Proof. For a proof of this result, see [11]. ◻

Proof. For the proof of this result, please see [14]. The proof relies on a theorem of Macdonald [15] relating \(v_\ell(\mathfrak{f}_\lambda)\) to the \(\ell\)-core tower of \(\pi_\lambda\). ◻

Proof. This is the principal result of loc.cit. We sketch the argument for the convenience of the reader.

Fix a prime \(q\). By the identity 3 , one has the lower bound \[v_q\!\left(\chi_\mu^\lambda\right) \ge v_q\!\left(\mathfrak{f}_\lambda\right) - v_q\!\left((n)_k\right).\] Applying Legendre’s formula, \[v_q(n!) = \frac{n - a_q(n)}{q-1},\] where \(a_q(n)\) denotes the sum of the base-\(q\) digits of \(n\), we obtain \[v_q\!\left((n)_k\right) = v_q\!\left(\frac{n!}{(n-k)!}\right) = \frac{k + a_q(n-k) - a_q(n)}{q-1}.\] Since \(a_q(n) \le (q-1)\log_q n\), it follows that \[v_q\!\left((n)_k\right) \le k + (q-1)\log_q n.\] Consequently, if \[v_q\!\left(\mathfrak{f}_\lambda\right) \ge m + (q-1)\log_q n,\] then \[v_q\!\left(\chi_\mu^\lambda\right) \ge m - k.\] Taking \(m = k + b\) and invoking Theorem 2, we deduce that \[\lim_{n \to \infty} \frac{ \#\left\{\lambda \vdash n \; \middle| \; v_q\!\left(\chi_\mu^\lambda\right) \le b \right\} }{p(n)} = 0.\] Since this holds for each prime divisor \(q\) of \(d\), the stated result follows. ◻

Proof. This result is well known and follows from how they are defined using Young tableaux. ◻

Proof. This result is [6]. ◻

Proof. The result is probably well known and follows from the construction of Deligne and Lusztig. Since we have been unable to find a reference, we give a brief sketch of a proof here following the notation and conventions in [9].

Let \(g\in G_n\) and let \[\alpha_1,\dots,\alpha_n\in\overline{\mathbb{F}}_q^\times\] be the eigenvalues of \(g\). Then the eigenvalues of \(g^{-1}\) are \(\alpha_1^{-1},\dots,\alpha_n^{-1}\). From Green’s definition [9] of the functions \[T_{1,d}(k;\alpha) = \theta_d^{k}(\alpha) + \theta_d^{kq}(\alpha) +\cdots+ \theta_d^{kq^{d-1}}(\alpha),\] one immediately obtains \[T_{1,d}(k;\alpha^{-1})=T_{1,d}(-k;\alpha).\] Consequently the class functions \(J_d(k)\) defined by \[J_d(k)(g) = \sum T_{1,d}(k;\alpha_{i_1})\cdots T_{1,d}(k;\alpha_{i_d})\] satisfy \[J_d(k)(g^{-1})=J_d(-k)(g).\] Since the basic characters \(B^\rho(h^\rho;\cdot)\) are obtained from the functions \(J_d(k)\) by Green’s induction product, it follows that \[B^\rho(h^\rho;g^{-1})=B^\rho((-h)^\rho;g),\] i.e. inversion replaces each parameter \(h_{d,i}\) by \(-h_{d,i}\).

Theorem 14 in loc. cit. expresses the symmetric–function class function \((\cdots g^{\mu(g)}\cdots)\) as a linear combination of the basic characters \(B^\rho(h^\rho;\cdot)\). Replacing \(\mu\) by \(\mu'\) with \(\mu'(f)=\mu(f^*)\) corresponds to replacing every root \(\alpha\) of \(f\) by \(\alpha^{-1}\), hence to replacing the parameters \(h\) by \(-h\) in the expansion. Comparing the resulting formulas shows that \[\chi_\mu(g^{-1})=\chi_{\mu'}(g) \qquad\text{for all }g\in G_n.\] Since the character of the dual representation satisfies \(\chi_{V^\vee}(g)=\chi_V(g^{-1})\), this implies \[(\chi_\mu)^\vee=\chi_{\mu'}.\] The final assertion is immediate from the definition of \(\mu'\). ◻

Proof. By [18], an irreducible representation of \(\operatorname{GL}_n(\mathbb{F}_q)\) is orthogonal if and only if it is self-dual and has Frobenius–Schur indicator \(+1\). For irreducible characters of \(\operatorname{GL}_n(\mathbb{F}_q)\), self-duality already forces the indicator to be \(+1\), and the claim follows. ◻

Proof. Since \(g^2=1\), the linear operator \(\pi(g)\) satisfies \(\pi(g)^2=I\). It follows that \(\pi(g)\) is diagonalizable over \(\mathbb{R}\), with eigenvalues contained in \(\{\pm 1\}\). Let \(V\) be the underlying real representation space of \(\pi\), and write \[V = V^{+} \oplus V^{-}\] for the decomposition into the \(+1\) and \(-1\) eigenspaces of \(\pi(g)\). Set \(d_+ = \dim V^{+}\) and \(d_- = \dim V^{-}\), then \[\dim \pi = d_+ + d_-.\] On the other hand, the character value is given by the trace: \[\chi_\pi(g) = \operatorname{tr}(\pi(g)) = d_+ - d_-.\] Solving these two equations for \(d_-\), we obtain \[d_- = \frac{\dim \pi - \chi_\pi(g)}{2}.\] We thus find that \[\det(\pi)(g) = (-1)^{d_-} = (-1)^{\frac{\dim \pi - \chi_\pi(g)}{2}},\] as claimed. ◻

Proof. The result follows from [4]. ◻

Proof. The above result is [19]. ◻

Proof. For further details, see [8]. ◻

Proof. This result is [8]. ◻

Proof. Fix \(\varepsilon>0\). By Theorem 2, there exists \(M=M(\varepsilon)\) such that for all \(m \ge M\), \[\frac{\#\{\lambda \vdash m : v_2(\mathfrak{f}_\lambda) < k + \log_2(k q^{k+1}) + \log_2 m\}}{p(m)} < \varepsilon.\] Set \(N = k q^{k+1} M\), and let \(n \ge N\). Fix \(F \in \mathscr{G}_n(\Phi)\), so that \(F(f)=F(f^*)\) for all \(f \in \Phi\). We consider the set \[\{\boldsymbol{\mu} \in \mathfrak{Y}_n : \boldsymbol{\mu} \mapsto F\}.\] Such \(\boldsymbol{\mu}\) are determined by choosing, for each orbit \(\{f,f^*\}\), a partition \(\lambda_f \vdash F(f)\), and setting \(\boldsymbol{\mu}(f)=\boldsymbol{\mu}(f^*)=\lambda_f\). Thus there is a bijection \[\{\boldsymbol{\mu} \in \mathfrak{Y}_n : \boldsymbol{\mu} \mapsto F\} \;\cong\; \prod_{\{f,f^*\}} \Lambda_{F(f)}.\]

