1549055384-Symplectic_Geometry_and_Topology__Eliashberg_

288 M. GRINBERG AND R. MACPHERSON, EULER CHARACTERISTICS

(a) Think of X as the complex line C. Take W 1 = X, and define n 1 : W1-+ X

by z r--+ z^2. Let F1 = (n1)*Z.

(b) Think of X as JR^2. Take W 2 = {(x, y , z) E JR^3 I x^2 + y^2 = z^2 }, and define

7r2 : W2-+ X by (x, y, z) r--+ (x, y). Let F2 = (n2)*Z.

Given two Fary functors F, F' on (X, S), a morphism f: F-+ F' is the obvious
thing: a collection of maps Hi(Y, Z; F) -+ Hi(Y, Z; F'), commuting with all the
structure maps. The set of all Fary functors on (X, S) forms an additive category.
This category of is not abelian, i.e., you can not define kernels and cokernels of
morphisms, so that the usual properties are satisfied.
In the rest of this lecture we will use the following shorthands. When (Y, 0)
is a standard pair, we let Hi (Y; F) = Hi (Y, 0; F). For p E X, we let Hi (p; F) =
Hi(B<,p; F), where B<,p is a small ball around p. The groups Hi(p; F) are called
the stalks of F.

5.2. Monodromy

Let F be a Fary functor on (X, S), and (yt, Zt), t E [O, l], be a smoothly varying
family of standard pairs. Then we can define a monodromy map

which is an isomorphism. Note that the structure of F does not directly give maps

Hi (yt, Zt; F) -+ Hi (yt1 , Zt'; F) for t < t'. The idea of defining the monodromy is to

impose a fast undulation on the family (yt, Zt), so that the homotopy axiom would
give isomorphisms between the groups for neighboring values oft.

Instead of a formal definition, we give an example. Let X = JR^2 , stratified

with one stratum. Take a family (yt, Zt), where yt is the unit disk around the
point (3t, 0), and Zt = 0. We can replace the discs yt by a family of ovals }t, with

Yo = Yo and Y 1 = Y 1 , as on Figure 3. The homotopy axiom gives isomorphisms

μ1 : Hi (Yo; F) -+ Hi W1;2; F) and μ2 : Hi (Y1; F) -+ Hi (Y 1 ;2; F). We then set

μ=μ2 - 1 0μ1.

We leave it to the reader to think through the general definition of monodromy.

0 :::; t :::; 1 /2 1 /2 :::; t :::; 1

Figure 3

Exercise. In the situation of Example 5.1.2, let p EX\ {O}. Show that H 0 (p; F) =

Z^2 , for both F = F 1 and F = F 2. Compute the monodromy, as p goes once around

the origin, and check that it is different for F = F 1 and F = F 2.