Traces of high powers of the Frobenius class in the moduli space of hyperelliptic curves

In the smooth models, the point at infinity will be replaced by 0, 1, or 2 points, accordingly, and we get that the number of points at infinity is

(4)

where \({{\mathrm{sgn}}}(Q)\) denotes the leading coefficient of Q. Note the relation between Eqs. (4) and (9); namely, the number of points at infinity is \(\lambda _Q+1\), with \(\lambda _Q\) as in (9).

In this paper, we study the traces of high powers of the Frobenius class of \(C\in {\mathscr {H}}_g\) over a fixed finite field \({\mathbb F}_q\), of odd cardinality q, as g tends to infinity. In particular, we concern ourselves with the expected values \(\langle {{\mathrm{tr}}}(\Theta _{C}^n) \rangle _{{\mathscr {H}}_g}\) as \(g \rightarrow \infty \) and we compare our work with the Random Matrix results of [3].

Once again, our results for odd n are exact and hold for all values of g. Although we continue to get deviations from the RMT results for even \(n\ge 4\), our results hold for \(n=2\) and our deviations are different from those obtained in [7].

Another approach to computing \(\langle \#C({\mathbb F}_{q^n}) \rangle _{{\mathscr {H}}_g}\) is the work of Alzahrani [1] which uses the distribution of the \({\mathbb F}_q\)-points of \({\mathscr {H}}_g\) in \({\mathbb F}_{q^n}\). Using these methods, the results of Alzahrani agree with the Corollary above (albeit with a larger error term).

Finally, we would like to mention that some of the computations done in Sect 2.1 through 3.2 were done independently by Lorenzo et al. in their study of statistics for biquadratic curves; their work is collected in [5].

In this section, we establish some notation, we introduce the main results of [7], and we prove some preliminary results. Since the majority of what follows is based off of the work in [7], we use the same notation and list important results for the convenience of the reader. We use [6] as a general reference.

Throughout this paper, \({\mathbb F}_q\) is a fixed finite field of odd cardinality q, P is solely used to represent monic irreducible polynomials in \({\mathbb F}_q[x]\), and Q will be used to denote squarefree polynomials of degree \(2g+1\) or \(2g+2\) with \(g\ge 1\). Unless stated otherwise, sums/products are over monic elements in \({\mathbb F}_q[x]\) (under the given restrictions). For example, sums/products indexed by P are over all monic irreducible polynomials in \({\mathbb F}_q[x]\), where as sums/products indexed by \(\deg (f)=n\) are over all monic polynomials of degree n in \({\mathbb F}_q[x]\). In the case where a sum involves elements \(B\in {\mathbb F}_q[x]\) that are not necessarily monic, we write the sum over B n.n.m..

The Zeta function associated to the hyperelliptic curve \(C_Q:y^2=Q(x)\) is then given by

$$\begin{aligned} {{\mathrm{Z}}}_{C_Q}(u)=L^*(u,\chi _Q)\zeta _q(u), \end{aligned}$$

where

$$\begin{aligned} \zeta _q(u):=\frac{1}{(1-u)(1-qu)} \end{aligned}$$

is the Zeta function of \({\mathbb F}_q(x)\) and where

$$\begin{aligned} L^*(u,\chi _Q):= & {} (1-\lambda _Q u)^{-1}\prod _{P}(1-\chi _Q(P) u^{\deg (P)})^{-1}\end{aligned}$$

(7)

$$\begin{aligned}= & {} \det ({\mathbb I}-u\sqrt{q} \Theta _{C_Q}), \end{aligned}$$

(8)

with

(9)

which relates to the count in Eq. (4).

In this section, we compute \(\langle {{\mathrm{tr}}}(\Theta _{C_Q}^n) \rangle _{{\mathcal F}_{2g+1} \cup {\mathcal F}_{2g+2}}\) by first computing \(\langle {{\mathrm{tr}}}(\Theta _{C_Q}^n) \rangle _{{\mathcal F}_{2g+2}}\) and then combining that result with Rudnick’s for \(\langle {{\mathrm{tr}}}(\Theta _{C_Q}^n) \rangle _{{\mathcal F}_{2g+1}}.\)

