Blobbed topological recursion of the quartic Kontsevich model II: Genus=0

We prove that the genus-0 sector of the quartic analogue of the Kontsevich model is completely governed by an involution identity which expresses the meromorphic differential $\omega_{0,n}$ at a reflected point $\iota z$ in terms of all $\omega_{0,m}$ with $m\leq n$ at the original point $z$. We prove that the solution of the involution identity obeys blobbed topological recursion, which confirms a previous conjecture about the quartic Kontsevich model.


Introduction and main result
1.1. Overview. This paper completes the solution of the genus-0 sector of the quartic analogue of the Kontsevich model. This is a model for N × N Hermitian matrices with the same covariance as the Kontsevich model [Kon92] but with quartic instead of cubic potential. The non-linear Dyson-Schwinger equation [GW09] for the planar 2-point function of the quartic Kontsevich model was solved in a special case in [PW20] and then in full generality in [GHW19,SW19]. Building on this foundation we identified in [BHW20a] three families of correlation functions and established interwoven loop equations between them. One family consists of meromorphic differential forms ω g,n labelled by genus g and number n of marked points of a complex curve. By a lengthy evaluation of residues the solution was found for ω 0,2 , ω 0,3 , ω 0,4 and ω 1,1 . It strongly suggested that the family ω g,n obeys blobbed topological recursion [BS17], a systematic extension of topological recursion [EO07] by additional terms holomorphic at ramification points of a covering x :Ĉ →Ĉ.
Recall that (blobbed) topological recursion (see e.g. [EO07,BS17] and references therein) starts from a spectral curve (x : Σ → Σ 0 , ω 0,1 = ydx, ω 0,2 ). Here y : Σ →Ĉ is regular at the ramification points of x and ω 0,2 is a symmetric bidifferential which extends (or is equal to) the Bergman kernel. We noticed in [BHW20a] that the two coverings x, y in the quartic Kontsevich model are related by y(z) = −x(−z) (already visible in [GHW19,SW19]) and that ω 0,2 (u, z) = −ω 0,2 (u, −z). We show in this paper that this observation is far more than a coincidence: the properties of x, y, ω 0,2 under reflection z → ιz := −z completely characterise the genus-0 sector of the quartic Kontsevich model. There is a single global equation (1.3) which describes the behaviour of the ω 0,n under reflection. This equation can be solved without connecting it to the matrix model, the solution is identical to the solution of the complicated system of loop equations in [BHW20a], and it obeys blobbed topological recursion [BS17] (restricted to genus g = 0). We observe that the reflection formula (1.3) is of similar form to the relations studied in the context of x-y symmetry in topological recursion (see for instance [EO13,Proposition 3.1]).
It is currently not known to us how to extend these results to genus g > 0. The loop equations of [BHW20a] bring in another structure which leads to poles of ω g>0,n at the fixed point of the involution ι. Nevertheless we speculate that the global involution z → ιz characterises the quartic Kontsevich model completely and that it describes a geometric structure of the moduli space of complex curves.
The proof is lengthy and will be divided into many steps. We start to prove uniqueness: if a consistent solution of (1.3) exists, it must be of the form (1.4)+(1.5). Then we prove that (1.4)+(1.5) implies consistency of (1.3).
In a second part we show that the loop equations [BHW20a] of the quartic analogue of the Kontsevich model lead for the choice ιz = −z and x(z) = R(z) = z −λ d k=1 k ε k +z found in [SW19] to exactly the same solution (1.4)+(1.5). Thereby we prove for genus 0 the main conjecture of [BHW20a] that the quartic Kontsevich model obeys blobbed topological recursion [BS17].
2. Proof of Theorem 1.2 2.1. Tools and conventions. Throughout this paper we denote by q, u, u k , w, z, z k ∈Ĉ complex numbers and by a, a , i, j, k, l, m, n, n 0 , ..., n s , p, r, s, s non-negative integers. By I = {u 1 , ..., u m } we understand a (multi-)set of length |I| = m of complex numbers, which are allowed to coincide. By I 1 ... Is=I we denote the sum over all partitions of the multiset I into disjoint non-empty subsets I 1 , ..., I s of any order. If we insist on a sum over ordered subsets we write I 1 ... Is=I I 1 <...<Is . We define {u k 1 , ..., u kn } < {u l 1 , ..., u lm } iff min(k i ) < min(l j ). This order is well-defined because the subsets {u k 1 , ..., u kn }, {u l 1 , ..., u lm } are disjoint. We will often write I u k or (I, u k ) for I {u k } and I \ u k for I \ {u k }. In the second part,q j for j ≥ 1 denotes another preimage of q under x, i.e. x(q j ) = x(q). We will often need the projection of a meromorphic 1-form ω to the principal part P w ω of its Laurent series about w ∈ C. This projection is obtained by the residue P w ω(z) = Res q→w ω(q)dz z − q . (2.1) The limit δ → 0 together with independence of all integrals from gives (2.4). The residue does not change under the local Galois involution, that is (2.5) Invariance of the term c −1 dq q−β i of the Laurent expansion follows from σ i (q) − β i = −(q − β i ) + O((q − β i ) 2 ). For poles of order n > 1 the term c −n dσ i (q) (σ i (q)−β i ) n = − 1 n d c −n (σ i (q)−β i ) n−1 does not have a residue. Of particular importance is the following residue: ∇ n ω 0,|I|+1 (I, q) := Res z→q ω 0,|I|+1 (I, z) (y(z) − y(q))(x(q) − x(z)) n (2.6) which is a function of q and a 1-form in every variable in I. In particular, ∇ 0 ω 0,|I|+1 (I, q) = where n 1 +...+ns=s−1 is the sum over all partitions of s − 1 into integers n i ≥ 0.
Proof. We evaluate the residue (1.3) as limit of a partial derivative: Leibniz's rule for a higher derivative of a product together with (2.6) give the assertion.
