10d. Bessel laws

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Let us go back to [math]H_N^+,K_N^+[/math], or rather to the whole series [math]H_N^{s+}[/math], with [math]s\in\{1,2,\ldots,\infty\}[/math] and work out the fusion rules, and probabilistic aspects. We first have:

Proposition

The algebra [math]C(H_N^{s+})[/math] has a family of [math]N[/math]-dimensional corepresentations [math]\{u_k|k\in\mathbb Z\}[/math], satisfying the following conditions:

• [math]u_k=(u_{ij}^k)[/math] for any [math]k\geq 0[/math].
• [math]u_k=u_{k+s}[/math] for any [math]k\in\mathbb Z[/math].
• [math]\bar{u}_k=u_{-k}[/math] for any [math]k\in\mathbb Z[/math].

Show Proof

Our claim is that all the above holds, with [math]u_k=(u_{ij}^k)[/math]. Indeed, all these results follow from the definition of [math]H_N^{s+}[/math]. See .

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Next, we have the following result, also from :

Theorem

With the convention [math]u_{i_1\ldots i_k}=u_{i_1}\otimes\ldots\otimes u_{i_k}[/math], for any [math]i_1,\ldots,i_k\in\mathbb Z[/math], we have the following equality of linear spaces,

[[math]] Hom(u_{i_1\ldots i_k},u_{j_1\ldots j_l})=span\left\{T_p\Big|p\in NC_s(i_1\ldots i_k,j_1\ldots j_l)\right\} [[/math]]

where the set on the right consists of elements of [math]NC(k,l)[/math] having the property that in each block, the sum of [math]i[/math] indices equals the sum of [math]j[/math] indices, modulo [math]s[/math].

Show Proof

This result is from , the idea of the proof being as follows:

(1) Our first claim is that, in order to prove [math]\supset[/math], we may restrict attention to the case [math]k=0[/math]. This follows indeed from the Frobenius duality isomorphism.

(2) Our second claim is that, in order to prove [math]\supset[/math] in the case [math]k=0[/math], we may restrict attention to the one-block partitions. Indeed, this follows once again from a standard trick. Consider the following disjoint union:

[[math]] NC_s=\bigcup_{k=0}^\infty\bigcup_{i_1\ldots i_k} NC_s(0,i_1\ldots i_k) [[/math]]

This is a set of labeled partitions, having property that each [math]p\in NC_s[/math] is noncrossing, and that for [math]p\in NC_s[/math], any block of [math]p[/math] is in [math]NC_s[/math]. But it is well-known that under these assumptions, the global algebraic properties of [math]NC_s[/math] can be checked on blocks.

(3) Proof of [math]\supset[/math]. According to the above considerations, we just have to prove that the vector associated to the one-block partition in [math]NC(l)[/math] is fixed by [math]u_{j_1\ldots j_l}[/math], when:

[[math]] s|j_1+\ldots+j_l [[/math]]

Consider the standard generators [math]e_{ab}\in M_N(\mathbb C)[/math], acting on the basis vectors by [math]e_{ab}(e_c)=\delta_{bc}e_a[/math]. The corepresentation [math]u_{j_1\ldots j_l}[/math] is given by the following formula:

[[math]] u_{j_1\ldots j_l}=\sum_{a_1\ldots a_l}\sum_{b_1\ldots b_l}u_{a_1b_1}^{j_1}\ldots u_{a_lb_l}^{j_l}\otimes e_{a_1b_1}\otimes\ldots\otimes e_{a_lb_l} [[/math]]

As for the vector associated to the one-block partition, this is [math]\xi_l=\sum_be_b^{\otimes l}[/math]. By using now several times the relations in Proposition 10.22, we obtain, as claimed:

[[math]] \begin{eqnarray*} u_{j_1\ldots j_l}(1\otimes\xi_l) &=&\sum_{a_1\ldots a_l}\sum_bu_{a_1b}^{j_1}\ldots u_{a_lb}^{j_l}\otimes e_{a_1}\otimes\ldots\otimes e_{a_l}\\ &=&\sum_{ab}u_{ab}^{j_1+\ldots+j_l}\otimes e_a^{\otimes l}\\ &=&1\otimes\xi_l \end{eqnarray*} [[/math]]

