2 Root function

Fix as modulus $\kappa \in (0, 1)$ . The rule

$$ \begin{align*} f(T) = \int_0^T F\,\bigg(\frac{1}{4}, \frac{3}{4}; \frac{1}{2}; \kappa^2 \sin^2 t\bigg) {\,d} t \end{align*} $$

defines an odd strictly increasing bijection from $\mathbb {R}$ to $\mathbb {R}$ . The evaluation

$$ \begin{align*} L := f\, (\tfrac{1}{2} \pi) = \tfrac{1}{2} \, \pi \, F\, (\tfrac{1}{4}, \tfrac{3}{4}; 1; \kappa^2) \end{align*} $$

is performed by termwise integration after expanding the hypergeometric series in the integrand.

Write $\phi : \mathbb {R} \to \mathbb {R}$ for the inverse to f; this inverse therefore satisfies

$$ \begin{align*}\phi (u + 2 L) = \phi (u) + \pi.\end{align*} $$

As an auxiliary function, introduce the composite $\psi : = \operatorname {arc}\sin (\kappa \sin \phi ).$ We now define the function $\sigma : \mathbb {R} \to \mathbb {R}$ by $\sigma = \sin \tfrac {1}{2} \psi $ with the warning that this is not a ‘sigma function’ in the traditional sense.

Recall (say, from [Reference Erdelyi, Magnus, Oberhettinger and Tricomi3, page 101]) the hypergeometric evaluation

$$ \begin{align*} F \bigg(\frac{1}{4}, \frac{3}{4}; \frac{1}{2}; \sin^2 \psi\bigg) = \frac{\cos \tfrac{1}{2} \psi}{\cos \psi}. \end{align*} $$

Proof. Here, $\sigma ^{\circ }$ denotes the derivative of $\sigma $ ; the traditional prime $'$ is used later for another (quite traditional) purpose. The initial condition is trivial; for the differential equation, differentiate. From the definition $\sigma = \sin \tfrac {1}{2} \psi $ , it follows that

$$ \begin{align*}\sigma^{\circ} = \tfrac{1}{2} \, (\cos \tfrac{1}{2} \psi) \, \psi^{\circ}; \end{align*} $$

from $\sin \psi = \kappa \, \sin \phi $ , it follows that

$$ \begin{align*}\psi^{\circ} = \kappa \, \frac{\cos \phi}{\cos \psi} \, \phi^{\circ};\end{align*} $$

while from

$$ \begin{align*}u = \int_0^{\phi (u)} F \bigg(\frac{1}{4}, \frac{3}{4}; \frac{1}{2}; \kappa^2 \sin^2 t\bigg) {\, d} t\end{align*} $$

together with the hypergeometric evaluation recalled before the theorem, it follows that

$$ \begin{align*} \phi^{\circ} = \frac{\cos \psi}{\cos \tfrac{1}{2} \psi}. \end{align*} $$

Mass cancellation produces $\sigma ^{\circ } = \tfrac {1}{2} \, \kappa \, \cos \phi ;$ finally, squaring and trigonometric duplication result in

$$ \begin{align*} 4 (\sigma^{\circ})^2 = \kappa^2 - \sin^2 \psi = \kappa^2 - 4 \sin^2 \tfrac{1}{2} \psi \, \cos^2 \tfrac{1}{2} \psi = \kappa^2 - 4 \sigma^2 (1 - \sigma^2).\\[-36pt] \end{align*} $$

This initial value problem has a second solution: namely, $- \sigma $ . The fact that the initial value problem has just these two solutions is evident upon further differentiation, leading to $\sigma ^{\circ \circ } = 2 \sigma ^3 - \sigma $ : as the right-hand side here is polynomial, specification of $\sigma (0)$ as zero and of $\sigma ^{\circ }(0)$ as a square root of $\tfrac {1}{4} \kappa ^2$ picks out a unique solution. The solution $\sigma = \sin \tfrac {1}{2} \psi $ is singled out by the specification $\sigma ^{\circ } (0) = \tfrac {1}{2} \kappa $ ; equivalently, by the requirement that $\sigma (t)> 0$ when $t> 0$ is small.

As the right-hand side of the differential equation in Theorem 2.3 is a quartic with simple zeros, its solutions are—or rather, extend to be—elliptic on the plane. We shall identify the specific elliptic extension of $\sigma = \sin \tfrac {1}{2} \psi $ in two alternative forms, each of which has its uses: in Section 3, we identify $\sigma $ in Weierstrassian terms; in Section 4, we identify $\sigma $ in Jacobian terms.

We end this section with some remarks on notation. In the Introduction, we gave the elliptic extension of $\sigma $ the name $\mathrm {rn}_2$ for two reasons: on the one hand, $\mathrm {rn}_2$ is a ‘root’ function from which the various functions in [Reference Shen7] may be derived; on the other hand, the proximity of $\mathrm {rn}$ to $\mathrm { sn}$ reflects the fact that $\mathrm {rn}_2$ is a signature four replacement for the Jacobian modular sine. Nevertheless, for largely typographical reasons, we continue to use the notation $\sigma $ for the elliptic extension of $\sin \tfrac {1}{2} \psi $ .

