6b. Circular variables

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Let us discuss now the unitary quantum group [math]U_N^+[/math]. We have 3 main topics to be discussed, namely the character law with [math]N\to\infty[/math], the representation theory at fixed [math]N\in\mathbb N[/math], and complexification, and the situation with respect to [math]U_N[/math] is as follows:

• The asymptotic character law appears as a “free complexification” of the Wigner law, the combinatorics being similar to the one in the classical case.
• The representation theory is definitely simpler, with the fusion rules being given by a “free complexification” of the Clebsch-Gordan rules, at any [math]N\geq2[/math].
• As for the complexification aspects, here the situation is extremely simple, with the passage [math]O_N^+\to U_N^+[/math] being a free complexification.

Obviously, some magic is going on here. Who would have imagined that, passed a few abstract things that can be learned, and we will learn indeed all this, the free quantum group [math]U_N^+[/math] is simpler than its classical counterpart [math]U_N[/math]. This might suggest for instance that quantum mechanics might be simpler than classical mechanics. And isn't this crazy. But hey, read Arnold first, and let me know if you find all that stuff simple.

Back to work now, let us first discuss the character problematics for [math]U_N^+[/math], or rather the difficulties that appear here. We have the following theoretical result, to start with, coming from the general [math]C^*[/math]-algebra theory developed in chapter 1 above:

Theorem

Given a [math]C^*[/math]-algebra with a faithful trace [math](A,tr)[/math], any normal variable,

[[math]] aa^*=a^*a [[/math]]

has a “law”, which is by definition a complex probability measure [math]\mu\in\mathcal P(\mathbb C)[/math] satisfying:

[[math]] tr(a^k)=\int_\mathbb Cz^kd\mu(z) [[/math]]

This law is unique, and is supported by the spectrum [math]\sigma(a)\subset\mathbb C[/math]. In the non-normal case, [math]aa^*\neq a^*a[/math], such a law does not exist.

Show Proof

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

(1) In the normal case, [math]aa^*=a^*a[/math], the Gelfand theorem, or rather the subsequent continuous functional calculus theorem, tells us that we have:

[[math]] \lt a \gt =C(\sigma(a)) [[/math]]

Thus the functional [math]f(a)\to tr(f(a))[/math] can be regarded as an integration functional on the algebra [math]C(\sigma(a))[/math], and by the Riesz theorem, this latter functional must come from a probability measure [math]\mu[/math] on the spectrum [math]\sigma(a)[/math], in the sense that we must have:

[[math]] tr(f(a))=\int_{\sigma(a)}f(z)d\mu(z) [[/math]]

We are therefore led to the conclusions in the statement, with the uniqueness assertion coming from the fact that the elements [math]a^k[/math], taken as usual with respect to colored integer exponents, [math]k=\circ\bullet\bullet\circ\ldots[/math]\,, generate the whole [math]C^*[/math]-algebra [math]C(\sigma(a))[/math].

(2) In the non-normal case now, [math]aa^*\neq a^*a[/math], we must show that such a law does not exist. For this purpose, we can use a positivity trick, as follows:

[[math]] \begin{eqnarray*} aa^*-a^*a\neq0 &\implies&(aa^*-a^*a)^2 \gt 0\\ &\implies&aa^*aa^*-aa^*a^*a-a^*aaa^*+a^*aa^*a \gt 0\\ &\implies&tr(aa^*aa^*-aa^*a^*a-a^*aaa^*+a^*aa^*a) \gt 0\\ &\implies&tr(aa^*aa^*+a^*aa^*a) \gt tr(aa^*a^*a+a^*aaa^*)\\ &\implies&tr(aa^*aa^*) \gt tr(aaa^*a^*) \end{eqnarray*} [[/math]]

Now assuming that [math]a[/math] has a law [math]\mu\in\mathcal P(\mathbb C)[/math], in the sense that the moment formula in the statement holds, the above two different numbers would have to both appear by integrating [math]|z|^2[/math] with respect to this law [math]\mu[/math], which is contradictory, as desired.