Iterating Lemma 2.2 we obtain a variant of with only a single term ∇ω: Lemma 2.3. The involution identity (1.3) can also be expressed as Proof. By induction on |I|, starting from the true statement for |I| = 1. For a partition I = I 1 ... I s of I = {u 1 , ..., u m } into s ≥ 2 subsets together with a given partition n 1 + ... + n s = s − 1 we let u µ = min( s k=1 n k >0 I k ) be the smallest element within those I k with n k > 0. Moving the subset which contains u µ to the first place allows us to get rid of the 1 s -factor in Lemma 2.2: By construction we have n 0 > 0. Take in the second line of (2.9) a term of the form X∇ 0 ω 0,|Ip|+1 (I p , q), for any product X which contains I 0 . Observe that (2.9) then contains for |I p | ≥ 2 also every term of the sum By induction hypothesis and with ∇ 0 ω 0,|Ip|+1 (I p , q) = ω 0,|Ip|+1 (Ip,q) dy(q) we have which is also true in case |I p | = 1. Repeat this procedure for the next X∇ 0 ω 0,|Ip|+1 (I p ; q) for which the product X does not yet contain a factor . At the end of this procedure, the second line of (2.9) is reduced to a sum of terms each containing a factor ω 0,|I j |+1 (I j ,ιq) −dy(q) . Now iterate the procedure for every X∇ 0 ω |Ip|+1 (I p , q) where the product X contains ∇ n 0 ω |I 0 |+1 (I 0 ; q) and precisely one factor . At the end of this step we have reduced the second line of (2.9) to terms of the form or with a double factor 2 j=1 ω 0,|I j |+1 (I j ,ιq) −dy(q) . Iterate again until all ∇ 0 in the second line of (2.9) are converted. This is the assertion. Since I 0 is anyway distinguished we can omit the condition u µ ∈ I 0 .
Lemma 2.4. The involution identity (1.3) together with convention (c) in Theorem 1.2 that ω 0,m+1 (u 1 , ..., u m , z) is for m≥2 holomorphic at z=u k imply that z → ω 0,m+1 (u 1 , ..., u m , ιz) has a pole at every z = u k . The principal part of the corresponding Laurent series is given by Equivalently, + terms which are holomorphic at z = ιu k .
In particular, the poles of ω 0,|I|+1 (I, z) at z = ιu k do not have a residue.
Proof. We divide the involution identity (1.3) by w − q and take the residue at q = u k . By convention (c) in Theorem 1.2 the term ω 0,|I|+1 (I, q) in the first line of (1.3) does not contribute to the residue. In the second line we commute the two residues via (2.2). Since the inner integrand is holomorphic at q = u k , we have We implemented the convention that there is only a pole at z = u k if a unique factor ω 0,2 (u k , z) is present. It can occur at all s places of the partition of I into s subsets, so that 1 s cancels. We shifted s − 1 → s. We write ω 0,2 (u k , z) according to the second line of (1.2) and commute the differential d u k according to (2.4) in front of the residues: This is the assertion (when renaming w → z).
We will derive an alternative formula: Proposition 2.5. The poles of z → ω 0,|I|+1 (I, ιz) at z = u k can also be evaluated by Equivalently, + terms which are holomorphic at z = ιu k .
Proof. We shift s → s − 1 in (2.10) and represent the term with j = 0 via Lemma 2.3 for q → ιu k : We have included the term ω |I 0 |+1 (I 0 , ιu k ) = dy(ιu k )∇ 0 ω |I 0 |+1 (I 0 , ιu k ) as n = 0. Implementing (1.1), i.e. −dy(ιu k ) = dx(u k ) suggests to change summation variables to s + n → s ∈ [0..|I| − 2]. Then we express the derivative with respect to y(q) as a residue: The term dy(q) w−u k added in the last step has vanishing residue (obvious before setting y(q) = −x(ιq)). It is added in order to extend the n-summation to any n ≥ 0, giving with (2.7) for q → ιu k , z → ιq and again (1.1) (2.12) The first two lines of (2.11) tell us that Here the inverse d −1 u of the exterior differential of a 1-form ω is its primitive, . Inserted into (2.12) we confirm with I ∪ u k = I 1 and symmetrisation in I 1 , I 0 the assertion.
2.3. Symmetry of the involution identity I: q → ιu k and q → u k . We consider the ι-reflection of (1.3), ω 0,|I|+1 (I, q) + ω 0,|I|+1 (I, ιq) (2.14) where (1.1) is used. We show that the rhs has the same pole at q = u k as the original equation (1.3), i.e. that the solution in Proposition 2.5 satisfies This is the same as where (2.2) has been used. Fixing u k ∈ I 1 gives a factor s. We write ω 0,|I 1 |+1 (I 1 , ιz) = d u k (d −1 u k ω 0,|I 1 |+1 (I 1 , ιz)), move d u k in front of the residues and ignore it below. Then we expand the denominator about x(z) = x(u k ): We will show that already the second line vanishes for every p ≥ 1. For p = 1 the equation to prove reduces to Res z→u k d −1 u k ω 0,|I|+1 (I, ιz) = 0, which is true by Lemma 2.4 (only higher order poles at z = ιu k ). Next for p = 2 we need to show ) . (2.17) Indeed by Proposition 2.5 the term in braces { } has at most a first-order pole at z = u k , which is removed by a prefactor (x(z) − x(u k )) n for any n ≥ 1. Hence (2.17) is true. Next for p = 3 we have to show By the argument employed to prove (2.17) this reduces to . (2.18) The sum in the first line will include the cases I 2 = I 3 and I 2 = I 2 . Using again the argument based on Proposition 2.5, in the first case d −1 u k ω 0,|I 1 |+1 (I 1 , ιz) + has at most a first-order pole at z = u k .

Multiplying this sum by
gives a regular term without residue. The same is true for I 2 ↔ I 3 . This proves (2.18). The same argument together with Pascal's triangle structure eventually shows that the second line of (2.16) vanishes identically for any p ≥ 1. In conclusion, (2.15) is proved, which means that the rhs of (1.3), minus its reflection q → ιq, is holomorphic at every q = ιu k (and then also at q = u k ).

Linear loop equation.
Let σ i be the local Galois involution defined in a neighbourhood of the ramification point β i . It satisfies x(z) = x(σ i (z)), σ i (z) = z for z = β i and lim z→β i σ i (z) = β i .
2.5. The recursion kernel. We start from Lemma 2.3 for q → ιq where (1.1) is taken into account: We introduce W a,s,s (I; q) := B a,a ,s,s (I; q, z) := (2.20) These are functions of q and 1-forms in every variable in I, and B a,a ,s,s (I; q, z) is also a 1-form in z. A lengthy calculation gives the following important tool: Lemma 2.8. Residues of W a,s,s satisfy for 0 < a ≤ s Res for any function f a,s,s meromorphic in a neighbourhood of β i .
Proof. We consider for a fixed partition I 0 I 1 ... I s I 1 ... I s = I and some 0 < a ≤ s the residue We have used (2.3) and the fact that the integrand is regular at z = β i . The residue at z = σ i (q) in the last line can be evaluated immediately and gives rise to the function − for which we insert (2.19) at q → σ i (q): We process the last two lines (*) in the same manner, i.e. commute the two residues according to (2.3). There is again no contribution of a residue at z = β i , but now an additional residue at z = q arises. The resulting term ω 0,|I j |+1 (I j , σ i (q)) (y(σ i (q)) − y(q)) a+a (dx(q)) s (dx(σ i (q))) s This is inserted back into the equation we started with. We sum over all partitions I 0 I 1 ... I s I 1 ... I s = I for fixed s, s , a and express the result in terms of W, B introduced in (2.20). The result is (2.21).
