# Show that $K=K'$ if and only if $k=(\sqrt{2}-1)^2$ (Ahlfors)

In Ahlfors' Complex Analysis text, page 240, he defines the following two integrals: $$K=\int_{-1}^1 \frac{dt}{\sqrt{(1-t^2)(1-k^2 t^2)}},$$ $$K'=\int_1^{1/k} \frac{dt}{\sqrt{(t^2-1)(1-k^2t^2)}}.$$ Here $0<k<1$ is a parameter. I want to prove that

$K=K'$ iff $k=(\sqrt{2}-1)^2$

My attempts:

• going "from left to right"

I've expanded $K$ as a power series in $k$:

$$K=\pi \sum_{n=0}^\infty \left[\frac{(2n)!}{2^{2 n} (n!)^2}\right]^2 k^{2n}$$

and I thought trying to do the same for $K'$. Then I realized that this probably won't get me anywhere...

I've also plotted both functions, and it appears that $K$ is increasing with $k$, while $K'$ is decreasing. Although I couldn't prove rigorously that $K'$ is decreasing, this shows that only one solution for $k$ is possible.

• going "from right to left"

I have no clue, I guess I should come up with a clever change of variables in the second integral, but I can't find it.

Any help is appreciated!

Ahlfors probably intended the reader to notice that $2K$ and $2iK'$ are the fundamental real and imaginary periods of the elliptic integral $$\int \frac{dt}{\sqrt{(1-t^2)(1-k^2 t^2)}},$$ and thus that $K=K'$ iff the period lattice is square, which in turn happens iff the branch points $t=\pm 1$ and $t=\pm 1/k$ have fourfold symmetry. Now there are various ways to finish; for instance, compute that the fractional linear transformation that takes $-1/k$ to $-1$ to $+1$ to $+1/k$ is $$t \mapsto \frac{(k+1)t - k + 3}{(3k-1)t + k+1}$$ and this map cycles $+1/k$ back to $-1/k$ iff $(k+1)(k^2-6k+1)=0$, whose only root in $0<k<1$ is $k = 3 - 2\sqrt{2} = (\sqrt{2} - 1)^2$. [The algebraic conjugate $k = (\sqrt{2} + 1)^2$ also gives rise to an elliptic integral with a square period lattice, while $k=-1$ is spurious.]

[added later] For any choice of $k>1$ the branch points $t=\pm1$ and $t=\pm1/k$ go to half-lattice points $0, K, iK', K+iK'$ modulo the period lattice $\Lambda = {\bf Z} K + {\bf Z} iK'$ (in some order depending on which of these branch points is chosen for the base point). Any 1:1 map of ${\bf C}/\Lambda$ that takes the set of half-lattice points to itself descends to an automorphism of the projective line (a.k.a. the Riemann sphere) with coordinate $t$, and that automorphism permutes the four branch points. The involution $z \leftrightarrow -z$ of ${\bf C}/\Lambda$ acts as $(t,u) \leftrightarrow (t,-u)$, and thus acts trivially on the $t$-line. Translations by the half-lattice points which descend to double transpositions: $t \leftrightarrow -t$ (from translation by $K$) and $t \leftrightarrow \pm 1 / kt$ (from translation by $iK'$ and $K+iK'$). For generic $k$ these translations and their compositions with $z \leftrightarrow -z$ are the only choices, but when $K=iK'$ there's the new map $z \mapsto iz$. This map fixes $0$ and $K+iK'$ and switches the other two half-lattice points, so descends to an involution that fixes two of the branch points, such as $1$ and $-1/k$, and switches the other two. Composing this map with translation by $K$ yields a 4-cycle $0 \mapsto K \mapsto K+iK' \mapsto iK' \mapsto 0$, and that's the 4-cycle I used in my answer.

