Collineation

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Collineation

Collineation is a transformation in projective geometry, which is a point-to-point and line-to-line mapping, and preservs incidence.
Let \( \pi \) be a projective plane, then a collineation on \( \pi \) is an isomorphism from \( \pi \) to itself.
A collineation from \( \pi \) to itself is denoted by \( f: \pi \to \pi \) where f is one-one-onto mapping and preserves collinearity.

In a collineation \( f: \pi \to \pi \)

  1. \( f:\pi \to \pi \) is one one
  2. \( f:\pi \to \pi \) is onto
  3. \( f:\pi \to \pi \) preserves collinearity
We say
  • \(f: \pi \to \pi \) fixes the point x if \(f(x)=x\)
  • \(f: \pi \to \pi \) fixes line L if \(f(L)=L\) (i.e., \( f(x) \in L \forall x \in L \) )
  • \(f : \pi \to \pi \) fixes line L pointwise if \( f(x)=x,\forall x \in L\)
  • \(f : \pi \to \pi \) fixes point x linewise if \( f(L)=L \forall L \in x \)

Let \( f:\pi \to \pi \) be a collineation, then following result holds.

  1. If f fixes two points p and q then f fixes the line pq
  2. If f fixes two lines L and M then f fixes the point \(L \cap M \)
  3. If f fixes two lines L and M point-wise, then f is identity
  4. If f fixes two points p and q line-wise, then f is identity.
Proof
  1. If \(f: \pi \to \pi \) fixes two points p and q, then f fixes line \(pq\).
    Given that, \(f: \pi \to \pi \) fixes two points \(p\) and \(q\)
    or \(p=p', q=q'\)
    By-P1: there exist a line \(pq\).
    Let, x is arbitrary point on \(pq\), then
    \(x,p,q\) are collinear.
    or \(x',p',q'\) are collinear
    or \(x’,p,q\) are collinear
    Hence,
    \(x'\) lies on \(pq\)
    Therefore
    \(f: \pi \to \pi \) fixes line \(pq\)
  2. The collineation \(f: \pi \to \pi \) fixes two lines Land M.
    i.e., f(L)=L, f(M)=M
    Let, \( p=L\cap M\), then
    \(p\in L\) and \(p\in M\)
    or \( p'\in L\) and \(p'\in M\)
    or \( p'\in (L\cap M) \)
    or p=p'
    Hence, f fixes \(L\cap M\)
  3. If f fixes two lines L and M point-wise, then f is identity
    Given, L and M are point-wise fixed by f.
    Let, x is arbitrary point in \( \pi \).
    Case 1: If \( x\in L\), then x is fixed
    Case 2: If \(x\in M\), then x is fixed
    Case 3: If x is neither on L nor on M, then
    There exist, \(L_1\) and \(L_2\) on x such that
    x,p,q are collinear.
    or \(L_1\cap L=p, L_1\cap M=q\)
    Then, \(p\in L\), thus p is fixed(i)
    Also, \( q\in M\), thus q is fixed(ii)
    Using (i) and (ii), line \(pq (L_1)\) is fixed (iii)
    Similarly, \(L_2\) is fixed.(iv)
    Using (iii) and (iv), point \(x (=L_1 \cap L_2 )\) is fixed
    From case 1, 2 and 3, f fixes every point in \( \pi \).
    Hence, f is an identity.
  4. If f fixes two points p and q line-wise, then f is identity.
    Given, two points p and q are line-wise fixed by f.
    Let, x is arbitrary point on \( \pi \)
    1. Case 1
      If x is not in pq, then lines px, qx are fixed
      Thus, \( px\cap qx=x\) is fixed
    2. Case 2
      If x is in pq, then consider L as arbitrary line on x other than pq.
      Since, every point which is not on pq is fixed, L is fixed.
      Here, pq and L is fixed, thus x is fixed
    From case 1 and 2, f fixes every point on \( \pi \)
Thus, f is an identity.

Let \( f:\pi \to \pi \) be a collineation, then following result holds.

  1. If f fixes two points p and q then f fixes the line pq
  2. If f fixes two lines L and M then f fixes the point \(L \cap M \)
  3. If f fixes two lines L and M point-wise, then f is identity
  4. If f fixes two points p and q line-wise, then f is identity.

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