Lemma 2.8 is our main tool to evaluate the polar part of (2.19) at q = β i . Taking condition (d) of Theorem 1.2 into account, we need to evaluate In a first (also very lengthy) step we show: Lemma 2.9.

= Res
Proof. We express |I|−1 s=1 Res q→β i dx(q)dw w−q W s,s,0 (I; q) via (2.21) at s = 0, a = s and f a,s,s ≡ 1. In the third line of (2.21), the case a = 1 of the second term cancels, when summing over s, every first term except for the single term with s = 1 and s = 0. This surviving term with s = a = 1 is the last term in the first line of (2.22). When subtracting the second line of (2.22), the term with s = 0 in (2.22) cancels directly, and then the term with s = 1 (and any s ≥ 1) cancels after reordering partial fractions. After renaming the parameters we arrive at In the second term of the last line we apply repeatedly the identity B a,a ,s,s (I; z, q) to express as linear combination of B a,0,s,s (I; q, z) and B 0,a ,s,s (I; q, z). The coefficient of B a,0,s,s (I; q, z) in this expansion is the number of paths made of steps up or right from (a, 0) to (s − 1, s ) with a first step right. This is the same as the number s+s −a−2 s −1 of words of s − 1 letters R and s − 1 − a letters U . Similarly, the coefficient of B 0,a ,s,s (I; q, z) in this expansion is the number of up-right paths from (0, a ) to (s − 1, s ) with a first step up. This is the same as the number s+s −a −2 s−2 of words of s − 2 letters U and s − a letters R. A right step comes with a factor (−1). We thus get The term with a = s cancels the first term of the last line of (2.23) so that we end up in the following equation in which k ≡ 0: (2.24) Next we process the line ( ‡) of (2.24) via (2.21). With the exception of one term the 'hockey-stick identity' s−k−1 a=1 s−a−1 k = s−1 k+1 occurs: (2.25) The following steps are performed: and the corresponding adjustments of the ranges for s, s we find that the first two lines ( †, ‡) of (2.24), where k = 0, equal the same two lines (2.24) †+ ‡ with k = 1, plus the iterated residue in the last three lines of (2.25), first for k = 0. Iterating this procedure until s ≥ 2 + k and s ≥ 1 + k becomes incompatible with the size |I| gives for the first two lines of (2.24) the identity Now we change the summation order and sum first over k. With Therefore, (2.24) and hence the rhs of (2.22) are identically zero.
We will prove by induction that the second line of (2.22) vanishes identically. This requires a rearrangement of the forms B. To simplify notation we introduce the split operator Sω 0,|I|+1 (I, q) := with Sω 0,2 (u, q) = 0. Then (2.22) can be written with (2.19) as .
We claim that this expression can be reordered intõ .
A p-fold product is then of the form .
2.6. Symmetry of the involution identity II: q → β i and q → ιβ i . Recall that the investigation of ω 0,|I|+1 (I, q) for q near a ramification point β i of x in Sections 2.4 and 2.5 started from the ι-reflection (2.14) of the involution identity (1.3). It thus remains to prove that the obtained solution is consistent with the original equation (1.3). This means we have to show This is the same as where (2.2) has been used. We expand 1 (y(q)−y(z)) s about y(z) = y(β i ) and then order into powers of y(β i ). Hence (2.31) holds iff We prove that the last line vanishes identically for any n = p − k: Proposition 2.11. For any family ω 0,|I|+1 (I, z) of 1-forms in z which satisfy the linear and quadratic loop equations 3 [BEO15, BS17] one has, for any n ≥ 1, In particular, (2.31) holds under these assumptions.
Proof. In Remark 2.12 below we indicate that the assertion would be an immediate consequence of existence of a loop insertion operator. Because we did not prove that such an operator exists in our case we give a direct combinatorial proof based on a technical Lemma A.1. We associate to the complex numbers y,ȳ, w,w in Lemma A.1 the functions (and forms in . We keep only those terms which give rise to an admissible partition of I (restrict to admissible products of t I k , then set t I k → 1).
Lemma A.1 together with e 2 := yw +ȳw + ww = y(w +w) + (ȳ − y)(w + ww y−y ) gives where Sω was defined in (2.26) and D n is a set of tuples specified in (A.3). The linear loop equation Proposition 2.6 and Remark 2.7 imply that is holomorphic at z = β i . Holomorphicity of the last line at z = β i follows from Proposition 2.10. Thus, (2.33) is regular at z = β i . The projection to admissible partitions of I guarantees that contributions to (2.33) with n − k > |I| are automatically zero.
We finish the proof of the proposition with the fact (2.5) that for any meromorphic 1-form ω(q) the residue does not change under the Galois involution, Since the residue of (2.33) vanishes 4 , this implies the assertion (2.32).
Remark 2.12. For topological recursion, the existence of a loop insertion operator D w is proved [EO07]. It is unclear whether the same holds for blobbed topological recursion in general as well. However, assuming that a loop insertion operator D w exists 5 for any blobbed topological recursion, we could prove (2.32) by induction in |I| → |I| + 1 with the following consideration: We have used that the sum I 1 ... Is=I w should be symmetric such that all terms with the form ω 0,2 (w, z) coming from the second line get a symmetry factor 1 s+1 so that n s n−s s+1 = n s+1 . 2.7. Finishing the proof of Theorem 1.2. We can now assemble the pieces into a proof of Theorem 1.2. In a first step we assume that the rhs of (1.3) and of (2.14) are the same. By induction these rhs have poles in the points q ∈ {β i , ιβ i , u k , ιu k }. Then conditions (b),(c),(d) imply that ω 0,|I|+1 (I, q) is meromorphic onĈ with poles only in q ∈ {β i , ιu k }. Therefore, Res q→ιu k ω 0,m+1 (u 1 , .., u m , q)dz z − q is a holomorphic 1-form on the Riemann sphereĈ z, hence identically zero. Inserting the residues from Proposition 2.5 and Proposition 2.10 represents ω 0,|I|+1 (I, z) as (1.4)+(1.5).