3.1 Computing \(\hbox {tr}(\Theta _{C_Q}^n)\) for \(Q\in \mathcal {F}_{2g+2}\)

For the time being, we restrict ourselves to \({\mathcal F}_{2g+2}\). Let \(Q\in {\mathcal F}_{2g+2}\) and consider the curve \(C_Q:y^2=Q(x)\). The trace of the powers of \(\Theta _{C_Q}\) is given by Eq. (10):

$$\begin{aligned} {{\mathrm{tr}}}(\Theta _{C_Q}^n)= & {} -\frac{1}{q^{\frac{n}{2}}}-\frac{1}{q^{\frac{n}{2}}}\sum _{\deg (f)=n}\Lambda (f)\chi _Q(f)\end{aligned}$$

(23)

$$\begin{aligned}= & {} -\frac{1}{q^{\frac{n}{2}}}-\frac{1}{q^{\frac{n}{2}}}\sum _{\begin{array}{c} {P,k}\\ {\deg (P^k)}=n \end{array}}\deg (P)\chi _Q(P^k)\end{aligned}$$

(24)

$$\begin{aligned}= & {} -\frac{1}{q^{\frac{n}{2}}}+{\mathcal P}_n+\square _n + {\mathbb H}_n, \end{aligned}$$

(25)

where \({\mathcal P}_n\) corresponds to \(k=1\), \(\square _n\) corresponds to the sum over all k even, and \({\mathbb H}_n\) corresponds to the sum over all odd \(k\ge 3\).

In the next three sections, we continue to use Rudnick’s methods in order to compute \({\mathcal P}_n\), \(\square _n\), and \({\mathbb H}_n\). Not surprinsingly, our results will only slightly differ from Rudnick’s. The addition of \(-1/q^\frac{n}{2}\) from (9) will be the main difference. We will also have different cut-off points for n when estimating \({\mathcal P}_n\) (see Sect. 5.3 of [7]).

3.1.1 Contribution of the primes: \(\mathcal {P}_n\)

The contribution of the primes in (23) is given by

$$\begin{aligned} {\mathcal P}_n=-\frac{1}{q^{\frac{n}{2}}}\sum _{\deg (P)=n}n\chi _Q (P). \end{aligned}$$

So,

$$\begin{aligned} \langle {\mathcal P}_n \rangle _{{\mathcal F}_{2g+2}}= & {} \frac{-n}{(q-1)q^{2g+1+\frac{n}{2}}}\sum _{\deg (P)=n}\sum _{2\alpha +\beta =2g+2}\sigma _n(\alpha )\sum _{\deg (B)=\beta }\Biggr (\frac{B}{P}\Biggr )\\= & {} \frac{-n}{(q-1)q^{2g+1+\frac{n}{2}}}\sum _{2\alpha +\beta =2g+2}\sigma _n(\alpha )S(\beta ;n). \end{aligned}$$

From Lemma 2.1, if \(n>g+1\), then

$$\begin{aligned} \sigma _n(\alpha )\ne 0\Rightarrow & {} (\alpha \equiv 0\, \hbox {mod}\, n \,or\, \alpha \equiv 1\, \hbox {mod}\, n)\\\Rightarrow & {} (\alpha =0 \text { or } \alpha =1), \end{aligned}$$

which follows from the fact that \(0\le \alpha \le g+1\). Since \(\sigma _n(0)=1\) and \(\sigma _n(1)=-q\), when \(n>g+1\), we have

$$\begin{aligned} \langle {\mathcal P}_n \rangle _{{\mathcal F}_{2g+2}}=\frac{-n}{(q-1)q^{2g+1 +\frac{n}{2}}} (S(2g+2;n)-qS(2g;n)). \end{aligned}$$

We now compute \(\langle {\mathcal P}_n \rangle _{{\mathcal F}_{2g+2}}\) by considering the case \(n\le g+1\) and the case \(n>g+1\), which we break into four (non-distinct) ranges:

1. (i)
   \(n\le g+1:\) If \(S(\beta ;n)\ne 0\), then \(\beta <n\); since \(\beta \) is even,

   $$\begin{aligned} S(\beta ;n)= & {} \pi _q(n)q^{\frac{\beta }{2}}+O \left( \frac{\beta }{n}q^{\beta +\frac{n}{2}}\right) \\= & {} \frac{q^{n+\frac{\beta }{2}}}{n}+O \Bigg (\frac{q^{\frac{n}{2}+\frac{\beta }{2}}}{n}\Bigg )+O \Bigg (\frac{\beta }{n}q^{\beta +\frac{n}{2}}\Bigg ). \end{aligned}$$