3 Weierstrassian representation

Our identification of $\sigma $ in Weierstrassian terms uses the result of [Reference Whittaker and Watson9, Example 2, page 454], which is a result attributed to Weierstrass himself. Explicitly, let f be the quartic polynomial given by

$$ \begin{align*}f(z) = a_0 z^4 + 4 a_1 z^3 + 6 a_2 z^2 + 4 a_3 z + a_4\end{align*} $$

with quadrinvariant

$$ \begin{align*}g_2 = a_0 a_4 - 4 a_1 a_3 + 3 a_2^2\end{align*} $$

and cubinvariant

$$ \begin{align*}g_3 = a_0 a_2 a_4 + 2 a_1 a_2 a_3 - a_2^3 - a_0 a_3^2 - a_1^2 a_4\end{align*} $$

and assume that the zeros of f are simple. Then the initial value problem

$$ \begin{align*}(w^{\circ})^2 = f(w),\quad w(0) = a\end{align*} $$

has solutions given by

$$ \begin{align*} w = a + \frac{A p^{\circ} + \tfrac{1}{2} f^{\circ} (a) [p - \tfrac{1}{24} f^{\circ \circ}(a)] + \tfrac{1}{24} f^{\circ \circ \circ} (a)}{2 [p - \tfrac{1}{24} f^{\circ \circ}(a)]^2 - \tfrac{1}{48} f(a) f^{\circ \circ \circ \circ}(a)}, \end{align*} $$

where A is a square root of $f(a)$ and where $p = \wp ( -; g_2, g_3)$ is the Weierstrass $\wp $ -function with $g_2$ and $g_3$ as invariants.

Theorem 3.1. The function $\sigma $ is given by

$$ \begin{align*} \sigma = \frac{\tfrac{1}{4} \kappa p^{\circ}}{(\tfrac{1}{4} \kappa)^2 - (\tfrac{1}{12} + p)^2} = \frac{\tfrac{1}{4} \kappa p^{\circ}}{(\tfrac{1}{4} \kappa - \tfrac{1}{12} - p)(\tfrac{1}{4} \kappa + \tfrac{1}{12} + p)}, \end{align*} $$

where $p = \wp (-; g_2, g_3)$ is the Weierstrass function with invariants

$$ \begin{align*}g_2 = \tfrac{1}{12} (1 + 3 \kappa^2), \quad g_3 = \tfrac{1}{216} (1 - 9 \kappa^2).\end{align*} $$

The Weierstrass function p has a fundamental pair of periods $(2 \omega , 2 \omega ')$ with $\omega> 0$ and $ - i \omega '> 0$ ; introduce the third half-period $\omega "$ by the symmetrical condition $\omega + \omega ' + \omega " = 0$ , as is customary. We present explicit evaluations of these fundamental periods in Section 5; their precise values are not important at present. The differential equation $(p^{\circ })^2 = 4 p^3 - g_2 p - g_3$ satisfied by p factorises as

$$ \begin{align*}(p^{\circ})^2 = 4 (p - \tfrac{1}{6}) (p + \tfrac{1}{12} - \tfrac{1}{4} \kappa) (p + \tfrac{1}{12} + \tfrac{1}{4} \kappa)\end{align*} $$

so that p has (stationary) midpoint values given in decreasing order by

$$ \begin{align*}e:= p(\omega) = \tfrac{1}{6}, \quad e" : = p(\omega") = - \tfrac{1}{12} + \tfrac{1}{4} \kappa, \quad e' : = p(\omega') = - \tfrac{1}{12} - \tfrac{1}{4} \kappa,\end{align*} $$

and, for $\sigma $ , we have the equivalent expression

$$ \begin{align*}\sigma = - \tfrac{1}{4} \kappa \, \frac{p^{\circ}}{(p - e")(p - e')} .\end{align*} $$

Such an expression for $\sigma $ (namely, $p^{\circ }$ times a rational function of p) is to be expected of an odd elliptic function in terms of a $\wp $ -function having the same periods. The following result enables us to confirm that $\sigma $ and p are, in fact, coperiodic.

Proof. The Weierstrass function p satisfies the familiar identity

$$ \begin{align*}(p(z + \omega) - e)(p(z) - e) = (e" - e) (e' - e) = : E\end{align*} $$

from which

$$ \begin{align*}p(z + \omega) = e + \frac{E}{p(z) - e}\end{align*} $$

and therefore

$$ \begin{align*}p^{\circ} (z + \omega) = - \, \frac{E \, p^{\circ} (z)}{(p(z) - e)^2}.\end{align*} $$

Introduce the function s by $\sigma = - \tfrac {1}{4} \kappa s$ so that

$$ \begin{align*}s = \frac{p^{\circ} }{(p - e')(p - e")}\end{align*} $$

and it follows, by direct calculation, that

$$ \begin{align*}s(z + \omega) = - \, \frac{E \, p^{\circ} (z)}{(E + (e - e')(p(z) - e))(E + (e - e")(p(z) - e))} ;\end{align*} $$

here,

$$ \begin{align*}E + (e' - e) (e) = e" (e' - e) \quad\mathrm{and} \quad E + (e" - e)(e) = e' (e" - e)\end{align*} $$

so that, after cancellation of E, it follows that $s(z + \omega ) = - s(z).$ It is perhaps of interest to note that s has this property when p is any Weierstrass function. All that remains of the proof is to reinstate the multiplier $ - \tfrac {1}{4} \kappa $ .

Here, we may note that the identity $\sigma (u + 2 L) = - \, \sigma (u)$ in the proof of Theorem 2.2 extends to the complex plane by analytic continuation.

We are also able to identify the zeros and poles of $\sigma $ .

In Theorem 3.2, we show that an $\omega $ shift reverses $\sigma $ . The effects on $\sigma $ of half-period shifts by $\omega '$ and $\omega "$ are as follows.

