298
14 Amplitude Factorization and Feynman Diagrams
The second condition is satisfied automatically because the Hilbert space is H −
when working with ˆ
P g,n (Sect. 13.3.1). However, the first condition further restricts
the states that propagate in internal lines. This leads to postulate that the external
states should also be taken to satisfy this condition
b
+
0 |V i = 0,
(14.50)
since external states are usually a subset of the internal states. This provides another
motivation of the statement in Sect. 3.2.2 that scattering amplitudes for the states not
annihilated by b
+
0 must be trivial. A field interpretation of this condition is given in
Chaps. 10 and 15.
Under these constraints on the states, the propagator can be inverted:
−1
= c
+
0 c
−
0 L
+
0 δ L
−
0 ,0 .
(14.51)
14.2.3 Fundamental Vertices
The vertices (14.33) can be constructed recursively assuming that all amplitudes are
known. The starting point is the tree-level cubic amplitude A 0,3 : since it does not
contain any internal propagator, it is equal to the fundamental vertex V 0,3 .
The first thing to extract from the recursion relations are the background
independent data. This amounts to find local coordinates and a characterization of
the subspaces V g,n ⊂ M g,n , starting with P 0,3 and iterating.
In the rest of this section, we show how this works schematically.
Recursive Definition: Tree-Level Vertices
The description of tree-level amplitudes A 0,n is the simplest since only the
separating plumbing fixture is used and Feynman graphs are trees. The possible
factorizations of the amplitude correspond basically to all the partitions of the set
{V i } into subsets.
Tree-Level Cubic Vertex Since M 0,3 = 0, the moduli space of the 3-punctured
sphere 0,3 reduces to a point, and so does the section S 0,3 of P 0,3 (Fig. 14.4a):
V 0,3 (V 1 , V 2 , V 3 ) := A 0,3 (V 1 , V 2 , V 3 ) = ω
0,3
0 (V 1 , V 2 , V 3 ).
(14.52)
The corresponding graph is indicated in Fig. 14.4b.
Tree-Level Quartic Vertex Part of the contributions to the 4-point amplitude A 0,4
with external states V i (i = 1, . . . , 4) comes from gluing two cubic vertices.
Because there are four external states, there are three different partitions 2|2 that
are described in Fig. 14.5 (see also Fig. 12.7). The sum of these three diagrams does
14 Amplitude Factorization and Feynman Diagrams
The second condition is satisfied automatically because the Hilbert space is H −
when working with ˆ
P g,n (Sect. 13.3.1). However, the first condition further restricts
the states that propagate in internal lines. This leads to postulate that the external
states should also be taken to satisfy this condition
b
+
0 |V i = 0,
(14.50)
since external states are usually a subset of the internal states. This provides another
motivation of the statement in Sect. 3.2.2 that scattering amplitudes for the states not
annihilated by b
+
0 must be trivial. A field interpretation of this condition is given in
Chaps. 10 and 15.
Under these constraints on the states, the propagator can be inverted:
−1
= c
+
0 c
−
0 L
+
0 δ L
−
0 ,0 .
(14.51)
14.2.3 Fundamental Vertices
The vertices (14.33) can be constructed recursively assuming that all amplitudes are
known. The starting point is the tree-level cubic amplitude A 0,3 : since it does not
contain any internal propagator, it is equal to the fundamental vertex V 0,3 .
The first thing to extract from the recursion relations are the background
independent data. This amounts to find local coordinates and a characterization of
the subspaces V g,n ⊂ M g,n , starting with P 0,3 and iterating.
In the rest of this section, we show how this works schematically.
Recursive Definition: Tree-Level Vertices
The description of tree-level amplitudes A 0,n is the simplest since only the
separating plumbing fixture is used and Feynman graphs are trees. The possible
factorizations of the amplitude correspond basically to all the partitions of the set
{V i } into subsets.
Tree-Level Cubic Vertex Since M 0,3 = 0, the moduli space of the 3-punctured
sphere 0,3 reduces to a point, and so does the section S 0,3 of P 0,3 (Fig. 14.4a):
V 0,3 (V 1 , V 2 , V 3 ) := A 0,3 (V 1 , V 2 , V 3 ) = ω
0,3
0 (V 1 , V 2 , V 3 ).
(14.52)
The corresponding graph is indicated in Fig. 14.4b.
Tree-Level Quartic Vertex Part of the contributions to the 4-point amplitude A 0,4
with external states V i (i = 1, . . . , 4) comes from gluing two cubic vertices.
Because there are four external states, there are three different partitions 2|2 that
are described in Fig. 14.5 (see also Fig. 12.7). The sum of these three diagrams does