Chapter 2 Preliminaries
2.5 Resurgence
2.5.6 Alien Calculus
We saw that sing0 is an inclusion of RES in SING. By definition, any element ofd d
RES can be continued on any path originating at 1 and that avoids 2πiZ. This
means that any element ofRES will have a singularityd 2 at any point ω ∈2πiZ, so 2
z1
z2
Figure 2.6: Continuation path inR1.
one could expect that by translating the origin to ω we get an element of SING. However translation on R1 is ambiguous. For this reason we need to consider the continuation of an element ofRES along a pathd γ, denoted by contγ. Let ˚f ∈RESd
and γ ∈ R1. Then there exists a unique analytic continuation of ˆf alongγ. Let n∈Zand γ : [0,1]→C\2πiZwith γ(0) =z1 not a negative real number and
γ(1) =z2, such that |z1|=ε1 and |z2−ω|=ε2 and ε1, ε2 < π, see Figure 2.6 for an example. Then the pathγ can be thought of as the concatenation of two paths. The first γ1 from 1 to z1 withγ1 in the principal sheet of R1 and γ2 from 1 to z2. So then γ is the concatenation of the reversed γ1 and γ2. It is worthy noting that in this procedure there is a hidden choice3, which is the value of the function on the principal sheet of R1. This choice is not unique but once it is made then the continuation depends only on γ.
The analytic continuation along γ gives a function analytic in ˜Dε2(γ2) and from
this we can define a function analytic on Dε2(z2) since ˜Dε2(γ2) is isomorphic to Dε2(z2). Now by a simple translation close to the origin to get a function analytic
inDε2(z2−ω). Then the function can be continued in the spiraling neighbourhood
of the origin V(C2ε2), withC2ε2 :R→ {2ε2}, thus we get an element of ANA. We 3
denote the procedure described above by
contγω:RESd →ANA.
Naturally we define
singγω :RESd →SING, with singγω := sing0◦contγω.
By the definition of RES, for anyd n ∈ Z and for any ˚f ∈ RES, the singularityd
singγωn( ˚f) can be extended into a function onR1. So we can use the map res to get an element of RES. We define thed alien operator associated with (ω, γ),
Aγω:RESd →RESd, with Aγω= res◦sing0◦contγω.
There is a particularly interesting class of alien operators denoted by ∆+ωn withωn=
2πin. This operator denotes the continuation to ωn by the path that circumvents
all ωm, 0< m < n, counterclockwise. In Figure 2.7 the path for ∆+ω4 is shown. For
the action of ∆+ωn on the algebra RES we have the following lemma.d
Lemma 2.20. Let n∈N andf˚,˚g∈RES. Thend
∆+ωn[ ˚f∗˚g] = ∆+ωn[ ˚f]∗˚g+
n−1
X
m=1
∆+ωm[ ˚f]∗∆+ωm−n[˚g] + ˚f∗∆+ωn[˚g].
A similar relation holds if n is a negative integer. An indication that this is true is given in Figure 2.8. The circles in these figures indicate what is considered to be close enough to a singularity so that we can use the convolution of SING. To get the contour in Figure 2.8, we begin with sclose to 0, Res > 0 and the second definition of convolution and then we continue to an sclose to ω3 along a straight line. Then we see that just by breaking path of the integralR
Γfˇ(σ)ˇg(s−σ)dσ into three, we get the relation. Notice that the function ˇf(σ) has singularities atωnand
the function ˇg(s−σ) has singularities ats−ωn.
Remark. Since[snis singular only at the origin we have ∆+ωn[[sn∗f˚] =[sn∗∆+ωn[ ˚f].
In particular we have ∆+ω1( ˚f ∗˚g) = ∆+ω1( ˚f)∗˚g+ ˚f∗∆+ω1(˚g), thus the operator ∆+ω1 acts on RES as a derivation. This motivates the definition of another set of lineard
ω4
ω3
ω2
ω1
0 ω1 ω2 s s-ω-1 s-ω-2 λ λeⅈ π+s 0 ω1 ω2 s s-ω1 s-ω2 λ λeⅈ π+s
Figure 2.8: On the left the convolution path for two elements of SING it is shown. On the rightsis moved on a straight line to get close toω2.
Definition 2.21. For each m∈N∗ we define ∆ωm = m X l=1 (−1)l l X m1,...,ml>1 m1+···+ml=m ∆+ωm 1· · ·∆ + ωml.
We define similarly ∆ωm form∈ −N
∗. Thus we have ∆ω1 = ∆ + ω1, ∆ω2 = ∆ + ω2 − 1 2∆ + ω1∆ + ω1, ∆ω3 = ∆ + ω3 − 1 2 ∆ + ω2∆ + ω1+ ∆ + ω1∆ + ω2 +13∆+ω1∆+ω1∆+ω1, .. .
These new alien operators are derivations on the ring RES and ∆d ω is called alien derivative at ω.
Knowing all the alien derivatives gives enough information to calculate any alien operator, which means that we can recover any singularity. Unfortunately the sub- ringsRESd
reg
and RESd
sim
are not invariant under the action of the alien derivatives. This is because a regular variation at the origin cannot force any other singularity it may have. This changes if ˚f satisfies some equation, see the following example.
Example 2.22. Let F :C→ C be analytic around the origin with F(0) = 0. In
other words, there are constants an such thatF(z) =Pn>1anzn in a neighbour-
hood of the origin. Assume that there exists ˜x∈RESg
reg such that ˜ x(t+ 1) =F(˜x(t)). (2.1) Let B[˜x](s) = ˆx(s) ∈ RESd reg
and assume that ˆx is of exponential type along any path that does not go to infinity vertically. Then we can apply the formal Borel transform to equation (2.1) by writing
F(˜x(t)) =X n>1 anx˜(t)n and F[ˆx](s) :=B[F(˜x)](s) = X n>1 anxˆ∗n(s). Since F0(z) =P n>0(n+ 1)an+1 zn we define F0[ˆx](s) :=B[F0(˜x)](s) =X n>1 (n+ 1)an+1 xˆ∗n(s).
The Borel transform of equation (2.1) is
e−sxˆ(s) =F[ˆx](s).
By applying the alien derivative ∆ω to the sum we get4
e−s∆ωxˆ(s) =F0[ˆx]∗∆ωxˆ(s).
The formal Laplace transform of this equation is
∆ωx˜(t+ 1) =F0(˜x(t) ∆ωx˜(t),
with ∆ωx˜the formal Laplace transform of ∆ωxˆ. This is a linear difference equation
and obviously ˜x0 is a solution, so we set ∆ωxˆ(t) = C(t) ˜x0(t) and from this we get
C(t+ 1) = C(t). This implies that if we look for a solution in C[[1t]]1 then C is a constant and we have
∆ωxˆ=Cωx˜0.
This indicates that even though in general the singularity at 0 of an element of
d
RESreg does not restrict any other singularities, if it is the solution of an equation, then the equation does restrict what singularities can appear. For an in depth treatment see [Sau13a, Sau08].