Collineations in Perspective
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Collineations in Perspective Now that we have a decent grasp of one-dimensional projectivities, we move on to their two di- mensional analogs. Although they are more complicated, in a sense, they may be easier to grasp because of the many applications to perspective drawing. Speaking of, let's return to the triangle on the window and its shadow in its full form instead of only looking at one line. Perspective Collineation In one dimension, a perspectivity is a bijective mapping from a line to a line through a point. In two dimensions, a perspective collineation is a bijective mapping from a plane to a plane through a point. To illustrate, consider the triangle on the window plane and its shadow on the ground plane as in Figure 1. We can see that every point on the triangle on the window maps to exactly one point on the shadow, but the collineation is from the entire window plane to the entire ground plane. We understand the window plane to extend inﬁnitely in all directions (even going through the ground), the ground also extends inﬁnitely in all directions (we will assume that the earth is ﬂat here), and we map every point on the window to a point on the ground. Looking at Figure 2, we see that the lamp analogy breaks down when we consider all lines through O. Although it makes sense for the base of the triangle on the window mapped to its shadow on 1 the ground (A to A0 and B to B0), what do we make of the mapping C to C0, or D to D0? C is on the window plane, underground, while C0 is on the ground. In this case, it appears that the lamp is acting like an eye, seeing the projection of the point on the window on the ground canvas. D is above the lamp, it can't cast a shadow on the ground... but if the lamp is acting like a camera obscura, then D projects through the aperture onto the ground at D0. 1. Is this a bijective mapping? In other words, can two points on the window get mapped to the same point on the ground (is it one-to-one)? Also, for every point on the ground, can we draw a point on the window that will cast a shadow onto that point (is it onto)? 2. If we know that there is a light source that is casting a shadow (that there is a perspective collineation), what is the minimum number of pairs of corresponding window and ground points will we need to ﬁgure out the location of the light source? 3. In the window-to-ground perspective collineation above, are there any ﬁxed points? Explain. Will perspective collineations mapping planes to diﬀerent planes always have ﬁxed points? Explain. Recall that given the deﬁnition of a one-dimensional perspectivity, the only perspectivity that maps a line to itself is the identity. This is because of the deﬁnition uses elementary correspondences. In two dimensions, if we consider not the real situation of the triangle and its shadow but instead, consider the image of the situation of the triangle and its shadow, it appears that it would be possible to deﬁne a perspective collineation such that the center of perspective is allowed to be in the plane (remember, with perspectivities, the center of perspective had to not be on the line). We notice a few properties: (1) A perspective collineation sends collinear points to collinear points. (2) Dually, a perspective collineation sends concurrent lines to concurrent lines. 2 (3) The center is linewise ﬁxed, i.e., every line through the center is ﬁxed (though the points along the line may be rearranged). (4) Dually, the axis is pointwise ﬁxed, i.e., every point along this line is ﬁxed (though the lines through these points may not be ﬁxed). So we will deﬁne a perspective collineation as follows. Deﬁnition: A perspective collineation is a bijective mapping from a projective plane to a projective plane (possibly the same plane) such that collinear points are sent to collinear points through a center. A perspective collineation that maps a projective plane to itself is called a perspective automorphism. 4. Use the deﬁnition to prove that a perspective collineation must preserve incidence. 5. Use the deﬁnition to prove that properties (3) and (4) are satisﬁed, i.e., there must always be a linewise ﬁxed point (a center) and a pointwise ﬁxed line (an axis). By exercise 5 above, we can deﬁne two types of collineations, Elation: center is on the axis. Homology: center is not on the axis. We will say that an ordered pair of distinct points is homologous if a collineation maps the ﬁrst to the second point. 6. Draw the center, axis and one homologous pair of points of an elation. Is this enough to determine the rest of the elation? Choose a random point, and ﬁgure out how to ﬁnd the point that it is mapped to by the collineation. Then generalize this and express the construction in words. 7. Draw the center, axis and one homologous pair of points of a homology. Is this enough to determine the rest of the homology? Choose a random point, and ﬁgure out how to ﬁnd the point that it is mapped to by the collineation. Then generalize this and express the construction in words. 3.