   Then \(S(\beta ;n)\ll \frac{\beta }{n}q^{n+\beta },\) which implies that

   $$\begin{aligned} \langle {\mathcal P}_n\rangle _{{\mathcal F}_{2g+2}}\ll & {} \frac{n}{q^{2g+\frac{n}{2}}}\sum _{\beta <n}\frac{\beta }{n}q^{\beta +n}\\\ll & {} \frac{n}{q^{2g+\frac{n}{2}}}q^{2n}=nq^{\frac{3n}{2}-2g}\ll gq^{\frac{-g}{2}}. \end{aligned}$$
    
2. (ii)
   \(g+1<n<2g+1:\) Since \(2g+2,2g\ge n\), \(S(2g+2;n)=S(2g;n)=0.\) Hence,

   $$\begin{aligned}\langle {\mathcal P}_n \rangle _{{\mathcal F}_{2g+2}}= \frac{-n}{(q-1)q^{2g+1+\frac{n}{2}}}(S(2g+2;n)-qS(2g;n))=0.\end{aligned}$$
    
3. (iii)
   \(n=2g+1:\) Since \(2g+2\ge n\), \(S(2g+2;n)=0\) and we get that

   $$\begin{aligned} \langle {\mathcal P}_n \rangle _{{\mathcal F}_{2g+2}}= & {} \frac{n}{(q-1)q^{2g+1+\frac{n}{2}}} q S(2g;n),\\= & {} \frac{2g+1}{(q-1)q^{3g+\frac{1}{2}}} S(2g;2g+1). \end{aligned}$$

   Using (18),

   $$\begin{aligned} \langle {\mathcal P}_n \rangle _{{\mathcal F}_{2g+2}}= & {} \frac{2g+1}{(q-1)q^{3g+\frac{1}{2}}} \pi _q(2g+1)q^{\frac{(2g+1)-1}{2}}. \end{aligned}$$

   By replacing \(\pi _q(2g+1)\) and simplifying, we obtain

   $$\begin{aligned} \langle {\mathcal P}_n \rangle _{{\mathcal F}_{2g+2}}= & {} \frac{2g+1}{(q-1)q^{3g+\frac{1}{2}}} \left( \frac{q^{2g+1}}{2g+1}+O\left( \frac{q^g}{2g+1}\right) q^g\right) \\= & {} \frac{q^{\frac{1}{2}}}{q-1}+O(q^{-g}). \end{aligned}$$
    
4. (iv)
   \(n=2g+2\): Similarly,

   $$\begin{aligned} \langle {\mathcal P}_n \rangle _{{\mathcal F}_{2g+2}}= & {} \frac{n}{(q-1)q^{2g+1+\frac{n}{2}}}q S(2g;n)\\= & {} \frac{2g+2}{(q-1)q^{3g+1}} S(2g;2g+2). \end{aligned}$$

   From Theorem 2.4,

   $$\begin{aligned} \langle {\mathcal P}_n \rangle _{{\mathcal F}_{2g+2}}= & {} \frac{2g+2}{(q-1)q^{3g+1}} \left( \left( \frac{q^{2g+2}}{2g+2}+O(\frac{q^g}{2g+2})\right) (q^g-q^{g-1})+O(q^{2g})\right) \\= & {} 1+O(q^{-g})+O(gq^{-g}). \end{aligned}$$
    
5. (v)
   \(n>2g+2:\) We apply Theorem 2.4 to get

   $$\begin{aligned} \langle {\mathcal P}_n \rangle _{{\mathcal F}_{2g+2}}= & {} \frac{-n}{(q-1)q^{2g+1+\frac{n}{2}}}\left( S(2g+2;n)-qS(2g;n)\right) \\= & {} \frac{-n}{(q-1)q^{2g+1+\frac{n}{2}}}\left( \pi _q(n)(q^{\frac{2g+2}{2}}-\eta _nq^{2g+2-\frac{n}{2}})\right. \\&\left. -q\pi _q(n)(q^{\frac{2g}{2}}-\eta _nq^{2g-\frac{n}{2}})+O(q^n)\right) . \end{aligned}$$

