TITLE: Trying to understand open (closed) subfunctors
QUESTION [13 upvotes]: I am trying to read about functor of points and I am struggling with the definition of open subfunctor.
The definition is the following. A subfunctor $\alpha\colon G\to F$, where $F,G\in \mathsf{Fct}(\mathsf{Rings},\mathsf{Sets})$ is called open (closed) if for any $\psi\colon h_{\mathrm{Spec}(R)}\to F$ the fibered product $G_\psi=G\times_F h_{\mathrm{Spec}(R)}$ yields a map $G_\psi\to h_{\mathrm{Spec}(R)}$ isomorphic to injection from the functor represented by some open (closed) subscheme of $\mathrm{Spec}(R)$.
Can you, please, explain why is it true that open subfunctors of $F=h_{\mathrm{Spec}(S)}$ are exactly the functors $G$ of the form $G(T)=\{\phi\in F(T)\mid \phi^*(I)T=T\}$ for some ideal $I\subset S$? This is the exercise $VI.6$ in Eisenbud-Harris "Geometry of Schemes". Can you, please, explain how to think about these subfunctors?
Thank you very much!

REPLY [5 votes]: $G$ is an open subfunctor of $F = h_{\operatorname{Spec}(S)}$ means that for any $T,$ $G(T) = \operatorname{Hom}(\operatorname{Spec}(T), U)$ where $U\subseteq\operatorname{Spec}(S)$ is an open subscheme. Thus $\varphi\in G(T)$ implies that $\phi = \iota\circ\varphi\in F(T)$ where $\iota:U\hookrightarrow \operatorname{Spec}(S)$ is the inclusion, and there is an obvious bijective correpondence between such $T$-valued points of $\operatorname{Spec}(S)$ and $T$-valued point of $U.$
Thus, we simply need to realize that the condition "$\phi\in F(T)$ factors through $U$" is equivalent with the condition "$\phi^*(I)T = T,$ where $\phi^*:S\to T$ is the dual map, and $I$ is the ideal of the complement of $U.$"
For example, in the case $U = \operatorname{Spec}(S_f)$ for some $f\in S,$ we are asking that $\phi^*$ factors through $S_f$ which is equivalent to $\phi^*(f)\in T^\times$ is invertible. The complement of $U$ is $V(f),$ so what this boils down to is that $\phi^*(I)T = T,$ where $I=(f).$ I leave the general case to you.


TITLE: Let f and g continuous, then if x contained in open ball with center a, and f(x)=g(x), then f(a)=g(a)
QUESTION [1 upvotes]: Let ($X, d$), ($Y, d'$) be two metric spaces, $ a \in X$, and $f, g : X \rightarrow Y$ two functions.
Suppose $f$ and $g$ continuous. Prove that if any open ball centered on $a$ contains a point $x$ such that $f(x) = g(x)$, then $ f(a) = g(a)$.
I am not sure how to approach this problem, but my intuition tells me that either $x$ is arbitrary close to $a$ (meaning that $a=x$), or $f = g$ at least where $x\in B(a,r)$, such that $f(B(a,r)) = g(B(a,r))$.
I have tried defining another ball $B$ with center on $x$ and same radius $r$, such that $a\in B(x,r)$. Because $f$ and $g$ are continuous, then $\exists \delta_1,\delta_2$ such that $f(B)\subset B(f(x),\delta_1)$ and $g(B) \subset B(g(x),\delta_2)$. Then $B(f(x),\delta_1)$ and $B(g(x),\delta_2)$ are concentric, meaning that they have the same center, $f(x)=g(x)$.
Now take the smallest of $\delta_1$or $\delta_2:\delta = min\{\delta_1,\delta_2\}$. If we make $\delta$ arbitrary small, then in order for $f(a)$ to be kept inside any of the balls, then we have to make $a$ arbitrary close to $x$. Therefore $a=x$ and $f(a)=f(x)=g(x)=g(a)$.
I am not sure of this proof, so I would assistance in solving this problem.
Thanks in advance!

REPLY [4 votes]: For each natural $n$, let $x_n\in B\left(a,\frac1n\right)$ such that $f(x_n)=g(x_n)$. Then $d(x_n,a)<\frac1n$. So, $\lim_{n\to\infty}x_n=a$ and, since $f$ and $g$ are continuous,$$f(a)=\lim_{n\to\infty}f(x_n)=\lim_{n\to\infty}g(x_n)=g(a).$$


TITLE: Evaluating $\lim_{x\to\infty}(\sqrt{x^2+1} - x)(x+1) $
QUESTION [4 upvotes]: The question is $$\lim_{x\to\infty}(\sqrt{x^2+1} - x)(x+1) $$ I know the answer is $\frac{1}{2}$ and I found it using this equality :
$$(\sqrt{x^2+1} - x)(x+1) = \frac{x+1}{\sqrt{x^2+1} + x}$$
But is there any other way to solve this? Any hints would be appreciated.

REPLY [4 votes]: Just for fun, try letting $x=\tan\theta=\sin\theta/\cos\theta$ with $\theta\to\pi/2^-$, and use
$$\sqrt{\tan^2\theta+1}=\sqrt{\sec^2\theta}=\sec\theta={1\over\cos\theta}$$
so that
$$\begin{align}
(\sqrt{x^2+1}-x)(x+1)
&=\left({1\over\cos\theta}-{\sin\theta\over\cos\theta}\right)\left({\sin\theta\over\cos\theta}+1\right)\\
&={(1-\sin\theta)(\sin\theta+\cos\theta)\over\cos^2\theta}\\
&={(1-\sin\theta)(\sin\theta+\cos\theta)\over1-\sin^2\theta}\\
&={\sin\theta+\cos\theta\over1+\sin\theta}
\end{align}$$
We get
$$\lim_{x\to\infty}(\sqrt{x^2+1}-x)(x+1)=\lim_{\theta\to\pi/2^-}{\sin\theta+\cos\theta\over1+\sin\theta}={1+0\over1+1}={1\over2}$$


TITLE: Is $\int_{1}^{\infty}\frac{dx}{\sin^{5}x}$ finite or infinite?
QUESTION [3 upvotes]: Is the definite integral given below finite or infinite? Why?
$$\int_{1}^{\infty}\frac{dx}{\sin^{5}x}$$

REPLY [0 votes]: The improper integral does not exist. Let
$$I_B=\int_1^B \frac{1}{\sin^5 x}\,dx$$
The improper integral exists if $\lim_{B\to\infty} I_B$ exists.
But $I_B$ itself fails to exist for arbitrarily large $B$, indeed already for $B=\pi$. We show that $\lim_{B\to \pi^-}I_B$ does not exist. Imagine calculating $I_{B-\epsilon}$ for small positive $\epsilon$. Make the change of variable $t=\pi-x$. Then
$$I_{B-\epsilon}=\int_{t=\epsilon}^{\pi-1} \frac{1}{\sin^5 t}\,dt.$$
Since $\sin t\le t$ in our interval,
$$\int_{t=\epsilon}^{\pi-1} \frac{1}{\sin t^5}\,dt\ge  \int_{t=\epsilon}^{\pi-1} \frac{1}{t^5}\,dt.$$
But $\int_{t=\epsilon}^{\pi-1} \frac{1}{t^5}\,dt$ can be calculated explicitly, and blows up as $\epsilon$ approaches $0$ from the right.
One reason is enough, but there are other reasons that the integral does not exist. For example, let $f(x)=\frac{1}{\sqrt{\sin x}}$ when $\sin x$ is positive, and let $f(x)=- \frac{1}{\sqrt{|\sin x|}}$ when $\sin x$ is negative. Then there is no existence issue about $J_B=\int_1^B f(x)\,dx$. However, $J_B$ has period $2\pi$, so does not converge.


TITLE: Understanding renormalizability and bare mass
QUESTION [6 upvotes]: I'm studying QFT on my own. I'm using the book Quantum Field Theory in a Nutshell by Zee.
I need some help with understanding and clarification with the following.
I'm going to quote a paragraph from Zee's book Quantum Field Theory in a Nutshell p.174. which I need help in understanding

Consider the following double integral $$
 I\equiv(-i\lambda)^2\int^\Lambda\int^\Lambda\frac{d^4p}{(2\pi)^4}\frac{d^4q}{(2\pi)^4}\frac{i}{p^2-m^2+i\epsilon}\frac{i}{q^2-m^2+i\epsilon}\frac{i}{(p+q+k)^2-m^2+i\epsilon}
 $$ Counting powers of $p$ and $q$ we see that the integral $\sim
 \int(d^8P/P^6)$ and so $I$ depends quadratically on the cutoff
  $\Lambda$. By Lorentz invariance $I$ is a function of $k^2$, which
  we can expand in a series $$ D+Ek^2+Fk^4+\ldots \tag{$\mathbf{1}$}
 $$ The quantity $D$ is just $I$ with external momentum $k$ set equal
  to zero so depends quadratically on the cutoff $\Lambda$. Next we get
  $E$ from differentiating $I$ with respect to $k$ twice and then set
  $k$ to zero. This clearly decreases the power of $p$ and $q$ in the
  integrand by 2 so it depends logarithmically on the cutoff $\Lambda$.
  Similarly we get $F$ from differentiating with respect to $k$ four
  times and setting $k$ to zero. This decreases the power of $p$ and $q$
  in the integrand by 4 and this F is given by an integral that goes as
  $\sim \int d^8P/P^{10}$ for large $P$. The integral is convergent and
  hence cutoff independent. Thus $F$ and all $(\ldots)$ terms are cutoff
  independent as the cutoff goes to infinity and we don't ahve to worry
  about them.
Putting it altogether, we have the inverse propagator $k^2-m^2+a+b+k^2$  The propagator is changed to $$
 \frac{1}{k^2-m^2}\rightarrow\frac{1}{(1+b)k^2-(m^2-a)}
 \tag{$\mathbf{2}$} $$ The pole $k^2$ is shifted to $m_\textbf{p}\equiv
 m+\delta m\equiv (m^2-a)(1+b)^{-1}$, which we identify as the physical
  mass.

Question
I have marked the parts that I have a hard time understanding with bold text. Those are some parts where I could need some clarification.

