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We would like to end this chapter with something still advanced, in relation with representations, but more analytic and refreshing, featuring some probability.

It is about formal graphs [math]X_N[/math] and their formal symmetry groups [math]G_N\subset S_N[/math] that we want to talk about, in the [math]N\to\infty[/math] limit, a bit as the physicists do. But, how to do this? Not clear at all, because while it is easy to talk about series of graphs [math]X=(X_N)[/math], in an intuitive way, the corresponding symmetry groups [math]G_N\subset S_N[/math] are not necessarily compatible with each other, as to form an axiomatizable object [math]G=(G_N)[/math].

Long story short, we are into potentially difficult mathematics here, and as a more concrete question that we can attempt to solve, we have: \begin{problem} What families of groups [math]G_N\subset S_N[/math] have compatible representation theory invariants, in the [math]N\to\infty[/math] limit? And then, can we use this in order to talk about “good” families of graphs [math]X_N[/math], whose symmetry groups [math]G_N=G(X_N)[/math] are of this type? \end{problem} But probably too much talking, let us get to work. The simplest graphs are undoubtedly the empty graphs and the simplices, with symmetry group [math]S_N[/math], so as a first question, we would like to know if the symmetric groups [math]S_N[/math] themselves have compatible representation theory invariants, with some nice [math]N\to\infty[/math] asymptotics to work out.

And here, surprise, or miracle, the answer is indeed yes, with the result, which is something very classical, remarkable, and beautiful, being as follows:

Theorem

The probability for a random [math]\sigma\in S_N[/math] to have no fixed points is

[[math]] P\simeq\frac{1}{e} [[/math]]

in the [math]N\to\infty[/math] limit, where [math]e=2.7182\ldots[/math] is the usual constant from analysis. More generally, the main character of [math]S_N[/math], which counts these permutations, given by

[[math]] \chi=\sum_i\sigma_{ii} [[/math]]

via the standard embedding [math]S_N\subset O_N[/math], follows the Poisson law [math]p_1[/math], in the [math]N\to\infty[/math] limit. Even more generally, the truncated characters of [math]S_N[/math], given by

[[math]] \chi=\sum_{i=1}^{[tN]}\sigma_{ii} [[/math]]

with [math]t \gt 0[/math], follow the Poisson laws [math]p_t[/math], in the [math]N\to\infty[/math] limit.

Show Proof

Obviously, many things going on here. The idea is as follows:

(1) In order to prove the first assertion, which is the key one, and probably the most puzzling one too, we will use the inclusion-exclusion principle. Let us set:

[[math]] S_N^k=\left\{\sigma\in S_N\Big|\sigma(k)=k\right\} [[/math]]

The set of permutations having no fixed points, called derangements, is then:

[[math]] X_N=\left(\bigcup_kS_N^k\right)^c [[/math]]

Now the inclusion-exclusion principle tells us that we have:

[[math]] \begin{eqnarray*} |X_N| &=&\left|\left(\bigcup_kS_N^k\right)^c\right|\\ &=&|S_N|-\sum_k|S_N^k|+\sum_{k \lt l}|S_N^k\cap S_N^l|-\ldots+(-1)^N\sum_{k_1 \lt \ldots \lt k_N}|S_N^{k_1}\cup\ldots\cup S_N^{k_N}|\\ &=&N!-N(N-1)!+\binom{N}{2}(N-2)!-\ldots+(-1)^N\binom{N}{N}(N-N)!\\ &=&\sum_{r=0}^N(-1)^r\binom{N}{r}(N-r)! \end{eqnarray*} [[/math]]

Thus, the probability that we are interested in, for a random permutation [math]\sigma\in S_N[/math] to have no fixed points, is given by the following formula:

[[math]] P =\frac{|X_N|}{N!} =\sum_{r=0}^N\frac{(-1)^r}{r!} [[/math]]

Since on the right we have the expansion of [math]1/e[/math], this gives the result.

