If $n$ is a natural number, let $\displaystyle \sigma_{11}(n) = \sum_{d \mid n} d^{11}$. The modular form $\Delta$ is defined by $\displaystyle \Delta(q) = q \prod_{n=1}^{\infty}(1 - q^n)^{24}$. Write $\tau(n)$ for the coefficient of $q^{n}$ in $\Delta(q)$.

I would like to know why $\tau(n) \equiv \sigma_{11}(n) \pmod{691}$. I think the proof may be somewhat difficult, so even just an outline of the argument would be much appreciated.

Thank you!

  • 1
    $\begingroup$ For searching purposes: $\tau(n)$ is Ramanujan's $\tau$ function. $\endgroup$ – J. M. isn't a mathematician Dec 28 '10 at 5:18
  • 2
    $\begingroup$ In any event, it is intimately related to the properties of Klein's invariant; see this (especially equations 18 and 19), this book by Apostol, and this article by D.H. Lehmer. $\endgroup$ – J. M. isn't a mathematician Dec 28 '10 at 5:33
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    $\begingroup$ This was first proved by Ramanujan, and a proof is available in his 1916 paper "On Some Arithmetical Functions" which is available for free in Google Books, "Collected Papers of Srinivasa Ramanujan". $\endgroup$ – Alon Amit Dec 28 '10 at 5:34

The proof is not at all obvious if you begin simply with the formula $$\Delta(q) = q \prod_{n=1}^{\infty} (1-q^n)^{24}.$$ However, as Derek Jennings explains in his answer, if you use the (absolutely crucial!) fact that $\Delta$ is a cusp form of weight twelve and level one, the proof is actually not very difficult.

As Derek explains, the ring of modular forms of level one is generated by two $q$-expansions, namely $$E_4 := 1 + 240 \sum_{n = 1}^{\infty} \sigma_3(n) q^n,$$ which has weight 4, and $$E_6 := 1 + 504 \sum_{n=1}^{\infty} \sigma_5(n) q^n,$$ which has weight 6. (The coefficients $240$ and $-504$ come from Bernoulli numbers, as Derek explains, but we don't need that at the moment.)

Now we see that we can make two monomials of weight 12 from these, namely $E_4^3$ and $E_6^2$. How do we get $\Delta$? Well, the constant term of $\Delta$ vanishes, while $E_4^3$ and $E_6^2$ have constant term $1$, so $\Delta$ must be proportional to $E_4^3 - E_6^2$. Since the coefficient of $q$ in $\Delta$ is $1$ (i.e. $\tau(1) = 1$) while the coefficient of $q$ in $E_4^3 - E_6^2$ is $1728,$ we find that $$E_4^3 - E_6^2 = 1728 \Delta.$$

It is useful to note that we can also use $E_4^3$ (say) and $\Delta$ as a basis for the weight $12$ modular forms. In fact, they are a very convenient basis, because if $f$ is any weight $12$ modular form, with constant term $a_0$, then we can subtract of $a_0 E_4^3$ to get rid of the constant term of $f$, and then $f - a_0 E_4^3$ must be a mulitple of $\Delta$.

To go further, we have to introduce another fact, also noted by Derek Jennings, namely that there is a weight 12 modular form $$E_{12} = 1 + \dfrac{65520}{691} \sum_{n=1}^{\infty} \sigma_{11}(n) q^n.$$ In fact, it is (for me) easier to work with $$691 E_{12} = 691 + 65520\sum_{n=1}^{\infty} \sigma_{11}(n) q^n,$$ which has integer coefficients.

Now we apply the above procedure to write $691 E_{12}$ in terms of $E_4^3$ and $\Delta$, to find that $$691 E_{12} = 691 E_4^3 + (65520 - 691\cdot 720) \Delta.$$ Now all the $q$-expansions in this formula have integral coefficients, and so what we find, looking at the coefficient of $q^n$, is that $65520\sigma_{11}(n) \equiv 65520\tau(n)$ for each $n \geq 1$.
Dividing by $65520$ (which is coprime to $691$) gives the desired formula.

The usual way this is summarized is to say that $$\dfrac{691}{65520} E_{12} = \dfrac{691}{65520} + \sum_{n=1}^{\infty} \sigma_{11}(n) q^n$$ is normalized (i.e. has coefficient of $q$ equal to $1$) and is a cuspform modulo $691$ (i.e. its constant term vanishes mod $691$). This forces it (as we have just seen) to be congruent to $\Delta$ modulo $691$.

Finally, let me note that by far the best place to read about the theory of modular forms used here is Serre's beautiful book A course in arithmetic.

Added: Since I was editing this anyway, let me add a cultural remark, namely that the study of congruences between Eisenstein series and cuspforms, of which the congruence between $\dfrac{691}{65520} E_{12}$ and $\Delta$ considered above is the first example, is a central topic in modern number theory. It lies at the heart of Mazur's determination of the possible torsion subgroups of elliptic curves over $\mathbb Q$, and is the basic method via which Ribet proved his "converse to Herbrand's theorem'' result giving a criterion for non-triviality of various $p$-power subgroup of the class group of the cyclotomic field $\mathbb Q(\zeta_p)$.

