Appendix A. Remaining Picard–Fuchs equations
In this appendix, we provide the details in the computation of the remaining three pencils \({\mathsf {F}}_2{\mathsf {L}}_2\), \({\mathsf {L}}_2{\mathsf {L}}_2\), and \({\mathsf {L}}_4\). We follow the same strategy as in Sects. 2.5–2.6.
A.1. The \({\mathsf {F}}_2{\mathsf {L}}_2\) pencil
Take the pencil
$$\begin{aligned} F_{\psi } :=x_0^4+x_1^4 + x_2^3 x_3 + x_3^3x_2 - 4\psi x_0x_1x_2x_3 \end{aligned}$$
that defines the pencil of projective hypersurfaces \(X_{\psi } = Z(F_\psi ) \subset {\mathbb {P}}^3\). There is a \({\mathbb {Z}}/8{\mathbb {Z}}\) scaling symmetry of this family generated by the element
$$\begin{aligned} g (x_0: x_1: x_2: x_3) = (\xi ^2 x_0: x_1: \xi x_2 : \xi ^5 x_3) \end{aligned}$$
where \(\xi \) is a primitive eighth root of unity. There are eight characters \(\chi _k: H \rightarrow {\mathbb {C}}^\times \), where \(\chi _k(g) = \xi ^k\). We can again decompose V into subspaces \(W_{\chi _k}\) and write their monomial bases. Note that the monomial bases for \(W_{\chi _1}, W_{\chi _3}, W_{\chi _5},\) and \(W_{\chi _7}\) are the same up to transpositions of \(x_0\) and \(x_1\) or \(x_2\) and \(x_3\) which leave the polynomial invariant; thus, they have the same Picard–Fuchs equations. The monomial bases for \(W_{\chi _2}\) and \(W_{\chi _6}\) are related by transposing \(x_0\) and \(x_1\), so they also have the same Picard–Fuchs equations. So, we are left with four types of monomial bases:
- (i)
\(W_{\chi _0}\) has monomial basis \(\{x_0x_1x_2x_3\}\);
- (ii)
\(W_{\chi _1}\) has monomial basis \(\{x_2^2x_0x_3, x_0^2x_1x_3\}\);
- (iii)
\(W_{\chi _2}\) has monomial basis \(\{x_0^2x_2x_3, x_1^2x_2^2, x_1^2x_3^2\}\); and
- (iv)
\(W_{\chi _4}\) has monomial basis \(\{x_0^2x_1^2,x_2^2x_3^2,x_2^2x_0x_1, x_3^2x_0x_1\}\).
Using (2.3.3), we compute the following period relations:
$$\begin{aligned} \begin{aligned} {\mathbf {v}} + (4,0,0,0)&= \frac{1+v_0}{4(\omega + 1)} {\mathbf {v}} + \psi ({\mathbf {v}}+(1,1,1,1)), \\ {\mathbf {v}} + (0,4,0,0)&= \frac{1+v_1}{4(\omega + 1)} {\mathbf {v}} + \psi ({\mathbf {v}}+(1,1,1,1)), \\ {\mathbf {v}} + (0,0,3,1)&= \frac{3(v_2+1)-(v_3+1)}{8(\omega + 1)} {\mathbf {v}} + \psi ({\mathbf {v}}+(1,1,1,1)), \\ {\mathbf {v}} + (0,0,1,3)&= \frac{-(v_2+1) + 3(v_3+1)}{8(\omega + 1)} {\mathbf {v}} + \psi ({\mathbf {v}}+(1,1,1,1)). \end{aligned} \end{aligned}$$
(A.1.1)
We can now use the diagram method to prove the following proposition.
Proposition A.1.2
For the \({\mathsf {F}}_2{\mathsf {L}}_2\) family, the primitive cohomology group \(H^2_{\text {prim}}(X_{{\mathsf {F}}_2{\mathsf {L}}_2, \psi }, {\mathbb {C}})\) has 15 periods whose Picard–Fuchs equations are hypergeometric differential equations as follows:
$$\begin{aligned} \begin{aligned} 3 \text { periods are annihilated by } D\left( \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4} ; 1, 1, 1 \,|\,\psi ^{-4}\right) , \\ 2 \text { periods are annihilated by } D\left( \tfrac{1}{4}, \tfrac{3}{4} ; 1, \tfrac{1}{2} \,|\,\psi ^{-4}\right) , \\ 2 \text { periods are annihilated by } D\left( \tfrac{1}{2} ; 1 \,|\,\psi ^{4}\right) , \\ 4 \text { periods are annihilated by } D\left( \tfrac{1}{8}, \tfrac{5}{8} ; 1, \tfrac{1}{4} \,|\,\psi ^{4}\right) , \text {and } \\ 4 \text { periods are annihilated by } D\left( \tfrac{-3}{8}, \tfrac{1}{8} ; 0, \tfrac{1}{4} \,|\,\psi ^{4}\right) . \end{aligned} \end{aligned}$$
We consider each of these in turn.
Lemma A.1.3
The Picard–Fuchs equation associated to the periods \(\psi (2,2,0,0)\) and \(\psi (0,0,2,2)\) is the hypergeometric differential equation \(D(\tfrac{1}{4}, \tfrac{3}{4} ; 1, \tfrac{1}{2} \,|\,\psi ^{-4})\).
Proof
For the periods (2, 2, 0, 0) and (0, 0, 2, 2), corresponding to the quartic monomials \(x_0^2x_1^2\) and \(x_2^2x_3^2\), we use the diagram
and get the relations
$$\begin{aligned} \begin{aligned} \eta (2,2,0,0) = \psi ^2(\eta +1)(0,0,2,2),\\ \eta (0,0,2,2) = \psi ^2(\eta +1)(2,2,0,0). \end{aligned} \end{aligned}$$
(A.1.4)
For the periods (2, 2, 0, 0) and (0, 0, 2, 2), corresponding to the quartic monomials \(x_0^2x_1^2\) and \(x_2^2x_3^2\), we get the same Picard–Fuchs equation
$$\begin{aligned} \begin{aligned} \left[ (\eta -2)\eta - \psi ^4(\eta +3)(\eta +1) \right] . \end{aligned} \end{aligned}$$
(A.1.5)
By multiplying by \(\psi \) and substituting \(t = \psi ^{-4}\) and \(\theta = t \displaystyle {\frac{\mathrm {d}}{\mathrm {d}t}} = -\eta /4\), we get
$$\begin{aligned} \psi \left[ (\eta -2)\eta - \psi ^4(\eta +3)(\eta +1) \right] (2,2,0,0)&= 0, \\ \left[ (\eta -3)(\eta -1) - \psi ^4(\eta +2)\eta \right] \psi (2,2,0,0)&= 0, \\ \left[ \left( \theta +\tfrac{3}{4}\right) \left( \theta +\tfrac{1}{4}\right) - t^{-1} \left( \theta -\tfrac{1}{2}\right) \theta \right] \psi (2,2,0,0)&= 0, \\ \left[ \left( \theta -\tfrac{1}{2}\right) \theta - t\left( \theta +\tfrac{3}{4}\right) \left( \theta +\tfrac{1}{4}\right) \right] \psi (2,2,0,0)&= 0, \end{aligned}$$
which is the hypergeometric differential equation \(D(\tfrac{1}{4}, \tfrac{3}{4} ; 1, \tfrac{1}{2} \,|\,\psi ^{-4})\). \(\square \)
Lemma A.1.6
The Picard–Fuchs equation associated to the periods (2, 0, 1, 1) is the hypergeometric differential equation \(D(\tfrac{1}{2} ; 1 \,|\,\psi ^{4})\).
Proof
For the period (2, 0, 1, 1), corresponding to the quartic monomial \(x_0^2x_2x_3\), we use the diagram
One can see that \((2,0,4,2) = \frac{1}{8}(2 + \eta ) (2,0,1,1)\), which one can then use to show that
$$\begin{aligned} \begin{aligned} \eta (2,0,1,1)&= 8 \psi ^2(0,2,3,3) \\&= 8 \psi ^3 (1,3,3,1) \\&= 8 \psi ^4 (2,0,4,2) \\&= \psi ^4 (\eta +2) (2,0,1,1). \end{aligned} \end{aligned}$$
(A.1.7)
Thus, the period (2, 0, 1, 1) corresponding to the quartic monomial \(x_0^2x_2x_3\) satisfies the differential equation:
$$\begin{aligned} \left[ \eta - \psi ^4(\eta +2)\right] (2,0,1,1)=0. \end{aligned}$$
By substituting \(u = \psi ^{4}\) and \(\sigma = u \displaystyle {\frac{\mathrm {d}}{\mathrm {d}u}} = 4\eta \), we get
$$\begin{aligned} \left[ \eta - \psi ^4(\eta +2)\right] (2,0,1,1)&=0, \\ \left[ \sigma - u(\sigma + \tfrac{1}{2})\right] (2,0,1,1)&=0. \end{aligned}$$
\(\square \)
Lemma A.1.8
The Picard–Fuchs equations associated to the periods \((2,1,0,1),\psi ^3(1,0,2,1)\) are the hypergeometric differential equations \(D(\tfrac{1}{8}, \tfrac{5}{8} ; 1, \tfrac{1}{4} \,|\,\psi ^{4}),D(\tfrac{1}{8}, \tfrac{-3}{8} ; 0, \tfrac{1}{4} \,|\,\psi ^{4})\), respectively.
Proof
For the period (2, 1, 0, 1), corresponding to the quartic monomial \(x_0^2x_1x_3\), we use the diagram
Note that:
$$\begin{aligned} \begin{aligned} \eta (2,1,0,1)&= 8\psi ^3(1,0,3,4),\\ (1,0,3,4)&= \tfrac{1}{8}\left( \eta +\tfrac{3}{2}\right) (1,0,2,1), \\ \eta (1,0,2,1)&= \psi \left( \eta +\tfrac{1}{2}\right) (2,1,0,1). \end{aligned} \end{aligned}$$
(A.1.9)
We can compute the two periods that satisfy each of the following Picard–Fuchs equations for the four sets of pairs:
$$\begin{aligned} \begin{aligned} \left[ (\eta -3) \eta - \psi ^4\left( \eta + \tfrac{5}{2}\right) \left( \eta + \tfrac{1}{2}\right) \right] (2,1,0,1)&= 0,\\ \left[ (\eta -1) \eta - \psi ^4\left( \eta + \tfrac{7}{2}\right) \left( \eta +\tfrac{3}{2}\right) \right] (1,0,2,1)&= 0. \end{aligned} \end{aligned}$$
(A.1.10)
With the first Picard–Fuchs equation, we can substitute \(u = \psi ^4\), \(\sigma = u\textstyle {\frac{\mathrm {d}}{\mathrm {d}u}} = 4 \eta \), and yield the equation:
$$\begin{aligned} \left[ (\eta -3) \eta - \psi ^4\left( \eta + \tfrac{5}{2}\right) \left( \eta + \tfrac{1}{2}\right) \right] (2,1,0,1)&= 0, \\ \left[ \left( \sigma -\tfrac{3}{4}\right) \sigma - u\left( \sigma + \tfrac{5}{8}\right) \left( \sigma + \tfrac{1}{8}\right) \right] (2,1,0,1)&= 0. \end{aligned}$$
which is the hypergeometric differential equation \(D(\tfrac{1}{8}, \tfrac{5}{8} ; 1, \tfrac{1}{4} \,|\,u)\). For the second Picard–Fuchs equation, we can multiply by \(\psi ^3\) and then substitute to find
$$\begin{aligned} \left[ \left( \eta -1\right) \eta - \psi ^4\left( \eta + \tfrac{7}{2}\right) \left( \eta +\tfrac{3}{2}\right) \right] \left( 1,0,2,1\right)&= 0,\\ \psi ^3 \left[ \left( \eta -1\right) \eta - \psi ^4\left( \eta + \tfrac{7}{2}\right) \left( \eta +\tfrac{3}{2}\right) \right] \left( 1,0,2,1\right)&= 0,\\ \left[ \left( \eta -4\right) \left( \eta -3\right) - \psi ^4\left( \eta + \tfrac{1}{2}\right) \left( \eta -\tfrac{3}{2}\right) \right] \psi ^3\left( 1,0,2,1\right)&= 0,\\ \left[ \left( \sigma -1\right) \left( \sigma -\tfrac{3}{4}\right) - u\left( \sigma + \tfrac{1}{8}\right) \left( \sigma -\tfrac{3}{8}\right) \right] \psi ^3\left( 1,0,2,1\right)&= 0, \end{aligned}$$
which is the hypergeometric function \(D(\tfrac{1}{8}, \tfrac{-3}{8} ; 0, \tfrac{1}{4} \,|\,\psi ^{4})\). \(\square \)
Proof of Proposition 2.7.1
The periods annihilated by \(D(\tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4} ; 1, 1, 1 \,|\,\psi ^{-4})\) are those corresponding to the holomorphic form. The 2 periods are annihilated by \(D(\tfrac{1}{4}, \tfrac{3}{4} ; 1, \tfrac{1}{2} \,|\,\psi ^{-4})\) and are provided by Lemma A.1.3. The period annihilated by \(D(\tfrac{1}{2} ; 1 \,|\,\psi ^{4})\) corresponds to a monomial in the basis for \(W_{\chi _2}\), which we compute in Lemma A.1.6. Since \(W_{\chi _2}\) and \(W_{\chi _6}\) are related by a transposition of \(x_0\) and \(x_1\), there are two periods annihilated by the hypergeometric differential equation computed here. The 4 periods annihilated by \(D(\tfrac{1}{8}, \tfrac{5}{8} ; 1, \tfrac{1}{4} \,|\,\psi ^{4})\) and the 4 periods annihilated by \(D(\tfrac{-3}{8}, \tfrac{1}{8} ; 0, \tfrac{1}{4} \,|\,\psi ^{4})\) correspond to the monomial bases for \(W_{\chi _1}, W_{\chi _3}, W_{\chi _5},\) and \(W_{\chi _7}\), which are the same up to transpositions. We compute in Lemma A.1.8 the Picard–Fuchs equations for \(W_{\chi _1}\) which then give us that each of those hypergeometric differential equations annihilates 4 periods. \(\square \)
A.2. The \({\mathsf {L}}_2{\mathsf {L}}_2\) pencil
Now, consider
$$\begin{aligned} F_{\psi } :=x_0^3x_1+x_1^3x_0 + x_2^3 x_3 + x_3^3x_2 - 4\psi x_0x_1x_2x_3 \end{aligned}$$
that defines the pencil of projective hypersurfaces \(X_{\psi } = Z(F_\psi ) \subset {\mathbb {P}}^3\). There is a \({\mathbb {Z}}/4{\mathbb {Z}}\) symmetry with generator
$$\begin{aligned} g(x_0:x_1:x_2:x_3)&= (\xi x_0:\xi ^5 x_1:\xi ^3 x_2:\xi ^7 x_3), \end{aligned}$$
where \(\xi \) is a primitive eighth root of unity. There are four characters \(\chi _{a}: H \rightarrow {\mathbb {G}}_m\), where \(\chi _{a}(g_1) = \sqrt{-1}^{a}\). We can again decompose V into subspaces \(W_{\chi _{a}}\). Out of the eight, the subspaces \(W_{\chi _{0}}, W_{\chi _{1}}, W_{\chi _{2}},\) and \(W_{\chi _{3}}\) are empty. The monomial bases for \(W_{\chi _{1}}\) and \(W_{\chi _{3}}\) are related by a transposition of the variables \(x_0\) and \(x_1\), so their Picard–Fuchs equations are the same. We have three types of monomial bases:
- (i)
\(W_{\chi _{0}}\) has monomial basis \(\{x_0x_1x_2x_3, x_0^2x_2^2, x_0^2x_3^2, x_1^2x_2^2, x_1^2x_3^2\}\);
- (ii)
\(W_{\chi _{1}}\) has monomial basis \(\{x_0^2x_1x_2, x_1^2x_0x_3, x_2^2x_1x_3, x_3^2x_0x_2\}\); and
- (iii)
\(W_{\chi _{2}}\) has monomial basis \(\{x_0^2x_1^2, x_2^2x_3^2, x_0^2x_2x_3, x_1^2x_2x_3, x_2^2x_0x_1, x_3^2x_0x_1 \}\).
