Free rotations

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

This article was automatically generated from a tex file and may contain conversion errors. If permitted, you may login and edit this article to improve the conversion.

5a. Gram determinants

We have seen that Tannakian duality allows us to get some substantial insight into the representation theory of [math]O_N^+,U_N^+[/math], with a free analogue of the classical Brauer theorem for [math]O_N,U_N[/math]. In this second part of the present book we discuss some concrete applications of this result, and some generalizations, following , then , and then .

Let us begin with a summary of the Brauer type results established in the previous chapter. The statement here, collecting what we have so far, is as follows:

Theorem

For the basic unitary quantum groups, namely

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

the intertwiners between the Peter-Weyl representations are given by

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

with the linear maps [math]T_\pi[/math] associated to the pairings [math]\pi[/math] being given by

[[math]] T_\pi(e_{i_1}\otimes\ldots\otimes e_{i_k})=\sum_{j_1\ldots j_l}\delta_\pi\begin{pmatrix}i_1&\ldots&i_k\\ j_1&\ldots&j_l\end{pmatrix}e_{j_1}\otimes\ldots\otimes e_{j_l} [[/math]]

and with the corresponding sets of pairings [math]D[/math] being as follows,

[[math]] \xymatrix@R=16mm@C=15mm{ \mathcal P_2\ar[d]&\mathcal{NC}_2\ar[l]\ar[d]\\ P_2&NC_2\ar[l]} [[/math]]

with calligraphic standing for matching, and with NC standing for noncrossing.

Show Proof

This is indeed a summary of the results that we have, established in the previous chapter, and coming from Tannakian duality, via some combinatorics.

■

In order to work out now some concrete applications, such as the classification of the irreducible representations of [math]O_N^+,U_N^+[/math], we must do some combinatorics. The problem indeed is that we do not know whether the linear maps [math]T_\pi[/math] in Theorem 5.1 are linearly independent or not, so we must solve this problem first. Things are quite tricky here, technically speaking, and we will solve this question as follows:

• By Frobenius duality, it is enough to examine the vectors [math]\xi_\pi=T_\pi[/math] associated to the pairings [math]\pi\in P_2(0,l)[/math], having no upper points.
• In order to decide whether these vectors [math]\xi_\pi[/math] are linearly independent or not, we will compute the determinant of their Gram matrix.
• We will actually compute the determinant of a bigger Gram matrix, that of the vectors [math]\xi_\pi=T_\pi[/math] coming from arbitrary partitions [math]\pi\in P(0,l)[/math], which is simpler.

What we have here is an accumulation of tricks, and some changes in notations too. By replacing [math]l\to k[/math] as well, for making things look better, we are led in this way to:

Proposition

To any partition [math]\pi\in P(k)[/math] we associate the vector

[[math]] \xi_\pi=\sum_{i_1\ldots i_k}\delta_\pi(i_1,\ldots,i_k)\,e_{i_1}\otimes\ldots\otimes e_{i_k} [[/math]]

with the Kronecker symbols being defined as usual, according to whether the indices fit or not. The Gram matrix of these vectors is then given by

[[math]] G_k(\pi,\sigma)=N^{|\pi\vee\sigma|} [[/math]]

where [math]\pi\vee\sigma\in P(k)[/math] is obtained by superposing [math]\pi,\sigma[/math], and [math]|\,.\,|[/math] is the number of blocks.

Show Proof

According to the formula of the vectors [math]\xi_\pi[/math], we have:

[[math]] \begin{eqnarray*} \lt \xi_\pi,\xi_\sigma \gt &=&\sum_{i_1\ldots i_k}\delta_\pi(i_1,\ldots,i_k)\delta_\sigma(i_1,\ldots,i_k)\\ &=&\sum_{i_1\ldots i_k}\delta_{\pi\vee\sigma}(i_1,\ldots,i_k)\\ &=&N^{|\pi\vee\sigma|} \end{eqnarray*} [[/math]]

Thus, we have obtained the formula in the statement.

■

Our goal in what follows will be that of computing [math]\det(G_k)[/math]. Which will actually take some time, to the point that you might start wondering, sometimes soon, if this is really the right thing to do, right now. To which I would say that yes, this is the right thing to do. Our goal is to understand the closed subgroups [math]G\subset U_N^+[/math], and 0 chances with that, until we know what this [math]U_N^+[/math] beast is. And for this, we need to compute [math]\det(G_k)[/math].

As an illustration now, at [math]k=2[/math] we have [math]P(2)=\{||,\sqcap\}[/math], and the Gram matrix is:

[[math]] G_2=\begin{pmatrix}N^2&N\\ N&N\end{pmatrix} [[/math]]

At [math]k=3[/math], we have [math]P(3)=\{|||,\sqcap|,\sqcap\hskip-3.2mm{\ }_|\,,|\sqcap,\sqcap\hskip-0.7mm\sqcap\}[/math], and the Gram matrix is:

[[math]] G_3=\begin{pmatrix} N^3&N^2&N^2&N^2&N\\ N^2&N^2&N&N&N\\ N^2&N&N^2&N&N\\ N^2&N&N&N^2&N\\ N&N&N&N&N \end{pmatrix} [[/math]]

These matrices might not look that bad, to the untrained eye, but in practice, their combinatorics can be fairly complicated. As an example here, the submatrix of [math]G_k[/math] coming from the usual pairings, that we are really interested in, according to Theorem 5.1, has as determinant a product of terms indexed by Young tableaux. This is actually why we use [math]G_k[/math], because, as we will soon discover, this matrix is something quite simple.

In order to compute the determinant of [math]G_k[/math], we will use a standard combinatorial trick, related to the Möbius inversion formula. Let us start with:

Definition

Given two partitions [math]\pi,\sigma\in P(k)[/math], we write

[[math]] \pi\leq\sigma [[/math]]

if each block of [math]\pi[/math] is contained in a block of [math]\sigma[/math].

Observe that this order is compatible with the previous convention for [math]\pi\vee\sigma[/math], in the sense that the [math]\vee[/math] operation is the supremum operation with respect to [math]\leq[/math]. At the level of examples, at [math]k=2[/math] we have [math]P(2)=\{||,\sqcap\}[/math], and the order relation is as follows:

[[math]] ||\leq\sqcap [[/math]]

At [math]k=3[/math] now, we have [math]P(3)=\{|||,\sqcap|,\sqcap\hskip-3.2mm{\ }_|\,,|\sqcap,\sqcap\hskip-0.7mm\sqcap\}[/math], and the order relation is:

[[math]] |||\leq\sqcap|,\sqcap\hskip-3.2mm{\ }_|\,,|\sqcap\leq\sqcap\hskip-0.7mm\sqcap [[/math]]

Summarizing, this order is very intuitive, and simple to compute. By using now this order, we can talk about the Möbius function of [math]P(k)[/math], as follows:

Definition

The Möbius function of any lattice, and so of [math]P(k)[/math], is given by

[[math]] \mu(\pi,\sigma)=\begin{cases} 1&{\rm if}\ \pi=\sigma\\ -\sum_{\pi\leq\tau \lt \sigma}\mu(\pi,\tau)&{\rm if}\ \pi \lt \sigma\\ 0&{\rm if}\ \pi\not\leq\sigma \end{cases} [[/math]]

with this construction being performed by recurrence.

