Proof. The odd-character projector at either modulus $q=b,b^2$ gives $$\sum_{\substack{\chi\bmod q\\\chi(-1)=-1}} \chi(u)\overline\chi(v) =\frac{\varphi(q)}2 \bigl(\mathbf1_{u\equiv v\;(q)}- \mathbf1_{u\equiv-v\;(q)}\bigr).$$ The imprimitive odd characters modulo $b^2$ are precisely the lifts of the odd characters modulo $b$. Subtract their projector. Setting $u=v=1$ also gives $N_b=(b-1)^2/2$. ◻

Proof. Expand $\sum_{\chi\in\mathcal P_b}|T_b(\chi)|^2$ and apply Lemma 2. Among $1\le u,v<b$, the positive congruence modulo $b^2$ holds exactly when $u=v$. The negative congruence never holds, since $0<u+v<b^2$. Modulo $b$, the $b-1$ positive pairs $u=v$ and the $b-1$ negative pairs $u+v=b$ cancel. Hence $$\sum_{\chi\in\mathcal P_b}|T_b(\chi)|^2 =\frac{b(b-1)^2}{2}=bN_b.$$

For an odd modulus $q$, the odd projector gives the complete Bernoulli square sum $$\sum_{\substack{\chi\bmod q\\\chi(-1)=-1}} \left|\frac1q\sum_{a\in\mathcal{U}_q}a\overline\chi(a)\right|^2 =\frac{\varphi(q)}{2q^2} \sum_{a\in\mathcal{U}_q}(2a^2-qa),$$ where $\mathcal{U}_q$ uses the representatives $1\le a<q$ coprime to $q$. At $q=b^2$, subtracting the multiples of $b$ from the full power sums yields $$\sum_{a\in\mathcal{U}_{b^2}}(2a^2-b^2a) =\frac{b^2(b-1)(b^3-2)}6.$$ The full odd-character Bernoulli square sum is therefore $$\frac{(b-1)^2(b^3-2)}{12b}.$$ If $\chi$ is lifted from an odd character $\psi$ modulo $b$, then $$\frac1{b^2}\sum_{r=1}^{b-1}\sum_{t=0}^{b-1} (r+bt)\overline\psi(r) =\frac1b\sum_{r=1}^{b-1}r\overline\psi(r).$$ Here the constant term vanishes because $\sum_r\psi(r)=0$. The imprimitive contribution is consequently the odd-character square sum at modulus $b$, namely $$\frac{(b-1)^2(b-2)}{12b}.$$ Subtracting gives $$\sum_{\chi\in\mathcal P_b}|B_b(\chi)|^2 =\frac{(b-1)^3(b+1)}{12}.$$ Divide by $N_b$ and apply (1). ◻

Proof. If such a constant existed, Proposition 3 and (2) would give $$c^2=\frac{\mathbf E_b|D_7|^2}{\mathbf E_b|B_7|^2} =\frac{28}{8}=\frac72.$$ The residue $3$ generates the units modulo $49$. Choose the character with $\overline\chi(3)=\zeta$, where $\zeta$ is a primitive fourteenth root of unity. It is odd, and it is primitive modulo $49$. Grouping the powers of $3$ by their exponents modulo $14$ gives $$T_7(\chi)=Q(\zeta),\qquad B_7(\chi)=H(\zeta),$$ where $$Q(z)=1+2z+z^{10}+z^{12}+z^{13}, \qquad H(z)=\sum_{j=3}^{9}z^j.$$ The required proportionality would imply $8|Q(\zeta)|^2-7|H(\zeta)|^2=0$. Use $\overline\zeta=\zeta^{-1}$ and reduce modulo $\Phi_{14}(z)=z^6-z^5+z^4-z^3+z^2-z+1$. Direct multiplication gives $$8|Q(\zeta)|^2-7|H(\zeta)|^2 =4\bigl(3-2\zeta^2+3\zeta^3-3\zeta^4+2\zeta^5\bigr).$$ The polynomial in parentheses is nonzero and has degree five. It cannot vanish at a primitive fourteenth root, whose minimal polynomial has degree six. This contradicts the proposed constant. ◻

Proof. With normalized Fourier coefficients, the spectrum factorization and Parseval identity of [5] give $$\widehat S_b^\circ(\chi) =-\frac{B_b(\chi)\overline{D_b(\chi)}}{b(b-1)}, \qquad \sum_{\chi\in\mathcal P_b}|\widehat S_b^\circ(\chi)|^2 =\frac{E_b}{b(b-1)}.$$ All omitted coefficients vanish by reflection or fiber centering. Using (1) and (2), we obtain $$\sum_{\chi\in\mathcal P_b}|T_b(\chi)|^2|L(1,\chi)|^2 =\frac{\pi^2(b-1)}{4b}E_b.$$ Division by $bN_b$ proves (7). Proposition 3 then gives (8). ◻