(4) Proof of [math]\subset[/math]. The spaces on the right in the statement form a Tannakian category in the sense of Woronowicz , so they correspond to a certain Woronowicz algebra [math]A[/math], which is by definition the maximal model for the Tannakian category. In other words, [math]A[/math] comes with a family of corepresentations [math]\{v_i\}[/math], such that:

[[math]] Hom(v_{i_1\ldots i_k},v_{j_1\ldots j_l})={\rm span}\left\{T_p\Big|p\in NC_s(i_1\ldots i_k,j_1\ldots j_l)\right\} [[/math]]

On the other hand, the inclusion [math]\supset[/math] that we just proved shows that [math]C(H_N^{s+})[/math] is a model for the category. Thus we have a quotient map [math]A\to C(H_N^{s+})[/math], mapping [math]v_i\to u_i[/math]. But this latter map can be shown to be an isomorphism, by suitably adapting the proof from the [math]s=1[/math] case, for the quantum permutation group [math]S_N^+[/math]. See , .

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Still following , we have the following result:

Theorem

Let [math]F= \lt \mathbb Z_s \gt [/math] be the monoid formed by the words over [math]\mathbb Z_s[/math], with involution [math](i_1\ldots i_k)^-=(-i_k)\ldots(-i_1)[/math], and with fusion product given by:

[[math]] (i_1\ldots i_k)\cdot (j_1\ldots j_l)=i_1\ldots i_{k-1}(i_k+j_1)j_2\ldots j_l [[/math]]

The irreducible representations of [math]H_N^{s+}[/math] can then be labeled [math]r_x[/math] with [math]x\in F[/math], such that

[[math]] r_x\otimes r_y=\sum_{x=vz,y=\bar{z}w}r_{vw}+r_{v\cdot w} [[/math]]

and [math]\bar{r}_x=r_{\bar{x}}[/math], and such that [math]r_i=u_i-\delta_{i0}1[/math] for any [math]i\in\mathbb Z_s[/math].

Show Proof

This basically follows from Theorem 10.23, the idea being as follows:

(1) Consider the monoid [math]A=\{a_x|x\in F\}[/math], with multiplication [math]a_xa_y=a_{xy}[/math]. We endow [math]\mathbb NA[/math] with fusion rules as in the statement, namely:

[[math]] a_x\otimes a_y=\sum_{x=vz,y=\bar{z}w}a_{vw}+a_{v\cdot w} [[/math]]

(2) The fusion rules on [math]\mathbb ZA[/math] can be then uniquely described by conversion formulae as follows, with [math]C[/math] being positive integers, and [math]D[/math] being integers:

[[math]] a_{i_1}\otimes\ldots\otimes a_{i_k}=\sum_l\sum_{j_1\ldots j_l}C_{i_1\ldots i_k}^{j_1\ldots j_l}a_{j_1\ldots j_l} [[/math]]

[[math]] a_{i_1\ldots i_k}=\sum_l\sum_{j_1\ldots j_l}D_{i_1\ldots i_k}^{j_1\ldots j_l}a_{j_1}\otimes\ldots\otimes a_{j_l} [[/math]]

(3) Now observe that there is a unique morphism of rings [math]\Phi:\mathbb ZA\to R[/math], such that [math]\Phi(a_i)=r_i[/math] for any [math]i[/math]. Indeed, consider the following elements of [math]R[/math]:

[[math]] r_{i_1\ldots i_k}=\sum_l\sum_{j_1\ldots j_l}D_{i_1\ldots i_k}^{j_1\ldots j_l}r_{j_1}\otimes\ldots\otimes r_{j_l} [[/math]]

In case we have a morphism as claimed, we must have [math]\Phi(a_x)=r_x[/math] for any [math]x\in F[/math]. Thus our morphism is uniquely determined on [math]A[/math], so it is uniquely determined on [math]\mathbb ZA[/math].

