Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

For the Dedekind eta function, defined as usual by $\eta(q) = q^{\frac1{24}} \prod\limits_{n=1}^{\infty} (1-q^{n})$, let for brevity $e_k:=\eta(q^k)$.

An eta product identity (or eta identity for short) is then defined as a homogenous polynomial in the $e_k$ with integer coefficients such that its Taylor series vanishes identically. There are already more than 6200 of them in Michael Somos' collection.

Asking bluntly like that how many eta identities exist, the answer is of course infinitely many, as they form a (graded) vector space.So we may try, as does Somos, to restrict to "irreducible" ones in an appropriate sense. Now it turns out that it is a bit tricky to come up with a good definition of irreducibility for an eta identity. Some requirements are obvious:

  • it should be irreducible as a polynomial, including that
  • not only the $\gcd$ of the coefficients (up to sign) should be $1$, but also the $\gcd$ of the eta exponents (i.e. the indices of the $e_k$'s) involved
  • it should not be expressible as the sum of two shorter eta identities
  • if there are several linear independant eta identities with given level, degree and rank, we should only count $k$ basis elements where $k$ is the dimension of the vector space generated by them, all others would be considered "reducible in terms of this basis".

A (minor) problem can occur if we add two identities whose monomials are not all disjoint in a way that some monomials cancel out. In particular, we can take any pair of 3-term identities, say $a+b+c$ and $d+e+f$, then $d(a+b+c)-a(d+e+f)=bd+cd-ae-af$ is a 4-term identity. Should we call it reducible? Some 4-term identities in Somos' collection can be obtained like that, but many can't (a priori). There are also pairs of 4-term identities of the form $a+b+c+d$ and $a+b+kc+e$ with $k\in\mathbb Q$, so their difference yields a 3-term identity, but as it is made up of two longer identities, there would be no reason anyway to call that one reducible.

So it seems best to stick to the four above criteria for an irreducible eta identity, and not to worry about some 4-term identities that would fall through.

There are still infinitely many irreducible eta identities, because e.g. all Schläfli type modular equations can be written in terms of eta products, so (irreducible) eta identities exist at least for all levels divisible by 4.

Now, many other types of modular equations, like e.g. theta functions that can be expressed as eta quotients, tend to come in finite quantities, and I think that we can impose feasible constraints on eta identities such that there are only finitely many fulfilling those constraints. So which constraints might that be? Some questions below are stronger than others.

  1. For a given level $N$, are there only finitely many irreducible eta identities? (Note that if length and degree are limited, that follows from the existence of the Sturm bound)

  2. (more or less equivalent to 1.) For a given level $N$, is there a maximal length and/or a maximal degree of an irreducible eta identity?
    It seems like for a given level, the identities of maximal degree tend to have more symmetries, more precisely, they are most often self-dual in the sense defined here, or sometimes, even stronger, each term is self-dual. (I am aware that the collection is not necessarily exhaustive even for rather small levels, so this may be a wrong impression).

  3. Are there only finitely many 3-term identities (all levels combined)?

  4. If so, then it makes also sense to ask: for a given length $k>3$, are there only finitely many irreducible $k$-term identities?

  5. Are there irreducible eta identities for each level $N$ that is not prime?

Concerning the last question above: it seems like for odd $N$'s, there are relatively few identities, and the bigger the minimal prime factor, the more intricate they are. E.g. for level $13^2$, according to Somos the "smallest example has 50 terms of degree 14 and rank 182", and for level $11^2$, the smallest example has 156 terms of degree 60 and rank 660. But it does exist!

On the other hand, if $N$ has only small prime factors, there are lots of identities. For $N=12$ alone, the collection lists over 1000, the vast majority of them with only four terms.

share|improve this question
add comment

1 Answer 1

up vote 3 down vote accepted

I feel like you had a little bit of trouble coming up with a good notion of "irreducible". A lot of the things you say suggest you want identities that "cannot be built from simpler ones." Here's a way of thinking about it that may be helpful.

Given a level $N$, consider the polynomial ring $R$ over $\mathbb{C}$ on $d(N)$ generators $u_{d}$. The map $\phi_{N} : R \to \oplus_{k} M_{k}(\Gamma_{0}(24N))$ that maps $u_{d}$ to $e_{24d}$ is a ring homomorphism. The kernel of $\phi$, is precisely the space of eta identities. It seems that you want to be asking for "generators" of this space.

