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In order to advance, we will need some standard Weingarten estimates for our quantum groups, which have their own interest, and that we will discuss now. So, consider the diagram formed by the main quantum permutation and quantum rotation groups:

[[math]] \xymatrix@R=15mm@C=15mm{ S_N^+\ar[r]&O_N^+\\ S_N\ar[r]\ar[u]&O_N\ar[u] } [[/math]]

Regarding the symmetric group [math]S_N[/math], the situation here is very simple, because we can explicitely compute the Weingarten function, and estimate it, as follows:

Proposition

For [math]S_N[/math] the Weingarten function is given by

[[math]] W_{kN}(\pi,\nu)=\sum_{\tau\leq\pi\wedge\nu}\mu(\tau,\pi)\mu(\tau,\nu)\frac{(N-|\tau|)!}{N!} [[/math]]

and satisfies the folowing estimate,

[[math]] W_{kN}(\pi,\nu)=N^{-|\pi\wedge\nu|}( \mu(\pi\wedge\nu,\pi)\mu(\pi\wedge\nu,\nu)+O(N^{-1})) [[/math]]

with [math]\mu[/math] being the Möbius function of [math]P(k)[/math].

Show Proof

The first assertion follows from the usual Weingarten formula, namely:

[[math]] \int_{S_N}v_{i_1j_1}\ldots v_{i_kj_k}=\sum_{\pi,\nu\in P(k)}\delta_\pi(i)\delta_\nu(j)W_{kN}(\pi,\nu) [[/math]]

Indeed, in this formula the integrals on the left are in fact known, from the explicit integration formula over [math]S_N[/math] that we established before, namely:

[[math]] \int_{S_N}g_{i_1j_1}\ldots g_{i_kj_k}=\begin{cases} \frac{(N-|\ker i|)!}{N!}&{\rm if}\ \ker i=\ker j\\ 0&{\rm otherwise} \end{cases} [[/math]]

But this allows the computation of the right term, via the Möbius inversion formula, explained before. As for the second assertion, this follows from the first one.

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The above result is of course something very special, coming from the fact that the Haar integration over the permutation group [math]S_N[/math], save for being just an averaging, this group being finite, is something very simple, combinatorially speaking.

Regarding now the quantum group [math]S_N^+[/math], that we are particularly interested in here, let us begin with some explicit computations. We first have the following simple and final result at [math]k=2,3[/math], directly in terms of the quantum group integrals:

Proposition

At [math]k=2,3[/math] we have the following estimate:

[[math]] \int_{S_N^+}u_{i_1j_1}\ldots u_{i_kj_k}=\begin{cases} 0&(\ker i\neq\ker j)\\ \simeq N^{-|\ker i|}&(\ker i=\ker j) \end{cases} [[/math]]

Show Proof

Since at [math]k\leq3[/math] we have [math]NC(k)=P(k)[/math], the Weingarten integration formulae for [math]S_N[/math] and [math]S_N^+[/math] coincide, and we obtain, by using the above formula for [math]S_N[/math]:

[[math]] \begin{eqnarray*} \int_{S_N^+}v_{i_1j_1}\ldots v_{i_kj_k} &=&\int_{S_N}v_{i_1j_1}\ldots v_{i_kj_k}\\ &=&\delta_{\ker i,\ker j}\frac{(N-|\ker i|)!}{N!} \end{eqnarray*} [[/math]]

Thus, we obtain the formula in the statement.