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All the above might look a bit abstract, so as an illustration here, consider the following matrix, which is the simplest example of a non-normal matrix:

[[math]] Z=\begin{pmatrix}0&1\\0&0\end{pmatrix} [[/math]]

We have then the following formulae, which show that [math]Z[/math] has no law, indeed:

[[math]] tr(ZZZ^*Z^*)=tr\begin{pmatrix}0&0\\0&0\end{pmatrix}=0 [[/math]]

[[math]] tr(ZZ^*ZZ^*)=tr\begin{pmatrix}1&0\\0&0\end{pmatrix}=\frac{1}{2} [[/math]]

Getting back now to [math]U_N^+[/math], its main character is not normal, so it does not have a law [math]\mu\in\mathcal P(\mathbb C)[/math]. Here is a concrete illustration for this phenomenon:

Proposition

The main character of [math]U_N^+[/math] satisfies, at [math]N\geq4[/math],

[[math]] \int_{U_N^+}\chi\chi\chi^*\chi^*=1 [[/math]]

[[math]] \int_{U_N^+}\chi\chi^*\chi\chi^*=2 [[/math]]

and so this main character [math]\chi[/math] does not have a law [math]\mu\in\mathcal P(\mathbb C)[/math].

Show Proof

This follows from the last assertion in Theorem 6.1, which tells us that the moments of [math]\chi[/math] are given by the following formula, valid at any [math]N\geq k[/math]:

[[math]] \int_{U_N^+}\chi^k=|\mathcal{NC}_2(k)| [[/math]]

Indeed, we obtain from this the following formula, valid at any [math]N\geq4[/math]:

[[math]] \begin{eqnarray*} \int_{U_N^+}\chi\chi\chi^*\chi^* &=&|\mathcal NC_2(\circ\circ\bullet\,\bullet)|\\ &=&|\Cap|\\ &=&1 \end{eqnarray*} [[/math]]

On the other hand, we obtain as well the following formula, once again at [math]N\geq4[/math]:

[[math]] \begin{eqnarray*} \int_{U_N^+}\chi\chi^*\chi\chi^* &=&|\mathcal NC_2(\circ\bullet\circ\,\bullet)|\\ &=&|\cap\cap\,,\Cap\,|\\ &=&2 \end{eqnarray*} [[/math]]

Thus, we have the formulae in the statement. Now since we cannot obtain both 1 and 2 by integrating [math]|z|^2[/math] with respect to a measure, our variable has no law [math]\mu\in\mathcal P(\mathbb C)[/math].

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Summarizing, we are a bit in trouble here, and we must proceed as follows:

Definition

Given a [math]C^*[/math]-algebra with a faithful trace [math](A,tr)[/math], the law of a variable [math]a\in A[/math] is the following abstract functional:

[[math]] \mu:\mathbb C \lt X,X^* \gt \to\mathbb C [[/math]]

[[math]] P\to tr(P(a)) [[/math]]

In particular two variables [math]a,b\in A[/math] have the same law, and we write in this case [math]a\sim b[/math], when all their moments coincide,

[[math]] tr(a^k)=tr(b^k) [[/math]]

with these moments being taken with respect to colored integers, [math]k=\circ\bullet\bullet\circ\ldots[/math]

Here the compatibility between the first and the second above conventions comes from the fact that, by linearity, the functional [math]\mu[/math] is uniquely determined by its values on the monomials [math]P(z)=z^k[/math], with [math]k=\circ\bullet\bullet\circ\ldots[/math] being a colored integer.

In the normal case, [math]aa^*=a^*a[/math], it follows from Theorem 6.5 that the law, as defined above, comes from a probability measure [math]\mu\in\mathcal P(\mathbb C)[/math], via the following formula:

[[math]] tr(P(a))=\int_\mathbb CP(z)d\mu(z) [[/math]]

In particular, in the case where we have two normal variables [math]a,b[/math], the equality [math]a\sim b[/math] tells us that the laws of [math]a,b[/math], taken in the complex measure sense, must coincide. In the general case, [math]aa^*\neq a^*a[/math], there is no such simple interpretation of the law, with this coming from the last assertion in Theorem 6.5, and also from the concrete example worked out in Proposition 6.6, and we must therefore use Definition 6.7 as it is.

