From Perspectivities to Perspective Collineations

Given distinct planes π and π′ in extended Euclidean space E3, and a point O on neither plane,the associated one-point projection with center O maps π bijectively onto π′. That’s a simplefact, but from the standpoint of real-world one-point projections such as perspective drawing orphotography, this map, called a perspectivity, is actually a bit strange. To see why, we first reviewthree types of practical, one-point projections in Figure 1.

Object

Center

Image

Center (sun)

Object

Image

Object Image Center Above left: Illustration of a camera obscura from

“Sketchbook on military art, including geometry,

fortifications, and pyrotechnics.” Seventeenth

century, origin unknown, possibly Italian.

Above: Hubble Telescope portrait of Jupiter and Io,

July 24, 1996. J. Spencer (Lowell Observatory) and

NASA/ESA.

Left: Illustration of the perspective drawing

process by Albrecht Dürer. Woodcut, 1525.

Figure 1: Three practical one-point projections.

In the upper left of Figure 1 is a diagram of a camera obscura, or pinhole camera. A light rayfrom an object (the spire of a domed building) passes through the center of projection (a hole ina wall of a darkened room) and meets the inverted image on the opposite wall of the room. The“photo itinerary” of the light ray is thus object-center-image. In the upper right of the figure,

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a light ray leaves the center of projection (the sun), passes by the object (Jupiter’s moon Io), andmeets its image (the shadow of Io on Jupiter). The “shadow itinerary” of the light ray is thuscenter-object-image. Finally, in the lower left of the figure, a light ray leaves an object (themodel’s hand), passes through the image (a perspective drawing of the hand), and meets the centerof projection (the artist’s eye). The “perspective drawing itinerary” is thus object-image-center.Figure 2 shows that not one, but all three types of practical one-point projections are involved ina plane-to-plane perspectivity.

OCEAN

PINHOLE PHOTOGRAPH

OF OCEAN

SHADOW

OF BEACH

BEACH

PRAIRIE

PERSPECTIVE DRAWING

OF PRAIRIEWest East

IMAGE OF THE FAR DISTANT,

EAST-SOUTHEASTERN OCEAN

IMAGE OF THE FAR DISTANT,

WEST-NORTHWESTERN PRAIRIE

π

π’

O

Figure 2: Schematic of a plane-to-plane perspectivity.

To highlight the different types of practical projections involved in the perspectivity, we dividethe horizontal plane π into three regions: two infinite half-planes called the “ocean” and the“prairie,” and a narrow strip dividing them called the “beach.” The center of projection O isdirectly above the “shoreline”—the boundary between the ocean and the beach. The perspectivityprojects the plane π onto the vertical plane π′, which meets π in the boundary between the oceanand the prairie. The entire plane π is paved with an avenue of initials “PG” for “perspectivegeometry.”

Looking at the dashed line through O connecting the ocean and its image, we see that theitinerary of the associated light ray is object-center-image: the photo type. Thus the image ofthe ocean is like a pinhole photograph: the initials PG are reversed, the ocean almost beneath Ois mapped very high up on π′, and the far eastern ocean is mapped just above the horizon line

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separating the images of the ocean and the prairie.From the dashed line through O connecting the beach and its image, we see that the itinerary

of the associated light ray is center-object-image—the shadow type. Thus the image of the beachis essentially an infinitely elongated shadow.

Finally, from the dashed line through O connecting the prairie and its image, we see that theitinerary of the associated light ray is object-image-center—the perspective drawing type. Thusthe image of the prairie is a perspective drawing on the vertical plane π′. The infinite road of PGinitials converges to a vanishing point on the horizon.

The strangest and most interesting part of the image is the neighborhood of the horizon line.The horizon line itself is the image of the line at infinity, or ideal line, in the plane π. Just abovethe horizon lies the image of the far eastern ocean, while just below it lies the image of the farwestern prairie! More generally, as indicated in Figure 2, points just above and below the horizoncome from opposite points of the compass (as applied to an infinite, flat Earth).