(2) Let us construct now the main character of [math]S_N[/math], as in the statement. The permutation matrices being given by [math]\sigma_{ij}=\delta_{i\sigma(j)}[/math], we have the following formula:

[[math]] \chi(\sigma) =\sum_i\delta_{\sigma(i)i} =\#\left\{i\in\{1,\ldots,N\}\Big|\sigma(i)=i\right\} [[/math]]

In order to establish now the asymptotic result in the statement, regarding these characters, we must prove the following formula, for any [math]r\in\mathbb N[/math], in the [math]N\to\infty[/math] limit:

[[math]] P(\chi=r)\simeq\frac{1}{r!e} [[/math]]

We already know, from (1), that this formula holds at [math]r=0[/math]. In the general case now, we have to count the permutations [math]\sigma\in S_N[/math] having exactly [math]r[/math] points. Now since having such a permutation amounts in choosing [math]r[/math] points among [math]1,\ldots,N[/math], and then permuting the [math]N-r[/math] points left, without fixed points allowed, we have:

[[math]] \begin{eqnarray*} \#\left\{\sigma\in S_N\Big|\chi(\sigma)=r\right\} &=&\binom{N}{r}\#\left\{\sigma\in S_{N-r}\Big|\chi(\sigma)=0\right\}\\ &=&\frac{N!}{r!(N-r)!}\#\left\{\sigma\in S_{N-r}\Big|\chi(\sigma)=0\right\}\\ &=&N!\times\frac{1}{r!}\times\frac{\#\left\{\sigma\in S_{N-r}\Big|\chi(\sigma)=0\right\}}{(N-r)!} \end{eqnarray*} [[/math]]

By dividing everything by [math]N![/math], we obtain from this the following formula:

[[math]] \frac{\#\left\{\sigma\in S_N\Big|\chi(\sigma)=r\right\}}{N!}=\frac{1}{r!}\times\frac{\#\left\{\sigma\in S_{N-r}\Big|\chi(\sigma)=0\right\}}{(N-r)!} [[/math]]

Now by using the computation at [math]r=0[/math], that we already have, from (1), it follows that with [math]N\to\infty[/math] we have the following estimate:

[[math]] P(\chi=r) \simeq\frac{1}{r!}\cdot P(\chi=0) \simeq\frac{1}{r!}\cdot\frac{1}{e} [[/math]]

Thus, we obtain as limiting measure the Poisson law of parameter 1, as stated.

(3) Finally, let us construct the truncated characters of [math]S_N[/math], as in the statement. As before in the case [math]t=1[/math], we have the following computation, coming from definitions:

[[math]] \chi_t(\sigma) =\sum_{i=1}^{[tN]}\delta_{\sigma(i)i} =\#\left\{i\in\{1,\ldots,[tN]\}\Big|\sigma(i)=i\right\} [[/math]]

Also before in the proofs of (1) and (2), we obtain by inclusion-exclusion that:

[[math]] \begin{eqnarray*} P(\chi_t=0) &=&\frac{1}{N!}\sum_{r=0}^{[tN]}(-1)^r\sum_{k_1 \lt \ldots \lt k_r \lt [tN]}|S_N^{k_1}\cap\ldots\cap S_N^{k_r}|\\ &=&\frac{1}{N!}\sum_{r=0}^{[tN]}(-1)^r\binom{[tN]}{r}(N-r)!\\ &=&\sum_{r=0}^{[tN]}\frac{(-1)^r}{r!}\cdot\frac{[tN]!(N-r)!}{N!([tN]-r)!} \end{eqnarray*} [[/math]]

Now with [math]N\to\infty[/math], we obtain from this the following estimate:

[[math]] P(\chi_t=0) \simeq\sum_{r=0}^{[tN]}\frac{(-1)^r}{r!}\cdot t^r \simeq e^{-t} [[/math]]

More generally, by counting the permutations [math]\sigma\in S_N[/math] having exactly [math]r[/math] fixed points among [math]1,\ldots,[tN][/math], as in the proof of (2), we obtain:

[[math]] P(\chi_t=r)\simeq\frac{t^r}{r!e^t} [[/math]]

Thus, we obtain in the limit a Poisson law of parameter [math]t[/math], as stated.