Next in line, we have to talk about freeness. For this purpose, let us recall that the independence of two subalgebras [math]B,C\subset A[/math] can be defined in the following way:

[[math]] tr(b)=tr(c)=0\implies tr(bc)=0 [[/math]]

In analogy with this, we have the following definition, due to Voiculescu :

Definition

Two subalgebras [math]B,C\subset A[/math] are called free when the following condition is satisfied, for any [math]b_i\in B[/math] and [math]c_i\in C[/math]:

[[math]] tr(b_i)=tr(c_i)=0\implies tr(b_1c_1b_2c_2\ldots)=0 [[/math]]

Also, two variables [math]b,c\in A[/math] are called free when the algebras that they generate,

[[math]] B= \lt b \gt \quad,\quad C= \lt c \gt [[/math]]

are free inside [math]A[/math], in the above sense.

In short, freeness appears as a kind of “free analogue” of independence, taking into account the fact that the variables do not necessarily commute. As a first result regarding this notion, in analogy with the basic theory of the independence, we have:

Proposition

Assuming that [math]B,C\subset A[/math] are free, the restriction of [math]tr[/math] to [math] \lt B,C \gt [/math] can be computed in terms of the restrictions of [math]tr[/math] to [math]B,C[/math]. To be more precise,

[[math]] tr(b_1c_1b_2c_2\ldots)=P\Big(\{tr(b_{i_1}b_{i_2}\ldots)\}_i,\{tr(c_{j_1}c_{j_2}\ldots)\}_j\Big) [[/math]]

where [math]P[/math] is certain polynomial in several variables, depending on the length of the word [math]b_1c_1b_2c_2\ldots[/math], and having as variables the traces of products of type [math]b_{i_1}b_{i_2}\ldots[/math] and [math]c_{j_1}c_{j_2}\ldots[/math]\,, with the indices being chosen increasing, [math]i_1 \lt i_2 \lt \ldots[/math] and [math]j_1 \lt j_2 \lt \ldots[/math]

Show Proof

This is something quite theoretical, so let us begin with an example. Our claim is that if [math]b,c[/math] are free then, exactly as in the case where we have independence:

[[math]] tr(bc)=tr(b)tr(c) [[/math]]

Indeed, we have the following computation, with the convention [math]a'=a-tr(a)[/math]:

[[math]] \begin{eqnarray*} tr(bc) &=&tr[(b'+tr(b))(c'+tr(c))]\\ &=&tr(b'c')+t(b')tr(c)+tr(b)tr(c')+tr(b)tr(c)\\ &=&tr(b'c')+tr(b)tr(c)\\ &=&tr(b)tr(c) \end{eqnarray*} [[/math]]

In general now, the situation is a bit more complicated, but the same trick applies. To be more precise, we can start our computation as follows:

[[math]] \begin{eqnarray*} tr(b_1c_1b_2c_2\ldots) &=&tr\big[(b_1'+tr(b_1))(c_1'+tr(c_1))(b_2'+tr(b_2))(c_2'+tr(c_2))\ldots\ldots\big]\\ &=&tr(b_1'c_1'b_2'c_2'\ldots)+{\rm other\ terms}\\ &=&{\rm other\ terms} \end{eqnarray*} [[/math]]

Observe that we have used here the freeness condition, in the following form:

[[math]] tr(b_i')=tr(c_i')=0\implies tr(b_1'c_1'b_2'c_2'\ldots)=0 [[/math]]

Now regarding the “other terms”, those which are left, each of them will consist of a product of traces of type [math]tr(b_i)[/math] and [math]tr(c_i)[/math], and then a trace of a product still remaining to be computed, which is of the following form, with [math]\beta_i\in B[/math] and [math]\gamma_i\in C[/math]:

[[math]] tr(\beta_1\gamma_1\beta_2\gamma_2\ldots) [[/math]]

To be more precise, the variables [math]\beta_i\in B[/math] appear as ordered products of those [math]b_i\in B[/math] not getting into individual traces [math]tr(b_i)[/math], and the variables [math]\gamma_i\in C[/math] appear as ordered products of those [math]c_i\in C[/math] not getting into individual traces [math]tr(c_i)[/math]. Now since the length of each such alternating product [math]\beta_1\gamma_1\beta_2\gamma_2\ldots[/math] is smaller than the length of the original alternating product [math]b_1c_1b_2c_2\ldots[/math], we are led into of recurrence, and this gives the result.

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As an illustration now, given two discrete groups [math]\Gamma,\Lambda[/math], the algebras [math]C^*(\Gamma),C^*(\Lambda)[/math] are independent inside [math]C^*(\Gamma\times\Lambda)[/math], are free inside [math]C^*(\Gamma*\Lambda)[/math]. More on this later.

With the above definitions in hand, we can now advance, in connection with our questions, in the following rather formal way:

Definition

The Voiculescu circular law [math]\Gamma_1[/math] is defined by

[[math]] \Gamma_1\sim\frac{1}{\sqrt{2}}(\alpha+i\beta) [[/math]]

with [math]\alpha,\beta[/math] being self-adjoint and free, each following the Wigner semicircle law [math]\gamma_1[/math].

Our goal in what follows will be that of proving that the main character law of [math]U_N^+[/math] becomes circular with [math]N\to\infty[/math], and in fact, more generally, with [math]N\geq2[/math]. In order to prove these results, we need first to study the Voiculescu circular law, a bit in the same way as we did with the Wigner semicircle law, in chapter 5. Let us start with:

Proposition

Consider the shift operator [math]S\in B(l^2(\mathbb N))[/math]. We have then

[[math]] S+S^*\sim\gamma_1 [[/math]]

with respect to the state [math]\varphi(T)= \lt T\delta_0,\delta_0 \gt [/math].

Show Proof

We must compute the moments of the variable [math]S+S^*[/math] with respect to the state [math]\varphi(T)= \lt T\delta_0,\delta_0 \gt [/math]. Our claim is that these moments are given by:

[[math]] \lt (S+S^*)^k\delta_0,\delta_0 \gt =|NC_2(k)| [[/math]]

Indeed, when expanding [math](S+S^*)^k[/math] and computing the value of [math]\varphi:T\to \lt T\delta_0,\delta_0 \gt [/math], the only contributions will come via the formula [math]S^*S=1[/math], which must succesively apply, as to collapse the whole product of [math]S,S^*[/math] variables into a 1 quantity. But these applications of [math]S^*S=1[/math] must appear in a non-crossing manner, and so the contributions, which are each worth 1, are parametrized by the partitions [math]\pi\in NC_2(k)[/math]. Thus, we obtain the above moment formula, which shows that we have [math]S+S^*\sim\gamma_1[/math], as claimed.

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The next step is that of taking a free product of the model found in Proposition 6.11 with itself. For this purpose, we can use the following construction:

Definition

Given a real Hilbert space [math]H[/math], we define the associated free Fock space as being the infinite Hilbert space sum

[[math]] F(H)=\mathbb C\Omega\oplus H\oplus H^{\otimes2}\oplus\ldots [[/math]]

and then we define the algebra [math]A(H)[/math] generated by the creation operators

[[math]] S_x:v\to x\otimes v [[/math]]

on this free Fock space.

At the level of examples, with [math]H=\mathbb R[/math] we recover the shift algebra [math]A= \lt S \gt [/math] on the Hilbert space [math]H=l^2(\mathbb N)[/math]. Also, with [math]H=\mathbb R^2[/math], we obtain the algebra [math]A= \lt S_1,S_2 \gt [/math] generated by the two shifts on the Hilbert space [math]H=l^2(\mathbb N*\mathbb N)[/math].