I'm onboard with that $I$ is a function of $k^2$ but why can we do a series expansion like that in (1)?
When he says putting it altogether, what is it that we are actually putting together and how do we obtain that expression?

REPLY [2 votes]: As I said in the comment, as long as you don't hit singularities for $k^2=0$, the function $I(k^2)$ is completely regular and you can perform the expansion. As well, you can take the one-dimensional integral (it may be a simplification of a more general case you can find in computing n-point Green functions)
$$
I(k) = \int_0^{+\infty} d q \frac{q^4}{(q+k)^2}
$$
where $k$ is the external momenta. This has degree of divergence $D=2$ but you can derive three-times with respect to the external momenta $k$ and you get I'''(k) which is completely convergent
$$
I'''(k) =\int_0^{+\infty} d q -6\frac{6q^2}{(k+q)^4} = -\frac{2}{k}
$$
Then, you can integrate back (integrate wrt to external momenta!!)
$$
I''(k) = -2\log(k) + A
$$
where A (and in the following all the upper case letters) is a divergent constant. Then,
$$
I'(k) = +2k -2k \log(k) + A k + B\rightarrow I(k) = Bk + \frac{1}{2}\left( 3+A\right)k^2 - k^2\log(k)
$$
Regarding your second question, you can express the n-point Green $G^{(n)}(p)$ function in terms of the amputated one $G^{(n)}_{amp}$
$$
G^{(n)}(p_1,...,p_n) = \Pi_{i=1}^n \left[G^{(2)}(p_i)\right] G^{(n)}_{amp}(p_1,...,p_n)
$$
The S-matrix is nothing else the amputated Green function in which you add the wave-function polarization and then put everything on shell. In the case of the scalar theory the wave-function polarization is trivial (i.e. it is 1). For $n=2$ you look at the full propagator $G^{(2)}(p)$
$$
G^{(2)}(p) =G^{(2)}(p)G^{(2)}(p) G^{(2)}_{amp}(p)
$$
and you see that
$$
 G^{(2)}(p) = \frac{1}{G^{(2)}_{amp}(p)} \,,\qquad \qquad (1)
$$
(EDITED)
In perturbation theory, you can set $G^{(2)}_{amp}(p) = k^2 - m^2 + a +b+k^2$. I guess the term $a +b+k^2$ include some power of the perturbative coupling constant. You can avoid higher order corrections because they are already taken into account in Eq. (1).
Notice that $G^{(2)}_{amp}(p)$ has that value because you are using the Feynman rules of the kinetic term seen as a vertex and the result of the loop-integral. If you want, the loop-integral provide you a contribution into the effective action proportional (roughly) to $(a+b)\phi^2 - (\partial\phi)^2$. Then, the term $a+b+k^2$ is the feynman rule associated to this vertex and it enters in $G^{(2)}_{amp}(p)$.
Notice I am not doing the sum like in the other answer. This is because I should do the same sum but with $1/(k^2-m^2)\rightarrow G^{(2)}(p)$ where $ G^{(2)}(p)$ is the full quantum propagator (the one which include quantum corrections). If you work in perturbation theory, that sum is equivalent to what I did here.


TITLE: Two different methods but not same answer
QUESTION [0 upvotes]: I wanted to work this problem 2 different ways, but am getting 2 different answers.  Can you assist in helping me understand why method 2 doesn't work?
Note:  There are 16 letters and 2 E's and 3 B's.
Method 1:  Prob of getting an E on the first draw + Prob getting a B on the second draw - minus the prob of getting EC (overlap).
$2/16 + 3/15 - (2/16 \cdot3/15) =3/10$ correct answer!
Method 2:  Prob of getting and E on the first draw and NOT a B on the second, PLUS Prob of getting NOT E on the first draw and a B on the second, PLUS Prob of getting an EB.  Add together all the possibilities.
Prob  of E on first  and NOT B on second = $2/16 \cdot13/15=26/240$
Prob of NOT E on first and B on second =   $14/16 \cdot3/15 = 42/240$
Prob of E on first and B on Second = $2/16 \cdot3/15 = 6/240 $
This Method 2 does not produce the $3/10$ answer.  I don't understand why not.

REPLY [1 votes]: If the first letter is not an E, it may be a B, in which case in the second draw the probability of drawing a B is reduced to $2/15$.  So, one needs to distinguish three disjoint events that amount to success:

The first letter drawn is an E.
The two letters are both Bs.
The first letter is neither E nor B, and the second letter is a B.

Adding the probabilities, we get:
$$ \frac{2}{16} + \frac{3}{16} \cdot \frac{2}{15} + \frac{11}{16}\cdot \frac{3}{15} = \frac{23}{80} \enspace. $$


TITLE: Arrows-only implication & disjunction in $\mathbf{Set}.$
QUESTION [4 upvotes]: Just before the truth-arrows in a topos subsection of Goldblatt's "Topoi: A Categorial Analysis of logic," descriptions of the truth functions $\Rightarrow$ and $\smallsmile$ are given in $\mathbf{Set}$ in terms of arrows. This is to motivate their abstraction to any given topos.

I'm having trouble understanding these descriptions thoroughly, so I would like some clarification, please :)

Definition 1: The disjunction truth-function $\smallsmile :2\times 2\to 2$ (for $2=\{0, 1\}$) is given by $1\smallsmile 1=0\smallsmile1=1\smallsmile0=1$ and $0\smallsmile0=0$, whereas the implication truth-function $\Rightarrow: 2\times 2\to 2$ is given by $1\Rightarrow0=0$ and $(0\Rightarrow1)=(0\Rightarrow0)=(1\Rightarrow1)=1$.
Definition 2: $\smallfrown$ can be taken as the characteristic function of the product map $\langle\operatorname{true}, \operatorname{true}\rangle$.

Implication.
Goldblatt states that $\Rightarrow$ is the characteristic function of $=\{\langle 0, 0\rangle,\langle 0, 1\rangle, \langle 1, 1\rangle\}$ with

a pullback square, which he then uses to write  as the equaliser of $\smallfrown:2\times 2\to 2$ and $pr_1$, where $pr_1(\langle x,y\rangle)=x$. I don't see how he does this.

Disjunction:
We take $\smallsmile$ as $\chi_D$ for $D=A\cup B$ for $A=\{\langle1, 1\rangle, \langle1, 0\rangle\}$ and $B=\{\langle1, 1\rangle,\langle0, 1\rangle\}$; identify $A$ with the product map $\langle\operatorname{true}_2, 1_2\rangle$, $B$ with $\langle1_2, \operatorname{true}_2\rangle$; form the coproduct map $f=[\langle\operatorname{true}_2, 1_2\rangle, \langle1_2, \operatorname{true}_2\rangle]$; identify $\operatorname{im} f=D$; epic-monic factor $f$ through $D$; then claim that all this specifies $D$ - and hence $\small\smile$ - uniquely up to isomorphism.

So yeah, I can't see the wood for the trees basically. Please help :)

REPLY [4 votes]: I was trying too hard.
Implication.
When Goldblatt used the lattice equivalence $$x\sqsubseteq y \iff x=x\sqcap y$$ to write  as $$\{\langle x, y\rangle\vert x\smallfrown y=x\},$$ he did so for a reason: it's so that  is the set for which $\smallfrown$ and $pr_1$ agree, which is then, by $\S 3.10$, the equaliser of the two.
Disjunction.
If you draw the commutative diagrams for the products $t_1=\langle\operatorname{true}_2=\operatorname{true}\circ\vert_2, 1_2\rangle$ and $t_2=\langle1_2,\operatorname{true}_2\rangle$ and write down all the equations (like $pr_2\circ t_1=1_2$), you'll see that $t_1$ and $t_2$ are nothing more than what Goldblatt said they were: $t_1$ sends $0$ to $\langle1, 0\rangle$ and $1$ to $\langle1, 1\rangle$; $t_2$, similarly; and we can identify them with $A$ and $B$, respectively, because $A$ and $B$ are two-element sets whose set-inclusions with respect to $2\times 2$ are $t_1$ and $t_2$, respectively.
The coproduct $2+2$ is just the disjoint union of $2$ with itself, so that we can write $$2+2=2^{(0)}\cup2^{(1)},$$ where $2^{(0)}=2\times\{0\}$ and $2^{(1)}=2\times\{1\}$, to see that $f$ is given by $$f(a, i)=\begin{cases}t_1(a) &: i=0 \\ t_{2}(a) &: i=1.\end{cases}$$
The rest is downhill from there (once you see that $\operatorname{im}f=\operatorname{im}t_1\cup\operatorname{im}t_2$).


TITLE: Potential energy $= mgh$, what is $h$?
QUESTION [3 upvotes]: NOTE: when I say potential energy I mean gravitational PE
The formula for potential energy is P.E = mgh.
What is h referring to? Height, obviously.
Consider the example: What is the potential energy of a 1kg mass lifted 2 metres off the ground?
m=1, g=9.8, h=2 => P.E=17.6J
My problem is this: why is h the height off the ground, this seems rather arbitrary. Would it not make more sense to have h as the height from the centre of gravity?
Suppose we repeat the experiment on top of a mountain, does the mass still have the same potential energy? I am pretty sure it doesn't.
I am foreseeing one or both of the following answers, so which one will it be?

It doesnt make sense to talk about potential energy in absolute terms, only in terms of gain in potential energy
Potential energy is defined in a way where h is the height off the ground (I dont buy this)

I am leaning towards the first one, but I am still generally uncomfortable with the idea that you can't have an absolute P.E

REPLY [1 votes]: "Why is h the height off the ground, this seems rather arbitrary?"
It is not arbitrary, it is useful and convenient because we live on the ground. But your intuition is still right.
P.E. = mgh, but this formula is only an approximation. It assumes that g is constant, which it is not, it depends on altitude. If you are at the height of the moon, g is very different. But if we are sticking with heights that are relatively close to the ground, this formula works well. If you want to be exact, you have to do the integral where you are using the correct value of g for each height.


TITLE: Deciding $\displaystyle o,\omega,\Theta$ notations
QUESTION [1 upvotes]: I have a question which I couldn't solve for about two hours. It goes like this:

Let $\displaystyle f(n)=\left(\frac{n+3\ln(n)}{n}\right)^n \ ; \ g(n)=27^{\ln(n)}$.
Fill the blank box with the correct notation ($o,\omega,\Theta):$ $f(n)=\Box{g(n)}$.