It remains to prove that the difference between the rhs of (1.3) and (2.14) is a holomorphic form onĈ q, hence zero. By induction it can have poles at most in q ∈ {β i , ιβ i , u k , ιu k }. In Section 2.3 we have shown that the difference is holomorphic at every q = u k and q = ιu k , and in Section 2.6 it is shown that the difference is holomorphic at every q = β i and q = ιβ i . This completes the proof of Theorem 1.2.
2.8. A particular symmetry under the involution ι. In the second part we prove that the planar sector (genus 0) of the quartic Kontsevich model is completely governed by the involution identity (1.3). In a decisive step of the proof we will need an intriguing symmetry resulting from (1.3) alone: (2.35) The residues in (2.35) can be expressed as limits of partial derivatives of with respect to x(z) and x(ιz). Using (2.6) we thus bring (2.35) into an equation that we need: (2.36) The main combinatorial tool to verify (2.35) is Conjecture A.2. Using Conjecture (A.2) we prove that the integrand in (2.35) is an exact 1-form in z: (2.37) In particular, the residue (2.35) at z = q is zero.
Proof. In the first line of (2.37), restricted to s ≥ 2, we write 1 (y(q)−y(z)) s as a multiple differential and integrate by parts: Hence the assertion is true if the 1-form in z is identically zero. The last line of f (I; z, q) is of the form of Conjecture A.2 and a constant x q = x(q). We thus get The derivatives in the last line are expressed as a residue: The case r = 1 combines to the involution identity (1.3) and is thus identified as the negative of the first line of (2.38). In the remainder we order the partitions of I: We change the order of the summations. The outer summation is a sum over ordered partitions I 1 ... I r given by I k = ∪ j∈J k I j , which is combined with an inner summation over ordered partitions of the individual I k . Renaming in the first line of (2.39) s → r and I j → I j , we arrive at The outcome is zero thanks to (1.3).

The quartic Kontsevich model
3.1. Summary of previous results. Let H N be the real vector space of selfadjoint N × N -matrices, H N be its dual and (e kl ) be the standard matrix basis in the complexification of H N . We define a measure dµ E,λ on H N by where dµ E,0 (Φ) is a Gaußian measure with covariance for some 0 < E 1 < · · · < E N . Moments or cumulants of dµ E,λ are viewed as general or connected correlation functions in a finite-dimensional approximation of a Euclidean quantum field theory. We call the objects resulting from (3.1)+(3.2) the Quartic Kontsevich Model because of its formal analogy with the Kontsevich model [Kon92] in which Tr(Φ 4 ) in (3.1) is replaced by Tr(Φ 3 ). The Gaußian measure dµ E,0 (Φ) is the same (3.2). Kontsevich proved in [Kon92] that (3.1) with Tr(Φ 3 )-term, viewed as function of the E k , is the generating function for intersection numbers of tautological characteristic classes on the moduli space M g,n of stable complex curves.
Derivatives |kl| was solved in a special case in [PW20] and then in [GHW19] in full generality. The solution introduces a ramified covering R :Ĉ →Ĉ of the Riemann sphereĈ = C ∪ {∞} given by (see [SW19]) Here (ε k , k ) are implicitly defined as solution of the system R(ε l ) = e l , l R (ε l ) = r l when assuming that (E 1 , ..., E N ) consists of d pairwise different values e 1 , ..., e d which arise with multiplicities r 1 , ..., r d . The planar 2-point function is then given by G with poles located at z + w = 0 and z, w ∈ { ε k j } for k, j ∈ {1, ..., d}.
The system of equations established in [BHW20a] permits to determine Ω (1) 1 in [BHW20a] gave strong support for the conjecture that the meromorphic forms ω g,n (z 1 , ..., z n ) := Ω (g) n (z 1 , ..., z n )dR(z 1 ) · · · dR(z n ) obey blobbed topological recursion [BS17] for the spectral curve (x :Ĉ → C, ω 0,1 = ydx, ω 0,2 ) with In the remainder of this paper we prove this conjecture for ω 0,n . More precisely, we prove that the solution of the system of equations given in [BHW20a] is identical to the solution of the involution identity (1.3) given in Theorem 1.2 for ιz = −z and x = R as in (3.4). In particular, the part of ω 0,n with poles at ramification points of x = R obeys exactly the universal formula of topological recursion [EO07], and the other part with poles along opposite diagonals z i +z j = 0 is described by a residue formula of very similar type.
3.3. Graphical description. We introduce in Table 1 weighted functions, vertices and edges. These are connected to chains which provide a graphical description for the terms (−1) p ∂ p ∂(R(z)) pt0,|I | (I z, q|) z=−q j andt 0,|I| (I u k , q|) and its constituents. We agree that arrow tips with label p = 0 are not shown. Also the surrounding circle segment indicating the n-th derivative with respect to R(z) is not shown for n = 0. Equation (3.12) has for |I| ≥ 1 the following graphical description (we keep the order of the last three lines of (3.12)):  (3.15) The integrand of the first line of (3.9) is now iteratively obtained by distinguishing int 0,|I| (I z, q|) the cases I = ∅ from I = ∅. We describe this iteration graphically. The integrand of the first line of (3.9) is the sum of weights of chains made of initial vertex v0, subsequent vertices v1, v2, v3 and edges in between. A vertex v3 can follow v2 or another v3, whereas v1,v2 can be placed anywhere. The edge to choose is governed by the type of vertices at both ends. One multiplies the weights given in Table 1 and sums for each order of vertices over partitions of I into subsets I 1 , ..., I s , u k 1 , ...., u kr at the vertices, over the v1-labels j, l, . . . (from 1 to d, but excluding the preceding label) and over the possible exponents n, p of the edges e1 p ,e3 p and e6 n . These exponents are not arbitrary; we discuss later their pattern.

Examples.
We write the first iteration in full details: The necessary sum over partitions of I and over ranges of labels j are obvious from the vertex labels. We therefore employ from now on a simplified notation were these summations are omitted. This means that instead of (3.16) we simply write For |I| = 2 only the first two chains contribute. The next iteration reads in simplified notation For |I| = 3 only the first three lines of the rhs are relevant. We give another iteration, but stop it at |I| = 4: 3.5. Cancellations between chains. It will be convenient to collect subchains of consecutive vertices v1 with the same upper label j:  A v1-block can terminate a chain iff the deficit is 0. A v1-block can be followed by an edge e1 p or e3 p ; the label p of such an edge is then given by the deficit p = s(n) − |n| of the v1-block before it. Since a v1-block is formed by repeatedly attaching a function f2 p labelled p ≥ 0, the condition on the deficit must hold at all intermediate steps. This amounts to a condition r i=1 n i ≤ r on any partial sum. For blocks of total deficit 0 (those which terminate a chain or are followed by edges e1 0 or e3 0 , this is equivalent to the opposite condition r i=0 n s−i > r for 0 ≤ r ≤ s − 1 and s i=0 n s−i = s when prepending n 0 = 0. This means that the reversely ordered tupleñ := (n s , n s−1 , ..., n 1 , 0) is a Catalan tuple in the sense of [dJHW19]. See Definition A.6. We consider the subset of chains which differ only in the degrees n of a v1-block of size s, but otherwise have identically labelled vertices. In this subset any degree n of the v1-block compatible with the deficit condition is produced, and precisely once. . . u I s starting with a vertex v2 of lower label u and several consecutive vertices v3 with the same inner label u, connected by edges e5. We let u be the label, s be the size and (I 1 , ..., I s ) be the partition distribution of a v2-block. A v2-block of size 0 is identified with a vertex v2 with lower label u.
