# Why does the definition of modularity demand weight 2?

Allow me to quote a definition from Gelbart in "Modular Forms and Fermat's Last Theorem":

Definition. Let $E/\mathbb{Q}$ be an elliptic curve. We say that $E$ is modular if there is some normalised eigenform

$$f(z) = \sum_{i=1}^{\infty} \ a_ne^{2\pi inz} \in S_2(\Gamma_0(N),\epsilon),$$

for some level $N$ and Nebentypus $\epsilon$, such that

$$a_q = q + 1 - \\#(E(\mathbb{F}_q))$$

for almost all primes $q$.

This is the basic question of the post:

Why is the weight of $f$ taken to be 2? Can I instead take 3, or 4, or 5, or even 19/2, without disturbing the peace?

I am aware of other definitions of modularity, some of which don't mention modular forms at all, but nonetheless I feel that weight 2 lurks beneath all of these.

I think one approach would involve differentials, and the construction of Eichler-Shimura, but I'm not so sure. Further, perhaps there are several reasons which fit together to tell a nice story.

Is it a corollary of this question that it doesn't matter what the weight is?

Finally, can I replace $E$ above with any abelian variety, and ask the same question?

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I'm sure someone will answer more fully, but in the meantime: it is explain very pleasantly and with the insight you require in Silverman's "Advanced Topics in the Arithmetic of Elliptic Curves". –  James D. Taylor Aug 14 '11 at 21:42
@James: Maybe you are referring to Theorem 11.4 in Ch 1; which says that the weight must be two to ensure the L-function of E satisfies the right functional equation, i.e, s goes to 2-s? –  Barinder Banwait Aug 14 '11 at 21:55
@Barinder: My response below should clarify the role of the weight. –  GH from MO Aug 16 '11 at 10:36
@GH: note that "the gamma factors determine the weight" is also in Kevin's comments. –  Dror Speiser Aug 16 '11 at 10:55
@Dror: Yes, but he talks about $L$-function associated to a Galois representation, while I talk about the $L$-function associated to a cusp form. I am emphasizing that if you use the correct "God given" normalization (actually Riemann's), i.e. the one in which $s$ is related to $1-s$, you can read off the weight from the gamma factor. If you use your favorite normalization depending on the weight (e.g. the one in which $s$ is related to $k-s$), then it is no surprise that you can read it off from the gamma factors. –  GH from MO Aug 16 '11 at 19:59


Q. What do the Galois representations attached to a variety know about the variety?

In order make this more precise, let us introduce some notation. Fix a prime $p$ and a non-negative integer $n$. Let $X$ be a proper smooth scheme over $\Q$, and let $V = H^n_{et}(X/\Qbar,\Q_p)$ denote the $n$th etale cohomology group of $X$. The basic and fundamental properties of etale cohomology tell us that:

1. $V$ is a vector space of dimension $H^n_{\Betti}(X(\C))$, where $H_{\Betti}$ denotes Betti (or singular) cohomology, and $X(\C)$ denotes the complex points of $X$ thought of as a topological manifold.

2. $V$ (with the $p$-adic topology) has a continuous action of $G_{\Q}:=\Gal(\Qbar/\Q)$.

Grothendieck and Serre further conjecture that the $G_{\Q}$-representation $V$ is semi-simple. The strongest possible conjecture one might make is to ask whether the functor from smooth projective varieties over $\Q$ to semi-simple $G_{\Q}$-representations (or the collection of all such representations for $n \le 2 \cdot \mathrm{dim}(X)$) is fully faithful. However, this is too much to ask, for the following reasons.

(i). The target category is semi-simple, but the category of varieties is far from semi-simple. (In particular, the existence of a map $X \rightarrow Y$ does not imply the existence of a non-trivial map $Y \rightarrow X$.)

(ii). Varieties built in a combinatorial way from projective spaces (think toric varieties) tend to have etale cohomology groups indistinguishable from products of projective spaces. This is because their cohomology groups are generated by geometric cycles, on which Galois acts in a well understood way (essentially by some power of the cyclotomic character).

These are - in some sense - manifestations of the same reason: A correspondence in $X \times Y$ gives rise to a cohomology class in $H^*(X \times Y)$; then by the Künneth formula, this leads to a relation between the cohomology of $X$ and $Y$ even when there is not necessarily any non-trivial map from $X$ to $Y$ (or vice versa). In order to account for this, one can try to take the quotient category of the category of algebraic varieties in which one is allowed to "break up" smooth proper varieties into pieces given the existence of certain correspondences on $X$. There are a variety of ways in which one might do this. Conjecturally, these constructions are all essentially the same, and the corresponding category is the category of pure motives. The Tate conjecture now says that etale cohomology is a fully faithful functor from pure motives to semi-simple $G_{\Q}$-representations.