• Would you mind explaining in a bit more detail? What is the period lattice, how is it related to the branch points of the integrand, and what is their fourfold symmetry? Why is it enough to compute the fractional linear transformation that permutes those points to conclude that $K=K'$. Is that obvious? How can I notice that $2K$ and $2iK'$ are periods of the elliptic integral? – Kirill Oct 16 '13 at 14:33
• Thanks Noam, but could you please elaborate about the "fourfold symmetry"? – user1337 Oct 17 '13 at 12:54
• Sorry I wasn't able to come back to this sooner. Since the problem was an exercise in Ahlfors, that text must explain the connection between these elliptic integrals and the period lattice of the differential $dt \Bigl/ \sqrt{(1-t^2)(1-k^2t^2)} \Bigr. = dt/u$ on the Riemann surface of $u^2 = (1-t^2)(1-k^2t^2)$. The fractional linear transformation gives an explicit change of variable that takes $K'$ to $K$ for $k = (\sqrt{2}-1)^2$. About the $4$-cycle I'll try to explain a bit more in an edit. – Noam D. Elkies Oct 22 '13 at 20:31

Here's one way of doing this. Set $t=\sqrt{u}=1/\sqrt{v}$, and $l=k^2$, so that the integrals become $$\int_0^1 \frac{du}{\sqrt{u(1-u)(1-l u)}}, \qquad \frac12\int_l^1 \frac{dv}{\sqrt{v(1-v)(v-l)}},$$ so look for a change of variable that will bring the second integral to the form of the first.

The integrand has three singular points, $v=0,1,l$, and the limits are $l$ and $1$, so first look at a fractional-linear transformation $v=\frac{a+bx}{c+dx}$ that will map $1\mapsto1$ and $l\mapsto0$. This is because every automorphism of the complex plane has this form, so this is a natural transformation to look at when trying to map specific points to specific points. The integrand becomes $$\frac{(bc-ad)\,dx}{\sqrt{(a+bx)(c+dx)((c-a)+(d-b)x)((a-cl)+(b-dl)x)}},$$ and the conditions $x(v=1)=1$, $x(v=l)=0$ become $$\frac{a-c}{-b+d} = 1, \qquad \frac{a-cl}{-b+dl} = 0.$$ There are four parameters, two equations, and one parameter will be cancelled, so we can impose one more condition by saying that the degree of the polynomial in $x$ inside the square root should be three, rather than four: $$b=0.$$

Thus we consider the change of variable $$v = \frac{l}{1+(l-1)x},$$ which makes the second integral $$\frac12 \int_0^1 \frac{dx}{\sqrt{x(1-x)(1+(l-1)x)}}.$$

Expressing both integrals in terms of complete elliptic integrals of the first kind, we get the equation $$2K(k) = K(\sqrt{1-k^2}),$$ where $\sqrt{1-k^2}$ is commonly denoted by $k'$. The equation $$K(k')/K(k) = \sqrt{r}$$ is solved by the elliptic lambda function, so setting $r=4$ gives $$k = \lambda^*(4) = 3-2\sqrt{2}.$$

Another way to find this is to use a result of Abel (equation 1 there), and p.525 of Whittaker and Watson, to find that $$k=\tan^2\frac\pi8, \quad\Rightarrow\quad K(k') = 2K(k).$$ This follows from Landen's transformation: $$K(k_1')/K(k_1) = 2K(k')/K(k), \quad\text{when}\quad k_1 = \frac{1-k'}{1+k'},$$ and $K(k)=K(k')$ when $k=k'=1/\sqrt{2}$, so $k_1 = 3-2\sqrt{2}$, and $k_1'=\sqrt{1-k_1^2}$.

Landen's transformation, in turn, is based on the AGM representation of the complete elliptic integral of the first kind. Let $M(a,b)$ be the AGM function, so that for any $a$ we have $$K(k) = \frac{\pi a}{2 M(a,ak')}.$$ Then with $k_1 = \frac{1-k'}{1+k'}$, we get $$K(k_1) = \frac{1+k'}{2}K(k),$$ and also $$K(k) = \frac{\pi a}{2 M(a(1+k),a(1-k))},$$ $$K(k_1') = (1+k')K(k'),$$ so that $K(k_1')/K(k_1) = 2K(k')/K(k).$

• Thanks Kirill, I think I understand your proof. – user1337 Oct 17 '13 at 12:54