6a. Gaussian laws

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We have seen that the Brauer type results for [math]O_N,O_N^+[/math] lead to some concrete and interesting consequences. In this chapter we discuss similar results for [math]U_N,U_N^+[/math]. The situation here is a bit more complicated than for [math]O_N,O_N^+[/math], and we will only do a part of the work, namely algebra and basic probability, following , with the other part, advanced probability, following , being left for later, in chapter 8 below.

Let us also mention that, probabilistically speaking, the basic probability theory that we used for [math]O_N,O_N^+[/math], while still applying to [math]U_N[/math], after some changes, will not apply to [math]U_N^+[/math], due to the fact that the main character here is not normal, [math]\chi\chi^*\neq\chi^*\chi[/math]. So, in order to deal with [math]U_N^+[/math] we will have to use something more advanced, namely Voiculescu's free probability theory . Which is something very beautiful. But more on this later.

Let us start with a summary of what we know about [math]U_N,U_N^+[/math], coming from the Brauer type results from chapter 4, and the partition computations from chapter 5:

Theorem

For the basic unitary quantum groups, namely

[[math]] U_N\subset U_N^+ [[/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 pairings [math]D[/math] being as follows, with calligraphic standing for matching:

[[math]] \mathcal P_2\supset\mathcal{NC}_2 [[/math]]

At the level of the moments of the main character, we have in both cases

[[math]] \int\chi^k\leq|D(k)| [[/math]]

with [math]D[/math] being the above sets of pairings, with equality happening at [math]N\geq k[/math].

Show Proof

This is a summary of the results that we have, established in the previous chapters, and coming from Tannakian duality, via some combinatorics. To be more precise, the Brauer type results are from chapter 4, the estimates for the moments follows from this and from Peter-Weyl, as explained in chapter 5, and finally the last assertion, regarding the equality at [math]N\geq k[/math], is something more subtle, explained in chapter 5.

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Let us first investigate the unitary group [math]U_N[/math]. As it was the case for the orthogonal group [math]O_N[/math], in chapter 5, the representation theory here is something quite complicated, related to Young tableaux, and we will not get into this subject. However, once again in analogy with [math]O_N[/math], there is one straightforward thing to be done, namely the computation of the law of the main character, in the [math]N\to\infty[/math] limit.

In order to do this, we will need a basic probability result, as follows:

Theorem

The moments of the complex Gaussian law, given by

[[math]] G_1\sim\frac{1}{\sqrt{2}}(a+ib) [[/math]]

with [math]a,b[/math] being independent, each following the real Gaussian law [math]g_1[/math], are given by

[[math]] M_k=|\mathcal P_2(k)| [[/math]]

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

Show Proof

This is something well-known, which can be done in several steps, as follows:

(1) We recall from chapter 5 that the moments of the real Gaussian law [math]g_1[/math], with respect to integer exponents [math]k\in\mathbb N[/math], are the following numbers:

[[math]] m_k=|P_2(k)| [[/math]]

Numerically, we have the following formula, explained as well in chapter 5:

[[math]] m_k=\begin{cases} k!!&(k\ {\rm even})\\ 0&(k\ {\rm odd}) \end{cases} [[/math]]

(2) We will show here that in what concerns the complex Gaussian law [math]G_1[/math], similar results hold. Numerically, we will prove that we have the following formula, where a colored integer [math]k=\circ\bullet\bullet\circ\ldots[/math] is called uniform when it contains the same number of [math]\circ[/math] and [math]\bullet[/math]\,, and where [math]|k|\in\mathbb N[/math] is the length of such a colored integer:

[[math]] M_k=\begin{cases} (|k|/2)!&(k\ {\rm uniform})\\ 0&(k\ {\rm not\ uniform}) \end{cases} [[/math]]

Now since the matching partitions [math]\pi\in\mathcal P_2(k)[/math] are counted by exactly the same numbers, and this for trivial reasons, we will obtain the formula in the statement, namely:

[[math]] M_k=|\mathcal P_2(k)| [[/math]]

(3) This was for the plan. In practice now, we must compute the moments, with respect to colored integer exponents [math]k=\circ\bullet\bullet\circ\ldots[/math]\,, of the variable in the statement:

[[math]] c=\frac{1}{\sqrt{2}}(a+ib) [[/math]]

As a first observation, in the case where such an exponent [math]k=\circ\bullet\bullet\circ\ldots[/math] is not uniform in [math]\circ,\bullet[/math]\,, a rotation argument shows that the corresponding moment of [math]c[/math] vanishes. To be more precise, the variable [math]c'=wc[/math] can be shown to be complex Gaussian too, for any [math]w\in\mathbb C[/math], and from [math]M_k(c)=M_k(c')[/math] we obtain [math]M_k(c)=0[/math], in this case.