This is something standard in combinatorics. As an illustration here, let us go back to the set of 2-point partitions, [math]P(2)=\{||,\sqcap\}[/math]. We have by definition:

[[math]] \mu(||,||)=\mu(\sqcap,\sqcap)=1 [[/math]]

Next in line, we know that we have [math]|| \lt \sqcap[/math], with no intermediate partition in between, and so the above recurrence procedure gives:

[[math]] \mu(||,\sqcap)=-\mu(||,||)=-1 [[/math]]

Finally, we have [math]\sqcap\not\leq||[/math], and so the last value of the Möbius function is:

[[math]] \mu(\sqcap,||)=0 [[/math]]

Thus, as a conclusion, we have computed the Möbius matrix [math]M_2(\pi,\sigma)=\mu(\pi,\sigma)[/math] of the lattice [math]P(2)=\{||,\sqcap\}[/math], the formula of this matrix being as follows:

[[math]] M_2=\begin{pmatrix}1&-1\\ 0&1\end{pmatrix} [[/math]]

The computation for [math]P(3)=\{|||,\sqcap|,\sqcap\hskip-3.2mm{\ }_|\,,|\sqcap,\sqcap\hskip-0.7mm\sqcap\}[/math] is similar, and leads to the following formula for the associated Möbius matrix:

[[math]] M_3=\begin{pmatrix} 1&-1&-1&-1&2\\ 0&1&0&0&-1\\ 0&0&1&0&-1\\ 0&0&0&1&-1\\ 0&0&0&0&1 \end{pmatrix} [[/math]]

In general, the Möbius matrix of [math]P(k)[/math] looks a bit like the above matrices at [math]k=2,3[/math], being upper triangular, with 1 on the diagonal, and so on. We will be back to this.

Back to the general case now, the main interest in the Möbius function comes from the Möbius inversion formula, which states that the following happens:

[[math]] f(\sigma)=\sum_{\pi\leq\sigma}g(\pi) \quad\implies\quad g(\sigma)=\sum_{\pi\leq\sigma}\mu(\pi,\sigma)f(\pi) [[/math]]

In linear algebra terms, the statement and proof of this formula are as follows:

Theorem

The inverse of the adjacency matrix of [math]P(k)[/math], given by

[[math]] A_k(\pi,\sigma)=\begin{cases} 1&{\rm if}\ \pi\leq\sigma\\ 0&{\rm if}\ \pi\not\leq\sigma \end{cases} [[/math]]

is the Möbius matrix of [math]P[/math], given by [math]M_k(\pi,\sigma)=\mu(\pi,\sigma)[/math].

Show Proof

This is well-known, coming for instance from the fact that [math]A_k[/math] is upper triangular. Indeed, when inverting, we are led into the recurrence from Definition 5.4.

■

As an illustration, for [math]P(2)=\{||,\sqcap\}[/math] the formula [math]M_2=A_2^{-1}[/math] appears as follows:

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

Also, for [math]P(3)=\{|||,\sqcap|,\sqcap\hskip-3.2mm{\ }_|\,,|\sqcap,\sqcap\hskip-0.7mm\sqcap\}[/math] the formula [math]M_3=A_3^{-1}[/math] reads:

[[math]] \begin{pmatrix} 1&-1&-1&-1&2\\ 0&1&0&0&-1\\ 0&0&1&0&-1\\ 0&0&0&1&-1\\ 0&0&0&0&1 \end{pmatrix}= \begin{pmatrix} 1&1&1&1&1\\ 0&1&0&0&1\\ 0&0&1&0&1\\ 0&0&0&1&1\\ 0&0&0&0&1 \end{pmatrix}^{-1} [[/math]]

Now back to our Gram matrix considerations, we have the following key result, based on this technology, which basically solves our determinant question:

Proposition

The Gram matrix is given by [math]G_k=A_kL_k[/math], where

[[math]] L_k(\pi,\sigma)= \begin{cases} N(N-1)\ldots(N-|\pi|+1)&{\rm if}\ \sigma\leq\pi\\ 0&{\rm otherwise} \end{cases} [[/math]]

and where [math]A_k=M_k^{-1}[/math] is the adjacency matrix of [math]P(k)[/math].

Show Proof

We have the following computation, using Proposition 5.2:

[[math]] \begin{eqnarray*} G_k(\pi,\sigma) &=&N^{|\pi\vee\sigma|}\\ &=&\#\left\{i_1,\ldots,i_k\in\{1,\ldots,N\}\Big|\ker i\geq\pi\vee\sigma\right\}\\ &=&\sum_{\tau\geq\pi\vee\sigma}\#\left\{i_1,\ldots,i_k\in\{1,\ldots,N\}\Big|\ker i=\tau\right\}\\ &=&\sum_{\tau\geq\pi\vee\sigma}N(N-1)\ldots(N-|\tau|+1) \end{eqnarray*} [[/math]]

According now to the definition of [math]A_k,L_k[/math], this formula reads:

[[math]] \begin{eqnarray*} G_k(\pi,\sigma) &=&\sum_{\tau\geq\pi}L_k(\tau,\sigma)\\ &=&\sum_\tau A_k(\pi,\tau)L_k(\tau,\sigma)\\ &=&(A_kL_k)(\pi,\sigma) \end{eqnarray*} [[/math]]

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

■

As an illustration for the above result, at [math]k=2[/math] we have [math]P(2)=\{||,\sqcap\}[/math], and the above decomposition [math]G_2=A_2L_2[/math] appears as follows:

[[math]] \begin{pmatrix}N^2&N\\ N&N\end{pmatrix} =\begin{pmatrix}1&1\\ 0&1\end{pmatrix} \begin{pmatrix}N^2-N&0\\N&N\end{pmatrix} [[/math]]

At [math]k=3[/math] now, we have [math]P(3)=\{|||,\sqcap|,\sqcap\hskip-3.2mm{\ }_|\,,|\sqcap,\sqcap\hskip-0.7mm\sqcap\}[/math], and the Gram matrix is:

[[math]] G_3=\begin{pmatrix} N^3&N^2&N^2&N^2&N\\ N^2&N^2&N&N&N\\ N^2&N&N^2&N&N\\ N^2&N&N&N^2&N\\ N&N&N&N&N \end{pmatrix} [[/math]]

