6c. Fusion rules

[math] \newcommand{\mathds}{\mathbb}[/math]

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With the above result in hand, we can now go ahead and do with [math]U_N^+[/math] exactly what we did with [math]O_N^+[/math] in chapter 5, with modifications where needed, namely constructing the irreducible representations by recurrence, using a Frobenius duality trick, computing the fusion rules, and concluding as well that we have [math]\chi\sim\Gamma_1[/math], at any [math]N\geq2[/math].

In practice, all this will be more complicated than for [math]O_N^+[/math], mainly because the fusion rules will be something new, in need of some preliminary combinatorial study. These fusion rules will be a kind of “free Clebsch-Gordan rules”, as follows:

[[math]] r_k\otimes r_l=\sum_{k=xy,l=\bar{y}z}r_{xz} [[/math]]

Following , let [math]W[/math] be the set of colored integers [math]k=\circ\bullet\bullet\circ\ldots[/math]\,, and consider the complex algebra [math]E[/math] spanned by [math]W[/math]. We have then an isomorphism, as follows:

[[math]] (\mathbb C \lt X,X^* \gt ,+,\cdot )\simeq(E,+,\cdot) [[/math]]

[[math]] X\to\circ\quad,\quad X^*\to\bullet [[/math]]

We define an involution on our algebra [math]E[/math], by antilinearity and antimultiplicativity, according to the following formulae, with [math]e[/math] being as usual the empty word:

[[math]] \bar{e}=e\quad,\quad\bar{\circ}=\bullet\quad,\quad \bar{\bullet}=\circ [[/math]]

With these conventions, we have the following result:

Proposition

The map [math]\times:W\times W\to E[/math] given by

[[math]] x\times y=\sum_{x=ag,y=\bar{g}b}ab [[/math]]

extends by linearity into an associative multiplication of [math]E[/math].

Show Proof

Observe first that [math]\times[/math] is well-defined, the sum being finite. Let us prove now that [math]\times[/math] is associative. Let [math]x,y,z\in W[/math]. Then:

[[math]] \begin{eqnarray*} (x\times y)\times z &=&\sum_{x=a\bar{g},y=gb} ab\times z\\ &=&\sum_{x=a\bar{g},y=gb,ab=ch,z=\bar{h}d}cd \end{eqnarray*} [[/math]]

Now observe that for [math]a,b,c,h\in W[/math] the equality [math]ab=ch[/math] is equivalent to [math]b=uh,c=au[/math] with [math]u\in W[/math], or to [math]a=cv,h=vb[/math] with [math]v\in W[/math]. Thus, we have:

[[math]] \begin{eqnarray*} (x\times y)\times z &=&\sum_{x=a\bar{g},y=guh,z=\bar{h}d}aud\\ &+&\sum_{x=cv\bar{g},y=gb,z=\overline{b}\bar{v}d}cd \end{eqnarray*} [[/math]]

A similar computation shows that [math]x\times (y\times z)[/math] is given by the same formula.

■

Still following , we have the following result:

Proposition

Consider the following morphism, with [math]S,T[/math] being the shifts,

[[math]] P:(E,+,\cdot )\to (B(l^2(W)),+,\circ ) [[/math]]

[[math]] \alpha\to S+T^* [[/math]]

and let [math]E_n\subset E[/math] be the linear space generated by the words of [math]W[/math] having length [math]\leq n[/math].

• If [math]J:E\to E[/math] is the map [math]f\to P(f)e[/math], then [math](J-Id)E_n\subset E_{n-1}[/math] for any [math]n[/math].
• [math]J[/math] is an isomorphism of [math]*[/math]-algebras [math](E,+,\cdot)\simeq(E,+,\times)[/math].

