ and $n\leq\frac{q^{t(n-\eta)}-1}{q^{t}-1}$, then there exists a $(k,k-t)$-SCID $\{\pi_{1},\ldots,\pi_{n}\}$ such that:

[TABLE]

with $S:=\langle\pi_{1},\ldots,\pi_{n}\rangle$ and $I:=\langle\pi_{i}\cap\pi_{j}\mid i\neq j\rangle$.

Proof.

We will construct a sunflower $\mathcal{C}$ meeting the conditions. Note that for a sunflower, $\dim I=k-t$. To have $\mathcal{C}$ fulfill the equality in the theorem, we must have that $\dim S=k+(n-1)t-\eta t$.

By Lemma 3.4, the existence of such a sunflower is equivalent to the existence of a partial $t$-spread in an $(n-\eta)t$-dimensional vector space $V((n-\eta)t,q)$, spanning that space. We can now use field reduction to guarantee the existence of $\mathcal{C}$.

Consider the vector space $V(n-\eta,q^{t})$. Choose $n$ lines, such that the last $n-\eta$ lines span the complete space $V(n-\eta,q^{t})$. Since $n-\eta<n\leq\frac{q^{t(n-\eta)}-1}{q^{t}-1}$, where $\frac{q^{t(n-\eta)}-1}{q^{t}-1}$ is the number of lines in $V(n-\eta,q^{t})$, this is always possible. By Lemma 3.1, this set of $n$ lines in $V(n-\eta,q^{t})$ corresponds to a partial $t$-spread in $V((n-\eta)t,q)$, spanning the whole space. ∎

Note that $\frac{q^{t(n-\eta)}-1}{q^{t}-1}$ is the cardinality of a $t$-spread in an $(n-\eta)t$-dimensional vector space over $\mathbb{F}_{q}$. By reversing the arguments used in the previous proof, it is clear that there cannot exist a $(k,k-t)$-sunflower with $\dim S+\dim I=2k+(n-2)t-\eta t$ if $n>\frac{q^{t(n-\eta)}-1}{q^{t}-1}$. However, this does not exclude the existence of an example of a $(k,k-t)$-SCID with these parameters.

Theorem 3.6**.**

If $1\leq\eta\leq n-2$, $0\leq\epsilon<t$, and $n\leq\frac{q^{t(n-\eta)}-1}{q^{t}-1}$, then there exists a $(k,k-t)$-SCID $\{\pi_{1},\ldots,\pi_{n}\}$ such that:

[TABLE]

with $S:=\langle\pi_{1},\ldots,\pi_{n}\rangle$ and $I:=\langle\pi_{i}\cap\pi_{j}\mid i\neq j\rangle$.

Proof.

For $\epsilon=0$, this is exactly the previous theorem.

For $\epsilon>0$, we will prove the existence of a $(k,k-t)$-sunflower meeting the conditions, in a similar way as in Theorem 3.5. Choose $n$ lines such that the last $n-\eta$ lines span the complete space $V(n-\eta,q^{t})$. Similarly to the previous proof, we have by Lemma 3.1 that this set of $n$ lines in $V(n-\eta,q^{t})$ corresponds to a partial $t$-spread $\{\pi_{1},\ldots,\pi_{n}\}$ in $V((n-\eta)t,q)$, spanning this whole space. Now embed this space $V=V((n-\eta)t,q)$ in a vector space $V^{\prime}=V((n-\eta)t+\epsilon,q)$. Choose an $\epsilon$-dimensional space $E$ in $V^{\prime}$, intersecting trivially with $V$. Then $\langle V,E\rangle=V^{\prime}$. Consider a $(t-\epsilon)$-dimensional subspace $U$ of $\pi_{1}$. Now replace $\pi_{1}$ by $\pi^{\prime}_{1}=\langle U,E\rangle$. Then $\{\pi^{\prime}_{1},\pi_{2},\ldots,\pi_{n}\}$ is a partial $t$-spread, spanning $V^{\prime}$. By Lemma 3.4, there exists a sunflower meeting the conditions. ∎

We have now proved that there exists a $(k,k-t)$-SCID (more precisely, a $(k,k-t)$-sunflower) with $\dim S+\dim I=N$, for any integer $N\in[2k,2k+(n-2)t]$, as long as the condition $n\leq\frac{q^{t(n-\eta)}-1}{q^{t}-1}$ holds. Note that a $(k,k-t)$-SCID with $\dim S+\dim I<2k$ cannot exist.

3.4 Summary

There exists a $(k,k-t)$-SCID with $n$ elements and with $\dim S+\dim I=N$,

• •

for any integer $N\in[2k+(n-2)t,nk]$, if $(n-1)(k-t)\leq k$.

• •

for any integer $N\in[2k,2k+(n-2)t]$, if $n\leq\frac{q^{t(n-\eta)}-1}{q^{t}-1}$.