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.
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3 Answers
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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.
answered Dec 30, 2009 at 16:30
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$\begingroup$ This is an easier proof, so I am going to select this one as the right answer. $\endgroup$– IdonealCommented Dec 31, 2009 at 4:34
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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.
answered Dec 16, 2009 at 16:47
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$\begingroup$ Neat! In fact, this proves much more. $\endgroup$– IdonealCommented Dec 17, 2009 at 4:53
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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.
answered Dec 16, 2009 at 18:25
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$\begingroup$ Yes, this is true. However, if L(E1)=0 then transcendence results of elliptic functions imply that the value determines E up to isogeny. $\endgroup$– IdonealCommented Dec 17, 2009 at 4:56
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$\begingroup$ I am guessing you mean L(E,1)!=0. $\endgroup$ Commented Dec 19, 2009 at 11:59
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$\begingroup$ @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? $\endgroup$ Commented Oct 23, 2011 at 23:18
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$\begingroup$ @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! $\endgroup$ Commented Oct 25, 2011 at 8:10