Regarding [math]L_3[/math], this can be computed by writing down the matrix [math]E_3(\pi,\sigma)=\delta_{\sigma\leq\pi}|\pi|[/math], and then replacing each entry by the corresponding polynomial in [math]N[/math]. We reach to the conclusion that the product [math]A_3L_3[/math] is as follows, producing the above matrix [math]G_3[/math]:

[[math]] A_3L_3=\begin{pmatrix} 1&1&1&1&1\\ 0&1&0&0&1\\ 0&0&1&0&1\\ 0&0&0&1&1\\ 0&0&0&0&1 \end{pmatrix} \begin{pmatrix} N^3-3N^2+2N&0&0&0&0\\ N^2-N&N^2-N&0&0&0\\ N^2-N&0&N^2-N&0&0\\ N^2-N&0&0&N^2-N&0\\ N&N&N&N&N \end{pmatrix} [[/math]]

In general, the formula [math]G_k=A_kL_k[/math] appears a bit in the same way, with [math]A_k[/math] being binary and upper triangular, and with [math]L_k[/math] depending on [math]N[/math], and being lower triangular.

We are led in this way to the following formula, due to Lindstöm :

Theorem

The determinant of the Gram matrix [math]G_k[/math] is given by

[[math]] \det(G_k)=\prod_{\pi\in P(k)}\frac{N!}{(N-|\pi|)!} [[/math]]

with the convention that in the case [math]N \lt k[/math] we obtain [math]0[/math].

Show Proof

If we order [math]P(k)[/math] as usual, with respect to the number of blocks, and then lexicographically, then [math]A_k[/math] is upper triangular, and [math]L_k[/math] is lower triangular. Thus, we have:

[[math]] \begin{eqnarray*} \det(G_k) &=&\det(A_k)\det(L_k)\\ &=&\det(L_k)\\ &=&\prod_\pi L_k(\pi,\pi)\\ &=&\prod_\pi N(N-1)\ldots(N-|\pi|+1) \end{eqnarray*} [[/math]]

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

■

We refer to for more on Gram determinants, and their conceptual meaning, from a modern perspective. Getting back now to quantum groups, or rather to the corresponding Tannakian categories, written as spans of diagrams, we have the following result:

Theorem

The vectors associated to the partitions, namely

[[math]] \left\{\xi_\pi\in(\mathbb C^N)^{\otimes k}\Big|\pi\in P(k)\right\} [[/math]]

and in particular the vectors associated to the pairings, namely

[[math]] \left\{\xi_\pi\in(\mathbb C^N)^{\otimes k}\Big|\pi\in P_2(k)\right\} [[/math]]

are linearly independent for [math]N\geq k[/math].

Show Proof

Here the first assertion follows from Theorem 5.7, the Gram determinant computed there being nonzero for [math]N\geq k[/math], and the second assertion follows from it.

■

In what follows, the above result will be all that we need, for deducing a number of interesting consequences regarding [math]O_N,U_N,O_N^+,U_N^+[/math]. Once these corollaries exhausted, we will have to go back to this, and work out some finer linear independence results.

5b. The Wigner law

We discuss here some applications of the above linear independence results, following , . As a first application, we can study the laws of characters. First, we have:

Proposition

For the basic unitary quantum groups, namely

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

the moments of the main character, which are the numbers [math]M_k=\int_G\chi^k[/math], depending on a colored integer [math]k[/math], are smaller than the following numbers,

[[math]] \xymatrix@R=16mm@C=15mm{ |\mathcal P_2(k)|\ar@{-}[d]&|\mathcal{NC}_2(k)|\ar@{-}[l]\ar@{-}[d]\\ |P_2(k)|&|NC_2(k)|\ar@{-}[l]} [[/math]]

and with equality happening in each case at [math]N\geq k[/math].

Show Proof

We have the following computation, based on Theorem 5.1, and on the character formulae from Peter-Weyl theory, for each of our quantum groups:

[[math]] \begin{eqnarray*} \int_G\chi^k &=&\dim(Fix(u^{\otimes k}))\\ &=&\dim\left(span\left(\xi_\pi\Big|\pi\in D(k)\right)\right)\\ &\leq&|D(k)| \end{eqnarray*} [[/math]]

Thus, we have the inequalities in the statement, coming from easiness and Peter-Weyl. As for the last assertion, this follows from Theorem 5.8.

■

In order to advance now, we must do some combinatorics and probability, first by counting the numbers in Proposition 5.9, and then by recovering the measures having these numbers as moments. We will restrict the attention to the orthogonal case, which is simpler, and leave the unitary case, which is more complicated, for later.

Since there are no pairings when [math]k[/math] is odd, we can assume that [math]k[/math] is even, and with the change [math]k\to 2k[/math], the partition count in the orthogonal case is as follows:

Proposition

We have the following formulae for pairings,

[[math]] |P_2(2k)|=(2k)!! [[/math]]

[[math]] |NC_2(2k)|=C_k [[/math]]

with the numbers involved, double factorials and Catalan numbers, being as follows:

[[math]] (2k)!!=(2k-1)(2k-3)(2k-5)\ldots [[/math]]

[[math]] C_k=\frac{1}{k+1}\binom{2k}{k} [[/math]]

Show Proof

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

(1) We must count the pairings of [math]\{1,\ldots,2k\}[/math]. Now observe that such a pairing appears by pairing 1 to a certain number, and there are [math]2k-1[/math] choices here, then pairing the next number, 2 if free or 3 if 2 was taken, to another number, and there are [math]2k-3[/math] choices here, and so on. Thus, we are led to the formula in the statement, namely:

[[math]] |P_2(2k)|=(2k-1)(2k-3)(2k-5)\ldots [[/math]]

(2) We must count the noncrossing pairings of [math]\{1,\ldots,2k\}[/math]. Now observe that such a pairing appears by pairing 1 to an odd number, [math]2a+1[/math], and then inserting a noncrossing pairing of [math]\{2,\ldots,2a\}[/math], and a noncrossing pairing of [math]\{2a+2,\ldots,2k\}[/math]. We conclude from this that we have the following recurrence for the numbers [math]C_k=|NC_2(2k)|[/math]:

[[math]] C_k=\sum_{a+b=k-1}C_aC_b [[/math]]

Consider now the generating series of these numbers:

[[math]] f(z)=\sum_{k\geq0}C_kz^k [[/math]]

In terms of this generating series, the recurrence that we found gives:

[[math]] \begin{eqnarray*} zf^2 &=&\sum_{a,b\geq0}C_aC_bz^{a+b+1}\\ &=&\sum_{k\geq1}\sum_{a+b=k-1}C_aC_bz^k\\ &=&\sum_{k\geq1}C_kz^k\\ &=&f-1 \end{eqnarray*} [[/math]]

Thus the generating series satisfies the following degree 2 equation:

[[math]] zf^2-f+1=0 [[/math]]

Now by solving this equation, using the usual degree 2 formula, and choosing the solution which is bounded at [math]z=0[/math], we obtain:

[[math]] f(z)=\frac{1-\sqrt{1-4z}}{2z} [[/math]]

By using now the Taylor formula for [math]\sqrt{x}[/math], we obtain the following formula:

[[math]] f(z)=\sum_{k\geq0}\frac{1}{k+1}\binom{2k}{k}z^k [[/math]]

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

■

Let us do now the second computation, which is probabilistic. We must find the real probability measures having the above numbers as moments, and we have here:

Theorem

The standard Gaussian law, and standard Wigner semicircle law

[[math]] g_1=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx [[/math]]

[[math]] \gamma_1=\frac{1}{2\pi}\sqrt{4-x^2}dx [[/math]]

have as [math]2k[/math]-th moments the numbers [math](2k)!![/math] and [math]C_k[/math], and their odd moments vanish.

