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One interesting question regarding the Weingarten calculus, where [math]W_{kN}=G_{kN}^{-1}[/math], is the computation of the determinant of [math]G_{kN}[/math]. Following Di Francesco , we discuss here this key question, for [math]O_N^+[/math]. Let us begin with something that we know, namely:

Theorem

The determinant of the Gram matrix of [math]P(k)[/math] is given by

[[math]] \det(G_{kN})=\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

This is indeed the Lindstöm formula that we established in chapter 5 above, by using a decomposition of type [math]G_{kN}=A_{kN}L_{kN}[/math].

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In what regards [math]O_N^+[/math], the set of partitions is here [math]NC_2(2k)[/math], and things are far more complicated than for [math]P(k)[/math]. However, we can use the Lindstöm formula at [math]k=1,2,3[/math], where [math]P(k)=NC(k)[/math], via the following fattening/shrinking trick:

Proposition

We have a bijection [math]NC(k)\simeq NC_2(2k)[/math], constructed by fattening and shrinking, as follows:

• The application [math]NC(k)\to NC_2(2k)[/math] is the “fattening” one, obtained by doubling all the legs, and doubling all the strings as well.
• Its inverse [math]NC_2(2k)\to NC(k)[/math] is the “shrinking” application, obtained by collapsing pairs of consecutive neighbors.

Show Proof

The fact that the two operations in the statement are indeed inverse to each other is clear, by computing the corresponding two compositions.

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At the level of the associated Gram matrices, the result is as follows:

Proposition

The Gram matrices of the sets of partitions

[[math]] NC_2(2k)\simeq NC(k) [[/math]]

are related by the following formula, where [math]\pi\to\pi'[/math] is the shrinking operation,

[[math]] G_{2k,n}(\pi,\sigma)=n^k(\Delta_{kn}^{-1}G_{k,n^2}\Delta_{kn}^{-1})(\pi',\sigma') [[/math]]

and where [math]\Delta_{kn}[/math] is the diagonal of [math]G_{kn}[/math].

Show Proof

It is elementary to see that we have the following formula:

[[math]] |\pi\vee\sigma|=k+2|\pi'\vee\sigma'|-|\pi'|-|\sigma'| [[/math]]

We therefore have the following formula, valid for any [math]n\in\mathbb N[/math]:

[[math]] n^{|\pi\vee\sigma|}=n^{k+2|\pi'\vee\sigma'|-|\pi'|-|\sigma'|} [[/math]]

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

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We can do now the computations for [math]O_N^+[/math] at [math]k=2,4,6[/math], as follows:

Proposition

The Gram matrices and determinants for [math]O_N^+[/math] are as follows,

[[math]] \det(N)=N [[/math]]

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

[[math]] \det\begin{pmatrix} N^3&N^2&N^2&N^2&N\\ N^2&N^3&N&N&N^2\\ N^2&N&N^3&N&N^2\\ N^2&N&N&N^3&N^2\\ N&N^2&N^2&N^2&N^3 \end{pmatrix}=N^5(N^2-1)^4(N^2-2) [[/math]]

at [math]k=2,4,6[/math], with the matrices written by using the lexicographic order on [math]NC_2(2k)[/math].

Show Proof

The formula at [math]k=2[/math], where [math]NC_2(2)=\{\sqcap\}[/math], is clear. The same goes for the formula at [math]k=4[/math], where [math]NC_2(4)=\{\sqcap\sqcap,\bigcap\hskip-4.9mm{\ }_\cap\,\}[/math]. At [math]k=6[/math] however, things are tricky, and we must use the Lindstöm formula. We have [math]NC(3)=\{|||,\sqcap|,\sqcap\hskip-3.2mm{\ }_|\,,|\sqcap,\sqcap\hskip-0.7mm\sqcap\}[/math], and the corresponding Gram matrix and its determinant are, according to Theorem 8.41:

[[math]] \det\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}=N^5(N-1)^4(N-2) [[/math]]

By using Proposition 8.43, the Gram determinant of [math]NC_2(6)[/math] is given by:

[[math]] \begin{eqnarray*} \det(G_{6N}) &=&\frac{1}{N^2\sqrt{N}}\times N^{10}(N^2-1)^4(N^2-2)\times\frac{1}{N^2\sqrt{N}}\\ &=&N^5(N^2-1)^4(N^2-2) \end{eqnarray*} [[/math]]

Thus, we have obtained the formula in the statement.

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In general, such tricks won't work, because [math]NC(k)[/math] is strictly smaller than [math]P(k)[/math] at [math]k\geq4[/math]. However, following Di Francesco , we have the following result:

Theorem

The determinant of the Gram matrix for [math]O_N^+[/math] is given by

[[math]] \det(G_{kN})=\prod_{r=1}^{[k/2]}P_r(N)^{d_{k/2,r}} [[/math]]

where [math]P_r[/math] are the Chebycheff polynomials, given by

[[math]] P_0=1\quad,\quad P_1=X\quad,\quad P_{r+1}=XP_r-P_{r-1} [[/math]]

and [math]d_{kr}=f_{kr}-f_{k,r+1}[/math], with [math]f_{kr}[/math] being the following numbers, depending on [math]k,r\in\mathbb Z[/math],

[[math]] f_{kr}=\binom{2k}{k-r}-\binom{2k}{k-r-1} [[/math]]

with the convention [math]f_{kr}=0[/math] for [math]k\notin\mathbb Z[/math].

Show Proof

This is something fairly heavy, obtained by using a decomposition as follows of the Gram matrix [math]G_{kN}[/math], with the matrix [math]T_{kN}[/math] being lower triangular:

[[math]] G_{kN}=T_{kN}T_{kN}^t [[/math]]

Thus, a bit as in the proof of the Lindstöm formula, we obtain the result, but the problem lies however in the construction of [math]T_{kN}[/math], which is non-trivial. See .

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The above result is interesting for countless reasons, and we refer here to Di Francesco , and also to , where a systematic study of the Gram determinants for the easy quantum groups was performed, following Lindstöm, Di Francesco, Zinn-Justin and others. Getting into all this, which is advanced mathematical physics, would be well beyond the purposes of the present book, so let us just mention here that:

(1) We will see in chapter 9 that [math]S_N,S_N^+[/math] are easy, coming from [math]P,NC[/math]. Thus the Lindstöm formula computes the determinant for [math]S_N[/math], and in what regards [math]S_N^+[/math], here the determinant can be computed by using Proposition 8.43 and Theorem 8.45.

(2) In what regards [math]O_N[/math], here the determinant is as follows, where [math]f^\lambda[/math] is the number of standard Young tableaux of shape [math]\lambda[/math], and [math]c_N(\lambda)=\prod_{(i,j)\in\lambda}(N+2j-i-1)[/math]:

[[math]] \det(G_{kN})=\prod_{|\lambda|=k/2}c_N(\lambda)^{f^{2\lambda}} [[/math]]

Obviously, some interesting mathematics is going on here. For more, you can check , , as suggested above, as well as the related literature, in mathematics and physics, which is considerable. And also, why not trying to compute one of these beasts yourself. There is no better introduction to advanced combinatorics than this.

General references

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