   Upon further simplification,

   $$\begin{aligned} \langle {\mathcal P}_n \rangle _{{\mathcal F}_{2g+2}}= & {} \frac{n}{(q-1)q^{2g+1+\frac{n}{2}}}\Biggr (\eta _n \pi _q(n) (q^{2g+2-\frac{n}{2}}-q^{2g+1-\frac{n}{2}})+O(q^n)\Biggr )\\= & {} \frac{n\eta _n \pi _q(n)}{q^n}+O(nq^{\frac{n}{2}-2g})\\= & {} \eta _n (1+O(q^{\frac{-n}{2}}))+O(nq^{\frac{n}{2}-2g}). \end{aligned}$$
    

Note 1 When \(n=2g+2\), (v) yields (iv).

3.1.2 Contribution of the squares: \(\square _n\)

For n even, we have the following contribution from the squares of prime powers:

$$\begin{aligned} \square _n= & {} -\frac{1}{q^{\frac{n}{2}}}\sum _{\deg (P^{2k})=n}\Lambda (P^{2k})\chi _Q(P^{2k})\\= & {} -\frac{1}{q^{\frac{n}{2}}}\sum _{\deg (P^{k})=\frac{n}{2}}\Lambda (P^{k})\chi _Q(P^{2k}) . \end{aligned}$$

Therefore,

$$\begin{aligned} \langle \square _n \rangle _{{\mathcal F}_{2g+2}}= & {} -\frac{1}{q^{\frac{n}{2}}}\sum _{\deg (P^{k})=\frac{n}{2}}\Lambda (P^{k})\langle \chi _Q(P^{2k})\rangle _{{\mathcal F}_{2g+2}}\\ \!= & {} \!-\frac{1}{q^{\frac{n}{2}}}\sum _{\deg (P^{k})=\frac{n}{2}}\deg (P)\frac{1}{(q\!-\!1)q^{2g+1}}\sum _{0\le \alpha \le g+1}\sum _{\begin{array}{c} {\deg (A)=\alpha }\\ {P\not \mid A} \end{array}}\mu (A)\sum _{\begin{array}{c} {\deg (B)=2g+2-2\alpha },\\ {P\not \mid B} \end{array}}1. \end{aligned}$$

Although Sect. 3.1.1 shows a slight deviation in \(\langle {\mathcal P}_n \rangle _{{\mathcal F}_{2g+2}}\) from \( \langle {\mathcal P}_n \rangle _{{\mathcal F}_{2g+1}}\), we will see that \(\langle \square _n \rangle _{{\mathcal F}_{2g+2}}=\langle \square _n \rangle _{{\mathcal F}_{2g+1}}.\)

Let \(m=\deg (P)\). Since

$$\begin{aligned} \#\{B\in {\mathbb F}_q[x]: B \; \mathrm monic, \deg (B)=\beta , P\not \mid B\}= q^{\beta }\cdot \left\{ \begin{array}{lll} 1 &{}\quad \text{ if } m>\beta , \\ 1 -\frac{1}{|P|} &{}\quad \text{ if } m\le \beta , \end{array}\right. \end{aligned}$$

$$\begin{aligned} \langle \square _n \rangle _{{\mathcal F}_{2g+2}} \!= & {} \!-\frac{1}{q^{\frac{n}{2}}}\sum _{\deg (P^{k})=\frac{n}{2}}\deg (P)\frac{1}{(q\!-\!1)q^{2g+1}}\sum _{0\le \alpha \le g+1}\sum _{\begin{array}{c} {\deg (A)=\alpha }\\ {P\not \mid A} \end{array}}\mu (A)\sum _{\begin{array}{c} {\deg (B)=2g+2-2\alpha }\\ {P\not \mid B} \end{array}}1\\= & {} -\frac{q}{(q-1)q^{\frac{n}{2}}}\sum _{\deg (P^{k})=\frac{n}{2}}\deg (P) \Biggr ( \Big (1-\frac{1}{|P|}\Big ) \sum _{\alpha \ge 0}\frac{\sigma _m(\alpha )}{q^{2\alpha }}+O(q^{-2g})\Biggr ). \end{aligned}$$