Show Proof

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

(1) Let [math]f\in E[/math]. We have then the following formula:

[[math]] P(\alpha )f=(S+T^*)f=\circ\times f [[/math]]

Thus, for any [math]g\in E[/math], we have the following formula:

[[math]] \begin{eqnarray*} J(\circ g) &=&P(\circ)J(g)\\ &=&\circ\times J(g)\\ &=&J(\circ)\times J(g) \end{eqnarray*} [[/math]]

The same argument shows that we have, for any [math]g\in E[/math]:

[[math]] J(\bullet g)=J(\bullet)\times J(g) [[/math]]

Now the algebra [math](E,+,\cdot )[/math] being generated by [math]\circ[/math] and [math]\bullet[/math], we conclude that [math]J[/math] is a morphism of algebras, as follows:

[[math]] J:(E,+,\cdot )\to (E,+,\times ) [[/math]]

We prove now by recurrence on [math]n\geq1[/math] that we have:

[[math]] (J-Id)E_n\subset E_{n-1} [[/math]]

At [math]n=1[/math] we have [math]J(\circ)=\circ[/math], [math]J(\bullet)=\bullet[/math] and [math]J(e)=e[/math], and since [math]E_1[/math] is generated by [math]e,\circ,\bullet[/math], we have [math]J=Id[/math] on [math]E_1[/math], as desired. Now assume that the above formula is true for [math]n[/math], and let [math]k\in E_{n+1}[/math]. We write, with [math]f,g,h\in E_n[/math]:

[[math]] k=\circ f+\bullet g+h [[/math]]

We have then the following computation:

[[math]] \begin{eqnarray*} (J-Id)k &=&J(\circ f+\bullet g+h)-(\circ f+\bullet g+h)\\ &=&[(S+T^*)J(f)+(S^*+T)J(g)+J(h)]-[Sf+Tg+h]\\ &=&S(J(f)-f)+T(J(g)-g)+T^*J(f)+S^*J(g)+(J(h)-h) \end{eqnarray*} [[/math]]

By using the recurrence assumption, applied to [math]f,g,h[/math] we find that [math]E_n[/math] contains all the terms of the above sum, and so contains [math](J-Id)k[/math], and we are done.

(2) Here we have to prove that [math]J[/math] preserves the involution [math]*[/math], and that it is bijective. We have [math]J*=*J[/math] on the generators [math]\{e,\circ,\bullet\}[/math] of [math]E[/math], so [math]J[/math] preserves the involution. Also, by (1), the restriction of [math]J-Id[/math] to [math]E_n[/math] is nilpotent, so [math]J[/math] is bijective.

■

Following , we can now formulate a main result about [math]U_N^+[/math], which is quite similar to the result for [math]O_N^+[/math] from chapter 5 above, as follows:

Theorem

For the quantum group [math]U_N^+[/math], with [math]N\geq2[/math], the main character follows the Voiculescu circular law,

[[math]] \chi\sim\Gamma_1 [[/math]]

and the irreducible representations can be labelled by the colored integers, [math]k=\circ\bullet\bullet\circ\ldots[/math]\,, with [math]r_e=1[/math], [math]r_\circ=u[/math], [math]r_\bullet=\bar{u}[/math], and with the involution and the fusion rules being

[[math]] \bar{r}_k=r_{\bar{k}} [[/math]]

[[math]] r_k\otimes r_l=\sum_{k=xy,l=\bar{y}z}r_{xz} [[/math]]

where [math]k\to\bar{k}[/math] is obtained by reversing the word, and switching the colors.

Show Proof

This is similar to the proof for [math]O_N^+[/math], as follows:

(1) In order to get familiar with the fusion rules, let us first work out a few values of the representations [math]r_k[/math], computed according to the formula in the statement:

[[math]] r_e=1 [[/math]]

[[math]] r_\circ=u [[/math]]

[[math]] r_\bullet=\bar{u} [[/math]]

[[math]] r_{\circ\circ}=u\otimes u [[/math]]

[[math]] r_{\circ\bullet}=u\otimes\bar{u}-1 [[/math]]

[[math]] r_{\bullet\circ}=\bar{u}\otimes u-1 [[/math]]

[[math]] r_{\bullet\bullet}=\bar{u}\otimes\bar{u} [[/math]]