We now consider two cases. First suppose that \(\max_{f \in \operatorname{supp}(F)} d(f) \le k\). Since there are at most \(q^{k+1}\) monic polynomials of degree at most \(k\), the number of orbits \(\{f,f^*\}\) in \(\operatorname{supp}(F)\) is at most \(q^{k+1}\). Therefore, \[\frac{1}{|\operatorname{supp}(F)|} \sum_{f \in \Phi} F(f) \;\ge\; \frac{\sum_{f \in \Phi} d(f) F(f)}{k q^{k+1}} = \frac{n}{k q^{k+1}} \;\ge\; M.\] Hence there exists \(f_0 \in \operatorname{supp}(F)\) such that \(m := F(f_0) \ge M\).

For any \(\boldsymbol{\mu} \mapsto F\), we have \[v_2(d_{\boldsymbol{\mu}}) \ge v_2\bigl(\mathfrak{f}_{\boldsymbol{\mu}(f_0)}\bigr),\] since \(d_{\boldsymbol{\mu}}\) contains the factor \(d_{\boldsymbol{\mu}(f_0)}(q^{d(f_0)})\), whose \(2\)-adic valuation is bounded below by that of \(\mathfrak{f}_{\boldsymbol{\mu}(f_0)}\).

It follows that \[\begin{align} &\frac{\# \{\boldsymbol{\mu}\in \mathfrak{Y}_n : \boldsymbol{\mu}\mapsto F,\;v_2(d_{\boldsymbol{\mu}})< v_2((n)_k)\}}{\# \{\boldsymbol{\mu}\in \mathfrak{Y}_n : \boldsymbol{\mu}\mapsto F\}} \\ &\qquad\le \frac{\#\{\lambda \vdash m : v_2(\mathfrak{f}_\lambda) < v_2((n)_k)\}}{p(m)}. \end{align}\]

Now observe that \[v_2((n)_k) = \sum_{i=0}^{k-1} v_2(n-i) \;\ge\; \log_2(n-k+1),\] so for \(n \ge k q^{k+1} M\) we have \[v_2((n)_k) \ge k + \log_2(k q^{k+1}) + \log_2 m.\] Therefore, \[\frac{\#\{\lambda \vdash m : v_2(\mathfrak{f}_\lambda) < v_2((n)_k)\}}{p(m)} < \varepsilon,\] by the choice of \(M\).

Next suppose that \(\max_{f \in \operatorname{supp}(F)} d(f) > k\). Let \(f_1 \in \operatorname{supp}(F)\) have maximal degree. Then \[\begin{align} n &= \sum_{f \in \Phi} d(f) F(f) \\ &> k F(f_1) + \sum_{f \ne f_1} F(f) \\ &= (k-1)F(f_1) + \sum_{f \in \Phi} F(f). \end{align}\] Since \(F(f_1)\ge 1\), it follows that \[n-k \ge \sum_{f \in \Phi} F(f).\] Thus \(\frac{n!}{\prod_f F(f)!}\) is divisible by \((n)_k\), and hence for any \(\boldsymbol{\mu} \mapsto F\) we have \[v_2(d_{\boldsymbol{\mu}}) \ge v_2\left(\frac{n!}{\prod_f F(f)!}\right) \ge v_2((n)_k).\] Therefore, \[\frac{\# \{\boldsymbol{\mu}\in \mathfrak{Y}_n : \boldsymbol{\mu}\mapsto F,\;v_2(d_{\boldsymbol{\mu}})< v_2((n)_k)\}}{\# \{\boldsymbol{\mu}\in \mathfrak{Y}_n : \boldsymbol{\mu}\mapsto F\}} = 0.\] Combining the two cases completes the proof. ◻

Proof. Fix \(\varepsilon>0\). Let \(N=N(\varepsilon)\) be as in Lemma 3. For all \(n \ge N\), we decompose according to the map \(\boldsymbol{\mu} \mapsto F\): \[\frac{ \#\{\boldsymbol{\mu} \in \mathfrak{Y}_n : v(d_{\boldsymbol{\mu}}) < v((n)_k)\} }{|\mathfrak{Y}_n|} = \sum_{F \in \mathscr{G}_n(\Phi)} \frac{ \#\{\boldsymbol{\mu} \in \mathfrak{Y}_n : \boldsymbol{\mu} \mapsto F,\;v(d_{\boldsymbol{\mu}}) < v((n)_k)\} }{|\mathfrak{Y}_n|}.\] For each \(F\), we write \[\begin{align}&\frac{ \#\{\boldsymbol{\mu} \in \mathfrak{Y}_n : \boldsymbol{\mu} \mapsto F,\;v(d_{\boldsymbol{\mu}}) < v((n)_k)\} }{|\mathfrak{Y}_n|}\\ = & \frac{ \#\{\boldsymbol{\mu} \in \mathfrak{Y}_n : \boldsymbol{\mu} \mapsto F,\;v(d_{\boldsymbol{\mu}}) < v((n)_k)\} }{\#\{\boldsymbol{\mu} \in \mathfrak{Y}_n : \boldsymbol{\mu} \mapsto F\}} \cdot \frac{ \#\{\boldsymbol{\mu} \in \mathfrak{Y}_n : \boldsymbol{\mu} \mapsto F\} }{|\mathfrak{Y}_n|}. \end{align}\] By Lemma 3, the first factor is \(<\varepsilon\) for all \(F \in \mathscr{G}_n(\Phi)\), provided \(n \ge N\). Hence \[\frac{ \#\{\boldsymbol{\mu} \in \mathfrak{Y}_n : v(d_{\boldsymbol{\mu}}) < v((n)_k)\} }{|\mathfrak{Y}_n|} < \varepsilon \sum_{F \in \mathscr{G}_n(\Phi)} \frac{ \#\{\boldsymbol{\mu} \in \mathfrak{Y}_n : \boldsymbol{\mu} \mapsto F\} }{|\mathfrak{Y}_n|} = \varepsilon.\] Since \(\varepsilon>0\) is arbitrary, the result follows. ◻