I tried to approximate LHS to get something that is similar to RHS, but it failed.
I also tried to evaluate $\displaystyle \lim_{n\to\infty}\frac{f(n)}{g(n)}$, but it also failed.
Unfortunately, I am not familiar with other methods, so I would appreciate if you could use these.
Thanks in advance for any kind of help!

REPLY [1 votes]: As $n$ is great, you have
$$
\ln f(n) =n\ln\left(1+\frac{3\ln(n)}{n}\right)=n\left( \frac{3\ln n}{n}+\mathcal{O}\left( \frac{\ln^2 n}{n^2}\right)\right)=\ln (n^3)+\mathcal{O}\left( \frac{\ln^2n}{n}\right)
$$ giving
$$
f(n)=n^3\times e^{\mathcal{O}\left( \frac{\ln^2n}{n}\right)}.
$$
On the other hand, you have
$$
g(n)=27^{\ln n}=e^{(\ln n) \ln (27)}=e^{\ln (n^{\ln27})}=n^{\ln 27}
$$
then $$
\frac{f(n)}{g(n)}=n^{3-\ln 27}e^{\mathcal{O}\left( \frac{\ln^2n}{n}\right)}.
$$ Observe that $3-\ln 27=3-3.2958\ldots<0$, since $\displaystyle e^{\mathcal{O}\left( \frac{\ln^2n}{n}\right)} \rightarrow 1$,
we deduce that
$$
\frac{f(n)}{g(n)}\sim n^{3-\ln 27} \rightarrow 0.
$$ We thus have
$$
f(n)=o(g(n))
$$


TITLE: Universal Property of Natural Transformations Proof Verification/ Proof tips
QUESTION [1 upvotes]: Let $\phi$ be a natural transformation between functors $F, F':\mathscr{C} \to \mathscr{D}$, $\tau :Arr\mathscr{D} \to \mathscr{D}$ be the identity natural transformation between the objects of the category of arrows of $\mathscr{D}$ to the morphisms of $\mathscr{D}$ by $\tau(\alpha)=\alpha$, and $\Phi :\mathscr{C} \to Arr\mathscr{D}$ be a functor which sends $c\in \mathscr{C}$ to $\phi(c):Fc \to F'c$
I am asked to show that $\phi=\tau \Phi$, and that for any other functor $\Psi:\mathscr{C} \to Arr\mathscr{D}$ which satisfies $\phi=\tau \Psi$, we must have $\Psi=\Phi$.
I am only asking for the verification of my attempt at the second portion of the proof because the first part is obvious.
Proof: Suppose BWOC there exists a $\Psi \neq \Phi$ such that $\Psi:\mathscr{C} \to Arr\mathscr{D}$ and $\phi=\tau \Psi$. Then for any $c\in \mathscr{C}$ we must have $\tau \Psi(c)=\phi(c)=\tau \Phi(c)$. Since $\tau$ is the identity natural transformation on objects of $Arr\mathscr{D}$, we must have that $\Psi=\Phi$.
Is there anything else that needs to be shown?
I still feel slightly uncomfortable with these kinds of proofs and any advice on attacking them would be much appreciated.
Thanks.

REPLY [1 votes]: The statement basically says that $\tau:\text{dom}\Rightarrow\text{cod}:\text{Arr}\mathscr D\to\mathscr D$ is a universal natural transformation between functors to $\mathscr D$, that is, given any natural transformation $\phi:F\Rightarrow F':\mathscr C\to\mathscr D$, there is a unique functor $\Phi:\mathscr C\to\text{Arr}\mathscr D$ such that $\phi=\tau\Phi$
If $\Phi$ is such a functor, then it must send an object $c$ to an object of $\text{Arr}\mathscr C$ at which $\tau$'s component is $\phi_c$, hence $\Phi(c)=\phi_c$. On an arrow $f:c\to c'$ we want $\Phi(f)$ to be a pair of arrows $Fc\to Fc'$ and $F'c\to F'c'$ whose image under $\text{dom}$ resp. $\text{cod}$ is $Ff$ resp. $F'f$, so we can say that $\Phi$, if it exists, must send an arrow $f:c\to c'$ to the commutative square
$$\begin{matrix}
Fc & \xrightarrow{Ff} & Fc' \\
\hspace{10pt} \downarrow\phi_c & & \hspace{10pt}\downarrow\phi_{c'} \\
F'c & \xrightarrow{F'f} & F'c'
\end{matrix}$$
But this obviously defines a functor such that $\tau\Phi=\phi$.
It is important to keep in mind that a natural transformation $\phi$ is not uniquely determined by its components. In fact, two transformations $\sigma$ and $\phi$ between functors from $\mathscr C$ to $\mathscr D$ may have exactly the same components while still being different because they go between different functors (with the same object functions, though). This is reminiscent to a function not being determined solely by a rule $x\mapsto f(x)$, you also need to specify what the domain and codomain are. So instead of just $f:x\mapsto x^2$ you should write $f:\Bbb R\to\Bbb R:x\mapsto x^2$, for example. This becomes evident when we see a natural transformation not just as an $\text{ob}\mathscr C$-indexed family of arrows, but as a morphism in the category of functors from $\mathscr C$ to $\mathscr D$, so this transformation comes together with source and target functors.
So altogether, what you forgot to check was that $\Phi$ and $\Psi$ are equal also on the arrows.


text:
\begin{document}

\begin{abstract}
We compute the variances of sums in arithmetic progressions of generalised $k$-divisor functions related to certain $L$-functions in $\mathbb{F}_q[t]$, in the limit as $q\rightarrow\infty$. This is achieved by making use of recently established equidistribution results for the associated Frobenius conjugacy classes.  The variances are thus expressed, when $q\rightarrow\infty$, in terms of matrix integrals, which may be evaluated.  Our results extend those obtained previously in the special case corresponding to the usual  $k$-divisor function, when the $L$-function in question has degree one.  They illustrate the role played by the degree of the $L$-functions; in particular, we find qualitatively new behaviour when the degree exceeds one.  Our calculations apply, for example, to elliptic curves defined over $\mathbb{F}_q[t]$, and we illustrate them by examining in some detail the generalised $k$-divisor functions associated with the Legendre curve.
\end{abstract}

\title[Variance of sums of divisor functions]
{Variance of sums in arithmetic progressions of divisor functions associated with higher degree $L$-functions in $\mathbb{F}_q[t]$}

\author{Chris Hall
\address{Department of Mathematics, Western University, London, ON, Canada, N6A 5B7}}

\author{Jonathan P.~Keating
\address{School of Mathematics, University of Bristol, Bristol BS8 1TW, UK}}

\author{Edva Roditty-Gershon
\address{School of Mathematics, University of Bristol, Bristol BS8 1TW, UK}}

\thanks{ We are pleased to acknowledge support under EPSRC Programme Grant EP/K034383/1
LMF: \textit{L}-Functions and Modular Forms.  JPK is also grateful for support from a Royal Society Wolfson Research Merit Award and ERC Advanced Grant 740900 (LogCorRM).}

\allowdisplaybreaks
\maketitle

\section{Introduction}\label{sec:introduction}

For a fixed integer $k\geq 2$ the generalized divisor function $\divid_k(n)$ denotes the number of ways of writing a (positive) integer as a product of $k$ factors:
$$
\divid_k(n)=\sum_{m_{1}\cdot m_{2}\cdots m_{k}=n}1
$$

It is related to the $k$th power of the Riemann zeta-function, defined by
\begin{equation}\label{zeta}
	\zeta(s)
	=
	\sum_{n=1}^\infty \frac{1}{n^s}
\end{equation}
when ${\rm Re}s>1$, in that
\begin{equation}\label{zetak}
	\zeta(s)^k
	=
	\sum_{n=1}^\infty \frac{\divid_k(n)}{n^s}
\end{equation}
for ${\rm Re}s>1$.

Define $\Delta_k(x)$ by
\begin{equation}
\Delta_k(x):=\sum_{n\leq x}\divid_k(n)-\Res_{s=1}\frac{x^{s}\zeta^{k}(s)}{s}=\sum_{n\leq x}\divid_k(n)-xP_{k-1}(\log x)
\end{equation}
where $P_{k-1}(u)$ is a certain polynomial of degree $k-1$; see, for example, \cite{Titch} Chapter XII.

The mean square of $\Delta_2(x)$ was computed by Cr\'amer \cite{Cramer} for $k=2$, and by Tong \cite{Tong} for $k\geq 3$ (assuming the Riemann Hypothesis (RH) if $k\geq 4$), to be
\begin{equation}\label{eq:Tong}
	\frac 1X\int_X^{2X} \Delta_k(x)^2 dx
	\sim
	c_k X^{1-\frac 1k},

\end{equation}
for a certain constant $c_k$. Heath-Brown \cite{HB1992} showed that $\Delta_k(x)/x^{\frac 12-\frac 1{2k}}$ has a limiting value distribution (for $k\geq 4$ one needs to assume RH); it is non-Gaussian.

Our main focus will be on  sums of  divisor functions over arithmetic progressions
\begin{equation}
	\ES_{\divid_k}(A)
	=
	\ES_{\divid_k;X;Q}(A)
	=
	\sum_{\substack{n\leq X\\n=A\bmod Q}}
	\divid_k(n)
\end{equation}
For the standard divisor function ($k=2$), it is known that if $Q<X^{2/3-\epsilon}$ then
\begin{equation}
	\ES_{\divid_2}(A)
	=
	\frac{X p_Q(\log X)}{\Phi(Q)} + O(X^{1/3+o(1)})
\end{equation}
for some linear polynomial $p_Q$. This is due to unpublished work of Selberg. For recent work on the asymptotics of sums of $\divid_3$ over arithmetic progressions, see \cite{FKM} and the literature cited therein.

