16
$\begingroup$

This is a problem I had a look at some years ago but always had the feeling that I was missing something behind its motivation.

D.H. Lehmer says in his 1947 paper, “The Vanishing of Ramanujan's Function τ(n),” that it is natural to ask whether τ(n)=0 for any n>0.

My question is: Why is it natural to wonder whether τ(n)=0 any n>0?

Are there any particular arithmetic properties among the many satisfied by τ(n) that would lead one to ponder its vanishing? The problem is mentioned here, where it's stated that it was a conjecture of Lehmer, although it's not actually presented as a conjecture in his paper, more a curiosity.

Maybe there is no deep reason to ponder the vanishing of τ(n), in which case that would be a satisfactory answer too.

$\endgroup$
3
  • 2
    $\begingroup$ From the viewpoint of Hecke eigenforms, the vanishing of $\tau(p)$ for prime $p$ seems more interesting than for general $n$; consider the analogy with elliptic curves over $\mathbb{Q}$ (for which $a_p = 0$ encodes supersingularity, except maybe for some issue when $p = 2, 3$). It ties in with the whole story of slopes of modular forms. But on a more concrete/classical level, doesn't $\tau(n)$ arise as the "error term" in one of those Ramanujan formulas for counting something related to a quadratic form, so vanishing means "no error" for that $n$. Perhaps that was Lehmer's motivation? $\endgroup$
    – Boyarsky
    Jul 8, 2010 at 15:15
  • 1
    $\begingroup$ @Boyarsky: If there exist any $n\geq1$ for which $\tau(n)$ vanishes, then the smallest such $n$ will be prime. This is not immediately obvious but is proved in Lehmer's paper IIRC. @DerekJ: because of this, one could look at Lehmer's question in the following way. CM elliptic curves have $a_p=0$ for 50% of primes. Non-CM elliptic curves have much sparser, but still infinitely many, $p$ with $a_p=0$. But by the time you get to weight 12 the $\Delta$ function is a candidate for a modular form with $a_p=0$ never happening at all! $\endgroup$ Jul 8, 2010 at 20:01
  • $\begingroup$ @Kevin: Yes, that's a good point about the least $n$ (if any) for which $\tau(n) = 0$. In view of the possible failure of $a_p = 0$ to hold when $p = 2, 3$ is a supersingular prime for an elliptic curve over $\mathbb{Q}$, it's nice that Lehmer's argument adapted to apply in general to eigenforms of any weight (using basics about quadratic fields generated by roots of unity in place of his trigonometric language) always requires a special calculation (which may fail...) for $p = 2, 3$. $\endgroup$
    – Boyarsky
    Jul 8, 2010 at 20:57

3 Answers 3

10
$\begingroup$

The key to your question is lacunarity in modular functions.

The tau function, as we know, occurs as the coefficient of the Discriminant function, which in turn is the 24th power of the Eta function. The Eta function was known to be lacunary (having gaps or zero coefficients). Therefore it was natural for Lehmer in 1947 to wonder if coefficients of powers of eta are also zero. See the opening passage of the following paper

MR0021027 (9,12b) Lehmer, D. H. The vanishing of Ramanujan's function $\tau(n)$. Duke Math. J. 14, (1947). 429--433. http://projecteuclid.org/euclid.dmj/1077474140

$\endgroup$
0
5
$\begingroup$

A simple reason: this is a function of $n$ satisfying significant congruences. If it vanishes, that is further congruence information.

$\endgroup$
1
  • 1
    $\begingroup$ I'll remark that Charles' observation can be used to yield fairly big lower bounds on the smallest $n$ for which $\tau(n)$ can possibly be zero. Known congruences for $\tau(n)$ modulo small powers of small primes imply that $n$ has to satisfy certain congruences if you want it to vanish. Serre's paper on lacunarity of powers of the eta function gives an explicit lower bound which is $O(10^{10})$ or so (I don't have his paper to hand). $\endgroup$ Jul 9, 2010 at 9:24
3
$\begingroup$

I don't know if this helps but you can put $D=24$ in (13) (14) of my paper to get an explicit formula for $\tau(n)$.

MR2218820 (2007c:17009) Westbury, Bruce W. Universal characters from the Macdonald identities. Adv. Math. 202 (2006), no. 1, 50--63.

doi:10.1016/j.aim.2005.03.013

Since $SL(5)$ is a simple Lie algebra of dimension 24 this also relates $\tau(n)$ to the affine root system of type $A_4$.

I doubt Lehmer would have had this in mind.

Addendum I started this project with the following problem. Let $\mathfrak{g}$ be a simple Lie algebra whose dimension is $D$. Normalise the Casimir so it acts as 1 on $\mathfrak{g}$. Now consider the subspace of the exterior power $\wedge^k \mathfrak{g}$ on which the Casimir acts by $k$. This is a representation of $\mathfrak{g}$ but obviously does not make sense for $k>D$. Taking $\mathfrak{g}=\mathfrak{sl}(5)$ we have $\tau(k)$ is the dimension of a representation for small $k$ (certainly no more than 24). I doubt this is interesting.

The conclusion of the project was that for all $k$ there is a complex of representations of $\mathfrak{g}$. Then the Euler characteristic is a virtual representation. This can be written as a sum (with signs) of representations of $\mathfrak{g}$ using the MacDonald identities for affine $\mathfrak{g}$. This gives $\tau(k)$ as the dimension of a virtual representation of $\mathfrak{sl}(5)$ for all $k$.

Because of the signs this does not give an immediate solution to Lehmer's question. However it is a different way of looking at the problem.

I also give an explicit formula for $\tau(k)$ in terms of partitions and hooklengths. I believe this is new.

$\endgroup$
1
  • $\begingroup$ Bruce, but what means "for each sufficiently large $N$" in your paper? Is $N=5$ sufficiently large? You interpret $\tau(n)$ as dimensions of certain representations. How does the sign of $\tau(n)$ appear there? What is wrong in having $\tau(n)=0$? I am wondering about the last question, since I am wondering whether your approach can be used in proving some partial results towards Lehmer's question (mathoverflow.net/questions/32620). $\endgroup$ Jul 22, 2010 at 0:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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