The kernel of $\phi_{N}$ is a prime ideal of $R$, and since every ideal of $R$ is finitely generated (by the Hilbert basis theorem), there will be "finitely many irreducible eta identities of level $N$," corresponding to the finitely many generators of $\ker \phi_{N}$, and hence an affirmative answer to question 1 and hence 2. (Of course the notion of irreducible in this context means "it cannot be built from the other identities that we've picked" and so labelling a given eta identity as irreducible or not is somewhat difficult to do - in the same way that choosing a single "best" generating set of an ideal is hard to do.)

Question 3 is quite a bit more subtle. For a given level $N$, an eta identity can be divided by one of the terms to give rise to an identity of the form $$ a_{1} + a_{2} + \cdots + a_{r} = 1 $$ where the $a_{i}$ are eta quotients that are weight zero modular functions. These modular functions have no zeroes or poles except at the cusps of $X_{0}(N)$, and there will only be finitely many such, by appealing to results about $S$-unit equations in global function fields (see the paper by R.C. Mason, "Norm form equations I" in J. Number Theory (1986), for example). However, this only applies for a fixed level $N$ - if the level is allowed to vary, there may be many more options. The same thing applies for 4. Every elliptic curve is modular, so if there are infinitely many elliptic curves $y^{2} + ay = x^{3}$ where the modular parametrization has $x$ and $y$ given by eta quotients (and this is a big if), then the answer is yes.

The issue with question 5 is the requirement that each identity not come from a relation of lower level. Given a positive integer $N$, if we take $\ker \phi_{d}$ for each divisor $d$ of $N$ and multiply each $k$ value by $N/d$ we get a subset of $\ker \phi_{N}$. Is that subset the whole ideal? I suspect in many cases the answer is yes - the field of functions on $X_{0}(N)$ is indeed generated by $j(z)$ and $j(Nz)$, and so it is plausible that there might not always be an irreducible identity of level $N$.

In general, it seems that the image of $\phi$ is a ring of Krull dimension $2$ (one dimension coming from the weight, one dimension coming from the dimension of $X_{0}(N)$). This would imply that $\ker \phi$ is a prime ideal of height $d(N) - 2$ and is hence non-trivial unless $N = 1$ or $N$ is prime. In the case that $N = p^{2}$, there will be an irreducible eta identity, and moreover such an identity will be unique, because a height $1$ prime ideal in a polynomial ring is principal.

share|improve this answer
Thank you very much! For the last paragraph, saying "it seems that", do you think that this should be easy to prove? And in the penultimate paragraph, "the function of functions on $X_0(N)$ ", is that a typo? Do you mean "the ring of functions"? –  Wolfgang Apr 19 at 17:13
It was a typo, and I meant "the field of functions" - the answer has been edited. –  Jeremy Rouse Apr 19 at 17:49
For the "it seems that", it should be easy to prove when $N > 1$. Since the image of $\phi$ is finitely generated, the Noether normalization lemma implies that the Krull dimension is the same as the transcendence degree of the fraction field. We can build this fraction field in three pieces: adjoin some non-constant modular function, adjoin all modular functions, and then adjoin all elements in the fraction field. The first extension will have transcendence degree $1$, the second is algebraic, and the third will also have transcendence degree $1$. –  Jeremy Rouse Apr 19 at 17:56
Nice! So your very last statement means that for each prime $p$, there is a unique irreducible eta identity for $N=p^2$, giving rise to a lot of parameters (degree, rank, length) with a presumably somewhat irregular distribution: $$\pmatrix{p&|&\partial&rk& \#\ terms\cr 3&|&12&36&4\cr5&|&6&30&6\cr 7&|&8&56&11\cr 11&|&60&660&156\cr 13&|&14&182&50}$$ It seems like the degree is always a multiple of $p+1$ (the first factors being $3, 1,1,5,1)$, moreover $rk=p\cdot \partial$. Might that be of interest to some number theorists?! –  Wolfgang Apr 19 at 19:59
This certainly is interesting. This eta identity may give a model for $X_{0}(p^{2})$, which is a degree $p(p+1)$ cover of $X_{0}(1)$. –  Jeremy Rouse Apr 19 at 20:19
show 5 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.