Using (2.3.3), we compute the following period relations:
$$\begin{aligned} \begin{aligned} {\mathbf {v}}+(3,1,0,0)&= \frac{3(v_0+1) -(v_1+1)}{8(\omega +1)}{\mathbf {v}} + \psi ({\mathbf {v}}+(1,1,1,1)) \\ {\mathbf {v}}+(1,3,0,0)&= \frac{-(v_0+1) +3(v_1+1)}{8(\omega +1)}{\mathbf {v}} + \psi ({\mathbf {v}}+(1,1,1,1)) \end{aligned} \end{aligned}$$
(A.2.1)
and the two symmetric relations replacing 0, 1 with 2, 3.
Proposition A.2.2
For the \({\mathsf {L}}_2{\mathsf {L}}_2\) family, the primitive cohomology group \(H^2_{\text {prim}}(X_{{\mathsf {L}}_2{\mathsf {L}}_2, \psi }, {\mathbb {C}})\) has 13 periods whose Picard–Fuchs equations are hypergeometric differential equations as follows:
$$\begin{aligned} \begin{aligned} 3 \text { periods are annihilated by } D\left( \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4} ; 1, 1, 1 \,|\,\psi ^{-4}\right) , \\ 8 \text { periods are annihilated by } D\left( \tfrac{1}{8}, \tfrac{3}{8}, \tfrac{5}{8}, \tfrac{7}{8} ; 0, \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4} \,|\,\psi ^{4}\right) , \text {and } \\ 2 \text { periods are annihilated by } D\left( \tfrac{1}{4}, \tfrac{3}{4} ; 1, \tfrac{1}{2} \,|\,\psi ^{4}\right) . \end{aligned} \end{aligned}$$
To prove Proposition A.2.2, we use the diagram method above in a few cases and then use symmetry. We first do two calculations.
Lemma A.2.3
The Picard–Fuchs equation associated to the periods (2, 1, 1, 0), (1, 0, 1, 2), (1, 2, 0, 1), and (0, 1, 2, 1) is the hypergeometric differential equation \(D(\tfrac{1}{8}, \tfrac{3}{8}, \tfrac{5}{8}, \tfrac{7}{8} ; 0, \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4} \,|\,\psi ^{4})\).
Proof
To find the Picard–Fuchs equations corresponding to all these cohomology pieces, we use the diagram
We obtain the following relations:
$$\begin{aligned} \begin{aligned} \eta (1,0,1,2)&= \psi \left( \eta +\tfrac{1}{2}\right) (2,1,1,0), \\ \eta (1,2,0,1)&= \psi \left( \eta +\tfrac{1}{2}\right) (1,0,1,2), \\ \eta (0,1,2,1)&= \psi \left( \eta +\tfrac{1}{2}\right) (1,2,0,1), \\ \eta (2,1,1,0)&= \psi \left( \eta +\tfrac{1}{2}\right) (0,1,2,1). \end{aligned} \end{aligned}$$
(A.2.4)
Using these relations, we can get the Picard–Fuchs equation:
$$\begin{aligned} \left[ (\eta -3) (\eta -2)(\eta -1) \eta - \psi ^4\left( \eta + \tfrac{7}{2}\right) \left( \eta + \tfrac{5}{2}\right) \left( \eta + \tfrac{3}{2} \right) \left( \eta + \tfrac{1}{2}\right) \right] (1,0,1,2) = 0. \end{aligned}$$
(A.2.5)
By substituting \(u = \psi ^4\) and \(\sigma = u \displaystyle {\frac{\mathrm {d}}{\mathrm {d}u}} = \tfrac{1}{4} \eta \), we obtain:
$$\begin{aligned}&\left[ (4\sigma -3) (4\sigma -2)(4\sigma -1) 4\sigma - u\left( 4\sigma + \tfrac{7}{2}\right) \left( \eta + \tfrac{5}{2}\right) \left( 4\sigma + \tfrac{3}{2} \right) \left( 4\sigma + \tfrac{1}{2}\right) \right] \\&\quad \cdot (1,0,1,2) = 0, \\&\left[ (\sigma -\tfrac{3}{4}) (\sigma -\tfrac{1}{2})(\sigma -\tfrac{1}{4}) \sigma - u\left( \sigma + \tfrac{7}{8}\right) \left( \sigma + \tfrac{5}{8}\right) \left( \sigma + \tfrac{3}{8} \right) \left( \sigma + \tfrac{1}{8}\right) \right] \\&\quad \cdot (1,0,1,2) = 0, \end{aligned}$$
which is the hypergeometric differential equation \(D(\tfrac{1}{8}, \tfrac{3}{8}, \tfrac{5}{8}, \tfrac{7}{8} ; 0, \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4} \,|\,\psi ^{4})\). The other three Picard–Fuchs equations are the same due to the symmetry in (A.2.4). \(\square \)
Lemma A.2.6
The Picard–Fuchs equation associated to the periods (2, 2, 0, 0) and (0, 0, 2, 2) is the hypergeometric differential equation \(D(\tfrac{1}{4}, \tfrac{3}{4} ; 1, \tfrac{1}{2} \,|\,\psi ^{4})\).
Proof
We use the diagram
We obtain the following relations:
$$\begin{aligned} \begin{aligned} \eta (0,0,2,2)&= \psi ^2(\eta + 1) (2,2,0,0),\\ \eta (2,2,0,0)&= \psi ^2(\eta + 1) (0,0,2,2), \end{aligned} \end{aligned}$$
(A.2.7)
giving the following Picard–Fuchs equations:
$$\begin{aligned} \begin{aligned} \left[ (\eta -2)\eta - \psi ^4 \left( \eta + 3\right) \left( \eta + 1\right) \right] (2,2,0,0)&= 0,\\ \left[ (\eta -2)\eta - \psi ^4 \left( \eta + 3\right) \left( \eta + 1\right) \right] (0,0,2,2)&= 0; \end{aligned} \end{aligned}$$
(A.2.8)
By substituting \(u = \psi ^4\) and \(\sigma = u\tfrac{\mathrm {d}}{\mathrm {d}u} = \tfrac{1}{4} \eta \), we obtain the hypergeometric form:
$$\begin{aligned} \left[ \left( \sigma -\tfrac{1}{2}\right) \sigma -u \left( \sigma + \tfrac{3}{4}\right) \left( \sigma + \tfrac{1}{4}\right) \right] \left( 2,2,0,0\right) = 0,\\ \left[ \left( \sigma -\tfrac{1}{2}\right) \sigma -u \left( \sigma + \tfrac{3}{4}\right) \left( \sigma + \tfrac{1}{4}\right) \right] \left( 0,0,2,2\right) = 0, \end{aligned}$$
which is the hypergeometric differential equation \(D(\tfrac{1}{4}, \tfrac{3}{4} ; 1, \tfrac{1}{2} \,|\,\psi ^{4})\). \(\square \)
Proof of Proposition A.2.2
The first three periods are the same for each family. Next, by Lemma A.2.3, all the monomial basis elements in \(W_{\chi _{1}}\) are annihilated by the hypergeometric differential equation \(D(\tfrac{1}{8}, \tfrac{3}{8}, \tfrac{5}{8}, \tfrac{7}{8} ; 0, \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4} \,|\,\psi ^{4})\). Since \(W_{\chi _{1}}\) and \(W_{\chi _{3}}\) are related by a transposition, we get 8 periods annihilated by it. The Picard–Fuchs equations for the last two periods are given by Lemma A.2.6. \(\square \)
A.3. The \({\mathsf {L}}_4\) pencil
Finally, we consider
$$\begin{aligned} F_{\psi } :=x_0^3x_1+x_1^3x_2 + x_2^3 x_3 + x_3^3x_0 - 4\psi x_0x_1x_2x_3, \end{aligned}$$
that defines the pencil of projective hypersurfaces \(X_{\psi } = Z(F_\psi ) \subset {\mathbb {P}}^3\). There is a \(H={\mathbb {Z}}/5{\mathbb {Z}}\) scaling symmetry on \(X_{\psi }\) generated by the element
$$\begin{aligned} g(x_0:x_1:x_2:x_3) = (\xi x_0:\xi ^2x_2:\xi ^4x_2: \xi ^3 x_3) \end{aligned}$$
where \(\xi \) is a fifth root of unity. There are five characters \(\chi _k: H\rightarrow {\mathbb {C}}^\times \) given by \(\chi _k(g) = \xi ^k\). We decompose V into five subspaces \(W_{\chi _k}\). The monomial bases for \(W_{\chi _1}, W_{\chi _2}, W_{\chi _3},\) and \(W_{\chi _4}\) are related by a rotation of the variables \(x_1, x_2, x_3,\) and \(x_4\), so their corresponding Picard–Fuchs equations are the same. We are then left with two types of monomial bases:
- (i)
\(W_{\chi _0}\) has monomial basis \(\{x_0x_1x_2x_3, x_0^2x_2^2, x_1^2x_3^2\}\); and
- (ii)
\(W_{\chi _1}\) has monomial basis \(\{x_0^2x_1^2, x_1^2x_2x_3, x_3^2x_0x_2, x_2^2x_0x_1\}\).
For this family, we can compute the period relations:
$$\begin{aligned} \begin{aligned} {\mathbf {v}} + (3,1,0,0) = \frac{27(1+v_0)- (1+v_1) +3(1+v_2) - 9(1+v_3)}{80(\omega +1)} {\mathbf {v}} + \psi ({\mathbf {v}}+(1,1,1,1)) \end{aligned} \end{aligned}$$
(A.3.1)
and its 4 cyclic permutations.
Proposition A.3.2
For the family \({\mathsf {L}}_4\), the primitive cohomology group \(H^2_{\text {prim}}(X_{{\mathsf {L}}_4, \psi },{\mathbb {C}})\) has 19 periods whose Picard–Fuchs equations are hypergeometric differential equations as follows:
$$\begin{aligned} \begin{aligned} 3 \text { periods are annihilated by } D\left( \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4} ; 1, 1, 1 \,|\,\psi ^{-4}\right) , \\ 4 \text { periods are annihilated by } D\left( \tfrac{1}{5}, \tfrac{2}{5}, \tfrac{3}{5}, \tfrac{4}{5}; 1, \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4} \,|\, \psi ^4\right) , \\ 4 \text { periods are annihilated by } D\left( \tfrac{-1}{5}, \tfrac{1}{5}, \tfrac{2}{5}, \tfrac{3}{5}; 0, \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4} \,|\,\psi ^4\right) , \\ 4 \text { periods are annihilated by } D\left( \tfrac{-2}{5}, \tfrac{-1}{5}, \tfrac{1}{5}, \tfrac{2}{5}; \tfrac{-1}{4}, 0, \tfrac{1}{4}, \tfrac{1}{2} \,|\,\psi ^4\right) , \text {and} \\ 4 \text { periods are annihilated by } D\left( \tfrac{-3}{5}, \tfrac{-2}{5}, \tfrac{-1}{5}, \tfrac{1}{5}; \tfrac{-1}{2}, \tfrac{-1}{4}, 0, \tfrac{1}{4} \,|\,\psi ^4\right) . \end{aligned} \end{aligned}$$
Proof
The period associated to the holomorphic form is found by the same strategy as before.
