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Proof of Proposition 6.2 and Lemma 6.3

6. Matrix integrals

6.2. Proof of Proposition 6.2 and Lemma 6.3

We first show that the free energy is a function of the correlators, and then that the correlators only depend on{LNi (x`)}`>0,16i6d andτBN. Finally, we deduce the largeN expansion of the free energy as well as its smoothness.

6.2.1. The free energy in terms of the correlators Recalling the definition of free energy, (6.5), and (6.4), we have

FβN,aV(A1, ..., Ad, B1, ..., Bm) := logIβN,aV

=

a

0

d

dulogIβN,uVdu

=N2

a

0

1

NrTr⊗rV dQN,uVβ du

=N2−r

a

0

(W1NuV)⊗r(V)du+r(r−1)N2−r

a

0

W2NuV⊗(W1NuV)r−2(V)du+RN, Z

Z Z

Z Z

whereRN has terms either with two cumulants of order 2, or a cumulant of order greater or equal to 3. By Lemma6.7(note that it applies uniformly inu∈[−a0, a0], for somea0

universally small), this latter term is at most of order 1/N, and is therefore negligible.

Moreover, using Corollary6.8and Lemmas6.12and 6.13, we find that FβN,aV(A1, ..., Ad, B1, ..., Bm) =N2

a

0

f0udu+N

a

0

f1udu+

a

0

f2udu+O 1

N

, with

f0u:= (τ10uV)⊗r(V),

f1u:=rτ11uV⊗(τ10uV)⊗(r−1)(V), f2u:=r(r−1)

11uV)⊗220uV

⊗(τ10uV)⊗(r−2)(V)+rτ12uV⊗(τ10uV)⊗(r−1)(V),

(6.29)

where we have used thatV is symmetric and such thatkVkξ1is finite forξ1big enough, so thatδξ1(uV)<(1+max{2, r})−1 providedu∈[−a0, a0] witha0 sufficiently small. In particular this implies that, fora0small enough and any 1<ξ4321,

δξi(uV)<(1+max{2, r})−1 for allu∈[−a0, a0], so that the previous lemmas apply. Hence, we deduce the following result.

Lemma 6.14. Let kVkξ11 be finite for some ξ1 large enough and let ζ1>1. Then, there existsa0>0so that,fora∈[−a0, a0],and uniformly on Hermitian matrices{Ai}di=1 and {Bi}16i6m whose operator norm is bounded by 1,we have

FβN,aV(A1, ..., Ad, B1, ..., Bm) =

2

X

l=0

N2−lFla+O 1

N

, with

Fla=

a

0

fludu andflu given by (6.29).

6.2.2. The correlators as functions of{LNi }di=1 and τBN Let us define the space

P:={Q(u1a1u−11 , ..., udadu−1d , b1, ..., bk) :Q∈Chx1, ..., xd, b1, ..., bmi}.

As the functionsFla only depend on the restriction toP ofτ10uV11uV12uV, andτ20uV for u∈[−a, a], we shall first prove that the latter only depend on

MA,B:=

1 N

N

X

j=1

(aij)`:`>0 and 16i6d

∪{τBN}.

Z Z Z

Z

• The restriction τ01aV|P depends only on MA,B. We start by showing thatτ01aV can be defined inductively, as is the case when V=0, since it depends analytically on the potentialV in the following sense.

Lemma6.15. Let p∈L and V be a potential such that,for some ξ >1 and ζ>1, δξ,ζ(V)< 1

1+max{2, r}. Then, for all a∈[−1,1],the solution τ10aV of

τ⊗τ(∂ip)+a(1+1β=1)τ(Di,τV p) = 0 (6.30) is uniquely defined. Moreover,we have the decomposition

τ10aV =X

n>0

anτnV,

with τnV∈Lξ,ζ satisfying kτnVkξ,ζ6CnDn, where {Cn}n>0 denote the Catalan numbers and D is a positive constant.

Proof. This result can be seen to be a consequence of the implicit function theorem.

However we will soon need additional information on theτnV, and therefore give a proof

“by hand”.

By uniqueness of solutions, it is enough to show that there exists a solution of (6.30), or more precisely of (6.13), which is analytic ina. Let us therefore look for such a solution and writeτaV(p):=P

n>0anτnV(p). We then find thatτaV satisfies (6.13) if and only if τnV(p)+ X

k∈{0,n}

τkV⊗τn−kV (Π ¯∆p)

=−

n−1

X

k=1

τkV⊗τn−kV (Π ¯∆p)

−X

hV, q1⊗...⊗qri

d

X

i=1 r

X

`=1

X

P

iki=n−1

Y

j6=`

τkVj(qj)

τkV`(Diq`·DiD−1p)

(6.31)

As ¯∆ splits monomialspinto simple tensorsq1⊗q2each of whose factors has degree strictly smaller than that ofp, we see that there exists a unique solution to this equation.