3-D letters

The exercise of drawing 3-D letters is instructive in relating plane-to-plane perspectivities withperspective collineations. Figure 3(a) is a perspective drawing of a 3-D letter Lying on the ground,along with a point X which lies in the plane of the upper, gray-shaded face of the L. Our goal is tomake a correct drawing of the point X ′ which lies in the ground plane directly below the point Xin space. As a hint, and for the purpose of our discussion, Figure 3(b) shows the vanishing pointO of the vertical edges of the L, and the vanishing point P of the retreating, horizontal edges ofthe L. Finding X ′ given X is like finding A′ given A, or B′ given B; it exemplifies the basic taskinvolved in drawing 3-D letters. The point P lies on the horizon line o. The line o is the image ofthe ideal line in E3 which is incident with both the upper and lower planes (front and back faces)of the L.

(b)(a)

X

A B

A’ B’

X

O

P o

C

C’

Figure 3: Setup of the problem of constructing the image X ′ of X.

As is typical when drawing 3-D letters, there are many correct ways to construct X ′ given X.

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Figure 4 illustrates a particular method which will be instructive later. Note in particular that thefigure depicts a plane-to-plane perspectivity in space with center O relating the upper and lowerplanes of the L; that is, the plane of the gray-shaded surface and the ground plane. The pictureplane of the figure—that is, the plane of the paper—is tilted slightly forward in space, so that wesee both vanishing points O and P , which are ideal points in space that represent perpendicularvertical and horizontal directions (see also Figure 7).

B

B’

X

O

P o

X’

Q

jk

l

j Draw the line BX until it meets the horizon line o at the point Q.

k Draw the line B’Q. Since BX and B’Q vanish at Q, they represent

parallel horizontal lines, like the top and bottom of a fence with a fence

post BB’.

l The desired point X’ lies at the bottom of the fence, that is, on B’Q. It

also lies on the vertical line OX, hence X’ = OX · B’Q.

Figure 4: Construction of X ′ given X, reasoning in 3-D.

There are other plane-to-plane perspectivities implied by Figure 4; for example the perspectivitythat maps the front plane of the letter L onto the plane of the paper, or the perspectivity thatmaps the back plane of the letter L (the ground plane) onto the plane of the paper . The centerof these perspectivities is the reader’s viewing eye E (not pictured in Figure 4, but see Figure 7),assuming it is located at the correct viewpoint with respect to the page. These perspectivites arelike the one in Figure 2 with center O. Since these mappings are onto the plane of the paper, anypoint on the page (say X) can represent a point in the plane of the gray front surface of the L, andlikewise any point on the page (say X ′) can represent a point in the ground plane behind the L.Thus we can think of the mapping X 7→ X ′ as representing a mapping from the plane of the paperto itself. This mapping is a perspective collineation.

However, because this perspective collineation involves the entire plane of the paper, there aresome point mappings of the type X 7→ X ′ from the front plane of the L to the back plane of theL that are a bit more challenging to the “artist’s intuition” method employed in Figure 4. Forexample, Figure 5(a) challenges us to find Y ′ when the point Y is given above the horizon lineo. The figure whimsically depicts this point as being in the “sky” of the picture. Recall fromFigure 2 that a complete plane-to-plane perspectivity involves shadow projection and perspectivedrawing and a reversed photographic negative. In particular, the point Y in Figure 5(a) is like thepoints in the “sky” of the plane π′ in Figure 2 which belong to the reversed photographic imageof the “ocean” part of the plane π. This makes 3-D intuition harder to apply when finding Y ′ inFigure 5(a).

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Rather than try to draw or imagine all three planes involved (the plane of the page and thefront and back planes of the L) it’s easier to just work with the plane of the page and some basicproperties of perspective collineations. (However, we discuss the three-planes point of view in thenext section.) Let’s jot down what we have to work with in Figure 5(b).

First, we have the point Y whose image Y ′ we wish to determine. We draw that in Figure 5(b).Second, a perspective collineation has a unique line of fixed points called the axis. In Figure 5(a)

the horizon line o is the image of the line at infinity that belongs to the (parallel) front and backplanes of the L, hence the points of o are fixed. Thus o in Figure 5(b) is the axis of the perspectivecollineation.

Third, a perspective collineation has a special fixed point called the center . Every point and itsimage (such as A 7→ A′, B 7→ B′, and C 7→ C ′ in Figure 3(b)) are collinear with the center, hencethe point O is the center, as drawn in Figure 5(b).

