We recall the axioms of incidence geometry:
(I1) Given \(2\) distinct points, there is a unique line incident with both of
them.
(I2) Given a line, there exist two distinct points incident with it.
(I3) There exists \(3\) non-collinear points.
(i)
Can you find two non-isomorphic models with only \(2\) points satisfying (I1)
and (I2) ?
(ii)
Assume that \((\mathcal {P},\mathcal {L},\mathcal {R})\) is an incidence geometry with \(3\) points only.
Show it is isomorphic to the model described in Example 4.
We say that there is a unique incidence geometry with three points up to
isomorphisms .
I1 and I2 Let
\[ \mathcal {P} = \{1,2\}, \qquad \mathcal {L} = \{\{1,2\}\}. \]
\[ P\,I\,\ell \iff P \in \ell \quad (\text {standard incidence}). \]
I1 and I2 are clearly satisfied.
Unique (up to isomorphism)?
Assume \((\mathcal {P}',\mathcal {L}',\mathcal {R}')\) is a model for axioms I1 and I2 and \(\mathcal {P}'\) has only two elements.
Call \(x\) and \(y\) the two elements of \(\mathcal {P}'\) .
By I1 , there exists \(\ell \in \mathcal {L}'\) such that
\[ x\,I\,\ell \quad \text {and} \quad y\,I\,\ell . \]
Are there any other lines?
Assume \(\ell ' \in \mathcal {L}'\) is a line.
By I2 , it is incident with at least \(2\) points.
Since we only have \(2\) points, it is incident with \(x\) and \(y\) .
But \(\ell \) is the only line incident with \(x\) and \(y\) .
Therefore \(\ell ' = \ell \) .
Conclusion
\[ \mathcal {P}' = \{x,y\}, \quad \mathcal {L}' = \{\ell \}, \quad \mathcal {R}' = \{(x,\ell ),(y,\ell )\}. \]
This model is isomorphic to \((\mathcal {P},\mathcal {L},\mathcal {R})\) .
Indeed define
\[ f(1)=x,\quad f(2)=y,\qquad g(\{1,2\})=\ell . \]
Name our three points \(x,y,z\) .
We need to determine \(\mathcal {L}\) and \(\mathcal {R}\) , i.e. the lines and incidences.
By I1 , there exists a unique line
\[ \ell _{xy} \text { incident with } x \text { and } y, \]
\[ \ell _{yz} \text { incident with } y \text { and } z, \]
\[ \ell _{xz} \text { incident with } x \text { and } z. \]
If two of these three lines were equal, then \(x,y,z\) are collinear, which contradicts
I3 .
For the same reason, none of these lines is incident to \(3\) points.
Therefore \(\ell _{xy},\ell _{yz},\ell _{xz}\) are distinct and only incident to \(2\) points each.
Are these all the lines?
Assume \(\ell \in \mathcal {L}\) .
Then \(\ell \) is incident with \(2\) points at least by I2 and cannot be incident with
\(3\) .
So \(\ell \in \{\ell _{xy},\ell _{yz},\ell _{xz}\}\) .
Conclusion
\[ \mathcal {L} = \{\ell _{xy},\ell _{yz},\ell _{xz}\}. \]
Now defining
\[ f(x)=A,\quad f(y)=B,\quad f(z)=C, \]
\[ g(\ell _{xy})=\{A,B\},\quad g(\ell _{yz})=\{B,C\},\quad g(\ell _{xz})=\{A,C\}, \]
gives an isomorphism. (Note that there are others!)
An affine plane is an incidence geometry
\((\mathcal {P},\mathcal {L},\mathcal {R})\) satisfying the Euclidean parallel property.
Example questions
(a)
Any affine planes with only \(3\) points?
(b)
An affine plane with \(4\) points? Example:
\(4\) points
\(6\) lines
each line incident to only \(2\) points
Definition
Let \(\mathbb {F}=\mathbb {R}\) or \(\mathbb {Q}\) (or \(\mathbb {C}\) , or any field).