[[math]] \vdots [[/math]]

(2) Equivalently, we want to decompose into irreducibles the Peter-Weyl representations, because the above formulae can be written as follows:

[[math]] u^{\otimes e}=r_e [[/math]]

[[math]] u^{\otimes\circ}=r_\circ [[/math]]

[[math]] u^{\otimes\bullet}=r_\bullet [[/math]]

[[math]] u^{\otimes\circ\circ}=r_{\circ\circ} [[/math]]

[[math]] u^{\otimes\circ\bullet}=r_{\circ\bullet}+r_e [[/math]]

[[math]] u^{\bullet\circ}=r_{\bullet\circ}+r_e [[/math]]

[[math]] u^{\bullet\bullet}=r_{\bullet\bullet} [[/math]]

[[math]] \vdots [[/math]]

(3) In order to prove the fusion rule assertion, let us construct a morphism as follows, by using the polynomiality of the algebra on the left:

[[math]] \Psi: (E,+,\times )\to C(U_N^+) [[/math]]

[[math]] \circ\to\chi (u) [[/math]]

[[math]] \bullet\to\chi(\bar{u}) [[/math]]

Our claim is that, given an integer [math]n\geq1[/math], assuming that [math]\Psi(x)[/math] is the character of an irreducible representation [math]r_x[/math] of [math]U_N^+[/math], for any [math]x\in W[/math] having length [math]\leq n[/math], then [math]\Psi(x)[/math] is the character of a non-null representation of [math]U_N^+[/math], for any [math]x\in W[/math] of length [math]n+1[/math].

(4) At [math]n=1[/math] this is clear. Assume [math]n\geq2[/math], and let [math]x\in W[/math] of length [math]n+1[/math]. If [math]x[/math] contains a [math]\geq 2[/math] power of [math]\circ[/math] or of [math]\bullet[/math], for instance if [math]x=z\circ\circ\,y[/math], then we can set:

[[math]] r_x=r_{z\circ}\otimes r_{\circ y} [[/math]]

Assume now that [math]x[/math] is an alternating product of [math]\circ[/math] and [math]\bullet[/math]. We can assume that [math]x[/math] begins with [math]\circ[/math]. Then [math]x=\circ\bullet\circ\,y[/math], with [math]y\in W[/math] being of length [math]n-2[/math]. Observe that [math]\Psi(\bar{z})=\Psi(z)^*[/math] holds on the generators [math]\{e,\circ,\bullet\}[/math] of [math]W[/math], so it holds for any [math]z\in W[/math]. Thus, we have:

[[math]] \begin{eqnarray*} \lt \chi (r_\circ\otimes r_{\bullet\circ y}),\chi (r_{\circ y}) \gt &=& \lt \chi (r_{\bullet\circ y}),\chi (r_\bullet\otimes r_{\circ y}) \gt \\ &=& \lt \chi (r_{\bullet\circ y}),\Psi(\bullet\times\circ y ) \gt \\ &=& \lt \chi (r_{\bullet\circ y}),\Psi(\bullet\circ y) +\Psi(y) \gt \\ &=& \lt \chi (r_{\bullet\circ y}),\chi (r_{\bullet\circ y})+\chi(r_y) \gt \\ &\geq&1 \end{eqnarray*} [[/math]]

Now since the corepresentation [math]r_{\circ y}[/math] is by assumption irreducible, we have [math]r_{\circ y}\subset r_\circ\otimes r_{\bullet\circ y}[/math]. Consider now the following quantity:

[[math]] \begin{eqnarray*} \chi (r_\circ\otimes r_{\bullet\circ y})-\chi (r_{\circ y}) &=&\Psi(\circ\times \bullet\circ y-\circ y)\\ &=&\Psi(x) \end{eqnarray*} [[/math]]

This is then the character of a representation, as desired.