Example If $E$ is an elliptic curve over $\Q$, and $n = 1$, then the etale cohomology group $V$ is the (dual) of the usual representation attached to the $p$-adic Tate module of $E$. Suppose that $E'$ is another elliptic curve over $\Q$ with first etale cohomology group $V'$. For curves, the theory of "motives" is essentially the theory of abelian varieties. (More generally, the theory of $H^1$ is essentially the theory of abelian varieties, since, for any proper variety $X$, there is an isomorphism $H^1(X) \simeq H^1(A(X))$, where $A(X)$ is the Albanese of $X$.) Tate's conjecture in this case says that $$\Hom(E,E') \otimes \Q_p \rightarrow \Hom_{G_{\Q}}(V,V')$$ is an isomorphism. This is how you will see the Tate conjecture stated for elliptic curves, for example, in AOEC. The Tate conjecture for abelian varieties is a theorem of Faltings. (Suggestion: to understand what the Tate conjecture really is about, and why it is hard, you should really think about the special case of Elliptic curves.)

If we now return to our question, we can (tautologically) say the following: assuming the Tate conjecture, the etale cohomology knows about the motive corresponding to the original variety. What does that really mean? One way of thinking about motives is as a universal cohomology theory''. In particular, we can recover from the motive not only the etale cohomology groups, but also the algebraic de Rham cohomology groups. Recall that de Rham cohomology is another cohomology theory that gives vector spaces of the "correct" dimension for a smooth proper variety $X/\Q$. The de Rham cohomology groups do not have associated Galois representations, but they do have a Hodge filtration. Over $\C$, if one takes the associated graded of the Hodge filtration, one recovers the Hodge decomposition: $$H^n_{dR}(X,\C) = \bigoplus_{p+q=n} H^{p}(\Omega^q_X).$$ The dimensions of the latter space are called the Hodge numbers $h^{pq}$. So, assuming the Tate conjecture, from $V$ we can recover the underlying motive, from which we may reconstruct the de Rham cohomology, and then the Hodge numbers. The Tate conjecture seems to be very hard. However, Grothendieck asked the following: given $V$, can we directly recover the (algebraic $p$-adic) de Rham cohomology along with its filtration without first constructing the motive? This was a great question, and the answer (yes!) constitutes one of the major achievements of $p$-adic Hodge Theory. I can do no more than give a cartoon description here. In order to do so, first recall the much more classical story connecting de Rham cohomology to Betti (singular) cohomology. These groups can both naturally be defined as vector spaces over $\Q$ (one has to define de Rham cohomology in the correct way), but the isomorphism relating these spaces comes from integrating forms over cycles. Yet these integrals are typically transcendental numbers, so to pass from Betti to de Rham cohomology one first has to tensor with a field bigger than $\Q$ which contains all these periods (usually, one simply tensors with $\C$). In order to pass from etale cohomology to algebraic de Rham cohomology, one might ask for a period ring in which we can compare both groups. In this refined setting, the period ring should both have a Galois action and a filtration. The most basic verion of a period ring is $B_{HT}$, specifically, $$B_{HT} := \bigoplus_{\Z} \C_p(n),$$ where $\C_p$ is the completion of $\Qbar_p$, and $\C_p(n)$ is $\C_p$ twisted (as a local Galois module) by the $n$th power of the cyclotomic character. The ring $B_{HT}$ has a natural filtration (indeed, it is even graded). Now we can consider $$D_{HT}(V) = (V \otimes B_{HT})^{\Gal(\Qbar_p/\Q_p)}.$$ The Galois group acts on both $V$ and $B_{HT}$. The result is a graded (and so filtered) module. On the other hand, one can also consider the ring $B_{HT} \otimes H^n_{dR}(X/\Q_p)$, where there is a natural way to make sense of the corresponding filtration. An important theorem of Faltings then says that $$H^{n}(X/\Qbar_p,\Q_p) \otimes B_{HT} = H^n_{dR}(X/\Q_p) \otimes B_{HT},$$ and $D_{HT}(V) = H^n_{dR}(X/\Q_p)$. In particular, from a geometric Galois representation, we can recover the Hodge filtration and the Hodge numbers.

Modular Forms. The Eichler-Shimura isomophism relates modular forms of weight $k \ge 2$ to $H^1(X_0(N),\Sym^{k-2}\Q)$. If $k = 2$, this is just $H^1(X_0(N),\Q)$. The Hecke algebra $\T$ acts on $H^1_{\Betti}(X_0(N),\Q)$, and (since it is constructed functorially) also on the etale cohomology $H^1(X_0(N),\Q_p)$. Now the Hodge decomposition of $H^1$ is $H^1 = H^{0,1} \oplus H^{1,0}$, where $h^{0,1} = h^{1,0}$ is the genus of $X_0(N)$. The Hecke algebra breaks up the cohomology into two dimensional pieces corresponding to the Galois representations associated to eigenforms; it turns out that each two dimensional piece contains one dimension from $H^{0,1}$ and one dimension from $H^{1,0}$. The result of Faltings above tells us that we can read off that $h^{0,1} = h^{1,0} = 1$ directly from the Galois representation.

