# How many Hecke operators span the level 1 Hecke algebra?

Let $k \ge 4$ be an even integer, and let $d$ be the dimension of the space $M_k(\operatorname{SL}_2(\mathbb{Z}))$ of modular forms of level 1 and weight $k$. Then the space of Hecke operators acting on $M_k$ also has dimension $d$. Is it spanned by $T_1, \dots, T_d$?

Equivalently (more explicitly but also more messily): if $f \in M_k(\operatorname{SL}_2(\mathbb{Z}))$ satisfies $a_i(f) = 0$ for $1 \le f \le d$, where $a_i(f)$ are the $q$-expansion coefficients of $f$, with no assumption on $a_0(f)$, then is it necessarily true that $f = 0$?

(Edit: See also this follow-up question which asks a related question for modular forms of higher level.)

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There's an algorithm for checking this, right? You take the Eisenstein series $E_k$ and then subtract an appropriate multiple of $\Delta.E_{k-12}$ to make the $q$ term equal to zero, and then use $\Delta^2.E_{k-24}$ to make the $q^2$ term equal to zero, and so on. And then when you run out of Eisenstein series you look at the next coefficient and see if it's zero and if it is then you have lost. For any given $k$ this will be super-quick so you could check it yourself for $k\leq 1000$ or whatever. What happens if you do this? Probably no pattern emerges... – Kevin Buzzard Oct 19 '10 at 17:28
Oh wait I can do this trivially because I have all this implemented in pari-gp. It's OK for $k\leq 300$. If you normalise things so all Eisenstein series are $1+\ldots$ then the coefficient of $q^d$ always seems to be an integer (probably not hard to prove) but it isn't always 1, which means that for mod~$p$ modular forms the corresponding result is false, which means that any geometric argument e.g. of Sturm Bound type needs to take this into account. The sequence isn't in Sloane. – Kevin Buzzard Oct 19 '10 at 17:41
The primes for which it fails are a bit random too, e.g. for $k=68$ it's false mod $51599$. So I don't know the answer. The numbers seem to be growing though. – Kevin Buzzard Oct 19 '10 at 17:49
I checked it's true for $k \le 1000$, in Sage, using roughly the algorithm you describe -- this is automated in Sage (and probably also Magma) because it's how you construct the Victor Miller basis. – David Loeffler Oct 19 '10 at 18:02
Random note: you get integers because you can replace $E_k$ with a suitable product of $E_4$ and $E_6$. – S. Carnahan Oct 19 '10 at 18:05

The answer is yes when $k$ is a multiple of $4$. There is a unique form of weight $k$ of the form $f_k=1+a_dq^d+\cdots$. When $k$ is a multiple of $4$ this is the theta series for a putative extremal even unimodular lattice of rank $2k$. Theorem 20 in chapter 7 of Conway and Sloane's Sphere Packings, Lattices and Groups asserts that $a_d>0$. They give several references for the proof, including a 1969 paper of Siegel.

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@Robin: computations for $k\leq 300$ suggest that the sign of $a_d$ alternates. So for $k=2$ mod 4 this trick isn't going to work :-/ – Kevin Buzzard Oct 19 '10 at 17:51
Unless Siegel's method (alas I haven't read the paper) gives $a_d<0$ for $k\equiv2$ (mod 4). – Robin Chapman Oct 19 '10 at 17:59
True ! – Kevin Buzzard Oct 19 '10 at 18:02

Write $k = 12\ell + k'$, where $k'$ is one of $0, 4, 6, 8, 10, 14$, and let $f_{k, m}$ be the unique weakly holomorphic modular form (poles allowed at cusps) of weight $k$ for $SL_2(\mathbb{Z})$ with Fourier expansion $f_{k, m} = q^{-m} + \sum_{n \geq \ell+1} a_k(m, n) q^n$. The duality of coefficients $a_k(m, n) = -a_{2-k}(n, m)$ between forms of weight $k$ and forms of weight $2-k$ holds (see http://www.math.ucla.edu/~wdduke/preprints/serre.pdf ), so the original question is equivalent to asking whether it is true that the coefficient $a_{k}(0, \ell+1)$ is never zero. By duality, this coefficient is the negative of the constant term in $f_{2-k, \ell+1} = \frac{E_{14-k'}}{\Delta^{\ell+1}} = q^{-\ell-1} + \sum_{n=-\ell}^\infty a_{2-k}(\ell+1, n) q^n$. Siegel's 1969 paper referenced in Robin Chapman's answer proves that this constant term is always nonzero (Theorem 2), so the answer to the original question is yes in all cases. An English translation of Siegel's paper appears as an appendix in his Advanced Analytic Number Theory book, available online at http://www.math.tifr.res.in/~publ/ln/tifr23.pdf .

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Wooah! That's pretty wild! So can you also prove things like $\tau(n)\not=0$ for all $n\geq1$ using this black magic? Somehow I felt the original question had a very geometric feel to it but this answer is somehow combinatorial plus analytic. – Kevin Buzzard Oct 20 '10 at 6:34
Very nice argument! I'm also delighted to learn that Siegel's wonderful lectures can be found online. – François Brunault Oct 20 '10 at 6:55
$\Delta$ is the basis element $f_{12, -1}$ using the notation above, so by duality, Lehmer's conjecture is equivalent to the nonvanishing, for all positive $n$, of the coefficient of $q^{-1}$ in the weight $-10$ form $f_{-10, n} = q^{-n} + \tau(n) q^{-1} + \ldots$. Unfortunately, it doesn't seem to be easy to show that this is never zero. – Paul Jenkins Oct 20 '10 at 20:20

It is known that $S_k(\operatorname{SL}_2(\mathbb{Z}))$ has a basis $(f_1,\ldots,f_d)$ satisfying $a_i(f_j) = \delta_{i,j}$ (see for example William Stein's book "Modular forms : a computational approach", Section 2.3, it is called Miller's basis). Thus the Hecke algebra of $S_k(\operatorname{SL}_2(\mathbb{Z}))$ is generated by $T_1,\ldots,T_d$.

I think one should be able to get the analogous result for $M_k(\operatorname{SL}_2(\mathbb{Z}))$ by using the fact that the Fourier coefficients of the Eisenstein series are so big with respect to cusp forms.

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That's not so clear to me. Some sort of growth argument would surely show that some large Fourier coefficient doesn't vanish (which is clear anyway), but this question seems much more delicate, the problem being that $\sigma_{k-1}(n)$ may actually be greater than $\sigma_{k-1}(n+1)$. – Kevin Buzzard Oct 19 '10 at 17:57
You're right, my argument seems incomplete. Numerically the Fourier coefficient $a_{d+1}(f_j)$ of the Miller basis elements seems not so big (especially as $j$ increases, because $f_j$ is divisible by $\Delta^j$). But after some thought I don't see how to use this here... – François Brunault Oct 19 '10 at 18:39