If several v2-blocks arise in a chain then its labels u k , u l , . . . are necessarily different.
We will prove that, after taking cancellations into account, also the labels j 1 , j 2 , . . . of v1-blocks in the surviving chains are pairwise different. These cancellations start with chains of 4 vertices: 0 which follows from the weights in Table 1 and with ∇ 0 . These identities reduce the set of graphs to a much simpler subset: Lemma 3.5. Let M be the set of chains generated by the loop equations for Then cancellations between weights remove all chains with edges e1 p and e3 p having a tip labelled p ≥ 1 and all chains with two or more identically labelled v1-blocks. The subset of surviving graphs is given by the set of chains made of v2-blocks and of v1-blocks which have deficit 0 and pairwise different labels, connected by appropriate edges without tip.
Proof. Consider a v1-block of label j, partition distribution I and degree n = (n 1 , n 2 , ..., n s ) with deficit p = s − n 1 − ... − n s ≥ 1. Its reverse degree (n s , n s−1 , ..., n 1 , 0) cannot be a Catalan tuple (see Definition A.6) for p ≥ 1. This means that either n s = 0, or there is a unique 2 ≤ t ≤ s such that (n s , n s−1 , ..., n t , 0) is a Catalan tuple but (n s , n s−1 , ..., n t , n t−1 , 0) is not. This necessarily means n t−1 = 0. We define a unique splitting into two v1-blocks of degrees n − and n + : By construction, n + has deficit 0 so that it can terminate a chain or is followed by edges e1 0 or e3 0 . The other label n − has deficit p − 1 and is followed by edges e1 p−1 or e3 p−1 . Conversely, two degrees n − of deficit p − 1 ≥ 0 and n + of deficit 0 can be joined to a unique degree n of deficit p. The weights given in Table 1 together with confirm the following identity: where the shaded circle stands for any identical subchain in both chains. The same cancellation arises if the v1-block labelled j 1 is replaced by a v2-block and e1 p by e3 p . Next for chains which extend by further blocks to the right, all with labels = j, we have 0 =  Again the shaded circle stands for any identical subchain. The same cancellation arises if any subset of v1-blocks (other than the one labelled j) is replaced by corresponding v2-blocks. After these preparations we prove that (3.17) and (3.18) provide the claimed reduction in the set of chains describing I 1 I 2 =I 0,|I 1 |+1 (I 1 ; q)v 0,|I 2 | (I 2 q).
(1) We start with the type of chains indicated by the left graph in (3.17), with p ≥ 1. Since the splitting of n into n − , n + is unique, it cancels against a unique chain indicated on the right of (3.17). Conversely, for any chain K terminating in a triple consisting of two v1-blocks of the same label j and any other block in between, there is a unique chain indicated on the left of (3.17) against which K cancels. As result we remove all chains with a single block after the last e1 p or e3 p edge (with p ≥ 1) and all those chains which terminate in a triple of blocks in which two v1-blocks are equally labelled.
(2) We pass to (3.18) for r = 2. The chain in the first line is only present for j 2 = j because the case j 2 = j was removed in step (1). According to (3.18) the chain K indicated in the first line cancels against two uniquely determined chains terminating in a quadruple of blocks two of which are labelled j, and conversely. After all we remove all chains with two blocks after the last e1 p or e3 p edge (with p ≥ 1) and all those chains terminating in a v1-block labelled j which is followed by three more blocks one of them also labelled j. (r) Continuing in this manner removes all chains with an e1 p or e3 p edge with p ≥ 1 and all chains with two or more identically labelled v1-blocks.
We are left with chains in which all blocks have different labels and are connected by edges e1 0 ,e3 0 , i.e. without tip.
All surviving v1-blocks have degrees of deficit 0, i.e. are reversals of Catalan tuples. In the next step we collect all v1-blocks which have the same union of their partition distribution (and deficit 0) to a v1-group: We have used that the leftmost vertex of every v1-block has weight 0,|I 0 |+1 (I 0 ; −q j ) = (−dR(q j ))∇ 0 0,|I 0 |+1 (I 0 ; −q j ). By C s we denote the set of Catalan tuples of length s, see Definition A.6.
Similarly we collect v2-blocks with the same union of their partition distribution to a v2-group: The summation .. Is=I is left out for |I| = ∅. For I = ∅ there is no contribution from s = 0. We summarise the previous simplifications and collections: Corollary 3.6. The integrand I 1 I 2 =I 0,|I 1 |+1 (I 1 ; q)v 0,|I 2 | (I 2 q) in the first line of (3.9) is the sum of weights of all different chains which meet the criteria: • The leftmost vertex is v0 with weight − 0,|I 1 |+1 (I 1 ; q).
• Any other vertex is a v1-group with weight (3.19) or a v2-group with weight (3.20). The labels j i of the v1-groups are pairwise different. • The union of all subsets I i at the initial vertex, the v1-groups and the v2-groups, together with the labels u k of the v2-groups, is I = {u 1 , ..., u m }. • The edges between the groups (and initial vertex) are given by e1 0 ,e2,e3 0 ,e4 depending on the groups they connect. Their weights are given in Table 1.
Proposition 3.8. Letq j i = σ i (q) be the preimage of q which corresponds to the local Galois involution near β i . Then for all I = {u 1 , ..., u m } with m ≥ 2 one has The application of d u 1 · · · d u k agrees with the restriction of (1.4) to poles at z = β i .
Proof. In the graphical representation of Corollary 3.6, the assertion amounts to (3.22) The rhs is one of the chains contributing to the residue at q = β i in the first line of (3.9). We have to prove that all other chains described in Corollary 3.6 sum up to expressions regular at q = β i .
