I start with definitions.

Definition 1. A linear space is a pair $(X,\mathcal L)$ consisting of a set $X$ and a family $\mathcal L$ of subsets of $X$ satisfying three axioms:

(L1) for any distinct points $x,y\in X$ there exists a unique line $L\in\mathcal L$ containing $x$ and $y$;

(L2) every set $L\in\mathcal L$ contains at least three points;

(L3) $X\notin\mathcal L$.

Elements of the family $\mathcal L$ are called lines.

For any distinct points $x,y\in X$ of a linear space $(X,\mathcal L)$, the unique line $L\in\mathcal L$ containing the points $x,y$ will be denoted by $\overline{xy}$.

Definition 2. A linear space $(X,\mathcal L)$ is called

$\ \bullet$ uniform  if all lines in $X$ have the same cardinality;

$\ \bullet$ hyperbolic  if for every points $\ x,o,y\in X$ and $p\in \overline{xy}\setminus(\overline{ox}\cup\overline{oy})\,\ $ the set $\ \{u\in\overline{oy}\,:\ \overline{up}\ \cap\overline{ox}\ = \varnothing\}\ $ contains more than one point.

Question. Is every uniform hyperbolic linear space infinite?

Added in Edit, after answers of @ihromant: It turns out that typical examples of finite uniform hyperbolic linear spaces (including those two in the second answer of @ihromant) are provided by classical unitals in finite projective planes of square order $q^2$ for some $q$ which is a power of a prime number. By Lemma 7.42 in the book "Unitals in projective planes", every classical unital $U$ in a projective plane $PG(2,q^2)$ contains no Pasch configurations, which means that for every points $o,x,y\in U$, $p\in\overline{oy}\setminus(\overline{ox}\cup\overline{oy})$, and $u\in\overline{oy}\setminus\{o,y\}$ the lines $\overline{up}$ and $\overline{ox}$ are disjoint. So $U$ is hyperbolic in a very strong sense. It is a longstanding conjecture that the absence of Pasch configurations characterizes classical unitals, see page 161 of the mentioned book "Unitals in projective planes". This conjecture is confirmed by Wilbrink under some additional conditions on the geometry of a unital, see Theorem 7.43, 7.44 in the book. I recall that a linear space $(X,\mathcal L)$ is a unital if $|X|=n^3+1$ for some number $n$ such that all lines $L\in\mathcal L$ have cardinality $n+1$. A classical unital in a projective plane $PG(2,q^2)$ is the subset defined by the equation $X^{q+1}+Y^{q+1}+Z^{q+1}=0$ in the homogeneous coordinates. Classical unitals have many exciting geometric properties, discussed in the book. The geometry of the classical unital in the projective plane $PG(2,9)$ is visualized in this MO-post.


5 Answers 5


The answer is "No".There exist a finite plane with required property.

First of all, let's rephrase your question. Your uniform linear space is a synonym to a BIBD (balanced incomplete block design) with λ = 1. There exists an example of a BIBD(217, 7, 1): v = 217, k = 7 so-called "Difference family" where blocks (in your case lines) are obtained by cyclic shifts by addition of some base sets.

So, there are 6 base sets for your family (please note that the last one produces only 31 lines):

  • [0,1,37,67,88,92,149]
  • [0,15,18,65,78,121,137]
  • [0,8,53,79,85,102,107]
  • [0,11,86,100,120,144,190]
  • [0,29,64,165,198,205,207]
  • [0,31,62,93,124,155,186]

This plane consists of 217 points, 217 * 5 + 31 = 1116 lines, there are 36 lines that go through every point. Through every point that does not lie on a given line you have 29 "parallel" lines. And last: hyperbolic index (number of non-crossing lines from your definition) is 2..5. This properties are checked by Java code present here:

  • Plane model
  • Test that proves required properties. Checking that it's indeed a plane is successful, but long-running, so I commented it.

First answer was just existence of such space. After routine computer search in large collection of BIBD collected by Vedran Krcadinac, I have two much smaller, nicer and more uniform examples.

First is BIBD(28,4,1). Lines of it are printed below in 28-base numeric system.


lines are printed vertically, total number of lines is 63. Its hyperbolic index is 2..2 (constant 2).

