A note on strong blocking sets
and higgledy-piggledy sets of lines

Let $H$ be an hyperplane containing all but at most $(n-1)$ of the sets $B_{i}$, and relabel the sets $B_{i}$ that are not contained in $H$ as $B_{1},\ldots,B_{s}$, where $s\leq n-1$. Since each $B_{i}$ is a blocking set, we find (at least) one point, say $P_{i}$ contained in $B_{i}\cap H$, $i=1,\ldots,s$. The $s$ points $P_{1},P_{2},\ldots,P_{s}$, span at most an $(n-2)$-dimensional space of $H$ so they are contained in some $n-2$-dimensional space $H^{\prime}$ which is a hyperplane of $H$. There are $q+1$ hyperplanes of $\mbox{\rm PG}(N,q)$ through $H^{\prime}$, one of which is $H$ and contains $B_{s+1},\ldots,B_{k}$. Of the other $q$ hyperplanes, at most $(N-1)(\sqrt{q}+1)$ contain further points of $\mathcal{S}$: there are at most $N-1$ blocking sets $B_{i}$ not contained in $H$, and the point $P_{i}$ lies on at most $\sqrt{q}+1$ lines in $B_{i}$ containing further points of $B_{i}$. Since $q\geq N^{2}$, we have that $N\leq\sqrt{q}$, hence $(N-1)(\sqrt{q}+1)\leq q-1<q$. It follows that at least one hyperplane $\pi$ through $H^{\prime}$ does not contain any points of $\mathcal{S}$ outside of $H^{\prime}$. Since $\pi$ meets $\mathcal{S}$ in a set not spanning $\pi$, $\mathcal{S}$ is not a strong blocking set. ∎

Let $B_{1},\ldots,B_{i}$ be the $i$ lines and $B_{i+1},\ldots,B_{i+k}$ be the $k$ Baer subplanes. Since $i$ lines span at most a $2i-1$-dimensional space, there is a hyperplane containing $B_{1},\ldots,B_{i}$. Since $k\leq N-1$, the result follows from Lemma 2.2.

∎

Assume to the contrary that $\mathcal{S}$ is a strong blocking set of size $S<Nq+c_{p}q^{2/3}-(N-1)(N-2)/2$. Then $\mathcal{S}$ is an $N$-fold blocking set. Let $B$ be the a minimal $N$-fold blocking set contained in $\mathcal{S}$. By Result 2.5, $B$ contains a set $B^{\prime}$ which is the disjoint union of $N$ lines and/or Baer subplanes. From Corollary 2.4, we know that there is a hyperplane $H$ meeting $B^{\prime}$ in points spanning at most an $(N-2)$-space. Let $\pi$ be that $(N-2)$-space. There are at most $Nq+c_{p}q^{2/3}-(N-1)(N-2)/2-N(q+1)<q+1$ points in $\mathcal{S}\setminus\mathcal{B^{\prime}}$. Hence, at least one hyperplane through $\pi$ contains no further points of $\mathcal{S}$, a contradiction since $\mathcal{S}$ is a strong blocking set.

∎

We will use induction on $n$ to show that there exist:

• •

  a set of $2n-2$ lines $\mathcal{L}$ of a Desarguesian spread $\mathcal{D}$ in $\mbox{\rm PG}(2n-1,q)$ such that the number of $(2n-3)$-spaces meeting all lines of $\mathcal{L}$ is at most $(n-1)(q+1)^{2n-2}$;

• •

  a line $M$ of $\mathcal{D}$ such that

  • –

    the number of $(2n-3)$-spaces meeting $M$ in exactly a point and meeting all lines of $\mathcal{L}$ is at most $2(n-1)(q+1)^{2n-3}$, and

  • –

    the number of $(2n-3)$-spaces containing $M$ and meeting all lines of $\mathcal{L}$ is at most $2(n-1)(q+1)^{2n-4}$.

Consider the base case $n=2$. Let $\mathcal{L}$ be any two distinct lines of a Desarguesian spread in $\mbox{\rm PG}(3,q)$. It is clear that the number of lines meeting both lines of $\mathcal{L}$ is precisely $(q+1)^{2}$. Furthermore, let $M$ be any line of the Desarguesian spread, not in $\mathcal{L}$. Then the lines meeting the three lines of $\mathcal{L}\cup\{M\}$ are precisely the $q+1\leq 2(q+1)$ lines of the opposite regulus determined by $\mathcal{L}$ and there is no line containing $M$ and meeting the two lines of $\mathcal{L}$.

