TITLE: PDF of $X = \max\{X_1,X_2\}$, being $X_1$ and $X_2$ independent Normal distributed random variables
QUESTION [11 upvotes]: Let $X_1 \sim \mathcal{N}(\mu_1,\sigma_1^2)$ and $X_2 \sim \mathcal{N}(\mu_2,\sigma_2^2)$, what's the CDF of $X = \max\{X_1,X_2\}$? Both variables are assumed to be independent.
I tried the following:
\begin{equation}
F_X(x) = \text{Prob}\{X<x\} = \text{Prob}\{X_1<x, X_2<x\} = \text{Prob}\{X_1<x\}\cdot \text{Prob}\{X_2<x\} = F_{X_1}(x)\cdot F_{X_2}(x)
\end{equation}
Since $X_1$ and $X_2$ are normal distributed variables, their CDFs are well known:
\begin{equation} \tag{1}
F_X(x) = F_{X_1}(x)\cdot F_{X_2}(x) = \bigg(1-\mathcal{Q}\Big(\frac{x-\mu_1}{\sigma_1}\Big)\bigg)\cdot \bigg(1-\mathcal{Q}\Big(\frac{x-\mu_2}{\sigma_2}\Big)\bigg)
\end{equation}
where $\mathcal{Q}\big(\frac{x-\mu}{\sigma}\big) = \int_x^{\infty} \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx$.
However, MATLAB simulation shows that this is not correct. The MATLAB code and the plot result comparing simulation and theoretical is shown below.
S=1e6;

% Define Normal Distributed RV #1
mu_1 = 1;
sigma_1 = 1;
X_1 = sigma_1/sqrt(2)*(randn(1,S))+mu_1;

% Define Normal Distributed RV #1
mu_2 = 1;
sigma_2 = 1;
X_2 = sigma_2/sqrt(2)*(randn(1,S))+mu_2;

% Define variable X = max{X_1,X_2}
X = max(X_1,X_2);
[X_pdf, X_var] = ecdf(X); % Obtain empirical CDF (similar to function hist)

X_pdf_th = (1-qfunc((X_var-mu_1)/sigma_1)).*(1-qfunc((X_var-mu_2)/sigma_2));

plot(X_var,X_pdf); hold on;grid on;
plot(X_var,X_pdf_th,'r');  legend('Simulation CDF', 'Theory CDF'); hold off;

REPLY [3 votes]: This is not really an answer, but notes for future reference about the distribution of $X = \max\{X_1,X_2\}$ when $X_1, X_2 \sim \mathcal{N}(0,1)$, independent, identically distributed.
As you stated, the cumulative distribution function (CDF) is
$$F_X(x) = P(X\le x) = P(X_1\le x\ \cap X_2 \le x) = P(X_1 \le x) \cdot P(X_2  \le x) = \Phi(x)^2,
$$
with $\Phi_{\mu,\sigma}(x) := \int_{-\infty}^x \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(t-\mu)^2}{2\sigma^2}} dt$, and $\Phi := \Phi_{0,1}$.
The density of $X$ is then $$f_X(x) = 2 \Phi(x) \Phi'(x) = \sqrt{\frac{2}{\pi}} \Phi(x)\  e^{-x^2/2}.$$
The expectation of $X$ is $\int_{\mathbb{R}} x f_X(x) d x = \frac{1}{\sqrt{\pi}}$ and its variance is $\int_{\mathbb{R}} x^2 f_X(x) d x - \Big(\frac{1}{\sqrt{\pi}}\Big)^2 = 1 - \frac{1}{\pi}.$
We can note that $f_X$ is very close to the density of a normal distribution $\mathcal{N}(1/\sqrt{\pi}, \sqrt{1-1/\pi})$:

Note: Does this distribution has a name? Any reference welcome.


TITLE: Rational cohomological dimension of a locally finite group
QUESTION [5 upvotes]: $\DeclareMathOperator\cd{cd}$Recall that the rational cohomological dimension of a group $G$ is the supremum of the set of integers $k$ such that there exists a $\mathbb{Q}[G]$-module $M$ with $H^k(G;M) \neq 0$.  Denote this by $\cd_{\mathbb{Q}}(G)$.
If $G$ is a finite group, then it is easy to see that $\cd_{\mathbb{Q}}(G) = 0$.
However, it is not clear to me what the rational cohomological dimension of a locally finite group is (where "locally finite" means that all finitely generated subgroups are finite).  Since homology commutes with direct limits, the rational homological dimension of these group are $0$.
So my question is what can be said about the rational cohomological dimension of a locally finite group $G$.  I mostly care about countable $G$.  In fact, the easiest example where I don't know the answer is
$$G = \bigoplus_{k=1}^{\infty} \mathbb{Z}/2\mathbb{Z}.$$

REPLY [4 votes]: Edited version. If $G$ is a countable infinite locally finite group, then the rational cohomological dimension is exactly $1$.  The rational cohomological dimension for an infinite group is never $0$ because if $\mathbb Q$ is a projective $\mathbb QG$-module, then you need an idempotent $e$ with $\mathbb QGe\cong \mathbb Q$ and that is impossible unless $G$ is finite and $e$ is the average of all elements of $G$.
Here are two proofs that the projective dimesion of $\mathbb QG$ is at most one for a countable locally finite group.  This amounts to showing the augmentation ideal is projective.  Then $\mathbb Q$ has projective dimension at most $1$ since $I\to \mathbb QG\to \mathbb Q\to 0$ is a projective resolution where $I$ is the augmentation ideal.
It follows from a result in Dicks book Groups, trees and projective modules (1980)  characterizing projectivity of the augmentation ideal of $RG$ that $\mathbb QG$ has a projective augmentation ideal whenever $G$ is countable and locally finite.   See this question for more details.
The second argument is that by a result of Connell, $\mathbb QG$ is von Neumann regular if $G$ is locally finite.  The augmentation ideal is countably generated as a left ideal if $G$ is countable and hence by a result of Kaplansky is projective (Kaplansky proved that a countably generated left ideal in a von Neumann regular ring is projective).
Thus for a countably infinite locally finite group the rational cohomological dimension is one.
It follows from Theorem 2 of Derek Holt's 1981 paper (DOI link) that if $G$ is a locally finite group of cardinality $\aleph_1$, then the rational cohomological dimension of $G$ is at least $2$.


TITLE: What shape of mug keeps a drink hottest longest?
QUESTION [6 upvotes]: I have a selection of mugs that are different shapes. Some are taller elongated cylinders, others are similar in height to their diameter, and some are flatter, wider vessels.
I was wondering, which shape of mug should I use if I want my tea or coffee to stay hottest for the longest time?
All are ceramic and have no lid (obviously a travel-style mug would beat other styles, but let's not worry about that; I'm just interested in "standard" mugs).
Do I want something that is closest in shape to a sphere to minimise the overall surface area of the drink? Or do I want something with the smallest area exposed to the air to minimise convection? Or is there some other consideration that wins out?

REPLY [1 votes]: To me, the answer is obvious by observation. I can't do the math, but if someone would care to dope it out, here is my theory.
If we eliminate designs that are specific to heat conservation, such as double walls and lids, and make it strictly a single-walled material with an open top:  a mug which angles inward to a narrow neck and top from a wider bases keep coffee warmer longer--a mug with a conical base incorporated into it.
This shape takes advantage of the common property of heat and steam-- they RISE.   In a standard cylinder, most of the steam and heat escape upwards into the air.  But the conical body allows for more steam and heat to make contact with the upper body of the mug and heat the material, which in turn warms the coffee.
As a standard cylinder mug's coffee is consumed, the same upward trajectory for heat remains constant because of the uniform circumference. More inner wall is exposed, but the wall is also making more contact with cool air and is ineffective at retaining and transferring heat back to the coffee, either thru transference to the deeper liquid layer or the surface layer.
But in a conical mug made of the same material, the steam/heat from the outer edge of the coffee is channeled along the acute angle of the inner wall.  More heat and steam is retained longer, utilized within a more contained chamber before eventually reaching the narrow neck and opening.  Some steam might also begin to condense into drops which return to the coffee.
Other advantages:  Less likely to tip because whether full or empty, it is always bottom heavy.
Disadvantages:  Harder to get clean, and most likely won't be cleaned at all in a dishwasher.


TITLE: For what $x$ does the series $\sum_{n=2}^{\infty} \frac{x^n}{(\ln(n))^2}$ converge?
QUESTION [0 upvotes]: The problem sounds as:

For what $x$ does the series $\sum_{n=2}^{\infty} \frac{x^n}{(\ln(n))^2}$ converge?

My solution:
First of all, we realize that since $1/(\ln(n))^2$ is decreasing for all $n \geq 2$, we have that:
$0 \leq \sum_{n=2}^{\infty} \frac{x^n}{(\ln(n))^2} \leq \frac{1}{\ln^2(2)}\sum_{n=2}^{\infty} x^n$. This holds as long as we work with positive x's. Since we have a geometric series, this converges for all $x \in (-1,1)$, and with the previous condition we have for all x's in [0,1).
Now, we have to check for the case in which x is negative. Let's substitute $x = -u$ into the sum.
$\sum_{n=2}^{\infty} \frac{(-u)^n}{(\ln(n))^2} = \sum_{n=2}^{\infty} (-1)^n\frac{u^n}{(\ln(n))^2}$
I'm thinking I can use the Leibniz convergence test here somehow, but I don't really know how to go on from here. Feel free to share any tips / ideas, especially if there's some general method solving problems like these. Are you supposed to divide it into two cases, one for positive and one for negative x's as I did?
Thanks.

REPLY [1 votes]: Compare
$$\sum_{n=2}^\infty \frac{x^n}{[\log(n)]^2}\tag1 $$
against both
$$\sum_{n=2}^\infty x^n\tag2 $$
and
$$\sum_{n=2}^\infty \frac{x^n}{n}\tag3. $$

$\underline{\textbf{Case 1 :} ~~|x| < 1}$
(2) above converges if and only if $|x| < 1.$
Since, for $n \geq 3, ~[\log(n)]^2 > 1,$ you therefore have that (1) above converges, for $|x| < 1.$

$\underline{\textbf{Case 2 :} ~~x \geq 1}$
For $x \geq 1$, it is sufficient to show that (1) above diverges for $x = 1$.
To show this, consider that

$[\log(10)]^2 < (10)$.

$f(n) = n - [\log(n)]^2 \implies $
$\displaystyle f'(n) = 1 - \frac{2\log(n)}{n}$,
which is positive for all $n \geq 10$.
Thus, $f(n)$ is a strictly increasing function, for $n \geq 10$.
This implies that when $x = 1$, you may conclude that (1) above is divergent, by comparing (1) with (3) above, which is known to be divergent for $x = 1.$

$\underline{\textbf{Case 3 :} ~~x  = -1}$
For consideration of $x = -1$, in (1) above, the alternating series test is relevant.

The terms are alternating in sign (positive - negative).

The absolute values of the terms are strictly decreasing.

The absolute values of the terms go to $(0)$ as $n \to \infty.$

Therefore, when $x = -1$, (1) above is convergent.

$\underline{\textbf{Case 4 :} ~~x  < -1}$
In consideration of (1) above, for $x < -1$, let $a_n = \displaystyle \frac{|x|^n}{[\log(n)]^2}.$
Then, $\displaystyle \frac{a_{n+1}}{a_n} =
|x| \times \left[\frac{\log(n)}{\log(n+1)}\right]^2.$
Using L'hopital's rule,
$$\lim_{n\to \infty} \frac{\log(n)}{\log(n+1)} = \lim_{n \to \infty} \frac{n+1}{n} = 1.$$
Therefore, for any $x < -1$, there clearly exists an $N \in \Bbb{Z^+}$ such that for all $n \geq N, ~a_n$ is strictly increasing.
Therefore, for $x < -1$, you have that after a finite number of terms, the absolute value of the terms in (1) above are strictly increasing.
Therefore, for $x < -1$, (1) above must be divergent.


TITLE: Why is it sufficient to consider 'forms' instead of polynomials in the projective case?
QUESTION [0 upvotes]: So I am reading through William Fulton's Algebraic Curves and currently I am reading that bit about projective curves.
Now, when the author discussed about the affine case (in the earlier chapters 1-3), for example, to define affine plane curves in Chapter 3, and all the discussion following he considered $F$ to be a polynomial.
In Chapter 2, he talked about how we can write a polynomial of n variables as a sum of forms and gave a few results related to them. Then when Projective varieties were introduced, he again gave a few result related to forms and from that point onwards, whenever a polynomial $F \in k(\mathbb{P^2}) $ was given it's always assumed to be a form.
Now, what I don't understand is, why is it sufficient to just consider forms in the projective case in contrary to considering polynomial(a sum of forms) in the affine case?
I would appreciate any kind of help/hint(s).
Thanks a lot!

REPLY [1 votes]: In projective space, everything is only defined up to scalar multiples. So if you had a non homogeneous polynomial, the question of whether or not a polynomial vanishes at a point of projective space isn't even well defined. For instance, $(0,1,0)$ is a solution to $x + y = 1$. This is the same point in the projective plane as $(0,2,0)$, which is not a solution.


TITLE: $H_1(U)=\mathbb{Z}^n$ if $U$ is complement of $n$ distinct points in $\mathbb{R}^2$
QUESTION [0 upvotes]: Let $U$ be the complement of $n$ distinct points in $\mathbb{R}^2.$ How can we show that $H_1(U)=\mathbb{Z}^n$?

REPLY [2 votes]: $U$ retracts to  a bouquet of $n$ circles whose fundamental group is a free group of $n$ generators so after abelianisation you get $Z^n$.


TITLE: Solving the matrix differential equation $\frac{d M} {dt} = \kappa \hspace{1mm} \max(0,(M_1-M) -\gamma I )$
QUESTION [3 upvotes]: I have a weird equation that I want to solve,
$$\frac{d M} {dt} = \kappa  \hspace{1mm} \max(0,(M_1-M) -\gamma I )$$
where, $M, M_1, I$ are all $2 \times 2$ matrices. With the $\max$ function I want to make sure the matrix $M$ does not shrink or become smaller, so the $\max$ condition I am imagining is that the condition for the argument matrix to be positive definite.

REPLY [2 votes]: We have a matrix differential equation in $\mathrm X \in \mathbb R^{m \times n}$
$$\dot{\mathrm X} = \kappa \cdot \max \left( \mathrm O_{m \times n}, \mathrm Y - \mathrm X\right)$$
where $\mathrm Y \in \mathbb R^{m \times n}$ is given. Thus, we have $m n$ ODEs of the form
$$\dot{x}_{ij} = \kappa \cdot \max \left( 0, y_{ij} - x_{ij} \right)$$
or,
$$\dot{x}_{ij} = \begin{cases} \kappa \left( y_{ij} - x_{ij} \right) & \text{if } x_{ij} < y_{ij}\\ 0 & \text{if } x_{ij} \geq y_{ij}\end{cases}$$
If $\color{blue}{x_{ij} (0) < y_{ij}}$, then
$$x_{ij} (t) = y_{ij} + \left( x_{ij} (0) - y_{ij} \right) \, \exp(- \kappa \,t)$$
If $\color{blue}{x_{ij} (0) \geq y_{ij}}$, then
$$x_{ij} (t) = x_{ij} (0)$$


TITLE: generating function of multinomial coefficient
QUESTION [0 upvotes]: How to express this series in closed form?
$$\sum_{i=1}^{\infty}\frac{(3i)!}{(i!)^3}x^{i}$$
Motive of the generating function is to evaluate the number of the paths from the $(0,0,0)$ to $(n,n,n)$ not passing through $(i,i,i)$ ($1\leq i\leq n-1)$. The answer is coefficient of $x^n$ in $$\sum_{k=1}^{n}(-1)^{k-1}\left(\sum_{i=1}^{n}\frac{(3i)!}{(i!)^3}x^{i}\right)^{k}.$$

REPLY [3 votes]: The summand $c_n = x^n (3n)!/(n!)^3$ is a hypergeometric term, meaning that $c_{n+1}/c_n$ is a rational function of $n$. Indeed:
$$
    \frac{c_{n+1}}{c_n} = \frac{x^{n+1}}{x_n} \frac{(3n+3)!}{(3n)!} \left(\frac{n!}{(n+1)!}\right)^3 = x \frac{(3n+3)(3n+2)(3n+1)}{(n+1)^3} = 27 x \frac{\left(n+\frac{1}{3}\right)}{(n+1)} \frac{\left(n+\frac{2}{3}\right)}{(n+1)}
$$
Such a recurrence equation implies that
$$
   c_n = c_0 \prod_{k=0}^{n-1} 27 x \frac{\left(k+\frac{1}{3}\right)}{(k+1)} \frac{\left(k+\frac{2}{3}\right)}{(k+1)} = c_0 \frac{(27 x)^n}{n!} \frac{\left(\frac{1}{3}\right)_n \left(\frac{2}{3}\right)_n}{(1)_n}
$$
Now the sum in question get's written as a defining series of the Gauss hypergeometric function:
$$
   \sum_{i=0}^\infty \binom{3i}{i,i,i} x^i = \sum_{n=0}^\infty \frac{(27 x)^n}{n!} \frac{\left(\frac{1}{3}\right)_n \left(\frac{2}{3}\right)_n}{(1)_n} = {}_2F_1\left(\frac{1}{3}, \frac{2}{3}; 1; 27 x\right)
$$
The result for summation from $i=1$ is obtained by subtracting the first term.


TITLE: If the family of functions $H$ is relatively compact in $C[K]$, then it is also relatively compact in $L^p(K)$ for all $p \in [1,\infty]$
QUESTION [0 upvotes]: I am working through the book Fundamentals of Applied Functional Analysis, and in the chapter where the Ascoli-Arzela theorem is introduced, the question is asked: show that if the family of functions $H$ is relatively compact in $C[K]$, then it is also relatively compact in $L^p(K)$ for all $p \in [1,\infty]$. How would I prove this?