Theorem 3.5.

$$ \begin{align*} \sigma(z + \omega') = \frac{\tfrac{1}{2} p^{\circ}(z)}{\tfrac{1}{6} - p(z)} \quad \mathrm{and} \quad \sigma (z + \omega") = \frac{\tfrac{1}{2} p^{\circ}(z)}{p(z) - \tfrac{1}{6}}. \end{align*} $$

In connection with the effect of an $\omega "$ shift, the devotee of [Reference Whittaker and Watson9] will be pleased to consult Miscellaneous Example 12 on page 457 for the case when $c = - 1/6$ and $e = \kappa / 2$ .

The following result makes more transparent the effect on $\sigma $ of an $\omega '$ shift.

4 Jacobian representation

The elliptic function $\sigma $ can also be given a manifestly Jacobian cast. Return to the differential equation

$$ \begin{align*}(\sigma^{\circ})^2 = \tfrac{1}{4} \kappa^2 - \sigma^2 + \sigma^4\end{align*} $$

of Theorem 2.3 and compare it with the differential equation

$$ \begin{align*}(\mathrm{sn}^{\circ})^2 = (1 - \mathrm{sn}^2) (1 - k^2 \mathrm{sn}^2)\end{align*} $$

that is satisfied by the Jacobian modular sine function $\mathrm {sn} = \mathrm {sn} (- , k)$ of modulus k. Alongside the modulus $\kappa \in (0, 1)$ we set the complementary modulus $\lambda = (1 - \kappa ^2)^{1/2} \in (0, 1)$ and introduce the abbreviations

$$ \begin{align*}\mu_{\pm} = \bigg(\frac{1 \pm \lambda}{2}\bigg)^{1/2} \in (0, 1)\end{align*} $$

so that

$$ \begin{align*} \mu_+ \mu_- = \bigg(\frac{1 - \lambda^2}{4}\bigg)^{1/2} = \frac{1}{2} \kappa \end{align*} $$

and

$$ \begin{align*}\mu_+^2 + \mu_-^2 = 1.\end{align*} $$

Theorem 4.1. The function $\sigma $ is given by $\sigma (z) = \mu _- \, \mathrm {sn} (\mu _+ z, k)$ , where

$$ \begin{align*}k = \frac{\mu_-}{\mu_+} = \bigg(\frac{1 - \lambda}{1 + \lambda} \bigg)^{1/2}.\end{align*} $$

Proof. With $s(z) : = \mu _- \, \mathrm {sn} (\mu _+ z, k)$ , we have

$$ \begin{align*}s^{\circ} (z) = \mu_- \mu_+ \, \mathrm{sn}^{\circ} (\mu_+ z, k) = \tfrac{1}{2} \kappa\, \mathrm{sn}^{\circ} (\mu_+ z, k)\end{align*} $$

from which the differential equation satisfied by $\mathrm {sn} (-, k)$ yields

$$ \begin{align*}s^{\circ}(z)^2 = \frac{1}{4} \kappa^2 \, \bigg(1 - \frac{s(z)^2}{\mu_-^2}\bigg) \bigg(1 -\bigg(\frac{\mu_-}{\mu_+}\bigg)^2 \, \frac{s(z)^2}{\mu_-^2}\bigg)\end{align*} $$

and mild simplification reveals that s satisfies the differential equation

$$ \begin{align*}(s^{\circ})^2 = \tfrac{1}{4} \kappa^2 - s^2 + s^4;\end{align*} $$

of course, $s(0) = 0$ and $s(t)> 0$ when $t> 0$ is small. Thus, s is none other than $\sigma $ .

Note that Theorem 3.2 embodies the property $\mathrm {sn} (u + 2 K, k) = - \,\mathrm {sn} (u, k)$ , while Theorem 3.6 reflects the property $\mathrm {sn} (u + i K', k) = k / \mathrm {sn} (u, k).$ As a curiosity, note further that the real Jacobi transformation

$$ \begin{align*}k \, \mathrm{sn} (u, k) = \mathrm{sn} (k u, 1/k)\end{align*} $$

assumes here the symmetrical form

$$ \begin{align*}\mu_- \, \mathrm{sn} \bigg(\mu_+ \, z,\, \frac{\mu_-}{\mu_+}\bigg) = \mu_+ \, \mathrm{sn} \bigg(\mu_- \, z, \, \frac{\mu_+}{\mu_-}\bigg).\end{align*} $$

Along with the modular sine $\mathrm {sn}$ , we have the modular cosine $\mathrm {cn}$ and the delta amplitude $\mathrm {dn}$ , all to the same modulus k. Accordingly, the elliptic function $\sigma $ belongs to a triple $(\sigma , \gamma , \delta )$ of ‘Jacobian’ elliptic functions defined by

$$ \begin{align*}\sigma (z) = \mu_- \, \mathrm{sn} (\mu_+ z, k), \quad \gamma (z) = \mu_- \, \mathrm{cn} (\mu_+ z, k), \quad \delta (z) = \mu_+ \, \mathrm{dn} (\mu_+ z, k).\end{align*} $$

Here, the choice of $\mu _+$ as the multiplier in $\delta $ leads to more elegant relationships, as follows.

Theorem 4.2. The functions $\sigma $ , $\gamma $ and $\delta $ satisfy the algebraic equations

$$ \begin{align*}\gamma^2 + \sigma^2 = \mu_-^2 \quad \mathrm{and} \quad \delta^2 + \sigma^2 = \mu_+^2.\end{align*} $$

By subtraction, we also have $\delta ^2 - \gamma ^2 = \lambda $ .