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In general now, the idea will be that of working out a “master estimate” for the Weingarten function, as above. Before starting, let us record the formulae at [math]k=2,3[/math], which will be useful later, as illustrations. At [math]k=2[/math], with indices [math]||,\sqcap[/math] as usual, and with the convention that [math]\approx[/math] means componentwise dominant term, we have:

[[math]] W_{2N}\approx\begin{pmatrix}N^{-2}&-N^{-2}\\-N^{-2}&N^{-1}\end{pmatrix} [[/math]]

At [math]k=3[/math] now, with indices [math]|||,|\sqcap,\sqcap|,\sqcap\hskip-3.2mm{\ }_|,\sqcap\hskip-0.8mm\sqcap[/math] as usual, and same meaning for [math]\approx[/math], we have:

[[math]] W_{3N}\approx\begin{pmatrix} N^{-3}&-N^{-3}&-N^{-3}&-N^{-3}&2N^{-3}\\ -N^{-3}&N^{-2}&N^{-3}&N^{-3}&-N^{-2}\\ -N^{-3}&N^{-3}&N^{-2}&N^{-3}&-N^{-2}\\ -N^{-3}&N^{-3}&N^{-3}&N^{-2}&-N^{-2}\\ 2N^{-3}&-N^{-2}&-N^{-2}&-N^{-2}&N^{-1} \end{pmatrix} [[/math]]

These formulae follow indeed from the plain formulae for the Weingarten matrix [math]W_{kN}[/math] at [math]k=2,3[/math] from and related papers, after rearranging the matrix indices as above.

Observe in particular, in the context of the above computations, that we have the following formula, which will be of interest in what follows:

[[math]] W_{3N}(|\sqcap,\sqcap|)\simeq N^{-3} [[/math]]

In order to deal now with the general case, let us start with some standard facts:

Proposition

The following happen, regarding the partitions in [math]P(k)[/math]:

• [math]|\pi|+|\nu|\leq|\pi\vee\nu|+|\pi\wedge\nu|[/math].
• [math]|\pi\vee\tau|+|\tau\vee\nu|\leq|\pi\vee\nu|+|\tau|[/math].
• [math]d(\pi,\nu)=\frac{|\pi|+|\nu|}{2}-|\pi\vee\nu|[/math] is a distance.

Show Proof

All this is well-known, the idea being as follows:

(1) This is well-known, coming from the fact that [math]P(k)[/math] is a semi-modular lattice.

(2) This follows from (1), as explained for instance in the paper .

(3) This follows from (2) above, which says that the following holds:

[[math]] \begin{eqnarray*} &&\frac{|\pi|+|\tau|}{2}-d(\pi,\tau)+\frac{|\tau|+|\nu|}{2}-d(\tau,\nu)\\ &\leq&\frac{|\pi|+|\nu|}{2}-d(\pi,\nu)+|\tau| \end{eqnarray*} [[/math]]

Thus, we obtain in this way the triangle inequality:

[[math]] d(\pi,\tau)+d(\tau,\nu)\geq d(\pi,\nu) [[/math]]

As for the other axioms for a distance, these are all clear.

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Actually in what follows we will only need (3) in the above statement. For more on this, and on the geometry and combinatorics of partitions, we refer to .

As a main result now regarding the Weingarten functions, we have:

Theorem

The Weingarten matrix [math]W_{kN}[/math] has a series expansion in [math]N^{-1}[/math],

[[math]] W_{kN}(\pi,\nu)=N^{|\pi\vee\nu|-|\pi|-|\nu|}\sum_{g=0}^\infty K_g(\pi,\nu)N^{-g} [[/math]]

where the various objects on the right are defined as follows:

• A path from [math]\pi[/math] to [math]\nu[/math] is a sequence as follows:
  [[math]] p=[\pi=\tau_0\neq\tau_1\neq\ldots\neq\tau_r=\nu] [[/math]]
• The signature of such a path is [math]+[/math] when [math]r[/math] is even, and [math]-[/math] when [math]r[/math] is odd.
• The geodesicity defect of such a path is:
  [[math]] g(p)=\sum_{i=1}^rd(\tau_{i-1},\tau_i)-d(\pi,\nu) [[/math]]
• [math]K_g[/math] counts the signed paths from [math]\pi[/math] to [math]\nu[/math], with geodesicity defect [math]g[/math].