The variance $\Var(\ES_{\divid_2;X;Q})$  of $\ES_{\divid_2}$

has been studied by Motohashi \cite{Motohashi},

Blomer \cite{Blomer}, Lau and Zhao \cite{Lau Zhao}, the result being \cite{Lau Zhao} (we assume $Q$ prime for simplicity):

\begin{enum}
\item If $1\leq Q<X^{1/2+\epsilon}$ then
\begin{equation}
	\Var(\ES_{\divid_2;X;Q}) \ll X^{1/2} + (\frac XQ)^{2/3+\epsilon}.
\end{equation}

\item For $X^{1/2}<Q<X$,
\begin{equation}\label{eq:Lau Zhao}
	\Var(\ES_{\divid_2;X;Q})
	=
	\frac XQ p_3(\log \frac{Q^2}X) + O((\frac XQ)^{5/6}(\log X)^3 )
\end{equation}
where $p_3$ is a polynomial of degree $3$ with positive leading coefficient.
\end{enum}

\noindent
For $k\geq 3$, Kowalski and Ricotta \cite{KowRic} considered smooth analogues of the divisor sums $\ES_{\divid_k;X;Q}(A)$, and among other things computed the variance for $Q^{k-1/2+\epsilon}<X<Q^{k-\epsilon}$.

In the function field setting, the analogous problem was studied by Rodgers, Rudnick and the second and third authors \cite{KRRR}.  Let $\MM_n\sub\Fq[t]$ be the set of monic polynomials of degree $n$ with coefficients in $\Fq$.  For a polynomial $Q \in F_{q}[x]$ of degree at least $2$, and $A$ co-prime to $Q$, set
\begin{equation}
	\ES_{\divid_k,n,Q}(A)
	:=
	\sum_{\substack{f\in \mathcal M_n\\f=A\bmod Q}} \divid_k(f) \;,
\end{equation}
Then for $n\leq k(\deg Q-1)$ the variance of this sum is given by
\begin{equation}
	\lim_{q\to \infty} \frac{\Var_Q(\ES_{\divid_k,n,Q})}{  q^n/|Q|}
	=
	I_k(n;\deg Q-1)

\end{equation}
where $I_k(n;\deg Q-1)$ is a certain matrix integral over the unitary group $U_{\deg Q -1}(\bbC)$ --- see \cite{KRRR} and (\ref{matrix integral def}) below.  This integral can be evaluated in a number of ways.  Some of the main results obtained in \cite{KRRR} are set out in Section 5 below.  It follows from these, for example that, for the classical divisor function $\divid=\divid_2$, if $\deg Q\geq 2$ and $n\leq 2(\deg Q-1)$, then
\begin{equation}
	\lim_{q\to \infty} \frac{\Var_Q(\ES_{\divid_2,n,Q})}{  q^n/|Q|} =
	\begin{cases}
		\mathrm{Pol}_3(n), & n\leq \deg Q-1 \\
		\mathrm{Pol}_3(2(\deg Q-1)-n),& \deg Q\leq n\leq 2(\deg Q-1)
	\end{cases}
\end{equation}
where
$$
	\mathrm{Pol}_3(x) = \binom{x+3}{3} = (x+1)(x+2)(x+3)/6.
$$
For general $k$, the variance may be expressed in terms of a lattice-point count and is a piecewise polynomial function of degree $k^2-1$.

The function-field expressions for the variance of the generalized divisor function lead immediately to conjectures in the standard number-field setting \cite{KRRR}.  The conjecture when $k=3$ has recently been proved by Rodgers and Soundararajan \cite{RodgersSound}.

Our aim here is to extend the line of research reviewed above to arithmetic functions associated with other $L$-functions in the same way that the generalized divisor function is related to the Riemann zeta-function. Our function field results (which we will soon state below; Theorem \ref{thm:Legendre-application} and Theorem \ref{thm:main-theorem}) can be used to motivate predictions in the number field setting.
In order to illustrate these predictions, we focus now on two representative examples:
elliptic curve $L$-functions and the Ramanujan $L$-function.

Let $E/\bbQ$ be an elliptic curve of conductor $N$ defined over $\bbQ$. The associated $L$-function $F(s)$ will be denoted by $L(s,E)$ and is given by
$$
L(s,E)=\prod_{p|N}(1-a_{p}p^{-s})^{-1}\prod_{p\nmid N}(1-a_{p}p^{-s}+p^{-2s+1})^{-1}
$$
where $a_{p}$ is the difference between $p+1$ and the number of points on the reduced curve mod $p$
$$
a_{p}=p+1-\#\tilde{E}(\bbF_{p}).
$$
When $p|N$, $a_{p}$ is either $1,$ $-1$, or $0$. In general, we have the Hasse bound on $a_{p},$ $|a_{p}| < 2\sqrt{p}$, hence the product converges and gives an analytic function for all $Re(s)>3/2.$
The $L$ function of $E$ expands as
$$
L(s,E)=\sum_{n=1}^{\infty}a_{n}n^{-s}
$$
where
$$
a_{p^{e}}=p^{e}+1-\#\tilde{E}(\bbF_{p^{e}}).
$$
and for $n=\prod_{i=1}^{r}p_{i}^{e_{i}}$ with $p_{1},\ldots,p_{r}$ distinct primes
$$
a_{n}=a_{p_{1}^{e_{1}}}\cdots a_{p_{r}^{e_{r}}}.
$$
For each positive integer $k$ consider the multiplicative function $d_{k,E}:\bbZ\rightarrow \bbZ$ given by
$$
d_{k,E}:=\prod_{n_{1}\cdots n_{k}=n}a_{n_{1}}\cdots a_{n_{k}}.
$$
Note that $d_{k,E}$ gives the coefficients in the Dirichlet series expansion
$$
L(s,E)^{k}=\sum_{n=1}^{\infty}d_{k,E}(n)n^{-s}.
$$

Our results in the function field setting are analogous to computing the variance of the sum of $d_{k,E}$ in arithmetic progressions
Define the sum over arithmetic progressions
$$
S_{x,c,E}(A):=\sum_{\substack{n\leq x \\ n=A \mod c}}d_{k,E}(n).
$$
Our function field result (see Theorem~\ref{thm:main-theorem} ) leads us to predict that for $x^{\epsilon}<c $, $\epsilon >0$, the following holds:
$$
\mathrm{Var}(S_{x,c,E}) \sim \frac{x}{\phi(c)}a_{k}(L(s,E))\gamma_k(\frac{\log x}{2\log c})(2\log c)^{k^{2}-1}.
$$
with
$$
	\gamma_k(c)
	=
	\frac{1}{k!\, G(1+k)^2}
	\int_{[0,1]^k}
		\delta_c(w_1 + \ldots + w_k)
		\prod_{i< j}(w_i-w_j)^2\,
		d^k w,
$$
with $\delta_c(x) =\delta(x-c)$ being the delta distribution translated by $c$, and $G$ is the Barnes $G$-function, so that for positive integers $k$, $G(1+k) = 1!\cdot 2! \cdot 3! \cdots (k-1)!$. Here $a_{k}(L(s,E))$ is an arithmetic factor that can be written explicitly by (2.5.9) in \cite{CFKRS}.

Note that we can detect the degree of the $L$-function in question as the coefficient of $\log c$.

Another example of a degree-two $L$-function is the Ramanujan $L$-function:
$$
L(s,\tau )=\prod _{p}\left(1-\tau (p)p^{-s}+p^{-2s+11}\right)^{-1} ,
$$
where $\tau$ is the Ramanujan tau function $\tau :\mathbb {N} \to \mathbb {Z} $ defined by the following identity:
$$
\sum _{n\geq 1}\tau (n)q^{n}=q\prod _{n\geq 1}(1-q^{n})^{24},
$$
where $q=\exp(2\pi iz)$.
Ramanujan conjectured (and his conjecture was proved by Deligne) that
$|\tau(p)| \leq 2p^{11/2}$ for all primes $p$,
hence the product converges and gives an analytic function for all $Re(s)>13/2.$
The $L$ function associated with $\tau$ expands as
$$
L(s,\tau )=\sum_{n=1}^{\infty}\tau(n)n^{-s}
$$
For each positive integer $k$ consider the multiplicative function $d_{k,\tau}:\bbZ\rightarrow \bbZ$ given by
$$
d_{k,\tau}:=\prod_{n_{1}\cdots n_{k}=n}\tau(n_{1})\cdots\tau(n_{k}).
$$
Note that $d_{k,E}$ gives the coefficients in the Dirichlet series expansion
$$
L(s,\tau )^{k}=\sum_{n=1}^{\infty}d_{k,\tau}(n)n^{-s}.
$$
Again we are led to speculate that for $x^{\epsilon}<c $, $\epsilon>0$,  if
$$S_{x,c,\tau}(A):=\sum_{\substack{n\leq x \\ n=A \mod c}}d_{k,\tau}(n) $$
then the following holds:
$$
\mathrm{Var}(S_{x,c,\tau}) \sim \frac{x}{\phi(c)}a_{k}(L(s,\tau))\gamma_k(\frac{\log x}{2\log c})(2\log c)^{k^{2}-1}..
$$
Where $a_{k}(L(s,\tau))$ is an arithmetic factor that can be written explicitly by (2.5.9) in \cite{CFKRS}.

We set out our main results below, but first, by way of illustration, we examine a specific example where the $L$-function in question take a relatively simple and explicit form, and can defined in a self-contained way.

\subsection{Divisors associated to Legendre curve L-function}

Let $q$ be a power of a prime $p$.  Let $\MM\sub\Fq[t]$ be the subset of monic polynomials and $\II\sub\MM$ be the subset of irreducibles.  For each $n\geq 1$, let $\MM_n\sub\MM$ and $\II_n\sub\II$ be the respective subsets of elements of degree $n$.

Suppose $q$ is odd, and let $E/\Fq(t)$ be the Legendre curve, that is, the elliptic curve with affine model
$$
	y^2 = x(x-1)(x-t).
$$

\ssni
Its $L$-function is given by an Euler product
$$
	L(T,E/\Fq(t))
	=
	\prod_{\pi\in\II}
	L(T^{\deg(\pi)},E/\Fpi)^{-1}
$$
where $\Fpi$ is the residue field $\Fq[t]/\pi\,\Fq[t]$.