Lastly, we use (A.3.1) to construct the diagram:
and consequently obtain the following relations:
$$\begin{aligned} \begin{aligned} \eta (1,1,2,0)&= \psi \left( \eta + \tfrac{2}{5}\right) (2,2,0,0), \\ \eta (1,0,1,2)&= \psi \left( \eta + \tfrac{1}{5}\right) (1,1,2,0), \\ \eta (0,2,1,1)&= \psi \left( \eta + \tfrac{4}{5}\right) (1,0,1,2), \\ \eta (2,2,0,0)&= \psi \left( \eta + \tfrac{3}{5}\right) (0,2,1,1). \end{aligned} \end{aligned}$$
(A.3.3)
We then cyclically use these relations to find a recursion which yields the following Picard–Fuchs equations:
$$\begin{aligned} \begin{aligned} \left[ \left( \eta + \tfrac{16}{5}\right) \left( \eta + \tfrac{12}{5}\right) \left( \eta + \tfrac{8}{5}\right) \left( \eta + \tfrac{4}{5}\right) - \tfrac{1}{\psi ^4} (\eta -3)(\eta -2)(\eta -1)\eta \right] (1,0,1,2)&=0, \\ \left[ \left( \eta + \tfrac{17}{5}\right) \left( \eta + \tfrac{13}{5}\right) \left( \eta + \tfrac{9}{5}\right) \left( \eta + \tfrac{1}{5}\right) - \tfrac{1}{\psi ^4} (\eta -3)(\eta -2)(\eta -1)\eta \right] (1,1,2,0)&= 0, \\ \left[ \left( \eta + \tfrac{18}{5}\right) \left( \eta + \tfrac{14}{5}\right) \left( \eta + \tfrac{6}{5}\right) \left( \eta + \tfrac{2}{5}\right) - \tfrac{1}{\psi ^4} (\eta -3)(\eta -2)(\eta -1)\eta \right] (2,2,0,0)&=0, \\ \left[ \left( \eta + \tfrac{19}{5}\right) \left( \eta + \tfrac{11}{5}\right) \left( \eta + \tfrac{7}{5}\right) \left( \eta + \tfrac{3}{5}\right) - \tfrac{1}{\psi ^4} (\eta -3)(\eta -2)(\eta -1)\eta \right] (0,2,1,1)&=0. \end{aligned} \end{aligned}$$
(A.3.4)
By multiplying these lines by 1, \(\psi \), \(\psi ^2\), and \(\psi ^3\), respectively, substituting \(u = \psi ^{4}\), \(\sigma = u \displaystyle {\frac{\mathrm {d}}{\mathrm {d}u}}\), and then multiplying by \(-u\), we obtain the following equations:
$$\begin{aligned}&\left[ \left( \sigma -\tfrac{3}{4}\right) \left( \sigma -\tfrac{1}{2}\right) \left( \sigma -\tfrac{1}{4}\right) \sigma - u\left( \sigma + \tfrac{4}{5}\right) \left( \sigma + \tfrac{3}{5}\right) \left( \sigma + \tfrac{2}{5}\right) \left( \sigma + \tfrac{1}{5}\right) \right] \\&\quad \cdot \left( 1,0,1,2\right) =0 \\&\left[ \left( \sigma -1\right) \left( \sigma -\tfrac{3}{4}\right) \left( \sigma -\tfrac{1}{2}\right) \left( \sigma -\tfrac{1}{4}\right) - u\left( \sigma + \tfrac{3}{5}\right) \left( \sigma + \tfrac{2}{5}\right) \left( \sigma + \tfrac{1}{5}\right) \left( \sigma - \tfrac{1}{5}\right) \right] \psi \\&\quad \cdot \left( 1,1,2,0\right) = 0 \\&\left[ \left( \sigma -\tfrac{5}{4}\right) \left( \sigma -1\right) \left( \sigma -\tfrac{3}{4}\right) \left( \sigma -\tfrac{1}{2}\right) - u\left( \sigma + \tfrac{2}{5}\right) \left( \sigma + \tfrac{1}{5}\right) \left( \sigma - \tfrac{1}{5}\right) \left( \sigma - \tfrac{2}{5}\right) \right] \psi ^2\\&\quad \cdot \left( 2,2,0,0\right) =0\\&\left[ \left( \sigma -\tfrac{3}{2}\right) \left( \sigma -\tfrac{5}{4}\right) \left( \sigma -1\right) \left( \sigma -\tfrac{3}{4}\right) - u\left( \sigma + \tfrac{1}{5}\right) \left( \sigma - \tfrac{1}{5}\right) \left( \sigma - \tfrac{2}{5}\right) \left( \sigma - \tfrac{3}{5}\right) \right] \psi ^3\\&\quad \cdot \left( 0,2,1,1\right) =0. \end{aligned}$$
These are the claimed hypergeometric differential equations. \(\square \)
Appendix B. Finite-field hypergeometric sums
In this Appendix, we write down the details of manipulations of hypergeometric sums.
B.1. Hybrid definition
In this section, we apply the argument of Beukers–Cohen–Mellit to show that the hybrid definition of the finite-field hypergeometric sum reduces to the classical one. We retain the notation from Sects. 3.1–3.2.
Lemma B.1.1
Suppose that q is good and splittable for \(\varvec{\alpha },\varvec{\beta }\). If \(\alpha _i q^{\times }, \beta _i q^{\times }\in {\mathbb {Z}}\) for all \(i=1,\dots ,d\), then Definitions 3.1.6 and 3.2.7 agree.
Proof
Our proof follows Beukers–Cohen–Mellit [3, Theorem 1.3]. We consider
$$\begin{aligned} G(m+\varvec{\alpha }'q^{\times },-m-\varvec{\beta }'q^{\times }) = \prod _{\alpha _i' \in \varvec{\alpha }'}\frac{g(m+ \alpha _i'q^{\times })}{g(\alpha _i q^{\times })} \prod _{\beta _i' \in \varvec{\beta }'}\frac{g(-m-\beta _i'q^{\times })}{g(-\beta _i q^{\times })}. \end{aligned}$$
We massage this expression, and for simplicity drop the subscripts 0. First,
$$\begin{aligned}&D\left( x\right) \prod _{\alpha _j \in \hat{\varvec{\alpha }}} \left( x- e^{2\pi \sqrt{-1} \alpha _j}\right) = \prod _{j=1}^r \left( x^{p_j} - 1\right) \quad \text {and} \\&D\left( x\right) \prod _{\beta _j \in \tilde{\varvec{\beta }}}\left( x- e^{2\pi \sqrt{-1} \beta _j}\right) = \prod _{j=1}^s \left( x^{q_j} - 1\right) . \end{aligned}$$
Write \(D(x) = \prod _{j=1}^\delta (x- e^{2\pi \sqrt{-1} c_j/q^{\times }})\). Then
$$\begin{aligned} \begin{aligned}&G(m+\varvec{\alpha }'q^{\times },-m-\varvec{\beta }'q^{\times }) \\&\quad = \left( \prod _{i=1}^r \prod _{j=0}^{p_i-1} \frac{g(m+ jq^{\times }\,/p_i)}{g(jq^{\times }\,/p_i)}\right) \left( \prod _{i=1}^s \prod _{j=0}^{q_i-1} \frac{g(-m-jq^{\times }\,/q_i)}{g(-jq^{\times }\,/q_i)}\right) \prod _{j=1}^\delta \frac{g(c_j)g(-c_j)}{g(m+c_j) g(-m-c_j)}. \end{aligned} \end{aligned}$$
(B.1.2)
Since \(p_i\) divides \(q^{\times }\), by the Hasse–Davenport relation (Lemma 3.1.3(c)) we have that
$$\begin{aligned} \prod _{j=0}^{p_i-1} \frac{g(m+ jq^{\times }\,/p_i)}{g(jq^{\times }\,/p_i)} = \frac{-g(p_i m)}{\omega (p_i)^{p_im}}. \end{aligned}$$
(B.1.3)
Analogously, since \(q_i\) divides \(q^{\times }\), we use Hasse–Davenport to find that
$$\begin{aligned} \prod _{j=0}^{q_i-1} \frac{g(-m-jq^{\times }\,/q_i)}{g(-jq^{\times }\,/q_i)} = -g(-q_i m) \omega (q_i)^{q_i m}. \end{aligned}$$
(B.1.4)
Note that if \(c_j \ne 0\), then \(g(c_j)g(-c_j) = (-1)^{c_j}q\) and 1 otherwise, hence
$$\begin{aligned} \prod _{i=1}^\delta g(c_j)g(-c_j) = (-1)^{\sum c_j} q^{\delta - s(0)}, \end{aligned}$$
where s(0) is the multiplicity of 1 in D(x), or, equivalently, the number of times \(c_j\) is 0. Now, note the number of times that \(m+c_j = 0\) is the multiplicity of the root \(e^{-2\pi \sqrt{-1} m / q^{\times }}\) in D(x), which, equivalently, is the multiplicity of \(e^{2\pi \sqrt{-1} m / q^{\times }}\) in D(x) as D(x) is a product of cyclotomic polynomials. This implies that
$$\begin{aligned} \prod _{j=1}^\delta g(m+c_j)g(-m-c_j) = (-1)^{m+c_j} q^{\delta -\lambda (m)}. \end{aligned}$$
We then have that
$$\begin{aligned} \prod _{j=1}^\delta \frac{g(c_j)g(-c_j)}{g(m+c_j) g(-m-c_j)} = \frac{ (-1)^{\sum _j c_j} q^{\delta - s(0)}}{ (-1)^{\sum _j(m+c_j)} q^{\delta -s(m)}} = (-1)^{-\delta m} q^{s(m) - s(0)}. \end{aligned}$$
(B.1.5)
Combining Eqs. (B.1.3), (B.1.4), and (B.1.5), we then have
$$\begin{aligned}&G(m+\varvec{\alpha }'q^{\times },-m-\varvec{\beta }'q^{\times }) \\&\quad = \left( \prod _{i=1}^r \frac{-g(p_i m)}{\omega (p_i)^{p_im}}\right) \left( \prod _{i=1}^s -g(-q_i m) \omega (q_i)^{q_i m}\right) \left( (-1)^{-\delta m} q^{s(m) - s(0)}\right) \\&\quad = (-1)^{r+s}q^{s(m) - s(0)} g(p_1m)\cdots g(p_rm) g(-q_1m) \cdots g(-q_sm) \\&\qquad \cdot \omega ((-1)^{\delta }p_1^{-p_1} \cdots p_r^{-p_r} q_1^{q_1} \cdots q_s^{q_s})^m\\&\quad = (-1)^{r+s}q^{s(m) - s(0)} g(p_1m)\cdots g(p_rm) g(-q_1m) \cdots g(-q_sm) \omega ((-1)^{\delta }M)^m. \end{aligned}$$
By plugging this equation into Definition 3.1.6 for the appropriate factors, we obtain the quantity given in Definition 3.2.7. \(\square \)
B.2. The pencil \({\mathsf {F}}_2{\mathsf {L}}_2\)
Proposition B.2.1
The number of \({\mathbb {F}}_q\)-points on \({\mathsf {F}}_2{\mathsf {L}}_2\) can be written in terms of hypergeometric functions, as follows:
- (a)
If \(q \equiv 3\pmod 4\), then
$$\begin{aligned} X_{{\mathsf {F}}_2{\mathsf {L}}_2,\psi }\left( {\mathbb {F}}_q\right)&=q^2-q+1+H_q\left( \tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4};0,0,0\,|\,\psi ^{-4}\right) - q H_q\left( \tfrac{1}{4},\tfrac{3}{4};0,\tfrac{1}{2} \,|\,\psi ^{-4}\right) . \end{aligned}$$
- (b)
If \(q\equiv 5\pmod 8\), then
$$\begin{aligned} X_{{\mathsf {F}}_2{\mathsf {L}}_2,\psi }\left( {\mathbb {F}}_q\right)&=q^2-q+1+H_q\left( \tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4};0,0,0\,|\,\psi ^{-4}\right) \\&\quad +qH_q\left( \tfrac{1}{4},\tfrac{3}{4};0,\tfrac{1}{2} \,|\,\psi ^{-4}\right) -2qH_q\left( \tfrac{1}{2};0\,|\,\psi ^{-4}\right) . \end{aligned}$$
- (c)
If \(q\equiv 1\pmod 8\), then
$$\begin{aligned} X_{{\mathsf {F}}_2{\mathsf {L}}_2,\psi }\left( {\mathbb {F}}_q\right)&= q^2+7q +1 +H_q\left( \tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4};0,0,0\,|\,\psi ^{-4}\right) +qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) \\&\quad + 2q H_q\left( \tfrac{1}{2}; 0 \,|\, \psi ^{-4}\right) + 2\omega \left( 2\right) ^{q^{\times }\,/4}qH_q\left( \tfrac{1}{8}, \tfrac{5}{8}; 0, \tfrac{1}{4} \,|\, \psi ^4\right) \\&\quad + 2\omega \left( 2\right) ^{q^{\times }\,/4}qH_q\left( \tfrac{3}{8}, \tfrac{7}{8}; 0, \tfrac{3}{4} \,|\, \psi ^4\right) . \end{aligned}$$
Remark B.2.2
Notice that the hypergeometric functions appearing in the point count correspond to the Picard–Fuchs equations in Proposition 2.7.1. We also see the appearance of six additional trivial factors.