Show Proof

We recall that the Weingarten matrix [math]W_{kN}[/math] appears as the inverse of the Gram matrix [math]G_{kN}[/math], which is given by the following formula:

[[math]] G_{kN}(\pi,\nu)=N^{|\pi\vee\nu|} [[/math]]

Now observe that the Gram matrix can be written in the following way:

[[math]] \begin{eqnarray*} G_{kN}(\pi,\nu) &=&N^{|\pi\vee\nu|}\\ &=&N^{\frac{|\pi|}{2}}N^{|\pi\vee\nu|-\frac{|\pi|+|\nu|}{2}}N^{\frac{|\nu|}{2}}\\ &=&N^{\frac{|\pi|}{2}}N^{-d(\pi,\nu)}N^{\frac{|\nu|}{2}} \end{eqnarray*} [[/math]]

This suggests considering the following diagonal matrix:

[[math]] \Delta=diag(N^{\frac{|\pi|}{2}}) [[/math]]

So, let us do this, and consider as well the following matrix:

[[math]] H(\pi,\nu)=\begin{cases} 0&(\pi=\nu)\\ N^{-d(\pi,\nu)}&(\pi\neq\nu) \end{cases} [[/math]]

In terms of these two matrices, the above formula for [math]G_{kN}[/math] simply reads:

[[math]] G_{kN}=\Delta(1+H)\Delta [[/math]]

Thus, the Weingarten matrix [math]W_{kN}[/math] is given by the following formula:

[[math]] W_{kN}=\Delta^{-1}(1+H)^{-1}\Delta^{-1} [[/math]]

In order to compute now the inverse of [math]1+H[/math], we will use the following formula:

[[math]] (1+H)^{-1}=1-H+H^2-H^3+\ldots [[/math]]

Consider indeed the set [math]P_r(\pi,\nu)[/math] of length [math]r[/math] paths between [math]\pi[/math] and [math]\nu[/math]. We have:

[[math]] \begin{eqnarray*} H^r(\pi,\nu) &=&\sum_{p\in P_r(\pi,\nu)}H(\tau_0,\tau_1)\ldots H(\tau_{r-1},\tau_r)\\ &=&\sum_{p\in P_r(\pi,\nu)}N^{-d(\pi,\nu)-g(p)} \end{eqnarray*} [[/math]]

Thus by using [math](1+H)^{-1}=1-H+H^2-H^3+\ldots[/math] we obtain:

[[math]] \begin{eqnarray*} (1+H)^{-1}(\pi,\nu) &=&\sum_{r=0}^\infty(-1)^rH^r(\pi,\nu)\\ &=&N^{-d(\pi,\nu)}\sum_{r=0}^\infty\sum_{p\in P_r(\pi,\nu)}(-1)^rN^{-g(p)} \end{eqnarray*} [[/math]]

It follows that the Weingarten matrix is given by the following formula:

[[math]] \begin{eqnarray*} W_{kN}(\pi,\nu) &=&\Delta^{-1}(\pi)(1+H)^{-1}(\pi,\nu)\Delta^{-1}(\nu)\\ &=&N^{-\frac{|\pi|}{2}-\frac{|\nu|}{2}-d(\pi,\nu)}\sum_{r=0}^\infty\sum_{p\in P_r(\pi,\nu)}(-1)^rN^{-g(p)}\\ &=&N^{|\pi\vee\nu|-|\pi|-|\nu|}\sum_{r=0}^\infty\sum_{p\in P_r(\pi,\nu)}(-1)^rN^{-g(p)} \end{eqnarray*} [[/math]]

Now by rearranging the various terms in the above double sum according to their geodesicity defect [math]g=g(p)[/math], this gives the following formula:

[[math]] W_{kN}(\pi,\nu)=N^{|\pi\vee\nu|-|\pi|-|\nu|}\sum_{g=0}^\infty K_g(\pi,\nu)N^{-g} [[/math]]

Thus, we are led to the conclusion in the statement.