Theorem 4.3. The functions $\sigma $ , $\gamma $ and $\delta $ satisfy the differential equations

$$ \begin{align*}\sigma^{\circ} = \gamma \, \delta, \quad \gamma^{\circ} = - \sigma \, \delta, \quad \delta^{\circ} = - \sigma \, \gamma.\end{align*} $$

Proof. These follow at once from the familiar Jacobian identities

$$ \begin{align*} \mathrm{sn}^{\circ} = \mathrm{cn} \cdot \mathrm{dn}, \quad \mathrm{cn}^{\circ} = -\, \mathrm{sn} \cdot \mathrm{dn}, \quad \mathrm{dn}^{\circ} = - \, k^2 \, \mathrm{sn} \cdot \mathrm{ cn}. \\[-33pt]\end{align*} $$

5 Elliptic periods

Here, we derive explicit expressions for the fundamental periods $2 \omega $ and $2 \omega '$ of the elliptic function $\sigma $ ; equivalently, of its coperiodic Weierstrass function p. We offer two such expressions: one in terms of the hypergeometric function $F(\tfrac {1}{2}, \tfrac {1}{2}; 1; -)$ that is appropriate to the classical theory of elliptic functions and one in terms of the signature four hypergeometric function $F(\tfrac {1}{4}, \tfrac {3}{4}; 1; -)$ . Before proceeding, it is helpful to dress the notation with the modulus on which our construction is based: thus, $2 \omega _{\kappa }$ and $2 \omega ^{\prime }_{\kappa }$ will be the fundamental periods of $\sigma _{\kappa }$ and $p_{\kappa }$ .

Expressions in terms of the classical hypergeometric function $F(\tfrac {1}{2}, \tfrac {1}{2}; 1; -)$ are provided at once by Theorem 4.1. Here, recall that $\lambda = (1 - \kappa ^2)^{1/2}$ is the modulus complementary to $\kappa $ and that $\mu _+ = (\tfrac {1}{2}(1 + \lambda ))^{1/2}$ .

Theorem 5.1. The fundamental half-periods $\omega _{\kappa }$ and $\omega _{\kappa }'$ of $\sigma _{\kappa }$ are given by

$$ \begin{align*}\mu_+ \, \omega_{\kappa} = \pi \, F\bigg(\frac{1}{2}, \frac{1}{2}; 1; \frac{1 - \lambda}{1 + \lambda}\bigg) \quad\mbox{and}\quad \mu_+\, \omega^{\prime}_{\kappa} = \frac{1}{2} \pi \, i \, F\bigg(\frac{1}{2}, \frac{1}{2}; 1; \frac{2 \lambda}{1 + \lambda}\bigg).\end{align*} $$

Proof. Recall (say, from [Reference Whittaker and Watson9, Ch. XXII]) that $\mathrm {sn} (- , k)$ has fundamental periods $4 K$ and $2 i K'$ , where

$$ \begin{align*} K = \tfrac{1}{2} \pi \, F (\tfrac{1}{2}, \tfrac{1}{2}; 1; k^2) \quad\mbox{and}\quad K' = \tfrac{1}{2} \pi \, F (\tfrac{1}{2}, \tfrac{1}{2}; 1; 1 - k^2). \end{align*} $$

It follows immediately from Theorem 4.1 that $\sigma $ has $4 K / \mu _+$ as real period and $2 i K' / \mu _+$ as imaginary period. Finally, recall from Theorem 4.1 the expression for k in terms of $\lambda $ .

An $F(\tfrac {1}{4}, \tfrac {3}{4}; 1; -)$ expression for the real period $2 \omega _{\kappa }$ is already in hand: reference to Theorem 2.2 procures the expression $\omega _{\kappa } =\pi \, F(\tfrac {1}{4}, \tfrac {3}{4}; 1; \kappa ^2)$ for the real half-period. Access to the imaginary half-period $\omega ^{\prime }_{\kappa }$ will be facilitated by means of an auxiliary Weierstrass function.

Explicitly, alongside the Weierstrass function

$$ \begin{align*}p_{\kappa} = \wp (-; \omega_{\kappa},\omega_{\kappa}') = \wp (-; g_2, g_3)\end{align*} $$

that is coperiodic with $\sigma _{\kappa }$ , we shall consider the Weierstrass function

$$ \begin{align*}q_{\kappa} = \wp (-; \omega_{\kappa} / 2, \omega_{\kappa}') = \wp (-; h_2, h_3)\end{align*} $$

that comes by halving the real (half-)period.

Theorem 5.2. The invariants $h_2$ and $h_3$ of $q_{\kappa }$ are given by

$$ \begin{align*}h_2 = \tfrac{4}{3} - \kappa^2 \quad \textit{and} \quad h_3 = \tfrac{8}{27} - \tfrac{1}{3} \kappa^2.\end{align*} $$

Proof. Reference to [Reference Lawden4, Section 9.8] provides us with expressions for the invariants of $q_{\kappa }$ in terms of those for $p_{\kappa }$ . Quite generally, such period dimidiation has the following effect:

$$ \begin{align*}h_2 = 60 p(\omega)^2 - 4 g_2, \quad h_3 = 56 p(\omega)^3 + 8 g_3.\end{align*} $$

With $p_{\kappa } (\omega ) = 1/6$ and with the invariants of $p_{\kappa }$ provided by Theorem 3.1, we calculate the invariants of $q_{\kappa }$ to be as advertised.