Each Euler factor of $L(T,E/\Fq(t))$ is the reciprocal of a polynomial in $\mathbb{Q}[T]$ and satisfies
$$
	T\frac{d}{dT}
	\log L(T,E/\mathbb{F}_\pi)^{-1}
	=
	\sum_{m=1}^\infty a_{\pi^m}T^m
	\in\mathbb{Z}[[T]].
$$
We regard the coefficients $a_{\pi^m}$ as values of the multiplicative function $f\mapsto a_f$ on $\MM$ given by the coefficients of the Dirichlet series expansion
$$
	L(T,E/\Fq(t))
	=
	\sum_{f\in\MM}
	a_f\,T^{\deg(f)}.
$$
Thus $a_1=1$, and if $f=\prod_{i=1}^r\pi_i^{e_i}$ with $\pi_1,\ldots,\pi_r$ distinct elements of $\II$, then
$$
	a_f = a_{\pi^{e_1}}\cdots a_{\pi^{e_r}}.
$$

For each positive integer $k$, consider the multiplicative function $\dkLeg\colon\MM\to\bbZ$ given by
$$
	\dkLeg(f)
	=
	\prod_{\substack{f_1,\ldots,f_k\in\MM\\[0.01in]f_1\cdots f_k=f}}
	a_{f_1}\cdots a_{f_k}.
$$
Equivalently, $\dkLeg$ gives the coefficients in the Dirichlet series expansion
$$
	L(T,E/\Fq(t))^k
	=
	\sum_{f\in\MM}\dkLeg(f)\,T^{\deg(f)}.
$$
It is easy to see that this is a generalization of the $k$th divisor function: replace $L(T,E/\Fq(t))$ by the zeta function $Z(T,\Aone/\Fq)$ so that $f\mapsto a_f$ becomes the constant function $f\mapsto 1$.

Let $\Q\in\MM$ be square free and $\BQ=\BQFq$.  For each $n\geq 1$ and $A\in\BQ$, consider the sum
$$
	S_{k,n,\Q}(A)
	\ :=
	\sum_{\substack{f\in\mathcal{M}_n\\f\equiv A\bmod\Q}}
	\dkLeg(f).
$$
Let $A$ vary uniformly over $\Gamma(\Q)$, and consider the expected value
$$
	\bbE_A[S_{k,n,\Q}(A)]
	:=
	\frac{1}{|\BQ|}
	\sum_{A\in\BQ}
	S_{k,n,\Q}(A)
$$
and the variance
$$
	\mathrm{Var}_A[S_{k,n,\Q}(A)]
	:=
	\frac{1}{|\BQ|}
	\sum_{A\in\BQ}
	|S_{k,n,\Q}(A)-\bbE_A[S_{k,n,\Q}(A)]|^2.
$$
These moments depend on $q$, so one can ask how they behave when we replace $\Fq$ by a finite extension, that is, let $q\to\infty$.  Our main result gives the variance  $\mathrm{Var}_A[S_{k,n,\Q}(A)]$, in the limit $q\to \infty$, in terms of a matrix integral.  Let $U$ be an $R\times R$ matrix.  Let $\std\colon U_R(\bbC)\to\GL_R(\bbC)$ be the representation given by the inclusion $U_R(\bbC)\sub\GL_R(\bbC)$ and
$$
	\wedge^j \std\colon U_R(\bbC)\to \GL_{R_j}(\wedge^j\bbC)
$$
be its $j$th exterior power where $R_j=\binom{R}{j}$; we define $(\wedge^j\,\std)(g)=0$ unless $0\leq j\leq R$.  Define the matrix integrals with respect to Haar measure $d\theta$ on $U_R(\bbC)$ the group of $R\times R$ unitary matrices
\begin{equation}\label{matrix integral def}
	I_{k}(n;R)
	=
	\int_{U_R(\bbC)}
	\Big|

		\sum_{n_1+\cdots +n_k=n}
		\,\,
		\Tr\Big(\big(\otimes_{i=1}^k (\wedge^{n_i}\,\std)\big)(\theta)\Big)
	\Big|^2
	d\theta .
\end{equation}
Then, as a special case of a general theorem we present in this paper, one can prove the following theorem:

\begin{theorem}\label{thm:Legendre-application}
If $\gcd(\Q,t(t-1))=t$ and if $\deg(\Q)\gg 1$, then
$$
	\lim_{q\to\infty}
	\frac{|\BQ|}{q^{2n}}
	\Var_A[S_{k,n,\Q}(A)]
	=
	I_{k}(n;2\deg (c)-1)
$$

\end{theorem}

See Remark~\ref{rmk:proof-of-legendre-theorem} for a brief explanation of how this follows from our general theorem.  The factor 2 multiplying $\deg(c)$ corresponds to the degree of the Legendre curve $L$-function.

\begin{cor}
In the case of $k=2$ we have for $n\leq 2\deg(c)-1$
$$
	\lim_{q\to\infty}
	\frac{|\BQ|}{q^{2n}}
	\Var_A[S_{2,n,\Q}(A)]
	=
	{n+3\choose 3}
$$
and for $2\deg(c)-1 < n < 4\deg(c)-2$
$$
	\lim_{q\to\infty}
	\frac{|\BQ|}{q^{2n}}
	\Var_A[S_{2,n,\Q}(A)]
	=
	{4 \deg c -n+1\choose 3}.
$$
\end{cor}

\ssni
The Corollary will follow from evaluating the matrix integral in \S\ref{integral}.

\section{Statement of Main Theorem}

Let $\Kbar$ be an algebraic closure of $K=\Fq(t)$ and $\GK$ be the Galois group $\Gal(\Kbar/K)$.  We regard $\PP=\II\cup\{\infty\}$ as the set of places of $K$, and for each $v\in\PP$, we write $\Fv$ for the residue field and $\deg(v)$ for its degree over $\Fq$.  We also fix inertia and decomposition groups $I(v)\sub D(v)\sub\GK$ respectively and write $\Gv$ for the quotient group $D(v)/I(v)$ and $\Frob_v\in\Gv$ for the Geometric Frobenius element.

For each finite subset $\SS\sub\PP$, let $\KS\seq\Kbar$ be the maximal subfield which is unramified over $\PP\ssm\SS$ and $\GKS$ be the Galois group $\Gal(\KS/K)$.  We abuse notation and write $I(v)\seq D(v)$ for the images of $I(v)\sub D(v)\sub\GK$ in $\GKS$ via the canonical quotient $\GK\onto\GKS$.  If $v\in\SS$, then $I(v)\seq\GKS$ is isomorphic to $I(v)\sub\GK$, and otherwise $I(v)\seq\GKS$ is the trivial group.

Let $\ell$ be a prime distinct from $p$ and $V$ be a finite-dimensional $\Qellbar$-vector space.  Consider a Galois representation
$$
	\rho\colon\GKS\to\GL(V),
$$

\ssni
that is, a continuous group homomorphism.  For each $v\in\PP$, let $V_v=V^{\rho(I(v))}$. We say that $s$ is the conductor of $\rho$ if $s$ is a square-free monic polynomial divisible by prime polynomials $v$ for which $V_v$ is strictly smaller than $V.$
Let
$$
	\rholv\colon\Gv\to\GL(V_v)
$$
be the composition of the restriction of $\rho$ to $D(v)$ and the quotient $D(v)\onto\Gv$, and let
$$
	L(T,\rho_v) := \det\left(1 - T\,\rho(\Frob_v)\mid V_v\right).
$$

We attach several $L$-functions to $\rho$.  One is the complete $L$-function and is given by the Euler product
\begin{equation}\label{eqn:euler-product-for-L}
	L(T,\rho) := \prod_{v\in\PP}L(T^{\deg(v)},\rho_v)^{-1}.
\end{equation}
Another is a partial $L$-function given by the Euler product
$$
	\Lf(T,\rho) := \prod_{\pi\in\II}L(T^{\deg(\pi)},\rho_\pi)^{-1}
$$
where now the Euler product is taken over the `finite' places of $K$.  If $L(T,\rho_\infty)$ is trivial, that is, if $V_\infty=\ZeroSpace$, then these $L$-functions coincide, but in general
$$
	\Lf(T,\rho) = L(T,\rho)\cdot L(T,\rho_\infty).
$$

One reason for considering the partial $L$-function $\Lf(T,\rho)$ is that we can expand its Euler product as a Dirichlet series
$$
	\Lf(T,\rho) = \sum_{f\in\MM}a_f T^{\deg(f)}
$$

\ssni
where $f\mapsto a_f$ is the multiplicative function $\MM\to\Qellbar$ given on prime powers by writing
$$
	T\frac{d}{dT}\log(L(T,\rho_\pi)^{-1})
	=
	\sum_{m=1}^\infty a_{\pi^m}T^m.
$$
Equivalently, for each $\pi\in\II$ and $m\geq 1$, we have
$$
	a_{\pi^m} = \Tr(\rho_v(\Frob_\pi)^m\mid V_v);
$$
and if $f=\prod_{i=1}^r\pi_i^{e_i}$ is a prime factorization in $\Fq[t]$ with $\pi_i\neq\pi_j$ for $i\neq j$, then
$$
	a_f = a_{\pi_1^{e_1}}\cdots a_{\pi_r^{e_r}}
$$
where $a_1=1$.

For each positive integer $k$, we define the \defi{$k$th divisor function of $\rho$} as follows: it is the multiplicative function $\dk\colon\MM\to\Qellbar$ given by
$$
	\dk(f)
	=
	\prod_{\substack{f_1,\ldots,f_k\in\MM\\[0.01in]f_1\cdots f_k=f}}
	a_{f_1}\cdots a_{f_k}.
$$
When $\rho$ is the trivial representation $\one$ (and thus $\dim(V)=1$), then this is the usual $k$ divisor function on $\MM$:
$$
	d_{k,\one}(f)
	=
	\left|\left\{\,
		f_1,\ldots,f_k\in\MM
		:
		f_1\cdots f_k=f
	\,\right\}\right|
$$

\ssni
In general, we have the identity
$$
	\Lf(T,\rho)^k
	=
	\sum_{f\in\MM} \dk(f)\,T^{\deg(f)}
	=
	\sum_{n=0}^\infty\left(\sum_{f\in\MM_n}\dk(f)\right)T^n.
$$

Let $\CC\sub\PP$ be a finite subset containing $\infty$ and
\begin{equation}\label{eqn:euler-product-of-L_C}
	\LC(T,\rho) := \prod_{\pi\not\in\CC}L(T^{\deg(\pi)},\rho_\pi)^{-1}
\end{equation}
where $\pi$ runs over $\II$ and let
\begin{equation}
R=\deg(\LC(T,\rho)).
\end{equation}
Thus $\Lf(T,\rho)=L_{\{\infty\}}(T,\rho)$, and in general,
\begin{equation}\label{eqn:L_CC-to-d_k}
	\LC(T,\rho)^{k}
	=
	\sum_{n=0}^\infty\Big(
		\sum_{\substack{f\in\MM_n\\[0.02in]\gcd(f,\Q)=1}}\dk(f)
	\Big)T^n
\end{equation}
where $\Q\in\MM$ is the product of all primes $\pi\in\II\cap\CC$.