Show Proof

There are several proofs here, depending on your calculus and probability knowledge. Normally the “honest”, white belt proof would be by trying to find centered measures [math]g_1,\gamma_1[/math] having as even moments the numbers [math](2k)!![/math] and [math]C_k[/math]. But this is something quite complicated, requiring the usage of the Stieltjes inversion formula, namely:

[[math]] d\mu (x)=\lim_{t\searrow 0}-\frac{1}{\pi}\,Im\left(G(x+it)\right)\cdot dx [[/math]]

Now the problem is that, assuming that you master this formula, you have certainly learned enough probability as to know about the solutions [math]g_1,\gamma_1[/math] to our problem. So, we will just cheat, assume that the laws [math]g_1,\gamma_1[/math] are found, and proceed as follows:

(1) The moments of the normal law [math]g_1[/math] in the statement are given by:

[[math]] \begin{eqnarray*} M_k &=&\frac{1}{\sqrt{2\pi}}\int_\mathbb Rx^ke^{-x^2/2}dx\\ &=&\frac{1}{\sqrt{2\pi}}\int_\mathbb R(x^{k-1})\left(-e^{-x^2/2}\right)'dx\\ &=&\frac{1}{\sqrt{2\pi}}\int_\mathbb R(k-1)x^{k-2}e^{-x^2/2}dx\\ &=&(k-1)\times\frac{1}{\sqrt{2\pi}}\int_\mathbb Rx^{k-2}e^{-x^2/2}dx\\ &=&(k-1)M_{k-2} \end{eqnarray*} [[/math]]

Thus by recurrence we have [math]M_{2k}=(2k)!![/math], and we are done.

(2) The moments of the Wigner law [math]\gamma_1[/math] in the statement are given by:

[[math]] \begin{eqnarray*} N_k &=&\frac{1}{2\pi}\int_{-2}^2\sqrt{4-x^2}\,x^{2k}dx\\ &=&\frac{1}{2\pi}\int_0^\pi\sqrt{4-4\cos^2t}\,(2\cos t)^{2k}(2\sin t)dt\\ &=&\frac{2^{2k+1}}{\pi}\int_0^\pi\cos^{2k}t\sin^2tdt\\ &=&\frac{2^{2k+1}}{\pi}\cdot\frac{(2k)!!2!!}{(2k+3)!!}\cdot\pi\\ &=&2^{2k+1}\cdot\frac{3\cdot5\cdot7\ldots(2k-1)}{2\cdot 4\cdot6\ldots(2k+2)}\\ &=&2^{2k+1}\cdot\frac{(2k)!}{2^kk!2^{k+1}(k+1)!}\\ &=&\frac{(2k)!}{k!(k+1)!} \end{eqnarray*} [[/math]]

Here we have used an advanced calculus formula, but a routine computation based on partial integration works as well. Thus we have [math]N_k=C_k[/math], and we are done.

■

As a comment here, the advanced calculus formula used in (2) above is as follows, with [math]\varepsilon(p)=1[/math] if [math]p[/math] is even and [math]\varepsilon(p)=0[/math] if [math]p[/math] is odd, and with [math]m!!=(m-1)(m-3)(m-5)\ldots[/math], with the product ending at [math]2[/math] if [math]m[/math] is odd, and ending at [math]1[/math] if [math]m[/math] is even:

[[math]] \int_0^{\pi/2}\cos^pt\sin^qt\,dt=\left(\frac{\pi}{2}\right)^{\varepsilon(p)\varepsilon(q)}\frac{p!!q!!}{(p+q+1)!!} [[/math]]

This formula is something extremely useful, in everyday life, with the proof being by partial integration, and then a double recurrence on [math]p,q[/math]. With spherical coordinates and Fubini it is possible to generalize this into an integration formula over the arbitrary real spheres [math]S^{N-1}_\mathbb R[/math], in arbitrary dimension [math]N\in\mathbb N[/math], but more on this later.

Now back to our orthogonal quantum groups, by using the above we can formulate a clear and concrete result regarding them, as follows:

Theorem

For the quantum groups [math]O_N,O_N^+[/math], the main character

[[math]] \chi=\sum_iu_{ii} [[/math]]

follows respectively the standard Gaussian, and the Wigner semicircle law

[[math]] g_1=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx\quad,\quad \gamma_1=\frac{1}{2\pi}\sqrt{4-x^2}dx [[/math]]

in the [math]N\to\infty[/math] limit.

Show Proof

This follows by putting together the results that we have, namely Proposition 5.9 applied with [math]N \gt k[/math], and then Proposition 5.10 and Theorem 5.11.

■

The above result is quite interesting, making the link with the law of Wigner . Note also that this is the first application of our Tannakian duality methods, developed in chapter 4. We will see in what follows countless versions and generalizations of it, basically obtained by using the same method, Tannakian duality and easiness first, then combinatorics for linear independence, and then more combinatorics and probability.

5c. Clebsch-Gordan rules

Let us try now to work out some finer results, at fixed values of [math]N\in\mathbb N[/math]. In the case of [math]O_N[/math] the above result cannot really be improved, the fixed [math]N\in\mathbb N[/math] laws being fairly complicated objects, related to Young tableaux and their combinatorics.

In the case of [math]O_N^+[/math], however, we will see that some miracles happen, and the convergence in the above result is in fact stationary, starting from [math]N=2[/math]. Following , we have:

Theorem

For the quantum group [math]O_N^+[/math], the main character follows the standard Wigner semicircle law, and this regardless of the value of [math]N\geq 2[/math]:

[[math]] \chi\sim\frac{1}{2\pi}\sqrt{4-x^2}dx [[/math]]

The irreducible representations of [math]O_N^+[/math] are all self-adjoint, and can be labelled by positive integers, with their fusion rules being the Clebsch-Gordan ones,

[[math]] r_k\otimes r_l=r_{|k-l|}+r_{|k-l|+2}+\ldots+r_{k+l} [[/math]]

as for the group [math]SU_2[/math]. The dimensions of these representations are given by

[[math]] \dim r_k=\frac{q^{k+1}-q^{-k-1}}{q-q^{-1}} [[/math]]

where [math]q,q^{-1}[/math] are the solutions of [math]X^2-NX+1=0[/math].

