# Integral relations in Fricke and Klein

Can someone please explain how Fricke and Klein obtain the integral relationa stated at the top of p. 34 in this book? The entire book can be previewed on Google Books. It is an old book and I do not speak German. Also, is there a modern account of this anywhere in the modern literature?

• Page 34 is not on google books. If you post the relation, or at least an image, then perhaps this question will be answered faster, since many people may not have the german book in question. – user207766 Mar 26 '15 at 0:09
• @Andy You can preview the entire book. – glebovg Mar 26 '15 at 0:10
• Your right. My computer glitched, and it was not working. – user207766 Mar 26 '15 at 0:12
• I am a German native speaker and I can translate passages of the book if you wish. However, I cannot see any integrals on p. 34... – sranthrop Apr 3 '15 at 3:17
• @sranthrop I would really appreciate your help. There are three relations at the top of p. 34. The derivation starts at the bottom of p. 33 right after (3). – glebovg Apr 5 '15 at 3:11

Using the Weierstrass Normal form, let's try to understand the elliptic integral

$$\int \frac{dy}{\sqrt{4y^3 - g_2 y - g_3}}$$

In order to reduce the number of parameters, Fricke and Klein do a change of variables $y = \frac{g_2}{g_3}z$. The elliptic curve has two basic periods

$$\Omega = \int \frac{dz}{\sqrt{4z^3 + g(z+1)}} \text{ and } -H = \int \frac{z\,dz}{\sqrt{4z^3 + g(z+1)}}$$ This new parameter $g$ is basically the $J$-invariant in disguise. $\frac{27 J}{1-J} = g$. Let's just call the denominator $R = \sqrt{\dots}$.

What happens if we change $g$? Let's differentiate under the integral sign:

\begin{eqnarray} \frac{d\Omega}{dg} &=& - &\int \frac{dz}{2R^3} & - &\int \frac{z \, dz}{2R^3}\\ \frac{dH}{dg} &=& & \int \frac{z \, dz}{2R^3} &+& \int \frac{z^2 \, dz}{2R^3} \end{eqnarray}

These are called the Picard-Fuchs equations. In modern language, this is an instance of the Gauss-Manin connection.

The integral of any derivative along a closed path on the elliptic curve is going to be zero.

$$\int d\left(\frac{z^a}{R}\right) = \int \frac{a z^{a-1}}{R} - \int \frac{z^a(12 z^2 + g)}{2R^3} = 0 \tag{\ast}$$

The first step says the derivative and integral are inverse operations - the fundamental theorem of calculus. For a closed path the two endpoints are the same so the integral is zero. For the second step use the quotient rule for derivatives and the definition of R as the denominator.

The first relation comes from setting $a = 0$ without any changes. The second and third relations start from setting $a = 1,2$ with some extra manipulation.

EDIT Yeah, they fall out of $R^2 = 4z^3 + g(z+1)$ and the definition of the periods $\Omega, H$.

Fricke and Klein solve for the periods and ultimately find the differential equation the period $\Omega$ satisfies in terms of the elliptic curve invariant $J$:

$$\frac{d^2 \Omega}{dJ^2} + \frac{1}{J} \cdot \frac{d \Omega}{dJ} + \frac{\frac{31}{144}J- \frac{1}{36}}{J^2(J-1)^2}\Omega = 0$$

## When is $\oint \neq 0$ ?

The fundamental theorem of calculus says: $\int_a^b F'(x) \, dx = F(b) - F(a)$ so we integrate along a closed path, $a = b$ so the integral should be $F(b) - F(a) = 0$. Yet,

$$\int_{|z|=1} \frac{dz}{z} = 2\pi i \neq 0$$

What goes wrong is that $\frac{1}{z}$ is not holomorphic on $\mathbb{C}$, it is only holomorphic on $\mathbb{C} \backslash \{0\}$. By removing the one point we no longer have a plane, but a punctured plane. The circle $|z| = 1$ moves around this puncture and therefore integrating has a residue.

Another important issue is $\frac{1}{\sqrt{z}}$. The problem here is as we move around the unit circle from $1$ back to itself, the square roots go to $-1$.

$$\sqrt{1} = \color{blue}{\mathbf{1}} \to \sqrt{i} = \mathbf{\tfrac{\color{blue}{1+i}}{\color{blue}{\sqrt{2}}}} \to \sqrt{-1} = \mathbf{\color{blue}{i}}\to \sqrt{-i} = \mathbf{\tfrac{\color{blue}{1-i}}{\color{blue}{\sqrt{2}}}} \to \sqrt{1} = -\mathbf{\color{green}{1}}$$

Basically every number has two square roots except for $0$, therefore the riemann surface $\{ (z, \pm \sqrt{z}): z \in \mathbb{C} \}$ has branching at $0$.

Using the identity $(\ast)$ with $a=1$ we can solve the 2nd identity.

\begin{eqnarray} \frac{1}{2}\Omega = \frac{1}{2}\int \frac{dz}{R} = \frac{1}{2}\int \frac{R^2 }{R^3 } dz &=& \frac{1}{2}\int \frac{4z^3 + g(z+1) }{R^3 } dz \\ &=& \frac{1}{2}\int \frac{4z^3 + g(z+1) + (-4z^3 + g(z+2))}{R^3 } dz \\ &=& 2g \int \frac{ z \, dz}{2R^3 } + 3g \int \frac{ dz}{2R^3 } \end{eqnarray}

Using the identity $(\ast)$ with $a=2$ we can solve the 3rd identity.

\begin{eqnarray} \tfrac{1}{2}H = -\tfrac{1}{2}\int \frac{z}{R} \, dz = -\frac{1}{2}\int \frac{z R^2}{R^3} \, dz &=& - \frac{1}{2}\int \frac{z(4z^3 + g(z+1)) }{R^3 } dz \\ &=& -\frac{1}{2}\int \frac{ -g(3z^2 + 4z) + g(z+1)z }{R^3 } dz \\ &=& 2g \int \frac{ z^2 \,dz }{2R^3 } + 3g\int \frac{ z \,dz}{2R^3 } \end{eqnarray}

• Do you know why the integral of any derivative along a closed path on the elliptic curve is going to be zero? – glebovg May 13 '15 at 1:07
• @glebovg That's just the fundamental theorem of calculus, basically. (Stoke's theorem implies the integral of an exact form over a closed path is zero.) – Potato May 13 '15 at 1:46
• I still do not understand how to obtain the last two relations. They do not simply follow from the definitions of $\Omega$ and $H$. Any help will be appreciated. – glebovg Jun 6 '15 at 23:09
• @glebovg OK I included details about those two equations. Let me know if you need more. – cactus314 Jun 10 '15 at 22:52
• @johnmangual Thanks. We just add zero (of course). – glebovg Jun 11 '15 at 19:24