So now assume that there is a set $\mathcal{L}$ of $2n-4$ lines of a Desarguesian spread $\mathcal{D}_{\Sigma}$ in $\Sigma=\mbox{\rm PG}(2n-3,q)$ such that the number of $(2n-5)$-spaces meeting all lines of $\mathcal{L}$ is at most $(n-2)(q+1)^{2n-4}$, and a line $M$, not in $\mathcal{L}$ such that the number of $(2n-5)$-spaces meeting all lines of $\mathcal{L}$ meeting $M$ in a point is at most $2(n-2)(q+1)^{2n-5}$ and the number of $(2n-5)-$spaces containing $M$ and meeting all lines of $\mathcal{L}$ is at most $2(n-2)(q+1)^{2n-6}$.

Now embded $\Sigma$ in $\mbox{\rm PG}(2n-1,q)$ and extend the Desarguesian spread $\mathcal{D}_{\Sigma}$ in $\Sigma$ to a Desarguesian spread $\mathcal{D}$ in $\mbox{\rm PG}(2n-1,q)$. Let $M_{0}$ and $M_{1}$ be two lines of $\mathcal{D}$, not in $\Sigma$ such that the $3$-space $\mu=\langle M_{0},M_{1}\rangle$ meets $\Sigma$ precisely in the line $M$. We will first show that the number of $(2n-3)$-spaces meeting $\mathcal{L}\cup\{M_{0},M_{1}\}$ is at most $(n-1)(q+1)^{2n-2}$.

Any $(2n-3)$-space $S$ meeting all lines of $\mathcal{L}\cup\{M_{0},M_{1}\}$ meets $\Sigma$ in either a $(2n-4)$-dimensional or a $(2n-5)$-dimensional space.

First consider the $(2n-3)$-spaces meeting $\Sigma$ in a hyperplane of $\Sigma$. Note that in the case that $S$ meets $\Sigma$ in a hyperplane of $\Sigma$, it is impossible that $S$ contains $\mu$, since $M_{0}$ is contained in $\mu$ and disjoint from $\Sigma$.

Each of the $q^{2n-3}+q^{2n-4}$ hyperplanes $H$ of $\Sigma$ meeting $M$ in exactly a point give rise to exactly one $(2n-3)$-space meeting all lines of $\mathcal{L}\cup\{M_{0},M_{1}\}$, namely the $(2n-3)$-space spanned by $H$ and the unique transversal line to $M_{0},M_{1}$ through $H\cap M$. Each of the $\frac{q^{2n-4}-1}{q-1}$ hyperplanes of $\Sigma$ through $M$ gives rise to $q+1$ $(2n-3)$-spaces meeting all lines of $\mathcal{L}\cup\{M_{0},M_{1}\}$, namely one for each of the $q+1$ planes of $\mu$ through $M$. This gives us $q^{2n-3}+q^{2n-4}+\frac{q^{2n-4}-1}{q-1}(q+1)$ subspaces meeting all lines of $\mathcal{L}\cap\{M_{0},M_{1}\}$ and meeting $\Sigma$ in a hyperplane. This number is smaller than $2q^{2n-3}$ if $q\geq 3$.

Now consider spaces meeting all lines of $\mathcal{L}\cup\{M_{0},M_{1}\}$ and meeting $\Sigma$ in a $(2n-5)-$space. We will make a distinction depending on their intersection size with the line $M$.

First consider such spaces $S$ containing the line $M$. Each of the $(2n-5)$-spaces $\tau$ meeting all lines of $\mathcal{L}$ and containing $M$ lies on at most $(q+1)(q^{2}+q+1)$ $(2n-3)$-spaces meeting all lines of $\mathcal{L}\cup\{M_{0},M_{1}\}$ since each of the $q+1$ planes through $M$ in $\mu$ gives rise to a $(2n-4)$-space meeting all lines of $\mathcal{L}\cup\{M_{0},M_{1}\}$, and each of those $(2n-4)$-spaces lies on $q^{2}+q+1$ $(2n-3)$-spaces. There are at most $2(n-2)(q+1)^{2n-6}$ such $(2n-5)$-spaces $\tau$.