(4) Our claim is that [math]\Phi[/math] commutes with the linear forms [math]x\to\#(1\in x)[/math]. Indeed, by linearity we just have to check the following equality:

[[math]] \#(1\in a_{i_1}\otimes\ldots\otimes a_{i_k})=\#(1\in r_{i_1}\otimes\ldots\otimes r_{i_k}) [[/math]]

Now remember that the elements [math]r_i[/math] are defined as [math]r_i=u_i-\delta_{i0}1[/math]. So, consider the elements [math]c_i=a_i+\delta_{i0}1[/math]. Since the operations [math]r_i\to u_i[/math] and [math]a_i\to c_i[/math] are of the same nature, by linearity the above formula is equivalent to:

[[math]] \#(1\in c_{i_1}\otimes\ldots\otimes c_{i_k})=\#(1\in u_{i_1}\otimes\ldots\otimes u_{i_k}) [[/math]]

Now by using Theorem 10.23, what we have to prove is:

[[math]] \#(1\in c_{i_1}\otimes\ldots\otimes c_{i_k})=\#NC_s(i_1\ldots i_k) [[/math]]

In order to prove this formula, consider the product on the left:

[[math]] P=(a_{i_1}+\delta_{i_10}1)\otimes(a_{i_2}+\delta_{i_20}1)\otimes\ldots\otimes (a_{i_k}+\delta_{i_k0}1) [[/math]]

But this quantity can be computed by using the fusion rules on [math]A[/math], and the combinatorics leads to the conclusion that we have [math]\#(1\in P)=\# NC_s(i_1\ldots i_k)[/math], as claimed.

(5) Our claim now is that [math]\Phi[/math] is injective. Indeed, this follows from the result in the previous step, by using a standard positivity argument.

(6) Our claim is that we have [math]\Phi(A)\subset R_{irr}[/math]. This is the same as saying that [math]r_x\in R_{irr}[/math] for any [math]x\in F[/math], and we will prove it by recurrence. Assume that the assertion is true for all the words of length [math] \lt k[/math], and consider a length [math]k[/math] word, [math]x=i_1\ldots i_k[/math]. We have:

[[math]] a_{i_1}\otimes a_{i_2\ldots i_k}=a_x+a_{i_1+i_2,i_3\ldots i_k}+\delta_{i_1+i_2,0}a_{i_3\ldots i_k} [[/math]]

By applying [math]\Phi[/math] to this decomposition, we obtain:

[[math]] r_{i_1}\otimes r_{i_2\ldots i_k}=r_x+r_{i_1+i_2,i_3\ldots i_k}+\delta_{i_1+i_2,0}r_{i_3\ldots i_k} [[/math]]

We have the following computation, which is valid for [math]y=i_1+i_2,i_3\ldots i_k[/math], as well as for [math]y=i_3\ldots i_k[/math] in the case [math]i_1+i_2=0[/math]:

[[math]] \begin{eqnarray*} \#(r_y\in r_{i_1}\otimes r_{i_2\ldots i_k}) &=&\#(1,r_{\bar{y}}\otimes r_{i_1}\otimes r_{i_2\ldots i_k})\\ &=&\#(1,a_{\bar{y}}\otimes a_{i_1}\otimes a_{i_2\ldots i_k})\\ &=&\#(a_y\in a_{i_1}\otimes a_{i_2\ldots i_k})\\ &=&1 \end{eqnarray*} [[/math]]

Moreover, we know from the previous step that we have [math]r_{i_1+i_2,i_3\ldots i_k}\neq r_{i_3\ldots i_k}[/math], so we conclude that the following formula defines an element of [math]R^+[/math]:

[[math]] \alpha=r_{i_1}\otimes r_{i_2\ldots i_k}-r_{i_1+i_2,i_3\ldots i_k}-\delta_{i_1+i_2,0}r_{i_3\ldots i_k} [[/math]]

On the other hand, we have [math]\alpha=r_x[/math], so we conclude that we have [math]r_x\in R^+[/math]. Finally, the irreducibility of [math]r_x[/math] follows from [math]\#(1\in r_x\otimes\bar{r}_x)=1[/math].

(7) Summarizing, we have constructed an injective ring morphism [math]\Phi:\mathbb ZA\to R[/math], having the property [math]\Phi(A)\subset R_{irr}[/math]. The remaining fact to be proved, namely that we have [math]\Phi(A)=R_{irr}[/math], is something of abstract nature, which is clear. Thus, we are done.

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Regarding the probabilistic aspects, we will need some general theory. We have the following definition, extending the Poisson limit theory from chapter 9 above:

Definition

Associated to any compactly supported positive measure [math]\rho[/math], not necessarily of mass [math]1[/math], are the probability measures

[[math]] p_\rho=\lim_{n\to\infty}\left(\left(1-\frac{c}{n}\right)\delta_0+\frac{1}{n}\rho\right)^{*n} [[/math]]

[[math]] \pi_\rho=\lim_{n\to\infty}\left(\left(1-\frac{c}{n}\right)\delta_0+\frac{1}{n}\rho\right)^{\boxplus n} [[/math]]

where [math]c=mass(\rho)[/math], called compound Poisson and compound free Poisson laws.