Solving the recurrence relation in Lemma 2.1,

$$\begin{aligned} \langle \square _n \rangle _{{\mathcal F}_{2g+2}}= & {} -\frac{q}{(q-1)q^{\frac{n}{2}}}\sum _{\deg (P^{k})=\frac{n}{2}}\deg (P)\Biggr (\Big (1-\frac{1}{|P|}\Big ) \frac{1-\frac{1}{q}}{1-\frac{1}{|P|^2}}+O(q^{-2g})\Biggr )\\= & {} -\frac{1}{q^{\frac{n}{2}}}\sum _{\deg (P)|\frac{n}{2}} \deg (P)\Big (\frac{|P|}{|P|+1}+O(q^{-2g})\Big )\\= & {} -\frac{1}{q^{\frac{n}{2}}}\sum _{\deg (P)|\frac{n}{2}} \deg (P)\Big (1-\frac{1}{|P|+1}+O(q^{-2g})\Big )\\= & {} -\frac{1}{q^{\frac{n}{2}}}\sum _{\deg (P)|\frac{n}{2}} \deg (P) +\frac{1}{q^{\frac{n}{2}}}\sum _{\deg (P)|\frac{n}{2}} \deg (P) \Big (\frac{1}{|P|+1}+O(q^{-2g})\Big ). \end{aligned}$$

Since

$$\begin{aligned}&q^l=\sum _{\deg (h)=l} \Lambda (h)=\sum _{\deg (P^k)=l}\deg (P)=\sum _{\deg (P)|l}\deg (P),\\&\langle \square _n \rangle _{{\mathcal F}_{2g+2}} = -1+\frac{1}{q^{\frac{n}{2}}}\sum _{\deg (P)|\frac{n}{2}} \frac{\deg (P)}{|P|+1}+O(q^{-2g}). \end{aligned}$$

3.1.3 Contribution of the higher prime powers: \(\mathbb {H}_n\)

Using trivial bounds for \({\mathbb H}_n\), we obtain slightly different results from Rudnick in the case where \(n>6g\). This is due to the fact that our estimates of \({\mathbb H}_n\) involve \(S(\beta ;n)\) for \(\beta \) even, as opposed to \(\beta \) odd, and, from Lemma 2.3,

$$\begin{aligned} S(\beta ;n)\ll \left\{ \begin{array}{lll} q^{n+\beta } &{} \quad \text {if }\beta \, \text {is even}, \\ q^{\frac{n}{2}+\beta } &{} \quad \text {if }\beta \, \text { is odd.} \end{array}\right. \end{aligned}$$

We shall see that our bounds for \(n>6g\) are absorbed in the error term from \(\langle {\mathcal P}_n \rangle _{{\mathcal F}_{2g+2}}\) when \(n>2g+1\).

The contribution to \({{\mathrm{tr}}}(\Theta _{C_Q}^n)\) from the higher odd prime powers in (23) is:

$$\begin{aligned} {\mathbb H}_n= & {} -\frac{1}{q^{\frac{n}{2}}}\sum _{\begin{array}{c} {d|n}\\ {3 \le d: \mathrm{odd}} \end{array}}\sum _{\deg (P)=\frac{n}{d}}\frac{n}{d}\chi _Q(P^d)\\= & {} -\frac{1}{q^{\frac{n}{2}}}\sum _{\begin{array}{c} {d|n}\\ {3 {\le d: \mathrm{odd}}} \end{array}}\sum _{\deg (P)=\frac{n}{d}}\frac{n}{d}\chi _Q(P), \end{aligned}$$

where the last equality follows from the fact that \(\chi _Q(P^d)=\chi _Q(P)\) for odd d.

This implies that

$$\begin{aligned} \langle {\mathbb H}_n \rangle _{{\mathcal F}_{2g+2}}= & {} -\frac{1}{(q-1)q^{2g+1+\frac{n}{2}}}\sum _{\begin{array}{c} {d|n}\\ {3 {\le d: \mathrm{odd}}} \end{array}}\frac{n}{d} \sum _{\deg (P)=\frac{n}{d}} \sum _{2\alpha + \beta = 2g+2} \sigma _{\frac{n}{d}}(\alpha ) \sum _{\deg (B)=\beta }\Big (\frac{B}{P}\Big )\\= & {} -\frac{1}{(q-1)q^{2g+1+\frac{n}{2}}}\sum _{\begin{array}{c} {d|n}\\ {3 {\le d: \mathrm{odd}}} \end{array}}\frac{n}{d}\sum _{2\alpha + \beta = 2g+2} \sigma _{\frac{n}{d}}(\alpha )S (\beta ;\frac{n}{d}). \end{aligned}$$

