178 M. Monastyrsky
Fori≥3wehave
0 →πi(S^3 )→πi(S^2 )→ 0.
In particularπ 3 (S^2 )=Z.
The homotopy classes of mapsS^3 →S^2 are characterized by the group of
integersZ. What is their geometrical meaning?
Let us consider a special mapp:S^3 →S^2 , which is calledHopf fibration
(Fig. 10.1). We representS^3 as a pair of complex numbers (z 1 ,z 2 ) with the
pointz 1 /z 2 =w(wis a point in the complex planeC). The map (z 1 ,z 2 )→w
is extended to the completion ofCwith the pointz 2 = 0 at infinity. We thus
obtained a mapS^3 →S^2 ∼C=C∪∞=CP(1) – Riemann sphere. It is
obvious that under this map the points (exp(iφ)z 1 ,exp(iφ)z 2 ) are sent into
the same pointw. Therefore the fiber of the bundleS^3 →S^2 is the set of
pointsλ=exp(iφ)∼S^1. We obtained a fiber mapp:S^3 →S^2 with the
fiberS^1. It is easy to see thatpis not trivial, i.e., not equivalent to the direct
product
S^3 =S^2 ×S^1.The simplest way to see this is to calculate the homotopy groups (e.g.,π 1
orπ 2 ) of both sides, namely,
(a)π 2 (S^3 )=0,π 2 (S^2 ×S^1 )=Z, (b)π 1 (S^3 )=0,π 1 (S^2 ×S^1 )=Z.The geometrical meaning of that any circle inS^3 can be contracted to a
point but not inS^1 ×S^2.
It is possible to prove the following proposition:
Proposition 1.The set of classes of homotopy mapsS^3 →S^2 is the compo-
sition of homotopy mapsf:S^3 →S^3 and the Hopf fibrationp:S^3 →S^2.
Fig. 10.1.Hopf fibration