In what follows we will be interested in the case where [math]\rho[/math] is discrete, as is for instance the case for [math]\rho=t\delta_1[/math] with [math]t \gt 0[/math], which produces the Poisson and free Poisson laws. The following result allows one to detect compound Poisson/free Poisson laws:

Proposition

For a discrete measure, written as

[[math]] \rho=\sum_{i=1}^sc_i\delta_{z_i} [[/math]]

with [math]c_i \gt 0[/math] and [math]z_i\in\mathbb R[/math], we have the formulae

[[math]] F_{p_\rho}(y)=\exp\left(\sum_{i=1}^sc_i(e^{iyz_i}-1)\right) [[/math]]

[[math]] R_{\pi_\rho}(y)=\sum_{i=1}^s\frac{c_iz_i}{1-yz_i} [[/math]]

where [math]F,R[/math] are respectively the Fourier transform, and Voiculescu's [math]R[/math]-transform.

Show Proof

Let [math]\mu_n[/math] be the measure appearing in Definition 10.25, under the convolution signs. In the classical case, we have the following computation:

[[math]] \begin{eqnarray*} &&F_{\mu_n}(y)=\left(1-\frac{c}{n}\right)+\frac{1}{n}\sum_{i=1}^sc_ie^{iyz_i}\\ &\implies&F_{\mu_n^{*n}}(y)=\left(\left(1-\frac{c}{n}\right)+\frac{1}{n}\sum_{i=1}^sc_ie^{iyz_i}\right)^n\\ &\implies&F_{p_\rho}(y)=\exp\left(\sum_{i=1}^sc_i(e^{iyz_i}-1)\right) \end{eqnarray*} [[/math]]

In the free case now, we use a similar method. The Cauchy transform of [math]\mu_n[/math] is:

[[math]] G_{\mu_n}(\xi)=\left(1-\frac{c}{n}\right)\frac{1}{\xi}+\frac{1}{n}\sum_{i=1}^s\frac{c_i}{\xi-z_i} [[/math]]

Consider now the [math]R[/math]-transform of the measure [math]\mu_n^{\boxplus n}[/math], which is given by:

[[math]] R_{\mu_n^{\boxplus n}}(y)=nR_{\mu_n}(y) [[/math]]

The above formula of [math]G_{\mu_n}[/math] shows that the equation for [math]R=R_{\mu_n^{\boxplus n}}[/math] is as follows:

[[math]] \begin{eqnarray*} &&\left(1-\frac{c}{n}\right)\frac{1}{y^{-1}+R/n}+\frac{1}{n}\sum_{i=1}^s\frac{c_i}{y^{-1}+R/n-z_i}=y\\ &\implies&\left(1-\frac{c}{n}\right)\frac{1}{1+yR/n}+\frac{1}{n}\sum_{i=1}^s\frac{c_i}{1+yR/n-yz_i}=1 \end{eqnarray*} [[/math]]

Now multiplying by [math]n[/math], rearranging the terms, and letting [math]n\to\infty[/math], we get:

[[math]] \begin{eqnarray*} &&\frac{c+yR}{1+yR/n}=\sum_{i=1}^s\frac{c_i}{1+yR/n-yz_i}\\ &\implies&c+yR_{\pi_\rho}(y)=\sum_{i=1}^s\frac{c_i}{1-yz_i}\\ &\implies&R_{\pi_\rho}(y)=\sum_{i=1}^s\frac{c_iz_i}{1-yz_i} \end{eqnarray*} [[/math]]

This finishes the proof in the free case, and we are done.

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We also have the following result, providing an alternative to Definition 10.25, and which is an extension of the classical and free Poisson limiting theorems (PLT) that we know from chapter 9, called Compound Poisson Limiting Theorem (CPLT):

Theorem

For a discrete measure, written as

[[math]] \rho=\sum_{i=1}^sc_i\delta_{z_i} [[/math]]

with [math]c_i \gt 0[/math] and [math]z_i\in\mathbb R[/math], we have the formulae

[[math]] p_\rho/\pi_\rho={\rm law}\left(\sum_{i=1}^sz_i\alpha_i\right) [[/math]]

where the variables [math]\alpha_i[/math] are Poisson/free Poisson[math](c_i)[/math], independent/free.

