eq. (1.18), it is then possible to rewriteeq. (2.21) as a recursive identity for Bell poly-nomials. Then, the terms are split into one part porportional to m2 and another one proportional topipj,i6=j which is possible due to symmetry in the onshell momenta.
After some manipulations and using lemma A.15, the individual terms boil down to lemma A.11.
By theorem 2.1, the objects bn are purely combinatorial. They do not depend on the involved momenta and masses. The tree sum with n+ 1 external edges where all propagators are amputed arises from bn by multiplying the inverse upper propagator
−ix1+2+...+n,
−i
p1+p
2+. . .+p
n
2
−m2
bn.
If the upper edge is onshell as well, the parenthesis vanishes and so does the whole amplitude. This means that at treelevel the diffeomorphism does not contribute to the S-matrix.
Compare [KV13, sec. 4] where this result is established without using theorem 2.1 by a discussion similar to the one in section 2.3which will be useful several times in our study.
An internal edge has two adjacent diffeomorphism vertices. So it can be cancelled either twice, or once, or not at all. We will consider these cases one at a time.
The first case is what eliminates them2contributions of vertices through a mechanism likelemma A.1. Of course, there are also contributions of double-cancelled edges which are proportional not to m2 but xj for some j. These contributions only change the proportionality constant for some already existing term, but not the fact if edges are cancelled or not. So, in terms of cancellation, a double-cancelled edge is an object which reduces to single-cancellation of its outer edges (and the double-cancelled edge itself is gone, so the double-cancelled edge together with its two adjacent vertices forms one new vertex which cancels one of its outer edges).
Similarly, the edges cancelled once are already included intheorem 2.1, they do not do anything new here. In terms of Graphs, they cause their two adjacent vertices to fuse.
A tree withvvertices hasv−1 internal edges by the Euler characteristicsection 1.5.2.
Since each of the diffeomorphism vertices in tree sums cancels precisely one edge, hav-ing one external edge offshell (and cancelled) in bn implies that all internal edges are cancelled. Consequently, if more than one external edge is cancelled, an internal edge will remain uncancelled. So it is uncancelled internal edges which make the difference betweenbn and some Ajn+1 withj >1 external offshell edges.
Example 5: Four external edges
We explicitly compute the Feynman amplitudeA44 of a tree sum with four external edges, all of which may be offshell. There are two types of contributions, 1. one 4-valent vertexv4 and 2. two 3-valent verticesv3, connected with an internal edge.
The latter involves three different ways of distributing the four external edges onto the two vertices.
Let the external momenta bep1, p2, p3, p4. The 4-valent byeq. (2.15)vertex has amplitude
A44,1=v4= 4im2a21+i 4a21+ 6a2
(x1+x2+x3+x4).
For the trees involving two vertices, first fix x1 and x2 to be the variables at one of the vertices and add the contributions with indices exchanged later.
A44,2a= 2ia1(x1+x2+x1+2) i
x1+22ia1(x1+2+x3+x4)
=−4ia21
x1+x2
x1+2
+ 1
(x1+2+x3+x4)
=−4ia21
x1+x2+x1+2+(x1+x2)(x3+x4)
x1+2 +x3+x4
=−4ia21(x1+x2+x3+x4)−4ia21x1+2−4ia21(x1+x2)(x3+x4) x1+2
.
Summing over all three ways and using lemma A.1eliminates the xi+j and yields A44,2=−16ia21(x1+x2+x3+x4)−4im2a21−4ia21
(x1+x2)(x3+x4)
x1+2 + 2 more
. The total amplitude of a tree sum with four external edges is
A44 =A44,1+A44,2 (2.22)
=i −12a21+ 6a2
(x1+x2+x3+x4)−4ia21
(x1+x2)(x3+x4) x1+2
+ 2 more
. Setting allxj except one to zero and including an external propagator ineq. (2.22) reproduces bn fromtheorem 2.1,
i x1
·A44
x2=x3=x4=0 = i x1
·A14 = 12a21−6a2 =b3.
There is an important connection between an edge with momentump being offshell and being cancelled:
Lemma 2.1.
Letebe an edge adjacent to Γ whereAnis a sum over all possible trees withnexternal edges . If An contains vertices capable of cancelling adjacent edges, then either there are terms cancellingeoreis onshell.