We prove this regularity by induction on the length of chains (with v1/v2groups as vertices). Byj we denote a label different from j i . There are two remaining chains of length 2, namely 0 (in the first chain summation overj = j i and over partitions I 1 I 2 = I, in the second chain summation over partitions I 1 u I 2 = I). Edges, v1-groups with labelj and v2-groups are regular at q = β i . The initial vertex v0 is regular for |I 1 | = 1 so that these chains only contribute to P i q for |I 1 | ≥ 2. In that case we can, up to terms holomorphic at β i , replace the initial vertex by (3.22) for I → I 1 , which is true by induction hypothesis. We thus have This identity removes all chains of length 3 with a v1-group labelled j i at any position. There remain only the chains of length 3 without v1-group labelled j i . For |I 1 | = 1 these are holomorphic at q = β i and can be discarded in the projection P i q . The only poles come from initial v0-vertices with |I 1 | ≥ 2 multiplied by regular expressions. We can thus use (3.22) for I → I 1 again and express by the same mechanism as (3.23) the survived length-3 chains as −P i q of all length-4 chains which have a v1-group labelled j i at any position. These cancel in the graphical representation. Since the v1-group labelled j i can occur only once by Lemma 3.5, only the length-4 chains without any v1-group labelled j i survive the cancellation.
We repeat this procedure up to chains of length |I|. The surviving ones have an initial v0-vertex and otherwise v1/v2-groups with other labels than j i . Now because the initial v0-vertex necessarily has |I 1 | = 1, it is also regular at q = β i . Therefore, all chains survived up to this point project with P i q to 0. 3.8. Poles of 0,|I|+1 (I; z) at z = −u k . We prove in Section 4: We can thus focus on poles of second or higher order captured by the projection for some 1-form ω(u k , z) in z (which may depend on further variables). We prove: Proposition 3.10. Let I = {u 1 , ..., u m } with m ≥ 2. The projection H k q of 0,|I|+1 (I; q) is recursively given by in the graphical description or explicitly by
Proof. Since the second line of (3.9) only has a first-order pole at z = −u k , the projection of (3.9) to poles of higher order reads The rhs of (3.25) is one of the chains contributing to the rhs of (3.27). We prove by induction on the chain length (with v1/v2-groups as vertices) that all other chains sum to expressions which at q = −u k are holomorphic or have at most a first-order pole. Byū we denote any u l = u k . At length 2 we have in addition to the rhs of (3.25) the chains 0 I 1 j I 2 + 0 I 1 I 2 ;ū . In the case u k ∈ I 2 these chains are holomorphic at q = −u k and can be discarded under H k q . Remains u k ∈ I 1 . If I 1 = {u k }, then the initial vertex v0 has weight − 0,2 (u k ; q) = dq u k −q + dq u k +q and thus only a first-order pole at q = −u k which does not contribute to H k q . The only contributions are thus from u k ∈ I 1 with |I 1 | ≥ 2. Here we can use the induction hypothesis (3.25) for I → I 1 → I 2 → I 3 so that This identity removes all chains of length 3 with a v2-group labelled u k at any position. The remaining length-3 chains have edges and v1/v2-groups which are holomorphic at = −u k . Poles arise only if u k ∈ I 1 in the initial vertex, and poles of second and higher order require |I 1 | ≥ 2. Here the induction hypothesis is available, so that the same mechanism removes all chains of length 4 with a v2group labelled u k . We repeat this construction until the initial vertex necessarily has |I 1 | = 1 and also projects to 0 under H k q . This finishes the proof of (3.26). As discussed in the remark directly after Proposition 3.10, the proof of Theorem 3.2 is complete provided that Assumption 3.9 holds.
4. Proof of Assumption 3.9 4.1. The residue. The recursion formula (3.9) generates, a priori, also a firstorder pole at z = −u k with residue Our goal is to prove Assumption 3.9, i.e. that the residue (4.1) vanishes for |I| ≥ 2.
Of particular importance will be the functions These arise as follows: Lemma 4.1. Let I = {u 1 , ..., u m } for m ≥ 2. Suppose Assumption 3.9 holds for 0,|I |+1 with u k ∈ I and 2 ≤ |I | < |I| (there is no condition for m = 2). Then Res Proof. We evaluate the residue on the rhs of (4.1). In the graphical representation we have a contribution from the chain (in which I = ∅ is allowed; the weight of the v2-group is given in (3.20)) where the definition (2.6) for ω → at q → −u k and z → q has been used. This provides the third term on the rhs of (4.3). The first term is copied from (4.1). We investigate the residues at q = −u k of all other chains. The remaining chains of length 2 (with v1/v2-groups as vertices) with residue at q = −u k are the same as in the first line of (3.28) with u k ∈ I 1 . Two cases are to distinguish. For I 1 = u k we have a purely first-order pole and Amputation of the initial v0-vertex gives the chains contributing to −v 0,|I | (I q). Hence, the rhs of the above equation is the restriction of −v 0,|I|−1 (I\u k −u k ) to chains of length 1. The other case is u k ∈ I 1 but |I 1 | ≥ 2. Since |I 1 | < |I|, Assumption 3.9 holds for the initial vertex − |I 1 |+1 (I 1 ; q) whose poles at q = −u k are thus of purely higher order. They are thus given by −H k q 0,|I 1 |+1 (I 1 ; q) , which can be expressed by (3.25): In the step from the second to third line we have used that only the whole projection H k q + dq q+u k Res q→−u k to the principal part of a Laurent series, but not H k q alone, is the identity operator under the residue. According to (4.4), the second line from below equals −∆ 0,|I 0 |+|I 1 |+1 (I 0 ∪ I 1 ; −u k ) times the restriction of −v 0,|I 3 | (I 3 −u k ) to chains of length 1, here with I 0 I 1 I 3 = I\u k . The last line removes from the residue all chains of length 3 with a v2-group labelled u k .
Thus only those length-3 chains for which u k ∈ I 1 is located at the initial vertex v0 contribute to the remaining residue. Again the case I 1 = u k produces the restriction of −v 0,|I|−1 (I\u k −u k ) to chains of length 2. For |I 1 | ≥ 2 we use Assumption 3.9 that − 0,|I 1 |+1 (I 1 ; q) has at q = −u k a pole of second or higher order given by (3.25). The same argument as before produces on one hand (4.4) times the restriction of −v 0,|I 3 | (I 3 ; −u k ) to chains of length 2, on the other hand removes from the residue all chains of length 4 with a v2-group labelled u k . Continuing this strategy until I 1 = u k is the only choice shows that the residue of all chains other than (4.4) evaluates to −v 0,|I|−1 (I\u k −u k ) plus (4.4) times −v 0,|I | (I −u k ), summed over partitions of I\u k . Assumption 3.9 for 0,|I|+1 , that the rhs of (4.3) evaluates to 0, is thus a condition on 0,|I |+1 or ω 0,|I |+1 for |I | < I. Here Theorem 3.2 is the induction hypothesis. Its proof is complete (following the previous considerations) if Theorem 3.2 implies (4.5) We are going to prove that the rhs of (4.5) is an entire holomorphic function on C, i.e. a constant, equal to its value 0 for q → ∞. This implies (4.5).
