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Projective geometry vs. affine geometry

As explained at the end of chapter one, an affine transformation on $ {\Bbb R}^n$ is a pair $ (A,y)$ consisting of an invertible $ n \times n$ matrix $ A$ and a vector $ y$. If $ x$ is a point in $ {\Bbb R}^n$, the action of $ (A,y)$ on $ x$ is $ Ax + y$.

Exercise 2.1 (10)   Let $ (A,y)$ be an affine transformation on $ {\Bbb R}^n$ and let $ B$ be the $ (n+1) \times (n+1)$ matrix

$\displaystyle \begin{pmatrix}
a_{11} & \cdots & a_{1n} & y_{1} \\
\cdot & \cd...
... \\
a_{n1} & \cdots & a_{nn} & y_{n} \\
0 & \cdots & 0 & 1
\end{pmatrix} .

Show that upon identifying $ U_{n+1}$ with $ {\Bbb R}^n$ via the chart $ \varphi_{n+1}$ the projective tranformation defined by $ B$ equals the affine transformation defined by $ (A,y)$.

Exercise 2.2 (15)   Use the preceding exercise to justify the following claim: the group of affine transformations on $ {\Bbb R}^n$ is the subgroup of projective transformations that fix the hyperplane at infinity.

The notion of parallelism is typical of affine geometry. It can be easily interpreted in projective geometry if we fix the hyperplane at infinity.

Exercise 2.3 (05)   Use figure 2 to show that two staight lines on the plane $ z = 1$ are parallel if and only if the corresponding projective lines intersect at infinity.

Figure 2.

Exercise 2.4 (00)   Show that two affine subspaces in $ {\Bbb R}^n$ are parallel if and only if all the points in their intersection lie on the hyperplane at infinity.

Part of the philosophy of projective geometry is to think of $ {\Bbb R}^n$ as a subset of $ {\Bbb R}P^n$. Equivalently, we may say that $ {\Bbb R}P^n$ is a compactification of $ {\Bbb R}^n$. This point of view is helpful in the solution of many geometric problems. Here is a pretty example:

Exercise 2.5 (25)   A convex set in $ {\Bbb R}^n$ is one that contains every line segment joining any two of its points. For example a disc is convex, but a croissant is not. Show that any unbounded convex subset on the plane contains a half line.

Likewise, some theorems of projective geometry are easier to prove if we reduce them to theorems in affine geometry. The classical examples are the theorems of Pappus and Desargues, which we permit ourselves to write in the form of experiments.

Note. In the geometric constructions that follow, we will denote points by uppercase letters and lines by lowercase letters. The symbol $ AB$ denotes the line passing through the points $ A$ and $ B$, while $ a \cdot b$ denotes the point of intersection of the lines $ a$ and $ b$.

Theorem 2.1 (Pappus)   Draw two lines on the projective plane and three points on each line. Denote the points on the first line by $ A,B, C$, and the points on the second line by $ A',B',C'$. Draw the lines that join points denoted by different letters (i.e., we do not draw the lines $ AA', BB'$, or $ CC'$). The points $ AB'\cdot BA', AC' \cdot CA'$, and $ BC' \cdot CB'$ are collinear.

Figure 3.

The proof is given in the following two exercises, the first of which is to prove the affine version of the theorem.

Exercise 2.6 (10)   Draw two lines on the plane and draw points $ A,B, C$ on the first line, and points $ A',B',C'$ on the second line such that $ AB'$ is parallel to $ BA'$, and $ BC'$ is parallel to $ CB'$. Show that $ AC'$ is necessarily parallel to $ CA'$.

Figure 4.

Now comes the reduction of the projective result to the affine result.

Exercise 2.7 (10)   Let $ {\cal T}$ be any projective transformation that sends the points $ AB'\cdot BA'$ and $ BC' \cdot CB'$ to infinity. Apply $ {\cal T}$ to the points and lines in the configuration of theorem 2.1 and verify that the new configuration is like the one of the previous exercise. Use this to prove Pappus' theorem.

Theorem 2.2 (Desargues)   Draw the lines $ a,b$, and $ c$ on the projective plane, and draw points $ A,A' \in a$, $ B,B' \in b$, and $ C,C' \in c$. The points $ AB \cdot A'B', AC \cdot A'C'$, and $ BC \cdot B'C'$ are collinear if and only if the lines $ a,b$, and $ c$ are concurrent.

Figure 5.

Exercise 2.8 (20)   Find and prove an affine version of Desargues' theorem. Reduce Desargues' theorem to its affine version by using a well-chosen projective transformation.

next up previous
Next: The fundamental theorems Up: The Real Projective Plane Previous: Basic concepts
Juan Carlos Alvarez 2001-01-22