18 0. Prequel

By Example 0.2.4, π1(Y ) has presentation α, β :

α5β−3,β3(αβ)−2

.

Thus π1(Y ) has two generators α and β and two relations

α5

=

β3

=

(αβ)2.

This is the well-known binary icosahedral group, so called because there is

an order-two homomorphism of π1(Y ) onto the group G of rigid motions of

the icosahedron.

In order to complete the proof we need to show that π1(Y ) is nontrivial.

We will do that by exhibiting a homomorphism of π1(Y ) onto a nonabelian

subgroup of G. To determine such a homomorphism, send α to the coun-

terclockwise rotation through 2π/5 radians about the point x indicated in

Figure 0.3 and send β to the counterclockwise rotation through 2π/3 radians

about the point y. Then αβ is the rotation through π radians about the

point z, so all three of

α5, β3,

and

(αβ)2

represent the identity motion. As a

result, this assignment extends to a homomorphism of π1(Y ) to G. Since βα

is rotation through an angle of π radians about z , we see that αβ = βα.

x

y

z

z

Figure 0.3. The icosahedron

In the preceding argument it was not necessary to compute π1(Y ) explic-

itly to determine its non-triviality, but it is known that G has 60 elements,

and that the order of π1(Y ) is 120.

Example 0.10.3. There exists a compact, contractible n-dimensional ∂-

manifold in Sn, n ≥ 5, that is not a ball.

Proof. Start with an acyclic 2-complex P (such as that in the preceding

example) and embed it in

Sn,

n ≥ 5. Name a regular neighborhood N of P

and set M =

Sn

Int N. General position considerations yield that

Sn

P

is 1-connected; the same holds for M, which is a (deformation) retract of

Sn

P , since N P

∼

=

∂N × [0, 1). Like P , N is acyclic; more importantly,

so is M, by duality or a simple Mayer-Vietoris argument (it helps to know

∂N = ∂M is orientable, due to §0.3). Hence M is contractible. Note that

∂M need not be a sphere. In particular, ∂M is 1-connected if and only if P

is, for general position implies that the arrow in the line below,

π1(∂M = ∂N)

∼

=

π1(∂N × [0, 1))

∼

=

π1(N P ) → π1(N)

∼

=

π1(P ),