Show Proof

There are several proofs for this fact, the simplest one being via purely algebraic methods, based on the easiness property of [math]O_N^+[/math] from Theorem 5.1:

(1) In order to get started, let us first work out the first few values of the representations [math]r_k[/math] that we want to construct, computed by recurrence, according to the Clebsch-Gordan rules in the statement, which will be useful for various illustrations:

[[math]] r_0=1 [[/math]]

[[math]] r_1=u [[/math]]

[[math]] r_2=u^{\otimes 2}-1 [[/math]]

[[math]] r_3=u^{\otimes 3}-2u [[/math]]

[[math]] r_4=u^{\otimes 4}-3u^{\otimes 2}+1 [[/math]]

[[math]] r_5=u^{\otimes 5}-4u^{\otimes 3}+3u [[/math]]

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

(2) We can see that what we want to do is to split the Peter-Weyl representations [math]u^{\otimes k}[/math] into irreducibles, because the above formulae can be written as well as follows:

[[math]] u^{\otimes0}=r_0 [[/math]]

[[math]] u^{\otimes1}=r_1 [[/math]]

[[math]] u^{\otimes2}=r_2+r_0 [[/math]]

[[math]] u^{\otimes3}=r_3+2r_1 [[/math]]

[[math]] u^{\otimes4}=r_4+3r_2+2r_0 [[/math]]

[[math]] u^{\otimes5}=r_5+4r_3+5r_1 [[/math]]

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

(3) In order to get fully started now, our claim, which will basically prove the theorem, is that we can define, by recurrence on [math]k\in\mathbb N[/math], a sequence [math]r_0,r_1,r_2,\ldots[/math] of irreducible, self-adjoint and distinct representations of [math]O_N^+[/math], satisfying:

[[math]] r_0=1 [[/math]]

[[math]] r_1=u [[/math]]

[[math]] r_k+r_{k-2}=r_{k-1}\otimes r_1 [[/math]]

(4) Indeed, at [math]k=0[/math] this is clear, and at [math]k=1[/math] this is clear as well, with the irreducibility of [math]r_1=u[/math] coming from the embedding [math]O_N\subset O_N^+[/math]. So assume now that [math]r_0,\ldots,r_{k-1}[/math] as above are constructed, and let us construct [math]r_k[/math]. We have, by recurrence:

[[math]] r_{k-1}+r_{k-3}=r_{k-2}\otimes r_1 [[/math]]

In particular we have an inclusion of representations, as follows:

[[math]] r_{k-1}\subset r_{k-2}\otimes r_1 [[/math]]

Now since [math]r_{k-2}[/math] is irreducible, by Frobenius reciprocity we have:

[[math]] r_{k-2}\subset r_{k-1}\otimes r_1 [[/math]]

Thus, there exists a certain representation [math]r_k[/math] such that:

[[math]] r_k+r_{k-2}=r_{k-1}\otimes r_1 [[/math]]

(5) As a first observation, this representation [math]r_k[/math] is self-adjoint. Indeed, our recurrence formula [math]r_k+r_{k-2}=r_{k-1}\otimes r_1[/math] for the representations [math]r_0,r_1,r_2,\ldots[/math] shows that the characters of these representations are polynomials in [math]\chi_u[/math]. Now since [math]\chi_u[/math] is self-adjoint, all the characters that we can obtain via our recurrence are self-adjoint as well.

(6) It remains to prove that [math]r_k[/math] is irreducible, and non-equivalent to [math]r_0,\ldots,r_{k-1}[/math]. For this purpose, observe that according to our recurrence formula, [math]r_k+r_{k-2}=r_{k-1}\otimes r_1[/math], we can now split [math]u^{\otimes k}[/math], as a sum of the following type, with positive coefficients:

[[math]] u^{\otimes k}=c_kr_k+c_{k-2}r_{k-2}+\ldots [[/math]]

We conclude by Peter-Weyl that we have an inequality as follows, with equality precisely when [math]r_k[/math] is irreducible, and non-equivalent to the other summands [math]r_i[/math]:

[[math]] \sum_ic_i^2\leq\dim(End(u^{\otimes k})) [[/math]]

(7) Now let us use the easiness property of [math]O_N^+[/math]. This gives us an upper bound for the number on the right, that we can add to our inequality, as follows:

[[math]] \sum_ic_i^2 \leq\dim(End(u^{\otimes k})) \leq C_k [[/math]]

The point now is that the coefficients [math]c_i[/math] come straight from the Clebsch-Gordan rules, and their combinatorics shows that [math]\sum_ic_i^2[/math] equals the Catalan number [math]C_k[/math], with the remark that this follows as well from the known theory of [math]SU_2[/math]. Thus, we have global equality in the above estimate, and in particular we have equality at left, as desired.

(8) In order to finish the proof of our claim, it still remains to prove that [math]r_k[/math] is non-equivalent to [math]r_{k-1},r_{k-3},\ldots[/math] But these latter representations appear inside [math]u^{\otimes k-1}[/math], and the result follows by using the embedding [math]O_N\subset O_N^+[/math], which shows that the even and odd tensor powers of [math]u[/math] cannot have common irreducible components.

(9) Summarizing, we have proved our claim, made in step (3) above.

(10) In order now to finish, since by the Peter-Weyl theory any irreducible representation of [math]O_N^+[/math] must appear in some tensor power of [math]u[/math], and we have a formula for decomposing each [math]u^{\otimes k}[/math] into sums of representations [math]r_k[/math], as explained above, we conclude that these representations [math]r_k[/math] are all the irreducible representations of [math]O_N^+[/math].

(11) In what regards now the law of the main character, we obtain here the Wigner law [math]\gamma_1[/math], as stated, due to the fact that the equality in (7) gives us the even moments of this law, and that the observation in (8) tells us that the odd moments vanish.

(12) Finally, from the Clebsch-Gordan rules we have in particular:

[[math]] r_kr_1=r_{k-1}+r_{k+1} [[/math]]

We obtain from this, by recurrence, with [math]q^2-Nq+1=0[/math]:

[[math]] \dim r_k=q^k+q^{k-2}+\ldots+q^{-k+2}+q^{-k} [[/math]]

But this gives the dimension formula in the statement, and we are done.

■

Let us discuss now the relation with [math]SU_2[/math]. This group is the most well-known group in mathematics, and there is an enormous quantity of things known about it. For our purposes, we need a functional analytic approach to it. This can be done as follows:

Theorem

The algebra of continuous functions on [math]SU_2[/math] appears as

[[math]] C(SU_2)=C^*\left((u_{ij})_{i,j=1,2}\Big|u=F\bar{u}F^{-1}={\rm unitary}\right) [[/math]]

where [math]F[/math] is the following matrix,

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

called super-identity matrix.