REPLY [0 votes]: Given a sequence in $H$, find a $C(K)$-convergent subsequence. Then it is an $L^\infty(K)$ convergent subsequence. (The two norms are the same for continuous functions.) Since $K$ has finite measure, convergence in $L^q(K)$ implies convergence in $L^p(K)$ whenever $q>p$. Hence the subsequence converges in $L^p(K)$ for $p \in [1,\infty]$.


TITLE: Do the signs in Puppe sequences matter?
QUESTION [32 upvotes]: A basic construction in homotopy is Puppe sequences.  Given a map $A \stackrel{f}{\to} X$, its homotopy cofiber is the map $X\to X/A=X \cup_f CA$ from $X$ to the mapping cone of $f$.  If we then take the cofiber again, something remarkable happens: $(X/A)/X$ is naturally homotopy equivalent to the suspension $\Sigma A$ of $A$.  This isn't hard to see geometrically; a nice picture and discussion can be found in pages 397-8 of Hatcher.  If we iterate this, we end up with a sequence $$A \to X \to X/A \to \Sigma A \to \Sigma X \to \Sigma(X/A) \to \Sigma^2 A \to \cdots$$
in which each map is the homotopy cofiber of the previous map.  If we then apply a functor which sends cofiber sequences to exact sequences, we get a long exact sequence.  This can be understood as the origin of long exact sequences of cofibrations in (co)homology, using the fact that $H^n(X)=H^{n+1}(\Sigma X)$.
One subtlety of this construction is that under the natural identifications of $(X/A)/X$ and $((X/A)/X)/(X/A)$ with $\Sigma A$ and $\Sigma X$, the map $\Sigma A\to\Sigma X$ is not the suspension of the original map $f$, but rather its negative (i.e., $-1\wedge f: S^1\wedge A=\Sigma A \to \Sigma X=S^1\wedge X$, where -1 is a map of degree -1).  The geometric explanation for this can neatly be seen in Hatcher's picture, where the cones are successively added on opposite sides, so the suspension dimensions are going in opposite directions.
However, you usually don't need to worry about this sign issue.  First, since a map of degree -1 is a self-homotopy equivalence (even homeomorphism) of $\Sigma A$, we could just change our identification of $(X/A)/X$ with $\Sigma A$ by such a map and then we would just have $\Sigma f:\Sigma A \to \Sigma X$ (note though that then we are not changing how we identify the next space in the sequence with $\Sigma X$, which breaks some of the symmetry of the picture).  Alternatively, if we only care about the Puppe sequence because of the long exact sequences it gives us, we could note that an exact sequence remains exact if you change the sign of one of its maps.
My question is: is there any situation where these signs really do matter and have interesting consequences?  Might they be somehow connected to the signs that show up in graded commutative objects in topology?

REPLY [19 votes]: There is a specific sort of situation I know about where that sign matters.  Suppose you have $f:X \rightarrow Y$ and $g:Z \rightarrow W$ cofibrations (if the maps are not cofibrations, all the same things work - you just replace the quotient spaces by mapping cones).  You extend both maps to their Dold-Puppe sequences, so you get the sequences
$X \rightarrow Y \rightarrow Y/X \rightarrow \Sigma X \rightarrow \Sigma Y \ldots$
and
$Z \rightarrow W \rightarrow W/Z \rightarrow \Sigma Z \rightarrow \Sigma W \ldots$
Now suppose you have maps $a: X \rightarrow W$ and $b: Y \rightarrow W/Z$ making the obvious square commute up to homotopy.  You can then extend these to make a commutative ladder from the first Dold-Puppe sequence to the second.  (Notice that the sequences are deliberately offset from each other by one spot.)
Using the usual parameters and the obvious choices of homotopies you will get a square involving $Y/X, \Sigma X, \Sigma Z, \Sigma W$.    This square will commute if it includes the map $-\Sigma g: \Sigma Z \rightarrow \Sigma W$, but not generally with the map $\Sigma g$.  (To check all this, I recommend doing the Dold-Puppe sequences with mapping cones rather than quotient spaces but keeping the homotopy equivalences with the quotient spaces in mind, which is the only way I know to calculate what the right maps should be.)
At this point, if you were feeling stubborn, you could replace the map in your ladder $\Sigma X \rightarrow \Sigma Z$ with $-1$ times that map, and that would allow you to have used $\Sigma g$ in the square I mention in the above paragraph, but that creates other issues;  if you choose not to simply use suspensions of your original maps to go from one Dold-Puppe sequence to the other then you run into problem when you are mapping between Dold-Puppe sequences without the shift of this example.
I hope this helps unravel Greg's answer (which is correct - you need the sign to get good mapping properties).
Of course one sees exactly the same phenomenon in the category of chain complexes of abelian groups (where homotopy is chain homotopy) and other such categories.  I agree with Theo and Mark that one thinks about the suspension as "odd" (in the sense of parity not the sense of peculiar).
The published paper that Mark refers to that has an error of exactly this sort (which is unfortunately fundamental to the paper) is by Lin Jinkun in Topology v. 29, no. 4, pp. 389-407.  I read this paper in preprint form in 1988 and missed this error, but discovered it in 1992 when reading another paper by the same author with the same error.  In the Topology paper the error is made in diagram 4.4 on the right hand square (proof of Lemma 4.3).


text:
\section{Fluctuation moments}\label{sec.proof.main.thm.2}

The proofs of Theorem \ref{thm.second.order.asymptotics} and Theorem \ref{thm.second.order.freeness} are very similar, and thus, in order to avoid redundancies, this section is devoted to prove only Theorem \ref{thm.second.order.asymptotics}; nonetheless, what has to be modified to obtain the conclusions from Theorem \ref{thm.second.order.freeness} is pointed out in the next section.

Let $X_{N,1}$ and $X_{N,2}$ be as in Theorem \ref{thm.second.order.asymptotics}.

Assume $Y_{N}=Y_{N,1} Y_{N,2} \cdots Y_{N,2m_{1}}$ and $Z_{N} = Z_{N,1} Z_{N,2} \cdots Z_{N,2m_{2}}$ where $Y_{N,k}$ and $Z_{N,l}$ are given by \eqref{yz.centered.matrices} for some polynomials $\mathrm{p}_{1}, \mathrm{p}_{2},\ldots, \mathrm{p}_{2m_{1}}, \mathrm{q}_{1},   \mathrm{q}_{2}, \ldots , \mathrm{q}_{2m_{2}}\in \mathbb{C}[\mathrm{x}]$ and some indexes $i_1, i_2, \ldots, \allowbreak i_{2m_{_1}}, j_{1},j_{2},\ldots,j_{2m_{_2}} \in \{1,2\}$ satisfying $i_{1}=j_{1}$ and (\ref{cyclic.alternating.indexes}).

Note that

\[
Y_{N} = \big( U^{}_{N,i_{1}}A^{}_{N,1}U^{*}_{N,i_{1}} \big)
\big( U^{}_{N,i_{2}}A^{}_{N,2}U^{*}_{N,i_{2}} \big) \cdots
\big( U^{}_{N,i_{2m_{1}}}A^{}_{N,2m_{1}}U^{*}_{N,i_{2m_{1}}} \big)
\]

and

\[
Z_{N} =\big( U^{}_{N,j_{1}}B^{}_{N,1}U^{*}_{N,j_{1}} \big)
\big( U^{}_{N,j_{2}}B^{}_{N,2}U^{*}_{N,j_{2}} \big) \cdots
\big( U^{}_{N,j_{2m_{1}}}B^{}_{N,2m_{1}}U^{*}_{N,j_{2m_{1}}} \big)
\]

with $A_{N,k}$ and $B_{N,l}$ defined as in \eqref{ab.centered.matrices.alternative}; moreover, we have $ i_{2k-1} = j_{2l-1} = i_{1}  \neq i_{2} = i_{2k} = j_{2l}$.

Thus, following similar arguments to those in the proof of Lemma \ref{lemma.bounded.cumulant.property}, we obtain

\begin{align}\label{eqn.cov.YN.ZN.1}
\covt{Y_{N}}{Z_{N}}
=&
\sum_{\theta \in P_{\chi}(\pm {2m})}
\left(\sum_{\substack{
		\pi \in P_{\text{even}}(\pm {2m}) \\
		\pi \leq \theta}}
\mathfrak{c}_{2} \left[ \pi \right]
\mu(\pi, \theta) \right)
\sum_{\substack{
		\mathbf{j}:[\pm {2m}]\rightarrow[N] \\ \mathrm{ker}(\mathbf{j}) \geq \theta}}
\mathbf{a}(\mathbf{j})
\end{align}

where $P_{\chi}(\pm {2m})$ denotes the set of all even partitions of $[\pm {2m}]$ with no blocks of the form $\{k,-k\}$,

$\mu:P(\pm {2m} ) \times P( \pm {2m})\rightarrow \mathbb{C}$ is the M\"{o}bius inversion function,

$\mathbf{a(j)}$ is given by

\begin{align*}

\mathbf{a(j)} = \prod_{k=1}^{2m_{1}} A_{N,k}(j_{-k},j_{+k}) \cdot
\prod_{l=1}^{2m_{2}} B_{N,l}(j_{-2m_{1}-l},j_{+2m_{1}+l}),

\end{align*}

for function each $\mathbf{j} : [\pm {2m} ] \rightarrow [N]$, and if $\mathbf{j} : [\pm {2m} ] \rightarrow [N]$ satisfies $\kernel{\mathbf{j}} = \pi $, then

\begin{align}\label{eqn.normalized.fluct.c[pi]}
\mathfrak{c}_{2} \left[ \pi \right] & =
\cov{ \prod_{k =1}^{2m_{1}}
	V_{k}(j_{\sigma(-k)},j_{\sigma(k)})}{
	\prod_{k = 	2m_{1}+1}^{2m_{1}+2m_{2}}						V_{k}(j_{\sigma(-k)},j_{\sigma(k)})}
\end{align}

with $V^{}_{2k-1} =  V^{*}_{2k} = U^{*}_{N,{i}_{1}} U^{}_{N,{i}_{2}}  $ for $k = 1,2,\ldots, m$ and $\sigma:[\pm 2m] \rightarrow [\pm 2m]$ is the cyclic permutation given by

\[
\sigma = (-1,1,-2,2,\ldots,-2m_{1},2m_{1})(-2m_{1}-1,2m_{1}+1,\ldots,-2m_{1}-2m_{2} ,2m_{1}+2m_{2}).
\]

It turns out that (\ref{eqn.cov.YN.ZN.1}) becomes

\begin{align}\label{eqn.fluct.moments.c2.chi.chi.1}
\covt{Y_{N} }{Z_{N} }
=&
\sum_{\theta \in P_{\chi\chi}(\pm {2m})}
\left(\sum_{\substack{
		\pi \in P_{\text{even}}(\pm {2m}) \\
		\pi \leq \theta}}
	\mathfrak{c}_{2} \left[ \pi \right]
	\mu(\pi, \theta) \right)
\sum_{\substack{
		\mathbf{j}:[\pm {2m}]\rightarrow[N] \\ \mathrm{ker}(\mathbf{j}) \geq \theta}}
\mathbf{a}(\mathbf{j})
+ O(N^{-1})
\end{align}

where $P_{\chi\chi}(\pm {2m})$ denotes the set of all partitions $\theta \in P_{\chi}(\pm {2m})$ such that the graph sum exponent $\tau_{\theta}$, defined in Section \ref{sec.general.graph.sums}, equals $m$.

Indeed, if we are given partitions $\pi \in P_{\text{even}}(\pm {2m})$ and $\theta \in P_{\chi}(\pm {2m}) $ satisfying $\theta \geq \pi$, then Theorem \ref{mingo-speicher} and Proposition \ref{prop.graph.sum.exp.bound1} imply

\begin{align}\label{eqn.bound.fluct.moments.c2.chi.chi.graph.sum}
\abs*{
	\mathfrak{c}_{2} \left[ \pi \right]
	\mu(\pi, \theta)
	\sum_{\substack{
			\mathbf{j}:[\pm 2m]\rightarrow[N] \\ \mathrm{ker}(\mathbf{j}) \geq \theta}}
	\mathbf{a}(\mathbf{j})  }
\leq
\abs*{	\mathfrak{c}_{2} \left[ \pi \right]	}
\abs*{	\mu(\pi, \theta) }
N^{\tau_{\theta}}
\prod_{k=1}^{2m_{1}} \norm*{ A_{N,k} }
\prod_{l=1}^{2m_{2}} \norm*{ B_{N,l} }

\end{align}

where $\tau_{\theta}$ is the number of connected components of the graph $\mathcal{G}_{\theta}$.

Now, by hypothesis, $\sup_{N} \norm*{D_{N,i}} < \infty$ and $\lim_{N\rightarrow \infty}\mathrm{tr}( D^{k}_{N,i} )$ exists for every  $k\geq 1$ and any $i \in \{1,2\}$, so we have

\[\sup_{N} \prod_{k=1}^{2m_{1}} \norm*{ A_{N,k} }
\prod_{l=1}^{2m_{2}} \norm*{ B_{N,l} } < \infty . \]

Moreover, every connected component of $\mathcal{G}_{\theta}$ contains at least two edges since $\theta$ is even and has no blocks of the form $\{-k,k\}$, and hence, the graph sum exponent $\tau_{\theta}$ satisfies

\[
\tau_{\theta} \leq m .
\]

Additionally, since the unitary ensemble $\{\{U_{N,1},U_{N,2}\}\}_{N=1}^{\infty}$ satisfies
\textit{(\ref{pnorms-uniformly-bounded})} from Lemma \ref{lemma.bounded.cumulant.property}, it follows from the proof of Lemma \ref{lemma.bounded.cumulant.property} that there is a constant $C_{2}$ independent from $N$ satisfying

\[
\abs*{\mathfrak{c}_{2} \left[ \pi \right] }  \leq C_{2} N^{-m}.
\]

Therefore, from (\ref{eqn.bound.fluct.moments.c2.chi.chi.graph.sum}) we obtain

\begin{align}
	\mathfrak{c}_{2} \left[ \pi \right]
	\mu(\pi, \theta)
	\sum_{\substack{
			\mathbf{j}:[\pm 2m]\rightarrow[N] \\ \mathrm{ker}(\mathbf{j}) \geq \theta}}
	\mathbf{a}(\mathbf{j})
=
O(N^{-1})

\end{align}
unless the graph sum exponent $\tau_{\theta} = m$, and, consequently, we get (\ref{eqn.fluct.moments.c2.chi.chi.1}).

Note that the condition $\tau_{\theta} = m$, for an even partition $\theta\in P(\pm 2m)$ with no blocks of the form $\{+k,-k\}$, forces each component of the undirected graph ${\mathcal{G}}_{\theta}$ to have exactly two edges.

Thus, for any partition $\theta \in P_{\chi\chi}(\pm {2m})$, each component of the directed graph $\vec{\mathcal{G}}_{\theta}$ has one of the following forms:

\begin{center}\includegraphics[scale=1]{graph_sums_surviving_edges}\end{center}

And therefore, as illustrated it at the end of Section \ref{sec.general.graph.sums}, each graph sum $\sum_{\substack{ \mathrm{ker}(\mathbf{j}) \geq \theta}}\mathbf{a}(\mathbf{j})$ appearing in (\ref{eqn.fluct.moments.c2.chi.chi.1})  can be written as a product of traces of matrices where each trace is of the followings forms: $\Trn{C_{N,k}C_{N,l}}$,
$\Trn{C^{}_{N,k}C^{T}_{N,l}}$, or  $\Trn{C_{N,k} \circ C_{N,l}}$ where $C_{N,k}$ and $C_{N,l}$ belong to the set $\{A_{N,1},  \ldots, \allowbreak A_{N,2m_{1}},B_{N,1},\ldots,  B_{N,2m_{2}}\}$.

Hence, letting $\mathfrak{c}[\pi]$ be given by (\ref{eqn.normalized.fluct.c[pi]}) for each partition $\pi \in P(\pm 2m)$, the conclusions in Theorem \ref{thm.second.order.asymptotics} will follow from (\ref{eqn.fluct.moments.c2.chi.chi.1}) once we determine the order of

\begin{align}\label{sum.c2.pi.mu.pi.theta}
\sum_{\substack{
		\pi \in P_{\text{even}}(\pm 2m)	\\
		\theta \geq \pi}}
\mathfrak{c}_{2} \left[ \pi \right]
\mu(\pi, \theta)  .
\end{align}

Now, recall the values of M\"{o}bius inversion function are determined by (\ref{eqn.defining.mobius.func.r}), and given explicitly by (\ref{mobius.inversion.function}).

Thus, to determine the order of (\ref{sum.c2.pi.mu.pi.theta}), it is enough to compute $\mathfrak{c}_{2}[\pi]$ for even partitions $\pi\in P(\pm 2m)$ satisfying  $\pi  \leq \theta$ for some another partition $\theta$ in the set $P_{\chi\chi}(\pm {2m} )$.

\begin{proposition}\label{prop.fluc}

	Suppose $\theta$ is a partition in $P_{\chi\chi}(\pm {2m} )$.