(4) In the uniform case now, where [math]k=\circ\bullet\bullet\circ\ldots[/math] consists of [math]p[/math] copies of [math]\circ[/math] and [math]p[/math] copies of [math]\bullet[/math]\,, the corresponding moment can be computed as follows:

[[math]] \begin{eqnarray*} M_k &=&\int(c\bar{c})^p\\ &=&\frac{1}{2^p}\int(a^2+b^2)^p\\ &=&\frac{1}{2^p}\sum_s\binom{p}{s}\int a^{2s}\int b^{2p-2s}\\ &=&\frac{1}{2^p}\sum_s\binom{p}{s}(2s)!!(2p-2s)!!\\ &=&\frac{1}{2^p}\sum_s\frac{p!}{s!(p-s)!}\cdot\frac{(2s)!}{2^ss!}\cdot\frac{(2p-2s)!}{2^{p-s}(p-s)!}\\ &=&\frac{p!}{4^p}\sum_s\binom{2s}{s}\binom{2p-2s}{p-s} \end{eqnarray*} [[/math]]

(5) In order to finish now the computation, let us recall that we have the following formula, coming from the generalized binomial formula, or from the Taylor formula:

[[math]] \frac{1}{\sqrt{1+t}}=\sum_{k=0}^\infty\binom{2k}{k}\left(\frac{-t}{4}\right)^k [[/math]]

By taking the square of this series, we obtain the following formula:

[[math]] \begin{eqnarray*} \frac{1}{1+t} &=&\sum_{ks}\binom{2k}{k}\binom{2s}{s}\left(\frac{-t}{4}\right)^{k+s}\\ &=&\sum_p\left(\frac{-t}{4}\right)^p\sum_s\binom{2s}{s}\binom{2p-2s}{p-s} \end{eqnarray*} [[/math]]

Now by looking at the coefficient of [math]t^p[/math] on both sides, we conclude that the sum on the right equals [math]4^p[/math]. Thus, we can finish the moment computation in (4), as follows:

[[math]] M_p=\frac{p!}{4^p}\times 4^p=p! [[/math]]

(6) As a conclusion, if we denote by [math]|k|[/math] the length of a colored integer [math]k=\circ\bullet\bullet\circ\ldots[/math]\,, the moments of the variable [math]c[/math] in the statement are given by:

[[math]] M_k=\begin{cases} (|k|/2)!&(k\ {\rm uniform})\\ 0&(k\ {\rm not\ uniform}) \end{cases} [[/math]]

On the other hand, the numbers [math]|\mathcal P_2(k)|[/math] are given by exactly the same formula. Indeed, in order to have matching pairings of [math]k[/math], our exponent [math]k=\circ\bullet\bullet\circ\ldots[/math] must be uniform, consisting of [math]p[/math] copies of [math]\circ[/math] and [math]p[/math] copies of [math]\bullet[/math], with [math]p=|k|/2[/math]. But then the matching pairings of [math]k[/math] correspond to the permutations of the [math]\bullet[/math] symbols, as to be matched with [math]\circ[/math] symbols, and so we have [math]p![/math] such matching pairings. Thus, we have the same formula as for the moments of [math]c[/math], and we are led to the conclusion in the statement.

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We should mention that the above proof is just one proof among others. There is a lot of interesting mathematics behind the complex Gaussian variables, whose knowledge can avoid some of the above computations, and we recommend some reading here.

By getting back now to the unitary group [math]U_N[/math], with the above results in hand we can formulate our first concrete result about it, as follows:

Theorem

For the unitary group [math]U_N[/math], the main character

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

follows the standard complex Gaussian law

[[math]] \chi\sim G_1 [[/math]]

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

Show Proof

This follows by putting together the results that we have, namely Theorem 6.1 applied with [math]N \gt k[/math], and then Theorem 6.2.

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As already mentioned, as it was the case for the orthogonal group [math]O_N[/math], in chapter 5, the representation theory for [math]U_N[/math] at fixed [math]N\in\mathbb N[/math] is something quite complicated, related to the combinatorics of Young tableaux, and we will not get into this subject here.

