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We have established two results for scalar quantum fields with interaction:

1. For arbitrary interaction polynomials of the underlying field, diffeomorphisms do not alter onshell tree sums (theorem 4.3). In kinematic renormalization this implies they do not alter the S-matrix and are therefore not observable in exper-iments. In this sense, all quantum fields which are related by diffeomorphisms form an equivalence class. This opens the opportunity to pick out of this equiv-alence class any representative with desirable properties for the problem under consideration.

2. If only one interaction monomial (of arbitrary order) is present, then there is a diffeomorphism which for specific fixed offshell external momenta posesses a free time-ordered 2-point function and it is explicitly given by theorem 4.2.

Still, a number of questions remains to be adressed in future work:

1. What are the conditions for the functionρ(φ) to exist in a mathematically rigorous way? Especially, if there are constraints on the fields as intheorem 3.2, are they fulfilled by quantum fields?

2. φs-theory has s-valent vertices in Feynman graphs. On the other hand, the adiabatic diffeomorphism by lemma 4.4 is given by the Fuss-Catalan numbers An(s−1,1). These numbers in turn count the planar trees built of s-valent ver-tices. Is there a deeper combinatoric meaning in the adiabatic diffeomorphism which is not yet understood?

3. What is the status of ρ? Does passing from φ to ρ effectively just implement kinematic renormalization at the renormalization point p? What does this even mean for the non-renormalizable theoryρ?

4. Is it possible to map from an interacting field φto a completely free one by the nonlocal transformationeq. (5.3)?

5. Are the identities noted in section 5.2 meaningful in any way? Are they even correct?

6. Is it actually possible to use the adiabatic diffeomorphism to cure at least in part the ill-definedness of perturbative quantum field theory, and if so, how?

7. If any set of (possibly infinitely many) vertex Feynman rules is given, is there a way to determine whether this is just a diffeomorphism of some simpler theory?

Equation (4.21)gives the most general form of vertices arising from a diffeomor-phism, but given some vertices, one has to determine both the values of λs and the diffeomorphism parameters aj with the boundary condition that “as few as possibleλsare6= 0”. Is there an “invariant” which can be easily computed for any given quantum field theory and determines, which equivalence class this theory lies in?

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A Lemmas

This section is rather a table than a didactic derivation and provides all the statements that have been used in the main text.

A.1 Sums of partitions of momenta

Consider a set ofn four-momentan

p1, . . . , pn o

which fulfill overall momentum conser-vation

n

X

j=1

pj = 0. (A.1)

Symmetric sums of offshell variables of two momenta (according toeq. (1.4)) decompose into summands proportional toxj and tom2.

Lemma A.1.

If momentum conservationeq. (A.1)holds, Xn:=

n

X

k=2 k−1

X

j=1

xj+k=x1+2+x1+3+. . .+x1+n+. . .+x(n−1)+n (A.2)

= (n−2)

n

X

k=1

xk+n(n−3) 2 m2.

Proof. Note that there are n(n−1)2 summands ineq. (A.2). For anyn≥2,

Xn=

n

X

k=2 k−1

X

j=1

pj+pk

2

−m2

=

n

X

k=2 k−1

X

j=1

p2j+p2k+ 2pjpk

−n(n−1) 2 m2

Rearranging summation indices and using momentum conservation yields for the

indi-vidual parts

n

X

k=2 k−1

X

j=1

p2

j +

n

X

k=2 k−1

X

j=1

p2

k=

n

X

j=1

p2

j n

X

k=j+1

1 +

n

X

k=2

p2

k k−1

X

j=1

1

=

n

X

k=1

(n−k)p2k+

n

X

k=2

(k−1)p2k= (n−1)

n

X

k=1

p2k

2

n

X

k=2

pk

k−1

X

j=1

pj =

n

X

k=2

pk

k−1

X

j=1

pj+

n

X

j=1

pj

n

X

k=j+1

pk

=

n

X

k=1

pkX

j6=k

pj =

n

X

k=1

pk

−pk

, so

Xn= (n−2)

n

X

k=1

p2k−(n−2)nm2+

(n−2)n−n(n−1) 2

m2

Lemma A.2.

If in in the sum from lemma A.1 still momentum conservationeq. (A.1) between alln momenta but all variables involving pn are left out, then

Xn0 :=

n−1

X

k=2 k−1

X

j=1

xj+k=x1+2+x1+3+. . .+x1+(n−1)+. . .+x(n−2)+(n−1) (A.3)

= (n−3)

n−1

X

i=1

xi+xn+(n−2)(n−3)

2 m2.

Proof.

Xn0 =Xn

n−1

X

j=1

p2j +p2n+ 2pjpn−m2

=Xn

n−1

X

j=1

xj−(n−1)xn−2p

n n−1

X

j=1

pj−(n−1)m2

=Xn

n

X

j=1

xj−(n−2)xn−2pn

−pn

−(n−1)m2

= (n−2)

n

X

k=1

xk+n(n−3) 2 m2

n

X

j=1

xj−(n−2)xn+ 2xn−(n−3)m2

= (n−3)

n

X

k=1

xk−(n−4)xn+n(n−3)

2 m2−(n−3)m2.

Note that this is not the same as Xn−1 because the latter involves momentum con-servation only between the first (n−1) momenta, which effectively impliespn= 0.

Example 21: n= 4

For n= 4, the lemmas yield

X4 = 2(x1+x2+x3+x4) + 2m2 X40 =x1+x2+x3+x4+m2.

By chance, 2X40 = X4. This is true only for n = 4 because in this case, due to momentum conservation, the offshell variables in X4 coincide pairwise, e.g. p1+ p2=−p

3−p4 implies x1+2=x3+4.