Fourth, a perspective collineation is completely determined by its center, its axis, and the imageof any non-fixed point (such as B 7→ B′ in Figure 5(b)). That’s because a perspective collineationmaps points to points, lines to lines, and preserves intersections. Now we’re ready to solve for Y ′.

B

B’

O

P

o

Y

B

B’

Y

O

o

(a) (b)

Figure 5: Two ways of envisioning the problem.

Figure 6 shows how we can use the above properties to find Y ′. Comparison with the stepsin Figure 4 shows that the two methods are analogous, but the intuitive reasoning is different. InFigure 4 we reasoned “artistically” in terms of spatial visualization, while the method of Figure 6depends strictly upon the properties of perspective collineations. There is no need to think in termsof real objects in space, but the connection still remains, as we illustrate in the next section.

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B

B’

Y

O

o

Y’

R

j

k

l

j Draw the line BY; it meets the horizon line o at the point R.

k Draw the line B’R. Since R is fixed (it lies on the axis), the line

BY maps to the line B’R, hence Y’ lies on B’R.

l But Y’ also lies on OY, hence Y’ = OY · B’R.

Figure 6: Construction of Y ′ given Y , using the properties of perspective collineations.

Thinking in 3-D

We can interpret the solution steps in Figure 6 in terms of 3-D letters in space. Figure 7 illustratesthis way of looking at the problem. A young reader looks at a version of Figure 5, which lies in aslanted plane α. The reader’s viewing eye is at the point E (points and lines outside α are denotedwith boldface letters). The 3-D letter L lies on a horizontal plane λ′, and the upper (front) face ofthe L lies in the horizontal plane λ. The ideal point O belongs to all vertical lines, such as the lineconnecting the vertices B and B′ of the L. The ideal line o belongs to the horizontal planes λ andλ′. The viewer sees the ideal point O as the point O in α, and viewer sees the ideal line o as thehorizon line o in α. The viewer sees the point P in Figure 5(a) as the image of an ideal point P ono, and the viewer sees the points B and B′ as the respective points B and B′ in α.

Before discussing the points Y and Y ′ it will be helpful to introduce some function notation.We denote three perspectivities as follows, one with center O and two with center E:

PO : λ→ λ′, PE : λ→ α, QE : λ′ → α.

In particular we have

PO(B) = B′, PE(B) = B, QE(B′) = B′.

We are after the perspective collineation PO : α→ α, where

PO = QE ◦PO ◦P−1E ,

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and in particular we want to find

Y ′ = PO(Y ) = QE ◦PO ◦P−1E (Y ).

Now as shown in Figure 7, P−1E (Y ) = Y, where Y ∈ λ and Y, E, and Y are collinear. Similarly,

PO(Y) = Y′, where Y′ ∈ λ′ and Y, Y′ and O are collinear. Finally, QE(Y′) = Y ′, where Y ′ ∈ αand Y′, Y ′, and E are collinear.

O

o

Y

B

B’

Y ’

α (plane of Figure 5)

reader’s eye E

to ideal

point O

λ

λ’

Y

Y’

B

B’

to ideal

point O to ideal

point O

to ideal point P

and ideal line o

P

Figure 7: Visualizing the three planes in space.

In Figure 7 we can think of a light ray (dashed line) leaving the vertex B of the L, piercing theplane α in the point B, and striking the reader’s eye E. This is the object-image-center itinerary ofa perspective drawing setup, like the one in the lower left of Figure 1. In this case it is reasonable tothink of E as the viewing eye of an onlooker. On the other hand, Figure 7 shows a “light ray” leavingthe point Y, passing through the back of the “reader’s eye” E, coming out the pupil, and strikingthe plane α at the point Y ! Obviously the characterization of E as a person’s eye is not appropriatein this case. In fact, this is the object-center-image itinerary of a pinhole camera, like the picturein the upper left of Figure 1. Another drawback of Figure 7 is that it is a rather complicatedmethod of locating Y ′ given Y , by comparison with the simple method of Figure 6. Nevertheless,the idea of Figure 7 suggests an interesting interpretation of the perspective collineation PO, andit reminds us that even simple exercises like drawing 3-D letters can have deeper implications.

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