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As an illustration for all this, we have the following explicit estimates:

Theorem

Consider an easy quantum group [math]G=(G_N)[/math], coming from a category of partitions [math]D=(D(k))[/math]. For any [math]\pi\leq\nu[/math] we have the estimate

[[math]] W_{kN}(\pi,\nu)=N^{-|\pi|}(\mu(\pi,\nu)+O(N^{-1})) [[/math]]

and for [math]\pi,\nu[/math] arbitrary we have

[[math]] W_{kN}(\pi,\nu)=O(N^{|\pi\vee\nu|-|\pi|-|\nu|}) [[/math]]

with [math]\mu[/math] being the Möbius function of [math]D(k)[/math].

Show Proof

We have two assertions here, the idea being as follows:

(1) The first estimate is clear from the general expansion formula established in Theorem 15.19, namely:

[[math]] W_{kN}(\pi,\nu)=N^{|\pi\vee\nu|-|\pi|-|\nu|}\sum_{g=0}^\infty K_g(\pi,\nu)N^{-g} [[/math]]

(2) In the case [math]\pi\leq\nu[/math] it is known that [math]K_0[/math] coincides with the Möbius function of [math]NC(k)[/math], as explained for instance in , so we obtain once again from Theorem 15.19 the fine estimate in the statement as well, namely:

[[math]] W_{kN}(\pi,\nu)=N^{-|\pi|}(\mu(\pi,\nu)+O(N^{-1}))\qquad\forall\pi\leq\nu [[/math]]

Observe that, by symmetry of [math]W_{kN}[/math], we obtain as well that we have:

[[math]] W_{kN}(\pi,\nu)=N^{-|\nu|}(\mu(\nu,\pi)+O(N^{-1}))\qquad\forall\pi\geq\nu [[/math]]

Thus, we are led to the conclusions in the statement.

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When [math]\pi,\nu[/math] are not comparable by [math]\leq[/math], things are quite unclear. The simplest example appears at [math]k=3[/math], where we have the following formula, which is elementary:

[[math]] W_{3N}(|\sqcap,\sqcap|)\simeq N^{-3} [[/math]]

Observe that the exponent [math]-3[/math] is precisely the dominant one, and this because:

[[math]] \Big||\sqcap\vee\sqcap|\Big|-\Big||\sqcap\Big|-\Big|\sqcap|\Big|=1-2-2=-3 [[/math]]

As for the corresponding coefficient, [math]K_0(|\sqcap,\sqcap|)=1[/math], this is definitely not the Möbius function, which vanishes for partitions which are not comparable by [math]\leq[/math]. According to Theorem 15.19, this is rather the number of signed geodesic paths from [math]|\sqcap[/math] to [math]\sqcap|[/math].

In relation to all this, observe that geometrically, [math]NC(5)[/math] consists of the partitions [math]|\sqcap,\sqcap|,\sqcap\hskip-3.2mm{\ }_|[/math], which form an equilateral triangle with edges worth 1, and then the partitions [math]|||,\sqcap\hskip-0.8mm\sqcap[/math], which are at distance 1 apart, and each at distance [math]1/2[/math] from each of the vertices of the triangle. It is not obvious how to recover the formula [math]K_0(|\sqcap,\sqcap|)=1[/math] from this.

Finally, also following , we will need as well the following result:

Proposition

We have the following results:

• If [math]D=NC,NC_2[/math], then [math]\mu_{D(k)}(\pi,\nu)=\mu_{NC(k)}(\pi,\nu)[/math].
• If [math]D=P,P_2[/math] then [math]\mu_{D(k)}(\pi,\nu)=\mu_{P(k)}(\pi,\nu)[/math].

Show Proof

Let [math]Q=NC,P[/math] according to the cases (1,2) above. It is easy to see in each case that [math]D(k)[/math] is closed under taking intervals in [math]Q(k)[/math], in the sense that if [math]\pi_1,\pi_2\in D(k)[/math], [math]\nu\in Q(k)[/math] and [math]\pi_1 \lt \nu \lt \pi_2[/math] then [math]\nu\in D(k)[/math]. With this observation in hand, the result now follows from the definition of the Möbius function. See .

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General references

Banica, Teo (2024). "Calculus and applications". arXiv:2401.00911 [math.CO].