The Weierstrass functions $q_{\kappa }$ and $p_{\lambda }$ are, therefore, related as follows.

Proof. Quite generally, a Weierstrass function $\wp (-; G_2, G_3)$ satisfies the homogeneity relation: $\wp (z; \mu ^4 G_2, \mu ^6 G_3) = \mu ^2 \wp (\mu \, z; G_2, G_3).$ Here, take $G_2 = \tfrac {1}{12} (1 + 3 \lambda ^2)$ and $G_3 = \tfrac {1}{216}(1 - 9 \lambda ^2)$ ; with $\mu = i \sqrt 2$ ,

$$ \begin{align*}\mu^4 G_2 = \tfrac{1}{3} (1 + 3 \lambda^2) = \tfrac{4}{3} - \kappa^2 \quad\mbox{and}\quad \mu^6 G_3 = \tfrac{1}{27} (9 \lambda^2 - 1) = \tfrac{8}{27} - \tfrac{1}{3} \kappa^2.\end{align*} $$

Finally, invoke Theorem 3.1 for the modulus $\kappa $ and for the complementary modulus $\lambda $ .

We may now make the imaginary period of $\sigma _{\kappa }$ explicit in $F(\tfrac {1}{4}, \tfrac {3}{4}; 1; -)$ terms; for reference, we record also the real period, as previously addressed.

Theorem 5.4. The fundamental half-periods $\omega _{\kappa }$ and $\omega _{\kappa }'$ of $\sigma _{\kappa }$ are given by

$$ \begin{align*}\omega_{\kappa} = \pi \, F\, (\tfrac{1}{4}, \tfrac{3}{4}; 1; \kappa^2) \quad\mbox{and}\quad \omega_{\kappa}' = i \, \pi \, F\, (\tfrac{1}{4}, \tfrac{3}{4}; 1 - \kappa^2)/{\sqrt2}.\end{align*} $$

Proof. Compare expressions for the real half-period of $q_{\kappa }$ : on the one hand, it is $\omega _{\kappa } / 2$ by definition; on the other hand, it is $- i \omega _{\lambda }' / \sqrt 2$ on account of Theorem 5.3. Switching moduli, we deduce that $\omega _{\kappa }' = i \omega _{\lambda } / \sqrt 2$ ; recalling our identification of the real half-period, we conclude that

$$ \begin{align*} \omega_{\kappa}' = {i} \pi F\,(\tfrac{1}{4}, \tfrac{3}{4}; 1; \lambda^2)/\sqrt2 = {i} \pi F(\tfrac{1}{4}, \tfrac{3}{4}; 1; 1 - \kappa^2)/\sqrt2.\\[-34pt] \end{align*} $$

These concrete expressions for the fundamental periods of $\sigma _{\kappa }$ at once furnish transformation laws relating the signature four hypergeometric function $F(\tfrac {1}{4}, \tfrac {3}{4}; 1; -)$ to the ‘classical’ hypergeometric function $F(\tfrac {1}{2}, \tfrac {1}{2}; 1; -)$ .

Theorem 5.5. If $0 < \lambda < 1$ , then

$$ \begin{align*} F\bigg(\!\frac{1}{4},\! \frac{3}{4};\! 1;\! 1 \!-\! \lambda^2\!\bigg) = \sqrt{\!\frac{2}{1 \!+\! \lambda}} \! F\bigg(\!\frac{1}{2}, \frac{1}{2};\! 1;\! \frac{1 \!-\! \lambda}{1 \!+\! \lambda}\!\bigg)\!\quad\mbox{and}\quad\! F\bigg(\!\frac{1}{4}, \frac{3}{4}; 1; \lambda^2\!\bigg) = \sqrt{\!\frac{1}{1\! +\! \lambda}} F\bigg(\!\frac{1}{2},\! \frac{1}{2};\! 1;\! \frac{2 \lambda}{1 \!+\! \lambda}\!\bigg).\end{align*} $$

These identities also appear in [Reference Berndt, Bhargava and Garvan2]: the first appears as Theorem 9.2 and is derived by manipulation of an entry in the second notebook of Ramanujan; the second appears as Theorem 9.1 and is precisely an entry in the notebook. In our approach, these identities fall directly from the periods of the elliptic function $\sigma _{\kappa }$ and thereby cement the place of this function in the signature four elliptic theory.

As in [Reference Berndt, Bhargava and Garvan2], these transformation laws engender a relationship between the signature four base

$$ \begin{align*} q_4 (x) = \exp \bigg[-\sqrt2 \, \pi \frac{F(\tfrac{1}{4}, \tfrac{3}{4}; 1; 1 - x)}{F(\tfrac{1}{4} , \tfrac{3}{4}; 1; x)}\bigg] \end{align*} $$

and the classical base

$$ \begin{align*} q(y) = \exp \bigg[ - \pi \, \frac{F(\tfrac{1}{2} , \tfrac{1}{2}; 1; 1 - y)}{F(\tfrac{1}{2} , \tfrac{1}{2}; 1; y)}\bigg]; \end{align*} $$

in fact, with $x = \lambda ^2$ and $y = {2 \lambda }/{(1 + \lambda )}$ , we see from Theorem 5.5 that

$$ \begin{align*} q_4(\lambda^2) = q\,\bigg(\frac{2 \lambda}{1 + \lambda}\bigg)^2. \end{align*} $$

Again as in [Reference Berndt, Bhargava and Garvan2], this relationship between bases supports a transfer principle whereby classical elliptic results yield counterparts in signature four. Section 9 of [Reference Berndt, Bhargava and Garvan2] contains several such instances of passage from classical results to signature four counterparts. In our next section, we demonstrate how some such signature four identities may be derived directly from the elliptic function $\sigma _{\kappa }$ .