Step 1: Computing and clustering the characters
To use Theorem 3.3.3 we compute the subset \(S\subset ({\mathbb {Z}}/q^{\times }{\mathbb {Z}})^r\) given by the constraints in (3.3.1).
- (a)
If \(q\equiv 3 \pmod 4\), then \(S\) can be clustered in the following way:
- (i)
the set \(S_1 = \{ k(1,1,1,1,-4) : k \in {\mathbb {Z}}/q^{\times }{\mathbb {Z}}\}\) and
- (ii)
the set \(S_4 = \{ k(1,1,1,1,-4) + \tfrac{q^{\times }}{2}(0,0,1,1,0): k \in {\mathbb {Z}}/q^{\times }{\mathbb {Z}}\}\).
- (b)
If \(q\equiv 5 \pmod 8\), then \(S\) contains the two sets above and:
- (i)
the set \(S_5 = \{ k(1,1,1,1,-4) + \tfrac{q^{\times }}{4}(0,2,1,1,0): k \in {\mathbb {Z}}/q^{\times }{\mathbb {Z}}\}\) and
- (ii)
the set \(S_6 = \{ k(1,1,1,1,-4) + 3\tfrac{q^{\times }}{4}(0,2,1,1,0): k \in {\mathbb {Z}}/q^{\times }{\mathbb {Z}}\}\).
- (c)
If \(q\equiv 1 \pmod 8\), then \(S\) contains the four sets above and
- (i)
two sets of the form \(S_{10} =\{ k(1,1,1,1,-4) + \tfrac{q^{\times }}{8}(0,2,1,5,0): k \in {\mathbb {Z}}/q^{\times }{\mathbb {Z}}\}\) and
- (ii)
two sets of the form \(S_{11} =\{ k(1,1,1,1,-4) + \tfrac{q^{\times }}{8}(0,6,7,3,0): k \in {\mathbb {Z}}/q^{\times }{\mathbb {Z}}\}\).
Step 2: Counting points on the open subset with nonzero coordinates
Lemma B.2.3
Suppose \(\psi \in {\mathbb {F}}_q^{\times }\).
- (a)
If \(q \equiv 3 \pmod 4\), then
$$\begin{aligned} \#U_{{\mathsf {F}}_2{\mathsf {L}}_2, \psi }\left( {\mathbb {F}}_q\right) = q^2-3q+1+H_q\left( \tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4};0,0,0\,|\,\psi ^{-4}\right) - qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) . \end{aligned}$$
(B.2.4)
- (b)
If \(q \equiv 5 \pmod 8\), then
$$\begin{aligned} \begin{aligned} \#U_{{\mathsf {F}}_2{\mathsf {L}}_2, \psi }\left( {\mathbb {F}}_q\right)&= q^2-3q+3+H_q\left( \tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4};0,0,0\,|\,\psi ^{-4}\right) + qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) \\&\quad -2q H_q\left( \tfrac{1}{2}; 0 \,|\, \psi ^{-4}\right) - 2 \left( \frac{g\left( \tfrac{q^{\times }}{4}\right) ^2 + g\left( \tfrac{3q^{\times }}{4}\right) ^2}{g\left( \tfrac{q^{\times }}{2}\right) }\right) . \end{aligned} \end{aligned}$$
(B.2.5)
- (c)
If \(q \equiv 1 \pmod 8\), then
$$\begin{aligned} \begin{aligned} \#U_{{\mathsf {F}}_2{\mathsf {L}}_2, \psi }\left( {\mathbb {F}}_q\right)&= q^2-3q+7+H_q\left( \tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4};0,0,0\,|\,\psi ^{-4}\right) +qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) \\&\quad + 2q H_q\left( \tfrac{1}{2}; 0 \,|\, \psi ^{-4}\right) + 2\omega \left( 2\right) ^{q^{\times }\,/4}qH_q\left( \tfrac{1}{8}, \tfrac{5}{8}; 0, \tfrac{1}{4} \,|\, \psi ^4\right) \\&\quad + 2\omega \left( 2\right) ^{q^{\times }\,/4}qH_q\left( \tfrac{3}{8}, \tfrac{7}{8}; 0, \tfrac{3}{4} \,|\, \psi ^4\right) \\&\quad - 2\left( \frac{g\left( \tfrac{q^{\times }}{4}\right) ^2 + g\left( \tfrac{3q^{\times }}{4}\right) ^2}{g\left( \tfrac{q^{\times }}{2}\right) }\right) - \tfrac{4}{q}g\left( \tfrac{q^{\times }}{8}\right) g\left( \tfrac{5q^{\times }}{8}\right) g\left( \tfrac{q^{\times }}{4}\right) \\&\quad - \tfrac{4}{q}g\left( \tfrac{3q^{\times }}{8}\right) g\left( \tfrac{7q^{\times }}{8}\right) g\left( \tfrac{3q^{\times }}{4}\right) . \end{aligned} \end{aligned}$$
(B.2.6)
Proof
If \(q \equiv 3\pmod 4\), then by Lemmas 3.4.4 and 3.4.7
$$\begin{aligned} \begin{aligned} \#U_{{\mathsf {F}}_2{\mathsf {L}}_2, \psi }\left( {\mathbb {F}}_q\right)&= \sum _{s \in S_1} \omega \left( a\right) ^{-s} c_s + \sum _{s \in S_4} \omega \left( a\right) ^{-s} c_s \\&= \left( q^2-3q+3+H_q\left( \tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4};0,0,0\,|\,\psi ^{-4}\right) \right) + \\&\qquad -2 - qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) \\&= q^2-3q+1+H_q\left( \tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4};0,0,0\,|\,\psi ^{-4}\right) - qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) . \end{aligned} \end{aligned}$$
(B.2.7)
If \(q \equiv 5 \pmod 8\), then \(q\equiv 1 \pmod 4 \) but \(q\not \equiv 1 \pmod 8\), so by Lemmas 3.4.4, 3.4.7, 3.4.12, and 3.4.21
$$\begin{aligned} \begin{aligned} \#U_{{\mathsf {F}}_2{\mathsf {L}}_2, \psi }\left( {\mathbb {F}}_q\right)&= \sum _{s \in S_1} \omega \left( a\right) ^{-s} c_s + \sum _{s \in S_4} \omega \left( a\right) ^{-s} c_s + \sum _{s \in S_5} \omega \left( a\right) ^{-s} c_s + \sum _{s \in S_6} \omega \left( a\right) ^{-s} c_s \\&= \left( q^2-3q+3+H_q\left( \tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4};0,0,0\,|\,\psi ^{-4}\right) \right) \\&\quad + \left( 2 + qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) \right) \\&\quad + \left( -1\right) ^{q^{\times }\,/4} q H_q\left( \tfrac{1}{2}; 0 \,|\, \psi ^{-4}\right) + \left( -1\right) ^{q^{\times }\,/4} - \left( \frac{g\left( \tfrac{q^{\times }}{4}\right) ^2 + g\left( \tfrac{3q^{\times }}{4}\right) ^2}{g\left( \tfrac{q^{\times }}{2}\right) }\right) \\&\quad + \left( -1\right) ^{q^{\times }\,/4} q H_q\left( \tfrac{1}{2}; 0 \,|\, \psi ^{-4}\right) + \left( -1\right) ^{q^{\times }\,/4} - \left( \frac{g\left( \tfrac{q^{\times }}{4}\right) ^2 + g\left( \tfrac{3q^{\times }}{4}\right) ^2}{g\left( \tfrac{q^{\times }}{2}\right) }\right) \\&= q^2-3q+3+H_q\left( \tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4};0,0,0\,|\,\psi ^{-4}\right) + qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) )\\&\quad -2q H_q\left( \tfrac{1}{2}; 0 \,|\, \psi ^{-4}\right) - 2 \left( \frac{g\left( \tfrac{q^{\times }}{4}\right) ^2 + g\left( \tfrac{3q^{\times }}{4}\right) ^2}{g\left( \tfrac{q^{\times }}{2}\right) }\right) . \end{aligned} \end{aligned}$$
(B.2.8)
If \(q\equiv 1 \pmod 8\), then by Lemmas 3.4.4, 3.4.7, 3.4.12, 3.4.21, B.2.12, and B.2.17, we have:
$$\begin{aligned} \begin{aligned} \#U_{{\mathsf {F}}_2{\mathsf {L}}_2, \psi }\left( {\mathbb {F}}_q\right)&= \sum _{s \in S_1} \omega \left( a\right) ^{-s} c_s + \sum _{s \in S_4} \omega \left( a\right) ^{-s} c_s + \sum _{s \in S_5} \omega \left( a\right) ^{-s} c_s + \sum _{s \in S_6} \omega \left( a\right) ^{-s} c_s \\&\quad + \sum _{s \in S_{10}} \omega \left( a\right) ^{-s} c_s + \sum _{s \in S_{11}} \omega \left( a\right) ^{-s} c_s \\&= \left( q^2-3q+3+H_q\left( \tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4};0,0,0\,|\,\psi ^{-4}\right) \right) \\&\quad + \left( 2 + qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) \right) \\&\quad + 2q H_q\left( \tfrac{1}{2}; 0 \,|\, \psi ^{-4}\right) + 2 - 2\left( \frac{g\left( \tfrac{q^{\times }}{4}\right) ^2 + g\left( \tfrac{3q^{\times }}{4}\right) ^2}{g\left( \tfrac{q^{\times }}{2}\right) }\right) \\&\quad + 2\omega \left( 2\right) ^{q^{\times }\,/4} qH_q\left( \tfrac{1}{8}, \tfrac{5}{8}; 0, \tfrac{1}{4} \,|\, \psi ^4\right) + 2\omega \left( 2\right) ^{q^{\times }\,/4}qH_q\left( \tfrac{3}{8}, \tfrac{7}{8}; 0, \tfrac{3}{4} \,|\, \psi ^4\right) \\&\quad - \tfrac{4}{q}g\left( \tfrac{q^{\times }}{8}\right) g\left( \tfrac{5q^{\times }}{8}\right) g\left( \tfrac{q^{\times }}{4}\right) - \tfrac{4}{q}g\left( \tfrac{3q^{\times }}{8}\right) g\left( \tfrac{7q^{\times }}{8}\right) g\left( \tfrac{3q^{\times }}{4}\right) \\&= q^2-3q+7+H_q\left( \tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4};0,0,0\,|\,\psi ^{-4}\right) +qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) \\&\quad + 2q H_q\left( \tfrac{1}{2}; 0 \,|\, \psi ^{-4}\right) + 2\omega \left( 2\right) ^{q^{\times }\,/4} qH_q\left( \tfrac{1}{8}, \tfrac{5}{8}; 0, \tfrac{1}{4} \,|\, \psi ^4\right) \\&\quad + 2\omega \left( 2\right) ^{q^{\times }\,/4}qH_q\left( \tfrac{3}{8}, \tfrac{7}{8}; 0, \tfrac{3}{4} \,|\, \psi ^4\right) - 2\left( \frac{g\left( \tfrac{q^{\times }}{4}\right) ^2 + g\left( \tfrac{3q^{\times }}{4}\right) ^2}{g\left( \tfrac{q^{\times }}{2}\right) }\right) \\&\quad - \tfrac{4}{q}g\left( \tfrac{q^{\times }}{8}\right) g\left( \tfrac{5q^{\times }}{8}\right) g\left( \tfrac{q^{\times }}{4}\right) - \tfrac{4}{q}g\left( \tfrac{3q^{\times }}{8}\right) g\left( \tfrac{7q^{\times }}{8}\right) g\left( \tfrac{3q^{\times }}{4}\right) \end{aligned} \end{aligned}$$
(B.2.9)
as claimed. \(\square \)
Before proving the lemmas that associate the quantities \(\sum _{s \in S_{10}}\omega (a)^{-s} c_s\) and \(\sum _{s \in S_{11}}\omega (a)^{-s} c_s\) to hypergeometric sums, we need the following lemma.