	If  $\pi$ is an even partition such that $\pi \leq \theta$ and $\mathfrak{c}_{2}[\pi]$  is given by (\ref{eqn.normalized.fluct.c[pi]}),

	then the following holds:

	\begin{enumerate}

		\item\label{prop.fluc.h} for $U^{*}_{N,{i}_{1}} U^{}_{N,{i}_{2}} = \frac{1}{\sqrt{N}} W^{*} H^{} W$, we have
		\begin{align*}
		N^{m}\mathfrak{c}_{2}[ \pi ]
		& =
		\left\{\begin{array}{cl}
		1 +  O\left( N^{-1}\right)
		& \text{if there symmetric pairing partition } \hat\theta \leq \pi \text{ satisfying either (\ref{prop.minimal.special.partitions.1}) or
			(\ref{prop.minimal.special.partitions.2})}
		\\ & \text{from Proposition \ref{prop.minimal.special.partitions}} \\
		O\left( N^{-1/2}\right)
		& \text{otherwise. }
		\end{array} \right.
		\end{align*}

		\item\label{prop.fluc.xh} for $U^{*}_{N,{i}_{1}} U^{}_{N,{i}_{2}} = \frac{1}{\sqrt{N}} W^{*} X H^{} W$, we obtain
		\begin{align*}
		N^{m}\mathfrak{c}_{2}[ \pi ]
		& =
		\left\{\begin{array}{cl}
		1 +  O\left( N^{-1}\right)
		& \text{if there symmetric pairing partition } \hat\theta \leq \pi \text{ satisfying (\ref{prop.minimal.special.partitions.1}) from Proposi-}
		\\ & \text{tion \ref{prop.minimal.special.partitions}} \\
		1 +  O\left( N^{-1}\right)
		& \text{if there symmetric pairing partition } \hat\theta \leq \pi \text{ satisfying (\ref{prop.minimal.special.partitions.2}) from Proposi-}
		\\ & \text{tion \ref{prop.minimal.special.partitions} and the graph } \mathcal{G}_{\pi} \text{ has only double-loops as components} \\
		O\left( N^{-1/2}\right)
		& \text{otherwise. }
		\end{array} \right.
		\end{align*}

		\item\label{prop.fluc.xhx} for $U^{*}_{N,{i}_{1}} U^{}_{N,{i}_{2}} = \frac{1}{N} W^{*} H^{*}X H^{} W$, we get
		\begin{align*}
		N^{m}\mathfrak{c}_{2}[ \pi ]
		& =
		\left\{\begin{array}{cl}
		1 +  O\left( N^{-1}\right)
		& \text{if } m_{1} = m_{2} \text{ and there are integers } 1 \leq k \leq 2m_{1} < l \leq 2m_{1} + 2m_{2}	\text{ so} 	\\ &
		\text{that } \sigma^{t}(-k) \sim_{\pi} \sigma^{-t}(l) \text{ for every integer } t \geq 0	\\
		2 +  O\left( N^{-1}\right)
		&\text{if there are integers } 1 \leq k \leq 2m_{1} < l \leq 2m_{1} + 2m_{2} \text{ so that for each} \\ &
		\text{integer } t \geq 0  \text{ we have } \sigma^{-t}(k)	\sim_{\pi} 	\sigma^{t+1}(k)	\text{ and }
		\sigma^{-t}(l)	\sim_{\pi}	\sigma^{t+1}(l)  \\
		O\left( N^{-1/2}\right)
		& \text{otherwise. }
		\end{array} \right.
		\end{align*}

	\end{enumerate}
\end{proposition}

The proof of Proposition \ref{prop.fluc} is based on the expected value of products of entries from $X$ and $W$, see relations (\ref{dist.entries.signature.matrix}) and (\ref{dist.entries.signed.perm.matrix}), and  the results on graph sums of the Discrete Fourier Transform from Section \ref{subsec.graph.sums.dft}, however, it requires some technical intermediate steps, so we will omit it for now and leave it to the end of this section.

Nonetheless, the computation of (\ref{sum.c2.pi.mu.pi.theta}) up to a term of order $N^{-m-1/2}$ is quite simple assuming Proposition \ref{prop.fluc} holds.

\begin{lemma}\label{lemma.fluc.mobius}
	Suppose $\theta$ is a partition in $P_{\chi\chi}(\pm {2m} )$ and let $\mathfrak{c}[\pi]$  be given by (\ref{eqn.normalized.fluct.c[pi]}) for each partition $\pi \in P(\pm 2m)$.

	Then the following holds:

	\begin{enumerate}
		\item for $ U^{*}_{N,{i}_{1}} U^{}_{N,{i}_{2}}  = \frac{1}{\sqrt{N}} W^{*} H W$, we have

		\begin{align*}
		\sum_{\substack{
				\pi \in P_{\text{even}}(\pm {2m})	\\
				\pi \leq \theta }}
		N^{m}
		\mathfrak{c}_{2} \left[ \pi \right]
		\mu(\pi, \theta)
		=&
		\left\{\begin{array}{cl}
	1  + 	O\left( N^{-1/2} \right)&
		\text{if  } \theta \text{ is a pairing partition satisfying either  (\ref{prop.minimal.special.partitions.1}) or} \\
		& \text{(\ref{prop.minimal.special.partitions.2}) from  Proposition \ref{prop.minimal.special.partitions}}, \\
	O\left( N^{-1/2}\right)	&
		\text{ otherwise }
		\end{array}\right.
		\end{align*}

		\item for $ U^{*}_{N,{i}_{1}} U^{}_{N,{i}_{2}}  = \frac{1}{\sqrt{N}} W^{*} XH W^{*}$, we obtain

		\begin{align*}
		\sum_{\substack{
				\pi \in P_{\text{even}}(\pm 2m)	\\
				\pi \leq \theta }}
		N^{m}
		\mathfrak{c}_{2} \left[ \pi \right]
		\mu(\pi, \theta)
		=&
	\left\{\begin{array}{cl}
	1	+	O\left( N^{-1/2} \right)  &
		\text{if  }
		\theta \text{ is a pairing partition satisfying 			(\ref{prop.minimal.special.partitions.1})  from Pro-} \\ & \text{position \ref{prop.minimal.special.partitions}}, \\
	1	+	O\left( N^{-1/2} \right) &
		\text{if there exists a pairing partition } \hat\theta \leq \theta  \text{ satisfying} \\
		& \text{(\ref{prop.minimal.special.partitions.2}) from  Proposition \ref{prop.minimal.special.partitions} and } \theta  \text{ has only blocks of} \\
		& \text{the form } \{k,-k,l,-l\}, \\
	O\left( N^{-1/2}\right)	&
		\text{otherwise }
	\end{array}\right.
		\end{align*}

		\item for $ U^{*}_{N,{i}_{1}} U^{}_{N,{i}_{2}}  = \frac{1}{N}  W^{*} H^{*}XH  W^{}$, we get
		\begin{align*}
		\sum_{\substack{
				\pi \in P_{\text{even}}(\pm 2m)	\\
				\pi \leq \theta }}
		N^{m}
		\mathfrak{c}_{2} \left[ \pi \right]
		\mu(\pi, \theta)
		& =	\left\{\begin{array}{cl}
1	+	O\left( N^{-1/2} \right)
		& \text{if } m_{1} = m_{2} \text{ and there is an integer }  \\
		& 2m_{1}+1 \leq l \leq  2m_{1} + 2m_{2} \text{ such that } \\
		& \theta= \{ \sigma^{t}(-1) , \sigma^{-t-1}(-l) \} \mid t \geq 0
		\} ,  	\\
2	+	O\left( N^{-1/2} \right)
		& \text{if there are integers }  1 \leq l_{1} \leq 2m_{1} \text{ and } \\
		& 2m_{1}+1 \leq l_{2} \leq  2m_{1} + 2m_{2} \text{ so that }  \theta= \theta_{1} \sqcup \theta_{2} \\
		& \text{where } \theta_{1}=\{ \{ \sigma^{t}(-l_{1}) , \sigma^{-t-1}(-l_{1}) \}  \mid t \geq 0 \}  \\
		&  \theta_{2}=\{ \{ \sigma^{t}(-l_{2}) , \sigma^{-t-1}(-l_{2}) \} \mid t \geq 0 \} ,	\\
	O\left( N^{-1/2}\right)
		& \text{otherwise. }
		\end{array} \right.
		\end{align*}
	\end{enumerate}
\end{lemma}

\begin{proof}

	Suppose $U^{*}_{N,{i}_{1}} U^{}_{N,{i}_{2}}=W^{*}XHW/\sqrt{N}$. Proposition \ref{prop.minimal.special.partitions} and Proposition \ref{prop.fluc} imply that

	\begin{align*}
	\sum_{\substack{	\pi \in P_{\text{even}}(\pm 2m)	\\
			\pi \leq \theta }}
	N^{m}\mathfrak{c}_{2} \left[ \pi \right]
	\mu(\pi, \theta)
	=
	O\left( N^{-1/2}\right)
	\end{align*}
	unless there are integers $ 1\leq k \leq 2m_{1} < l \leq 2m_{1} + 2m_{2}$ satisfying one of the following:

	\begin{enumerate}[(i)]
		\item\label{lemma.fluc.mobius.1.case1} $k+l$ is even and $\sigma^{t}(-k) \sim_{\theta} \sigma^{-t}(l)$  for all integers  $t \geq 0$ or
		\item\label{lemma.fluc.mobius.1.case2} $k+l$ is odd, $\sigma^{t}(-k) \sim_{\theta} \sigma^{t}(-l)$  for all integers  $t \geq 0$ and $\mathcal{G}_{\theta}$ has only double-loops as	 components
	\end{enumerate}

	Assuming (\ref{lemma.fluc.mobius.1.case1}) above holds, consider the pairing partition

	$\hat\theta = \{ \{ \sigma^{t}(-k) , \sigma^{-t}(l) \}\mid t \geq 0\}$

	and note that Proposition \ref{prop.fluc} implies that

	\begin{align}\label{cumulant.reduction}
	\sum_{\substack{
			\pi \in P_{\text{even}}(\pm 2m)	\\	\pi \leq \theta }}
	N^{m}\mathfrak{c}_{2} \left[ \pi \right]	\mu(\pi, \theta)
	& =
	\sum_{\substack{	\pi \in P_{\text{even}}(\pm 2m)	\\
			\hat\theta \leq \pi \leq \theta }}
	N^{m}\mathfrak{c}_{2} \left[ \pi \right]
	\mu(\pi, \theta)
	+
	O\left( N^{-1/2}\right) .
	\end{align}

	Moreover, since $N^{m}\mathfrak{c}_{2} \left[ \pi \right] = 1 + O(N^{-1})$ for any partition $\pi$ satisfying $\hat\theta \leq \pi \leq \theta$, we get

	\begin{align*}
	\sum_{\substack{
			\pi \in P_{\text{even}}(\pm 2m)	\\	\pi \leq \theta }}
	N^{m}\mathfrak{c}_{2} \left[ \pi \right]	\mu(\pi, \theta)
	& =
	\sum_{\substack{	\pi \in P_{\text{even}}(\pm 2m)	\\
			\hat\theta \leq \pi \leq \theta }}
	\mu(\pi, \theta)
	+
	O ( N^{-1/2} )
	=
	\left\{\begin{array}{cl}
	1 + O\left( N^{-1/2}\right) & \text{if } \hat\theta = \theta \\
	O\left( N^{-1/2}\right)  & \text{if } \hat\theta < \theta
	\end{array}\right.
	\end{align*}

	from equations in (\ref{eqn.defining.mobius.func.r}) defining the M\"{o}bius inversion function.

	On the other hand, if (\ref{lemma.fluc.mobius.1.case2}) above holds, consider $\hat\theta = \{ \{ \sigma^{t}(-k) , \sigma^{t}(-l) \}\mid t \geq 0\}$ instead and note that (\ref{cumulant.reduction}) above holds also in this case.

	Hence, since $N^{m}\mathfrak{c}_{2} \left[ \theta \right] = 1 + O(N^{-1})$ and $N^{m}\mathfrak{c}_{2} \left[ \pi \right] = O(N^{-1/2})$ for any partition $\hat\theta \leq \pi < \theta$, we obtain

	\begin{align*}
	\sum_{\substack{
			\pi \in P_{\text{even}}(\pm 2m)	\\	\pi \leq \theta }}
	N^{m}\mathfrak{c}_{2} \left[ \pi \right]	\mu(\pi, \theta)
	=
	N^{m}\mathfrak{c}_{2} \left[ \theta \right]
	\mu(\theta, \theta)
	+
	\sum_{\substack{	\pi \in P_{\text{even}}(\pm 2m)	\\
			\hat\theta \leq \pi < \theta }}
	N^{m}\mathfrak{c}_{2} \left[ \pi \right]
	\mu(\pi, \theta)
	=
	1 +	O( N^{-1/2})
	\end{align*}

	The other cases, namely, $U^{*}_{N,{i}_{1}} U^{}_{N,{i}_{2}} = W^{*}HW /\sqrt{N} $ and $U^{*}_{N,{i}_{1}} U^{}_{N,{i}_{2}} = W^{*} H^{*}XH W/N$, are proved in the same way, one chooses a suitable pairing partition $\hat\theta$ such that  (\ref{cumulant.reduction}) holds, and then the corresponding conclusion follows from Proposition \ref{prop.fluc} and the equations in  (\ref{eqn.defining.mobius.func.r}) defining the M\"{o}bius function.
\end{proof}

As mentioned earlier, the proof of Theorem \ref{thm.second.order.asymptotics} is complete once we apply Lemma \ref{lemma.fluc.mobius} to the relation  (\ref{eqn.fluct.moments.c2.chi.chi.1}).

For instance, suppose $ U^{*}_{N,{i}_{1}} U^{}_{N,{i}_{2}}  = W^{*} H^{*}XH  W^{} / N$.

Then, Theorem \ref{mingo-speicher}, Lemma \ref{lemma.fluc.mobius}, and (\ref{eqn.fluct.moments.c2.chi.chi.1}) imply that

\begin{align}\label{rel.cov.YN.ZN.last}
\covt{Y_{N} }{Z_{N} }

= &

\delta_{m_{1},m_{2}}
\sum_{ \theta \in \mathcal{P}_{1} }
\sum_{\substack{
		\mathbf{j}:[\pm {2m}]\rightarrow[N]
		\\ \mathrm{ker}(\mathbf{j}) \geq \theta}}
\mathbf{a}(\mathbf{j})

 +

\sum_{\theta \in \mathcal{P}_{2} }
2 \sum_{\substack{
		\mathbf{j}:[\pm {2m}]\rightarrow[N]
		\\ \mathrm{ker}(\mathbf{j}) \geq \theta}}
\mathbf{a}(\mathbf{j})

+

O\left(N^{-\frac{1}{2}}\right)
\end{align}

where $\mathcal{P}_{1}$ and $\mathcal{P}_{2}$ are subsets of $P_{\chi\chi}(\pm {2m})$ given by

\[
\mathcal{P}_{1} = \{
\{ \{ \sigma^{t}(-1) , \sigma^{-t-1}(-l) \} \mid t = 0,1, 2, \ldots, 4m_{}  \}  \mid  l \in [2m] \setminus [2m_{1}]  \}
\]

and

\[
\mathcal{P}_{2} = \{	\{ \{ \sigma^{t}(-l_{1}) , \sigma^{-t-1}(-l_{1}) \} ,
\{ \sigma^{t}(-l_{2}) , \sigma^{-t-1}(l_{2}) \} \mid t \geq 0 \}	\mid  l_{1} \in [2m_{1}] ,  l_{2} \in [2m] \setminus [2m_{1}] \}.
\]

Now, note the set $\mathcal{P}_{1}$ has cardinality $2m_{1}$ provided $m_{1} = m_{2}$. Moreover,  $m_{1} = m_{2}$ implies that a partition $\theta \in P(\pm 2m)$ belongs to the set $\mathcal{P}_{1}$ if and only if for some integer $1\leq l \leq 2m_{1}$ the directed graph $\vec{\mathcal{G}}_{\theta}$ can be represented as

\begin{center}\includegraphics[scale=1]{thm5_pairings_1b}\end{center}

where $F_{k}$ denotes the edge $	E_{2m_{1}+k}$ and $l-k$ is taken module $2m_{1}$ for $k=1,2,\ldots, 2m_{1}$.

Thus, for each integer $1\leq l \leq 2m_{1}$, there exists a unique $\theta  \in  \mathcal{P}_{1}$ so that

\begin{align*}
\sum_{\substack{\mathbf{j}:[\pm {m}]\rightarrow[N] \\
		\mathrm{ker}(\mathbf{j}) \geq \theta}}
\mathbf{a}(\mathbf{j})

=\prod_{k=1}^{2m_{1}} \Tr{A_{N,k} B_{N,l-k}}
,
\end{align*}

and hence, we obtain

\begin{align*}
\covt{ Y_{N} }{ Z_{N} }
=&
 \delta_{m_{1},m_{2}} \sum_{l=1}^{2m_{1}}
\left( \prod_{k=1}^{2m_{1}} \tr{A_{N,k}B_{N,l-k}} \right)

+
\sum_{\theta \in \mathcal{P}_{2} }
2 \sum_{\substack{
		\mathbf{j}:[\pm {2m}]\rightarrow[N]
		\\ \mathrm{ker}(\mathbf{j}) \geq \theta}}
\mathbf{a}(\mathbf{j})
+
	O \left( N^{-\frac{1}{2	}}\right) .
\end{align*}

On the other hand, the set $\mathcal{P}_{2}$ has cardinality $m_{1} \cdot m_{2}$ since

\[
		\{ \{ \sigma^{t}(-l_{1}) , \sigma^{-t-1}(-l_{1}) \} \mid t \geq 0 \}
	=
		\{ \{ \sigma^{t}(-m_{1}-l_{1}) , \sigma^{-t-1}(-m_{1}-l_{1}) \} \mid t \geq 0 \}
\]

and

\[
\{ \{ \sigma^{t}(-l_{2}) , \sigma^{-t-1}(-l_{2}) \} \mid t \geq 0 \}
=
\{ \{ \sigma^{t}(-m_{2}-l_{2}) , \sigma^{-t-1}(-m_{2}-l_{2}) \} \mid t \geq 0 \}
\]

for $l_{1}=1,2,\ldots, m_{1}$ and $l_{2}=1,2,\ldots, m_{2}$. Moreover, a partition $\theta \in P(\pm 2m)$ belongs to the set $\mathcal{P}_{2}$ if and only if for some integers $1\leq l \leq m_{1}$ and  $1\leq l_{2} \leq m_{2}$ the directed graph $\vec{\mathcal{G}}_{\theta}$ can be represented as

\begin{center}\includegraphics[scale=1]{thm5_pairings_3b_2}\end{center}

where $l_{1}-k_{1}$ and $l_{2}-k_{2}$ are taken modulo $2m_{1}$ and $2m_{2}$, respectively, for $k_{1}=1,2,\ldots,m_{1}$ and $k_{2} = 1,2,\ldots,m_{2}$.