(5) We know from easiness that we have the following estimate:

[[math]] \dim(Fix(u^{\otimes k}))\leq|\mathcal{NC}_2(k)| [[/math]]

By identifying as usual [math](\mathbb C \lt X,X^* \gt ,+,\cdot )=(E,+,\cdot )[/math], the noncommutative monomials in [math]X,X^*[/math] correspond to the elements of [math]W\subset E[/math]. Thus, we have, on [math]W[/math]:

[[math]] h\Psi J\leq\tau J [[/math]]

(6) We prove now by recurrence on [math]n\geq 0[/math] that for any [math]z\in W[/math] having length [math]n[/math], [math]\Psi(z)[/math] is the character of an irreducible representation [math]r_z[/math].

(7) At [math]n=0[/math] we have [math]\Psi_G(e)=1[/math]. So, assume that our claim holds at [math]n\geq 0[/math], and let [math]x\in W[/math] having length [math]n+1[/math]. By Proposition 6.19 (1) we have, with [math]z\in E_n[/math]:

[[math]] J(x)=x+z [[/math]]

Let [math]E^N\subset E[/math] be the set of functions [math]f[/math] such that [math]f(x)\in\mathbb N[/math] for any [math]x\in W[/math]. Then [math]J(\alpha ),J(\beta )\in E^N[/math], so by multiplicativity [math]J(W)\subset E^N[/math]. In particular, [math]J(x)\in E^N[/math]. Thus there exist numbers [math]m(z)\in\mathbb N[/math] such that:

[[math]] J(x)=x+\sum_{l(z)\leq n}m(z)z [[/math]]

(8) It is clear that for [math]a,b\in W[/math] we have [math]\tau(a\times\bar{b})=\delta_{a,b}[/math]. Thus:

[[math]] \begin{eqnarray*} \tau J(x\bar{x}) &=&\tau\left(\left(x+\sum m(z)z\right)\times\left(\bar{x}+\sum m(z)\bar{z}\right)\right)\\ &=&1+\sum m(z)^2 \end{eqnarray*} [[/math]]

(9) By recurrence and by (3), [math]\Psi(x)[/math] is the character of a representation [math]r_x[/math]. Thus [math]\Psi J(x)[/math] is the character of [math]r_x+\sum_{l(z)\leq n}m(z)r_z[/math], and we obtain from this:

[[math]] h\Psi J(x\bar{x})\geq h(\chi (r_x)\chi (r_x)^*)+\sum m(z)^2 [[/math]]

(10) By using (5), (8), (9) we conclude that [math]r_x[/math] is irreducible, which proves (6).

(11) The fact that the [math]r_x[/math] are distinct comes from (5). Indeed, [math]W[/math] being an orthonormal basis of [math]((E,+,\times ),\tau )[/math], for any [math]x,y\in W[/math], [math]x\neq y[/math] we have [math]\tau (x\times\bar{y})=0[/math], and so:

[[math]] \begin{eqnarray*} h(\chi (r_x\otimes\bar{r_y})) &=&h\Psi J(x\bar{y})\\ &\leq&\tau J(x\bar{y})\\ &=&\tau (x\times\bar{y})\\ &=&0 \end{eqnarray*} [[/math]]

(12) The fact that we obtain all the irreducible representations is clear too, because we can now decompose all the tensor powers [math]u^{\otimes k}[/math] into irreducibles.

(13) Finally, since [math]W[/math] is an orthonormal system in [math]((E,+,\times ),\tau )[/math], the set [math]\Psi(W)=\{\chi (r_x)|x\in W\}[/math] is an orthonormal system in [math]C(U_N^+)[/math], and so we have:

[[math]] h\Psi J=\tau_0P [[/math]]

Now since the distribution of [math]\chi (u)\in (C(G),h)[/math] is the functional [math]h\Psi_GJ[/math], and the distribution of [math]S+T^*\in(B(l^2(\mathbb N*\mathbb N)),\tau_0)[/math] is the functional [math]\tau_0P[/math], we have [math]\chi\sim\Gamma_1[/math], as claimed.

■

The above proof, from , is the original proof, still doing well after all these years, but there are some alternative proofs as well, to be discussed in the next section.

General references

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