Proof. Let \(g \in G_{n_0}\) be fixed. From Lemma 2, we have that \[\frac{\chi_{\boldsymbol{\mu}}(g)}{d_{\boldsymbol{\mu}}} \cdot [G_n : G_{n-n_0}]\] is an algebraic integer. It follows that if \[v_2(d_{\boldsymbol{\mu}}) \;\ge\; r + v_2\!\left(\prod_{i=0}^{n_0-1} (q^{n-i}-1)\right),\] then \(2^r \mid \chi_{\boldsymbol{\mu}}(g)\). Set \[r_0 := r + n_0\, v(q^2 - 1).\] As in the proof of [8], we have that \[\{\boldsymbol{\mu} \in \mathfrak{Y}_n : 2^r \nmid \chi_{\boldsymbol{\mu}}(g)\} \subset \{\boldsymbol{\mu} \in \mathfrak{Y}_n : v(d_{\boldsymbol{\mu}}) < v((n)_{n_0 + 2r_0})\}.\] Applying Proposition 14 with \(k = n_0 + 2r_0\), we obtain \[\lim_{n\to\infty} \frac{ \#\{\boldsymbol{\mu} \in \mathfrak{Y}_n : 2^r \nmid \chi_{\boldsymbol{\mu}}(g)\} }{|\mathfrak{Y}_n|} = 0,\] as required. ◻

Proof. The result follows directly from 4 , Theorem 10 and Proposition 15. ◻

Proof. Recall that \[\mathrm{Ind}_H^G V = \{ f: G \to V \mid f(hg)=\sigma(h)f(g)\;\text{for all } h\in H,\, g\in G \},\] has \(G\)-action given by \[(g\cdot f)(x)=f(xg), \quad\text{for}\quad g,x\in G.\] Similarly, \[\mathrm{Ind}_H^G V^\vee = \{ \lambda: G \to V^\vee \mid \lambda(hg)=\sigma^\vee(h)\lambda(g)\;\text{for all } h\in H,\, g\in G \},\] with the same right translation action.

Define a pairing \[\langle\cdot,\cdot\rangle : \mathrm{Ind}_H^G V \times \mathrm{Ind}_H^G V^\vee \longrightarrow \mathbb{C}\] by \[\langle f,\lambda\rangle := \sum_{x \in H\backslash G} \lambda(x)\bigl(f(x)\bigr).\] For \(g\in G\), \[\langle g\cdot f,\lambda\rangle = \sum_{x\in H\backslash G} \lambda(x)\bigl(f(xg)\bigr).\] Making the change of variables \(y=xg\), which permutes the cosets \(H\backslash G\), we obtain \[\langle g\cdot f,\lambda\rangle = \sum_{y\in H\backslash G} \lambda(yg^{-1})\bigl(f(y)\bigr) = \langle f, g^{-1}\cdot \lambda\rangle,\] since \((g^{-1}\cdot \lambda)(y)=\lambda(yg^{-1})\). Thus the pairing is \(G\)-invariant.

This pairing is clearly bilinear, and nondegenerate. If \(f\neq 0\), then there exists \(x\) such that \(f(x)\neq 0\), and one may choose \(\lambda\) supported on the coset \(Hx\) so that \(\lambda(x)(f(x))\neq 0\). Thus, \(\langle f, \lambda\rangle \neq 0\). Similarly, if \(\lambda\neq 0\), one may choose \(f\) such that \(\langle f, \lambda\rangle \neq 0\).

Therefore the pairing identifies \(\mathrm{Ind}_H^G V^\vee\) with \((\mathrm{Ind}_H^G V)^\vee\) as \(G\)-representations, yielding the desired isomorphism \[(\mathrm{Ind}_H^G \sigma)^\vee \simeq \mathrm{Ind}_H^G(\sigma^\vee).\] ◻

Proof. Let \(B\subset G\) be the Borel subgroup of upper triangular matrices, and consider the (unnormalized) induced representation \[\pi := \mathrm{Ind}_B^G \mathbf{1}.\] It is well known that \(\pi\) contains the trivial representation \(\mathbf{1}\) as a subrepresentation (via constant functions), and that the Steinberg representation is defined as the quotient \[\mathrm{St}:= \pi / \mathbf{1}.\] We first observe that \(\pi\) is self-dual. Indeed, by Lemma 4, \[\pi^\vee = (\mathrm{Ind}_B^G \mathbf{1})^\vee \simeq \mathrm{Ind}_B^G(\mathbf{1}) = \pi.\] Realize \(\pi\) as \[\pi = \{ f: G \to \mathbb{C} \mid f(bg)=f(g)\;\text{for all } b\in B \}.\] Define \[\langle f_1, f_2 \rangle := \sum_{x \in B\backslash G} f_1(x) f_2(x).\] As in the proof of Lemma 4, \(\pi \simeq \pi^\vee\) via this pairing.

Consider the subrepresentation \(\mathbf{1} \subset \pi\) consisting of constant functions. We have that \[\mathbf{1}^\perp = \left\{ f \in \pi \;\middle|\; \sum_{x\in B\backslash G} f(x)=0 \right\}.\] This subspace has codimension one, hence \[\mathbf{1}^\perp \simeq \mathrm{St}.\] Since the pairing on \(\pi\) is nondegenerate and \(G\)-invariant, it induces a nondegenerate \(G\)-invariant pairing on \(\mathbf{1}^\perp\). Therefore \(\mathbf{1}^\perp\) is self-dual, and thus, \(\mathrm{St}^\vee \simeq \mathrm{St}\), as claimed. ◻