With the above notions in hand, we have the following freeness result, from :

Proposition

Given a real Hilbert space [math]H[/math], and two orthogonal vectors [math]x,y\in H[/math],

[[math]] x\perp y [[/math]]

the corresponding creation operators [math]S_x[/math] and [math]S_y[/math] are free with respect to

[[math]] tr(T)= \lt T\Omega,\Omega \gt [[/math]]

called trace associated to the vacuum vector.

Show Proof

In standard tensor notation for the elements of the free Fock space [math]F(H)[/math], the formula of a creation operator associated to a vector [math]x\in H[/math] is as follows:

[[math]] S_x(y_1\otimes\ldots\otimes y_n)=x\otimes y_1\otimes\ldots\otimes y_n [[/math]]

As for the formula of the adjoint of this creation operator, this is as follows:

[[math]] S_x^*(y_1\otimes\ldots\otimes y_n)= \lt x,y_1 \gt \otimes y_2\otimes\ldots\otimes y_n [[/math]]

We obtain from this the following formula, valid for any two vectors [math]x,y\in H[/math]:

[[math]] S_x^*S_y= \lt x,y \gt id [[/math]]

With these formulae in hand, the result follows by doing some elementary computations, a bit similar to those in the proof of Proposition 6.11.

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In order now to model the circular variables, still following Voiculescu , we can use the following key observation, coming from Proposition 6.13 via a rotation trick:

Proposition

Given two polynomials [math]f,g\in\mathbb C[X][/math], consider the variables

[[math]] R^*+f(R)\quad,\quad S^*+g(S) [[/math]]

where [math]R,S[/math] are two creation operators, or shifts, associated to a pair of orthogonal norm [math]1[/math] vectors. These variables are then free, and their sum has the same law as

[[math]] T^*+(f+g)(T) [[/math]]

with [math]T[/math] being the usual shift on [math]l^2(\mathbb N)[/math].

Show Proof

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

(1) The freeness assertion comes from the general freeness result from Proposition 6.13, via the various identifications coming from the previous results.

(2) Regarding now the second assertion, the idea is that this comes from a [math]45^\circ[/math] rotation trick. Let us write indeed the two variables in the statement as follows:

[[math]] X=R^*+a_0+a_1R+a_2R^2+\ldots [[/math]]

[[math]] Y=S^*+b_0+b_1S+a_2S^2+\ldots [[/math]]

Now let us perform the following [math]45^\circ[/math] base change, on the real span of the vectors [math]r,s\in H[/math] producing our two shifts [math]R,S[/math]:

[[math]] t=\frac{r+s}{\sqrt{2}}\quad,\quad u=\frac{r-s}{\sqrt{2}} [[/math]]

The new shifts, associated to these vectors [math]t,u\in H[/math], are then given by:

[[math]] T=\frac{R+S}{\sqrt{2}}\quad,\quad U=\frac{R-S}{\sqrt{2}} [[/math]]

By using now these new shifts, which are free as well according to Proposition 6.13, we obtain the following equality of distributions:

[[math]] \begin{eqnarray*} X+Y &=&R^*+S^*+\sum_ka_kR^k+b_kS^k\\ &=&\sqrt{2}T^*+\sum_ka_k\left(\frac{T+U}{\sqrt{2}}\right)^k+b_k\left(\frac{T-U}{\sqrt{2}}\right)^k\\ &\sim&\sqrt{2}T^*+\sum_ka_k\left(\frac{T}{\sqrt{2}}\right)^k+b_k\left(\frac{T}{\sqrt{2}}\right)^k\\ &\sim&T^*+\sum_ka_kT^k+b_kT^k \end{eqnarray*} [[/math]]

To be more precise, here in the last two lines we have used the freeness property of [math]T,U[/math] in order to cut [math]U[/math] from the computation, as it cannot bring anything, and then we did a rescaling at the end. Thus, we are led to the conclusion in the statement.