Proof. The cubic law [6] gives $E_b=b^3+O(b^2(\log b)^2)$. The secondary-term theorem [7] sharpens this to $$E_b=b^3-\frac1{\pi^2}b^2(\log b)^2 +O\!\left(\frac{b^2(\log(2b))^2}{\log\log(3b)}\right).$$ Insert this formula into (7) and (8). Replacing $b-1$ by $b$ in the lower-order denominators changes the answers by quantities absorbed in the stated errors. Finally, $$\operatorname{Cov}_b(X_b^2,Y_b^2)=\mathbf E_b(X_b^2Y_b^2)-\nu_b,$$ and $\nu_b\to\pi^2/6$. ◻

Proof. Hölder’s inequality with exponents $3/2$ and $3$ gives $$d_b\le H_b^{2/3}M^{1/3}.$$ The monotonicity of probability-space moments gives $a_b\le H_b^{2/3}$, and Cauchy–Schwarz gives $c_b\le\sqrt{\nu_b}$. Therefore $$|d_b-a_bc_b| \le (M^{1/3}+\sqrt{\nu_b})H_b^{2/3}.$$ Also $1-a_b^2\ge1-H_b^{4/3}$, while the assumed lower bound controls the other variance. Substitution into (11) proves (13). Its right side tends to zero under (12). ◻

Proof. The first identity follows from Lemma 2 with $u=2$ and $v=1$. Neither $2\equiv1$ nor $2\equiv-1$ holds modulo $b$ or $b^2$.

For an odd modulus $q$, write $[2c]_q$ for the representative of $2c$ in $\{1,\ldots,q-1\}$ and put $$J(q)=\sum_{c=1}^{q-1}c\bigl(2[2c]_q-q\bigr).$$ Splitting at $(q-1)/2$ and summing the first two powers gives $$\begin{aligned} J(q) &=4\sum_{c=1}^{q-1}c^2-q\sum_{c=1}^{q-1}c -2q\sum_{c=(q+1)/2}^{q-1}c\\ &=\frac{q(q-1)(q-5)}{12}. \end{aligned}$$ Odd-character orthogonality gives $$\begin{aligned} &\sum_{\substack{\psi\bmod q\\\psi(-1)=-1}} \left|\frac1q\sum_{a\in\mathcal{U}_q}a\overline\psi(a)\right|^2\psi(2)\\ &\hspace{15mm}= \frac{\varphi(q)}{2q^2} \sum_{c\in\mathcal{U}_q}c\bigl(2[2c]_q-q\bigr). \end{aligned}$$ At modulus $b^2$, removing the multiples of $b$ changes the inner sum to $$J(b^2)-b^2J(b)=\frac{b^2(b-1)(b^3-5)}{12}.$$ Thus the full odd-character twisted Bernoulli moment is $$\frac{(b-1)^2(b^3-5)}{24b}.$$ The lifted odd characters contribute the corresponding moment modulo $b$. The Bernoulli sum is unchanged by lifting, as established in Proposition 3, and the value at $2$ is unchanged as well. This contribution is $$\frac{(b-1)^2(b-5)}{24b}.$$ Subtracting and dividing by $N_b$ gives $$\mathbf E_b\bigl(|B_b(\chi)|^2\chi(2)\bigr)=\frac{b^2-1}{12}.$$ Multiplication by $\pi^2/b^2$ proves the second identity. ◻

Proof. Put $Z(\chi)=\Re\chi(2)$ and $c=\mathbf E_bY_b$. Then $|Z|\le1$, $\mathbf E_bZ=0$, and Proposition 8 gives $$\frac{\nu_b}{2} =\mathbf E_b\bigl((Y_b-c)(Y_b+c)Z\bigr).$$ By Cauchy–Schwarz, $$\begin{aligned} \frac{\nu_b}{2} &\le\sqrt{\mathop{\mathrm{Var}}_b(Y_b)} \sqrt{\mathbf E_b((Y_b+c)^2Z^2)}\\ &\le\sqrt{\mathop{\mathrm{Var}}_b(Y_b)}\sqrt{\nu_b+3c^2} \le2\sqrt{\nu_b\mathop{\mathrm{Var}}_b(Y_b)}. \end{aligned}$$ Squaring proves the first bound. The second follows by inserting $b\ge5$ in the expression for $\nu_b$. ◻

Proof. Put $q=b^2$. For every nonprincipal character modulo $q$, periodicity and the vanishing of a complete-period sum give $$\left|\sum_{n\le t}\chi(n)\right|\le q.$$ Partial summation therefore gives, uniformly in such characters, $$L(1,\chi)=A_q(\chi)+O(q^{-1}), \qquad A_q(\chi)=\sum_{n\le q^2}\frac{\chi(n)}n.$$ Indeed, the tail after an integer $T$ has magnitude at most $2q/T$.