Lemma B.2.10
Suppose \(q \equiv 1 \pmod 8\) and \(q = p^r\) for some natural number r and prime p. Then
$$\begin{aligned} \frac{g\left( \tfrac{3q^{\times }}{4}\right) g\left( \tfrac{q^{\times }}{8}\right) g\left( \tfrac{5q^{\times }}{8}\right) }{g\left( \tfrac{q^{\times }}{2}\right) } = \omega \left( 2\right) ^{q^{\times }\,/4} q. \end{aligned}$$
Proof
Since \(q\equiv 1 \pmod 8\), we can use Hasse–Davenport with \(N=2\) and \(m = \tfrac{q^{\times }}{8}\) to get that
$$\begin{aligned} g\left( \tfrac{q^{\times }}{4}\right) = \omega \left( 2\right) ^{q^{\times }/4} \frac{g\left( \tfrac{q^{\times }}{8}\right) g\left( \tfrac{5q^{\times }}{8}\right) }{g\left( \tfrac{q^{\times }}{2}\right) }. \end{aligned}$$
(B.2.11)
By multiplying both sides by \(g(\tfrac{3q^{\times }}{4})\), and dividing by \(\omega (2)^{q^{\times }/4}\), we have
$$\begin{aligned} \omega \left( 2\right) ^{-q^{\times }/4}g\left( \tfrac{q^{\times }}{4}\right) g\left( \tfrac{3q^{\times }}{4}\right) =\frac{g\left( \tfrac{q^{\times }}{8}\right) g\left( \tfrac{5q^{\times }}{8}\right) g\left( \tfrac{3q^{\times }}{4}\right) }{g\left( \tfrac{q^{\times }}{2}\right) }. \end{aligned}$$
We obtain the identity above after noting that \(g(\tfrac{q^{\times }}{4})g(\tfrac{3q^{\times }}{4}) = (-1)^{q^{\times }/4} q = q\), since \(q\equiv 1 \pmod 8\) and that \(\omega (2)^{q^{\times }\,/4} = \pm 1\) when \(q\equiv 1\pmod 8\); hence, \(\omega (2)^{q^{\times }\,/4} = \omega (2)^{-q^{\times }\,/4}\). \(\square \)
Lemma B.2.12
Suppose \(q \equiv 1 \pmod 8\). Then for
$$\begin{aligned} S_{10} = \left\{ k(1,1,1,1,-4) + \tfrac{q^{\times }}{8}(0,2,1,5,0) : k \in {\mathbb {Z}}/q^{\times }{\mathbb {Z}}\right\} \end{aligned}$$
we have
$$\begin{aligned}&\sum _{s \in S_{10}} \omega \left( a\right) ^{-s} c_s = \omega \left( 2\right) ^{q^{\times }\,/4}qH_q\left( \tfrac{1}{8}, \tfrac{5}{8}; 0, \tfrac{1}{4} \,|\, \psi ^4\right) - \frac{1}{q}g\left( \tfrac{q^{\times }}{8}\right) g\left( \tfrac{5q^{\times }}{8}\right) g\left( \tfrac{q^{\times }}{4}\right) \\&\quad - \frac{1}{q}g\left( \tfrac{3q^{\times }}{8}\right) g\left( \tfrac{7q^{\times }}{8}\right) g\left( \tfrac{3q^{\times }}{4}\right) . \end{aligned}$$
Proof
First, we take the definition of the sum and take out all terms that are of the form \(\ell \tfrac{q^{\times }}{4}\) to obtain the equality:
$$\begin{aligned} \begin{aligned} \sum _{s \in S_{10}} \omega \left( a\right) ^{-s} c_s&= \frac{1}{qq^{\times }} g\left( \tfrac{q^{\times }}{4}\right) g\left( \tfrac{q^{\times }}{8}\right) g\left( \tfrac{5q^{\times }}{8}\right) - \frac{1}{qq^{\times }} g\left( \tfrac{q^{\times }}{4}\right) g\left( \tfrac{q^{\times }}{2}\right) g\left( \tfrac{3q^{\times }}{8}\right) g\left( \tfrac{7q^{\times }}{8}\right) \\&\quad - \frac{1}{qq^{\times }} g\left( \tfrac{q^{\times }}{2}\right) g\left( \tfrac{3q^{\times }}{4}\right) g\left( \tfrac{q^{\times }}{8}\right) g\left( \tfrac{5q^{\times }}{8}\right) \\&\quad + \frac{1}{qq^{\times }} g\left( \tfrac{3q^{\times }}{4}\right) g\left( \tfrac{7q^{\times }}{8}\right) g\left( \tfrac{q^{\times }}{8}\right) \\&\quad + \frac{1}{qq^{\times }} \sum _{{\mathop {4k\not \equiv 0~({\text {mod}}~{q^{\times }})}\limits ^{k=0}}}^{q-2}\omega \left( 4\psi \right) ^{4k}g\left( k\right) \\&\quad \cdot g\left( k+\tfrac{q^{\times }}{4}\right) g\left( k + \tfrac{q^{\times }}{8}\right) g\left( k+\tfrac{5q^{\times }}{8}\right) g\left( -4k\right) . \end{aligned} \end{aligned}$$
(B.2.13)
Next, we use the Hasse–Davenport relationship to expand \(g(-4k)\) and then use relation from 3.1.3(b) to cancel out the \(g(k+\tfrac{q^{\times }}{4})\) factor in the summation. Through this, we obtain:
$$\begin{aligned} \begin{aligned} \sum _{s \in S_{10}} \omega \left( a\right) ^{-s} c_s&= \frac{1}{qq^{\times }} g\left( \tfrac{q^{\times }}{4}\right) g\left( \tfrac{q^{\times }}{8}\right) g\left( \tfrac{5q^{\times }}{8}\right) - \frac{1}{qq^{\times }} g\left( \tfrac{q^{\times }}{4}\right) g\left( \tfrac{q^{\times }}{2}\right) g\left( \tfrac{3q^{\times }}{8}\right) g\left( \tfrac{7q^{\times }}{8}\right) \\&\quad - \frac{1}{qq^{\times }} g\left( \tfrac{q^{\times }}{2}\right) g\left( \tfrac{3q^{\times }}{4}\right) g\left( \tfrac{q^{\times }}{8}\right) g\left( \tfrac{5q^{\times }}{8}\right) \\&\quad + \frac{1}{qq^{\times }} g\left( \tfrac{3q^{\times }}{4}\right) g\left( \tfrac{7q^{\times }}{8}\right) g\left( \tfrac{q^{\times }}{8}\right) \\&\quad + \frac{1}{q^{\times }} \sum _{{\mathop {4k\not \equiv 0~({\text {mod}}~{q^{\times }})}\limits ^{k=0}}}^{q-2} \omega \left( \psi \right) ^{4k} \\&\quad \cdot \frac{g\left( k+\tfrac{q^{\times }}{8}\right) g\left( k+\tfrac{5q^{\times }}{8}\right) g\left( -k+\tfrac{q^{\times }}{4}\right) g\left( -k+\tfrac{q^{\times }}{2}\right) }{g\left( \tfrac{q^{\times }}{2}\right) }. \end{aligned} \end{aligned}$$
(B.2.14)
Here, we reindex the summation by \(m = k+ \tfrac{q^{\times }}{2}\) to obtain:
$$\begin{aligned} \begin{aligned} \sum _{s \in S_{10}} \omega \left( a\right) ^{-s} c_s&= \frac{1}{qq^{\times }} g\left( \tfrac{q^{\times }}{4}\right) g\left( \tfrac{q^{\times }}{8}\right) g\left( \tfrac{5q^{\times }}{8}\right) - \frac{1}{qq^{\times }} g\left( \tfrac{q^{\times }}{4}\right) g\left( \tfrac{q^{\times }}{2}\right) g\left( \tfrac{3q^{\times }}{8}\right) g\left( \tfrac{7q^{\times }}{8}\right) \\&\quad - \frac{1}{qq^{\times }} g\left( \tfrac{q^{\times }}{2}\right) g\left( \tfrac{3q^{\times }}{4}\right) g\left( \tfrac{q^{\times }}{8}\right) g\left( \tfrac{5q^{\times }}{8}\right) \\&\quad + \frac{1}{qq^{\times }} g\left( \tfrac{3q^{\times }}{4}\right) g\left( \tfrac{7q^{\times }}{8}\right) g\left( \tfrac{q^{\times }}{8}\right) \\&\quad + \frac{1}{q-1} \sum _{{\mathop {4m\not \equiv 0~({\text {mod}}~{q^{\times }})}\limits ^{m=0}}}^{q-2} \omega \left( \psi \right) ^{4m} \\&\quad \cdot \frac{g\left( m+\tfrac{q^{\times }}{8}\right) g\left( m+\tfrac{5q^{\times }}{8}\right) g\left( -m-\tfrac{q^{\times }}{4}\right) g\left( -m\right) }{g\left( \tfrac{q^{\times }}{2}\right) }. \end{aligned} \end{aligned}$$
(B.2.15)
We now multiply the final form by the expression
$$\begin{aligned} \frac{g\left( \tfrac{3q^{\times }}{4}\right) g\left( \tfrac{q^{\times }}{8}\right) g\left( \tfrac{5q^{\times }}{8}\right) }{g\left( \tfrac{3q^{\times }}{4}\right) g\left( \tfrac{q^{\times }}{8}\right) g\left( \tfrac{5q^{\times }}{8}\right) }=1. \end{aligned}$$
We put the denominator of this factor into the summation to relate the summation to a hypergeometric function but factor out the numerator along with a factor of \(g(\tfrac{q^{\times }}{2})\). We then apply Lemma B.2.10 to this factor ahead of the summation. We thus obtain:
$$\begin{aligned} \sum _{s \in S_{10}} \omega \left( a\right) ^{-s} c_s&= \frac{1}{qq^{\times }} g\left( \tfrac{q^{\times }}{4}\right) g\left( \tfrac{q^{\times }}{8}\right) g\left( \tfrac{5q^{\times }}{8}\right) - \frac{1}{qq^{\times }} g\left( \tfrac{q^{\times }}{4}\right) g\left( \tfrac{q^{\times }}{2}\right) g\left( \tfrac{3q^{\times }}{8}\right) g\left( \tfrac{7q^{\times }}{8}\right) \nonumber \\&\quad - \frac{1}{qq^{\times }} g\left( \tfrac{q^{\times }}{2}\right) g\left( \tfrac{3q^{\times }}{4}\right) g\left( \tfrac{q^{\times }}{8}\right) g\left( \tfrac{5q^{\times }}{8}\right) \nonumber \\&\quad + \frac{1}{qq^{\times }} g\left( \tfrac{3q^{\times }}{4}\right) g\left( \tfrac{7q^{\times }}{8}\right) g\left( \tfrac{q^{\times }}{8}\right) \nonumber \\&\quad - \frac{ \omega \left( 2\right) ^{q^{\times }\,/4} q}{q^{\times }} \sum _{{\mathop {4m\not \equiv 0~({\text {mod}}~{q^{\times }})}\limits ^{m=0}}}^{q-2} \omega \left( \psi \right) ^{4m} \nonumber \\&\quad \cdot \frac{g\left( m+\tfrac{q^{\times }}{8}\right) g\left( m+\tfrac{5q^{\times }}{8}\right) g\left( -m-\tfrac{q^{\times }}{4}\right) g\left( -m\right) }{g\left( \tfrac{q^{\times }}{8}\right) g\left( \tfrac{5q^{\times }}{8}\right) g\left( -\tfrac{q^{\times }}{4}\right) g\left( 0\right) }. \end{aligned}$$
(B.2.16)
By comparing terms of the summations above and the hypergeometric function itself, we obtain the desired result. \(\square \)
Lemma B.2.17
Suppose \(q \equiv 1 \pmod 8\). Then for
$$\begin{aligned} S_{11} = \left\{ k(1,1,1,1,-4) +\tfrac{q^{\times }}{8}(0,6,7,3,0) : k \in {\mathbb {Z}}/q^{\times }{\mathbb {Z}}\right\} \end{aligned}$$
we have
$$\begin{aligned}&\sum _{s \in S_{11}} \omega \left( a\right) ^{-s} c_s = \omega \left( 2\right) ^{q^{\times }\,/4} qH_q\left( \tfrac{3}{8}, \tfrac{7}{8}; 0, \tfrac{3}{4} \,|\, \psi ^4\right) \\&\quad - \frac{1}{q}g\left( \tfrac{q^{\times }}{8}\right) g\left( \tfrac{5q^{\times }}{8}\right) g\left( \tfrac{q^{\times }}{4}\right) - \frac{1}{q}g\left( \tfrac{3q^{\times }}{8}\right) g\left( \tfrac{7q^{\times }}{8}\right) g\left( \tfrac{3q^{\times }}{4}\right) . \end{aligned}$$
Proof
The proof is analogous to the proof in Lemma B.2.12 except we substitute \(m = k + \tfrac{q^{\times }}{2}\). Alternatively, apply complex conjugation to the conclusion of Lemma B.2.12, negating indices as in the proof of Lemma 3.5.6. \(\square \)
Step 3: Count points when at least one coordinate is zero
Lemma B.2.18
The following statements hold.
- (a)
If \(q\equiv 3 \pmod 4\), then
$$\begin{aligned} \#X_{{\mathsf {F}}_2{\mathsf {L}}_2, \psi }({\mathbb {F}}_q) - \#U_{{\mathsf {F}}_2{\mathsf {L}}_2, \psi }({\mathbb {F}}_q) = 2q. \end{aligned}$$
- (b)
If \(q\equiv 5 \pmod 8\), then
$$\begin{aligned} \#X_{{\mathsf {F}}_2{\mathsf {L}}_2, \psi }\left( {\mathbb {F}}_q\right) - \#U_{{\mathsf {F}}_2{\mathsf {L}}_2, \psi }\left( {\mathbb {F}}_q\right) = 2q-2 + 2\left( \frac{g\left( \tfrac{q^{\times }}{4}\right) ^2 + g\left( \tfrac{3q^{\times }}{4}\right) ^2}{g\left( \tfrac{q^{\times }}{2}\right) }\right) . \end{aligned}$$
- (c)
If \(q \equiv 1 \pmod 8\), then
$$\begin{aligned} \#X_{{\mathsf {F}}_2{\mathsf {L}}_2, \psi }\left( {\mathbb {F}}_q\right) - \#U_{{\mathsf {F}}_2{\mathsf {L}}_2, \psi }\left( {\mathbb {F}}_q\right)&= 10q - 6 + 2\left( \frac{g\left( \tfrac{q^{\times }}{4}\right) ^2 + g\left( \tfrac{3q^{\times }}{4}\right) ^2}{g\left( \tfrac{q^{\times }}{2}\right) }\right) \\&\quad + \tfrac{4}{q}g\left( \tfrac{q^{\times }}{4}\right) g\left( \tfrac{q^{\times }}{8}\right) g\left( \tfrac{5q^{\times }}{8}\right) \\&\quad + \tfrac{4}{q}g\left( \tfrac{3q^{\times }}{4}\right) g\left( \tfrac{3q^{\times }}{8}\right) g\left( \tfrac{7q^{\times }}{8}\right) . \end{aligned}$$
Proof
We do this case by case. If \(x_1\) is the only variable equaling zero, then we must count the number of solutions in the open torus for the hypersurface \(Z(x^4 + y^3z+ z^3y)\subset {\mathbb {P}}^2\). We can see by using Theorem 3.3.3 that this depends on q. Here, in case (a) we get \(q-1\) points, in case (b) we get \(q-3 + (g(\tfrac{q^{\times }}{4})^2 + g(\tfrac{3q^{\times }}{4})^2)g(\tfrac{q^{\times }}{2})^{-1}\), and in case (c) we get \(q-3 + (g(\tfrac{q^{\times }}{4})^2 + g(\tfrac{3q^{\times }}{4})^2)g(\tfrac{q^{\times }}{2})^{-1} + 2q^{-1}(g(\tfrac{q^{\times }}{4})g(\tfrac{q^{\times }}{8})g(\tfrac{5q^{\times }}{8}) + g(\tfrac{3q^{\times }}{4})g(\tfrac{3q^{\times }}{8})g(\tfrac{7q^{\times }}{8}))\). There are two such cases, when either \(x_1\) or \(x_2\) is the only variable equaling 0.