Each of the $(2n-5)$-spaces $\tau$ meeting all lines of $\mathcal{L}$ and meeting $M$ in a point lies on at most $(q^{2}+q+1)$ $(2n-3)$-spaces meeting all lines of $\mathcal{L}\cup\{M_{0},M_{1}\}$ namely one for each of the $(q^{2}+q+1)$ $(2n-3)$-spaces through the unique $(2n-4)$-space spanned by the $(2n-5)$-space $\tau$ and the unique transversal line through the point $\tau\cap\mu$ to the regulus defined by $M_{0},M_{1},M$. There are at most $2(n-2)(q+1)^{2n-5}$ such $(2n-5)$-spaces $\tau$.

Finally, each of the $(2n-5)$-spaces $\tau$ meeting all lines of $\mathcal{L}$ but not intersecting $M$ lie on $(q+1)^{2}-(q+1)$ $(2n-3)$-spaces intersecting all lines of $\mathcal{L}\cup\{M_{0},M_{1}\}$, namely those spanned by $\tau$ and a line meeting both $M_{0}$ and $M_{1}$, but not $M$ (if the subspace $S$ would meet $M$, then $S$ would meet $\Sigma$ in $(2n-4)$-space). There are at most $(n-2)(q+1)^{2n-4}$ such $(2n-5)$-spaces $\tau$.

We find that the total number of $(2n-3)$-spaces $S$ meeting all lines of $\mathcal{L}\cup\{M_{0},M_{1}\}$ is at most

since $q\geq 4n$.

We will now show that there is a line $N$ such that the number of $(2n-3)$-spaces meeting $N$ in a point and meeting the lines $\mathcal{L}\cup\{M_{0},M_{1}\}$ is at most $2(n-1)(q+1)^{2n-3}$, while the number of $(2n-3)$-spaces containing $N$ and meeting the lines $\mathcal{L}\cup\{M_{0},M_{1}\}$ is at most $2(n-1)(q+1)^{2n-4}$.

We already know that there is a set $\mathcal{S}$ of at most $(n-1)(q+1)^{2n-2}$ $(2n-3)$-spaces $S$ meeting the lines $\mathcal{L}\cup\{M_{0},M_{1}\}$. Each of those spaces covers $\frac{q^{2n-2}-1}{q-1}$ points. Since there are $\frac{q^{2n}-1}{q-1}-(2n-2)(q+1)$ points not on the lines $\mathcal{L}$, it follows that there exists a point $P$, not on a line of $\mathcal{L}\cup\{M_{0},M_{1}\}$ such that the number of spaces of $\mathcal{S}$ through it is at most

Let $N$ be the spread element of $\mathcal{D}$ through $P$ and let $\mathcal{S}_{P}$ be the subset of spaces of $\mathcal{S}$ containing $P$. Let $x$ be the number of spaces of $\mathcal{S}_{P}$ containing $N$, and $y$ is the number of spaces of $\mathcal{S}_{P}$ not containing $N$; we have just seen that $x+y\leq 2(n-1)(q+1)^{2n-4}$ Since the lines of $\mathcal{L}\cup\{M_{0},M_{1}\}$, as well as $N$ are elements of $\mathcal{D}$, we find that the the number of $(2n-3)$-spaces meeting all lines of $\mathcal{S}$ is given by $x+(q+1)y.$ The number of $(2n-3)$-spaces meeting all lines of $\mathcal{L}\cup\{M_{0},M_{1}\}$ and meeting $N$ exactly in a point is $(q+1)y\leq(q+1)(x+y)\leq 2(n-1)(q+1)^{2n-4}(q+1)=2(n-1)(q+1)^{2n-3}.$

Finally, observe that, since $q>\frac{4n}{ln(2)}$, $2n(q+1)^{2n-3}>2(n-1)(q+1)^{2n-3}+2(n-1)(q+1)^{2n-4}$.