Show Proof

This can be done in several steps, as follows:

(1) Let us first compute [math]SU_2[/math]. Consider an arbitrary [math]2\times2[/math] complex matrix:

[[math]] U=\begin{pmatrix}a&b\\c&d\end{pmatrix} [[/math]]

Assuming [math]\det U=1[/math], the unitarity condition [math]U^{-1}=U^*[/math] reads:

[[math]] \begin{pmatrix}d&-b\\-c&a\end{pmatrix} =\begin{pmatrix}\bar{a}&\bar{c}\\\bar{b}&\bar{d}\end{pmatrix} [[/math]]

Thus we must have [math]d=\bar{a}[/math], [math]c=-\bar{b}[/math], and we obtain the following formula:

[[math]] SU_2=\left\{\begin{pmatrix}a&b\\-\bar{b}&\bar{a}\end{pmatrix}\Big|\ |a|^2+|b|^2=1\right\} [[/math]]

(2) With the above formula in hand, the fundamental corepresentation of [math]SU_2[/math] is:

[[math]] u=\begin{pmatrix}a&b\\-\bar{b}&\bar{a}\end{pmatrix} [[/math]]

Now observe that we have the following equality:

[[math]] \begin{pmatrix}a&b\\ -\bar{b}&\bar{a}\end{pmatrix} \begin{pmatrix}0&1\\ -1&0\end{pmatrix} =\begin{pmatrix}-b&a\\-\bar{a}&-\bar{b}\end{pmatrix} =\begin{pmatrix}0&1\\ -1&0\end{pmatrix} \begin{pmatrix}\bar{a}&\bar{b}\\ -b&a\end{pmatrix} [[/math]]

Thus, with [math]F[/math] being as in the statement, we have [math]uF=F\bar{u}[/math], and so:

[[math]] u=F\bar{u}F^{-1} [[/math]]

We conclude that, if [math]A[/math] is the universal algebra in the statement, we have:

[[math]] A\to C(SU_2) [[/math]]

(3) Conversely now, let us compute the universal algebra [math]A[/math] in the statement. For this purpose, let us write its fundamental corepresentation as follows:

[[math]] u=\begin{pmatrix}a&b\\c&d\end{pmatrix} [[/math]]

We have [math]uF=F\bar{u}[/math], with these quantities being respectively given by:

[[math]] uF=\begin{pmatrix}a&b\\c&d\end{pmatrix} \begin{pmatrix}0&1\\ -1&0\end{pmatrix} =\begin{pmatrix}-b&a\\-d&c\end{pmatrix} [[/math]]

[[math]] F\bar{u}=\begin{pmatrix}0&1\\ -1&0\end{pmatrix} \begin{pmatrix}a^*&b^*\\c^*&d^*\end{pmatrix} =\begin{pmatrix}c^*&d^*\\-a^*&-b^*\end{pmatrix} [[/math]]

Thus we must have [math]d=a^*[/math], [math]c=-b^*[/math], and we obtain the following formula:

[[math]] u=\begin{pmatrix}a&b\\-b^*&a^*\end{pmatrix} [[/math]]

We also know that this matrix must be unitary, and we have:

[[math]] uu^*=\begin{pmatrix}a&b\\-b^*&a^*\end{pmatrix} \begin{pmatrix}a^*&-b\\b^*&a\end{pmatrix} =\begin{pmatrix}aa^*+bb^*&ba-ab\\a^*b^*-b^*a^*&a^*a+b^*b\end{pmatrix} [[/math]]

[[math]] u^*u=\begin{pmatrix}a^*&-b\\b^*&a\end{pmatrix} \begin{pmatrix}a&b\\-b^*&a^*\end{pmatrix} =\begin{pmatrix}a^*a+bb^*&a^*b-ba^*\\b^*a-ab^*&aa^*+b^*b\end{pmatrix} [[/math]]

Thus, the unitarity equations for [math]u[/math] are as follows:

[[math]] aa^*=a^*a=1-bb^*=1-b^*b [[/math]]

[[math]] ab=ba,a^*b=ba^*,ab^*=a^*b,a^*b^*=b^*a^* [[/math]]

It follows that [math]a,b,a^*,b^*[/math] commute, so our algebra is commutative. Now since this algebra is commutative, the involution [math]*[/math] becomes the usual conjugation [math]-[/math], and so:

[[math]] u=\begin{pmatrix}a&b\\-\bar{b}&\bar{a}\end{pmatrix} [[/math]]

But this tells us that we have [math]A=C(X)[/math] with [math]X\subset SU_2[/math], and so we have a quotient map [math]C(SU_2)\to A[/math], which is inverse to the map constructed in (2), as desired.

■

Now with the above result in hand, we can see right away the relation with [math]O_N^+[/math], and more specifically with [math]O_2^+[/math]. Indeed, this latter quantum group appears as follows:

[[math]] C(O_2^+)=C^*\left((u_{ij})_{i,j=1,2}\Big|u=\bar{u}={\rm unitary}\right) [[/math]]

Thus, [math]SU_2[/math] appears from [math]O_2^+[/math] by replacing the identity with the super-identity, or perhaps vice versa. In any case, these two quantum groups are related by some “twisting” operation, so they should have similar representation theory. This is indeed the case:

Theorem

For the group [math]SU_2[/math], the main character follows the Wigner law:

[[math]] \chi\sim\frac{1}{2\pi}\sqrt{4-x^2}dx [[/math]]

The irreducible representations of [math]SU_2[/math] are all self-adjoint, and can be labelled by positive integers, with their fusion rules being the Clebsch-Gordan ones,

[[math]] r_k\otimes r_l=r_{|k-l|}+r_{|k-l|+2}+\ldots+r_{k+l} [[/math]]

as for the quantum group [math]O_N^+[/math]. The dimensions of these representations are given by

[[math]] \dim r_k=k+1 [[/math]]

exactly as for the quantum group [math]O_2^+[/math].

Show Proof

This result is as old as modern mathematics, with many proofs available, all instructive. Here is our take on the subject, in connection with what we do here:

(1) A first proof, which is straightforward but rather long, is by taking everything that has been said so far about [math]O_N^+[/math], starting from the middle of chapter 4, setting [math]N=2[/math], and then twisting everything with the help of the super-identity matrix:

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

What happens then is that a Brauer theorem for [math]SU_2[/math] holds, involving the set [math]D=NC_2[/math] as before, but with the implementation of the partitions [math]\pi\to T_\pi[/math] being twisted by [math]F[/math]. In particular, we obtain in this way, as before, inequalities as follows:

[[math]] \dim(End(u^{\otimes k}))\leq C_k [[/math]]

But with such inequalities in hand, the proof of Theorem 5.13 applies virtually unchanged, and gives the result, with of course [math]q=1[/math] in the dimension formula.