Let $\BQ$ be the finite abelian group $\BQFq$, and for each $A\in\BQ$, let
$$
	\SnAQ := \sum_{f\in\MM_n(A)}\dk(f)
$$
where
$$
	\MM_n(A) := \{\,f\in\MM_n : f\equiv A\bmod\Q\,\}.
$$
We regard $A$ as a random variable uniformly distributed over $\BQ$ and define its expected value
$$
	\bbE_A[\SnAQ]
	:=
	\frac{1}{|\BQ|}\sum_{A\in\BQ}\SnAQ
$$
and variance
$$
	\Var_A[\SnAQ]
	:=
	\frac{1}{|\BQ|}\sum_{A\in\BQ}\left|\SnAQ-\bbE_A[\SnAQ]\right|^2
$$
accordingly.

It follows easily the definition and \eqref{eqn:L_CC-to-d_k} that
$$
	\LC(T,\rho)
	=
	|\BQ|\cdot
	\sum_{n=0}^\infty
	\bbE_A[\SnAQ]\,T^n,
$$
so each expected value is the coefficient of an $L$-series.  The main goal of this paper is to analyze the asymptotic behavior of $|\Var_A[\SnAQ]|^2$ as $q\to\infty$, that is, as we replace $q$ by a power $q^r$ and take $r\to\infty$.  To do so, we must impose some hypotheses on $\rho$, e.g., we suppose that $\rho$ is punctually pure of weight $w$ (see section $6$ in \cite{HKRG}).  We also impose hypotheses on the Mellin transform of $\rho$, but before doing so we need some additional notation and terminology.

Let $\PhiQ$ be the finite abelian group $\Hom(\BQ,\Qellbar^\times)$.  For each $\dc\in\PhiQ$, there is a corresponding Dirichlet character
$$
	\dc\colon\GKC\to\GL(\Qellbar)
$$

\ssni
which we regard as a representation of $\GKR$ for $\RR=\CC\cup\SS$ by composing $\dc$ with the quotient $\GKR\onto\GKC$.  We also regard $\rho$ as a representation of $\GKR$ via the quotient $\GKR\onto\GKS$, and we define the tensor-product representation
$$
	\rhochi\colon\GKR\to\GL(V_\varphi)
$$
where $V_\varphi=V$ and $f\mapsto\rho(f)\varphi(f)$.

Let $L(T,\rhochi)$ and $\LC(T,\rhochi)$ be the $L$-functions respectively defined by the Euler products in \eqref{eqn:euler-product-for-L} and \eqref{eqn:euler-product-of-L_C} with $\rhochi$ in lieu of $\dc$.  A priori each, of these is a power series with coefficients in $\Qellbar$, but Grothendieck showed both are rational functions in $\Qellbar(T)$, compare (1.4.7) of \cite{Deligne:WeilII}.  We say that $\dc\in\PhiQ$ is \defi{good for $\rho$} iff it lies in the set
\begin{equation}\label{good}
\PhiQGood\rho
	:=
	\{\,
		\dc\in\PhiQ
		:
		L(T,\rhochi) = \LC(T,\rhochi)\in\Qellbar[T]
	\,\},
\end{equation}
and otherwise we say that $\rho$ is \defi{bad for $\rho$}.

The hypothesis that $\rho$ is punctually pure implies that both $L$-functions lie in $\Qbar(T)$, and $\dc$ is good for $\rho$ iff every zero $\alpha\in\Qbar$ of $\LC(T,\rhochi)$ satisfies $|\iota(\alpha)|^2=1/q^{1+w}$ for every field embedding $\iota\colon\Qbar\to\bbC$.  Equivalently, $\dc$ is good for $\rho$ iff the `unitarized' $L$-function
$$
	\LCStar(T,\rhochi) := \LC(T/(\sqrt{q})^{1+w},\rhochi)
$$

\ssni
is the characteristic polynomial of a unitary conjugacy class $\trhochi\sub U_R(\bbC)$ for $R=\deg(\LC(T,\rho))$.

We further distinguish bad characters by saying that $\rho\in\PhiQ$ is \defi{mixed for $\rho$} iff it lies in the set
\begin{equation}\label{mixed}
\PhiQMixed\rho
	:=
	\{\,
		\dc\in\PhiQ\ssm\PhiQGood\rho
		:
		\LC(T,\rhochi)\in\Qellbar[T]
	\,\},
\end{equation}
and otherwise we say that the elements of
\begin{equation}\label{heavy}
	\PhiQHeavy\rho
	:=
	\PhiQ\ssm(\PhiQGood\rho\cup\PhiQMixed\rho)
\end{equation}
are \defi{heavy for $\rho$}.  The mixed characters are those for which $\LCStar(T,\rhochi)$ is not the characteristic polynomial of a unitary matrix, and the heavy characters are those for which $\LCStar(T,\rhochi)$ is not even a polynomial.

We are now in a position to state our main theorem:

\begin{theorem}\label{thm:main-theorem}
Suppose that $\rho$ is punctually pure of weight $w$, that its Mellin transform has big monodromy (see section \ref{Big Monodromy}), and that $\PhiQHeavy\rho\seq\{\one\}$ for all $q$. Then
$$
	\lim_{q\to\infty}\frac{|\PhiQ|}{q^{n(1+w)}}\cdot \Var_A[S_{k,n,\Q}(A)]
	=
	I_{k}(n;R)
$$
for each $n\geq 1$.
\end{theorem}

\begin{remark}\label{rmk:proof-of-legendre-theorem}
If $\rho$ is the representation associated to the $\ell$-adic Tate module of the Legendre curve, then the hypotheses on $\Q$ in Theorem~\ref{thm:Legendre-application} imply that the Mellin transform of $\rho$ has big monodromy (see \cite[\S8]{HKRG}).  One can also show that $R=2\deg(\Q)-1$ (cf.~\loccit).  Thus Theorem~\ref{thm:Legendre-application} follows from Theorem~\ref{thm:main-theorem}.
\end{remark}

\begin{cor}
In the case of $k=2$ we have for $n\leq R$
$$
	\lim_{q\to\infty}
	\frac{|\BQ|}{q^{2n}}
	\Var_A[S_{2,n,\Q}(A)]
	=
	{n+3\choose 3}
$$
and for $R<n<2R$
$$
	\lim_{q\to\infty}
	\frac{|\BQ|}{q^{2n}}
	\Var_A[S_{2,n,\Q}(A)]
	=
	{2R-n+3\choose 3}
$$
\end{cor}

\section{Big Monodromy and Equidistribution}\label{Big Monodromy}

In \cite[\S10]{HKRG}, we defined a subgroup $\PhiUNu\seq\PhiQ$, and for each coset $\dc\PhiUNu$, we defined a monodromy group
$$
	\Ggeom{\rho}{\dc\PhiUNu}
	\seq
	\GL_R(\Qellbar)
$$
generated by Frobenii corresponding to the good characters in $\dc\PhiUNu$, where $R:=\deg(\LC(T,\rho))=\deg(\LC(T,\rhochi))$ (see \cite[Prop.~4.3.1]{HKRG}). More precisely, the Frobenius attached to a character $\dc\alpha^\nu\in\dc\PhiUNu$ is an element of $\GL_R(\Qellbar)$ with characteristic polynomial $\LC(T,\rho\otimes\chi\alpha^\nu)$, and $\Ggeom{\rho}{\dc\PhiUNu}$ is the Zariski closure of all such elements when one takes $q\to\infty$.

We say that $\dc\in\PhiQ$ is \defi{big for $\rho$} iff it lies in the set
$$
	\PhiQBig\rho
	:=
	\{\,
		\dc\in\PhiQ
		:
		\Ggeom{\rho}{\dc\PhiUNu}
		=
		\GL_R(\Qellbar)
	\,\},
$$
and we say that \defi{the Mellin transform of $\rho$ has big monodromy} iff
$$
	|\PhiQ|
	\sim
	|\PhiQBig\rho|
	\text{ as }
	q\to\infty.
$$

\begin{lemma}\label{size}
If the Mellin transform of $\rho$ has big monodromy, then
$$
	|\PhiQ|
	\sim
	|\PhiQGood\rho|
	\sim
	|\PhiQBig\rho|
	\text{ as }
	q\to\infty
$$
\end{lemma}

\begin{proof}
See \cite[Corollary .~10.4.3]{HKRG}.
\end{proof}
If the Mellin transform of $\rho$ has big monodromy it implies that $\trhochi$ become equidistributed in $U_R(\bbC)$ (for reference see Theorem 10.0.4 combined with remark 8.2.4 in \cite{HKRG}):
\begin{theorem}\label{equidistribution}
Suppose that $\rho$ is punctually pure of weight $w$, that its Mellin transform has big monodromy, and that $\PhiQHeavy\rho\seq\{\one\}$ for all $q$. Then for any continuous function $F:U_R(\bbC)\rightarrow \bbC$
\begin{equation}
	\lim_{q\to\infty}
	\frac{1}{|\PhiQGood\rho|}
	\sum_{\dc\in\PhiQGood\rho}
	F(\trhochi)
	\ =
	\int_{U_R(\bbC)}
	F(\theta)
	d\theta
\end{equation}
with respect to Haar measure $d\theta$ on $U_R(\bbC)$.
\end{theorem}
In the following Theorem we present a sufficient criteria for the Mellin transform of $\rho$ to have big monodromy (explicit example for representations meeting this criteria can be found in \cite[\S 12]{HKRG}).
\begin{theorem}
Let $s$ be the conductor of $\rho$ and suppose that $\gcd(\s,c)=t$ and that $\deg c\geq 3$.  Suppose moreover that $V(0)$ has a unique unipotent block of exact multiplicity one and that $\rho$ is geometrically simple and pointwise pure.  If $r:=\dim(V)$ and $\deg c$ satisfy
\begin{equation}
\deg c >\frac{1}{r}(72(r^2+1)^2 - r - \deg(L(T,\rho)) +\sum_{\nu \in C}\deg \nu(r-\deg L(T,\rho_{\nu
}))
\end{equation}
then the Mellin transform of $\rho$ has big monodromy.
\end{theorem}

