MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $f$ and $g$ are two newforms of certain levels, weights etc. If we know that L(f,n)=L(g,n) for all sufficiently large $n$, can we conclude that $f=g$?

Same question when the forms have the same weight and $n$ runs over critical points.

share|cite|improve this question
up vote 5 down vote accepted

The answer to the first question is "yes". The standard proof of the uniqueness of a Dirichlet series expansion actually generalizes to show the following.

Theorem. Suppose that $A(s) = \sum_n a_n n^{-s}$ and $B(s) = \sum_n b_n n^{-s}$ are Dirichlet series with coefficients $a_n, b_n$ bounded by a polynomial. If there exists a sequence of complex numbers $s_k$ with real part approaching infinity such that $A(s_k) = B(s_k)$ for all $k$, then $a_n = b_n$ for all $n$.

Proof (sketch). Proceed by induction. For $k$ big we have $A(s_k) = a_1 + O(2^{-\sigma_k})$ where $\sigma_k$ is the real part of $s_k$. Similarly, $B(s_k) = b_1 + O(2^{-\sigma_k})$. Since $A(s_k) = B(s_k)$, we conclude that $a_1 = b_1$. A similar argument shows $a_2 = b_2$, $a_3 = b_3$, etc.

share|cite|improve this answer
This is an easier proof, so I am going to select this one as the right answer. – Idoneal Dec 31 '09 at 4:34

I think the answer to your first question is "yes." Suppose $L(f,s) = \sum_{m} a(m)m^{-s}$ and $L(g,s) = \sum_{m} b(m) m^{-s}$, and that $L(f,n) = L(g,n)$ for $n \geq n_0$, with $n_0$ large enough that the sums converge absolutely. Then pick an integer $M \geq n_0$ and weights $C_M(n)$ so that $\sum_{n \geq M} C_M(n) m^{-n}$ is $1$ if $m=M$, and $0$ otherwise. One can surely come up with such weights without too much trouble. Then $a(M) = \sum_{n \geq M} C_M(n) L(f,n) = \sum_{n \geq M} C_M(n) L(g,n) = b(M)$. It's not too hard to see that if two modular forms eventually have the same Fourier coefficients, then they are the same.

edit: After some further thought, I'm having trouble justifying the existence of those weights. I found a different solution that I'm posting as a separate answer.

share|cite|improve this answer
Neat! In fact, this proves much more. – Idoneal Dec 17 '09 at 4:53

I think the answer to your second question is "no". For example if $k=2$ and $f$ and $g$ correspond to elliptic curves over $Q$ with positive rank, then the only critical point is $s=1$ and (at least conjecturally, and in sufficiently many cases provably) both $L$-functions will vanish at this point.

share|cite|improve this answer
Yes, this is true. However, if L(E1)=0 then transcendence results of elliptic functions imply that the value determines E up to isogeny. – Idoneal Dec 17 '09 at 4:56
I am guessing you mean L(E,1)!=0. – Kevin Buzzard Dec 19 '09 at 11:59
Yes, indeed i was. – Idoneal Dec 31 '09 at 4:32
@idoneal: so, is it then conjectured that the above question, replaced with "first non-zero derivatives at critical points", is true?? If the rank of the elliptic curve is 1 then, if I understand correctly, it is also known, and Birch-Swinnerton-Dyer implies it for all abelian varieties? – Dror Speiser Oct 23 '11 at 23:18
@Dror: my understanding of the system is that adding a comment to a two-year-old question doesn't bounce it to the front (it's only adding a new answer that will work). Hence if I'm right we can be pretty confident that only you (who wrote the comment) and I (who was notified about it because it's a comment to my answer) are currently reading this. I'd ask a new question if you have something more to ask. Note: of course my understanding of the system could be wrong! – Kevin Buzzard Oct 25 '11 at 8:10

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.