(2) Here is as well a second proof, which is quite instructive. With [math]a=x+iy[/math], [math]b=z+it[/math], the formula for [math]SU_2[/math] that we found in the proof of Theorem 5.14 reads:

[[math]] SU_2=\left\{\begin{pmatrix}x+iy&z+it\\-z+it&x-iy\end{pmatrix}\Big|\ x^2+y^2+z^2+t^2=1\right\} [[/math]]

Thus, [math]SU_2[/math] is isomorphic to the real unit sphere [math]S^3_\mathbb R\subset\mathbb R^4[/math]. The point now is that the uniform measure on [math]SU_2[/math] corresponds in this way to the uniform measure on [math]S^3_\mathbb R[/math], and so in this picture, the moments of the main character of [math]SU_2[/math] are given by:

[[math]] M_k=\int_{S^3_\mathbb R}(2x)^kd(x,y,z,t) [[/math]]

In order to compute now such integrals, we can use the following advanced calculus formula, valid for any exponents [math]k_i\in2\mathbb N[/math], which at [math]N=2[/math] corresponds to the advanced calculus formula mentioned after Theorem 5.11, and at [math]N\geq3[/math] comes as well from that advanced calculus formula, via spherical coordinates and Fubini:

[[math]] \int_{S^{N-1}_\mathbb R}x_1^{k_1}\ldots x_N^{k_N}\,dx=\frac{(N-1)!!k_1!!\ldots k_N!!}{(N+\Sigma k_i-1)!!} [[/math]]

Indeed, by using this formula at [math]N=4[/math], we obtain:

[[math]] \begin{eqnarray*} \int_{S^3_\mathbb R}x_1^{2k}\,dx &=&\frac{3!!(2k)!!}{(2k+3)!!}\\ &=&2\cdot\frac{3\cdot5\cdot7\ldots (2k-1)}{2\cdot4\cdot6\ldots (2k+2)}\\ &=&2\cdot\frac{(2k)!}{2^kk!2^{k+1}(k+1)!}\\ &=&\frac{C_k}{4^k} \end{eqnarray*} [[/math]]

Thus the even moments of our character [math]\chi=2x_1[/math] are the Catalan numbers, [math]M_{2k}=C_k[/math], and since the odd moments vanish via [math]x\to-x[/math], we conclude that we have [math]\chi\sim\gamma_1[/math]. But this formula, or rather the moment formula [math]M_{2k}=C_k[/math] it comes from, gives:

[[math]] \dim(End(u^{\otimes k}))=C_k [[/math]]

Thus we can conclude as in the above first proof (1), by arguing that the recurrence construction of [math]r_k[/math] from the proof of Theorem 5.13 applies virtually unchanged, and gives the result, with of course [math]q=1[/math] in the dimension formula.

■

As a conclusion, we have two fringe proofs for the [math]SU_2[/math] result, one by crazy algebraists, and one by crazy probabilists. We recommend, as a complement, any of the proofs by geometers or physicists, which can be found in any good mathematical book. In fact [math]SU_2[/math], and also [math]SO_3[/math], are cult objects in geometry and physics, and there are countless things that can be said about them, and the more such things you know, the better your mathematics will be, no matter what precise mathematics you are interested in.

5d. Symplectic groups

Let us discuss now the unification of the [math]O_N^+[/math] and [math]SU_2[/math] results. In view of Theorem 5.14, and of the comments made afterwards, the idea is clear, namely that of looking at compact quantum groups appearing via relations of the following type:

[[math]] u=F\bar{u}F^{-1}={\rm unitary} [[/math]]

In order to clarify what exact matrices [math]F\in GL_N(\mathbb C)[/math] we can use, we must do some computations. Following , , , we first have the following result:

Proposition

Given a closed subgroup [math]G\subset U_N^+[/math], with irreducible fundamental corepresentation [math]u=(u_{ij})[/math], this corepresentation is self-adjoint, [math]u\sim\bar{u}[/math], precisely when

[[math]] u=F\bar{u}F^{-1} [[/math]]

for some unitary matrix [math]F\in U_N[/math], satisfying the following condition:

[[math]] F\bar{F}=\pm 1 [[/math]]

Moreover, when [math]N[/math] is odd we must have [math]F\bar{F}=1[/math].

Show Proof

Since [math]u[/math] is self-adjoint, [math]u\sim\bar{u}[/math], we must have [math]u=F\bar{u}F^{-1}[/math], for a certain matrix [math]F\in GL_N(\mathbb C)[/math]. We obtain from this, by using our assumption that [math]u[/math] is irreducible:

[[math]] \begin{eqnarray*} u=F\bar{u}F^{-1} &\implies&\bar{u}=\bar{F}u\bar{F}^{-1}\\ &\implies&u=(F\bar{F})u(F\bar{F})^{-1}\\ &\implies&F\bar{F}=c1\\ &\implies&\bar{F}F=\bar{c}1\\ &\implies&c\in\mathbb R \end{eqnarray*} [[/math]]

Now by rescaling we can assume [math]c=\pm1[/math], so we have proved so far that:

[[math]] F\bar{F}=\pm 1 [[/math]]

In order to establish now the formula [math]FF^*=1[/math], we can proceed as follows:

[[math]] \begin{eqnarray*} (id\otimes S)u=u^* &\implies&(id\otimes S)\bar{u}=u^t\\ &\implies&(id\otimes S)(F\bar{u}F^{-1})=Fu^tF^{-1}\\ &\implies&u^*=Fu^tF^{-1}\\ &\implies&u=(F^*)^{-1}\bar{u}F^*\\ &\implies&\bar{u}=F^*u(F^*)^{-1}\\ &\implies&\bar{u}=F^*F\bar{u}F^{-1}(F^*)^{-1}\\ &\implies&FF^*=d1 \end{eqnarray*} [[/math]]

We have [math]FF^* \gt 0[/math], so [math]d \gt 0[/math]. On the other hand, from [math]F\bar{F}=\pm 1[/math], [math]FF^*=d1[/math] we get:

[[math]] |\det F|^2=\det(F\bar{F})=(\pm1)^N [[/math]]

[[math]] |\det F|^2=\det(FF^*)=d^N [[/math]]

Since [math]d \gt 0[/math] we obtain from this [math]d=1[/math], and so [math]FF^*=1[/math] as claimed. We obtain as well that when [math]N[/math] is odd the sign must be 1, and so [math]F\bar{F}=1[/math], as claimed.