Points are elements of \(\mathbb {F}^2\) , i.e. ordered pairs \((x,y)\) where \(x,y\in \mathbb {F}\) .
Lines are subsets of \(\mathbb {F}^2\) of the form
\[ \ell =\{(x,y)\in \mathbb {F}^2 : ax+by+c=0\} \]
for some \((a,b,c)\in \mathbb {F}^3\) with \((a,b)\neq (0,0)\) .
\[ P\,I\,\ell \iff P\in \ell . \]
\((\mathbb {F}^2,\mathcal {L},\mathcal {R})\) is an affine plane.
(I1) Assume \((x_1,y_1)\neq (x_2,y_2)\) .
Finding a line
\[ \ell =\{ax+by+c=0\} \]
such that
\(P_1,P_2\in \ell \) amounts to solving:
\[ \begin{cases} ax_1+by_1+c=0 \\ ax_2+by_2+c=0 \end{cases} \]
This system can be rewritten in matrix form as
\[ \begin{pmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. \]
The matrix
\[ M= \begin{pmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end{pmatrix} \]
has rank
\(2\) since
\((x_1,y_1)\neq (x_2,y_2)\) .
Therefore the homogeneous system
\[ M \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \]
has a nontrivial solution
\((a,b,c)\neq (0,0,0)\) .
We claim that necessarily \((a,b)\neq (0,0)\) . Indeed, if \(a=b=0\) , then the equations reduce to
\[ c=0, \]
which
contradicts
\((a,b,c)\neq (0,0,0)\) .
Thus we obtain a line
\[ \ell =\{ax+by+c=0\} \]
passing through
\(P_1\) and
\(P_2\) .
Moreover, since \(\mathrm {rank}(M)=2\) , the space of solutions is \(1\) -dimensional.
Thus any two solutions \((a,b,c)\) and \((a',b',c')\neq (0,0,0)\) are proportional:
\[ (a',b',c')=\lambda (a,b,c) \]
for some
\(\lambda \neq 0\) .
But proportional coefficients define the same line, so all solutions correspond to the
same line.
Therefore there exists a unique line incident with \(P_1\) and \(P_2\) .
(I2) Next HW or section.
(I3) Homework 4.
Euclidean parallel property
Let
\[ \ell =\{ax+by+c=0\}, \quad P=(x_0,y_0)\notin \ell . \]
Consider another line
\[ m=\{dx+ey+f=0\}. \]
We analyze the system:
\[ \begin{cases} ax+by+c=0 \\ dx+ey+f=0 \end{cases} \]
Number of solutions:
\(0\) solutions \(\Rightarrow \ell \parallel m\)
\(1\) solution \(\Rightarrow \ell \not \parallel m\)
more than \(1\) solution \(\Rightarrow \ell = m\)
Compute the determinant:
\[ \Delta = \begin{vmatrix} a & b \\ d & e \end{vmatrix} = ae - bd. \]
If \(\Delta =0\) , then \((a,b)\) and \((d,e)\) are proportional.
Thus
\[ m=\{ax+by+d=0\} \]
for some
\(d\) .
Hence lines parallel to \(\ell \) take the form:
\[ \{ax+by+d=0\}, \quad d\neq c, \]
and one can show with similar arguments that
such a line is indeed parallel to
\(\ell \) .
Now we want a line \(m\parallel \ell \) passing through \(P\) .
Solve:
\[ ax_0+by_0+d=0. \]
Thus
\[ d=-ax_0-by_0. \]
Since \(P\notin \ell \) , we have \(d\neq c\) .
Thus \(m\parallel \ell \) and it is unique.
For every \(n\geq 1\) and prime \(p\) , there exists a field \(\mathbb {F}_{p^n}\) .
Thus there exists an affine plane with exactly
\[ (p^n)^2 = p^{2n} \]
elements.
MATH 128A