Thus, for each partition $\theta  \in  \mathcal{P}_{2}$ there are integers $1 \leq l_{1} \leq m_{1}$ and $1 \leq l_{2} \leq m_{2}$ satisfying

\begin{align*}
\sum_{\substack{\mathbf{j}:[\pm {m}]\rightarrow[N] \\
		\mathrm{ker}(\mathbf{j}) \geq \theta}}
\mathbf{a}(\mathbf{j})

=& \prod_{k_{1}=1}^{m_{1}} \Tr{ A_{N,l_{1}+k_{1}-1} A_{N,l_{1}-k_{1}}} \cdot \prod_{k_{2}=1}^{m_{2}} \Tr{B_{N,l_{2}+k_{2}-1} B_{N,l_{2}-k_{2}}}	.
\end{align*}

Therefore, we have

\begin{align*}
\covt{ Y_{N} }{ Z_{N} }
=&
2
\sum_{l_{1}=1}^{m_{1}} \sum_{l_{2}=1}^{m_{2}}
\prod_{k_{1}=1}^{m_{1}}
\tr{A_{N,l_{1}+k_{1}-1} A_{N,l_{1}-k_{1}}}
\prod_{k_{2}=1}^{m_{2}}
\tr{B_{N,l_{2}+k_{2}-1} B_{N,l_{2}-k_{2}}}
\\	&
+ \delta_{m_{1},m_{2}} \sum_{l=1}^{2m_{1}}
\left( \prod_{k=1}^{2m_{1}} \tr{A_{N,k}B_{N,l-k}} \right)
+
O ( N^{ -1/2 } ) .
\end{align*}

The other cases, namely, $U^{*}_{N,{i}_{1}} U^{}_{N,{i}_{2}} = W^{*}HW /\sqrt{N} $ and $U^{*}_{N,{i}_{1}} U^{}_{N,{i}_{2}} = W^{*} XH W/N$, are proved in the same way, applying Lemma \ref{lemma.fluc.mobius} to the relation  (\ref{eqn.fluct.moments.c2.chi.chi.1}) we obtain similar relations to that in (\ref{rel.cov.YN.ZN.last}) that lead to (\ref{thm.second.order.asymptotics.1}) and (\ref{thm.second.order.asymptotics.2}) in Theorem \ref{thm.second.order.asymptotics}.

The remaining of this section is devoted to the proof of Proposition \ref{prop.fluc}.

For clarity, we have considered two cases: 	$ U^{*}_{N,{i}_{1}} U^{}_{N,{i}_{2}} = W^{*} H^{*} X H^{}  W / N $  and $ U^{*}_{N,{i}_{1}} U^{}_{N,{i}_{2}} =  W^{*} Y H^{}  W /\sqrt{N} $ where $Y$ is either the identity matrix $I_{N}$ or an $N$-by-$N$ uniformly distributed signature matrix $X$.

But first, let us introduce some more notation for partitions.

Given a partition $\pi \in P(\pm 2m)$, we let $\pi_{\text{even}}$ and $\pi_{\text{odd}}$ denote the restriction of $\pi$ to the sets
$\{ k \in [\pm 2m] \mid k \text{ is even}  \}$ and $\{ k \in [\pm 2m] \mid k \text{ is odd}  \}$, respectively.

Moreover, we let $\pi^{\text{even}}$ and $\pi^{\text{odd}}$ denote the partitions of $\{ k \in [\pm 4m] \mid k \text{ is even }  \} $ and $\{ k \in [\pm 4m] \mid k \text{ is odd} \}$, respectively, given by $\pi^{\text{even}} = \{\{ 2k \mid k \in B \} \mid B \in \pi \}$ and
$\pi^{\text{odd}} = \{\{ 2k-\sign{(k)} \mid k \in B \} \mid B \in \pi \}$ where $\sign(k) = 1$, if $k$ is positive, and $\sign(k) = -1$,  otherwise.

For instance, if $\pi$ is the partition in $P(\pm 6)$ given by

\[\pi=\{\{-1,4\},\{1,-3\},\{-2,3\},\{2,-4\}  \}\]

then

\begin{align*}
\pi_{\text{even}} & =  \{ \{4\}, \{-2\},\{2,-4\} \},
	& \pi_{\text{odd}} =& \{\{-1\}, \{1,-3\}, \{3\} \},\\

\pi^{\text{even}}  & =  \{\{-2,8\},\{2,-6\},\{-4,6\},\{4,-8\} \},
\quad \text{and} \quad
		& \pi^{\text{odd}} = &\{\{-1,7\},\{1,-5\},\{-3,5\},\{3,-7\} \}.
\end{align*}

\subsection*{Case $\bm{ U^{*}_{N,{i}_{1}} U^{}_{N,{i}_{2}}=\frac{1}{\sqrt{N}} W^{*} Y H^{}  W}$ }

Let $Y$ be an $N$-by-$N$ diagonal random matrix independent from $W$.

Given a function $\mathbf{i}: [\pm 2m] \rightarrow [N]$, we let
$$\mathbf{h}(\mathbf{i}) = \mathbf{h}_{1}(\mathbf{i}) \mathbf{h}_{2}(\mathbf{i}) \quad  \text{and} \quad
\mathbf{y}(\mathbf{i}) = \mathbf{y}_{1}(\mathbf{i}) \mathbf{y}_{2}(\mathbf{i})
$$ where  $\mathbf{h_{1}}(\mathbf{i})$, $\mathbf{h_{2}}(\mathbf{i})$, $\mathbf{y_{1}}(\mathbf{i})$, and $\mathbf{y_{2}}(\mathbf{i})$ are given by

\begin{align*}

\mathbf{h_{1}}(\mathbf{i}) &=
\prod_{k=1}^{m_{1}}
H(i_{-2k+1},i_{2k-1}) H^{*}(i_{-2k},i_{2k}) ,

&\mathbf{h}_{2}(\mathbf{i})=&
\prod_{k=m_{1}+1}^{m_{1}+m_{2}}
H(i_{-2k+1},i_{2k-1}) H^{*}(i_{-2k},i_{2k}) , \\

\mathbf{y}_{1}(\mathbf{i}) &=
\prod_{k=1}^{m_{1}}
Y(i_{-2k+1},i_{-2k+1}) Y(i_{2k-1},i_{2k-1}),

&\mathbf{y}_{2}(\mathbf{i})=&
\prod_{k=m_{1}+1}^{m_{1}+m_{2}}
Y(i_{-2k+1},i_{-2k+1}) Y(i_{2k-1},i_{2k-1});
\end{align*}

additionally, if we are given a function $\mathbf{j}: [\pm 2m] \rightarrow [N]$,  we put $$\mathbf{w}(\mathbf{i,j}) = \mathbf{w}_{1}(\mathbf{i,j}) \mathbf{w}_{2}(\mathbf{i,j})$$ where $\mathbf{w}_{1}(\mathbf{i,j})$ and  $\mathbf{w}_{2}(\mathbf{i,j})$ are given by

\begin{align*}
\mathbf{w}_{1}(\mathbf{i,j}) &=
\prod_{k=1}^{2m_{1}}
W(i_{k},j_{k}) W(i_{-k},j_{-k})

\quad \text{ and }

&\mathbf{w}_{2}(\mathbf{i,j})=&
\prod_{k=2m_{1}+1}^{2m_{1}+2m_{2}}
W(i_{k},j_{k}) W(i_{-k},j_{-k}).
\end{align*}

Now, for every partition $\pi \in P_{}(\pm {2m})$ and any function $\mathbf{j}:[\pm {2m}] \rightarrow [N]$ satisfying $ \kernel{\mathbf{j}} = \pi$, we define $\mathfrak{C}_{2} \left[ \pi \right]$ by

\begin{align}\label{eqn.unnormalized.flucts}
&
\sum_{\substack{ \mathbf{i} : [\pm 2m] \rightarrow [N] \\ \kernel{\mathbf{i}}=\pi}}
\mathbf{h}(\mathbf{i} \circ \bm{\sigma} )
\exptr{\mathbf{w}(\mathbf{i,j})}\exptr{\mathbf{y}(\mathbf{i})}
-
\sum_{\substack{ \mathbf{i} : [\pm 2m] \rightarrow [N] \\  \kernel{\mathbf{i}}=\pi_{1}\sqcup\pi_{2}  }}
\mathbf{h}(\mathbf{i}  \circ \bm{\sigma}  )
\exptr{\mathbf{w}_{1}(\mathbf{i,j})}
\exptr{\mathbf{w}_{2}(\mathbf{i,j})}
\exptr{\mathbf{y}_{1}(\mathbf{i})}
\exptr{\mathbf{y}_{2}(\mathbf{i})}  \nonumber \\
& =
\mathfrak{C}_{2} \left[ \pi \right]
\end{align}

where ${\pi}_1$ and ${\pi}_2$ denote the restrictions of ${\pi}$ to $[\pm 2m_1]$ and $[\pm 2m ] \setminus [\pm 2m_1]$, respectively, and $\sigma$ is the cycle permutation given by

\[
\sigma = (-1,1,-2,2,\ldots,-2m_{1},2m_{1})(-2m_{1}-1,2m_{1}+1,\ldots,-2m_{1}-2m_{2}, 2m_{1}+2m_{2}).
\]

\begin{proposition}\label{prop.cpi-and-Cpi}
	Let $\pi$ be an even partition in $P(\pm {2m})$.

	Suppose $U^{*}_{N,{i}_{1}} U^{}_{N,{i}_{2}}  = W^{*} Y H^{} W / \sqrt{N}$ where $Y$ is an $N$-by-$N$ diagonal matrix independent from $W$ so that each entry $Y(i,i)$ takes values in the set $\{-1,1\}$.

	If $\mathfrak{c}_{2}[\pi]$ and $\mathfrak{C}_{2} \left[ \pi \right]$ are given by (\ref{eqn.normalized.fluct.c[pi]}) and (\ref{eqn.unnormalized.flucts}), respectively, then

	\begin{align}\label{eqn.relating.c[pi].C[pi]}
	N^{m}\mathfrak{c}_{2} \left[ \pi \right]
	=
	\mathfrak{C}_{2} \left[ \pi \right] 	+	O(N^{-1})
	\end{align}

\end{proposition}

\begin{proof}

	Fix a function $\mathbf{j}:[\pm {2m}] \rightarrow [N]$ satisfying  $\kernel{\mathbf{j}} = \pi$.

	Note that the $(j_{-2k+1},j_{2k-1})$-entry of $U^{*}_{N,{i}_{1}} U^{}_{N,{i}_{2}}$ and the $(j_{-2k},j_{2k})$-entry of $U^{*}_{N,{i}_{2}} U^{}_{N,{i}_{1}}$ are given by

	\begin{align*}

	\sum_{i_{-2k+1},i_{2k+1}=1}^{N}
	\frac{1}{\sqrt{N}}  W^{*}(j_{-2k+1},i_{-2k+1}) Y (i_{-2k+1},i_{-2k+1}) H^{}(i_{-2k+1},i_{2k-1})  W(i_{2k-1},j_{2k-1})
	\end{align*}

	and

	\begin{align*}

	\sum_{i_{-2k},i_{2k}=1}^{N}
	\frac{1}{\sqrt{N}}	W^{*}(j_{-2k},i_{-2k}) H^{*} (i_{-2k},i_{2k}) Y(i_{2k},i_{2k}),  W(i_{2k},j_{2k}),
	\end{align*}

	respectively.

	Thus, from (\ref{eqn.normalized.fluct.c[pi]}) and the linearity of the covariance, we have that

	\begin{align*}
	N^{m} \mathfrak{c}_{2} \left[ \pi \right]
	\quad =
	\sum_{\mathbf{i}:[\pm {2m}]\rightarrow[N]}
	{\mathbf{h}}(\mathbf{i}) \cdot
	\text{cov} \big[ {\mathbf{w}}_{1}\mathbf{(i, j \circ \bm{{\sigma}})} \cdot \mathbf{y}_{1}\mathbf{(i)} , \mathbf{w}_{2}\mathbf{(i, j \circ \bm{{\sigma}})} \cdot \mathbf{y}_{2}\mathbf{(i )} \big]
	\end{align*}

	where ${\mathbf{h}}(\mathbf{i})$, ${\mathbf{w}}_{1}\mathbf{(i,j)}$, ${\mathbf{w}}_{2}\mathbf{(i,j)}$, $\mathbf{y}_{1}\mathbf{(i)}$, $\mathbf{y}_{2}\mathbf{(i)}$, and $\bm{\sigma}$ are defined as above.

	But, for every function $\mathbf{i}:[\pm {2m}]\rightarrow[N]$, we have that

	\[
	{\mathbf{w}}_{k}\mathbf{(i,j)} =
	{\mathbf{w}}_{k}\mathbf{(i\circ \bm{ {\sigma}}, j \circ \bm{{\sigma}})}
	\quad \text{ and } \quad
	\mathbf{y}_{k}\mathbf{(i)} =	\mathbf{y}_{k}\mathbf{(i\circ \bm{{\sigma}})}
	\quad \text{ for }	\quad k=1,2,
	\]

	so we obtain

	\begin{align}\label{eqn.c2[pi].1}
	N^{m}\mathfrak{c}_{2} \left[ \pi \right]
	\quad =
	\sum_{\mathbf{i}:[\pm {2m}]\rightarrow[N]}
	{\mathbf{h}}(\mathbf{i}\circ \bm{{\sigma}}) \cdot
	\cov{ {\mathbf{w}}_{1}\mathbf{(i, j )} \cdot \mathbf{y}_{1}\mathbf{(i)} }{\mathbf{w}_{2}\mathbf{(i, j )} \cdot \mathbf{y}_{2}\mathbf{(i )} } .
	\end{align}

	Moreover, from (\ref{dist.entries.signed.perm.matrix}) we get that

	\[
	\exptr{{\mathbf{w}}_{1}(\mathbf{i},\mathbf{j}) {\mathbf{w}}_{2}(\mathbf{i},\mathbf{j})}= 0
	\quad  \text{and} \quad
	\exptr{{\mathbf{w}}_{1}(\mathbf{i},\mathbf{j})}\exptr{{\mathbf{w}}_{2}(\mathbf{i},\mathbf{j})}=0
	\]

	provided  $\kernel{\mathbf{i}} \neq \pi$ and $\kernel{\mathbf{i}} \not\geq \pi_{1} \sqcup \pi_{2} $, respectively.

	And hence, equality in (\ref{eqn.c2[pi].1}) becomes

	\begin{align*}

	N^{m}\mathfrak{c}_{2} \left[ \pi \right]
	\quad = &
	\sum_{\substack{ 	\mathbf{i}:[\pm {2m}]\rightarrow[N] \\
			\mathrm{ker}(\mathbf{i}) = \pi}}
	\mathbf{h}(\mathbf{i \circ \bm{\sigma}})
	\exptr{\mathbf{w}_{1}(\mathbf{i,j}) \mathbf{w}_{2}(\mathbf{i,j})} \exptr{\mathbf{y}_{1}(\mathbf{i})\mathbf{y}_{2}(\mathbf{i})} \\
	& -
	\sum_{\substack{	\theta \in P(\pm {2m})  \\
			{\theta} \geq \pi_{1} \sqcup \pi_{2}}}
	\sum_{\substack{	\mathbf{i}:[\pm {2m}]\rightarrow[N] \\
			\mathrm{ker}(\mathbf{i}) = \theta}}
	\mathbf{h}(\mathbf{i \circ \bm{\sigma}})
	\exptr{\mathbf{w}_{1}(\mathbf{i,j})} 	\exptr{\mathbf{w}_{2}(\mathbf{i,j})}
	\exptr{\mathbf{y}_{1}(\mathbf{i})}		\exptr{\mathbf{y}_{2}(\mathbf{i})}.

	\end{align*}

	To obtain (\ref{eqn.relating.c[pi].C[pi]}), it only remains to show that for ${\theta} \gneq \pi_{1} \sqcup \pi_{2} $, i.e., $\theta \geq \pi_{1} \sqcup \pi_{2}$ but $\theta \neq \pi_{1} \sqcup \pi_{2}$, implies

	\begin{align*}
	&
	\sum_{\substack{  \mathbf{i}:[\pm {2m}]\rightarrow[N] \\ \kernel{\mathbf{i}}={\theta}   }}
	{\mathbf{h}}(\mathbf{i}\circ \bm{{\sigma}})
	\exptr{{\mathbf{w}}_{1}(\mathbf{i},\mathbf{j})}
	\exptr{{\mathbf{w}}_{2}(\mathbf{i},\mathbf{j})}
	\exptr{{\mathbf{y}}_{1}(\mathbf{i})}
	\exptr{{\mathbf{y}}_{2}(\mathbf{i})}
	=
	O\left(N^{ -1  }\right) .
	\end{align*}

	Suppose ${\theta} \in P(\pm {2m})$ satisfies ${\theta} \gneq \pi_{1} \sqcup \pi_{2} $.

	Then, we must have $\#(\theta) < \#(\pi_{1} \sqcup \pi_{2})  = \#(\pi_{1}) +  \#(\pi_{2})$, or, equivalently,

	\[
	\#(\theta) - \#(\pi_{1}) -  \#(\pi_{2}) \leq -1.
	\]

	Now, $\mathbf{h}(\mathbf{i}\circ \bm{{\sigma}})$ has absolute value $1$ for any function $\mathbf{i}:[\pm {2m}]\rightarrow[N]$ and $\abs*{\exptr{{\mathbf{y}}_{1}(\mathbf{i})}
		\exptr{{\mathbf{y}}_{2}(\mathbf{i})}}\leq 1$, so	(\ref{dist.entries.signed.perm.matrix}) implies

	\begin{align*}
	& \abs*{
		\sum_{\substack{  \mathbf{i}:[\pm {2m}]\rightarrow[N] \\ \kernel{\mathbf{i}}={\theta}   }}
		{\mathbf{h}}(\mathbf{i}\circ \bm{{\sigma}})
		\exptr{{\mathbf{w}}_{1}(\mathbf{i},\mathbf{j})}
		\exptr{{\mathbf{w}}_{2}(\mathbf{i},\mathbf{j})}
		\exptr{{\mathbf{y}}_{1}(\mathbf{i})}
		\exptr{{\mathbf{y}}_{2}(\mathbf{i})}				} \\
	& \leq
	\frac{(N-\#(\pi_{1}))!}{N!} \cdot \frac{(N-\#(\pi_{2}))!}{N!}  \cdot
	\abs*{\sum_{\substack{  \mathbf{i}:[\pm {2m}]\rightarrow[N] \\ \kernel{\mathbf{i}}={\theta}   }}
		{\mathbf{h}}(\mathbf{i}\circ \bm{{\sigma}})			}
	=
	O\left(N^{- \#(\pi_{1}) -\#(\pi_{2}) + \#({\theta})  }\right)
	=
	O\left(N^{	-1	}\right).
	\end{align*}

\end{proof}

\begin{proof}[\textbf{Proof of (\ref{prop.fluc.h}) from Proposition \ref{prop.fluc}}]

	Let $Y$ be the identity matrix $I_{N}$.

