# criterion for a near-linear space being a linear space

Suppose $\mathcal{S}$ is near-linear space with $v$ points and $b$ lines, and ${s}_{i}$ is the number of points in the $i$th line, for $i=1,\mathrm{\dots},b$. Then

$$\sum _{i=1}^{b}{s}_{i}({s}_{i}-1)\le v(v-1),$$ |

and equality holds if and only if $\mathcal{S}$ is a linear space.

## Proof

Let $N$ be the number of ordered pairs of points that are joined by a line. Clearly $N$ can be no more than $v(v-1)$, and $N=v(v-1)$ if and only if every pair of points are joined by a line. Since two points in a near-linear space are on at most one line, we can label each pair by the line to which the two points belong to. We thus have a partition^{} of the $N$ pairs into $b$ groups, and each group is associated with a distinct line. The group corresponding to the line consisting of ${s}_{i}$ points contributes ${s}_{i}({s}_{i}-1)$ to the total sum. Therefore

$$\sum _{i=1}^{b}{s}_{i}({s}_{i}-1)=N\le v(v-1).$$ |

Title | criterion for a near-linear space being a linear space |
---|---|

Canonical name | CriterionForANearlinearSpaceBeingALinearSpace |

Date of creation | 2013-03-22 14:32:47 |

Last modified on | 2013-03-22 14:32:47 |

Owner | kshum (5987) |

Last modified by | kshum (5987) |

Numerical id | 9 |

Author | kshum (5987) |

Entry type | Theorem |

Classification | msc 05B25 |