■

Escalating difficulties, let us discuss now the hyperoctahedral group [math]H_N[/math]. Here the result is more technical, getting us into more advanced probability, as follows:

Theorem

For the hyperoctahedral group [math]H_N\subset O_N[/math], the law of the truncated character [math]\chi=g_{11}+\ldots +g_{ss}[/math] with [math]s=[tN][/math] is, in the [math]N\to\infty[/math] limit, the measure

[[math]] b_t=e^{-t}\sum_{r=-\infty}^\infty\delta_r\sum_{p=0}^\infty \frac{(t/2)^{|r|+2p}}{(|r|+p)!p!} [[/math]]

called Bessel law of parameter [math]t \gt 0[/math].

Show Proof

We regard [math]H_N[/math] as being the symmetry group of the graph [math]I_N=\{I^1,\ldots ,I^N\}[/math] formed by [math]n[/math] segments. The diagonal coefficients are then given by:

[[math]] u_{ii}(g)=\begin{cases} \ 0\ \mbox{ if $g$ moves \ltmath\gtI^i[[/math]]

}\\ \ 1\ \mbox{ if [math]g[/math] fixes [math]I^i[/math]}\\ -1\mbox{ if [math]g[/math] returns [math]I^i[/math]} \end{cases} </math>

Let us denote by [math]F_g,R_g[/math] the number of segments among [math]\{I^1,\ldots ,I^s\}[/math] which are fixed, respectively returned by an element [math]g\in H_N[/math]. With this notation, we have:

[[math]] u_{11}+\ldots+u_{ss}=F_g-R_g [[/math]]

Let us denote by [math]P_N[/math] the probabilities computed over [math]H_N[/math]. The density of the law of the variable [math]u_{11}+\ldots+u_{ss}[/math] at a point [math]r\geq 0[/math] is then given by the following formula:

[[math]] D(r) =P_N(F_g-R_g=r) =\sum_{p=0}^\infty P_N(F_g=r+p,R_g=p) [[/math]]

Assume first that we are in the case [math]t=1[/math]. We have the following computation:

[[math]] \begin{eqnarray*} \lim_{N\to\infty}D(r) &=&\lim_{N\to\infty}\sum_{p=0}^\infty(1/2)^{r+2p}\binom{r+2p}{r+p}P_N(F_g+R_g=r+2p)\\ &=&\sum_{p=0}^\infty(1/2)^{r+2p}\binom{r+2p}{r+p}\frac{1}{e(r+2p)!}\\ &=&\frac{1}{e}\sum_{p=0}^\infty\frac{(1/2)^{r+2p}}{(r+p)!p!} \end{eqnarray*} [[/math]]

The general case [math]0 \lt t\leq 1[/math] follows by performing some modifications in the above computation. Indeed, the asymptotic density can be computed as follows:

[[math]] \begin{eqnarray*} \lim_{N\to\infty}D(r) &=&\lim_{N\to\infty}\sum_{p=0}^\infty(1/2)^{r+2p}\binom{r+2p}{r+p}P_N(F_g+R_g=r+2p)\\ &=&\sum_{p=0}^\infty(1/2)^{r+2p}\binom{r+2p}{r+p}\frac{t^{r+2p}}{e^t(r+2p)!}\\ &=&e^{-t}\sum_{p=0}^\infty\frac{(t/2)^{r+2p}}{(r+p)!p!} \end{eqnarray*} [[/math]]

Together with [math]D(-r)=D(r)[/math], this gives the formula in the statement.