Next is when both \(x_1\) and \(x_2\) are zero and the other two variables are nonzero. Here, the number of solutions is \(1 + (-1)^{q^{\times }\,/2}\).
Next is when \(x_3\) is zero but the rest are nonzero. Here, this is \((q-1)\) times the number of solutions of \(Z(x^4+y^4)\) in the open torus of \({\mathbb {P}}^1\). We then get that the number of solutions is 0 if \(q \not \equiv 1 \pmod 8\) and 4 if \(q\equiv 1 \pmod 8\), hence \(4q-4\) points. There are two such cases, when \(x_3\) or \(x_4\) are uniquely zero.
The next case is when \(x_3\) and \(x_4\) are both zero. Then the number of nonzero solutions is exactly the number of solutions of \(Z(x^4+y^4)\) in the open torus of \({\mathbb {P}}^1\), i.e., 0 if \(q \not \equiv 1 \pmod 8\) and 4 if \(q\equiv 1 \pmod 8\).
There are no rational points where \(x_1\) and \(x_3\) are both zero and \(x_2\) nonzero and the same when you swap \(x_1\) with \(x_2\) or \(x_3\) with \(x_4\). Finally, there are two more solutions: (0 : 0 : 1 : 0) and (0 : 0 : 0 : 1). We now count.
- (a)
If \(q\not \equiv 1 \pmod 4\) and q is odd, then
$$\begin{aligned} \#X_{{\mathsf {F}}_2{\mathsf {L}}_2, \psi }({\mathbb {F}}_q) - \#U_{{\mathsf {F}}_2{\mathsf {L}}_2, \psi }({\mathbb {F}}_q) = 2(q-1) + 0 + 0 + 0 + 2 = 2q. \end{aligned}$$
- (b)
If \(q\equiv 5 \pmod 8\), then
$$\begin{aligned} \#X_{{\mathsf {F}}_2{\mathsf {L}}_2, \psi }\left( {\mathbb {F}}_q\right) - \#U_{{\mathsf {F}}_2{\mathsf {L}}_2, \psi }\left( {\mathbb {F}}_q\right)&= 2\left( q-3 +\frac{g\left( \tfrac{q^{\times }}{4}\right) ^2 + g\left( \tfrac{3q^{\times }}{4}\right) ^2}{g\left( \tfrac{q^{\times }}{2}\right) }\right) \\&\quad + 2 + 0 + 0 + 2 \\&= 2q-2 + 2\left( \frac{g\left( \tfrac{q^{\times }}{4}\right) ^2 + g\left( \tfrac{3q^{\times }}{4}\right) ^2}{g\left( \tfrac{q^{\times }}{2}\right) }\right) . \end{aligned}$$
- (c)
If \(q\equiv 1 \pmod 8\), then
$$\begin{aligned}&\#X_{{\mathsf {F}}_2{\mathsf {L}}_2, \psi }\left( {\mathbb {F}}_q\right) - \#U_{{\mathsf {F}}_2{\mathsf {L}}_2, \psi }\left( {\mathbb {F}}_q\right) \\&\quad = 2\left( q-3 +\frac{g\left( \tfrac{q^{\times }}{4}\right) ^2 + g\left( \tfrac{3q^{\times }}{4}\right) ^2}{g\left( \tfrac{q^{\times }}{2}\right) } + \frac{2}{q}g\left( \tfrac{q^{\times }}{4}\right) g\left( \tfrac{q^{\times }}{8}\right) g\left( \tfrac{5q^{\times }}{8}\right) \right. \\&\qquad \left. + \frac{2}{q}g\left( \tfrac{3q^{\times }}{4}\right) g\left( \tfrac{3q^{\times }}{8}\right) g\left( \tfrac{7q^{\times }}{8}\right) \right) + 2 + 2\left( 4\left( q-1\right) \right) + 4 + 2 \\&= 10q - 6 + 2\frac{g\left( \tfrac{q^{\times }}{4}\right) ^2 + g\left( \tfrac{3q^{\times }}{4}\right) ^2}{g\left( \tfrac{q^{\times }}{2}\right) }+ \tfrac{4}{q}g\left( \tfrac{q^{\times }}{4}\right) g\left( \tfrac{q^{\times }}{8}\right) g\left( \tfrac{5q^{\times }}{8}\right) \\&\qquad + \tfrac{4}{q}g\left( \tfrac{3q^{\times }}{4}\right) g\left( \tfrac{3q^{\times }}{8}\right) g\left( \tfrac{7q^{\times }}{8}\right) . \end{aligned}$$
\(\square \)
Step 4: Combine Steps 2 and 3 to reach the conclusion
Proof of Proposition 3.6.1
We now combine Lemmas B.2.3 and B.2.18 as follows. For (a), for \(q \equiv 3 \pmod {4}\),
$$\begin{aligned} \#X_{{\mathsf {F}}_2{\mathsf {L}}_2, \psi }\left( {\mathbb {F}}_q\right)&= q^2-3q+1+H_q\left( \tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4};0,0,0\,|\,\psi ^{-4}\right) - qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) + 2q \\&= q^2-q+1+H_q\left( \tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4};0,0,0\,|\,\psi ^{-4}\right) - qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) . \end{aligned}$$
For (b), for \(q\equiv 5 \pmod 8\),
$$\begin{aligned} \#X_{{\mathsf {F}}_2{\mathsf {L}}_2, \psi }\left( {\mathbb {F}}_q\right)&=q^2-3q+3+H_q\left( \tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4};0,0,0\,|\,\psi ^{-4}\right) + qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) \\&\quad -2q H_q\left( \tfrac{1}{2}; 0 \,|\, \psi ^{-4}\right) - 2 \left( \frac{g\left( \tfrac{q^{\times }}{4}\right) ^2 + g\left( \tfrac{3q^{\times }}{4}\right) ^2}{g\left( \tfrac{q^{\times }}{2}\right) }\right) \\&\quad + 2q-2 + 2\left( \frac{g\left( \tfrac{q^{\times }}{4}\right) ^2 + g\left( \tfrac{3q^{\times }}{4}\right) ^2}{g\left( \tfrac{q^{\times }}{2}\right) }\right) \\&= q^2-q+1+H_q\left( \tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4};0,0,0\,|\,\psi ^{-4}\right) \\&\quad + qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) -2q H_q\left( \tfrac{1}{2}; 0 \,|\, \psi ^{-4}\right) \end{aligned}$$
Finally, for (c) with \(q\equiv 1 \pmod 8\),
$$\begin{aligned}&\#X_{{\mathsf {F}}_2{\mathsf {L}}_2, \psi }\left( {\mathbb {F}}_q\right) = q^2-3q+5+H_q\left( \tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4};0,0,0\,|\,\psi ^{-4}\right) +qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) \\&\qquad + 2q H_q\left( \tfrac{1}{2}; 0 \,|\, \psi ^{-4}\right) + 2\omega \left( 2\right) ^{q^{\times }\,/4}qH_q\left( \tfrac{1}{8}, \tfrac{5}{8}; 0, \tfrac{1}{4} \,|\, \psi ^4\right) \\&\qquad + 2\omega \left( 2\right) ^{q^{\times }\,/4}qH_q\left( \tfrac{3}{8}, \tfrac{7}{8}; 0, \tfrac{3}{4} \,|\, \psi ^4\right) \\&\qquad - 2\left( \frac{g\left( \tfrac{q^{\times }}{4}\right) ^2 + g\left( \tfrac{3q^{\times }}{4}\right) ^2}{g\left( \tfrac{q^{\times }}{2}\right) }\right) \\&\qquad - \tfrac{4}{q}g\left( \tfrac{q^{\times }}{8}\right) g\left( \tfrac{5q^{\times }}{8}\right) g\left( \tfrac{q^{\times }}{4}\right) - \tfrac{4}{q}g\left( \tfrac{3q^{\times }}{8}\right) g\left( \tfrac{7q^{\times }}{8}\right) g\left( \tfrac{3q^{\times }}{4}\right) \\&\qquad + 10q - 6 + 2\left( \frac{g\left( \tfrac{q^{\times }}{4}\right) ^2 + g\left( \tfrac{3q^{\times }}{4}\right) ^2}{g\left( \tfrac{q^{\times }}{2}\right) }\right) + \tfrac{4}{q}g\left( \tfrac{q^{\times }}{4}\right) g\left( \tfrac{q^{\times }}{8}\right) g\left( \tfrac{5q^{\times }}{8}\right) \\&\qquad + \tfrac{4}{q}g\left( \tfrac{3q^{\times }}{4}\right) g\left( \tfrac{3q^{\times }}{8}\right) g\left( \tfrac{7q^{\times }}{8}\right) \\&\quad = q^2+7q +1 +H_q\left( \tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4};0,0,0\,|\,\psi ^{-4}\right) +qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) \\&\qquad + 2q H_q\left( \tfrac{1}{2}; 0 \,|\, \psi ^{-4}\right) + 2\omega \left( 2\right) ^{q^{\times }\,/4}qH_q\left( \tfrac{1}{8}, \tfrac{5}{8}; 0, \tfrac{1}{4} \,|\, \psi ^4\right) \\&\qquad +2\omega \left( 2\right) ^{q^{\times }\,/4}qH_q\left( \tfrac{3}{8}, \tfrac{7}{8}; 0, \tfrac{3}{4} \,|\, \psi ^4\right) . \end{aligned}$$
\(\square \)
B.3. The pencil \({\mathsf {L}}_2{\mathsf {L}}_2\)
Proposition B.3.1
The number of \({\mathbb {F}}_q\)-points on \({\mathsf {L}}_2{\mathsf {L}}_2\) can be written in terms of hypergeometric functions, as follows:
- (a)
If \(q\equiv 3\pmod 4\), then
$$\begin{aligned} \#X_{{\mathsf {L}}_2{\mathsf {L}}_2,\psi }\left( {\mathbb {F}}_q\right) = q^2 + q+1 + H_q\left( \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4}; 0,0,0 \,|\, \psi ^{-4}\right) - qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) . \end{aligned}$$
(B.3.2)
- (b)
If \(q\equiv 1\pmod 4\), then
$$\begin{aligned} \begin{aligned} \#X_{{\mathsf {L}}_2{\mathsf {L}}_2,\psi }\left( {\mathbb {F}}_q\right)&= q^2+9q+1 + H_q\left( \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4}; 0,0,0 \,|\, \psi ^{-4}\right) + qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) \\&\quad + 2 \left( -1\right) ^{q^{\times }\,/4} \omega \left( \psi \right) ^{q^{\times }\,/2} q H_q\left( \tfrac{1}{8},\tfrac{3}{8},\tfrac{5}{8},\tfrac{7}{8};0,\tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4}\,|\,\psi ^{-4}\right) . \end{aligned} \end{aligned}$$
(B.3.3)
Remark B.3.4
Again, notice that the hypergeometric functions appearing in the point count correspond to exactly one of the Picard–Fuchs equations in Proposition A.2.2. We also see the appearance of eight additional trivial factors.
Step 1: Computing and clustering the characters
As with all the previous families, we first compute the set \(S\) of solutions to the system of congruences given by Theorem 3.3.3:
- (a)
If \(q\not \equiv 1\pmod 4\), and q is odd, then \(S\) consists of
- (i)
the set \(S_1 = \{k(1,1,1,1,-4) : k \in {\mathbb {Z}}/q^{\times }{\mathbb {Z}}\}\) and
- (ii)
the set \(S_4 = \{k(1,1,1,1,-4) + \tfrac{q^{\times }}{2}(0,0,1,1,0) : k \in {\mathbb {Z}}/q^{\times }{\mathbb {Z}}\}. \)
- (b)
If \(q\equiv 1 \pmod 4\), then
- (i)
the set \(S_1 = \{k(1,1,1,1,-4) : k \in {\mathbb {Z}}/q^{\times }{\mathbb {Z}}\},\)
- (ii)
the set \(S_4 = \{k(1,1,1,1,-4) + \tfrac{q^{\times }}{2}(0,0,1,1,0) : k \in {\mathbb {Z}}/q^{\times }{\mathbb {Z}}\}, \) and
- (iii)
two sets of the form \(S_{12} = \{k(1,1,1,1,-4) + \tfrac{q^{\times }}{4}(0,2,3,1,2) : k \in {\mathbb {Z}}/q^{\times }{\mathbb {Z}}\}\).
