Cubic Singular Values

An interesting test case for the application of integer relation methods to the task of finding minimal polynomials as in Section 2.3 arises in the setting of modular or theta functions, see [9], [23]. Consider

$$a:=a(q) = \sum_{m,n=-{\infty}}^{\infty} q^{m^2+mn+n^2},\qquad... ...q) = \sum_{m,n=-{\infty}}^{\infty}\omega^{n-m}q^{n^2 + mn+m^2},$$

where $\omega = e^{2\pi i/3}$ and

$$c:=c(q) = \sum_{m,n=-{\infty}}^{\infty} q^{(n+1/3)^2 + (n+1/3)(m+1/3) + (m+1/3)^2}.$$

These three functions lie on the Fermat curve a³=b³+c³and one has the lovely parameterization of a hypergeometric function:

$$_2F_1(\frac 13,\frac 23; 1; \frac{c^3} {a^3}) = a.$$

If we set $q:= e^{-2\pi\sqrt {N/3}}$ it follows from the theory in [9], [23] that, for N rational, $s_N:= \frac ca$ is expressible in terms of radicals, and a fortiori is an algebraic number. For a positive integer N, s_N is called the N-th cubic singular value--in analogy to the classical case of the complete elliptic integral of the first kind and quadratic singular values k_N ([7], p. 139). While with more work it is possible to be more explicit about the exact form of s_N, it is interesting to ask what may be determined entirely by our computational methods.

In terms of the classical theta functions

$$\theta_3(q):=\sum_{n=-\infty}^\infty q^{n^2} \qquad \theta_2(q):=\sum_{n=-\infty}^\infty q^{{(n+\frac 12)}^2}$$

one has the following highly lacunary representations for a, b and c

$$a(q)= \theta_3(q)\theta_3(q^3)+\theta_2(q)\theta_2(q^3)$$

$$b(q)=(3a(q^3)-a(q))/2 \qquad c(q)=(a(q^{1/3})-a(q))/2$$

which allows one to compute many hundreds or thousands of digits of s_Nalmost instantly.

This then provides a rather good test of the ability of a program to determine the algebraic value at hand. For example:

$$\begin{eqnarray*}s_1& =& 1/{\root{3} \of {2}}\\ s_2&=&0.52710011025237060710\cd... ...4& =& 0.26625264629019611364 \cdots = (1/2-5/18\sqrt{3})^{1/3}. \end{eqnarray*}$$

From the first few cases, and in analogy with the classical (quadratic) case¹, we were led to look instead at

$$G_N:=(1/2-s_N^3)^2 \qquad \mbox{ or } \qquad g_N:=\frac{3s_N}{{1-s_N^3}}$$ (5)

for $N \equiv \pm 1$ or $N \equiv 0$ mod 3 respectively. This has the effect of reducing the degree of the polynomial sought significantly and so made it practical--with refinements to PSLQ--to obtain the first 100 values of s_N. Thus, we obtain $s_{14}=(1/2-1/250\sqrt{14}-99/500\sqrt{6})^{1/3}$. In addition, suppose we compute g₁₄ to 100 places and then obtain the potential polynomial from, say, a 50 digit computation. If the putative root is then verified to near full precision we are almost certainly right.

To extract the radical representation of s_N for a given N, we compute P_N, the minimal polynomial for G_N or g_N (depending on $N\bmod 3$). Then we try factoring P_N over different quadratic number fields until we get a factor which is (essentially) of degree 4 or less, solve it in radicals and use a transformation inverse to (5) to get s_N in radical form. The least value of N for which this approach does not seem to work is N=53. Computing G₅₃ to 200 places and using 100 digits in Maple's `minpoly', we obtain

$$\begin{eqnarray*}P_{53}(x):=-1468655558161-9100986507260x +23965205326688x^2\\ +672242305831040x^3-3033916937098496x^4+3679773552653312x^5 \end{eqnarray*}$$

as a polynomial we hope satisfies P₅₃ (G₅₃)=0. We also verify that $P_{53}(G_{53})=-1.9\cdot 10^{-185}$. Extra comfort is provided by `galois(P₅₃(x))' which returns

+D5, 10, (2 5)(3 4), (1 2 3 4 5)

telling us that the polynomial indeed has a solvable Galois group of order 10. Now, Daniel Lazard has written Maple code which returns a radical for any solvable quintic. The procedure, which works well on tame examples, returns a radical with 7508 symbols. Kevin Hare at CECM has greatly simplified the radical expression to one with only 860 symbols which, as Maple can now happily verify symbolically, solves P₅₃ and which ultimately will become something attractive.

Most reassuringly, our evaluations of s_N can often be verified via the equation

 

J₂(k_3N)=J₃(s_N) (6)

which holds for any rational N. Here, k and s are as above, while

$$J_2(x)={4\over{27}} {{(1-x^2(1-x^2))^3}\over{x^4(1-x^2)^2}}$$

is Klein's absolute invariant ([7], p. 115) and

$$J_3(x)={1\over{64}} {{(9-8x^3)^3}\over{x^9(1-x^3)}}$$

its cubic counterpart².

The identity (6) follows from Proposition 5.8 in [7], p. 185 (equation (5.5.26)) and [8]. Using known values of k_3N (see [7], pp. 139, 162 and many other places) the corresponding s_N can be verified. Maple easily checks the identity (6) symbolically using its `radnormal' procedure for $N\le 10$; for larger N the complexity of intermediate results becomes a limiting factor and the computation requires human guidance. Still, after adequate effort we were able to verify our value of s₇₀symbolically, using the evaluation of k₂₁₀ (see [7], p. 141) that Hardy celebrated as ``one of the most striking of Ramanujan's results.'' (See [28], p. 229.)

It is interesting to note that the s_N values can be also verified by a mixture of symbolic and numerical computing. For example, using a Liouville type bound on the minimal polynomials for k₂₁₀and our minpoly-guessed value of s₇₀ (let us call it S), we can prove that either (i)  J₂(k₂₁₀)=J₃(S) or (ii)  |J₂(k₂₁₀)-J₃(S)|>10^-6400. It takes just 9 minutes of CPU time on an SGI workstation to prove numerically that (ii) cannot hold, again using Maple. Thus, G₇₀=

$${\frac {735540276593}{3996969003000}}+{\frac {3413048639}{222053833500 }}\,\sqrt {2}\sqrt {3}$$

$$+{\frac {357407303}{55513458375}}\,\sqrt {3} \sqrt {5}\sqrt {... ...rac {8992317139}{1998484501500}}\,\sqrt {2} \sqrt {5}\sqrt {7}.$$

We have found similar proofs and bounds for other larger values of N, such as 110 and 154.