We close this section by pointing out that the identities in Theorem 5.5 may, of course, be derived otherwise; as an illustration, the real quarter-period

$$ \begin{align*}L = \tfrac{1}{2} \, \pi \, F (\tfrac{1}{4}, \tfrac{3}{4}; 1; \kappa^2) \end{align*} $$

may be cast in $F(\tfrac {1}{2}, \tfrac {1}{2}; 1; -)$ terms by integration, as follows. Let $\theta = \operatorname {arc}\sin \kappa $ be the modular angle, so that $\cos \theta = \lambda $ and $\sigma (L) = \sin \tfrac {1}{2} \psi (L) = \sin \tfrac {1}{2} \theta .$ From

$$ \begin{align*} \sin^2 \tfrac{1}{2} \theta= \tfrac{1}{2} (1 - \cos \theta) = \tfrac{1}{2} (1 - \lambda), \end{align*} $$

it follows that $\sin \tfrac {1}{2} \theta = \mu _-$ and, similarly, $\cos \tfrac {1}{2} \theta = \mu _+$ in the notation set up ahead of Theorem 4.1. From the differential equation in Theorem 2.3, we now deduce that

$$ \begin{align*}L = \int_0^{\mu_-} \frac{{\,d} \sigma}{\sqrt{ \sigma^4 - \sigma^2 + \frac{1}{4} \kappa^2}}, \end{align*} $$

where $\kappa ^2 = 4 \sin ^2 \tfrac {1}{2} \theta \, \cos ^2 \tfrac {1}{2} \theta = 4 \mu _-^2 \, \mu _+^2$ and, therefore,

$$ \begin{align*}L = \int_0^{\mu_-} \frac{{\, d} \sigma} {\sqrt{ (\mu_-^2 - \sigma^2) (\mu_+^2 - \sigma^2)}} \,. \end{align*} $$

Put $\sigma = \mu _- \sin t$ for $0 \leqslant t \leqslant \tfrac {1}{2} \pi $ to see that

$$ \begin{align*} L = \int_0^{\frac{1}{2} \pi} \frac{{\, d} t}{\sqrt{\mu_+^2 - \mu_-^2 \sin^2 t}} \,\end{align*} $$

and conclude that

$$ \begin{align*} \mu_+ \, L = \int_0^{\frac{1}{2} \pi} \frac{{\, d} t}{\sqrt{1 - k^2 \sin^2 t}} = \frac{1}{2} \pi \, F\,\bigg(\frac{1}{2}, \frac{1}{2}; 1; k^2\bigg) \end{align*} $$

by the familiar evaluation of a complete elliptic integral of the first kind.

6 Eisenstein series

We open this section by recalling the familiar connection between Weierstrass functions and Eisenstein series. For additional details, see [Reference Serre6, Ch. VII] or [Reference Armitage and Eberlein1, Ch. 7].

Let $p = \wp (-; \omega , \omega ') = \wp (-; g_2, g_3)$ be a quite arbitrary Weierstrass function, with $2 \omega $ and $2 \omega '$ as fundamental periods and with $g_2$ and $g_3$ as invariants; take the period ratio $\tau : = \omega ' / \omega $ to have positive imaginary part as usual. The quadrinvariant $g_2$ is given by

$$ \begin{align*}g_2 = \frac{60}{(2 \omega)^4} {\sum}' \frac{1}{(m + n \tau)^4}\end{align*} $$

and the Eisenstein series $E_4 (\tau )$ by

$$ \begin{align*}E_4 (\tau) = \frac{45}{\pi^4} {\sum}' \frac{1}{(m + n \tau)^4},\end{align*} $$

where, in each case, the prime $'$ indicates that summation takes place over all pairs of integers $(m, n)$ other than $(0, 0)$ ; thus,

$$ \begin{align*}E_4(\tau) = \frac{3}{4}\, \bigg(\frac{2 \omega}{\pi}\bigg)^4 \, g_2.\end{align*} $$

Similarly, the cubinvariant $g_3$ is given by

$$ \begin{align*}g_3 = \frac{140}{(2 \omega)^6} {\sum}' \frac{1}{(m + n \tau)^6}\end{align*} $$

and the Eisenstein series $E_6 (\tau )$ by

$$ \begin{align*}E_6 (\tau) = \frac{945}{2 \cdot \pi^6} {\sum}' \frac{1}{(m + n \tau)^6}\end{align*} $$

so that

$$ \begin{align*}E_6 (\tau) = \frac{27}{8} \, \bigg(\frac{2 \omega}{\pi}\bigg)^6 \, g_3.\end{align*} $$

It will now be convenient to employ the notational abbreviation $F_4 = F(\tfrac {1}{4}, \tfrac {3}{4}; 1; -)$ : thus, $F_4(x) : = {F(\tfrac {1}{4}, \tfrac {3}{4}; 1; x)}.$ In [Reference Berndt, Bhargava and Garvan2], this is written $z(4; x)$ or just $z(4)$ when x is understood.