	By Proposition \ref{prop.cpi-and-Cpi}, we only need to show that

	\begin{align*}
	\mathfrak{C}_{2}[ \pi ]
	& =
	\left\{\begin{array}{cl}
	1 +  O\left( N^{-1}\right)
	& \text{if there symmetric pairing partition } \hat\theta \leq \pi \text{ satisfying either (\ref{prop.minimal.special.partitions.1}) or
		(\ref{prop.minimal.special.partitions.2}) }
	\\ & \text{from Proposition \ref{prop.minimal.special.partitions}}, \\
	O\left( N^{-1/2}\right)
	& \text{otherwise. }
	\end{array} \right.
	\end{align*}

	where $\mathfrak{C}_{2} \left[ \pi \right]$ is given by (\ref{eqn.unnormalized.flucts}).

	Note that from (\ref{dist.entries.signed.perm.matrix}) and (\ref{eqn.unnormalized.flucts}) we obtain the inequality

	\begin{align*}
	\abs*{\mathfrak{C}_{2}[\pi]  }
	\leq
	\frac{(N-\#(\pi))!}{N!}
	\abs*{ \sum_{\substack{
				\mathbf{i}:[\pm {2m} ] \rightarrow [N] \\
				\kernel{\mathbf{i}}=\pi}}
		\mathbf{h}(\mathbf{i}\circ\bm{\sigma}) }
	+
	\frac{(N-\#(\pi_{1}))!}{N!}
	\frac{(N-\#(\pi_{2}))!}{N!}
	\abs*{ \sum_{\substack{
				\mathbf{i}:[\pm {2m} ] \rightarrow [N] \\
				\kernel{\mathbf{i}}=\pi_{1}\sqcup\pi_{2} }}
		\mathbf{h}(\mathbf{i}\circ\bm{\sigma}) } .
	\end{align*}

	But then, if $\mathrm{p}_{\sigma^{-1} \circ \pi}$ is a non-zero polynomial, so is $\mathrm{p}_{\sigma^{-1} \circ \pi_{1} \sqcup \pi_{2} }$ by Proposition \ref{minimal.special.polynomials}, and therefore, the last inequality and Corollary \ref{cor.bound.gauss.sum} would imply

	\(
	\mathfrak{C}_{2} \left[ \pi \right] = O(N^{-1/2}) . \text{ }
	\)

	And so, we can assume $\mathrm{p}_{\sigma^{-1} \circ \pi}$ is the zero polynomial without loss of generality.

	Now, for every function $\mathbf{i}:[\pm {2m} ] \rightarrow [N]$ satisfying $\kernel{\mathbf{i}}=\pi$ we have $\mathbf{h}(\mathbf{i}\circ\bm{\sigma}) = 1$ since $\mathrm{p}_{\sigma^{-1} \circ \pi}$ is the zero polynomial; additionally, (\ref{dist.entries.signed.perm.matrix}) gives

	\(
	\exptr{\mathbf{w}(\mathbf{i,j})} =
	\frac{(N-\#(\pi_{}))!}{N!}
	\)

	since $\pi$ is an even partition.

	Thus, from (\ref{eqn.unnormalized.flucts}) we obtain

	\begin{align}\label{C2pi.sigma-pi-zero.case.1}
	\mathfrak{C}_{2}[\pi] = 1 -
	\sum_{\substack{
			\mathbf{i}:[\pm {2m} ] \rightarrow [N] \\ \kernel{\mathbf{i}}=\pi_1 \sqcup \pi_2 }}
	\mathbf{h}(\mathbf{i}\circ\bm{\sigma})  \exptr{\mathbf{w}_1(\mathbf{i,j})} \exptr{\mathbf{w}_2(\mathbf{i,j})} .
	\end{align}

	Moreover, by Proposition \ref{minimal.special.polynomials}, there is a symmetric pairing partition $\hat\theta \leq \pi$ such that $\mathrm{p}_{\sigma^{-1} \circ \hat\theta}$ is also the zero polynomial, and hence,

	the partition $\hat\theta$ must satisfy one of the conditions \textit{(1)-(6)} from Proposition \ref{prop.minimal.special.partitions}.

	Notice $\hat\theta = \hat\theta_{1} \sqcup \hat\theta_{2}$, where  $\hat\theta_1$ and $\hat\theta_2$ denote the restrictions of $\hat\theta$ to $[\pm 2m_1]$ and $[\pm 2m ] \setminus [\pm 2m_1]$, respectively, implies

	\begin{align}\label{C2pi.pi=pi1_pi2.case.1}
	\mathfrak{C}_{2}[\pi]
	=
	1 -
	\frac{(N-\#(\pi_1))!}{N!}	\cdot \frac{(N-\#(\pi_2))!}{N!} \cdot \frac{N!}{(N-\#(\pi_1\sqcup\pi_2))!}
	=
	O\left( N^{-1}\right).
	\end{align}

	Indeed, if $\hat\theta = \hat\theta_{1} \sqcup \hat\theta_{2}$, then
	 $\hat\theta_{1}$ and  $\hat\theta_{2}$ must be even partitions, and so are $\pi_{1}$ and $\pi_{2}$ since  $\hat\theta \leq \pi$ implies $\hat\theta_{1} \leq \pi_{1}$ and  $\hat\theta_{2} \leq \pi_{2}$, so (\ref{dist.entries.signed.perm.matrix}) gives

	\begin{align*}
	\exptr{\mathbf{w}_{1}(\mathbf{i,j})} =
	\frac{(N-\#(\pi_{1}))!}{N!}
	\quad\text{and}\quad
	\exptr{\mathbf{w}_{2}(\mathbf{i,j})} =
	\frac{(N-\#(\pi_{2}))!}{N!}
	\end{align*}

	for every function $\mathbf{i}:[\pm {2m} ] \rightarrow [N]$ satisfying $\kernel{\mathbf{i}}=\pi_{1} \sqcup \pi_{2}$; moreover, Proposition \ref{minimal.special.polynomials} implies the polynomial  $\mathrm{p}_{\sigma^{-1} \circ \pi_1 \sqcup \pi_2}$ is also zero since $\hat\theta = \hat\theta_{1} \sqcup \hat\theta_{2}  \leq \pi_1 \sqcup \pi_2$, and thus, we obtain

	$$\mathbf{h}(\mathbf{i}\circ\bm{\sigma}) = 1 .$$

	Hence, (\ref{C2pi.pi=pi1_pi2.case.1}) follows from (\ref{C2pi.sigma-pi-zero.case.1}) provided $\hat\theta$ satisfies either
	\textit{(\ref{prop.minimal.special.partitions.3})},
	\textit{(\ref{prop.minimal.special.partitions.4})},
	\textit{(\ref{prop.minimal.special.partitions.5})}, or
	\textit{(\ref{prop.minimal.special.partitions.6})} from Proposition \ref{prop.minimal.special.partitions}.

 	Assume now $\hat\theta$ satisfies either \textit{(\ref{prop.minimal.special.partitions.1})} or \textit{(\ref{prop.minimal.special.partitions.2})} from Proposition \ref{prop.minimal.special.partitions}.

	Then, either $\pi_1 \sqcup \pi_2$ contains some singletons, if $\{k,l\} \in \pi$ or $\{k,-l\} \in \pi$ for some integers $1\leq k \leq 2m_{1} < l \leq 2m_{1}+2m_{2}$, or $\pi_1 \sqcup \pi_2 = \{\{-k,k\} \mid k \in [\pm 2m ]\}$, otherwise.

	In any case, the graph $\vec{\mathcal{G}}_{\pi_1 \sqcup \pi_2}$ does not satisfy none of the conditions \textit{(1)-(6)} from Remark \ref{rmk.single.loops} since  $m_{1}+m_{2} > 2$, and hence,  the polynomial $\mathrm{p}_{\sigma^{-1} \circ \pi_1 \sqcup \pi_2}$ is non-zero.

	Thus, by (\ref{dist.entries.signed.perm.matrix})  and Corollary \ref{cor.bound.gauss.sum}, we have

	\begin{align}\label{ineq.h.w1.w2}
	\abs*{
		\sum_{\substack{
				\mathbf{i}:[\pm {2m} ] \rightarrow [N] \\ \kernel{\mathbf{i}}=\pi_1 \sqcup \pi_2 }}
		\mathbf{h}(\mathbf{i}\circ\bm{\sigma})  \exptr{\mathbf{w}_1(\mathbf{i,j})} \exptr{\mathbf{w}_2(\mathbf{i,j})}
	}
	\leq
	C N^{\#( \pi_1 \sqcup \pi_2 ) - \frac{1}{2} } \cdot	\frac{(N-\#(\pi_1))!}{N!}	\cdot \frac{(N-\#(\pi_2))!}{N!}
	\end{align}

	for some constant $C > 0$ independent from $N$.

	Therefore,  from  (\ref{C2pi.sigma-pi-zero.case.1}) we get that

	\[
	\mathfrak{C}_{2}[\pi]
	=
	1 +	O( N^{-1/2}).
	\]

\end{proof}

\begin{proof}[\textbf{Proof of (\ref{prop.fluc.xh}) from Proposition \ref{prop.fluc}}]

	Let $Y$ be a random $N$-by-$N$ signature matrix independent from $W$.

	Similar to the previous case, $U^{*}_{N,{i}_{1}} U^{}_{N,{i}_{2}} = W^{*} H^{}  W / \sqrt{N}$,
	we can assume  $\mathrm{p}_{\sigma^{-1} \circ \pi}$ is the zero polynomial and it suffices to show that

	\begin{align*}
	\mathfrak{C}_{2}[ \pi ]
	& =
	\left\{\begin{array}{cl}
	1 +  O\left( N^{-1}\right)
	& \text{if there symmetric pairing partition } \hat\theta \leq \pi \text{ satisfying (\ref{prop.minimal.special.partitions.1}) from }
	\\ & \text{Proposition \ref{prop.minimal.special.partitions}}, \\
	1 +  O\left( N^{-1}\right)
	& \text{if there symmetric pairing partition } \hat\theta \leq \pi \text{ satisfying (\ref{prop.minimal.special.partitions.2}) from }
	\\ & \text{Proposition \ref{prop.minimal.special.partitions} and the graph } \mathcal{G}_{\pi} \text{ has only double-loops as components}, \\
	O\left( N^{-1/2}\right)
	& \text{otherwise. }
	\end{array} \right.
	\end{align*}

	Let $\pi_{\text{odd},1}$ and  $\pi_{\text{odd},2}$ denote the restrictions of $\pi_{\text{odd}}$ to $[\pm 2m_1]$ and $[\pm 2m ] \setminus [\pm 2m_1]$, respectively.

	Note that if $\pi_{\text{odd}}$ is not an even partition, then either  $\pi_{\text{odd},1}$ or  $\pi_{\text{odd},2}$ is not even, and hence, we obtain $\mathfrak{C}_{2}[\pi]=0$ since (\ref{dist.entries.signature.matrix}) would imply

	$\exptr{\mathbf{y}_1(\mathbf{i})} \exptr{\mathbf{y}_2(\mathbf{i})} =\exptr{\mathbf{y}(\mathbf{i})}
	=0
	$

	for every function $\mathbf{i}:[\pm {2m} ] \rightarrow [N]$ satisfying $\kernel{\mathbf{i}}=\pi$.

	Thus, we can further assume $\pi_{\text{odd}}$ is even.

	It then follows from (\ref{dist.entries.signature.matrix}) and (\ref{dist.entries.signed.perm.matrix}) that

	\[
	\mathbf{h}(\mathbf{i}\circ\bm{\sigma}) = 1,
	\quad
	\exptr{\mathbf{y}(\mathbf{i})} =1
	\quad \text{and} \quad
	\exptr{\mathbf{w}(\mathbf{i,j})} = \frac{(N-\#(\pi))!}{N!}
	\]

	for $\mathbf{i}:[\pm {2m} ] \rightarrow [N]$ satisfying $\kernel{\mathbf{i}}=\pi$, and hence, we obtain

	\begin{align}\label{C2pi.sigma-pi-zero.case.2}
	\mathfrak{C}_{2}[\pi] = 1 -
	\sum_{\substack{
			\mathbf{i}:[\pm {2m} ] \rightarrow [N] \\ \kernel{\mathbf{i}}=\pi_1 \sqcup \pi_2 }}
	\mathbf{h}(\mathbf{i}\circ\bm{\sigma})
	\exptr{\mathbf{y}_1(\mathbf{i})} \exptr{\mathbf{y}_2(\mathbf{i})}  \exptr{\mathbf{w}_1(\mathbf{i,j})} \exptr{\mathbf{w}_2(\mathbf{i,j})} .
	\end{align}

	By Proposition \ref{minimal.special.polynomials}, there is a symmetric pairing partition $\hat\theta \leq \pi$ such that $\mathrm{p}_{\sigma^{-1} \circ \hat\theta}$ is also the zero polynomial, and thus,

	the partition $\hat{\theta}$ must satisfy one of the conditions \textit{(1)-(6)} from Proposition \ref{prop.minimal.special.partitions}.

	However, if $\hat{\theta}$ satisfies either
	\textit{(\ref{prop.minimal.special.partitions.3})},
	\textit{(\ref{prop.minimal.special.partitions.4})},
	\textit{(\ref{prop.minimal.special.partitions.5})}, or
	\textit{(\ref{prop.minimal.special.partitions.6})}, then

	\begin{align}\label{C2pi.pi=pi1_pi2.case.2}
	\mathfrak{C}_{2}[\pi]
	=
	1 -	\frac{N!}{(N-\#(\pi_1\sqcup\pi_2))!} \cdot
	\frac{(N-\#(\pi_1))!}{N!}	\cdot
	\frac{(N-\#(\pi_2))!}{N!}
	=
	O\left( N^{-1}\right).
	\end{align}

	Indeed, suppose $\hat{\theta}$ satisfies either
	\textit{(\ref{prop.minimal.special.partitions.3})},
	\textit{(\ref{prop.minimal.special.partitions.4})},
	\textit{(\ref{prop.minimal.special.partitions.5})}, or
	\textit{(\ref{prop.minimal.special.partitions.6})} from Proposition \ref{prop.minimal.special.partitions}, let $\hat{\theta}_1$ and $\hat{\theta}_2$ denote the restrictions of $\hat{\theta}$ to $[\pm 2m_1]$ and $[\pm 2m ] \setminus [\pm 2m_1]$, respectively, and let $\mathbf{i}:[\pm {2m} ] \rightarrow [N]$ be a function satisfying $\kernel{\mathbf{i}}=\pi_{1} \sqcup \pi_{2}$.

	Note that $\hat{\theta} = \hat{\theta}_1 \sqcup \hat{\theta}_2  \leq  \pi_{1} \sqcup \pi_{2}$ since $\hat{\theta} \leq \pi$ implies $\hat{\theta}_{1} \leq \pi_{1}$ and  $\hat{\theta}_{2} \leq \pi_{2}$, and thus, by Proposition \ref{minimal.special.polynomials},  the polynomial $\mathrm{p}_{\sigma^{-1} \circ \pi_{1} \sqcup \pi_{2}}$ is zero, and hence, we get

	\begin{align*}
	\mathbf{h}(\mathbf{i}\circ\bm{\sigma}) = 1.
	\end{align*}

	Moreover, $\pi_{1}$ and $\pi_{2}$ are even partitions since
	$\hat{\theta}$ is even and $\hat{\theta} = \hat{\theta}_1 \sqcup \hat{\theta}_2 \leq \pi$, so, from (\ref{dist.entries.signed.perm.matrix}),  we get

	\begin{align*}
	\exptr{\mathbf{w}_{1}(\mathbf{i,j})} =
	\frac{(N-\#(\pi_{1}))!}{N!}
	\quad\text{and}\quad
	\exptr{\mathbf{w}_{2}(\mathbf{i,j})} =
	\frac{(N-\#(\pi_{2}))!}{N!}  .
	\end{align*}

	The partitions $\pi_{\text{odd},1}$ and $\pi_{\text{odd},2}$ are also even since $\hat{\theta}_{\text{odd}}$ is even and   $\hat{\theta} = \hat{\theta}_1 \sqcup \hat{\theta}_2 \leq \pi$ implies $\hat{\theta}_{\text{odd}} = \hat{\theta}_{\text{odd},1} \sqcup \hat{\theta}_{\text{odd},2}$, $\hat{\theta}_{\text{odd},1} \leq \pi_{\text{odd},1}$, and  $\hat{\theta}_{\text{odd},2} \leq \pi_{\text{odd},2}$ where  $\hat{\theta}_{\text{odd},1}$ and  $\hat{\theta}_{\text{odd},2}$ denote the restrictions of $\hat{\theta}_{\text{odd}}$ to $[\pm 2m_1]$ and $[\pm 2m ] \setminus [\pm 2m_1]$, respectively.

	Thus, from (\ref{dist.entries.signature.matrix}), we have

	\[
	\exptr{\mathbf{y}_1(\mathbf{i})} \exptr{\mathbf{y}_2(\mathbf{i})} = 1  .
	\]

	Consequently, we obtain (\ref{C2pi.pi=pi1_pi2.case.2}) from (\ref{C2pi.sigma-pi-zero.case.2}).