Proof. We dualize the defining expression term by term. For any subgroup \(H\subset G_q\) and any finite-dimensional representation \(\sigma\) of \(H\), we have \[(\mathrm{Ind}_H^{G_q} \sigma)^\vee \simeq \mathrm{Ind}_H^{G_q}(\sigma^\vee).\] Applying this to each term gives \[\pi_\theta^\vee = \mathrm{Ind}_{ZN}^{G_q}(\theta_\psi^\vee) - \mathrm{Ind}_E^{G_q}(\theta^\vee).\] Since \(\theta\) is one-dimensional, \[\theta^\vee = \theta^{-1}.\] Thus \[(\mathrm{Ind}_E^{G_q} \theta)^\vee \simeq \mathrm{Ind}_E^{G_q}(\theta^{-1}).\] Recall that \(\theta_\psi\) is defined on \(ZN\) by \[\theta_\psi(zn) = \theta(z)\,\psi(n),\] where \(\psi:N\to\mathbb{C}^\times\) is a nontrivial additive character (via the standard identification \(N\simeq \mathbb{F}_q\)). Thus \[\theta_\psi^\vee(zn) = \theta(z)^{-1}\,\psi(n)^{-1}.\] Since \(\psi^{-1}\) is again a nontrivial additive character of \(N\), it differs from \(\psi\) by conjugation in \(G_q\): more precisely, there exists \(g\in G_q\) such that \[\psi(n)^{-1} = \psi(gng^{-1}) \quad \text{for all } n\in N.\] It follows that \(\theta_\psi^\vee\) is conjugate (as a representation of \(ZN\)) to \((\theta^{-1})_\psi\). Therefore \[\mathrm{Ind}_{ZN}^{G_q}(\theta_\psi^\vee) \simeq \mathrm{Ind}_{ZN}^{G_q}((\theta^{-1})_\psi),\] since induction is invariant under conjugation of the inducing representation. Combining the above, we obtain \[\pi_\theta^\vee = \mathrm{Ind}_{ZN}^{G_q}((\theta^{-1})_\psi) - \mathrm{Ind}_E^{G_q}(\theta^{-1}) = \pi_{\theta^{-1}}.\] This proves the desired isomorphism. ◻

Proof. Part (1) is immediate.

For part (2), let \(\pi = \operatorname{Ind}_B^{G_q}(\chi_1 \otimes \chi_2)\). By Lemma 4, one has \[\left(\operatorname{Ind}_B^{G_q}(\chi_1 \otimes \chi_2)\right)^\vee \simeq \operatorname{Ind}_B^{G_q}(\chi_1^{-1} \otimes \chi_2^{-1})\] The Weyl group symmetry gives \[\operatorname{Ind}_B^{G_q}(\chi_1 \otimes \chi_2) \simeq \operatorname{Ind}_B^{G_q}(\chi_2 \otimes \chi_1).\] It follows that \(\pi \simeq \pi^\vee\) if and only if the unordered pair \(\{\chi_1,\chi_2\}\) is invariant under inversion, i.e.\(\{\chi_1,\chi_2\}=\{\chi_1^{-1},\chi_2^{-1}\}\).

For part (3), Proposition 4 asserts that \(\mathrm{St}^\vee \simeq \mathrm{St}\). Hence, \[(\mathrm{St}\otimes (\psi \circ \det))^\vee \simeq \mathrm{St}\otimes (\psi^{-1} \circ \det).\]

Part (4) follows from Lemma 5. ◻

Proof. Let \(\{e_i\}\) be the standard basis of \(\mathrm{Ind}_H^G \sigma\) corresponding to the cosets \(Hx_i\). By definition of induction, \[g\cdot e_i = \sigma\bigl(h_i(g)\bigr)\, e_{\rho(i)}.\] Thus the matrix of \(\pi(g)\) is a permutation matrix corresponding to \(\rho\), with nonzero entries \(\sigma(h_i(g))\) in positions \((i,\rho(i))\). The determinant of such a matrix is the sign of the permutation multiplied by the product of the nonzero entries, hence \[\det(\pi)(g) = \operatorname{sgn}(\rho)\prod_i \sigma\bigl(h_i(g)\bigr),\] as claimed. ◻

Proof. The action of \(G_q\) on \(\mathbb{P}^1(\mathbb{F}_q)\) factors through \(\operatorname{PGL}_2(\mathbb{F}_q)\), since scalar matrices act trivially. Thus the map \[g \longmapsto \operatorname{sgn}(\rho_{\mathbb{P}^1}(g))\] factors through \(\operatorname{PGL}_2(\mathbb{F}_q)\) and defines a group homomorphism \[\operatorname{PGL}_2(\mathbb{F}_q)\longrightarrow \{\pm1\}.\]

For \(q\) odd, \(\operatorname{PSL}_2(\mathbb{F}_q)\) is a normal subgroup of index \(2\) in \(\operatorname{PGL}_2(\mathbb{F}_q)\), and one has a canonical identification \[\operatorname{PGL}_2(\mathbb{F}_q)/\operatorname{PSL}_2(\mathbb{F}_q)\;\simeq\; \mathbb{F}_q^\times/(\mathbb{F}_q^\times)^2,\] via the determinant map. It follows that any homomorphism \(\operatorname{PGL}_2(\mathbb{F}_q)\to\{\pm1\}\) is determined by a quadratic character of \(\mathbb{F}_q^\times\), and hence there exists a character \(\chi:\mathbb{F}_q^\times\to\{\pm1\}\) such that \[\operatorname{sgn}(\rho_{\mathbb{P}^1}(g))=\chi(\det g).\]

It remains to identify \(\chi\). For this, consider the element \[g=\begin{pmatrix}a & 0 \\ 0 & 1\end{pmatrix}.\] Then \(g\) acts on \(\mathbb{P}^1(\mathbb{F}_q)=\mathbb{F}_q\cup\{\infty\}\) by \[x\mapsto ax, \qquad \infty\mapsto\infty.\] Thus \(\rho_{\mathbb{P}^1}(g)\) fixes \(\infty\) and permutes \(\mathbb{F}_q\) via multiplication by \(a\). Since \(0\) is also fixed, the sign of \(\rho_{\mathbb{P}^1}(g)\) is equal to the sign of the permutation of \(\mathbb{F}_q^\times\) given by \(x\mapsto ax\).

Choose a generator \(\gamma\) of the cyclic group \(\mathbb{F}_q^\times\), so that \(\mathbb{F}_q^\times=\{\gamma^k : 0\le k\le q-2\}\). Writing \(a=\gamma^m\), the map \(x\mapsto ax\) corresponds to the permutation \(k\mapsto k+m\) of \(\mathbb{Z}/(q-1)\mathbb{Z}\). This permutation is a cycle of length \((q-1)/\gcd(m,q-1)\) repeated \(\gcd(m,q-1)\) times, and its sign is \[(-1)^{(q-1)-\gcd(m,q-1)}=(-1)^{\gcd(m,q-1)}=(-1)^m=\mu(a).\] Thus conclude that \[\operatorname{sgn}(\rho_{\mathbb{P}^1}(g))=\mu(a)=\mu(\det g)\] for all such diagonal elements, and hence for all \(g\in G_q\). ◻

Proof. We compute \(\det(\pi)\) in each case.