Proof. Assume the edge eis actually onshell. Then by definition 1,xe = 0. The parts of the integrand An which can cancel e are precisely the summands proportional to xe since cancellation amounts to multiplication with the inverse propagator, which is
−ixe. But since xe = 0, all these summands vanish. This means there exists no term in the integrand which can cancel the edge, hence it is uncancelled.
On the contrary, since there is at least one vertex in Angenerating a factorxj for its adjacent edgej, by symmetry also a termxe must appear in the sum over all trees. So if this term does not vanish then it will cancele.
Lemma 2.1seems trivial, it is nevertheless the core ingredient for constructing tree sums with uncancelled internal edges. Assume for simplicity there are two external offshell edges, so just one internal edge e is uncancelled. Then from the point of view of its adjacent subtrees this edgeeis an onshell edge. Nevertheless it is not actually onshell, this is, there is still a propagator xi
e withxe6= 0. So the whole tree disintegrates into two components which are connected via this edge. Summing over all trees especially includes summing over all ways to reorganize these components without changing the uncancelled edge. But these are precisely the sums which yieldbs bydefinition 7. This is, a tree sum A2n with two of its n external edges offshell consists of two tree sums bs1, bs2 such that s1+s2 =nand they are connected by an internal edge e. This edge has a propagator xi
e wherepeis the sum of all momenta entering either of the two parts.
In summing over all permutations of external edges, this becomes a symmetric function of the possible intermediate momenta as in eq. (2.22). Since A2n also is symmetric in
external edges, the remaining momentum-dependent function connectingbs1 and bs2 is symmetric in both the external momenta and the sums of external momenta which make up the uncancelled internal propagator. But since the latter appear in the denominator, it is not a symmetric polynomial inxj but rather a rational function.
One might be tempted to reject this procedure because, apart from edges being can-celled or not, a Feynman amplitude generally is a function of its external momentan and therefore it makes a difference if these momenta are onshell or not. This is true indeed but does not apply here: The “building blocks” of the tree sum are by con-struction precisely the parts where no internal edge is uncancelled. They are always tree sums where the sum is carried out without any restriction on the allowed internal cancellations, i.e. somebnaccording todefinition 7. But bytheorem 2.1, nobndepends on external momenta.
One might also object that for bn, all but one external momenta are onshell and in this sense the amplitude does depend on the specific value of the momenta. This is also true, but if more then one external edge were offshell, still the terms formingbnwould be present, there would just be additional summands which are proportional to somexj. In the end the whole argument reduces to recognizing that with diffeomorphism Feynman rules eq. (2.15) there cannot be any other dependence on external momenta than in the form of terms proportional to somexj. Consequently, if all terms proportional to (some power of) xj are zero then the whole tree sum is necessarily independent of xj and hence of the corresponding external momentum.
Consider for illustration the computation example 5where from the very beginning x4 = 0 is imposed. The result would be just the same as calculating A44 for arbitrary x4 and settingx4 = 0 in the end result, i.e. all terms which are not proportional to x4 are unchanged no matter ifx4 is onshell or not.
Example 6: Four external edges
The resulteq. (2.22)from the aboveexample 5 can be interpreted in terms of tree sumsbs connected by symmetric functions of the external momenta.
A4 =−i(x1+x2+x3+x4)·b3−i
(x1+x2)(x3+x4)
x1+2 + 2 more
·b2·b2. (2.23)
Example 7: Five external edges
A computation similar toexample 5yields for the tree sum with five external edges A5 =i −120a1a2+ 24a3+ 120a31
(x1+x2+x3+x4+x5) (2.24) +i 24a31−12a1a2
(x1+x2)(x3+x4+x5) x1+2
+ 9 more
+ 8ia31
(x1+x2)x3(x4+x5)
x1+2x4+5 + 14 more
.
Again, this expression consists of factors bs connected with internal propagators:
A5 =−i(x1+x2+x3+x4+x5)·b4
−i
(x1+x2)(x3+x4+x5)
x1+2 + 9 more
·b3·b2
−i
(x1+x2)x3(x4+x5) x1+2x4+5
+ 14 more
·b2·b2·b2.
In the last line, 2 + 2 + 2 = 66= 5. This is because this line represents terms where three external edges are offshell (the symmetric part is of order three in the xj), hence there are two uncancelled internal lines. That the valuessof bs add up to 6 then follows from the Euler characteristic eq. (1.34) regarding bs as meta-vertices with valences+ 1.