  • 2
    $\begingroup$ +1 What a beautiful answer. Just a small comment. You have a couple of typos, when you introduce $E_{12}$ the first two times you should write $\sigma_{11}(n)$ instead of $\sigma_{1} 1(n)$ and in the paragraph where you mention that all the $q$-expansions in the formula have integral coefficients, you're missing a $ symbol so the formulas don't show up well. $\endgroup$ – Adrián Barquero Dec 28 '10 at 18:33
  • $\begingroup$ @Adrian: Dear Adrian, Thanks for the compliment, and for pointing out the typos. Best wishes, $\endgroup$ – Matt E Dec 28 '10 at 19:55
  • $\begingroup$ @Mariano: Dear Mariano, Thanks for helping to clean up my typos! Best wishes, $\endgroup$ – Matt E Dec 28 '10 at 19:55

The modular discriminant $\Delta (z) \in M_{12,0},$ the linear space of cusp forms of weight $12.$ Now dimension $M_{12,0} = 1$ (see, for example, T.M. Apostol: Moldular Functions and Dirichlet Series in Number Theory)

Every modular form for the full modular group is a polynomial in $E_4$ and $E_6,$ where $E_{2k}$ are the Eisenstein series defined by

$$E_{2k}(z) = \frac{1}{2\zeta(2k)} \sum_{(m,n) \ne (0,0) } \frac{1}{(m+nz)^{2k}} \quad \textrm{for } Im(z) >0,$$

where the summation extends over all integral $m$ and $n$ not both equal to $0$. So if we use the Eisenstein series to construct cusp forms of weight $12,$ because $\textrm{dim } M_{12,0} = 1$ these will necessarily be constant multiples of $\Delta (z).$

Now we can show that

$$E_{2k}(z) = 1 - \frac{4k}{B_{2k}} \sum_{n=1}^\infty \sigma_{2k-1}(n)e^{2 \pi i n z} \quad \textrm{for } Im(z) > 0,$$

where $\sigma_k(n) = \sum_{d|n} d^k$ and the $B_{2k}$ are the Bernoulli numbers.

Therefore we can expect identities between the divisors functions $\sigma_{2k-1}(n)$ and $\tau(n)$ to exist. Ramanujan's paper "On Certain Arithmetical Functions" shows how to express

$$\phi_{r,s}(x) = \sum_{n=1}^\infty n^r \phi_{s-r}(n)x^n$$

for integer $r,s \ge 0$ and $ |x| < 1$ as polynomials in $E_2,E_4,$ and $E_6,$ from which many such identities follow.

EDIT: To cite a random example:

$$\tau(n) = n^2\sigma_7(n) - 540\sum_{k=1}^{n-1}k(n-k)\sigma_3(k)\sigma_3(n-k)$$

and so $\tau(n) \equiv n^2\sigma_7(n) \textrm{ mod } 540.$

EDIT2: $B_{12} = -691/2730.$

EDIT3: For completeness, using Ramanujan's notation

$$P = 1 - 24 \sum_{n=1}^\infty \frac{nx^n}{1-x^n}$$ $$Q = 1 + 240 \sum_{n=1}^\infty \frac{n^3x^n}{1-x^n}$$ $$R = 1 - 504 \sum_{n=1}^\infty \frac{n^5x^n}{1-x^n},$$

the relevant entries from his paper "On Certain Arithmetical Functions" are $(3)$ and $(6)$ from his Table I, given below. (Note that $Q$ is $E_4$ and $R$ is $E_6.$)

$$\begin{align} 1 - 24\phi_{0,1}(x) &= P \\ 1 + 240\phi_{0,3}(x) &= Q \\ 1 - 504\phi_{0,5}(x) &= R \quad (3) \\ 1 + 480\phi_{0,7}(x) &= Q^2 \\ 1 - 264\phi_{0,9}(x) &= QR \\ 691 + 65520\phi_{0,11}(x) &= 441Q^3+250R^2 \quad (6) \\ 1 - 24\phi_{0,13}(x) &= Q^2R \\ 3617 + 16320\phi_{0,15}(x) &= 1617Q^4+2000QR^2 \\ 43867 - 28728\phi_{0,17}(x) &= 38367Q^3R+5500R^3 \\ 174611 + 13200\phi_{0,19}(x) &= 53361Q^5+121250Q^2R^2 \\ 77683 - 552\phi_{0,21}(x) &= 57183Q^4R+20500QR^3 \\ 236364091 + 131040\phi_{0,23}(x) &= 49679091Q^6+176400000Q^3R^2 + 10285000R^4 \\ 657931 - 24\phi_{0,25}(x) &= 392931Q^5R+265000Q^2R^3 \\ 3392780147 + 6960\phi_{0,27}(x) &= 489693897Q^7+2507636250Q^4R^2 + 395450000QR^4 \\ 1723168255201 - 171864\phi_{0,29}(x) &= 815806500201Q^6R+88134070500Q^3R^3 + 26021050000R^5 \\ 7709321041217 + 32640\phi_{0,31}(x) &= 764412173217Q^8+5323905468000Q^5R^2 \\ &+ 1621003400000Q^2R^4 \end{align} $$

Equations $(3)$ and $(6)$, along with $1728 \sum_{n=1}^\infty \tau(n)x^n = Q^3 - R^2,$ give

$$691+65520 \sum_{n=1}^\infty \sigma_{11}(n) x^n = 441 \times 1728 \sum_{n=1}^\infty \tau(n) x^n + 691R^2.$$

The result follows by taking this equation modulo $691$ and noting that $566$ and $691$ are coprime.

  • $\begingroup$ Even though I happened to have Ramanujan's table already typed up in Tex, I no longer have a copy of his paper, "On Certain Arithmetical Functions," and cannot find it anywhere on the net. If anyone can point me to a copy, or can email me one I would be very grateful. My email is name(at)ntlworld.com where "name" is my name (no separating dot). Thanks. $\endgroup$ – Derek Jennings Dec 30 '10 at 14:57
  • $\begingroup$ See this: books.google.com/… $\endgroup$ – PrimeNumber Dec 30 '10 at 15:07
  • $\begingroup$ @Trevor: Thanks for that! $\endgroup$ – Derek Jennings Dec 30 '10 at 15:54

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