Linear Algebra/Topic: Projective Geometry/Solutions

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Problem 1

What is the equation of this point?

\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}

From the dot product

0=\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\cdot\begin{pmatrix} L_1 &L_2 &L_3 \end{pmatrix}

we get that the equation is L_1=0.

Problem 2
  1. Find the line incident on these points in the projective plane.
\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix},\,\begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix}
  2. Find the point incident on both of these projective lines.
\begin{pmatrix} 1 &2 &3 \end{pmatrix},\,\begin{pmatrix} 4 &5 &6 \end{pmatrix}
  1. This determinant
1  &4  &x \\
2  &5  &y \\
3  &6  &z
    shows that the line is L=\begin{pmatrix} -3 &6 &-3 \end{pmatrix}.
  2. \begin{pmatrix} -3 \\ 6 \\ -3 \end{pmatrix}
Problem 3

Find the formula for the line incident on two projective points. Find the formula for the point incident on two projective lines.


The line incident on

u=\begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix}
v=\begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}

can be found from this determinant equation.

u_1  &v_1  &x  \\
u_2  &v_2  &y  \\
u_3  &v_3  &z
=(u_2v_3-u_3v_2)\cdot x
+ (u_3v_1-u_1v_3)\cdot y
+ (u_1v_2-u_2v_1)\cdot z

The equation for the point incident on two lines is the same.

Problem 4

Prove that the definition of incidence is independent of the choice of the representatives of p and L. That is, if p_1, p_2, p_3, and q_1, q_2, q_3 are two triples of homogeneous coordinates for p, and L_1, L_2, L_3, and M_1, M_2, M_3 are two triples of homogeneous coordinates for L, prove that p_1L_1+p_2L_2+p_3L_3=0 if and only if q_1M_1+q_2M_2+q_3M_3=0.


If p_1, p_2, p_3, and q_1, q_2, q_3 are two triples of homogeneous coordinates for p then the two column vectors are in proportion, that is, lie on the same line through the origin. Similarly, the two row vectors are in proportion.

k\cdot\begin{pmatrix} p_1 \\ p_2 \\ p_3 \end{pmatrix}
=\begin{pmatrix} q_1 \\ q_2 \\ q_3 \end{pmatrix}
m\cdot\begin{pmatrix} L_1 &L_2 &L_3 \end{pmatrix}
=\begin{pmatrix} M_1 &M_2 &M_3 \end{pmatrix}

Then multiplying gives the answer (km)\cdot (p_1L_1+p_2L_2+p_3L_3)=q_1M_1+q_2M_2+q_3M_3=0.

Problem 5

Give a drawing to show that central projection does not preserve circles, that a circle may project to an ellipse. Can a (non-circular) ellipse project to a circle?


The picture of the solar eclipse — unless the image plane is exactly perpendicular to the line from the sun through the pinhole — shows the circle of the sun projecting to an image that is an ellipse. (Another example is that in many pictures in this Topic, the circle that is the sphere's equator is drawn as an ellipse, that is, is seen by a viewer of the drawing as an ellipse.)

The solar eclipse picture also shows the converse. If we picture the projection as going from left to right through the pinhole then the ellipse I projects through P to a circle S.

Problem 6

Give the formula for the correspondence between the non-equatorial part of the antipodal modal of the projective plane, and the plane z=1.


A spot on the unit sphere

\begin{pmatrix} p_1 \\ p_2 \\ p_3 \end{pmatrix}

is non-equatorial if and only if p_3\neq 0. In that case it corresponds to this point on the z=1 plane

\begin{pmatrix} p_1/p_3 \\ p_2/p_3  \\ 1 \end{pmatrix}

since that is intersection of the line containing the vector and the plane.

Problem 7

(Pappus's Theorem) Assume that T_0, U_0, and V_0 are collinear and that T_1, U_1, and V_1 are collinear. Consider these three points: (i) the intersection V_2 of the lines T_0U_1 and T_1U_0, (ii) the intersection U_2 of the lines T_0V_1 and T_1V_0, and (iii) the intersection T_2 of U_0V_1 and U_1V_0.

  1. Draw a (Euclidean) picture.
  2. Apply the lemma used in Desargue's Theorem to get simple homogeneous coordinate vectors for the T's and V_0.
  3. Find the resulting homogeneous coordinate vectors for U's (these must each involve a parameter as, e.g., U_0 could be anywhere on the T_0V_0 line).
  4. Find the resulting homogeneous coordinate vectors for V_1. (Hint: it involves two parameters.)
  5. Find the resulting homogeneous coordinate vectors for V_2. (It also involves two parameters.)
  6. Show that the product of the three parameters is 1.
  7. Verify that V_2 is on the T_2U_2 line..
  1. Other pictures are possible, but this is one.