Write $$A_q(\chi)^2=\sum_{u\le q^4}c_u\chi(u),$$ where $$0\le c_u=\frac{\#\{n_1n_2=u\mid n_1,n_2\le q^2\}}u \le\frac{d(u)}u.$$ Here $d(u)$ is the divisor function. Averaging over all characters modulo $q$ and using orthogonality yields $$\frac1{\varphi(q)}\sum_{\chi\bmod q}|A_q(\chi)|^4 =\sum_{\substack{u,v\le q^4\\u\equiv v\;(q)\\(uv,q)=1}}c_uc_v.$$ On the diagonal, the sum is at most $\sum_{u\ge1}d(u)^2/u^2<\infty$. For the off-diagonal contribution, use $d(u)\ll_\varepsilon u^\varepsilon$ with $\varepsilon=1/16$. This elementary bound follows from the prime-factorization formula for $d(u)$ by separating the finitely many small primes. Since $u,v\le q^4$, their two divisor factors have product $O(q^{1/2})$. Consequently $$\begin{aligned} \sum_{\substack{u,v\le q^4\\u\equiv v\;(q)\\u\ne v}}c_uc_v &\ll q^{1/2}\sum_{u\le q^4}\frac1u \sum_{1\le k\le q^3}\frac1{u+kq}\\ &\ll q^{-1/2}(\log(2q))^2. \end{aligned}$$ Thus the full-character fourth moment of $A_q$ is bounded.

The primitive odd family has proportion $(b-1)/(2b)$ among all characters modulo $b^2$. Positivity gives $$\mathbf E_b|A_q|^4 \le\frac{2b}{b-1} \frac1{\varphi(q)}\sum_{\chi\bmod q}|A_q(\chi)|^4 \ll1.$$ Every character in this family is nonprincipal, so (16) proves the fourth-moment bound for $Y_b$. Hölder’s inequality gives the third-moment bound. ◻

Proof. Set $m=b^2$, $x=b-1$, and denote the average over all characters modulo $m$ by $$\mathbf E_m^{\rm all}W =\frac1{\varphi(m)}\sum_{\chi\bmod m}W(\chi).$$ For units $u,v$ with $1\le u,v<m$, orthogonality gives $$\mathbf E_m^{\rm all}\chi(u)\overline\chi(v) =\mathbf1_{u=v} =\mathbf E f(u)\overline{f(v)},$$ where $f$ is a Steinhaus random multiplicative function. Its values at primes are independent and uniform on the unit circle, and its other values are obtained multiplicatively.

The unit restriction is the only change in the exact comparison used in [1]. To verify it, retain the notation for the short prime cutoff $P$ and use Harper’s parameters with moment exponent $3/4$. Choose $P$ as in [1], including the harmless rounding condition that $(\log P)^{0.01}$ is an integer, and put $$\begin{aligned} \log P&\asymp(\log x)^{1/6}, &K&=2(\log P)^{1.02},\\ J&=\lceil1.2\log\log P\rceil, &\delta&=(\log P)^{-1.3}. \end{aligned}$$ Here $K$ indexes the conditioning frequencies, and $J$ is the range parameter in the smooth partition of unity. Write $\ell=\log P$ and $A=2K+1$ for the number of partition factors. The required comparison condition is $$xP^{400(A/\delta)^2\log(J\ell)}<m.$$ The logarithm of the factor multiplying $x$ is $$O\bigl(\ell^{5.64}\log\ell\bigr) =O\bigl((\log x)^{0.94}\log\log x\bigr) =o(\log x).$$ Since $m/x\asymp x$, condition (19) holds for all sufficiently large $b$.

To check the full products in Harper’s Proposition 1, take its Taylor parameter $$S=100A\left\lfloor\delta^{-2}\log(J\ell)\right\rfloor.$$ Each partition function is approximated by a polynomial of degree $2S-1$. Its argument is a prime-and-prime-square sum, so each occurrence contributes an integer at most $P^2$. Expanding all $A$ polynomials and the accompanying short-sum square gives character pairings $\chi(u)\overline\chi(v)$ with $$u,v\le xP^{2A(2S-1)}<xP^{4SA}, \qquad 4SA\le400(A/\delta)^2\log(J\ell).$$ Thus (19) controls each entire product across all partition factors.