We start to discuss absence of poles at q = ±β i . Recall that (4.5) equals Res z→q 0,|I|+2 (I, −q; z). The projection of (4.5) to poles at q = β i is thus given by 2) into account. The final term gives zero because non of the chains contributing to 0,|I|+2 (I, −q; z) has a pole at −q = −β i . The other term is also zero because 0,|I|+2 (I, −q; z) has due to the kernel K i (z, q) in Theorem 3.2 at z = β i poles of purely higher order, without residue. We have established this fact in Proposition 3.8 without relying on Assumption 3.9. In summary, (4.5) is regular at q = β i . We will show in Subsection 4.2 that (4.5) is antisymmetric under q → −q. This means that (4.5) is also regular at q = −β i .
The same simple argument cannot be used to prove that (4.5) is regular at q = ±u k because this would need Assumption 3.9. We therefore give in Subsection 4.3 a direct proof which uses the antisymmetry of (4.5).
In principle, the functions v 0,|I| (I ±q) may (and do) have poles at the other Recalling that (4.5) equals Res z→q 0,|I|+2 (I, −q; z), the projection of (4.5) to a pole at such q = v is The first term in the last line is trivially zero, but the second term can indeed have a pole at −q = u k j coming from the edge e4 in Table 1. An edge with these labels can only occur once in a chain so that it is a first-order pole. Its residue Res q→v 0,|I|+2 (I,−q;z)dq w−q is a 1-form in z from which we take the residue at the same z = − u k j . But there are no such poles so that (4.5) is regular at any

4.2.
A necessary condition. Adding (4.5) and its copy for q → −q shows that necessary for (4.5) to be true is the identity (we apply d u 1 · · · d um to pass to ω) 0 = ∆ω 0,|I|+1 (I, q) + ∆ω 0,|I|+1 (I, −q) − I 1 I 2 =I ∆ω 0,|I 1 |+1 (I 1 , q)∆ω 0,|I 2 |+1 (I 2 , −q) . This logarithm can also be represented as follows: Proposition 4.2. and insert it into the product in (4.2). This shows that products of ∆ω 0,|I j |+1 (I j , q) expand into products of ∇ n j ω 0,|I j |+1 (I j , q) with the given condition on the sum of n j . That the prefactor reduces to 1 s is, however, by no means obvious. The first step of the proof is Lemma 4.3 below, which relies on a combinatorial Conjecture A.9 in the appendix. Then a discussion given after the proof of Lemma 4.3 completes the proof. It relies on the same Conjecture A.9. ∇ n j ω 0,|I j |+1 (I i , q) (4.10) in which C 1 n 1 ...ns is symmetric in all its arguments. To determine C 1 n 1 ...ns we can consider a subsector of the n j -summations where n 1 , ..., n p > 0 for some p and n p+1 = ... = n s = 0. Other sectors are then obtained by symmetry. We will count the contributions from (4.2) which contribute to C 1 n 1 ...np0...0 for given positive integers n 1 , ..., n p (which are followed by n 1 + ... + n p − p zeros). In a first step we show that the number of these contributions is C 1 s0...0 = (s − 1)! (which is clear) and for p ≥ 2 given by The number C 1 n 1 ...np0...0 is the sum over all j with n j ≥ 2 of specially ordered contributions from . (4.12) The factors are expressed via (4.9), but only contributions compatible with n p+1 = ... = n s = 0 are retained. The positive n 1 , ..., n p , excluding n j , arise from the part of (4.9) in which all factors ∇ 0 ω have a larger order than any ∇ r ω with r > 0. In particular, contributions ∇ r ω with r > 0 only arise from every of the first factors in (4.12), for some with 1 ≤ < p − 1 to sum over. From the last n j − 1 − factors in (4.12) we only take the special term ∇ 0 ω 0,|Ĩ i |+1 (Ĩ i , q).
In the fourth line we have used the binomial theorem for the falling factorial. The final line is obvious. For a general order of the n i we thus have C 1 n 1 ...ns = (s − 2)!#(n j = 0), where #(n j = 0) is the number of n j which equal zero. Relaxing the condition that the I j are ordered amounts to an additional factor 1 s! . This is the assertion. Lemma 4.3 is the starting point to evaluate the sum over r in the first line of (4.8). It is clear that this sum has a similar expansion as (4.10): − log(1 − ∆ω 0,|.|+1 ( . , q)) (I) (4.13) = I s=1 I 1 ... Is=I I 1 <...<Is n 1 +...+ns=s where C n 1 ...ns is symmetric. We first show that for an order n 1 , ..., n p ≥ 1 and n p+1 = ... = n s = 0 it is given by C n 1 ...np0...0 (4.14) The factor (r − 1)! combines the step from any partitions I 1 ... I r = I into r! ordered ones with the prefactor 1 r in (4.8). The subset J i corresponds to I k where the I k are taken from I p+1 , ..., I n 1 +...+np . There are (n 1 +...+np−p)! r i=1 (|n| J i −|J i |)! different distributions of these I k , which explains the corresponding factor above. The numerator (|n| J i − 2)!(|n| J i − |J i |) is the weight of C 1 n J i 0....0 found in Lemma 4.3. We write (4.14) in terms of rising factorials and insert the corollary (A.11) of Conjecture A.9, where x j → n j − 1 and → r: We have used (r − 1)! = 1 r−1 and then applied the binomial theorem. By symmetry we thus have C n 1 ...ns = (s − 1)! for any partition n 1 + ... + n s = s. Relaxing the condition that the I j are ordered has to be corrected with an additional factor 1 s! . This completes the proof of Proposition 4.2, provided that Conjecture A.9 is true.
Proposition 4.2 together with (4.7) give as necessary condition for (4.5) to be true the equality (2.36) which we have proved in Section 2.8. 4.3. Absence of poles of (4.5) at q = −u k . The function v 0,|I| (I q) is holomorphic at u k = q and has poles at u k = −q which exclusively come from v2-groups with label u k . There are two possibilities. Either this v2-group is the single vertex of a length-1 chain − 0 I ; u k I=I u k , or it is part of a chain of larger length. In the second case it can be collectively taken out of all other chains, the remnant is just another copy of −v 0,|I| (I q) of smaller length with We recall that indicates that for |I | = 0 the sum is omitted, whereas for |I | > 0 the case s = 0 is left out. The following lemma (which we formulate for q → −q) characterises the polar part of the second line at q = −u k : Proof. The lhs of (4.16) can be rewritten with (2.6) as Up to O((q − u k ) 0 )-contributions it coincides with its projection to the polar part in which we change with (2.2) the order of residues: We have used that only ω 0,2 has a pole at z = u k which is given by (1.2).