There is, however, one more interesting topic regarding [math]U_N[/math] to be discussed, namely its precise relation with [math]O_N[/math], and more specifically the passage [math]O_N\to U_N[/math].

Contrary to the passage [math]\mathbb R^N\to\mathbb C^N[/math], or to the passage [math]S^{N-1}_\mathbb R\to S^{N-1}_\mathbb C[/math], which are both elementary, the passage [math]O_N\to U_N[/math] cannot be understood directly. In order to understand this passage we must pass through the corresponding Lie algebras, a follows:

Theorem

The passage [math]O_N\to U_N[/math] appears via a Lie algebra complexification,

[[math]] O_N\to\mathfrak o_N\to\mathfrak u_n\to U_N [[/math]]

with the Lie algebra [math]\mathfrak u_N[/math] being a complexification of the Lie algebra [math]\mathfrak o_N[/math].

Show Proof

This is something rather philosophical, and advanced as well, that we will not really need here, the idea being as follows:

(1) The orthogonal and unitary groups [math]O_N,N_N[/math] are both Lie groups, in the sense that they are smooth manifolds, and the corresponding Lie algebras [math]\mathfrak o_N,\mathfrak u_N[/math], which are by definition the respective tangent spaces at 1, can be computed by differentiating the equations defining [math]O_N,U_N[/math], with the conclusion being as follows:

[[math]] \mathfrak o_N=\left\{ A\in M_N(\mathbb R)\Big|A^t=-A\right\} [[/math]]

[[math]] \mathfrak u_N=\left\{ B\in M_N(\mathbb C)\Big|B^*=-B\right\} [[/math]]

(2) This was for the correspondences [math]O_N\to\mathfrak o_N[/math] and [math]U_N\to\mathfrak u_N[/math]. In the other sense, the correspondences [math]\mathfrak o_N\to O_N[/math] and [math]\mathfrak u_N\to U_N[/math] appear by exponentiation, the result here stating that, around 1, the orthogonal matrices can be written as [math]U=e^A[/math], with [math]A\in\mathfrak o_N[/math], and the unitary matrices can be written as [math]U=e^B[/math], with [math]B\in\mathfrak u_N[/math].

(3) In view of all this, in order to understand the passage [math]O_N\to U_N[/math] it is enough to understand the passage [math]\mathfrak o_N\to\mathfrak u_N[/math]. But, in view of the above explicit formulae for [math]\mathfrak o_N,\mathfrak u_N[/math], this is basically an elementary linear algebra problem. Indeed, let us pick an arbitrary matrix [math]B\in M_N(\mathbb C)[/math], and write it as follows, with [math]A,C\in M_N(\mathbb R)[/math]:

[[math]] B=A+iC [[/math]]

In terms of [math]A,C[/math], the equation [math]B^*=-B[/math] defining the Lie algebra [math]\mathfrak u_N[/math] reads:

[[math]] A^t=-A [[/math]]

[[math]] C^t=C [[/math]]

(4) As a first observation, we must have [math]A\in\mathfrak o_N[/math]. Regarding now [math]C[/math], let us decompose it as follows, with [math]D[/math] being its diagonal, and [math]C'[/math] being the remainder:

[[math]] C=D+C' [[/math]]

The remainder [math]C'[/math] being symmetric with 0 on the diagonal, by switching all the signs below the main diagonal we obtain a certain matrix [math]C'_-\in\mathfrak o_N[/math]. Thus, we have decomposed [math]B\in\mathfrak u_N[/math] as follows, with [math]A,C'\in\mathfrak o_N[/math], and with [math]D\in M_N(\mathbb R)[/math] being diagonal:

[[math]] B=A+iD+iC'_- [[/math]]

(5) As a conclusion now, we have shown that we have a direct sum decomposition of real linear spaces as follows, with [math]\Delta\subset M_N(\mathbb R)[/math] being the diagonal matrices:

[[math]] \mathfrak u_N\simeq\mathfrak o_N\oplus\Delta\oplus\mathfrak o_N [[/math]]

Thus, we can stop our study here, and say that we have reached the conclusion in the statement, namely that [math]\mathfrak u_N[/math] appears as a “complexification” of [math]\mathfrak o_N[/math].

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As before with many other things, that we will not really need in what follows, this was just an introduction to the subject. More can be found in any Lie group book.

General references

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