Moreover, we prove by induction that there exists finite constantD>0 such that, ifCn

denote the Catalan numbers, then

nVkξ,ζ6CnDn.

Indeed, for n=0, we simply have the law of free variables bounded by 1, so that the result is clear. Using the inductive hypothesis until n−1 to bound the right-hand side in (6.31), and (6.18) to bound the second term in the left-hand side of (6.31), we deduce that

(1−δξ,ζ(V))kτnVkξ,ζ6 8 ξ−1Dn

n−1

X

k=1

CkCn−k+Dn−1X

|hV, q1⊗...⊗qri|

× r

X

i=1

degqi

ζPri=1degA,B(qi)ξPri=1degUqi X

Pr

i=1ki6n−1 r

Y

i=1

Cki.

Using the fact thatPn

k=0CkCn−k=Cn+164Cn, we find recursively that X

Pr

i=1ki6n−1 r

Y

i=1

Cki6Cn+r−164r−1Cn.

Thus we can bound the last term by 4r−1CnDn−1

|V|

ξ, which implies that kτnVkξ,ζ6CnDn

providedD is chosen sufficiently large. As Cn64n, this implies thatτaV=P

n>0anτnV is absolutely converging provided|a|<1/4D and it satisfies (6.30), so we get τaV01aV as desired.

We finally show thatτnV|P only depends onMA,B. Again, we can argue by induction.

As already mentioned, this is clear whenn=0 asτ0V is the law of free variables. Also, if p∈P and deg(p)=0 thenpdepends only onb1, ..., bk, and thereforeτnV(p) only depends onτBN for alln>0. Thus, by the inductive hypothesis, we can assume that the result is true forτkV(p) whenk6n−1 andp∈P, and forτnV(p) whenp∈P and deg(p)6`.

To show that this property propagates we shall use the fact that (6.31) can be seen as an induction relation where all monomials belong to P. To this end, first note that {τnV}n>0 are tracial, that is

τnV(pq) =τnV(qp) for allp, q∈ P.

Indeed this property is clear as it is satisfied byτaV, and{τnV}n>0are derivatives ofτaV with respect toa.

Next, observe thatD−1 keepsP stable. Moreover, if p=Q({uiaiu−1i }mi=1), where Q is a monomial, then

Dip= X

Q=q1xiq2

(aiu−1i q2q1ui−u−1i q2q1uiai),

so that, up to cyclic symmetry,Dip·Diq∈P for eachiandq⊂P. (Here and in the sequel, cyclic symmetry is just the action of exchangingpqintoqp.) We also show that ¯∆ maps P into P ⊗P up to cyclic symmetry. Indeed, it follows from (6.12) that, forp∈P,

∆¯ip= X

p=p1uiaiu−1i p2

X

aiu−1i p2p1ui=aiu−1i q1uiaiu−1i q2ui

aiu−1i q1ui⊗aiu−1i q2ui

− X

aiu−1i p2p1ui=aiu−1i q1uiaiu−1i q2ui

(aiu−1i q1uiai⊗q2−ai⊗p2p1−p2p1⊗ai)

− X

u−1i p2p1uiai=u−1i q1uiaiu−1i q2uiai

q1⊗aiu−1i q2uiai

+ X

u−1i p2p1uiai=u−1i q1uiaiu−1i q2uiai

u−1i q1uiai⊗u−1i q2uiai

,

so that, up to cyclic symmetry, ¯∆ip∈P ⊗P for alli∈{1, ..., m}andp∈P.

Hence, by induction we see thatτnV restricted toP only depends on the restriction of{τkV}k6n−1 toP, therefore only on the restriction of τ0V to P. Since we have already seen thatτ0V|P only depends onMA,B, the conclusion follows.

• τ11aV depends only on MA,B. A direct inspection shows that ˜∆ mapsP intoP up to cyclic symmetry. Indeed, ˜∆=P

i∆˜i with

∆˜ip= X

p=p1uiaiu−1i p2

X

aiu−1i p2p1ui=aiu−1i q1uiaiu−1i q2ui

u−1i q2uia2iu−1i q1ui

− X

aiu−1i p2p1ui=aiu−1i q1uiaiu−1i q2ui

(u−1i q2uiaiu−1i q1uiai−u−1i p1p2uiai−aiu−1i p2p1ui)

− X

u−1i p2p1uiai=u−1i q1uiaiu−1i q2uiai

aiu−1i q2uiaiu−1i q1ui

+ X

u−1i p2p1uiai=u−1i q1uiaiu−1i q2uiai

aiu−1i q2q1uiai

.

Moreover, the previous considerations showed that ΨaVτ

10 mapsP intoP forasmall, and therefore

τ11aV(p) =1β=1τ10aV( ˜∆(ΨaVτ10)−1(p))

only depends on τ10aV|P. Since we just checked that the latter only depends on MA,B, this proves the result.