(i) One-dimensional representations. If \(\pi=\chi\circ\det\), then \(\det(\pi)=\chi\circ\det\) by definition.

(ii) Principal series. Let \(\pi=\mathrm{Ind}_B^{G_q}(\chi_1\otimes\chi_2)\), and write \(\sigma=\chi_1\otimes\chi_2\). By the determinant formula for induced representations, fixing coset representatives \(\{x_i\}\) for \(B\backslash G_q\), we have \[\det(\pi)(g) = \operatorname{sgn}(\rho_g)\prod_i \sigma(h_i(g)), \quad\text{where } x_i g = h_i(g)\,x_{\rho_g(i)}.\] Since \(\sigma\) depends only on diagonal entries, we have \[\sigma(h_i(g)) = \chi_1(a_i(g))\chi_2(d_i(g)),\] where \(a_i(g)d_i(g)=\det(g)\). Hence \[\prod_i \sigma(h_i(g)) = \prod_i (\chi_1\chi_2)(\det g) = (\chi_1\chi_2)(\det g)^{q+1}.\] Since the representation is self dual, \(\chi_1\chi_2=1\). Hence by Lemma 7, we find that \[\det(\pi)=\mu\circ \det.\]

(iii) Steinberg twists. Using the virtual identity \[\mathrm{St}= \mathrm{Ind}_B^{G_q}\mathbf{1} - \mathbf{1},\] we compute determinants multiplicatively: \[\det(\mathrm{St}) = \frac{\det(\mathrm{Ind}_B^{G_q}\mathbf{1})}{\det(\mathbf{1})}=\mu\circ \det.\] Now if \(\pi=\mathrm{St}\otimes(\psi\circ\det)\), then \[\det(\pi) = \det(\mathrm{St})\cdot (\psi\circ\det)^{\dim \mathrm{St}} = (\mu\circ \det) (\psi\circ\det)^q=(\mu\circ \det) (\psi\circ\det)=\mu\psi\circ \det.\] ◻

Proof. We treat each case separately.

(1) Since \(\det(h_1)=-1\) and \(\det(h_2)=1\), we obtain \[\chi_\pi(h_1)=\psi(-1) \quad \text{and}\quad\chi_\pi(h_2)=\psi(1)=1.\]

(2) We use the standard formula for the character of an induced representation: \[\chi_{\mathrm{Ind}_B^{G_q}\sigma}(g) = \frac{1}{|B|} \sum_{\substack{x\in G_q \\ x^{-1}gx \in B}} \chi_\sigma(x^{-1}gx),\] where \(\sigma=\chi_1\otimes\chi_2\). The element \(h_1\) is semisimple with distinct eigenvalues. There are exactly two such \(B\)-conjugacy representatives, namely \[\begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix} \quad \text{and} \quad \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}.\]Thus \[\chi_\pi(h_1) = \chi_\sigma\!\left(\begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix}\right) + \chi_\sigma\!\left(\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\right),\] and we obtain \[\chi_\pi(h_1) = \chi_1(-1)\chi_2(1) + \chi_1(1)\chi_2(-1) = \chi_1(-1)+\chi_2(-1).\] For \(h_2=-I\), which is central in \(G_q\), we have \(x^{-1}h_2x=h_2\) for all \(x\in G_q\), and hence \[\chi_\pi(h_2) = \frac{1}{|B|} \sum_{x\in G_q} \chi_\sigma(h_2) = \frac{|G_q|}{|B|}\,\chi_\sigma(h_2).\] Since \(\dim(\pi)=[G_q:B]=q+1\), it follows that \[\chi_\pi(h_2) = (q+1)\chi_\sigma(-I) = (q+1)\chi_1(-1)\chi_2(-1).\]

(3) The computation for Steinberg representations follows along the same lines as part (2).

(4) We use the virtual representation \[\pi_\theta = \mathrm{Ind}_{ZN}^{G_q}(\theta_\psi) - \mathrm{Ind}_E^{G_q}(\theta),\] and compute characters using the standard formula for induced representations: \[\chi_{\mathrm{Ind}_H^{G_q}\sigma}(g) = \frac{1}{|H|} \sum_{\substack{x\in G_q \\ x^{-1}gx \in H}} \chi_\sigma(x^{-1}gx).\] The element \(h_1\) is split semisimple with distinct eigenvalues. We claim that neither \(ZN\) nor \(E\) contains any conjugate of \(h_1\). First, every element of \(ZN\) is unipotent up to scalar, hence has a single eigenvalue; thus no conjugate of \(h_1\) lies in \(ZN\). Therefore \[\chi_{\mathrm{Ind}_{ZN}^{G_q}(\theta_\psi)}(h_1)=0.\] Diagonalizable elements in \(E\simeq \mathbb{F}_{q^2}^\times\) are the elements in \(\mathbb{F}_q^\times\). Thus the only diagonalizable elements in \(E\) are scalar. Since \(h_1\) is non-scalar and diagonalizable over \(\mathbb{F}_q\), it is not conjugate to any element of \(E\). Hence, \(\chi_{\mathrm{Ind}_E^{G_q}(\theta)}(h_1)=0\) and we deduce that \[\chi_{\pi_\theta}(h_1)=0.\] Since \(h_2=-I\) is central, the induced character formula simplifies to \[\chi_{\mathrm{Ind}_H^{G_q}\sigma}(h_2) = \frac{|G_q|}{|H|}\,\chi_\sigma(h_2)\] provided \(h_2\in H\), and is \(0\) otherwise. First, \(h_2\in Z\subset ZN\), so \[\chi_{\mathrm{Ind}_{ZN}^{G_q}(\theta_\psi)}(h_2) = \frac{|G_q|}{|ZN|}\,\theta_\psi(h_2).\] Since \(\theta_\psi(zn)=\theta(z)\psi(n)\) and \(h_2=-I\in Z\), we have \[\theta_\psi(h_2)=\theta(-1).\] A direct computation shows \[\frac{|G_q|}{|ZN|} = q^2-1,\] so \[\chi_{\mathrm{Ind}_{ZN}^{G_q}(\theta_\psi)}(h_2)=(q^2-1)\theta(-1).\]Next, \(h_2\in E\), so similarly \[\chi_{\mathrm{Ind}_E^{G_q}(\theta)}(h_2) = \frac{|G_q|}{|E|}\,\theta(-1).\] Since \(|E|=q^2-1\), we have \[\frac{|G_q|}{|E|} = q(q-1),\] and hence \[\chi_{\mathrm{Ind}_E^{G_q}(\theta)}(h_2)=q(q-1)\theta(-1).\]Therefore \[\chi_{\pi_\theta}(h_2) = (q^2-1)\theta(-1) - q(q-1)\theta(-1) = (q-1)\theta(-1).\] ◻