Theorem 6.1. Let $0 < x < 1$ . If

$$ \begin{align*}\tau = \frac{i}{\sqrt2} \, \frac{F_4(1 - x)}{F_4 (x)},\end{align*} $$

then $E_4 (\tau ) = F_4(x)^4 \, (1 + 3 x)$ and $E_6 (\tau ) = F_4 (x)^6 \, (1 - 9 x).$

Proof. Let $x = \kappa ^2$ and apply the foregoing recollections to the Weierstrass function $p_{\kappa }$ that is coperiodic with $\sigma _{\kappa }$ . From Theorem 5.4,

$$ \begin{align*}\frac{2 \omega_k}{\pi} = 2 F_4 (\kappa^2),\end{align*} $$

and from Theorem 3.1, $g_2 = \tfrac {1}{12} (1 + 3 \kappa ^2)$ , from which it follows at once that

$$ \begin{align*}E_4(\tau) = \tfrac{3}{4} \cdot (2 \, F_4 (\kappa^2))^4 \cdot \tfrac{1}{12} (1 + 3 \kappa^2) = F_4 (\kappa^2)^4 \, (1 + 3 \kappa^2),\end{align*} $$

where

$$ \begin{align*}\tau = \frac{\omega_{\kappa}'}{\omega_{\kappa}} = \frac{i}{\sqrt2} \, \frac{F_4(1 - \kappa^2)}{F_4 (\kappa^2)},\end{align*} $$

again by Theorem 5.4. This handles $E_4$ ; $E_6$ is handled similarly.

These formulas appear as Theorems 9.5 and 9.6 in [Reference Berndt, Bhargava and Garvan2] as applications of the signature four transfer principle. They also appear as (6.3) and (6.4) in [Reference Shen7], where they are approached quite differently, via theta functions and the Landen transformation.

Theorem 6.2. Let $0 < x < 1$ . If

$$ \begin{align*}\tau = i \sqrt2 \, \frac{F_4(1 - x)}{F_4 (x)},\end{align*} $$

then $E_4 (\tau ) = F_4(x)^4 \, (1 - \tfrac {3}{4} x)$ and $E_6 (\tau ) = F_4 (x)^6 \, (1 - \tfrac {9}{8} x).$

Proof. Let $x = \kappa ^2$ and apply the recollections above to the Weierstrass function $q_{\kappa }$ that was obtained from $p_{\kappa }$ by halving the real period. For $E_6$ ,

$$ \begin{align*}E_6 (\tau) = \tfrac{27}{8}\cdot F_4 (\kappa^2)^6 \cdot (\tfrac{8}{27} - \tfrac{1}{3} \kappa^2) = F_4 (\kappa^2)^6 (1 - \tfrac{9}{8} \kappa^2)\end{align*} $$

using Theorems 5.4 and 5.2; for $E_4$ the calculations are similar.

Again, these formulas are derived using the signature four transfer principle in [Reference Berndt, Bhargava and Garvan2], where they appear as Theorems 9.7 and 9.8. They also appear in [Reference Shen7], where in Corollary 3.4 they are derived substantially as in the present paper. Theorems 6.1 and 6.2 serve to cement still further the place of $\sigma $ in signature four.

7 Shen functions

The elliptic function $\sigma $ originated in a study of the elliptic function $\mathrm {dn}_2$ that was introduced in [Reference Shen7] by Shen. Here, we set out the relationship of [Reference Shen7] to the present paper.

With minor notational differences, the construction in [Reference Shen7] is based on the same bijection $\phi : \mathbb {R} \to \mathbb {R}$ , according to which

$$ \begin{align*}u = \int_0^{\phi(u)} F\,\bigg(\frac{1}{4}, \frac{3}{4}; \frac{1}{2}; \kappa^2 \sin^2 t\bigg)\, d t,\end{align*} $$

and on the same auxiliary function

$$ \begin{align*}\psi : = \operatorname{arc}\sin (\kappa \sin \phi)\end{align*} $$

with which we began Section 2; the elliptic function $\mathrm {dn}_2$ is defined in [Reference Shen7] as the elliptic extension of the real function $\cos \psi $ . The motivation for this construction comes from a classical approach to the Jacobian elliptic functions, in which the hypergeometric integrand $F(\tfrac {1}{2}, \tfrac {1}{2}; \tfrac {1}{2}; -)$ is simply replaced by $F(\tfrac {1}{4}, \tfrac {3}{4}; \tfrac {1}{2}; -)$ ; this motivation suggested the name $\mathrm {dn}_2$ because use of the classical hypergeometric integrand, instead, leads directly to the Jacobian delta amplitude $\mathrm {dn}$ .

More explicitly, in [Reference Shen7], it is shown that the function $y = \cos \psi $ satisfies the differential equation

$$ \begin{align*}(y^{\circ})^2 = 2 (1 - y) (y^2 - \lambda^2),\end{align*} $$

while, of course, $y(0) = 1$ . The solution to this initial value problem is a little easier to identify than was the solution in Theorem 3.1, because the right-hand side of the differential equation for y has the initial value as a zero, so that the simpler arguments on [Reference Whittaker and Watson9, page 453] apply. The result is that

$$ \begin{align*}y = 1 - \frac{\tfrac{1}{2} \kappa}{\tfrac{1}{3} + \wp (-; h_2, h_3)},\end{align*} $$

where the indicated Weierstrass function has invariants

$$ \begin{align*}h_2 = \tfrac{4}{3} - \kappa^2 \quad \mathrm{and} \quad h_3 = \tfrac{8}{27} - \tfrac{1}{3} \kappa^2.\end{align*} $$

Notice that this Weierstrass function is precisely the function $q_{\kappa }$ that we introduced immediately before Theorem 5.2. The elliptic extension $\mathrm {dn}_2$ of $\cos \psi $ is, thus, given by

$$ \begin{align*}\mathrm{dn}_2 = 1 - \frac{\frac12 \kappa}{\frac13 + q_{\kappa}}\end{align*} $$

and $q_{\kappa }$ is plainly its coperiodic Weierstrass function. Incidentally, the proof of Theorem 6.2 that appears in [Reference Shen7] as Corollary 3.4 was based on the periods of $\mathrm {dn}_2$ . The absence of $\sigma $ from [Reference Shen7] necessitated use of the Landen transformation to deduce the equivalents (6.3) and (6.4) of our Theorem 6.1. That the Landen transformation should enter is, of course, not surprising, in view of the relationship between the Weierstrass functions $p_{\kappa }$ and $q_{\kappa }$ .