If \(S(\beta ;\frac{n}{d})\ne 0\), then \(\beta < \frac{n}{d}\); also, \(S(\beta ; \frac{n}{d})\ll q^{\frac{n}{d}+\beta }\). Hence,

If \(\frac{n}{3}\le 2g\), then \(\min (\frac{n}{d},2g)=\frac{n}{d}\) for all \(d\ge 3\); and so,

$$\begin{aligned} \langle {\mathbb H}_n \rangle _{{\mathcal F}_{2g+2}}\ll & {} \frac{n}{q^{2g+\frac{n}{2}}}\sum _{\begin{array}{c} {d|n}\\ {3 {\le d: \mathrm{odd}}} \end{array}}q^{\frac{2n}{d}}\ll \frac{n}{q^{2g+\frac{n}{2}}}q^\frac{2n}{3} = \frac{n}{q^{2g}}q^{\frac{n}{6}}\\\ll & {} \frac{g}{q^{2g}} q^g=gq^{-g}. \end{aligned}$$

If \(\frac{n}{3}>2g\), then \(\min (\frac{n}{d},2g)\le \frac{n}{3}\) for all \(d\ge 3\); therefore,

$$\begin{aligned} \langle {\mathbb H}_n \rangle _{{\mathcal F}_{2g+2}} \ll \frac{n}{q^{2g+\frac{n}{2}}} q^\frac{2n}{3} \ll nq^{\frac{n}{6}-2g}. \end{aligned}$$

As we mentioned in the introduction, averaging over monic squarefree polynomials of a fixed degree is not the same as averaging over the moduli space of hyperelliptic curves of genus g: in the latter case, we consider polynomials of degree \(2g+1\) and \(2g+2\). Also, by restricting ourselves to monic polynomials, we introduce a bias in the average value of the trace: the contribution of the point at infinity is related to the leading coefficient of Q, as seen by Eq. (10).

We now turn our attention to finding the average of \({{\mathrm{tr}}}(\Theta _{C}^n)\) over \({\mathscr {H}}_g\). If we let

$$\begin{aligned} \widetilde{{\mathcal F}}_g=\widehat{{\mathcal F}}_{2g+1} \cup \widehat{{\mathcal F}}_{2g+2}, \end{aligned}$$

then, from Eqs. (5), (10), and (13),

$$\begin{aligned} \langle {{\mathrm{tr}}}(\Theta _{C}^n) \rangle _{{\mathscr {H}}_g}= & {} \frac{1}{\#\widetilde{{\mathcal F}}_g} \sum _{Q\in \widetilde{{\mathcal F}}_g}\Big ( -\frac{\lambda _Q^n}{q^\frac{n}{2}}-\frac{1}{q^\frac{n}{2}}\sum _{\deg (f)=n}\Lambda (f) \chi _Q(f)\Big ), \end{aligned}$$

(26)

where

Given \(D\in {\mathbb F}_q[x]\) with \(\deg (D)=d\), we may write \(D=A^2 B\), where \(A,B\in {\mathbb F}_q[x]\) with A monic, B not necessarily monic, and \(\deg (A)=\alpha \), \(\deg (B)=\beta \), so that \(d=2\alpha + \beta \). From here, we can take the character sum above over all elements of degree \(2g+1,2g+2\) by sieving out the squarefree terms (as we did earlier):