Proof. Let \(h_1=\mathrm{diag}(-1,1)\) and \(h_2=-I\). For any representation \(\pi\), recall that \[m_\pi = \frac{\dim \pi - \chi_\pi(h_1)}{2}.\] By Theorem 10, we have \[w_2(\pi)=0 \iff m_\pi \equiv 0 \pmod{4},\] and \[w_4(\pi)|_D = \binom{m_\pi/2}{2}\sum_i t_i^2 + \frac{\dim \pi - \chi_\pi(h_2)}{8} \sum_{i<j} t_i t_j.\] Thus \(w_4(\pi)=0\) if and only if both coefficients vanish modulo \(2\).

Case 1: \(\pi\) one-dimensional. If \(\pi=\psi\circ\det\) with \(\psi^2=1\), then \(\psi(-1)=1\) since \(-1\) is a square in \(\mathbb{F}_q^\times\) (as \(q\equiv 1 \pmod{4}\)). Hence \(\chi_\pi(h_1)=1\), \(m_\pi=0\), so \(w_2(\pi)=0\). Also \(\chi_\pi(h_2)=1=\dim\pi\), so \(w_4(\pi)=0\).

Case 2: \(\dim \pi>1\).

(a) Principal series. Let \(\pi=\mathrm{Ind}_B^{G_q}(\chi_1\otimes\chi_1^{-1})\). Then \[\chi_\pi(h_1)=\chi_1(-1)+\chi_1(-1)^{-1}.\]

If \(\chi_1(-1)=1\), then \(\chi_\pi(h_1)=2\), and hence \[m_\pi=\frac{(q+1)-2}{2}=\frac{q-1}{2}.\]

If \(\chi_1(-1)=-1\), then \(\chi_\pi(h_1)=-2\), and hence \[m_\pi=\frac{(q+1)-(-2)}{2}=\frac{q+3}{2}.\]

Therefore \[w_2(\pi)=0 \iff \begin{cases} \frac{q-1}{2}\equiv 0 \pmod{4}, & \text{if } \chi_1(-1)=1,\\[6pt] \frac{q+3}{2}\equiv 0 \pmod{4}, & \text{if } \chi_1(-1)=-1, \end{cases}\] which is equivalent to \[w_2(\pi)=0 \iff \begin{cases} q\equiv 1 \pmod{8}, & \text{if } \chi_1(-1)=1,\\[6pt] q\equiv 5 \pmod{8}, & \text{if } \chi_1(-1)=-1. \end{cases}\]

(b) Steinberg twists. Let \(\pi=\mathrm{St}\otimes(\psi\circ\det)\) with \(\psi^2=1\). Then \(\psi(-1)=1\), and one computes \[\chi_\pi(h_1)=0,\qquad \dim \pi=q,\] so \[m_\pi=\frac{q-1}{2}.\] Thus \[w_2(\pi)=0 \iff q\equiv 1 \pmod{8}.\]

(c) Cuspidal representations. Let \(\pi=\pi_\theta\) be cuspidal and self-dual. Then \(\chi_\pi(h_1)=0\) and \(\dim \pi=q-1\), so \[m_\pi=\frac{q-1}{2},\] and hence \[w_2(\pi)=0 \iff q\equiv 1 \pmod{8}.\]

We now analyze the two coefficients in the expression for \(w_4(\pi)\). By Theorem 10, the class \(w_4(\pi)\) vanishes if and only if both \[\binom{m_\pi/2}{2} \quad \text{and} \quad \frac{\dim \pi - \chi_\pi(h_2)}{8}\] are zero modulo \(2\).

We first consider the binomial coefficient. A direct congruence calculation shows that \[\binom{m_\pi/2}{2}\equiv 0 \pmod{2} \iff m_\pi \equiv 0 , 2\pmod{8}.\] From the case-by-case analysis above, one has \(m_\pi \in \{(q-1)/2,\,(q+3)/2\}\). Checking these two possibilities separately and noting that \(q\equiv 1\pmod{4}\), we find \[m_\pi \equiv 0 \pmod{8} \iff \begin{cases} q \equiv 1 \pmod{16}, & \text{if } m_\pi=\frac{q-1}{2},\\[6pt] q \equiv 13 \pmod{16}, & \text{if } m_\pi=\frac{q+3}{2}. \end{cases}\]

We now turn to the second coefficient. For principal series and Steinberg representations, one has \[\chi_\pi(h_2)=\dim \pi,\] so this term vanishes. For cuspidal representations \(\pi=\pi_\theta\), one has \[\chi_\pi(h_2)=(q-1)\theta(-1).\] Since \(\pi\) is self-dual, the parameter \(\theta\) satisfies \(\theta^q=\theta^{-1}\), and hence \(\theta^{q+1}=1\). Restricting to \(\mathbb{F}_q^\times\), it follows that \(\theta(a)^2=1\) for all \(a\in \mathbb{F}_q^\times\), so \(\theta|_{\mathbb{F}_q^\times}\) is quadratic. As \(q\equiv 1 \pmod{4}\), the element \(-1\) is a square in \(\mathbb{F}_q^\times\), and therefore \(\theta(-1)=1\). It follows that \[\chi_\pi(h_2)=q-1=\dim \pi,\] and hence the second coefficient also vanishes.

Combining these two observations, we conclude that \(w_4(\pi)=0\) if and only if \(m_\pi \equiv 0 \pmod{8}\), which yields the stated congruence conditions on \(q\). ◻

Proof. Let \(h_1=\mathrm{diag}(-1,1)\). Recall that \[m_\pi=\frac{\dim \pi - \chi_\pi(h_1)}{2}, \quad w_2(\pi)=0 \iff m_\pi \equiv 0,1 \pmod{4}.\]

Since \(q\equiv 3 \pmod{4}\), the element \(-1\) is not a square in \(\mathbb{F}_q^\times\). Hence for any quadratic character \(\chi\) of \(\mathbb{F}_q^\times\), one has \(\chi(-1)=-1\). We compute \(m_\pi\) using Proposition 20.