The origin of $\mathrm {dn}_2$ as a signature four version of the Jacobian function $\mathrm {dn}$ invites consideration of the functions $\cos \phi $ and $\sin \phi $ , as these yield $\mathrm { cn}$ and $\mathrm { sn}$ in the classical theory. It is shown in [Reference Shen7] that, in signature four, $\cos \phi $ does extend as an elliptic function $\mathrm { cn}_2$ on the plane; in contrast, $\sin \phi $ does not even extend to a meromorphic function on the plane, as we show shortly.

In one significant respect, the elliptic function $\mathrm {dn}_2$ is imperfect as a signature four counterpart to the classical Jacobian function $\mathrm {dn}$ : the classical function $\mathrm {dn}$ has two simple poles in each of its period parallelograms and is, therefore, of generally Jacobian type; but the signature four function $\mathrm {dn}_2$ is of Weierstrassian type, as it has a double pole in each period parallelogram. A similar comment applies to the elliptic function $\mathrm {cn}_2$ ; and $\mathrm {sn}_2$ simply fails to exist as an elliptic function. These are the circumstances that prompted our search for a set of truly Jacobian functions in signature four; particularly, a signature four counterpart to the odd function $\mathrm {sn}$ .

The precise relationships between our functions $\sigma $ , $\gamma $ and $\delta $ and the functions in [Reference Shen7] are readily determined. First, the trigonometric duplication identity

$$ \begin{align*}\cos \psi = 1 - 2 \sin^2 \tfrac{1}{2} \psi\end{align*} $$

on $\mathbb {R}$ extends at once by analytic continuation to the identifications

$$ \begin{align*}\mathrm{dn}_2 = 1 - 2 \sigma^2 = 2 \gamma^2 + \lambda = 2 \delta^2 - \lambda\end{align*} $$

with the aid of Theorem 4.2. We extract from the proof of Theorem 2.3 the identification

$$ \begin{align*}\tfrac{1}{2} \kappa \, \mathrm{cn}_2 = \sigma^{\circ}\end{align*} $$

to which Theorem 4.3 at once contributes the identification

$$ \begin{align*}\tfrac{1}{2} \kappa \, \mathrm{cn}_2 = \gamma \, \delta.\end{align*} $$

As noted above, the case of $\mathrm {sn}_2$ is different: from

$$ \begin{align*}\kappa^2 \sin^2 \phi = \sin^2 \psi = 4 \sin^2 \tfrac{1}{2} \psi \, \cos^2 \tfrac{1}{2} \psi,\end{align*} $$

it follows that

$$ \begin{align*}\tfrac{1}{4} \kappa^2 \sin^2 \phi = \sigma^2 (1 - \sigma^2)\end{align*} $$

on $\mathbb {R}$ . The elliptic function $\sigma ^2 (1 -\sigma ^2)$ has simple zeros where $\sigma = \pm 1$ because $(\sigma ^{\circ })^2 = \tfrac {1}{4} \kappa ^2 \neq 0$ there; this prevents $\sigma ^2 (1 - \sigma ^2)$ from having meromorphic square roots and so prevents $\sin \phi $ from having meromorphic extensions.

The preceding formulas for $\mathrm {dn}_2$ and $\mathrm {cn}_2$ have versions in terms of the classical Jacobian functions to modulus $k = \sqrt {(1 - \lambda ) / (1 + \lambda )}$ : explicitly, Theorem 4.1 at once gives

$$ \begin{align*}\mathrm{dn}_2 (z) = 1 - (1 - \lambda) \, \mathrm{sn}^2 (\mu_+ z, k)\end{align*} $$

and by differentiation gives

$$ \begin{align*}\mathrm{cn}_2 (z) = \mathrm{cn} (\mu_+ z, k) \, \mathrm{dn} (\mu_+ z, k).\end{align*} $$

Finally, in [Reference Shen8], Shen presented elliptic or hyperelliptic functions of interest in each of the three standard alternative signatures. In each case, the proposed function satisfies a differential equation of the form

$$ \begin{align*}(y^{\circ})^2 = T_n (y) - (1 - 2 \kappa^2),\end{align*} $$

where $T_n$ is the degree n Chebyshev polynomial of the first kind. The case appropriate to signature four has $n = 4$ : the corresponding function $y_4$ satisfies

$$ \begin{align*}(y^{\circ})^2 = 8 y^4 - 8 y^2 + 2 \kappa^2\end{align*} $$

and is required to have as its initial value one of the four zeros of the quartic on the right; in fact, it is required that $y_4(0) = \cos \tfrac {1}{2} \theta $ , where $\theta $ is the modular angle, as above. This function $y_4$ is plainly related to $\sigma $ : indeed, $y_4(z) = \sigma (2 \sqrt 2 z + c)$ for a suitable choice of the shift c.