Show Proof

Let [math]\alpha[/math] be the sum of Poisson/free Poisson variables in the statement. We will show that the Fourier/[math]R[/math]-transform of [math]\alpha[/math] is given by the formulae in Proposition 10.26. Indeed, by using some well-known Fourier transform formulae, we have:

[[math]] \begin{eqnarray*} F_{\alpha_i}(y)=\exp(c_i(e^{iy}-1)) &\implies&F_{z_i\alpha_i}(y)=\exp(c_i(e^{iyz_i}-1))\\ &\implies&F_\alpha(y)=\exp\left(\sum_{i=1}^sc_i(e^{iyz_i}-1)\right) \end{eqnarray*} [[/math]]

Also, by using some well-known [math]R[/math]-transform formulae, we have:

[[math]] \begin{eqnarray*} R_{\alpha_i}(y)=\frac{c_i}{1-y} &\implies&R_{z_i\alpha_i}(y)=\frac{c_iz_i}{1-yz_i}\\ &\implies&R_\alpha(y)=\sum_{i=1}^s\frac{c_iz_i}{1-yz_i} \end{eqnarray*} [[/math]]

Thus we have indeed the same formulae as those in Proposition 10.26.

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We can go back now to quantum reflection groups, and we have:

Theorem

The asymptotic laws of truncated characters are as follows, where [math]\varepsilon_s[/math] with [math]s\in\{1,2,\ldots,\infty\}[/math] is the uniform measure on the [math]s[/math]-th roots of unity:

• For [math]H_N^s[/math] we obtain the compound Poisson law [math]b_t^s=p_{t\varepsilon_s}[/math].
• For [math]H_N^{s+}[/math] we obtain the compound free Poisson law [math]\beta_t^s=\pi_{t\varepsilon_s}[/math].

These measures are in Bercovici-Pata bijection.

Show Proof

This follows from easiness, and from the Weingarten formula. To be more precise, at [math]t=1[/math] this follows by counting the partitions, and at [math]t\in(0,1][/math] general, this follows in the usual way, for instance by using cumulants. See .

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The above measures are called Bessel and free Bessel laws. This is because at [math]s=2[/math] we have [math]b_t^2=e^{-t}\sum_{k=-\infty}^\infty f_k(t/2)\delta_k[/math], with [math]f_k[/math] being the Bessel function of the first kind:

[[math]] f_k(t)=\sum_{p=0}^\infty \frac{t^{|k|+2p}}{(|k|+p)!p!} [[/math]]

The Bessel and free Bessel laws have particularly interesting properties at the parameter values [math]s=2,\infty[/math]. So, let us record the precise statement here:

Theorem

The asymptotic laws of truncated characters are as follows:

• For [math]H_N[/math] we obtain the real Bessel law [math]b_t=p_{t\varepsilon_2}[/math].
• For [math]K_N[/math] we obtain the complex Bessel law [math]B_t=p_{t\varepsilon_\infty}[/math].
• For [math]H_N^+[/math] we obtain the free real Bessel law [math]\beta_t=\pi_{t\varepsilon_2}[/math].
• For [math]K_N^+[/math] we obtain the free complex Bessel law [math]\mathfrak B_t=\pi_{t\varepsilon_\infty}[/math].

Show Proof

This follows indeed from Theorem 10.28 above, at [math]s=2,\infty[/math].

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In addition to what has been said above, there are as well some interesting results about the Bessel and free Bessel laws involving the multiplicative convolution [math]\times[/math], and the multiplicative free convolution [math]\boxtimes[/math]. For details, we refer here to .

As a conclusion to all this, work that we did in chapter 9 and here, things in the discrete setting are often more complicated than in the continuous setting, although when restricting the attention to [math]H_N,H_N^+[/math] and [math]K_N,K_N^+[/math], everything is after all quite similar to what we knew from chapters 5-6, regarding [math]O_N,O_N^+[/math] and [math]U_N,U_N^+[/math]. We will keep building in chapter 11 below, with this kind of philosophy, with the idea in mind of unifying the theory of [math]O_N,O_N^+[/math] and [math]U_N,U_N^+[/math] with the theory of [math]H_N,H_N^+[/math] and [math]K_N,K_N^+[/math].

General references

Banica, Teo (2024). "Introduction to quantum groups". arXiv:1909.08152 [math.CO].