Step 2: Counting points on the open subset with nonzero coordinates
Lemma B.3.5
Suppose \(\psi \in {\mathbb {F}}_q^\times \). For q, we have:
- (a)
If \(q\equiv 3\pmod 4\), then
$$\begin{aligned} \#U_{{\mathsf {L}}_2{\mathsf {L}}_2, \psi }\left( {\mathbb {F}}_q\right) = q^2 -3q+1 + H_q\left( \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4}; 0,0,0 \,|\, \psi ^{-4}\right) - qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) . \end{aligned}$$
(B.3.6)
- (b)
If \(q\equiv 1\pmod 4\), then
$$\begin{aligned} \begin{aligned} \#U_{{\mathsf {L}}_2{\mathsf {L}}_2, \psi }\left( {\mathbb {F}}_q\right)&= q^2-3q+5 + H_q\left( \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4}; 0,0,0 \,|\, \psi ^{-4}\right) + qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) \\&\qquad + 2 \left( -1\right) ^{q^{\times }\,/4} \omega \left( \psi \right) ^{q^{\times }\,/2} q H_q\left( \tfrac{1}{8},\tfrac{3}{8},\tfrac{5}{8},\tfrac{7}{8};0,\tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4}\,|\,\psi ^{-4}\right) . \end{aligned} \end{aligned}$$
(B.3.7)
Proof
We do this by cases. For (a), where \(q\equiv 3 \pmod 4\), then by using Theorem 3.3.3 with Lemmas 3.4.4 and 3.4.7 we have that
$$\begin{aligned} \begin{aligned} \#U_{{\mathsf {L}}_2{\mathsf {L}}_2, \psi }\left( {\mathbb {F}}_q\right)&= \sum _{s \in S_1} \omega \left( a\right) ^{-s}c_s + \sum _{s \in S_4} \omega \left( a\right) ^{-s}c_s \\&= q^2-3q+3 + H_q\left( \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4}; 0,0,0 \,|\, \psi ^{-4}\right) -2 - qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) \\&= q^2 -3q+1 + H_q\left( \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4}; 0,0,0 \,|\, \psi ^{-4}\right) - qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) . \end{aligned} \end{aligned}$$
(B.3.8)
For (b) with \(q\equiv 1 \pmod 4\), by using Theorem 3.3.3 with Lemmas 3.4.4, 3.4.7, and B.3.10 below, we have that
$$\begin{aligned} \#U_{{\mathsf {L}}_2{\mathsf {L}}_2, \psi }\left( {\mathbb {F}}_q\right)&= \sum _{s \in S_1} \omega \left( a\right) ^{-s}c_s + \sum _{s \in S_4} \omega \left( a\right) ^{-s}c_s +2\sum _{s \in S_{12}} \omega \left( a\right) ^{-s}c_s \nonumber \\&= q^2-3q+3 + H_q\left( \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4}; 0,0,0 \,|\, \psi ^{-4}\right) + 2 + qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) \nonumber \\&\quad + 2 \left( -1\right) ^{q^{\times }\,/4} \omega \left( \psi \right) ^{q^{\times }\,/2} q H_q\left( \frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8};0,\frac{1}{4},\frac{1}{2},\frac{3}{4}\,|\,\psi ^{-4}\right) \nonumber \\&= q^2-3q+5 + H_q\left( \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4}; 0,0,0 \,|\, \psi ^{-4}\right) + qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) \nonumber \\&\quad + 2 \left( -1\right) ^{q^{\times }\,/4} \omega \left( \psi \right) ^{q^{\times }\,/2} q H_q\left( \tfrac{1}{8},\tfrac{3}{8},\tfrac{5}{8},\tfrac{7}{8};0,\tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4}\,|\,\psi ^{-4}\right) . \end{aligned}$$
(B.3.9)
\(\square \)
Lemma B.3.10
Suppose that \(q \equiv 1 \pmod 4\). Then
$$\begin{aligned} \sum _{s \in S_{12}} \omega \left( a\right) ^{-s} c_s = \left( -1\right) ^{q^{\times }\,/4} \omega \left( \psi \right) ^{q^{\times }\,/2} q H_q\left( \tfrac{1}{8},\tfrac{3}{8},\tfrac{5}{8},\tfrac{7}{8};0,\tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4}\,|\,\psi ^{-4}\right) . \end{aligned}$$
Proof
We start with the definition, factor out \(\omega (\psi )^{q^{\times }\,/2}\), and then use the Hasse–Davenport relation (3.1.4) with \(N=4\) with respect to \(m=k\) to obtain:
$$\begin{aligned} \begin{aligned} \sum _{s \in S_{12}} \omega \left( a\right) ^{-s} c_s&= \frac{1}{qq^{\times }} \sum _{k=0}^{q-2} \omega \left( -4\psi \right) ^{4k + \tfrac{q^{\times }}{2}} g\left( k\right) g\left( k+ \tfrac{q^{\times }}{4}\right) g\left( k+\tfrac{q^{\times }}{2}\right) \\&\quad \cdot g\left( k+\tfrac{3q^{\times }}{4}\right) g\left( -4k+\tfrac{q^{\times }}{2}\right) \\&= \frac{\omega \left( \psi \right) ^{q^{\times }\,/2}}{qq^{\times }} \sum _{k=0}^{q-2} \omega \left( -4\psi \right) ^{4k} g\left( 4k\right) \omega \left( 4\right) ^{-4k} g\left( \tfrac{q^{\times }}{2}\right) \\&\quad \cdot g\left( \tfrac{q^{\times }}{4}\right) g\left( \tfrac{3q^{\times }}{4}\right) g\left( -4k+\tfrac{q^{\times }}{2}\right) . \end{aligned} \end{aligned}$$
(B.3.11)
Simplify with Lemma 3.1.3(b) to get
$$\begin{aligned} \sum _{s \in S_{12}} \omega \left( a\right) ^{-s} c_s = \frac{\left( -1\right) ^{q^{\times }\,/4}\omega \left( \psi \right) ^{q^{\times }\,/2}}{q-1} \sum _{k=0}^{q-2} \omega \left( -\psi \right) ^{4k} g\left( 4k\right) g\left( \tfrac{q^{\times }}{2}\right) g\left( -4k+\tfrac{q^{\times }}{2}\right) . \end{aligned}$$
(B.3.12)
Now, we use the Hasse–Davenport relation again with \(N=2\) and \(m=-4k\) to find:
$$\begin{aligned} \sum _{s \in S_{12}} \omega \left( a\right) ^{-s} c_s= & {} \frac{\left( -1\right) ^{q^{\times }\,/4}\omega \left( \psi \right) ^{q^{\times }\,/2}}{q-1} \sum _{k=0}^{q-2} \omega \left( -\psi \right) ^{4k} g\left( 4k\right) g\left( \tfrac{q^{\times }}{2}\right) \nonumber \\&\cdot \left( \frac{g\left( -8k\right) g\left( \tfrac{q^{\times }}{2}\right) \omega \left( 2\right) ^{8k}}{g\left( -4k\right) }\right) . \end{aligned}$$
(B.3.13)
We now simplify using Lemma 3.1.3(b) again and then expand the summation to get:
$$\begin{aligned} \begin{aligned} \sum _{s \in S_{12}} \omega (a)^{-s} c_s&= (-1)^{q^{\times }\,/4}\omega (\psi )^{q^{\times }\,/2}q \\&\quad \cdot \left( -\frac{4}{q^{\times }} + \frac{1}{q^{\times }} \sum _{{\mathop {4k \not \equiv 0 ~({\text {mod}}~{q^{\times }})}\limits ^{k=0}}}^{q-2} \omega (4\psi )^{4k} q^{-1}g(4k)^2g(-8k) \right) . \end{aligned} \end{aligned}$$
(B.3.14)
We finally reindex the sum with \(m=-k\), yielding
$$\begin{aligned} \begin{aligned} \sum _{s \in S_{12}} \omega (a)^{-s} c_s&= (-1)^{q^{\times }\,/4}\omega (\psi )^{q^{\times }\,/2}\\&\quad \cdot q\left( -\frac{4}{q^{\times }} + \frac{1}{q^{\times }} \sum _{{\mathop {4m \not \equiv 0 ~({\text {mod}}~{q^{\times }})}\limits ^{m=0}}}^{q-2} \omega (4\psi )^{-4m} q^{-1}g(-4m)^2g(8m) \right) \\&= (-1)^{q^{\times }\,/4}\omega (\psi )^{q^{\times }\,/2}q H_q\left( \tfrac{1}{8},\tfrac{3}{8},\tfrac{5}{8},\tfrac{7}{8};0,\tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4}\,|\,\psi ^{-4}\right) \end{aligned} \end{aligned}$$
(B.3.15)
relating back to the finite-field hypergeometric sum. \(\square \)
Step 3: Count points when at least one coordinate is zero
Lemma B.3.16
Let q be an odd prime that is not 7. Then
- (a)
If \(q\equiv 3 \pmod 4\), then
$$\begin{aligned} \#X_{{\mathsf {L}}_2{\mathsf {L}}_2, \psi }({\mathbb {F}}_q) - \#U_{{\mathsf {L}}_2{\mathsf {L}}_2, \psi }({\mathbb {F}}_q) = 4q. \end{aligned}$$
- (b)
If \(q\equiv 1 \pmod 4\), then
$$\begin{aligned} \#X_{{\mathsf {L}}_2{\mathsf {L}}_2, \psi }({\mathbb {F}}_q) - \#U_{{\mathsf {L}}_2{\mathsf {L}}_2, \psi }({\mathbb {F}}_q) = 12q - 4. \end{aligned}$$
Proof
Suppose that \(x_1=0\) and the rest are nonzero. Then, by using Theorem 3.3.3, we can see that there are \((q-1)((-1)^{q^{\times }\,/2} +1)\) such points. Since there are four choices of one coordinate being zero, this counts \(4(q-1)((-1)^{q^{\times }\,/2} +1)\) points.
Suppose now that \(x_1=x_2=0\) and the rest nonzero, then by Theorem 3.3.3 again, we have \(((-1)^{q^{\times }\,/2} +1)\) points. By symmetry, this is the same as the case where \(x_3=x_4=0\) and the rest nonzero, so we now count \(2((-1)^{q^{\times }\,/2} +1)\).
Next, suppose \(x_1=x_3=0\) and the rest nonzero. Automatically, the polynomial vanishes, hence there are \(q-1\) such points. There are 4 such cases from choosing one of \(x_1\) and \(x_2\) and another from \(x_3\) and \(x_4\) to equal zero, hence we count \(4(q-1)\) points. Finally, the four points where three coordinates are zero are all solutions, hence we count 4 more points. Thus,
$$\begin{aligned} \#X_{{\mathsf {L}}_2{\mathsf {L}}_2, \psi }\left( {\mathbb {F}}_q\right) - \#U_{{\mathsf {L}}_2{\mathsf {L}}_2, \psi }\left( {\mathbb {F}}_q\right)= & {} 4\left( q-1\right) \left( \left( -1\right) ^{q^{\times }\,/2} +1\right) + 2\left( \left( -1\right) ^{q^{\times }\,/2} +1\right) \\&+ 4\left( q-1\right) + 4. \end{aligned}$$
If \(q\equiv 3 \pmod 4\), then \((-1)^{q^{\times }\,/2} = -1\), so \( \#X_{{\mathsf {L}}_2{\mathsf {L}}_2, \psi }({\mathbb {F}}_q) - \#U_{{\mathsf {L}}_2{\mathsf {L}}_2, \psi }({\mathbb {F}}_q) = 4q\). If \(q\equiv 1 \pmod 4\), then \((-1)^{q^{\times }\,/2} = 1\), so \( \#X_{{\mathsf {L}}_2{\mathsf {L}}_2, \psi }({\mathbb {F}}_q) - \#U_{{\mathsf {L}}_2{\mathsf {L}}_2, \psi }({\mathbb {F}}_q) = 12q-4\). \(\square \)
Step 4: Combine Steps 2 and 3 to reach the conclusion
Proof of Proposition 3.6.2
If \(q \not \equiv 1 \pmod 4\), then by Lemmas B.3.5 and B.3.16, we have that
$$\begin{aligned} \#X_{{\mathsf {L}}_2{\mathsf {L}}_2,\psi }\left( {\mathbb {F}}_q\right)= & {} \left( q^2 -3q+1 + H_q\left( \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4}; 0,0,0 \,|\, \psi ^{-4}\right) - qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) \right) \nonumber \\&+ 4q= q^2 + q+1 + H_q\left( \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4}; 0,0,0 \,|\, \psi ^{-4}\right) \nonumber \\&- qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) . \end{aligned}$$
(B.3.17)
If \(q \equiv 1 \pmod 4\), then by Lemmas B.3.5 and B.3.16, we have that
$$\begin{aligned} \#X_{{\mathsf {L}}_2{\mathsf {L}}_2,\psi }\left( {\mathbb {F}}_q\right)&= q^2-3q+5 + H_q\left( \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4}; 0,0,0 \,|\, \psi ^{-4}\right) + qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) \nonumber \\&\quad + 2 \left( -1\right) ^{q^{\times }\,/4} \omega \left( \psi \right) ^{q^{\times }\,/2} q H_q\left( \tfrac{1}{8},\tfrac{3}{8},\tfrac{5}{8},\tfrac{7}{8};0,\tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4}\,|\,\psi ^{-4}\right) + 12q-4 \nonumber \\&= q^2+9q+1 + H_q\left( \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4}; 0,0,0 \,|\, \psi ^{-4}\right) + qH_q\left( \tfrac{1}{4}, \tfrac{3}{4}; 0, \tfrac{1}{2} \,|\, \psi ^{-4}\right) \nonumber \\&\quad + 2 \left( -1\right) ^{q^{\times }\,/4} \omega \left( \psi \right) ^{q^{\times }\,/2} q H_q\left( \tfrac{1}{8},\tfrac{3}{8},\tfrac{5}{8},\tfrac{7}{8};0,\tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4}\,|\,\psi ^{-4}\right) . \end{aligned}$$
(B.3.18)
\(\square \)
B.2. The pencil \({\mathsf {L}}_4\)
Proposition B.4.1
The number of \({\mathbb {F}}_q\) points on \({\mathsf {L}}_4\) for q odd is given in terms of hypergeometric functions as follows.
