Lecture 7 43
opposite them, which have been painted all three colours. (We neglect
the boundaries of the wedges, since they have area zero.) Hence if we
add up the areas of the double wedges, we obtain
areas of wedges = blue area + yellow area + red area
= (area of sphere) + 4 × (area of triangle),
which allows us to write an equation for the area A of the triangle:
4(α + β +
γ)R2
=
4πR2
+ 4A.
Solving, we see that
(1.6) A =
R2(α
+ β + γ π).
Thus the area of the triangle is directly proportional to its angular
excess; this result has no analogue in planar geometry, due to the
flatness of the Euclidean plane. As we will see later on in the course,
it does have an analogue in the hyperbolic plane, where the angles of
a triangle add up to less than π, and the area is proportional to the
angular defect.
Exercise 1.20. Express the area of a geodesic polygon on the sphere
in terms of its angles.
Lecture 7
a. Spaces with lots of isometries. In our discussion of the isome-
tries of R2, S2, and RP 2, we have observed a number of differences
between the various spaces, as well as a number of similarities. One of
the most important similarities is the high degree of symmetry each
of these spaces possesses, as evidenced by the size of their isometry
groups.
We can make this a little more concrete by observing that the
isometry group acts transitively on each of these spaces; given any
two points a and b in the plane, on the sphere, or in the projective
plane, there is an isometry I of the space such that Ia = b.
In fact, we can make the stronger observation that the group acts
transitively on the set of unit tangent vectors. That is to say, if v is
a unit tangent vector at a, which can be thought of as indicating a
particular direction along the surface from the point a, and w is a
Previous Page Next Page