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Discrete Holomorphic Local Dynamical Systems

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94 Eric Bedford<br />

which corresponds to the matrix<br />

⎛ ⎞<br />

2 0 0 1<br />

⎜<br />

⎜−1<br />

10−1 ⎟<br />

⎝−2<br />

01−2⎠<br />

−3 00−2<br />

As expected, the square of this matrix is the identity.<br />

Exercise: Define a new ordered basis 〈H,E1,E2,E3〉 for Pic(X3) by setting E1 =<br />

F1 + F2 + F3, E2 = F2 + F3, E3 = F3. Show that with respect to this basis, L∗ is<br />

represented by the matrix (6).<br />

Exercise: Perform an analysis on the maps L j(x,y)=(x,−y + x j ) for j = 1,2,3,...<br />

2.5 Linear Fractional Recurrences<br />

Here we consider the possibility of producing a space X and an automorphism<br />

f ∈ Aut(X) by the following procedure. That is, we consider a birational map f<br />

of projective space P 2 , and we would like to perform some blowups π : X → P 2<br />

such that the induced map fX will be an automorphism. Now if the original map<br />

f fails to be an automorphism, then there will be an exceptional curve C which is<br />

blown down to a point p1. We can try to fix this by blowing up the point p1 to produce<br />

a space π1 : X1 → P2 . If we are lucky, then the induced map fX1 will not map<br />

C to a point but will map it to the whole curve P1. However, this is only a temporary<br />

success if p1 is not a point of indeterminacy, because fX1 will map the fiber P1 to<br />

the point p2 := fp1. Thus we see that the blowing-up procedure cannot work unless<br />

we have the property that any exceptional curve C in P2 is eventually mapped to a<br />

point of indeterminacy.<br />

We spend this section looking at the example of linear fractional recurrences:<br />

� �<br />

y + a<br />

f (x,y)= y, . (9)<br />

x + b<br />

See [BK3] for further discussion of this family. In projective coordinates this map<br />

takes the form<br />

f [x0 : x1 : x2]=[x0(bx0 + x1) : x2(bx0 + x1) : x0(ax0 + x2)].<br />

We explain that to obtain the homogeneous representation, we write<br />

and so<br />

�<br />

f = 1:y :<br />

[x0 : x1 : x2]=[1:x1/x0 : x2/x0]=[1:x : y]<br />

� �<br />

y + a<br />

= 1:<br />

x + b<br />

x2<br />

:<br />

x0<br />

x2<br />

x0<br />

x1<br />

x0<br />

�<br />

+ a<br />

+ b<br />

�<br />

= 1: x2<br />

:<br />

x0<br />

x2<br />

�<br />

+ ax0<br />

.<br />

x1 + bx0<br />

(8)

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