■

It is convenient to diagonalize [math]F[/math]. Once again following Bichon-De Rijdt-Vaes , up to an orthogonal base change, we can assume that our matrix is as follows, where [math]N=2p+q[/math] and [math]\varepsilon=\pm 1[/math], with the [math]1_q[/math] block at right disappearing if [math]\varepsilon=-1[/math]:

[[math]] F=\begin{pmatrix} 0&1\ \ \ \\ \varepsilon 1&0_{(0)}\\ &&\ddots\\ &&&0&1\ \ \ \\ &&&\varepsilon 1&0_{(p)}\\ &&&&&1_{(1)}\\ &&&&&&\ddots\\ &&&&&&&1_{(q)} \end{pmatrix} [[/math]]

We are therefore led into the following definition, from :

Definition

The “super-space” [math]\mathbb C^N_F[/math] is the usual space [math]\mathbb C^N[/math], with its standard basis [math]\{e_1,\ldots,e_N\}[/math], with a chosen sign [math]\varepsilon=\pm 1[/math], and a chosen involution on the set of indices,

[[math]] i\to\bar{i} [[/math]]

with [math]F[/math] being the “super-identity” matrix, [math]F_{ij}=\delta_{i\bar{j}}[/math] for [math]i\leq j[/math] and [math]F_{ij}=\varepsilon\delta_{i\bar{j}}[/math] for [math]i\geq j[/math].

In what follows we will usually assume that [math]F[/math] is the explicit matrix appearing above. Indeed, up to a permutation of the indices, we have a decomposition [math]n=2p+q[/math] such that the involution is, in standard permutation notation:

[[math]] (12)\ldots (2p-1,2p)(2p+1)\ldots (q) [[/math]]

Let us construct now some basic compact quantum groups, in our “super” setting. Once again following , let us formulate:

Definition

Associated to the super-space [math]\mathbb C^N_F[/math] are the following objects:

• The super-orthogonal group, given by:
  [[math]] O_F=\left\{U\in U_N\Big|U=F\bar{U}F^{-1}\right\} [[/math]]
• The super-orthogonal quantum group, given by:
  [[math]] C(O_F^+)=C^*\left((u_{ij})_{i,j=1,\ldots,n}\Big|u=F\bar{u}F^{-1}={\rm unitary}\right) [[/math]]

As explained in , it it possible to considerably extend this list, but for our purposes here, this is what we need for the moment. We have indeed the following result, from , making the connection with our unification problem for [math]O_N^+[/math] and [math]SU_2[/math]:

Theorem

The basic orthogonal groups and quantum groups are as follows:

• At [math]\varepsilon=-1[/math] we have [math]O_F=Sp_N[/math] and [math]O_F^+=Sp_N^+[/math].
• At [math]\varepsilon=-1[/math] and [math]N=2[/math] we have [math]O_F=O_F^+=SU_2[/math].
• At [math]\varepsilon=1[/math] we have [math]O_F=O_N[/math] and [math]O_F^+=O_N^+[/math].

Show Proof

These results are all elementary, as follows:

(1) At [math]\varepsilon=-1[/math] this follows from definitions, because the symplectic group [math]Sp_N\subset U_N[/math] is by definition the following group:

[[math]] Sp_N=\left\{U\in U_N\Big|U=F\bar{U}F^{-1}\right\} [[/math]]

(2) Still at [math]\varepsilon=-1[/math], the equation [math]U=F\bar{U}F^{-1}[/math] tells us that the symplectic matrices [math]U\in Sp_N[/math] are exactly the unitaries [math]U\in U_N[/math] which are patterned as follows:

[[math]] U=\begin{pmatrix} a&b&\ldots\\ -\bar{b}&\bar{a}\\ \vdots&&\ddots \end{pmatrix} [[/math]]

In particular, the symplectic matrices at [math]N=2[/math] are as follows:

[[math]] U=\begin{pmatrix} a&b\\ -\bar{b}&\bar{a} \end{pmatrix} [[/math]]

Thus we have [math]Sp_2=U_2[/math], and the formula [math]Sp_2^+=Sp_2[/math] is elementary as well, via an analysis similar to the one in the proof of Theorem 5.14 above.

(3) At [math]\varepsilon=1[/math] now, consider the root of unity [math]\rho=e^{\pi i/4}[/math], and set:

[[math]] J=\frac{1}{\sqrt{2}}\begin{pmatrix}\rho&\rho^7\\ \rho^3&\rho^5\end{pmatrix} [[/math]]

This matrix [math]J[/math] is then unitary, and we have:

[[math]] J\begin{pmatrix}0&1\\1&0\end{pmatrix}J^t=1 [[/math]]

Thus the following matrix is unitary as well, and satisfies [math]KFK^t=1[/math]:

[[math]] K=\begin{pmatrix}J^{(1)}\\&\ddots\\&&J^{(p)}\\&&&1_q\end{pmatrix} [[/math]]

Thus in terms of the matrix [math]V=KUK^*[/math] we have:

[[math]] U=F\bar{U}F^{-1}={\rm unitary} \quad\iff\quad V=\bar{V}={\rm unitary} [[/math]]

We obtain in this way an isomorphism [math]O_F^+=O_N^+[/math] as in the statement, and by passing to classical versions, we obtain as well [math]O_F=O_N[/math], as desired.

■

With the above formalism and results in hand, we can now formulate the unification result for [math]O_N^+[/math] and [math]SU_2[/math], which in complete form is as follows:

Theorem

For the quantum group [math]O_F^+\in\{O_N^+,Sp_N^+\}[/math] with [math]N\geq2[/math], the main character follows the standard Wigner semicircle law,

[[math]] \chi\sim\frac{1}{2\pi}\sqrt{4-x^2}dx [[/math]]

the irreducible representations are all self-adjoint, and can be labelled by positive integers, with their fusion rules being the Clebsch-Gordan ones,

[[math]] r_k\otimes r_l=r_{|k-l|}+r_{|k-l|+2}+\ldots+r_{k+l} [[/math]]

and the dimensions of these representations are given by

[[math]] \dim r_k=\frac{q^{k+1}-q^{-k-1}}{q-q^{-1}} [[/math]]

where [math]q,q^{-1}[/math] are the solutions of [math]X^2-NX+1=0[/math]. Also, we have [math]Sp_2^+=SU_2[/math].

Show Proof

This is a straightforward unification of the results that we already have for [math]O_N^+[/math] and [math]SU_2[/math], the technical details being all standard. See .

■

We will be back to [math]O_N^+[/math] and [math]O_F^+[/math] later on, first in chapter 7 below, with a number of more advanced algebraic considerations, in relation with super-structures and twists, and then in chapter 8 below, with a number of advanced probabilistic computations.

Finally, as the saying in geometry and physics goes, there is no [math]SU_2[/math] without [math]SO_3[/math]. We will construct in chapter 9 below a kind of “[math]SO_3[/math] companion” for [math]O_N^+[/math]. This companion will be something quite unexpected, namely the quantum permutation group [math]S_N^+[/math].

General references

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