• τ20aV depends only on MA,B. By Lemma6.12,

τ20aVaVτ10p, q) =−(1+1β=110aV(PqτaV

10

p),

and recalling thatτ10aV expands in a convergent series ina, we see that so doesτ20aV. We only need to check that the operators which appear in the equation defining τ20aV keep P stable. But we have already seen that both the operators ¯∆ andPV keep P stable, and henceτ20aV(p, q) only depends onMA,B and it is in fact a convergent series in such elements.

• τ12aV depends only on MA,B. By Lemma6.13, τ12aVaVτ

10p) =τ11aV( ˜∆p)−[τ20aV11aV⊗τ11aV]( ¯∆p+SaVβp),

from which we see thatτ12aV(p) is a convergent series in a(recall that we already proved thatτ10aV(p), τ11aV(p) andτ20aV(p) are convergent series ina). So the main point is to prove that, up to cyclic symmetry, ¯∆p+SaVβp∈P ⊗P wheneverp∈P.

We already proved that this is the case for ¯∆p, so we focus onSaVβp. We notice that it is the sum of two parts. One part is linear over tensors of two monomials appearing in the decomposition of aV, and as aV∈P⊗r this part clearly belongs to P⊗2. The other part is linear over tensors of one monomial appearing in the decomposition ofaV (which therefore belongs toP) andDip·Diqj withqj appearing in the decomposition of aV (which we have seen belongs to P up to cyclic symmetry). Hence also this second part satisfies the desired property, which concludes the proof.

6.2.3. Smoothness of the functionsF2, F1, and F0

By Lemma6.14and the discussion in the previous subsection, we know that

FβN,aV =

2

X

l=0

N2−lFla(LN1, ..., LNd, τBN)+O 1

N

,

where the functionalsF0a,F1aandF2adepend on{LNi }di=1and onτBN through the asymp-totic correlators{τ1guV}2g=0 andτ20uV. We finally prove that they are smooth functions of these measures.

Recall the notation introduced in (6.3). We show the following.

Lemma 6.16. There exists ξ0>1 large enough such that the following holds: let V have finite k · kξ,ζ norm for some ξ >ξ0 and ζ>1. Then there exists a0>0 such that, for all a∈[−a0, a0], Fla is Fr´echet differentiable ` times for all `∈N, and if νj= (ν1j, ..., νdj, τj)∈P([−1,1])d×T(B), we have

|D`Fla(LN1 , ..., LNd, τBN)[ν1, ..., ν`]|6C`|a| kν1kζ...kν`kζ.

Moreover,the derivativeDkF0a(LN1, ..., LNd, τBN)=DF0a(LN1, ..., LNd , τBN)[0, ...,0, δx,0, ...,0]

of F in the direction of the measure LNk is a function on the real line with finite k · kζ

norm for any k∈{1, ..., d}. As a consequence,it is of class C in an open neighborhood of [−1,1].

Proof. First, fix ξ0 sufficiently large so that all previous results apply. By the previous section it is enough to show that {τ1guV}2g=0 and τ20uV depend smoothly on ({LNi }di=1, τBN), uniformly with respect to u∈[−a, a]. Indeed, by (6.29), F0a is the in-tegral of (τ10uV)⊗r(V) over u∈[0, a]. We have seen in Lemma 6.8 that τABN 7!τuV,τ

N AB

01 is

` times Fr´echet differentiable. Moreover, we have also seen that, once restricted to P, it depends only on {LNi }di=1 and τBN, and not the full distribution τABN . As a conse-quence, the smoothness ofτuV,τ

N AB

01 as a function of τABN reduces to the smoothness as a function of the probability measures {LNi }di=1 andτBN. The fact that DF0a is C is a direct consequence of formulas (6.25) and (6.26). For instance, if we denote by Dk the derivative along LNk , and Π0k is the projection onto the algebra generated by {ak}, for anyp∈Lξ,ζ∩P we have

Dkτ0,1V,τ1[p] =−Π0k[ TτΠ+τ1Π0+PVτΠ+τβ

1Π0+ QVτΠ+τβ

1Π0](Id+ΞV

τ01V,τ1)−1p∈ P, (6.32) where we use the fact (see Lemma6.2) that

[ TτΠ+τ1Π0+PVτΠ+τβ

1Π0+ QVτΠ+τβ

1Π0](Id+ΞV

τ01V,τ1)−1(P)⊂ P

so that once we project it onA B we get only polynomials either in theai or in thebi, and hence differentiating in the direction ofLNk we only keep those inak.

The same argument holds forF1u andF0u, since alsoτ10uV11uV, andτ20uV are smooth and only depend on{LNi }di=1 andτBN.

7. Law of polynomials of random matrices