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Still following Voiculescu , we can now formulate an explicit and very useful modelling result for the semicircular and circular variables, as follows:

Theorem

Let [math]H[/math] be the Hilbert space having as basis the colored integers [math]k=\circ\bullet\bullet\circ\ldots[/math]\,, and consider the shift operators [math]S:k\to\circ k[/math] and [math]T:k\to\bullet k[/math]. We have then

[[math]] S+S^*\sim\gamma_1 [[/math]]

[[math]] S+T^*\sim\Gamma_1 [[/math]]

with respect to the state [math]\varphi(T)= \lt Te,e \gt [/math], where [math]e[/math] is the empty word.

Show Proof

This is standard free probability, the idea being as follows:

(1) The formula [math]S+S^*\sim\gamma_1[/math] is something that we already know, in a slightly different formulation, from Proposition 6.11 above.

(2) The formula [math]S+T^*\sim\Gamma_1[/math] follows from this, by using the freeness result in Proposition 6.13, and the rotation trick in Proposition 6.14.

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At the combinatorial level now, we have the following result, which is in analogy with the moment theory of the Wigner semicircle law, developed in chapter 5:

Theorem

A variable [math]a\in A[/math] is circular when its moments are given by

[[math]] tr(a^k)=|\mathcal{NC}_2(k)| [[/math]]

for any colored integer [math]k=\circ\bullet\bullet\circ\ldots[/math]

Show Proof

By using Theorem 6.15, it is enough to do the computation in the model there. With [math]S:k\to\circ k[/math] and [math]T:k\to\bullet k[/math], our claim is that we have:

[[math]] \lt (S+T^*)^ke,e \gt =|\mathcal{NC}_2(k)| [[/math]]

In order to prove this formula, we can proceed as in the proof of Proposition 6.11. Indeed, let us expand the quantity [math](S+T^*)^k[/math], and then apply the state [math]\varphi[/math]. With respect to the previous computation, from Proposition 6.11, what happens is that the contributions will come this time via the formulae [math]S^*S=1[/math], [math]T^*T=1[/math], which must succesively apply, as to collapse the whole product of [math]S,S^*,T,T^*[/math] variables into a 1 quantity.

As before, in the proof of Proposition 6.11, these applications of the rules [math]S^*S=1[/math], [math]T^*T=1[/math] must appear in a noncrossing manner, but what happens now, in contrast with the computation from the proof of Proposition 6.11, where [math]S+S^*[/math] was self-adjoint, is that at each point where the exponent [math]k[/math] has a [math]\circ[/math] entry we must use [math]T^*T=1[/math], and at each point where the exponent [math]k[/math] has a [math]\bullet[/math] entry we must use [math]S^*S=1[/math]. Thus the contributions, which are each worth 1, are parametrized by the partitions [math]\pi\in\mathcal{NC}_2(k)[/math], and we are done.

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We will be back to this in chapter 8 below. For our purposes here, the above theory is all we need. Getting back now to [math]U_N^+[/math], following , we can reformulate the main result that we have so far about it, by using the above notions, as follows:

Theorem

For the quantum group [math]U_N^+[/math] with [math]N\geq2[/math] we have

[[math]] Hom(u^{\otimes k},u^{\otimes l})=span\left(T_\pi\Big|\pi\in D(k,l)\right) [[/math]]

and at the level of the moments of the main character we have

[[math]] \int_{U_N^+}\chi^k\leq|\mathcal{NC}_2(k)| [[/math]]

with equality at [math]N\geq k[/math], the numbers on the right being the moments of [math]\Gamma_1[/math].

Show Proof

This is something that we already know. To be more precise, the Brauer type result is from chapter 4, the estimate for the moments follows from this and from Peter-Weyl, as explained in chapter 5, the equality at [math]N\geq k[/math] is something more subtle, explained in chapter 5, and the last statement comes from the above discussion.

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Summarizing, with a bit of abstract probability theory, of free type, we are now on our way into the study of [math]U_N^+[/math], paralleling the previous study of [math]O_N^+[/math].

General references

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