    Linalg pappus.png

    The intersections 
T_0U_1\,\cap T_1U_0=V_2
T_0V_1\,\cap T_1V_0=U_2
, and 
U_0V_1\,\cap U_1V_0=T_2
are labeled so that on each line is a T, a U, and a V.

  2. The lemma used in Desargue's Theorem gives a basis B with respect to which the points have these homogeneous coordinate vectors.
{\rm Rep}_{B}(\vec{t}_0)=\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}
{\rm Rep}_{B}(\vec{t}_1)=\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}
{\rm Rep}_{B}(\vec{t}_2)=\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}
{\rm Rep}_{B}(\vec{v}_0)=\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}
  3. First, any U_0 on T_0V_0
{\rm Rep}_{B}(\vec{u}_0)=a\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}
+b\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}
=\begin{pmatrix} a+b \\ b \\ b \end{pmatrix}
    has homogeneous coordinate vectors of this form
\begin{pmatrix} u_0 \\ 1 \\ 1 \end{pmatrix}
    (u_0 is a parameter; it depends on where on the T_0V_0 line the point U_0 is, but any point on that line has a homogeneous coordinate vector of this form for some u_0\in\mathbb{R}). Similarly, U_2 is on T_1V_0
{\rm Rep}_{B}(\vec{u}_2)=c\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}
+d\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}
=\begin{pmatrix} d \\ c+d \\ d \end{pmatrix}
    and so has this homogeneous coordinate vector.
\begin{pmatrix} 1 \\ u_2 \\ 1 \end{pmatrix}
    Also similarly, U_1 is incident on T_2V_0
{\rm Rep}_{B}(\vec{u}_1)=e\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}
+f\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}
=\begin{pmatrix} f \\ f \\ e+f \end{pmatrix}
    and has this homogeneous coordinate vector.
\begin{pmatrix} 1 \\ 1 \\ u_1 \end{pmatrix}
  4. Because V_1 is T_0U_2\,\cap\,U_0T_2 we have this.
g\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}+h\begin{pmatrix} 1 \\ u_2 \\ 1 \end{pmatrix}
=i\begin{pmatrix} u_0 \\ 1 \\ 1 \end{pmatrix}+j\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}
g+h  &= iu_0 \\
hu_2 &= i    \\
h    &= i+j
    Substituting hu_2 for i in the first equation
\begin{pmatrix} hu_0u_2 \\ hu_2 \\ h \end{pmatrix}
    shows that V_1 has this two-parameter homogeneous coordinate vector.
\begin{pmatrix} u_0u_2 \\ u_2 \\ 1 \end{pmatrix}
  5. Since V_2 is the intersection T_0U_1\,\cap\,T_1U_0
k\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}+l\begin{pmatrix} 1 \\ 1 \\ u_1 \end{pmatrix}
=m\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}+n\begin{pmatrix} u_0 \\ 1 \\ 1 \end{pmatrix}
k+l  &= nu_0 \\
l    &= m+n    \\
lu_1 &= n
    and substituting lu_1 for n in the first equation
\begin{pmatrix} lu_0u_1 \\ l \\ lu_1 \end{pmatrix}
    gives that V_2 has this two-parameter homogeneous coordinate vector.
\begin{pmatrix} u_0u_1 \\ 1 \\ u_1 \end{pmatrix}
  6. Because V_1 is on the T_1U_1 line its homogeneous coordinate vector has the form
p\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}+q\begin{pmatrix} 1 \\ 1 \\ u_1 \end{pmatrix}
=\begin{pmatrix} q \\ p+q \\ qu_1 \end{pmatrix}
    but a previous part of this question established that V_1's homogeneous coordinate vectors have the form
\begin{pmatrix} u_0u_2 \\ u_2 \\ 1 \end{pmatrix}
    and so this a homogeneous coordinate vector for V_1.
\begin{pmatrix} u_0u_1u_2 \\ u_1u_2 \\ u_1 \end{pmatrix}
    By (*) and (**), there is a relationship among the three parameters: u_0u_1u_2=1.
  7. The homogeneous coordinate vector of V_2 can be written in this way.
\begin{pmatrix} u_0u_1u_2 \\ u_2 \\ u_1u_2 \end{pmatrix}
=\begin{pmatrix} 1 \\ u_2 \\ u_1u_2 \end{pmatrix}
    Now, the T_2U_2 line consists of the points whose homogeneous coordinates have this form.
r\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}+s\begin{pmatrix} 1 \\ u_2 \\ 1 \end{pmatrix}
=\begin{pmatrix} s \\ su_2 \\ r+s \end{pmatrix}
    Taking s=1 and r=u_1u_2-1 shows that the homogeneous coordinate vectors of V_2 have this form.