\section{Proof of Theorem \ref{thm:main-theorem}}

We express the variance of the arithmetic progressions sums $S_{k,n,c}(A)$ in terms of sums of divisor functions twisted by Dirichlet characters.  Let $k$ be a positive integer and let
\begin{equation}\label{twisted-d-sum}
	\ckn{\rhochi}
	=
	\sum_{\substack{f\in\MM_n\\[0.02in]\gcd(f,\Q)=1}} d_{k,\rhochi}(f)
\end{equation}
be the coefficients in the expansion of the k-th power of the partial L-function \eqref{eqn:L_CC-to-d_k}
$$
	\LC(T,\rhochi)^k
	=
	\sum_{n=0}^\infty
	\ckn{\rhochi}
	\,T^n
$$
For each $\dc\in\PhiQ$, we extend $\dc$ to a multiplicative map $\chiz\colon\MM\to\Qellbar$ by defining:
$$
	\chiz(f)
	=
	\begin{cases}
		\dc(f+\Q\,\Fq[t]) & \mbox{if }
		\gcd(f,\Q)=1 \\
		0 & \mbox{otherwise}.
	\end{cases}
$$
It is multiplicative and satisfies
\begin{equation*}\label{eq:chis.vs.chiz}
	\chiz(\pi)
	=
	\begin{cases}
		\dc(\Frob_{v(\pi)}) &  \mbox{if }\pi\nmid\Q \\
		0 & \mbox{otherwise}
	\end{cases}
	\mbox{ for }\pi\in\II.
\end{equation*}

The Orthogonality relations for Dirichlet characters are:
\begin{enum}
\item for each $A_1,A_2\in\BQ$
\begin{equation}\label{eq:orth:A}
	\frac{1}{|\PhiQ|}\sum_{\dc\in\PhiQ}\chiz(A_1)\chibarz(A_2)
	=
	\begin{cases}
	1 & \mbox{if }A_1=A_2 \mbox{ mod } c \\
	0 & \mbox{if }otherwise,
	\end{cases}
\end{equation}
\item for every $\chione,\chitwo\in\PhiQ$
\begin{equation}\label{eq:orth:chi}
	\frac{1}{|\PhiQ|}\sum_{A\in\BQ}\chionez(A)\chitwobarz(A)
	=
	\begin{cases}
	1 & \mbox{if }\chione=\chitwo \\
	0 & \mbox{if }\chione\neq\chitwo.
	\end{cases}
\end{equation}
\end{enum}

We use \eqref{eq:orth:A} and \eqref{twisted-d-sum} to express the sum of $d_{k,\rho}(f)$ over arithmetic progressions in terms of $\ckn{\rhochi}:$
\begin{equation*}\label{eqn:sum-of-bs}
	\SnAQ
	=
	\frac{1}{|\PhiQ|}
	\sum_{f\in\MM_n}
		d_{k,\rho}(f)
	\sum_{\dc\in\PhiQ}
		\chiz(A)
		\bar{\chiz}(f)
	=
	\frac{1}{|\PhiQ|}
	\sum_{\dc\in\PhiQ}
		\ckn{\rhochi}\cdot\bar{\chiz}(A)
\end{equation*}
Therefore, if we write $\chinot\in\PhiQ$ for the trivial character and by using the second orthogonality relation \eqref{eq:orth:chi} then expected value of $\SnAQ$ equals
\begin{equation}
	\begin{split}
    	\bbE_A[\SnAQ]
    	& := \frac{1}{|\PhiQ|}\sum_{A\in\BQ}\SnAQ \\
    	& = \frac{1}{|\PhiQ|^2}
    		\sum_{\dc\in\PhiQ}
        	\ckn{\rhochi}
        	\sum_{A\in\BQ}
    	    	\bar\chiz(A) \\
        & = \frac{1}{|\PhiQ|} \ckn{\rho\otimes\chinot}
	\end{split}
\end{equation}

In particular, we have the identity
\begin{equation}\label{eq:s-e}
	\SnAQ - \bbE_A[\SnAQ]
	=
	\frac{1}{|\PhiQ|}
	\sum_{\substack{\dc\in\PhiQ \\ \dc\neq\chinot}}
		\ckn{\rhochi}\cdot\bar\dc(A).
\end{equation}

Now consider the variance
$$
	\Var_A[\SnAQ]
	=
	\frac{1}{|\PhiQ|}
	\sum_{A\in\BQ}
		\left|\SnAQ - \bbE_A[\SnAQ]\right|^2.
$$
If we apply identities \eqref{eq:orth:chi} and \eqref{eq:s-e}, then the right side equals
$$
	\frac{1}{|\PhiQ|^3}
	\sum_{A\in\BQ}
	\sum_{\substack{\chione,\chitwo\in\PhiQ\\\chionez,\chitwoz\neq\chinot}}
		\ckn{\rho\otimes\chione}\overline{\ckn{\rho\otimes\chitwo}}
		\cdot
		\chionebarz(A)\chitwoz(A)
	=
	\frac{1}{|\PhiQ|^2}
	\sum_{\substack{\dc\in\PhiQ\\\dc\neq\chinot}}
		|\ckn{\rhochi}|^2.
$$

\medskip
In summary, the function $\SnAQ$ of the random variable $A$ satisfies
\begin{equation}\label{eqn:E-and-V-formulae}
	\bbE_A[\SnAQ]
	=
	\frac{1}{|\PhiQ|}\ckn{\rho\otimes \chinot},
	\quad
	\Var_A[\SnAQ]
	=
	\frac{1}{|\PhiQ|^2}
	\sum_{\substack{\dc\in\PhiQ\\\dc\neq\chinot}}
		|\ckn{\rhochi}|^2.
\end{equation}

Next, we break $\ckn\rhochi$ into smaller pieces which we will express in terms of the associated Frobenius matrices:

\begin{lemma}\label{coefficient}
$$
	\ckn\rhochi
	=
	\sum_{n_1+\cdots+n_k=n}
	\,\,
	\prod_{i=1}^k
	\,
	c_{1,\rhochi,n_i}
$$
\end{lemma}

\begin{proof}
This follows immediately from the definition of $\ckn\rhochi$ as the coefficients in the expansion of $\LC(T,\rhochi)^k$:
\begin{equation}
	\LC(T,\rhochi)^k
	=
	(\sum_{n=0}^{\infty}c_{1,\rhochi,n}T^{n})^{k}
	=
	\sum_{n=0}^{\infty}\sum_{n_{1}+\ldots+n_{k}=n}
	\prod_{i=1}^k
		c_{1,\rhochi,n_{i}}T^{n}
\end{equation}
\end{proof}

Recall that $\std\colon U_R(\bbC)\to\GL_R(\bbC)$ is the representation given by the inclusion $U_R(\bbC)\sub\GL_R(\bbC)$ and
$$
	\wedge^j \std\colon U_R(\bbC)\to \GL_{R_j}(\wedge^j\bbC)
$$
is its $j$th exterior power where $R_j=\binom{R}{j}$; and we defined $(\wedge^j\,\std)(g)=0$ unless $0\leq j\leq R$.

\begin{lemma}\label{eval coeff good}
Let $\dc\in\PhiQGood\rho$, then
$$
	c_{1,\rhochi,n}
	=
	(-1)^n
	\cdot
	(\sqrt{q})^{n(1+w)}
	\cdot
	\Tr((\wedge^n\std)(\trhochi))
$$
\end{lemma}

\begin{proof}
A Dirichlet character $\dc$ is good for $\rho$ iff the `unitarized' $L$-function
$$
	\LCStar(T,\rhochi) := \LC(T/(\sqrt{q})^{1+w},\rhochi)
$$

\ssni
is the characteristic polynomial of a unitary conjugacy class $\trhochi\sub U_R(\bbC)$ for $R=\deg(\LC(T,\rho))$.

The coefficients of the characteristic polynomial of an $N\times N$ matrix with eigenvalues $\lambda_{1},\ldots, \lambda_{n}$ are the elementary symmetric functions $\sum_{1\leq i_{1}<\ldots<i_{r}}\lambda_{i_{1}}\cdots \lambda_{i_{r}}$ which give the character of the exterior power representation. Thus we may write
$$
	\LCStar(T,\rhochi)
	=
	\sum_{i=0}^{R}
		(-1)^i
		\cdot
		(\sqrt{q})^{i(1+w)}
		\cdot
		\Tr((\wedge^i\std)(\trhochi)) T^{i}
$$
\end{proof}

\begin{lemma}\label{eval coeff}
If $\dc\in\PhiQ$, then
$$
	|c_{1,\rhochi,n}|^2
	=
	\begin{cases}
    	O\left(q^{n(2+w)}\right) & \text{ if }\dc\in\PhiQHeavy\rho \\[0.05in]
    	O\left(q^{n(1+w)}\right) & \text{otherwise}
	\end{cases}
$$
The implied constants depend only on $\rho$ and $n$.
\end{lemma}

\begin{proof}
Fix a field embedding $\iota\colon\Qbar\to\bbC$ and identify $\Qbar$ with its image.  Recall that, for some integer $s$ depending on $\dc$ and satisfying $s\leq\dim(V)$, we can express $\LC(T,\rhochi)$ as a ratio
$$
	\prod_{i=1}^{s+R}(1-\alpha_iT)
	\,/
	\prod_{j=1}^s(1-\beta_jT)
$$
where the $\alpha_i$ and $\beta_j$ lie in $\Qbar$ and satisfy
$$
	|\alpha_i|^2\leq q^{1+w},
	\quad
	|\beta_j|^2\leq q^{2+w}.
$$
For reference see equation $(3.4.2)$ and Theorem $6.2.1$ in \cite{HKRG}.