	Now, similar to the case $U^{*}_{N,{i}_{1}} U^{}_{N,{i}_{2}} = W^{*} H^{}  W / \sqrt{N}$, if $\hat{\theta}$ satisfies either  \textit{(\ref{prop.minimal.special.partitions.1})} or \textit{(\ref{prop.minimal.special.partitions.2})} from Proposition \ref{prop.minimal.special.partitions}, then $\mathrm{p}_{\sigma^{-1} \circ \pi_1 \sqcup \pi_2}$ is a non-zero polynomial and (\ref{ineq.h.w1.w2}) holds, so, from (\ref{C2pi.sigma-pi-zero.case.2}), we obtain

	\[
	\mathfrak{C}_{2}[\pi]
	=
	1 +	O( N^{-1/2})
	\]

	since we have $\abs*{\exptr{\mathbf{y}_1(\mathbf{i})} \exptr{\mathbf{y}_2(\mathbf{i})} = 1 } \leq 1$ for any function $\mathbf{i}:[\pm {2m} ] \rightarrow [N]$.

	It only remains to show that the undirected graph $\mathcal{G}_{\pi}$ must have only double-loops as connected components if  $\hat{\theta}$ satisfies \textit{(\ref{prop.minimal.special.partitions.2})} from Proposition \ref{prop.minimal.special.partitions}.

	So, suppose $\hat{\theta}$ satisfies \textit{(\ref{prop.minimal.special.partitions.2})} from Proposition \ref{prop.minimal.special.partitions}.

	Note that if $\pi$ has a block of the form $\{k,l\}$, then  $k+l$ is odd, and hence, we must have either $\{k\}$ or $\{l\}$ is a block of $\pi_{\text{odd}}$, contradicting the assumption that $\pi_{\text{odd}}$ is an even partition.

	Thus, $\pi$ has only blocks of the form $\{k,l,-k,-l\}$, or, equivalently, the undirected graph $\mathcal{G}_{\pi}$ has only double loops as connected components.

\end{proof}

\subsection*{Case $\bm{ U^{*}_{N,{i}_{1}} U^{}_{N,{i}_{2}} =\frac{1}{N} W^{*} H^{*} X H^{}  W}$ }

For each function $\mathbf{i}:[\pm {4m}] \rightarrow [N]$, we let $\widehat{\mathbf{h}}_{}(\mathbf{i})$, $\widehat{\mathbf{g}}_{}(\mathbf{i})$, $\widehat{\mathbf{x}}_{1}(\mathbf{i})$, and $\widehat{\mathbf{x}}_{2}(\mathbf{i})$ be given by

\begin{align*}

\widehat{\mathbf{h}}_{}(\mathbf{i})	=&
\prod_{k=1}^{2m}
H^{*}(i_{-2k+1},i_{2k-1}) H(i_{-2k},i_{2k}) ,

&\widehat{\mathbf{g}}_{}(\mathbf{i})=&
\prod_{k=1}^{2m}
H^{*}(i_{-2k+1},i_{-2k}) H(i_{2k},i_{2k-1})  ,  \\

\widehat{\mathbf{x}}_{1}(\mathbf{i}) = &
\prod_{k=1}^{2m_{1}}
X(i_{-2k},i_{2k}), \qquad \quad \text{and}

&\widehat{\mathbf{x}}_{2}(\mathbf{i}) =&
\prod_{k=2m_{1}+1}^{2m_{1}+2m_{2}}
X(i_{-2k},i_{2k});

\end{align*}

additionally, if we are given a function $\mathbf{j}:[\pm {2m}] \rightarrow [N]$, we take

\begin{align}\label{eqn.variable.t}
\mathbf{t}  =
(t_{-1},t_{1},t_{-3},t_{3},\ldots,t_{4m_{1}+4m_{2}-1})
=
(j_{-1},j_{1},j_{-2},j_{2},\ldots,j_{-2m},j_{2m})
\end{align}

and let $\widehat{\mathbf{w}}_{1}(\mathbf{i,t})$ and $\widehat{\mathbf{w}}_{2}(\mathbf{i,t})$, also denoted  $\widehat{\mathbf{w}}_{1}(\mathbf{i})$ and $\widehat{\mathbf{w}}_{2}(\mathbf{i})$, respectively, be defined by

\begin{align*}
\widehat{\mathbf{w}}_{1}(\mathbf{i,t})
=&
\prod_{k=1}^{2m_{1}}
W(i_{2k-1},t_{2k-1}) W(i_{-2k+1},t_{-2k+1})  \quad \text{and}\\

\widehat{\mathbf{w}}_{2}(\mathbf{i,t})
=&
\prod_{k=2m_{1}+1}^{2m_{1}+2m_{2}}
W(i_{2k+1},t_{2k+1}) W(i_{-2k+1},t_{-2k+1}) .

\end{align*}

Now, given partitions $\pi \in P(\pm 2m)$ and ${\alpha} \in P_{2}(2m)$ and a function $\mathbf{j}:[\pm {2m}] \rightarrow [N]$ satisfying $ \kernel{\mathbf{j}} = \pi$, we define $\mathfrak{C}_{2} \left[ \pi, {\alpha} \right]$ by

\begin{align}\label{eqn.unnormalized.fluct.HXH}

&
\sum_{\substack{ \mathbf{i}:[\pm {4m} ] \rightarrow [N] \\  \kernel{\mathbf{i}}= \widehat{\alpha} \sqcup \pi^{\text{odd}} }}
\widehat{\mathbf{h}}(\mathbf{i} \circ \bm{\widehat{\sigma}} )
\exptr{\widehat{\mathbf{w}}(\mathbf{i})}
\exptr{\widehat{\mathbf{x}}(\mathbf{i})}  \nonumber

-\sum_{\substack{ \mathbf{i}:[\pm {4m} ] \rightarrow [N] \\  \kernel{\mathbf{i}}=\widehat{\alpha} \sqcup \pi_{1}^{\text{odd}} \sqcup \pi_{2}^{\text{odd}}  }}
\widehat{\mathbf{h}}(\mathbf{i} \circ \bm{\widehat{\sigma}})
\exptr{\widehat{\mathbf{w}}_{1}(\mathbf{i})}
\exptr{\widehat{\mathbf{w}}_{2}(\mathbf{i})}
\exptr{\widehat{\mathbf{x}}_{1}(\mathbf{i})}
\exptr{\widehat{\mathbf{x}}_{2}(\mathbf{i})}  \\

&= N^{m}\mathfrak{C}_{2} \left[ \pi, {\alpha} \right]

\end{align}

where $\pi^{\text{odd}}_{1}$ and $\pi^{\text{odd}}_{2}$ denote the restrictions of $\pi^{\text{odd}}$ to the sets $[\pm 4m_{1}]$ and $[\pm (4m_{1}+4m_{2}) ] \setminus [\pm 4m_{1}]$, respectively, $\widehat\alpha$ is the partition given by $\widehat\alpha=\{\{-2k,2k,-2l,2l\} \mid \{k,l\} \in \alpha \}$, and  $\widehat{\sigma} : [\pm {4m}] \rightarrow [\pm {4m}]$ is the permutation with cycle decomposition

\begin{align*}
\widehat{\sigma} = 	(-1,1,-2,2,\ldots,-4m_{1},4m_{1})
(-4m_{1}-1,4m_{1}+1,-4m_{2}-2,\ldots,4m_{1}+4m_{2}).
\end{align*}

\begin{proposition}\label{prop.cpi-and-Cpi-nu}

	Let $\pi$ be an even partition in $P(\pm {2m})$.

	Suppose $U^{*}_{N,{i}_{1}} U^{}_{N,{i}_{2}} =  W^{*} H^{*}X H^{} W / N$.

	If $\mathfrak{c}_{2} \left[ \pi \right] $ is given by (\ref{eqn.normalized.fluct.c[pi]}) and  $\mathfrak{C}_{2} \left[ \pi, {\alpha} \right] $ is given by (\ref{eqn.unnormalized.fluct.HXH}) for every pairing partition ${\alpha} \in P_{2}(2m)$, then

	\begin{align}\label{eqn.c[pi].And.C[pi,eta]}
	N^{2m}\mathfrak{c}_{2} \left[ \pi \right]
	=	&
	\left(\sum_{\substack{ {\alpha} \in {P}_{2}( 2m )   }}
	\mathfrak{C}_{2} \left[ \pi, {\alpha} \right]  \right)
	+
	O\left( N^{m-1} \right)
	\end{align}

\end{proposition}

\begin{proof}

	Fix a function $\mathbf{j}:[\pm {2m}] \rightarrow [N]$ satisfying   $\kernel{\mathbf{j}} = \pi$ and let $\mathbf{t}$ be as in (\ref{eqn.variable.t}).

	The $(j_{-k},j_{k})$-entry of $U^{*}_{N,{i}_{1}} U^{}_{N,{i}_{2}}$ is then given by the sum

	\begin{align*}
	\sum_{i_{-2k+1},i_{2k+1},i_{-2k},i_{2k}=1}^{N}
	W^{*}(t_{-2k+1},i_{-2k+1}) H^{*}(i_{-2k+1},i_{-2k}) X(i_{-2k},i_{2k}) H^{}(i_{2k},i_{2k-1})  W(i_{2k-1},t_{2k-1}),
	\end{align*}

	and hence, by Equation (\ref{eqn.normalized.fluct.c[pi]}) and the linearity of the covariance, we get

	\begin{align*}
	N^{2m} \mathfrak{c}_{2} \left[ \pi \right]
	& =
	\sum_{\mathbf{i}:[\pm {4m}]\rightarrow[N]}
	\widehat{\mathbf{g}}(\mathbf{i}) \cdot
	\text{cov} \big[ \widehat{\mathbf{w}}_{1}\mathbf{(i, t \circ \bm{\widetilde{\sigma}})} \cdot \widehat{\mathbf{x}}_{1}\mathbf{(i)} , \mathbf{w}_{2}\mathbf{(i, t \circ \bm{\widetilde{\sigma}})} \cdot \widehat{\mathbf{x}}_{2}\mathbf{(i )} \big]
	\end{align*}

	where $\widehat{\mathbf{g}}(\mathbf{i})$, $\widehat{\mathbf{w}}_{1}\mathbf{(i,t)}$, $\widehat{\mathbf{w}}_{2}\mathbf{(i,t)}$, $\widehat{\mathbf{x}}_{1}\mathbf{(i)}$, and $\widehat{\mathbf{x}}_{2}\mathbf{(i)}$ are defined as above and $\widetilde\sigma : [\pm 4m] \rightarrow [\pm 4m] $ is the permutation with cycle decomposition

	\[
	\widetilde\sigma = 	(-1,1,-3,3,\ldots,-4m_{1}+1,4m_{1}-1)
	(-4m_{1}-1,4m_{1}+1,\ldots,4m_{1}+4m_{2}-1).
	\]

	Note that for every function $\mathbf{i}:[\pm {4m}]\rightarrow[N]$ we have

	\[
	\widehat{\mathbf{h}}(\mathbf{i}\circ \bm{\widehat{\sigma}}) =
	\widehat{\mathbf{g}}(\mathbf{i}\circ \bm{\widetilde{\sigma}}), \quad
	\widehat{\mathbf{w}}_{k}\mathbf{(i,t)} =
	\widehat{\mathbf{w}}_{k}\mathbf{(i\circ \bm{\widetilde{\sigma}}, t \circ \bm{\widetilde{\sigma}})} ,\quad \text{ and } \quad
	\widehat{\mathbf{x}}_{k}\mathbf{(i)} =	\widehat{\mathbf{x}}_{k}\mathbf{(i\circ \bm{\widetilde{\sigma}})}
	\]

	for $k=1,2$, so we get

	\begin{align*}
	N^{2m}\mathfrak{c}_{2} \left[ \pi \right]
	&=
	\sum_{\mathbf{i}:[\pm {4m}]\rightarrow[N]}
	\widehat{\mathbf{h}}(\mathbf{i}\circ \bm{\widehat{\sigma}}) \cdot
	\cov{ \widehat{\mathbf{w}}_{1}\mathbf{(i, t )} \cdot \widehat{\mathbf{x}}_{1}\mathbf{(i)} }{\mathbf{w}_{2}\mathbf{(i, t )} \cdot \widehat{\mathbf{x}}_{2}\mathbf{(i )} } .
	\end{align*}

	Now, suppose $\theta = \kernel{\mathbf{i}}$ for a function $\mathbf{i}:[\pm {4m}]\rightarrow[N]$.

	Since $\pi^{\text{odd}} = \kernel{\mathbf{t}}$, from (\ref{dist.entries.signed.perm.matrix}) we have that

	\[
	\exptr{\widehat{\mathbf{w}}_{1}(\mathbf{i},\mathbf{t}) \widehat{\mathbf{w}}_{2}(\mathbf{i},\mathbf{t})}= 0
	\quad  \text{and} \quad
	\exptr{\widehat{\mathbf{w}}_{1}(\mathbf{i},\mathbf{t})}\exptr{\widehat{\mathbf{w}}_{2}(\mathbf{i},\mathbf{t})}=0
	\]

	provided  ${\theta}_{\text{odd}} \neq \pi^{\text{odd}}$ and ${\theta}_{\text{odd}} \not\geq (\pi_{1} \sqcup \pi_{2})^{\text{odd}} = \pi_{1} ^{\text{odd}}  \sqcup \pi_{2}^{\text{odd}} $, respectively;

	moreover, (\ref{dist.entries.signature.matrix}) implies that

	\[
	\exptr{\widehat{\mathbf{x}}_{1}(\mathbf{i})\widehat{\mathbf{x}}_{2}(\mathbf{i})}=\exptr{\widehat{\mathbf{x}}_{1}(\mathbf{i})}\exptr{\widehat{\mathbf{x}}_{2}(\mathbf{i})}=0
	\]

	if ${\theta}_{\text{even}}$ is not an even partition, ${\theta}_{\text{even}}$ has a block of the form $\{ 2k, -2k\}$, or $2k \not\sim_{{\theta}} -2k$ for some $k \in [2m]$.

	Thus, we obtain

	\begin{align*}
	N^{2m} \mathfrak{c}_{2} \left[ \pi \right]
	= &
	\sum_{\substack{ {\theta} \in \widetilde{P}_{\pi}( \pm 4m )   }}
	\sum_{\substack{  \mathbf{i}:[\pm {4m}]\rightarrow[N] \\ \kernel{\mathbf{i}}={\theta}   }}

	\widehat{\mathbf{h}}(\mathbf{i}\circ \bm{\widehat{\sigma}})	\exptr{\widehat{\mathbf{w}}_{1}(\mathbf{i},\mathbf{t}) \widehat{\mathbf{w}}_{2}(\mathbf{i},\mathbf{t})}
	\exptr{\widehat{\mathbf{x}}_{1}(\mathbf{i})\widehat{\mathbf{x}}_{2}(\mathbf{i})}  \\ 	& -
	\sum_{\substack{ {\theta} \in \widetilde{P}_{\pi_{1} \sqcup \pi_{2}}( \pm 4m )   }}
	\sum_{\substack{  \mathbf{i}:[\pm {4m}]\rightarrow[N] \\ \kernel{\mathbf{i}}={\theta}   }}

	\widehat{\mathbf{h}}(\mathbf{i}\circ \bm{\widehat{\sigma}})
	\exptr{\widehat{\mathbf{w}}_{1}(\mathbf{i},\mathbf{t})}
	\exptr{\widehat{\mathbf{w}}_{2}(\mathbf{i},\mathbf{t})}
	\exptr{\widehat{\mathbf{x}}_{1}(\mathbf{i})}
	\exptr{\widehat{\mathbf{x}}_{2}(\mathbf{i})}
	\end{align*}

	where $\widetilde{P}_{\beta}(\pm 4m)$ denotes the set of all partitions ${\theta} \in P(\pm 4m)$ such that ${\theta}_{\text{odd}} \geq \beta^{\text{odd}}$ and for every integer $k \in [2m]$ there exists  $l \in [2m]\setminus\{k\}$ such that $2k \sim_{{\theta}} -2k \sim_{{\theta}} -2l \sim_{{\theta}} 2l$.

	Now, letting $\widehat{P}_{\beta}(\pm 4m)$ denote the set of partitions ${\theta}  \in \widetilde{P}_{\beta}(\pm 4m)$ so that ${\theta} = {\theta}_{\text{even}} \sqcup {\theta}_{\text{odd}}$, ${\theta}_{\text{odd}} = \beta^{\text{odd}}$, and every block of ${\theta}_{\text{even}}$ is of the form $\{2k,-2k,2l,-2l\}$ with $k,l \in [2m]$ and $k\neq l$,

	note the mapping $${\alpha} \mapsto \widehat{\alpha} \sqcup \beta^{\text{odd}}$$ with $\widehat{{\alpha}} = \left\{ \{2k,-2k,2l,-2l\} \mid \{k,l\} \in {\alpha} \right\}$ gives a bijection between the set of pairing partitions $P_{2}(2m)$ and the set $\widehat{P}_{\beta}(\pm 4m)$ for any partition $\beta \in P(\pm 2m)$.

	Thus, to get (\ref{eqn.c[pi].And.C[pi,eta]}), it only remains to show that

	\begin{align*}
	N^{2m} \mathfrak{c}_{2} \left[ \pi \right]
	= &

	\sum_{\substack{ {\theta} \in \widehat{P}_{\pi}( \pm 4m )   }}
	\sum_{\substack{  \mathbf{i}:[\pm {4m}]\rightarrow[N] \\ \kernel{\mathbf{i}}={\theta}   }}
	\widehat{\mathbf{h}}(\mathbf{i}\circ \bm{\widehat{\sigma}})
	\exptr{\widehat{\mathbf{w}}(\mathbf{i},\mathbf{t})}
	\exptr{\widehat{\mathbf{x}}(\mathbf{i})}  \\
	& -

	\sum_{\substack{ 	{\theta} \in \widehat{P}_{\pi_{1} \sqcup \pi_{2} } ( \pm 4m ) }}
	\sum_{\substack{  \mathbf{i}:[\pm {4m}]\rightarrow[N] \\ \kernel{\mathbf{i}}={\theta}   }}
	\widehat{\mathbf{h}}(\mathbf{i}\circ \bm{\widehat{\sigma}})
	\exptr{\widehat{\mathbf{w}}_{1}(\mathbf{i},\mathbf{t})}
	\exptr{\widehat{\mathbf{w}}_{2}(\mathbf{i},\mathbf{t})}
	\exptr{\widehat{\mathbf{x}}_{1}(\mathbf{i})}
	\exptr{\widehat{\mathbf{x}}_{2}(\mathbf{i})}
	\\ & 	+
	O\left( N^{m-1} \right).
	\end{align*}

	Suppose ${\theta} \in \widetilde{P}_{\pi_{1} \sqcup \pi_{2}}( \pm 4m )$.