The statement now follows by induction. ∎

Let $\mathcal{D}$ denote the Desarguesian line spread of $\Sigma$, and let $\mathcal{L}=\{\ell_{1},\dots,\ell_{2n-1}\}$ be a set of lines as obtained in Lemma 3.1. We will show that we can add lines $\ell_{2n},\ell_{2n+1},\ldots,\ell_{4n-4}$ to $\mathcal{L}$ such that there is no subspace of dimension $(2n-3)$ meeting all lines $\ell_{1},\ldots,\ell_{4n-4}$. Since it is possible that $\ell_{i}=\ell_{j}$ for some $i,j$, we will eventually obtain a set of at most $4n-4$ lines which is a higgledy-piggledy set $\mathcal{S}$: if $K$ is a hyperplane, it meets $\mathcal{S}$ in at least $2n-1$ points arising from the intersections of $K$ with the lines of $\mathcal{L}$, and by construction it is impossible that all points of $\mathcal{S}\cap K$ are contained in a hyperplane of $K$.

We double count the number of pairs

such that $Q\in H$ and $H$ is an $(2n-3)$-space with $H\cap\ell\neq\emptyset$ for any $\ell\in\mathcal{L}$.

By Lemma 3.1, this number is at most

where $\theta_{2n-3}=(q^{2n-2}-1)/(q-1)$ is the number of points of $H$.
On the other hand, the number of such pairs is also

where $M_{1}$ is the average number of $(2n-3)$-dimensional spaces $H$ meeting each of the lines $\ell_{i}$, $i=1,\dots,2n-1$ in at least a point and through a point $Q\in\Sigma$.

As a consequence

In particular, there exists a point $Q$ with the property that the number of $(2n-3)$-spaces through $Q$ meeting each of the lines $\ell_{i}$, $i=1,\dots,2n-1$ in at least a point is at most $M_{1}$. By Result 1.2, the unique line $\ell_{2n}$ of the spread $\mathcal{D}$ passing through $Q$ has the property that there are at most $(q+1)M_{1}$ $(2n-3)$-spaces meeting $\ell_{2n}$ and each of the lines $\ell_{i}$, $i=1,\dots,2n-1$ in at least a point. Note that $Q$ might be a point of one of the lines $\ell_{i}$, namely in such a case $\ell_{2n}\in\mathcal{L}$. The number of pairs

such that $H\cap\ell_{i}\neq\emptyset$ for any $i=1,\dots,2n$ and $Q$ belongs to $H$, is now at most

On the other hand, this number is also

where $M_{2}$ is the average number of $(2n-3)$-dimensional spaces $H$ through a fixed point $Q$ and meeting each of the lines $\ell_{i}$, $i=1,\dots,2n$ in at least a point. As a consequence

We can iterate the reasoning, taking each time a new line $\ell_{2n-1+i}$ contained in at most $M_{i}$ $(n-2)$-dimensional spaces $H$ meeting each of the previous lines, and double counting the pairs $(H,Q)$ such that $H\cap S_{i}\neq\emptyset$ and $Q$ belongs to $H$. We obtain

Combining with $M_{1}<\frac{2n(q+1)^{2n-3}\theta_{2n-3}}{\theta_{2n-1}}$, we obtain

In particular,

Observe that

holds for any $q$ such that

Indeed $(q+1)^{4n-7}<q^{4n-7}+(4n-6)q^{4n-8}$ is equivalent to $(1+1/q)^{4n-7}<1+\frac{4n-6}{q}$. Assuming $q>4n-6$, the right hand side of this inequality is at most $2$. Isolating $q$, we obtain the claim. Therefore,

Finally, $2n\left(\frac{1}{q}+\frac{4n-6}{q^{2}}\right)<1$ is equivalent to $q^{2}-2nq-4n+6>0$, which is always true under our hypothesis. It can be checked that if $q>4n/ln(2)$ and $n\geq 2$, then $q>\frac{1}{e^{ln(2)/(4n-7)}-1}$ as well. Since $M_{2n-3}<1$, there must be a point not contained in any $(2n-3)$-subspace meeting all the lines $\ell_{i}$, $i=1,\dots,4n-5$. Again using Result 1.2, we see that the unique line $\ell_{4n-4}$ through that point is such that there is no $(2n-3)$-subspace meeting all the lines $\ell_{i}$, $i=1,\dots,4n-4$. ∎