Timeline for The vanishing of Ramanujan's Function tau(n)
Current License: CC BY-SA 4.0
9 events
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Jan 30, 2022 at 11:12 | history | edited | YCor | CC BY-SA 4.0 |
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Jul 9, 2010 at 7:23 | vote | accept | Derek Jennings | ||
Jul 8, 2010 at 20:57 | comment | added | Boyarsky | @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 $p = 2, 3$. | |
Jul 8, 2010 at 20:01 | comment | added | Kevin Buzzard | @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! | |
Jul 8, 2010 at 18:24 | answer | added | Charles Matthews | timeline score: 5 | |
Jul 8, 2010 at 16:18 | answer | added | SandeepJ | timeline score: 10 | |
Jul 8, 2010 at 15:31 | answer | added | Bruce Westbury | timeline score: 3 | |
Jul 8, 2010 at 15:15 | comment | added | Boyarsky | 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? | |
Jul 8, 2010 at 15:01 | history | asked | Derek Jennings | CC BY-SA 2.5 |