(i) One-dimensional representations. If \(\pi=\psi\circ\det\) with \(\psi\neq 1\) and \(\psi^2=1\), then \(\psi(-1)=-1\), so \[m_\pi=\frac{1-(-1)}{2}=1.\] Thus \(w_2(\pi)= 0\).

(ii) Principal series. We have \[m_\pi= \begin{cases} \frac{q-1}{2} & \text{if } \chi_1(-1)=1,\\[6pt] \frac{q+3}{2} & \text{if } \chi_1(-1)=-1. \end{cases}\] If \(q\equiv 3 \pmod{8}\), then \[\frac{q-1}{2}\equiv 1 \pmod{4} \quad\text{and}\quad \frac{q+3}{2}\equiv 3 \pmod{4}.\] Therefore if \(q\equiv 3\pmod{8}\), we have that \(w_2(\pi)=0\) if and only if \(\chi_1(-1)=1\).

On the other hand, if \(q\equiv 7 \pmod{8}\), then \[\frac{q-1}{2}\equiv 3 \pmod{4}, \qquad \frac{q+3}{2}\equiv 1 \pmod{4},\] so again \(w_2(\pi)=0\). if and only if \(\chi_1(-1)=-1\). Thus for principal series, one has \(w_2(\pi)=0\) if and only if:

• \(q\equiv 3\pmod{8}\) and \(\chi_1(-1)=1\),

• \(q\equiv 7\pmod{8}\) and \(\chi_1(-1)=-1\).

(iii) Steinberg twists. Here \[m_\pi=\frac{q-\psi(-1)}{2}=\frac{q+1}{2}.\] If \(\psi(-1)=1\), then \(m_\pi=\frac{q-1}{2}\); if \(\psi(-1)=-1\), then \(m_\pi=\frac{q+1}{2}\). If \(q\equiv 3 \pmod{8}\), then \[\frac{q-1}{2}\equiv 1,\quad \frac{q+1}{2}\equiv 2 \pmod{4},\] and hence \(w_2(\pi)\neq 0\) in this case.

If \(q\equiv 7 \pmod{8}\), then \[\frac{q-1}{2}\equiv 3,\quad \frac{q+1}{2}\equiv 0 \pmod{4},\] so again \(w_2(\pi)=0\) occurs if and only if \(\psi\neq 1\).

(iv) Cuspidal representations. Here \[m_\pi=\frac{q-1}{2}.\] If \(q\equiv 3 \pmod{8}\), then \(m_\pi\equiv 1 \pmod{4}\), while if \(q\equiv 7 \pmod{8}\), then \(m_\pi\equiv 3 \pmod{4}\). Thus \(w_2(\pi)\neq 0\) for all cuspidal representations.

Combining the above, we see that \(w_2(\pi)=0\) occurs precisely in the Steinberg case, and this happens if and only if \(q\equiv 7 \pmod{8}\) and \(\psi\neq 1\). This completes the proof. ◻

Proof. We estimate the sizes of the relevant families. We recall that self-dual representations of \(G_q\) are orthogonal. The self-dual irreducible representations of \(G_q\) consist of:

• \(\sim \frac{q}{2}\) principal series,

• \(\sim \frac{q}{2}\) cuspidal representations,

• \(2\) one-dimensional representations,

• \(2\) Steinberg twists.

There are exactly two one-dimensional self-dual representations, namely the trivial and the quadratic character. Thus \[\frac{\#\{\text{1-dimensional}\}}{|\mathcal{O}_q|} \;\ll\; \frac{1}{q}\;\longrightarrow\;0.\] There are exactly two Steinberg twists \(\mathrm{St}\otimes(\psi\circ\det)\) with \(\psi^2=1\). Hence \[\frac{\#\{\text{Steinberg twists}\}}{|\mathcal{O}_q|} \;\ll\; \frac{1}{q}\;\longrightarrow\;0.\] ◻

Proof. We give explicit counts of irreducible orthogonal representations of \(G_q=\operatorname{GL}_2(\mathbb{F}_q)\) and analyze the condition \(w_2(\pi)=0\) in each case. Irreducible orthogonal representations of \(G_q\) fall into three families.

(i) Principal series. These are representations of the form \[\pi=\mathrm{Ind}_B^{G_q}(\chi\otimes \chi^{-1})\] with \(\chi\neq \chi^{-1}\). The number of such representations is \[\frac{(q-1)-2}{2}=\frac{q-3}{2},\] since there are \(q-1\) characters of \(\mathbb{F}_q^\times\), of which exactly two (the trivial and quadratic characters) satisfy \(\chi=\chi^{-1}\), and we identify \(\chi\sim\chi^{-1}\).

(ii) Cuspidal representations. Self-dual cuspidal representations correspond to characters \(\theta\) of \(\mathbb{F}_{q^2}^\times\) satisfying \(\theta^{q+1}=1\) and \(\theta\neq \theta^q\), modulo \(\theta\sim\theta^q\). There are \(q+1\) characters with \(\theta^{q+1}=1\), of which two satisfy \(\theta=\theta^q\), so the number of self-dual cuspidal representations is \[\frac{q-1}{2}.\]

(iii) One-dimensional and Steinberg twists. There are exactly two one-dimensional orthogonal representations (the trivial and quadratic determinant characters), and exactly two Steinberg twists \[\mathrm{St}\otimes(\psi\circ\det)\] with \(\psi^2=1\).

Thus \[|\mathcal{O}_q| = \frac{q-3}{2}+\frac{q-1}{2}+O(1) = q+O(1).\]

We now analyze the vanishing of \(w_2(\pi)\) according to the residue class of \(q \pmod{8}\).

(i) \(q\equiv 1 \pmod{8}\). From Proposition 22, all orthogonal representations with \(\dim\pi>1\) satisfy \(w_2(\pi)=0\), and the same holds for the one-dimensional representations. Hence \[\#\{\pi : w_2(\pi)=0\} = |\mathcal{O}_q|,\] so \[\frac{\#\{\pi : w_2(\pi)=0\}}{|\mathcal{O}_q|} \to 1.\]

(ii) \(q\equiv 5 \pmod{8}\). In this case, \(w_2(\pi)=0\) occurs precisely for principal series representations \[\pi=\mathrm{Ind}_B(\chi\otimes\chi^{-1})\] with \(\chi(-1)=-1\). Among the \(q-1\) characters of \(\mathbb{F}_q^\times\), exactly \(\frac{q-1}{2}\) satisfy \(\chi(-1)=-1\). Since the trivial and quadratic characters both satisfy \(\chi(-1)=1\), all such characters are admissible. Passing to unordered pairs \(\{\chi,\chi^{-1}\}\) yields \[\frac{q-1}{4}\] principal series representations with \(w_2(\pi)=0\).