The same condition controls the Taylor remainders. A product of $j\le A$ remainder majorants reduces by Hölder’s inequality to even moments of order $2Sj$ of the prime-and-prime-square sums. In Harper’s Lemma 1, the integers on each side of the corresponding pairing are bounded by $$x(P^2)^{Sj}\le xP^{2SA}<xP^{4SA}<m.$$ Every factor is either an integer $n\le x=b-1$ or a prime at most $P<b$. Hence all the products are units modulo $b^2$, and (18) applies term by term to both the polynomial comparisons and the remainder moments. The smooth-divisor bound used in that lemma is unchanged. If $\widetilde d(n)$ counts the $P$-smooth divisors of $n$, then $$\sum_{n\le x}\widetilde d(n) \le x\prod_{p\le P}(1-p^{-1})^{-1}\ll x\ell.$$ Taking $x=1$ in the comparison gives the partition-weight estimate of Harper’s Proposition 2.

The removal of smooth integers in [1] also uses only a second moment on integers at most $x$. Set $y=x^{1/\log\log x}$. These integers are units, and the contribution of the $y$-smooth terms to the $3/2$ moment is at most $\Psi(x,y)^{3/4}$, where $\Psi(x,y)$ counts $y$-smooth integers up to $x$. Harper’s smooth-number estimate makes this $o(x^{3/4}(\log\log x)^{-3/8})$. Also $P<y<b$ eventually.

It remains to check that the comparison errors stay small after summing all conditioning boxes, as in [1]. Put $$\alpha=\frac34,\qquad \eta=(J\ell)^{-A\delta^{-2}},\qquad \mathcal B=(2J+2)^A.$$ The polynomial comparison error is $O(x\eta)$ per box, and the partition-weight error is $O(\eta)$. The two accumulated losses are bounded by constant multiples of $$x^\alpha(\mathcal B\eta)^\alpha \quad\hbox{and}\quad x^\alpha(\mathcal B\eta)^{1-\alpha}.$$ Both are $o(x^{3/4}(\log\log x)^{-3/8})$, since $$\log(\mathcal B\eta) =A\{\log(2J+2)-\delta^{-2}\log(J\ell)\} \le-\tfrac12 A\delta^{-2}\log(J\ell)$$ for all sufficiently large $b$. In the weighted conditioning inequalities the partition weights sum to one. Keeping their powers inside Hölder’s inequality gives the displayed losses without requiring a positive lower bound for an individual weight.

Finally, the strongest precision restriction in [1] is $$\delta\le\frac1{J\ell^{1.2}\sqrt{\log\ell}}.$$ It holds because $\delta J\ell^{1.2}\sqrt{\log\ell} =O(\ell^{-0.1}(\log\ell)^{3/2})\to0$. The weaker precision restriction follows, and $J\ge1.2\log\ell$ by construction. The frequency range $|k|\le K=2\ell^{1.02}$ covers the translated windows used in the random stage.

The comparisons yield Harper’s equation (3.2), a random conditional expectation at exponent $\alpha=3/4$, with negligible errors. Sections 3.4–3.6 then contain no character modulus. Their random Euler-product estimates apply without a further support restriction. Hence $$\mathbf E_m^{\rm all} \left|\sum_{n\le x}\chi(n)\right|^{3/2} \ll\left(\frac{x}{1+\tfrac14\sqrt{\log\log P}}\right)^{3/4} \ll x^{3/4}(\log\log x)^{-3/8}.$$ Finally, positivity permits restriction to the primitive odd family, $$\mathbf E_b\left|\sum_{n<b}\chi(n)\right|^{3/2} \le\frac{2b}{b-1} \mathbf E_m^{\rm all} \left|\sum_{n<b}\chi(n)\right|^{3/2}.$$ Divide by $b^{3/4}$. Complex conjugation does not change the magnitude of the short sum, so this proves (17) for the normalization of $X_b$. ◻

Proof of Theorem 1. We have $H_b\ll(\log\log b)^{-3/8}$ by Proposition 11. Proposition 10 supplies an absolute upper bound for $\mathbf E_bY_b^3$, and Corollary 9 supplies the lower bound $\mathop{\mathrm{Var}}_b(Y_b)\ge\pi^2/100$. Also $$\mathbf E_bX_b\le H_b^{2/3}\ll(\log\log b)^{-1/4},$$ so $\mathop{\mathrm{Var}}_b(X_b)=1-(\mathbf E_bX_b)^2\to1$. Both variances are therefore positive for all sufficiently large $b$. Proposition 7 now gives $$|r_b|\ll H_b^{2/3}\ll(\log\log b)^{-1/4}.$$ ◻

Proof. Markov’s inequality bounds the displayed proportion by $\varepsilon^{-3/2}H_b$, which tends to zero. On $X_b\le\varepsilon$, $$\mathbf E_b\bigl(X_b^2\mathbf1_{X_b\le\varepsilon}\bigr) \le\varepsilon^{1/2}H_b\longrightarrow0.$$ Subtract from $\mathbf E_bX_b^2=1$. The final two limits are Corollary 6. ◻