Second is BIBD(65,5,1). Its hyperbolic index is 3..3 (constant 3). It's first in the list of designs Designs S(2,5,65). Common property of them is that both of them are unitals.


Other answer that satisfies your criteria. It's corrected $\operatorname{BIBD}(175, 7, 1)$ from Handbook of combinatorial designs VI.16.2 chapter. Corrected it is because this example is incorrect in the book.

Consider following difference family over $\mathbb Z_7\times\mathbb Z_5\times\mathbb Z_5$:

  • $\{\{0, 0, 0\}, \{1, 0, 0\}, \{2, 0, 0\}, \{3, 0, 0\}, \{4, 0, 0\}, \{5, 0, 0\}, \{6, 0, 0\}\}$,
  • $\{\{0, 0, 0\}, \{1, 1, 3\}, \{1, 4, 2\}, \{2, 2, 2\}, \{2, 3, 3\}, \{4, 2, 0\}, \{4, 3, 0\}\}$,
  • $\{\{0, 0, 0\}, \{1, 3, 4\}, \{1, 2, 1\}, \{2, 2, \boldsymbol{3}\}, \{2, 3, 2\}, \{4, 0, 2\}, \{4, 0, 3\}\}$,
  • $\{\{0, 0, 0\}, \{1, 1, 2\}, \{1, 4, 3\}, \{2, 1, \boldsymbol{1}\}, \{2, 4, 4\}, \{4, 0, 1\}, \{4, 0, 4\}\}$,
  • $\{\{0, 0, 0\}, \{1, 3, 1\}, \{1, 2, 4\}, \{2, 4, 1\}, \{2, 1, 4\}, \{4, 1, 0\}, \{4, 4, 0\}\}$

Misprints in the book are marked bold. First line gives $25$ lines, other give $175$. This difference family forms $\operatorname{BIBD}(175,7,1)$ that has hyperbolic index 2..5. Also it is first example that is not unital.


There is very simple example of hyperbolic $\textrm{BIBD}(91,7,1)$. It's following difference family:

  • $[0, 8, 29, 51, 54, 61, 63]$
  • $[0, 11, 16, 17, 31, 35, 58]$
  • $[0, 13, 26, 39, 52, 65, 78]$

over cyclic group of size $91$. First two blocks give $91$ line by cyclic shifts, second gives only $13$ lines ($195$ lines in total). It's hyperbolic and also has hyperbolic index 2..5.

For now it's very probable candidate for smallest hyperbolic plane that is not unital.


Obtained very interesting example (actually set of examples, but want to emphacise this one). It's classical (?) example of maximal $\{2^{m+n} - 2^n + 2^m;2^m\}$-arc in $PG(2,2^n)$ projective plane.

Definition. An $\{k;n\}$-arc in a projective plane $P$ of order $q$ is a nonempty set $K$ of $k$ points with the property that $n$ is the maximum number of points of $K$ which are collinear. An $\{k;n\}$-arc is maximal when it is not subset of any other $\{k';n\}$-arc.

Theorem (Denniston [1971]). Let $P=PG(2,2^n)$ be a Desarguesian projective plane. Then for any $m \le n$, $P$ contains a maximal $\{2^{m+n} - 2^n + 2^m;2^m\}$-arc.

Proof (surprisingly simple and contains constructive definition). Consider corresponding affine plane of the point set $\{(x, y):x, y \in \operatorname{GF}(2^n)\}$. Choose a nondegenerate quadratic form $\Phi$ over $\operatorname{GF}(2^n)$. Since $2^m$ divides $2^n$, there is subgroup $H$ of order $2^m$ of the additive group of $\operatorname{GF}(2^n)$. Then $K:=\{(x,y):\Phi(x,y)\in H\}$ is the desired maximal arc.

When we take $m = 4$ and $n = 5$. This gives us maximal $\{496;16\}$-arc or $\operatorname{BIBD}(496,16,1)$. This design has hyperbolic index $5..11$ from possible $0..14$ values. This is very surprising assuming that all other designs have bounds of their hyperbolic index either $0$ or $k - 2$.

If we take $m = 3$ and $n = 4$, we obtain maximal $\{120;8\}$-arc or $\operatorname{BIBD}(120,8,1)$. This design is also hyperbolic with index $2..6$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.