	Then, since ${\theta}_{\text{odd}} \geq (\pi_{1} \sqcup \pi_{2})^{\text{odd}} = \pi_{1}^{\text{odd}} \sqcup  \pi_{2}^{\text{odd}}$ and each block of ${\theta}_{\text{even}}$ has at least 4 elements, we get the inequality

	\[
	\#({\theta})  \leq
	\#({\theta}_{\text{odd}})+\#({\theta}_{\text{even}})
	\leq \#(\pi_{1}) + \#(\pi_{2}) + m
	\]

	with equality only if ${\theta} = {\theta}_{\text{even}} \sqcup {\theta}_{\text{odd}}$, ${\theta}_{\text{odd}} =  \pi_{1}^{\text{odd}} \sqcup  \pi_{2}^{\text{odd}}$,
	and each block of ${\theta}_{\text{even}}$ has exactly 4 elements, i.e.,  ${\theta} \in \widehat{P}_{\pi_{1} \sqcup  \pi_{2} }( \pm 4m )$;

	moreover, (\ref{dist.entries.signature.matrix}) and (\ref{dist.entries.signed.perm.matrix}) imply that

	\begin{align*}
	& \abs*{
		\sum_{\substack{  \mathbf{i}:[\pm {4m}]\rightarrow[N] \\ \kernel{\mathbf{i}}={\theta}   }}
		\widehat{\mathbf{h}}(\mathbf{i}\circ \bm{\widehat{\sigma}})
		\exptr{\widehat{\mathbf{w}}_{1}(\mathbf{i},\mathbf{t})}
		\exptr{\widehat{\mathbf{w}}_{2}(\mathbf{i},\mathbf{t})}
		\exptr{\widehat{\mathbf{x}}_{1}(\mathbf{i})}
		\exptr{\widehat{\mathbf{x}}_{2}(\mathbf{i})}				} \\
	& \leq
	\frac{(N-\#(\pi_{1}))!}{N!} \cdot \frac{(N-\#(\pi_{2}))!}{N!}  \cdot
	\abs*{\sum_{\substack{  \mathbf{i}:[\pm {4m}]\rightarrow[N] \\ \kernel{\mathbf{i}}={\theta}   }}
		\widehat{\mathbf{h}}(\mathbf{i}\circ \bm{\widehat{\sigma}})			}
	=
	O\left(N^{- \#(\pi_{1}) -\#(\pi_{2}) + \#({\theta})  }\right).
	\end{align*}

	Hence, if ${\theta} \in \widetilde{P}_{\pi_{1} \sqcup \pi_{2}}( \pm 4m ) \setminus \widehat{P}_{\pi_{1} \sqcup  \pi_{2} }( \pm 4m )$,  we have

	\begin{align*}
	&
	\sum_{\substack{  \mathbf{i}:[\pm {4m}]\rightarrow[N] \\ \kernel{\mathbf{i}}={\theta}   }}
	\widehat{\mathbf{h}}(\mathbf{i}\circ \bm{\widehat{\sigma}})
	\exptr{\widehat{\mathbf{w}}_{1}(\mathbf{i},\mathbf{t})}
	\exptr{\widehat{\mathbf{w}}_{2}(\mathbf{i},\mathbf{t})}
	\exptr{\widehat{\mathbf{x}}_{1}(\mathbf{i})}
	\exptr{\widehat{\mathbf{x}}_{2}(\mathbf{i})}
	=
	O\left(N^{ m-1  }\right) .
	\end{align*}

	Similar arguments show that $ \#({\theta}) \leq m + \#(\pi_{})$ for every partition ${\theta} \in \widetilde{P}_{\pi_{}}( \pm 4m )$ with equality only if ${\theta} \in \widehat{P}_{\pi}( \pm 4m )$, and hence, we get

	\[
	\sum_{\substack{  \mathbf{i}:[\pm {4m}]\rightarrow[N] \\ \kernel{\mathbf{i}}={\theta}   }}
	\widehat{\mathbf{h}}(\mathbf{i}\circ \bm{\widehat{\sigma}})
	\exptr{\widehat{\mathbf{w}}_{}(\mathbf{i},\mathbf{t})}
	\exptr{\widehat{\mathbf{x}}_{}(\mathbf{i})}

	=
		O\left(N^{ m-1  }\right)
	\]

	for any ${\theta} \in  \widetilde{P}_{\pi_{}}( \pm 4m ) \setminus \widehat{P}_{\pi}( \pm 4m )$.

\end{proof}

\begin{proposition}\label{prop.fluc.xhx.C[p,a]}
	Suppose $\pi \in P_{\text{}}(\pm 2m )$ and ${\alpha} \in P_{2}(2m)$ and let  ${\alpha}_{1}$ and  ${\alpha}_{2}$ denote the restrictions of  ${\alpha}$ to the sets $[2m_{1}]$ and $[ 2m_{1} + 2m_{2} ] \setminus [2m_{1}]$, respectively.

	If 	 $\mathfrak{C}_{2}[\pi,{\alpha}] $ is given by (\ref{eqn.unnormalized.fluct.HXH}), then

	\[
	\mathfrak{C}_{2}[\pi,{\alpha}]
	=
	\left\{\begin{array}{cl}
1 +  O\left( N^{-1}\right)
	& \text{if there is a symmetric pairing } \eta \in P(\pm {4m}) \text{ satisfying (\ref{prop.minimal.special.partitions.1}) from Proposi-}  \\
	& \text{tion \ref{prop.minimal.special.partitions} and such that }   \eta \leq \widehat{\alpha} \sqcup \pi^{\text{odd}} , \\
1 +  O\left( N^{-1}\right)
	& \text{if } {\alpha} \neq {\alpha}_{1} \sqcup {\alpha}_{2} \text{ and there is a symmetric pairing }   \eta \in P(\pm {4m}) \text{ satisfying}	\\
	&  \text{(\ref{prop.minimal.special.partitions.3}) from Proposition \ref{prop.minimal.special.partitions}} \text{ with } $k$ \text{ and } $l$ \text{ even and such that } \eta \leq \widehat{\alpha} \sqcup \pi^{\text{odd}} , \\
O( N^{-\frac{1}{2}} )
	& \text{otherwise. } \\
	\end{array} \right.
	\]
\end{proposition}

\begin{proof}

	Note that if the polynomial $\mathrm{p}_{\widehat\sigma^{-1} \circ (\widehat{\alpha} \sqcup \pi^{\text{odd}})}$ is non-zero,
	then $\mathfrak{C}_{2}[\pi,{\alpha}] =  O(N^{-1/2})$.

	Indeed, if $\mathrm{p}_{\widehat\sigma^{-1} \circ (\widehat{\alpha} \sqcup \pi^{\text{odd}})}$ is a non-zero polynomial, so is  $\mathrm{p}_{\widehat\sigma^{-1}  \circ (\widehat{\alpha} \sqcup \pi_{1}^{\text{odd}} \sqcup \pi_{2}^{\text{odd}} ) }$ by Proposition \ref{minimal.special.polynomials},

	and thus, Corollary \ref{cor.bound.gauss.sum} implies there is a constant $C$ independent from $N$ such that

	\begin{align*}
	\abs*{ \sum_{\substack{
			\mathbf{i}:[\pm {4m} ] \rightarrow [N] \\
			\kernel{\mathbf{i}} = {\widehat{\sigma}}^{-1}  \circ (  \widehat{\alpha} \sqcup \beta^{\text{odd}}  ) } }
			\widehat{\mathbf{h}}(\mathbf{i}  ) }
	=
	\abs*{ \sum_{\substack{ \mathbf{i}:[\pm {4m} ] \rightarrow [N] \\  \kernel{\mathbf{i}}= \widehat{\alpha} \sqcup \beta^{\text{odd}} }} 		\widehat{\mathbf{h}}(\mathbf{i} \circ \bm{\widehat{\sigma}} ) }
	\leq

	C_{} N ^{ m + \#(\beta) - \frac{1}{2}}
	\end{align*}

	for $\beta = \pi$ and  $\beta =\pi_{1} \sqcup \pi_{2}$.

	But then, we get that $\mathfrak{C}_{2}[\pi,{\alpha}] =  O(N^{-1/2})$ since from (\ref{dist.entries.signature.matrix}),  (\ref{dist.entries.signed.perm.matrix}), and (\ref{eqn.unnormalized.fluct.HXH}) we have

	\begin{align*}
	\Big|\mathfrak{C}_{2}[\pi, {\alpha}]  \Big|
	\leq
	C \cdot	\frac{(N-\#(\pi))!}{N!} \cdot N ^{ \#(\pi_{})  - \frac{1}{2}}
	+
	C \cdot \frac{(N-\#(\pi_{1}))!}{N!} \cdot \frac{(N-\#(\pi_{2}))!}{N!}
	\cdot N ^{ \#(\pi_{1}) + \#(\pi_{2}) - \frac{1}{2}}.

	\end{align*}

	Assume  $\mathrm{p}_{\widehat\sigma^{-1} \circ (\widehat{\alpha} \sqcup \pi^{\text{odd}})}$ is the zero polynomial.

	Then, by Proposition \ref{minimal.special.polynomials}, there is a symmetric pairing partition $\eta \leq \widehat{\alpha} \sqcup \pi^{\text{odd}}$ such that $\mathrm{p}_{\widehat\sigma^{-1} \circ \eta}$ is also the zero polynomial, and hence, the partition $\eta$ must satisfy one of the conditions \textit{(1)-(6)} from Proposition \ref{prop.minimal.special.partitions}.

	However, we have $\eta=\eta_{\text{odd}} \sqcup \eta_{\text{even}}$, since $\eta \leq \widehat{\alpha} \sqcup \pi^{\text{odd}}$, and neither $2m_1$ or $2m_2$ is odd, so conditions \textit{(\ref{prop.minimal.special.partitions.2})} and \textit{(4)-(6)} can not hold.

	Now, note that if $\mathrm{p}_{\widehat\sigma^{-1} \circ (\widehat{\alpha} \sqcup \beta^{\text{odd}})}$ is zero polynomial for some partition $\beta \in P(\pm 2m)$, then

	\begin{align*}
	\sum_{\substack{ \mathbf{i}:[\pm {4m} ] \rightarrow [N] \\  \kernel{\mathbf{i}}= \widehat{\alpha} \sqcup \beta^{\text{odd}} }} 		\widehat{\mathbf{h}}(\mathbf{i} \circ \bm{\widehat{\sigma}} )
	=
	\frac{N!}{	(	N	-	\#(\widehat{\alpha} \sqcup \beta^{\text{odd}})  + 1) !}
	=
	\frac{N!}{	(	N	-	m  - \#( \beta)  + 1) !}
	\end{align*}

	since we would have $\widehat{\mathbf{h}}(\mathbf{i} \circ \bm{\widehat{\sigma}} ) =1$ for any function $ \mathbf{i}:[\pm {4m} ] \rightarrow [N]$ satisfying $\kernel{\mathbf{i}}= \widehat{\alpha} \sqcup \beta^{\text{odd}}$.

	Hence, if ${\alpha}_{1}$, ${\alpha}_{2}$, $\pi_{1}$, and $\pi_{2} $ are all even partitions and the polynomial $\mathrm{p}_{\sigma^{-1} \circ \hat{\alpha} \sqcup \pi^{\text{odd}}_{1} \sqcup \pi^{\text{odd}}_{2} }$ is zero, from (\ref{dist.entries.signature.matrix}),  (\ref{dist.entries.signed.perm.matrix}), and (\ref{eqn.unnormalized.fluct.HXH}), we obtain

	\begin{align}\label{eqn.vanishing.pi.eta}
	&
	\frac{(N-\#(\pi))!}{ N!} \cdot
	\sum_{\substack{
			\mathbf{i}:[\pm {4m} ] \rightarrow [N]  \\ \kernel{\mathbf{i}} = \hat{{\alpha}} \sqcup \pi^{\text{odd}}} }
	\widehat{\mathbf{h}}(\mathbf{i}\circ \bm{\widehat{\sigma}}) -
	\frac{(N-\#(\pi_{1}))!}{ N!} \cdot
	\frac{(N-\#(\pi_{2}))!}{ N!}	\cdot
	\sum_{\substack{
			\mathbf{i}:[\pm {4m} ] \rightarrow [N]  \\ \kernel{\mathbf{i}}= \hat{{\alpha}}\sqcup \pi^{\text{odd}}_1 \sqcup \pi^{\text{odd}}_2}}
	\widehat{\mathbf{h}}(\mathbf{i}\circ \bm{\widehat{\sigma}})  \nonumber \nonumber \\

	& = N^{m}\mathfrak{C}_{2}[\pi,{\alpha}] = O\left( N^{m-1} \right)
	\end{align}

	On the other hand, if either ${\alpha}_{1}$ or ${\alpha}_{2}$ is not an even partition, from (\ref{dist.entries.signature.matrix})  and (\ref{eqn.unnormalized.fluct.HXH}),

	we get

	\begin{align}\label{eqn.non-vanishing.pi.eta}
	\mathfrak{C}_{2}[\pi,{\alpha}]
	=	&
	\frac{1}{N^{ m } } \cdot \frac{(N-\#(\pi))!}{ N!} \cdot
	\sum_{\substack{
			\mathbf{i}:[\pm {4m} ] \rightarrow [N]  \\ \kernel{\mathbf{i}} = \hat{{\alpha}} \sqcup \pi^{\text{odd}}} }
	\widehat{\mathbf{h}}(\mathbf{i}\circ \bm{\widehat{\sigma}})
	=
	1 + O\left(N^{-1}\right) .
	\end{align}

	Suppose ${\eta}$ satisfies \textit{(\ref{prop.minimal.special.partitions.3})}  from  Proposition \ref{prop.minimal.special.partitions} and let $1\leq k \leq 4m_{1}$ and $4m_{1}+1 \leq l \leq 4m_{1}+4m_{2}$ such that

	\(
	\eta = 	\left\{
	\{\widehat\sigma^{t_{1}}(-k),\widehat\sigma^{-t_{1}}(k)\},
	\{\widehat\sigma^{t_{2}}(-l),\widehat\sigma^{-t_{2}}(l)\}
	\mid
	t_{2}, t_{2} \geq 0	\right\}.
	\)

	We need to consider three cases: $k$ and $l$ are both odd, $k+l$ is odd, and $k$ and $l$ are both even.

	First, if $k$ and $l$ are both odd, then $\mathrm{p}_{\sigma^{-1} \circ \hat{\alpha} \sqcup \pi^{\text{odd}}_{1} \sqcup \pi^{\text{odd}}_{2} }$ is also the zero polynomial, $\pi_{1}$ and $\pi_{2} $ are both even partitions, and ${\alpha} = {\alpha}_{1} \sqcup {\alpha}_{2}$, so  (\ref{eqn.vanishing.pi.eta}) holds.

	Second, if $k+l$ is odd, then ${\alpha} = {\alpha}_{1} \sqcup {\alpha}_{2}$ and $\pi = \pi_{1} \sqcup \pi_{2} $, but then (\ref{eqn.vanishing.pi.eta}) holds too.

	Third, if $k$ and $l$ are both even, then $\pi = \pi_{1} \sqcup \pi_{2} $ and either ${\alpha} = {\alpha}_{1} \sqcup {\alpha}_{2}$ or ${\alpha} \neq {\alpha}_{1} \sqcup {\alpha}_{2}$.

	However, if  ${\alpha} = {\alpha}_{1} \sqcup {\alpha}_{2}$, we already know that  $\mathfrak{C}_{2}[\pi,{\alpha}]= O(N^{-1})$ from (\ref{eqn.vanishing.pi.eta}), and if ${\alpha} \neq {\alpha}_{1} \sqcup {\alpha}_{2}$, then  ${\alpha}_{1}$ and ${\alpha}_{2}$ are not even partitions, so (\ref{eqn.non-vanishing.pi.eta}) holds.

	Finally, if ${\eta}$ satisfies  \textit{(\ref{prop.minimal.special.partitions.1})}  from  Proposition \ref{prop.minimal.special.partitions}, we must have ${\alpha} \neq {\alpha}_{1}\sqcup{\alpha}_{2}$, so we obtain  $\mathfrak{C}_{2}[\pi,{\alpha}] = 1 + O\left(N^{-1}\right) $.

\end{proof}

\begin{proof}[\textbf{Proof of (\ref{prop.fluc.xhx}) from Proposition \ref{prop.fluc}}]

	Fix an even partition $\pi \in P(\pm 2m)$ such that $\pi \leq \theta$ for some partition $\theta \in P_{\chi\chi}(\pm 2m)$ and

	let $\mathfrak{C}_{2}[\pi,\alpha] $ be  given by (\ref{eqn.unnormalized.fluct.HXH}) for each pairing partition $\alpha \in P_{2}(2m)$.