$$\begin{aligned} \sum _{Q\in \widetilde{{\mathcal F}}_g}\chi _Q(f)= & {} \sum _{\begin{array}{c} {2\alpha +\beta =d}\\ {d=2g+1,2g+2} \end{array}} \sum _{\begin{array}{c} {\deg (B)=\beta }\\ B\, \text {n.n.m.} \end{array}} \sum _{\deg (A)=\alpha } \mu (A)\Big (\frac{A}{f}\Big )^2\Big (\frac{B}{f}\Big )\\= & {} \sum _{\begin{array}{c} {2\alpha +\beta =d}\\ {d=2g+1,2g+2} \end{array}}\sigma (f;\alpha ) \sum _{\begin{array}{c} {\deg (B)=\beta }\\ {B\, \text {n.n.m.}} \end{array}}\Big (\frac{B}{f}\Big )\\= & {} \sum _{\begin{array}{c} {2\alpha +\beta =d}\\ {d=2g+1,2g+2} \end{array}} \sigma (f;\alpha )\sum _{a\in {\mathbb F}_q^*}\sum _{\deg (B)=\beta }\Big (\frac{aB}{f}\Big )\\= & {} \sum _{\begin{array}{c} {2\alpha +\beta =d}\\ {d=2g+1,2g+2} \end{array}} \sigma (f;\alpha )\sum _{a\in {\mathbb F}_q^*}\sum _{\deg (B)=\beta }\Big (\frac{a}{f}\Big )\Big (\frac{B}{f}\Big )\\= & {} \sum _{a\in {\mathbb F}_q^*}\Big (\frac{a}{f}\Big ) \sum _{\begin{array}{c} {2\alpha +\beta =d}\\ {d=2g+1,2g+2} \end{array}} \sigma (f;\alpha )\sum _{\deg (B)=\beta }\Big (\frac{B}{f}\Big ). \end{aligned}$$

Hence,

$$\begin{aligned} \langle {{\mathrm{tr}}}(\Theta _{C}^n) \rangle _{{\mathscr {H}}_g}\!=\!-\frac{1}{q^\frac{n}{2} \#\widetilde{{\mathcal F}}_g}\Biggr ( \sum _{Q\in \widetilde{{\mathcal F}}_g}\lambda _Q^n+\! \sum _{\deg (f)=n} \Lambda (f)\sum _{a\in {\mathbb F}_q^*}\Big (\frac{a}{f}\Big ) \sum _{\begin{array}{c} {2\alpha +\beta =d}\\ {d=2g+1,2g+2} \end{array}} \sigma (f;\alpha )\sum _{\deg (B)=\beta }\Big (\frac{B}{f}\Big )\Biggr ). \end{aligned}$$

Since there are exactly \((q-1)/2\) squares and \((q-1)/2\) non-squares in \({\mathbb F}_q^*\), if n is odd,

$$\begin{aligned} \frac{1}{\#\widetilde{{\mathcal F}}_g}\sum _{Q\in \widetilde{{\mathcal F}}_g}\lambda _Q^n=\frac{1}{\#\widetilde{{\mathcal F}}_g}\sum _{Q\in \widetilde{{\mathcal F}}_g}\lambda _Q=0. \end{aligned}$$

On the other hand, if n is even,

$$\begin{aligned} \frac{1}{\#\widetilde{{\mathcal F}}_g}\sum _{Q\in \widetilde{{\mathcal F}}_g}\lambda _Q^n=\frac{1}{\#\widetilde{{\mathcal F}}_g}\sum _{Q\in \widetilde{{\mathcal F}}_g}|\lambda _Q|=\frac{\#\widehat{{\mathcal F}}_{2g+2}}{\#\widetilde{{\mathcal F}}_g}=\frac{q}{q+1}. \end{aligned}$$

Also, if f is a power of some prime in \({\mathbb F}_q[x]\), say \(f=P^k\), then for all \(a\in {\mathbb F}_q^*\) (see Proposition 3.2 of [6]),

$$\begin{aligned}\Big (\frac{a}{f}\Big )=\Big (\frac{a}{P}\Big )^k=\Big (a^{\frac{q-1}{2}\deg (P)}\Big )^k=a^{\frac{q-1}{2}\deg (f)}.\end{aligned}$$

If \(\deg (f)=\deg (P^k)=n\) is even, then \(\Big (\frac{a}{f}\Big )=1\) because \(|{\mathbb F}_q^*|=q-1\). This tells us that \(\sum _{a\in {\mathbb F}_q^*}\Big (\frac{a}{f}\Big )=q-1\). If \(\deg (f)=\deg (P^k)=n\) is odd, then we have that \(\sum _{a\in {\mathbb F}_q^*}\Big (\frac{a}{f}\Big )=0\) because there are exactly \((q-1)/2\) QR in \({\mathbb F}_q^*\) and exactly \((q-1)/2\) NQR in \({\mathbb F}_q^*\).

In particular,