- (a)
If \(q\not \equiv 1\pmod 5\), then
$$\begin{aligned} \#X_{{\mathsf {L}}_4,\psi }({\mathbb {F}}_q)=q^2+3q+1+H_q\left( \tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4};0,0,0\,|\,\psi ^{-4}\right) . \end{aligned}$$
- (b)
If \(q\equiv 1 \pmod 5\), then
$$\begin{aligned} \#X_{{\mathsf {L}}_4,\psi }({\mathbb {F}}_q)= & {} q^2+3q+1+H_q\left( \tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4};0,0,0\,|\,\psi ^{-4}\right) \\&+4qH_q\left( \tfrac{1}{5},\tfrac{2}{5}, \tfrac{3}{5},\tfrac{4}{5} ;0, \tfrac{1}{4},\tfrac{1}{2}, \tfrac{3}{4}\,|\,\psi ^{4}\right) . \end{aligned}$$
Remark B.4.2
As before, we can identify the parameters of the hypergeometric function
$$\begin{aligned} H_q\left( \tfrac{1}{5},\tfrac{2}{5}, \tfrac{3}{5},\tfrac{4}{5} ;0, \tfrac{1}{4},\tfrac{1}{2}, \tfrac{3}{4}\,|\,\psi ^{4}\right) \end{aligned}$$
with the parameters of the second Picard–Fuchs equation in Proposition A.3.2. If we use Theorem 3.4 of [3] again to shift parameters, then we see that in fact all of the Picard–Fuchs equations satisfied by the non-holomorphic periods correspond to this same hypergeometric motive over \({\mathbb {Q}}\).
Also notice that in the discussion following Proposition A.3.2, we see two periods that are “missed” by the Griffiths–Dwork method, and here they clearly correspond to the two additional trivial factors coming from the 3q term in the point count.
Step 1: Computing and clustering the characters
Again, we compute the solutions to the system of congruences given by Theorem 3.3.3. We obtain
- (a)
If \(q\not \equiv 1\pmod 5\), the solution set is
- (i)
the set \(S_1 = \{ k(1,1,1,1,-4) : k \in {\mathbb {Z}}/ q^{\times }{\mathbb {Z}}\}\).
- (b)
If \(q\equiv 1\pmod 5\), then clusters of solutions are
- (i)
the set \(S_1 = \{ k(1,1,1,1,-4) : k \in {\mathbb {Z}}/ q^{\times }{\mathbb {Z}}\}\) and
- (ii)
four sets of the form \(S_{13} = \{ k(1,1,1,1,-4) + \frac{q^{\times }}{5}(1,2,4,3,0) : k \in {\mathbb {Z}}/ q^{\times }{\mathbb {Z}}\}\).
Step 2: Counting points on the open subset with nonzero coordinates
Lemma B.4.3
Suppose \(\psi \in {\mathbb {F}}_q^\times \). For q odd, we have:
- (a)
If \(q\not \equiv 1\pmod 5\), then
$$\begin{aligned} \#U_{{\mathsf {L}}_4, \psi }({\mathbb {F}}_q) = q^2-3q+3 + H_q\left( \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4}; 0,0,0 \,|\, \psi ^{-4}\right) . \end{aligned}$$
(B.4.4)
- (b)
If \(q\equiv 1\pmod 5\), then
$$\begin{aligned} \#U_{{\mathsf {L}}_4, \psi }\left( {\mathbb {F}}_q\right) = q^2-3q+3 + H_q\left( \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4}; 0,0,0 \,|\, \psi ^{-4}\right) + 4q H_q\left( \tfrac{1}{5},\tfrac{2}{5}, \tfrac{3}{5},\tfrac{4}{5} ;0, \tfrac{1}{4},\tfrac{1}{2}, \tfrac{3}{4}\,|\,\psi ^{4}\right) . \end{aligned}$$
(B.4.5)
Proof
When \(q \not \equiv 1 \pmod 5\), we know that there is only one cluster of characters, \(S_1\). By Lemma 3.4.4, we know that
$$\begin{aligned} \#U_{{\mathsf {L}}_4, \psi }({\mathbb {F}}_q) = \sum _{s \in S_1} \omega (a)^{-s} c_s= q^2-3q+3 + H_q\left( \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4}; 0,0,0 \,|\, \psi ^{-4}\right) . \end{aligned}$$
(B.4.6)
When \(q\equiv 1 \pmod 5\), we have two types of clusters of characters. By Lemmas 3.4.4 and B.4.8,
$$\begin{aligned} \#U_{{\mathsf {L}}_4, \psi }\left( {\mathbb {F}}_q\right)&= \sum _{s \in S_1} \omega \left( a\right) ^{-s} c_s + 4 \sum _{s \in S_{13}} \omega \left( a\right) ^{-s} c_s \nonumber \\&= q^2-3q+3 + H_q\left( \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4}; 0,0,0 \,|\, \psi ^{-4}\right) \nonumber \\&\quad + 4q H_q\left( \tfrac{1}{5},\tfrac{2}{5}, \tfrac{3}{5},\tfrac{4}{5} ;0, \tfrac{1}{4},\tfrac{1}{2}, \tfrac{3}{4}\,|\,\psi ^{4}\right) . \end{aligned}$$
(B.4.7)
\(\square \)
We now just need a hypergeometric way to write the point count associated to the cluster \(S_{13}\).
Lemma B.4.8
If \(q \equiv 1 \pmod 5\) and q is odd, then
$$\begin{aligned} \sum _{s\in S_{13}} \omega (a)^{-s} c_s = qH_q\left( \tfrac{1}{5},\tfrac{2}{5}, \tfrac{3}{5},\tfrac{4}{5} ;0, \tfrac{1}{4},\tfrac{1}{2}, \tfrac{3}{4}\,|\,\psi ^{4}\right) . \end{aligned}$$
(B.4.9)
Proof
By using the hybrid hypergeometric definition, this equality is found quickly:
$$\begin{aligned} \sum _{s\in S_{13}} \omega \left( a\right) ^{-s} c_s&= \frac{1}{qq^{\times }} \sum _{k=0}^{q-2} \omega \left( -4^4\psi ^4\right) g\left( k+\tfrac{q^{\times }}{5}\right) g\left( k+\tfrac{2q^{\times }}{5}\right) \nonumber \\&\quad \cdot g\left( k+\tfrac{3q^{\times }}{5}\right) g\left( k + \tfrac{4q^{\times }}{5}\right) g\left( -4k\right) \nonumber \\&= \frac{1}{qq^{\times }} \sum _{k=0}^{q-2} \omega \left( 4^4\psi ^4\right) g\left( k+\tfrac{q^{\times }}{5}\right) g\left( k+\tfrac{2q^{\times }}{5}\right) \nonumber \\&\quad \cdot g\left( k+\tfrac{3q^{\times }}{5}\right) g\left( k + \tfrac{4q^{\times }}{5}\right) g\left( -4k\right) \nonumber \\&= \frac{q}{q^{\times }} \sum _{k=0}^{q-2} \omega \left( 4^4\psi ^4\right) \nonumber \\&\quad \cdot \frac{g\left( k+\tfrac{q^{\times }}{5}\right) g\left( k+\tfrac{2q^{\times }}{5}\right) g\left( k+\tfrac{3q^{\times }}{5}\right) g\left( k + \tfrac{4q^{\times }}{5}\right) }{q^2}g\left( -4k\right) \nonumber \\&= \frac{q}{q^{\times }} \sum _{k=0}^{q-2} \omega \left( 4^4\psi ^4\right) \nonumber \\&\quad \cdot \frac{g\left( k+\tfrac{q^{\times }}{5}\right) g\left( k+\tfrac{2q^{\times }}{5}\right) g\left( k+\tfrac{3q^{\times }}{5}\right) g\left( k + \tfrac{4q^{\times }}{5}\right) }{g\left( \tfrac{q^{\times }}{5}\right) g\left( \tfrac{2q^{\times }}{5}\right) g\left( \tfrac{3q^{\times }}{5}\right) g\left( \tfrac{4q^{\times }}{5}\right) } g\left( -4k\right) \nonumber \\&= qH_q\left( \tfrac{1}{5},\tfrac{2}{5}, \tfrac{3}{5},\tfrac{4}{5} ;0, \tfrac{1}{4},\tfrac{1}{2}, \tfrac{3}{4}\,|\,\psi ^{4}\right) . \end{aligned}$$
(B.4.10)
The last line uses the hybrid definition (Definition 3.2.7) of the hypergeometric function
$$\begin{aligned} H_q\left( \tfrac{1}{5},\tfrac{2}{5}, \tfrac{3}{5},\tfrac{4}{5} ;0, \tfrac{1}{4},\tfrac{1}{2}, \tfrac{3}{4}\,|\,\psi ^{4}\right) . \end{aligned}$$
Even though the hypergeometric function is defined over \({\mathbb {Q}}\), we can get to the relation much more quickly using the hybrid definition.
Step 3: Count points when at least one coordinate is zero.
Lemma B.4.11
If q is odd and not 7, then
$$\begin{aligned} \#X_{{\mathsf {L}}_4,\psi }({\mathbb {F}}_q) - \#U_{{\mathsf {L}}_4,\psi }({\mathbb {F}}_q) = 6q-2. \end{aligned}$$
Proof
First, we count the number of rational points when exactly variable equals zero. Without loss of generality, assume \(x_1=0\). Then we want solutions of
$$\begin{aligned} x_2^3 x_3 + x_3^3 x_4 = 0 \end{aligned}$$
which we can solve for \(x_4\). Since \(x_4\) is completely determined by \(x_2\) and \(x_3\), we can normalize \(x_2=1\) and see there are exactly \(q-1\) solutions when only \(x_1\) is zero. By symmetry, this shows that there are \(4q-4\) solutions when exactly one variable equals zero. If two consecutive variables are zero (say \(x_1=x_2=0\)), then we want solutions of the form \(x_3^3x_4 = 0\) which implies that a third variable equals zero. Thus, there are 4 solutions with 3 variables equaling zero and no solutions when exactly two variables equal zero and those variables are consecutive. Lastly, if two non-consecutive variables are zero then any other solution works. For any pair of non-consecutive variables (of which there are two), we then have \(q-1\) solutions. Therefore,
$$\begin{aligned} \#X_{{\mathsf {L}}_4,\psi }({\mathbb {F}}_q) - \#U_{{\mathsf {L}}_4,\psi }({\mathbb {F}}_q) = 4q-4 + 4 + 2(q-1) = 6q-2. \end{aligned}$$
\(\square \)
Step 4: Combine Steps 2 and 3 to find conclusion
We now prove Proposition 3.6.3.
Proof of Proposition 3.6.3
Combining Lemmas B.4.3 and B.4.11, we have
- (a)
If \(q\not \equiv 1\pmod 5\), then
$$\begin{aligned} \#X_{{\mathsf {L}}_4, \psi }\left( {\mathbb {F}}_q\right)= & {} \left( q^2-3q+3 + H_q\left( \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4}; 0,0,0 \,|\, \psi ^{-4}\right) \right) + \left( 6q-2\right) \nonumber \\= & {} q^2+3q+1 + H_q\left( \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4}; 0,0,0 \,|\, \psi ^{-4}\right) . \end{aligned}$$
(B.4.12)
- (b)
If \(q\equiv 1\pmod 5\), then
$$\begin{aligned} \#X_{{\mathsf {L}}_4, \psi }\left( {\mathbb {F}}_q\right)= & {} \left( q^2-3q+3 + H_q\left( \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4}; 0,0,0 \,|\, \psi ^{-4}\right) \right. \nonumber \\&\left. + 4q H_q\left( \tfrac{1}{5},\tfrac{2}{5}, \tfrac{3}{5},\tfrac{4}{5} ;0, \tfrac{1}{4},\tfrac{1}{2}, \tfrac{3}{4}\,|\,\psi ^{4}\right) \right) + \left( 6q-2\right) \nonumber \\= & {} q^2+3q+1 + H_q\left( \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4}; 0,0,0 \,|\, \psi ^{-4}\right) \nonumber \\&+ 4q H_q\left( \tfrac{1}{5},\tfrac{2}{5}, \tfrac{3}{5},\tfrac{4}{5} ;0, \tfrac{1}{4},\tfrac{1}{2}, \tfrac{3}{4}\,|\,\psi ^{4}\right) . \end{aligned}$$
(B.4.13)
This completes the proof. \(\square \)