■

In order to further discuss now all this, and extend the above results, we will need a number of standard probabilistic preliminaries. We have the following notion:

Definition

Associated to any compactly supported positive measure [math]\nu[/math] on [math]\mathbb C[/math], not necessarily of mass [math]1[/math], is the probability measure

[[math]] p_\nu=\lim_{n\to\infty}\left(\left(1-\frac{t}{n}\right)\delta_0+\frac{1}{n}\nu\right)^{*n} [[/math]]

where [math]t=mass(\nu)[/math], called compound Poisson law.

In what follows we will be mainly interested in the case where the measure [math]\nu[/math] is discrete, as is for instance the case for [math]\nu=t\delta_1[/math] with [math]t \gt 0[/math], which produces the Poisson laws. The following standard result allows one to detect compound Poisson laws:

Proposition

For [math]\nu=\sum_{i=1}^st_i\delta_{z_i}[/math] with [math]t_i \gt 0[/math] and [math]z_i\in\mathbb C[/math], we have

[[math]] F_{p_\nu}(y)=\exp\left(\sum_{i=1}^st_i(e^{iyz_i}-1)\right) [[/math]]

where [math]F[/math] denotes the Fourier transform.

Show Proof

Let [math]\mu_n[/math] be the Bernoulli measure appearing in Definition 11.32, under the convolution sign. We have then the following Fourier transform computation:

[[math]] \begin{eqnarray*} F_{\mu_n}(y)=\left(1-\frac{t}{n}\right)+\frac{1}{n}\sum_{i=1}^st_ie^{iyz_i} &\implies&F_{\mu_n^{*n}}(y)=\left(\left(1-\frac{t}{n}\right)+\frac{1}{n}\sum_{i=1}^st_ie^{iyz_i}\right)^n\\ &\implies&F_{p_\nu}(y)=\exp\left(\sum_{i=1}^st_i(e^{iyz_i}-1)\right) \end{eqnarray*} [[/math]]

Thus, we have obtained the formula in the statement.

■

Getting back now to the Poisson and Bessel laws, we have:

Theorem

The Poisson and Bessel laws are compound Poisson laws,

[[math]] p_t=p_{t\delta_1}\quad,\quad b_t=p_{t\varepsilon} [[/math]]

where [math]\delta_1[/math] is the Dirac mass at [math]1[/math], and [math]\varepsilon[/math] is the centered Bernoulli law, [math]\varepsilon=(\delta_1+\delta_{-1})/2[/math].

Show Proof

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

(1) The first assertion, regarding the Poisson law [math]p_t[/math], is clear from Definition 11.22, which for [math]\nu=t\delta_1[/math] takes the following form:

[[math]] p_\nu=\lim_{n\to\infty}\left(\left(1-\frac{t}{n}\right)\delta_0+\frac{t}{n}\,\delta_1\right)^{*n} [[/math]]

But this is a usual Poisson limit, producing the Poisson law [math]p_t[/math], as desired.

(2) Regarding the second assertion, concerning the Bessel law [math]b_t[/math], by further building on the various formulae from Theorem 11.31 and its proof, we have:

[[math]] F_{b_t}(y)=\exp\left(t\left(\frac{e^{iy}+e^{-iy}}{2}-1\right)\right) [[/math]]

On the other hand, the formula in Proposition 11.33 gives, for [math]\nu=t\varepsilon[/math], the same Fourier transform formula. Thus, with [math]\nu=t\varepsilon[/math] we have [math]p_\nu=b_t[/math], as claimed.

■

More generally now, and in answer to Problem 11.29, for the groups [math]H_N^s[/math] from chapter 9 the asymptotic truncated characters follow the law [math]b_t^s=p_{t\varepsilon_s}[/math], with [math]\varepsilon_s[/math] being the uniform measure on the [math]s[/math]-roots of unity, and if you are really picky about what happens with [math]N\to\infty[/math], these are the only solutions. For details on this, you can check my book .

General references

Banica, Teo (2024). "Graphs and their symmetries". arXiv:2406.03664 [math.CO].