With these preparations we control the polar part of (4.5) at q = ±u k : Proposition 4.5. Holomorphicity of (4.5) at q = −u k is a consequence of (2.36) proved in Proposition 2.13.
Proof. The first term v 0,|I| (I −q) in (4.5) is holomorphic at q = −u k . In the sum over I 1 I 2 = I we distinguish u k ∈ I 1 from u k ∈ I 2 . The function v 0,|I| (I q) is for u k ∈ I written as (4.15) with Lemma 4.4 used for the rhs. We thus find (4.5) = −A(I; q) +

Its iterative solution is
The factor 1 s arises by symmetrisation when dropping the condition u k ∈ I 1 . The consistency condition (2.36) of (4.5) together with Proposition 4.2 thus imply A(I; q) ≡ 0, which gives the assertion.
As result we have proved that (4.5) does not have any poles onĈ, it is thus a constant equal to its value 0 at q = ∞. This means that Assumption 3.9 is true and the proof of Theorem 3.2 complete.

Conclusion and outlook
We have proved for genus g = 0 the main conjecture of [BHW20a] that meromorphic forms ω g,n which naturally appear in the quartic analogue of the Kontsevich model follow blobbed topological recursion [BS17]. This makes the quartic Kontsevich model part of the growing family of structures in mathematics and physics governed by topological recursion [EO07,Eyn16]. Other examples include the combinatorics of the Kontsevich model [Kon92], the one-and two matrix models [CEO06], Hurwitz theory [BMn08], Gromov-Witten theory [BKMnP09], Weil-Petersson volumes of moduli spaces of hyperbolic Riemann surfaces [Mir06] and many more.
We consider as most important result of this paper the discovery that the quartic Kontsevich model is completely characterised by the behaviour of its objects ω g,n under the global involution ιz = −z. We showed how a single equation (1.3), To answer the first two of these critical questions we had to prove that the naïve solution is equivalent to the solution (1.4)+(1.5) given in Theorem 1.2. The generated material also allowed to affirm question (c), where a difficulty was to show that all poles are of purely higher order. This property is a consequence of a hidden symmetry (2.35) resulting from (1.3) alone. As result we have established for genus g = 0 a precise connection between (b) and (c), which was conjectured in [BHW20a]. But the statement is more general: in [BHW20a] it is shown that the poles of ω g≥1,n (z 1 , ..., z n ) are located, besides ramification points of x and diagonals z k = ιz l , at the fixed points z k = ιz k of the involution. The next step in our programme will be to extend (1.3) to higher genus. This should help to answer the exciting question whether the intersection numbers [BS17] encoded in the quartic Kontsevich model capture geometric information about a moduli space of curves equipped with an involution.

Appendix A. Combinatorial tools
We collect here combinatorial results which we need in the proof of Theorem 1.2 and Theorem 3.2. This includes two conjectures A.2 and A.9 for which we have not found a proof yet.
Shifting summation indices to n 1 = a+k−t, n 2 = b+k−t, n 4 = c+k+k−n 1 −n 2 = c+t+t−a−b leads to [y kȳk w twt ](A.4) =t  The second sum is over the S(n, r) (Stirling number of the second kind) partitions of a set of n objects into r parts.
Remark A.4. Trying to prove Conjecture A.2 by induction in n by sending c 1 (y) → c 0 (y)c 1 (y) and acting once more with b(y)∂ x + ∂ y leads for all ∂ r x a(x) to a simpler combinatorial relation, which consists of a sum of only two sets. This is achieved by factorising all the rest out. Let J be a finite index set with 0, 1 / ∈ J, then the simpler relation is given by where ∂ 0 y f (y) = f (y) and j∈∅ c j = 1. Therefore, Conjecture A.2 holds if (A.8) is true.
The set of Catalan tuples of length |ñ| := k is denoted by C k .
The cardinality of C k is the k th Catalan number OEIS A000108.
We prove a variant of the multinomial theorem 6 : Lemma A.7. Let T n be the set of rooted plane trees with n + 1 vertices. Let k i (t) be the number of children of the i-th vertex of t ∈ T n , for any chosen vertex order. Then for any x 0 , ..., x n in some commutative ring one has Alternatively, T n can be replaced by the set C n of Catalan tuples t = (k 0 (t), k 1 (t), ...., k n (t)).
Proof. For a partition n = s 0 + s 1 + ... + s n into integers s i ≥ 0, consider D s 0 ,s 1 ,...,sn := ∂ n ∂x s 0 0 ∂x s 1 1 ···∂x sn n . Any such operator applied to the lhs of (A.9) gives n!. Applied to the rhs it gives the number of labelled rooted plane trees with n + 1 vertices where the vertex labelled π −1 (i) has s i children. Now view the given partition n = s 0 + s 1 + ... + s n as consisting of α 0 times 0, α 1 times 1, . . . , α n times n, with n j=0 α j = n + 1 and n j=0 jα j = n. A rooted plane tree in which α j vertices have j children, for any j = 0, ..., n, has α 0 !α 1 ! · · · α n ! different labellings. This means that D s 0 ,s 1 ,...,sn applied to the rhs gives α 0 !α 1 ! · · · α n ! times the number of different rooted plane trees in which α j of the n + 1 vertices have j children. As proved by Tutte [Tut64], the number of such trees is n! α 0 !α 1 !···αn! . 6 We thank Abdelmalek Abdesselam for decisive help with the proof. The lemma can also be proved via the matrix-tree theorem [Abd04]. Consider the matrix A ij with i, j ∈ [0..n] given by A ij = −x i for i = j and A ii = j =i x j . The determinant of its principal minor is x 0 (x 0 + ... + x n ) n−1 . The matrix-tree theorem [Abd04, Thm. 1] expresses this determinant as a sum over oriented trees. On the other hand, start from the rhs of (A.9) but with permutations π ∈ S n+1 restricted to π(0) = 0 where 0 labels the root. Forgetting the planar structure removes the k i ! in the denominator. The remaining sum is of the form [Abd04, Thm. 1]. A final symmetrisation over all π ∈ S n+1 leads to (x 0 + ... + x n ) n .