12.3 Plumbing Fixture
257
If x s are given by (12.13), the vectors have support in only one circle.
There is, however, a redundancy in these vectors. Not all of them lead to a motion
in P g,n because some modifications can be absorbed with a reparametrization of
the z a . For example, if a given v (α) can be extended holomorphically outside the
circle C α in the neighbour sphere, then its effect can be undone by reparametrizing
the corresponding coordinate. A similar discussion holds for the other spaces,
and relations can be found by restricting the vector on subspaces. A non-trivial
vector (v (i) , C i ) of P g,n becomes trivial on M g,n if it can be cancelled with a
reparametrization of w i that leaves the origin fixed.
12.3 Plumbing Fixture
The plumbing fixture is a way to glue together two Riemann surfaces (separating
case) or two parts of the same surface (non-separating case), in order to build a
surface with a higher number of holes and punctures. This geometric operation will
correspond precisely to the concept of gluing two Feynman graphs with a propagator
in Siegel gauge.
The plumbing fixture depends on a (complex) one parameter, which leads to
a family of surfaces. This provides the correct number of moduli for the surface
obtained after gluing. This brings to the question of describing the moduli spaces
M g,n in terms of the moduli spaces with lower genus and number of punctures.
12.3.1 Separating Case
Consider two Riemann surfaces g 1 ,n 1 and g 2 ,n 2 with local coordinates w
(1)
1 , . . . ,
w
(1)
n 1 and w
(2)
1 , . . . , w
(2)
n 2 .
The first step is to cut two disks D
(1)
q and D
(2)
q of radius |q| 1/2 around a puncture
on each surface, taken to be the n 1 -th and n 2 -th for definiteness:
D
(1)
q =
|w
(1)
n 1
| ≤ |q|
1/2
,
D
(2)
q =
|w
(2)
n 2
| ≤ |q|
1/2
,
(12.28)
where q ∈ C is fixed (Fig. 12.3). 2 Then, both surfaces can be glued (indicated by
the binary operation #) together into a new surface
g,n = g 1 ,n 1 # g 2 ,n 2 ,
g = g 1 + g 2 ,
n = n 1 + n 2 − 2
(12.29)
2 The disks D
(i)
q should be equal or smaller than the disks D
(1)
n 1 and D
(2)
n 2 .
257
If x s are given by (12.13), the vectors have support in only one circle.
There is, however, a redundancy in these vectors. Not all of them lead to a motion
in P g,n because some modifications can be absorbed with a reparametrization of
the z a . For example, if a given v (α) can be extended holomorphically outside the
circle C α in the neighbour sphere, then its effect can be undone by reparametrizing
the corresponding coordinate. A similar discussion holds for the other spaces,
and relations can be found by restricting the vector on subspaces. A non-trivial
vector (v (i) , C i ) of P g,n becomes trivial on M g,n if it can be cancelled with a
reparametrization of w i that leaves the origin fixed.
12.3 Plumbing Fixture
The plumbing fixture is a way to glue together two Riemann surfaces (separating
case) or two parts of the same surface (non-separating case), in order to build a
surface with a higher number of holes and punctures. This geometric operation will
correspond precisely to the concept of gluing two Feynman graphs with a propagator
in Siegel gauge.
The plumbing fixture depends on a (complex) one parameter, which leads to
a family of surfaces. This provides the correct number of moduli for the surface
obtained after gluing. This brings to the question of describing the moduli spaces
M g,n in terms of the moduli spaces with lower genus and number of punctures.
12.3.1 Separating Case
Consider two Riemann surfaces g 1 ,n 1 and g 2 ,n 2 with local coordinates w
(1)
1 , . . . ,
w
(1)
n 1 and w
(2)
1 , . . . , w
(2)
n 2 .
The first step is to cut two disks D
(1)
q and D
(2)
q of radius |q| 1/2 around a puncture
on each surface, taken to be the n 1 -th and n 2 -th for definiteness:
D
(1)
q =
|w
(1)
n 1
| ≤ |q|
1/2
,
D
(2)
q =
|w
(2)
n 2
| ≤ |q|
1/2
,
(12.28)
where q ∈ C is fixed (Fig. 12.3). 2 Then, both surfaces can be glued (indicated by
the binary operation #) together into a new surface
g,n = g 1 ,n 1 # g 2 ,n 2 ,
g = g 1 + g 2 ,
n = n 1 + n 2 − 2
(12.29)
2 The disks D
(i)
q should be equal or smaller than the disks D
(1)
n 1 and D
(2)
n 2 .