If $\dc\not\in\PhiQHeavy\rho$, then $s=0$ and $|\alpha_i|^2\leq q^{1+w}$, and thus
$$
	|c_{1,\rhochi,n}|^2
	=
	\left|
    	\sum_{i_1<\cdots<i_n}
    	\alpha_{i_1}\cdots\alpha_{i_n}
	\right|^2
	\leq
 	\sum_{i_1<\cdots<i_n}
	\left|
    	\alpha_{i_1}\cdots\alpha_{i_n}
	\right|^2
	\leq
	\binom{R}{k}q^{1+w}.
$$
When $\rho$ is heavy we get

\begin{eqnarray*}
	|c_{1,\rhochi,n}|^2
	& = &
	\left|
		\sum_{n_1+n_2=n}
		(-1)^{n_1}
    	\sum_{i_1<\cdots<i_{n_1}}
    	\alpha_{i_1}\cdots\alpha_{i_{n_1}}
     	\sum_{j_1\leq\cdots\leq\,j_{n_2}}
       	\beta_{j_1}\cdots\beta_{j_{n_2}}
	\right|^2 \\[0.1in]
	& \leq &
	\sum_{n_1+n_2=n}\,
   	\sum_{i_1<\cdots<i_{n_1}}\,
   	\sum_{j_1\leq\cdots\leq\,j_{n_2}}
	\left|
    	\alpha_{i_1}\cdots\alpha_{i_{n_1}}
       	\beta_{j_1}\cdots\beta_{j_{n_2}}
	\right|^2 \\[0.05in]
	& = &
	O(q^{n(2+w)}).
\end{eqnarray*}
\end{proof}

\begin{cor}\label{eval coeff k}
If $\dc\in\PhiQ$, then
$$
	|\ckn\rhochi|^2
	=
	\begin{cases}
    	O\left(q^{n(2+w)}\right) & \text{ if }\dc\in\PhiQHeavy\rho \\[0.05in]
    	O\left(q^{n(1+w)}\right) & \text{otherwise}
	\end{cases}
$$
The implied constants depend only on $\rho$ and $n$.
\end{cor}

\begin{proof}
Combine Lemma \ref{coefficient} and Lemma \ref{eval coeff}.
\end{proof}

\begin{lemma}\label{just good}
If the Mellin transform of $\rho$ has big monodromy and if $\PhiQHeavy\rho\seq\{\one\}$, then
$$
	\sum_{\substack{\dc\in\PhiQ\\[0.02in]\varphi\neq\one}}
	|\ckn\rhochi|^2
	\ \ \sim
	\sum_{\substack{\dc\in\PhiQGood\rho\\[0.02in]\varphi\neq\one}}
	|\ckn\rhochi|^2
	\ \
	\text{ as }
	q\to\infty
$$

\end{lemma}

\begin{proof}
We break the sum over all Dirichlet characters mod $c$ to sums over "good", "heavy" and "mixed" characters (see definitions \eqref{good}, \eqref{heavy}, \eqref{mixed})
$$
	\sum_{\substack{\dc\in\PhiQ\\[0.02in]\varphi\neq\one}}
		|\ckn\rhochi|^2
	\ \ =
	\sum_{\substack{\dc\in\PhiQGood\rho\\[0.02in]\varphi\neq\one}}
		|\ckn\rhochi|^2
	+
	\sum_{\substack{\dc\in\PhiQHeavy\rho\\[0.02in]\varphi\neq\one}}
		|\ckn\rhochi|^2
	+
	\sum_{\substack{\dc\in\PhiQMixed\rho\\[0.02in] \varphi\neq\one}}
		|\ckn\rhochi|^2
$$
which give by Corollary \ref{eval coeff k}
$$
	\sum_{\substack{\dc\in\PhiQGood\rho\\[0.02in]\varphi\neq\one}}
		|\ckn\rhochi|^2
	+
	|\PhiQHeavy\rho\ssm\{1\}|\cdot O(q^{n(2+w)})
	+
	|\PhiQMixed\rho\ssm\{1\}|\cdot O(q^{n(1+w)})
$$
Lemma~\ref{size} and having $\PhiQHeavy\rho\seq\{\one\}$ conclude the proof.
\end{proof}

\begin{prop}
If the Mellin transform of $\rho$ has big monodromy and if $\PhiQHeavy\rho\seq\{\one\}$, then
$$
	\lim_{q\to\infty}
	\frac{|\PhiQ|}{q^{n(1+w)}}
	\cdot
	\Var_A[\SnAQ]
	\ =
	\int_{U_R(\bbC)}
	\left|\sum_{\substack{n_1+\cdots +n_k=n\\0\leq n_1,\ldots,n_k\leq R}}\,\,
		\Tr\Big(\big(\otimes_{i=1}^k (\wedge^{n_i}\,\std)\big)(\theta)\Big)
	\right|^2
	d\theta.
$$
\end{prop}

\begin{proof}
In \eqref{eqn:E-and-V-formulae} we found the following expression for the variance
$$
	\Var_A[\SnAQ]
	=
	\frac{1}{|\PhiQ|^2}
	\sum_{\substack{\dc\in\PhiQ\\\dc\neq\chinot}}
		|\ckn{\rhochi}|^2.
$$
In the limit of $q\rightarrow\infty$ we have from Lemma \ref{just good}
$$
	\Var_A[\SnAQ]
	\sim
	\frac{1}{|\PhiQ|^2}
	\sum_{\substack{\dc\in\PhiQGood\rho\\\dc\neq\chinot}}
		|\ckn{\rhochi}|^2
$$
which Lemma \ref{coefficient} and Lemma \ref{eval coeff good} equals to
$$
	\frac{1}{|\PhiQ|^2}
	\sum_{\substack{\dc\in\PhiQGood\rho\\\dc\neq\chinot}}
    	\left|
    		\sum_{\substack{n_1+\cdots +n_k=n\\0\leq n_1,\ldots,n_k\leq R}}
				\Tr\Big(
					\big(\otimes_{i=1}^k (\wedge^{n_i}\,\std)\big)(\trhochi)
				\Big)
    	\right|^2
$$
Now note that $|\PhiQ|\sim|\PhiQGood\rho|$ (Lemma~\ref{size}) and apply the equidistribution result Theorem \ref{equidistribution} to conclude the proof.
\end{proof}

\section{Matrix Integral}\label{integral}

In this section we state a few results evaluating the matrix integral in certain ranges and its asymptotic behaviour.  Proofs can be found in \cite{KRRR}.

In certain ranges the matrix integral evaluates to a very simple expression. For
$n<R$ or $(k-1)R<n<kR$, we obtain the following formulae:
\begin{theorem}\label{Thm: Ik(m,N) for large m}
Let $I_k(n;R)$ be the matrix integral defined in \eqref{matrix integral def}.  Then
\begin{enum}
\item for $(k-1)R<n<kR$,
$$
I_k(n;R) = \binom{kR-n+k^2-1}{k^2-1}\;.
$$
\item for $n<R$
$$
I_k(n;R) = \binom{n+k^2-1}{k^2-1}\;.
$$
\end{enum}
\end{theorem}

There is also a closed-form formula for the matrix integral for any range of the parameters, in terms of a lattice point count:

\begin{theorem}
\label{lattice_count}
$I_k(m;N)$ is equal to the count of lattice points $x= (x_i^{(j)})\in\mathbb{Z}^{k^2}$ satisfying each of the following relations:
\begin{enum}
\item $0 \leq x_i^{(j)} \leq N$ for all $1\leq i,j \leq k$;
\item $x_1^{(k)} + x_2^{(k-1)} + \cdots + x_k^{(1)} = kN - m$, and
\item $x$ is a $k \times k$ matrix whose entries satisfy the following system of inequalities,
$$
    \begin{matrix}
    	x_1^{(1)} & \leq & x_1^{(2)} & \leq & \cdots & \leq & x_1^{(k)} \\
    	  \vleq   &      &   \vleq   &      &        &      & \vleq \\
        x_2^{(1)} & \leq & x_2^{(2)} & \leq & \cdots & \leq & x_2^{(k)} \\
    	  \vleq   &      &   \vleq   &      &        &      & \vleq \\
          \vdots  &      &   \vdots  &      & \ddots &      & \vdots \\
    	  \vleq   &      &   \vleq   &      &        &      & \vleq \\
        x_k^{(1)} & \leq & x_k^{(2)} & \leq & \cdots & \leq & x_k^{(k)}
    \end{matrix}
$$
\end{enum}
\end{theorem}

The asymptotic behavior of $I_k(n;R)$ for $n\approx R$ is given in the following theorem:
\begin{theorem}\label{thm:Asymp of I}
Let $c := n/R$.  Then for $c \in [0,k]$,
\begin{equation}\label{thm:poly_approx}
	I_k(n;R)
	=
	\gamma_k(c) R^{k^2-1} + O_k(R^{k^2-2}),
\end{equation}
with
$$
	\gamma_k(c)
	=
	\frac{1}{k!\, G(1+k)^2}
	\int_{[0,1]^k}
		\delta_c(w_1 + \ldots + w_k)
		\prod_{i< j}(w_i-w_j)^2\,
		d^k w,
$$
with $\delta_c(x) =\delta(x-c)$ being the delta distribution translated by $c$, and $G$ is the Barnes $G$-function, so that for positive integers $k$, $G(1+k) = 1!\cdot 2! \cdot 3! \cdots (k-1)!$.

\end{theorem}

\bibliographystyle{amsplain.bst}
\bibliography{main}

\end{document}

config: arxiv


TITLE: Why is the following subset of $\mathbb{C}$ simply connected.
QUESTION [1 upvotes]: Let $E=\mathbb{C} \setminus \lbrace te^{it} \mid t\in \mathbb{R}_{\geq{0}} \rbrace$. Then we can define a branch of logarithm on $E$. My complex analysis instructor said that we can do it for any simply connected domain of $\mathbb{C}$.
How do we show using basic algebraic topology that $E$ is simply connected.

REPLY [4 votes]: The map $z\mapsto ze^{-i|z|}$ is a homeomorphism of $E$ onto $E':=\mathbb{C}\,\backslash\,\mathbb{R}_{\geq 0}$, which is star convex (for every $z\in E'$, the line segment connecting $z$ and $-1$ is contained in $E'$). This shows $E'$ is contractible (and in particular simply connected), hence so is $E$.