All other representations (cuspidal, Steinberg, and one-dimensional) satisfy \(w_2(\pi)\neq0\). Hence \[\frac{\#\{\pi : w_2(\pi)=0\}}{|\mathcal{O}_q|} = \frac{(q-1)/4}{q+O(1)} \to \frac{1}{4}.\]

(iii) \(q\equiv 3 \pmod{8}\). In this case, \(w_2(\pi)=0\) occurs precisely for the one-dimensional representations and for principal series representations \[\pi=\mathrm{Ind}_B(\chi\otimes\chi^{-1})\] with \(\chi(-1)=1\). Among the \(q-1\) characters of \(\mathbb{F}_q^\times\), exactly \(\frac{q-1}{2}\) satisfy \(\chi(-1)=1\). Since \(q\equiv3\pmod4\), the quadratic character satisfies \(\chi(-1)=-1\), whereas the trivial character satisfies \(\chi(-1)=1\). Removing the trivial character leaves \[\frac{q-3}{2}\] admissible characters. Passing to unordered pairs \(\{\chi,\chi^{-1}\}\) yields \[\frac{q-3}{4}\] principal series representations with \(w_2(\pi)=0\). Together with the two one-dimensional orthogonal representations, this gives \[\frac{q+5}{4}\] representations satisfying \(w_2(\pi)=0\).

All other representations (cuspidal and Steinberg twists) satisfy \(w_2(\pi)\neq0\). Hence \[\frac{\#\{\pi : w_2(\pi)=0\}}{|\mathcal{O}_q|} = \frac{(q+5)/4}{q+O(1)} \to \frac{1}{4}.\]

(iv) \(q\equiv 7 \pmod{8}\). In this case, \(w_2(\pi)=0\) occurs precisely for the nontrivial Steinberg twist \[\mathrm{St}\otimes(\psi\circ\det), \qquad \psi^2=1,\;\psi\neq1.\] Thus only \(O(1)\) representations satisfy \(w_2(\pi)=0\), and therefore \[\frac{\#\{\pi : w_2(\pi)=0\}}{|\mathcal{O}_q|} \to 0.\]

It remains to compute the global average over all odd prime powers. For \(a\in\{1,3,5,7\}\), define \[\mathcal{Q}_a(X) = \{q\le X : q \text{ odd prime power and } q\equiv a\pmod8\}.\] We first show that these four residue classes are asymptotically equidistributed.

Restrict first to odd primes. By Dirichlet’s theorem on primes in arithmetic progressions, \[\pi_a(X) := \#\{p\le X : p\equiv a\pmod8\} \sim \frac{1}{\varphi(8)}\frac{X}{\log X} = \frac{1}{4}\frac{X}{\log X}.\] Hence \[\pi_a(X) \sim \pi_b(X)\] for all \(a,b\in\{1,3,5,7\}\).

We now analyze the contribution of higher prime powers. Let \[R(X) = \#\{p^k\le X : p \text{ odd prime},\;k\ge2\}.\] If \(p^k\le X\) with \(k\ge2\), then necessarily \(p\le X^{1/2}\). Therefore \[R(X) \le \sum_{2\le k\le \log_2 X}\pi(X^{1/k}),\] where \(\pi(y)\) denotes the number of primes at most \(y\). Using the estimate \(\pi(y)\ll y/\log y\), we obtain \[R(X) \ll \sum_{2\le k\le \log_2 X}X^{1/k} = O(X^{1/2}).\] Since the number of odd primes up to \(X\) is asymptotic to \(X/\log X\), it follows that \[R(X) = o\!\left(\frac{X}{\log X}\right).\] Thus non-prime prime powers contribute negligibly to asymptotic density, and consequently \[\#\mathcal{Q}_a(X) \sim \frac{1}{4} \sum_{b\in\{1,3,5,7\}} \#\mathcal{Q}_b(X).\]

Now set \[N(q) = \#\{\pi\in\mathcal{O}_q : w_2(\pi)=0\} \quad\text{and}\quad D(q) = |\mathcal{O}_q|.\] From the preceding analysis, for \(q\equiv a\pmod8\) we have \[\frac{N(q)}{D(q)} = c_a+o(1),\] where \[c_1=1, \qquad c_3=c_5=\frac{1}{4}, \qquad c_7=0.\] Moreover, \[D(q)=q+O(1).\]

We decompose the denominator according to residue classes: \[\sum_{\substack{q\le X\\ q\text{ odd}}}D(q) = \sum_{a\in\{1,3,5,7\}} \sum_{q\in\mathcal{Q}_a(X)}(q+O(1)).\] Since the residue classes are equidistributed and the error term is of lower order, each congruence class contributes asymptotically one quarter of the total. More precisely, \[\sum_{q\in\mathcal{Q}_a(X)} q \sim \frac{1}{4} \sum_{\substack{q\le X\\ q\text{ odd}}} q.\]

Similarly, \[\sum_{\substack{q\le X\\ q\text{ odd}}}N(q) = \sum_{a\in\{1,3,5,7\}} \sum_{q\in\mathcal{Q}_a(X)} \bigl(c_a q+o(q)\bigr).\] Hence \[\sum_{\substack{q\le X\\ q\text{ odd}}}N(q) \sim \sum_{a\in\{1,3,5,7\}} c_a \sum_{q\in\mathcal{Q}_a(X)} q.\] Using the equidistribution just proved, we obtain \[\sum_{\substack{q\le X\\ q\text{ odd}}}N(q) \sim \frac{1}{4}(c_1+c_3+c_5+c_7) \sum_{\substack{q\le X\\ q\text{ odd}}} q.\]

Since \[\sum_{\substack{q\le X\\ q\text{ odd}}}D(q) \sim \sum_{\substack{q\le X\\ q\text{ odd}}} q,\] it follows that \[\lim_{X\to\infty} \frac{\sum_{\substack{q\le X\\ q\text{ odd}}}N(q)}{\sum_{\substack{q\le X\\ q\text{ odd}}}D(q)} = \frac{1}{4}(c_1+c_3+c_5+c_7).\] Substituting the values of the constants gives \[\frac{1}{4}\left(1+\frac{1}{4}+\frac{1}{4}+0\right) = \frac{3}{8}.\] This proves the theorem. ◻