	By Proposition \ref{prop.fluc.xhx.C[p,a]}, we have that

	\[

	\sum_{\alpha\in P_{2}(2m)} \mathfrak{C}_{2} \left[ \pi, \alpha \right]
	=
	\abs*{E_{\pi}} + \abs*{F_{\pi}} - \abs*{E_{\pi} \cap F_{\pi}}
	+ O ( N^{-1/2}  )
	\]

	where $E_{\pi}$ and $F_{\pi}$ are the subsets of $P_{2}(2m)$ given by

	\[
	E_{\pi} = \left\{ \alpha\in P_{2}(2m) \left| \begin{array}{c}
	\eta \leq \widehat{\alpha} \sqcup \pi^{\text{odd}} \text{ for some symmetric pairing } \eta \in P(\pm 4m)   \\
	\text{satisfying (\ref{prop.minimal.special.partitions.1}) from Proposition \ref{prop.minimal.special.partitions} }
	\end{array} \right. \right\}
	\]

	and

	\[
	F_{\pi} = \left\{ \alpha\in P_{2}(2m) \left| \begin{array}{c}
	\alpha \neq \alpha_{1} \sqcup \alpha_{2} \text{ and }\eta \leq \widehat{\alpha} \sqcup \pi^{\text{odd}} \text{ for some symmetric pairing } \eta \in P(\pm 4m)   \\
	\text{satisfying (\ref{prop.minimal.special.partitions.3}) from Proposition \ref{prop.minimal.special.partitions} with } k \text { and } l \text{ even}
	\end{array} \right. \right\}.
	\]

	Thus, by Proposition \ref{prop.cpi-and-Cpi-nu}, we only need to show that  $E_{\pi} \neq \emptyset$ implies $\abs*{E_{\pi}} =1$, $F_{\pi} \neq \emptyset$ implies $\abs*{F_{\pi}} = 2$, and $E_{\pi} \cap F_{\pi}$ is empty.

	Let $\alpha$ and $\beta$ be pairing partitions in $P_{2}(2m)$ and suppose there are symmetric pairings $\eta_{\alpha}, \eta_{\beta} \in P(\pm {4m})$ satisfying (\ref{prop.minimal.special.partitions.1}) from Proposition \ref{prop.minimal.special.partitions}, $\eta_{\alpha} \leq \widehat{\alpha} \sqcup \pi^{\text{odd}}$,  and  $\eta_{\beta} \leq \widehat{\beta} \sqcup \pi^{\text{odd}}$.

	Then, there are integers $ 2m_{1} + 1  \leq l_{\alpha} , l_{\beta} \leq  2m_{1} + 2m_{2}$ so that

	\[
	\widehat{\sigma}^{-t}(2l_{\alpha}-1)
	\sim_{\eta_{\alpha}}
	\widehat{\sigma}^{t}(-1)
	\sim_{\eta_{\beta}}
	\widehat{\sigma}^{-t}(2l_{\beta}-1)

	\quad \forall  t\geq 0
	\]

	where $\widehat{\sigma}$ is the permutation given by

	\[
	\widehat{\sigma} = 	(-1,1,-2,2,\ldots,-4m_{1},4m_{1})
	(-4m_{1}-1,4m_{1}+1,-4m_{1}-2,\ldots,-4m_{1}-4m_{2},4m_{1}+4m_{2}).
	\]

	But then, we must have

	\begin{align}\label{rel.sigma.hat.alpha}
	\widehat{\sigma}^{-t}(2l_{\alpha}-1)
	\sim_{\widehat\alpha \sqcup \pi^{\text{odd}}}
	\widehat{\sigma}^{t}(-1)
	\sim_{\widehat\beta \sqcup \pi^{\text{odd}}}
	\widehat{\sigma}^{-t}(2l_{\beta}-1)

	\quad \forall  t\geq 0
	\end{align}

	since $\eta_{\alpha} \leq \widehat{\alpha} \sqcup \pi^{\text{odd}}$ and  $\eta_{\beta} \leq \widehat{\beta} \sqcup \pi^{\text{odd}}$,

	and thus, we get that

	\[
	{\sigma}^{-t}(l_{\alpha})
	\sim_{ \pi }
	{\sigma}^{t}(-1)
	\sim_{ \pi }
	{\sigma}^{-t}(l_{\beta})

	\quad \forall  t\geq 0
	\]

	where ${\sigma}$ is the permutation given by

	\[{\sigma} = 	(-1,1,-2,2,\ldots,-2m_{1},2m_{1})
	(-2m_{1}-1,2m_{1}+1,-2m_{1}-2,\ldots,-2m_{1}-2m_{2},2m_{1}+2m_{2}) . \]

	In particular, for $t=0$, we obtain $l_{\alpha} \sim_{ \pi } -1 \sim_{ \pi } l_{\beta}$, and thus, we have  $l_{\alpha}=l_{\beta}$ since $\pi$ is an even partition with only blocks of the form  $\{-k,+k,-l,l\}$ and $\{+k,-l\}$.

	Therefore, it follows from (\ref{rel.sigma.hat.alpha}) that  $\widehat{\alpha}=\widehat{\beta}$, or, equivalently, ${\alpha}={\beta}$.

	This shows that  $E_{\pi} \neq \emptyset$ implies $\abs*{E_{\pi}} =1$.

	Suppose now there are symmetric pairings $\eta_{\alpha}, \eta_{\beta} \in P(\pm {4m})$ satisfying (\ref{prop.minimal.special.partitions.3}) from Proposition \ref{prop.minimal.special.partitions} with $k$ and $l$ even, $\eta_{\alpha} \leq \widehat{\alpha} \sqcup \pi^{\text{odd}}$,  and  $\eta_{\beta} \leq \widehat{\beta} \sqcup \pi^{\text{odd}}$.

	Then, there exists integers $ 1  \leq k_{\alpha}  \leq  k_{\beta} \leq  2m_{1}$ so that

	\[
	\widehat{\sigma}^{t}(-2 k_{\alpha}) \sim_{\eta_{\alpha}} \widehat{\sigma}^{-t}(2 k_{\alpha})
	\quad \text{and} \quad
	\widehat{\sigma}^{t}(-2 k_{\beta}) \sim_{\eta_{\beta}} \widehat{\sigma}^{-t}(2 k_{\beta})
	\quad \forall  t\geq 0,
	\]

	and hence, we get

	\begin{align}\label{rel.sigma.hat.alpha.pi.odd}
	\widehat{\sigma}^{t}(-2 k_{\alpha}) \sim_{ \widehat{\alpha} \sqcup \pi^{\text{odd}}} \widehat{\sigma}^{-t}(2 k_{\alpha})
	\quad \text{and} \quad
	\widehat{\sigma}^{t}(-2 k_{\beta}) \sim_{\widehat{\beta} \sqcup \pi^{\text{odd}}} \widehat{\sigma}^{-t}(2 k_{\beta})
	\quad \forall  t\geq 0
	\end{align}

	since $\eta_{\alpha} \leq \widehat{\alpha} \sqcup \pi^{\text{odd}}$ and  $\eta_{\beta} \leq \widehat{\beta} \sqcup \pi^{\text{odd}}$;

	in particular, we must have

	\[
	\widehat{\sigma}^{4t}(2 k_{\alpha}) \sim_{ \widehat{\alpha} } \widehat{\sigma}^{-4t}(-2 k_{\alpha})
	\quad \text{and} \quad
	\widehat{\sigma}^{4t}(2 k_{\beta}) \sim_{\widehat{\beta} } \widehat{\sigma}^{-4t}(-2 k_{\beta})
	\quad \forall  t\geq 0.
	\]

	Now, since ${\alpha}$ and ${\beta}$ are pairing partitions of $[2m_{1}+2m_{2}]$,  $\widehat{\alpha} = \{\{+2k,-2k,+2l,-2l\} \mid \{k,l\} \in \alpha \} $, and  $\widehat{\beta} = \{\{+2k,-2k,+2l,-2l\} \mid \{k,l\} \in \beta \} $, we get

	\[
	{\sigma}^{2t}( k_{\alpha} )	\sim_{\alpha} 	{\sigma}^{-2t}( k_{\alpha} )

	\quad \text{and} \quad

	{\sigma}^{2t}( k_{\beta} )	\sim_{\beta} 	{\sigma}^{-2t}( k_{\beta} )

	\quad \forall  t\geq 0,
	\]

	where ${\sigma}$ is the permutation defined above; hence, we obtain

	\[
	\alpha_{1} 	= \{\{k_{\alpha}\},\{k_{\alpha}+1,k_{\alpha}-1\},\ldots,\{k_{\alpha}+m_{1}-1,k_{\alpha}-m_{1}+1\},\{k_{\alpha}+m_{1}\}\}

	\]

	and

	\[
	\beta_{1} = \{\{k_{\beta}\},\{k_{\beta}+1,k_{\beta}-1\},\ldots,\{k_{\beta}+m_{1}-1,k_{\beta}-m_{1}+1\},\{k_{\beta}+m_{1}\}\}
	\]

	where $\alpha_{1}$ and $\beta_{1}$ denote the restrictions of $\alpha$ and $\beta$, respectively, to the set $[\pm 2m_{1}]$.

	Let us show that $\alpha_{1}=\beta_{1}$.

	From (\ref{rel.sigma.hat.alpha.pi.odd}), we also have that

	\begin{align*}
	\widehat{\sigma}^{4t}( - 2 k_{\alpha}-1) \sim_{ \pi^{\text{odd}} } \widehat{\sigma}^{-4t}( 2 k_{\alpha}-1), & &
	\widehat{\sigma}^{4t}( 2 k_{\alpha}+1) \sim_{\pi^{\text{odd}} } \widehat{\sigma}^{-4t}( -2 k_{\alpha}+1), \\
	\widehat{\sigma}^{4t}( - 2 k_{\beta}-1) \sim_{ \pi^{\text{odd}} } \widehat{\sigma}^{-4t}( 2 k_{\beta}-1), &
	\quad \text{and}  &
	\widehat{\sigma}^{4t}( 2 k_{\beta}+1) \sim_{\pi^{\text{odd}} } \widehat{\sigma}^{-4t}( -2 k_{\beta}+1)
	\end{align*}

	for every integer $t \geq 0$, and thus, since $\pi^{\text{odd}} = \{\{ 2k-\sign{(k)} \mid k \in B \} \mid B \in \pi \}\}$, we obtain

	\[
	{\sigma}^{t+1}( k_{\alpha} )	=

	{\sigma}^{t}( - k_{\alpha} - 1  )
			\sim_{ \pi }
	{\sigma}^{-t}( k_{\alpha} )
	\quad \text{and} \quad
	{\sigma}^{t+1}( k_{\beta} )		=

	{\sigma}^{t}( -k_{\beta} - 1 )
			\sim_{ \pi }
	{\sigma}^{-t}( k_{\beta} )

	\quad \forall  t\geq 0.
	\]

	Let $t=k_{\beta}-k_{\alpha}$ and note that

	\[
	{\sigma}^{2t}(	k_{\alpha} )
	\sim_{\pi}
	{\sigma}^{-2t+1}(	k_{\alpha} )
	\sim_{\pi}
	{\sigma}^{2t+1}(	k_{\alpha} )
	\]
	since

	\[
	k_{\beta} =
	{\sigma}^{2t}(	k_{\alpha} )
	\sim_{\pi}
	{\sigma}^{-2t+1}(	k_{\alpha} )

	\quad	\text{and}	\quad

	{\sigma}^{2t+1}(k_{ \alpha } ) = \sigma^{}(k_{\beta}) \sim_{\pi}  k_{\beta};
	 \]

	moreover, since ${\sigma}^{2t}(k_{\alpha} ) > 0$, ${\sigma}^{-2t+1}(	k_{\alpha} ), {\sigma}^{2t+1}(	k_{\alpha} ) < 0$, and $\pi$ is a partition with only blocks of the form  $\{-k,+k,-l,l\}$ and $\{+k,-l\}$ with $k,l > 0$, we must have

	\[
	{\sigma}^{2t}(	k_{\alpha} ) 	=	-{\sigma}^{-2t+1}(	k_{\alpha} ) ,

	\quad
	{\sigma}^{2t}(	k_{\alpha} ) =	-{\sigma}^{2t+1}(	k_{\alpha} ),
	\quad	\text{or}	\quad
	{\sigma}^{-2t+1}(	k_{\alpha} )  = {\sigma}^{2t+1}(	k_{\alpha} ),
	\]

	or, equivalently,

	\begin{align*}
	{\sigma}^{4t-2}(	k_{\alpha} ) 	=		k_{\alpha} ,
	\quad
	{\sigma}^{-2}(	k_{\alpha} )		=		k_{\alpha} ,
	\quad	\text{or}	\quad
	{\sigma}^{-4t}(	k_{\alpha} )  		=		k_{\alpha}.
	\end{align*}

	But the equality ${\sigma}^{s}(k_{\alpha}) = k_{\alpha}$ holds if only if $s \equiv 0 \mod 4m_{1}$, so only ${\sigma}^{-4t}( k_{\alpha} ) = k_{\alpha}$ can hold, and thus, we get 	$k_{\alpha} = k_{\beta}$ or $k_{\beta} = k_{\alpha} + m_{1}$ since $0 \leq t = k_{\beta} - k_{\alpha} \leq 2m_{1}-1$.

	Therefore, $\alpha_{1} = \beta_{1}$ and there is an integer $1\leq k=k_{\alpha} \leq 2m_{1}$ so that

	\begin{align*}
	\alpha_{1}
	& =
	\{ \{k\}, \{k+1,k-1\}, \ldots, \{k+m_{1}-1,k-m_{1}+1\}, \{k+m_{1}\} \}
	=
	\beta_{1} \\
	& =
	\{
	\{ \sigma^{2t}(k) ,\sigma^{-2t}(k)    \}
	\mid 0 \leq t \leq m_{1}  \} .
	\end{align*}

	Similarly, letting $\alpha_{2}$ and $\beta_{2}$ denote the restrictions of $\alpha$ and $\beta$, respectively, to the set $[\pm (2m_{1}+2m_{2})] \setminus [\pm 2m_{1}]$, we have $\alpha_{2} = \beta_{2}$ and there is an integer $2m_{1}+1 \leq \l_{} \leq 2m_{1}+2m_{2}$ so that

	\begin{align*}
	\alpha_{2}
			&=
	\{ \{l\}, \{l+1,l-1\}, \ldots, \{l+m_{2}-1,l-m_{2}+1\}, \{l+m_{2}\} \}
			=
	\beta_{2} \\
			&=
	\{	\{ \sigma^{2t}(l) ,\sigma^{-2t}(l)    \}
			\mid 0 \leq t \leq m_{2}  \}.
	\end{align*}

	This shows that $\abs*{F_{\pi}} = 2$ provided $F_{\pi} \neq \emptyset$ since  $\gamma \in F_{\pi}$ implies

	\[
	\gamma  =
	\{\{k, l\} , \{k+m_{1},l+m_{2}\}\} 	\cup \widetilde{\alpha}

	\quad \text{ or } \quad

	\gamma  =
	\{\{k, l+m_{2}\}, \{k+m_{1},l\}\} \cup \widetilde{\alpha}
	\]

	where  $\widetilde{\alpha} =\{\{ \sigma^{2t}(k) ,\sigma^{-2t}(k)    \} \mid 1 \leq t \leq m_{1}-1  \}
	\cup \{\{ \sigma^{2t}(l) ,\sigma^{-2t}(l)    \} \mid 1 \leq t \leq m_{2}-1  \} $.

	Finally, $E_{\pi} \cap F_{\pi}$ is empty since $E_{\pi} \neq \emptyset$ implies $\pi \neq \pi_{1} \sqcup \pi_{2}$, and, on the other hand, $ F_{\pi} \neq \emptyset $ implies $\pi=\pi_{1} \sqcup \pi_{2}$.
\end{proof}

config: arxiv


TITLE: How to extend Galois character?
QUESTION [8 upvotes]: Let $D_p$ be a decomposition group of $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ above $p$ (for all $p$) and $I_p$ the inertia group. Let $\chi_p$ be a character of $D_p$, such that $\chi_p$ is trivial for almost all $p$.
Question : Why does there exist a character $\chi$ of $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ such that $\chi_{|I_p}={\chi_p}_{|I_p}$ for all $p$ ?
This result is used in : Serre, 'Modular forms of weight one and Galois representations'. I suspect that it is a consequence of class field theory, but I don't know so much about it.

REPLY [5 votes]: I guess you meant to write that $\chi_p|_{I_p}$ is trivial for almost all $p$.
Notice that this is not the same as taking characters of $I_p$, since the natural morphism $I_p^{\mathrm{ab}} \rightarrow D_p^{\mathrm{ab}}$ is not injective (the Frobenius "twists" elements of $I_p$).
Local class field theory gives us an isomorphism between $\widehat{\mathbb{Q}_p^{\times}}$ and $D_p^{\mathrm{ab}}$, and the image of $I_p$ in $D_p^{\mathrm{ab}}$ corresponds to $\mathbb{Z}_p^{\times}$, so by taking restrictions to $I_p$ of characters $\chi_p$ of $D_p$ we just give ourselves characters $\eta_p$ of $\mathbb{Z}_p^{\times}$.
Global class field theory gives you an isomorphism between $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})^{\mathrm{ab}}$ and $\mathbb{A}^{\times}/\mathbb{Q}^{\times} \mathbb{R}_{>0} \simeq \prod_p \mathbb{Z}_p^{\times}$ (this last isomorphism is particular to the case where the number field is $\mathbb{Q}$, and it is possible to prove the result you want without appealing to it, but it is very convenient here).
We also have the obvious commutative diagram linking the local and global reciprocity maps, where $D_p^{\mathrm{ab}} \rightarrow \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})^{\mathrm{ab}}$ is parallel to $\widehat{\mathbb{Q}_p^{\times}} \rightarrow \mathbb{A}^{\times}/\mathbb{Q}^{\times} \mathbb{R}_{>0}$ (this last arrow is the one obtained from $x_p \mapsto (\ldots, 1, x_p , 1 , \ldots)$), and conviniently, the composition $\mathbb{Z}_p^{\times} \hookrightarrow \widehat{\mathbb{Q}_p^{\times}} \rightarrow \mathbb{A}^{\times}/\mathbb{Q}^{\times} \mathbb{R}_{>0} \simeq \prod_q \mathbb{Z}_q^{\times}$ is the obvious thing.
We now define $\eta$ on $\prod_p \mathbb{Z}_p^{\times}$ by $\eta ( (x_p)_p) = \prod_p \eta_p(x_p)$ (well-defined and continuous thanks to the hypothesis).