For $k > 2$, recall that (technical issues aside) there is a universal elliptic curve $\E \rightarrow X_0(N)$. The Kuga-Sato variety is (again, roughly) The $k-1$ dimensional variety $K = \E \times_X \E \ldots \times_X \E$ where $X = X_0(N)$. There is a natural map $\pi: K \rightarrow X$. The local system $\Sym^{k-2}(\Q^2_p)$ is trivialized over $K$, and so, using the proper base change theorem, Deligne shows that $H^1(X_0(N),\Sym^{k-2}\Q_p)$ is a sub-quotient of the cohomology group $H^{k-1}(K,\Q_p)$. (Warning: this requires more than simply a formal cohomological argument, it also requires some trickiness with weights to show that terms on different diagonals the Leray spectral sequence don't "mix", and hence the sequence degenerates.) The Galois representation associated to a modular form is now a two-dimensional piece of $H^{k-1}(K,\Q_p)$. Faltings proves that the corresponding "piece" of de Rham cohomology seen by this representation is $H^{0,k-1} \oplus H^{k-1,0}$. In particular, the representation has Hodge numbers $h^{0,k-1} = 1$ and $h^{k-1,0} = 1$.

Given a Galois representation $V$, one can twist $V$ by the cyclotomic character. How does this effect the Hodge decomposition? One can compute this on the Hodge side by seeing what happens to the cohomology of $X \times \mathbf{G}^1_m$ and comparing with the Künneth formula. It turns out that $h^{p,q}(V(n)) = h^{p-n,q-n}$. Thus, if only know $V$ up to twist, we still recover some information about the Hodge numbers.

Returning to modular forms. The coefficients $a_p$ determine the Galois representation, by Cebotarev. A modular form of weight $k$ has Hodge numbers $h^{0,k-1} = h^{k-1,0} = 1$. The determinant of the representation is the $k-1$th power of the cyclotomic character (up to a finite character) which can be read off from the "degree". By twisting, we can easily change the determinant, and change the Hodge numbers to $h^{-d,k-d-1} = h^{k-d-1,-d} = 1$. Yet, it is clear that we cannot twist so that $h^{1,0} = h^{0,1} = 1$ unless $k = 2$. Thus, given a modular form of weight $k > 2$, it cannot be associated to an elliptic curve even after twisting. This is Kevin's answer.

Secondly, any motive has (conjecturally) an $L$-function. The recipe of building this $L$-function breaks up into two parts. The first involves the factors at finite primes, which give rise to the Euler product. The second involves the infinite primes, which give rise to Gamma factors. The information at $\infty$, however, (by Tate's conjecture) can be read off from the Galois representation, and the recipe of Deligne shows that it will exactly depend on the Hodge numbers of the motive, and visa versa. Moreover, twisting by $\epsilon^k$ some power of the cyclotomic factor has the effect of replacing $L(s)$ by $L(s+k)$ (and shifting the corresponding central value) In particular, given an elliptic curve, one knows the Gamma factors (because one knows the Hodge decomposition of $E$), and one sees that even after twisting one cannot get Gamma factors that "look like" the Gamma factors associated to a modular form of weight different from $2$. This is GH's answer.

More generally, arithmetic conjectures of Langlands type imply that all motives should be "automorphic", and that the Hodge structure of the motive determines the infinity type of the automorphic form, which in turn determines the Gamma factors. So, at least morally, given a pure irreducible motive $V$, we know that if it is automorphic, it must be automorphic of a particular weight determined by the underlying geometry of $V$.

Of course, even before the result of Faltings, one had enough faith in terms of how these things were connected to be very confident that Elliptic curves over $\Q$ should correspond exactly to weight two forms - GH's remark that "It was an experimental fact that the gamma factors are always the same, hence the precise form of the modularity conjecture was formulated, which then turned out to be right, namely it was proved by great efforts of great mathematicians" seems spot on.

Related problems. Given a modular eigenform $f = \sum a_n q^n$ of weight four (in the arithmetic normalization), one can ask: is it possible to show that there does not exist a weight two modular form $g = \sum b_n q^n$ where $a_p = p b_p$ for all primes $p$ without using $p$-adic Hodge theory? I think this is not so easy. For example:

(i) The arithmetic approach: The weight $4$ form $f$ would have the property that it is not ordinary at every prime, since clearly $a_p = p b_p \equiv 0 \mod p$. One conjectures that a set of primes of density one are modular (or $1/2$ if $f$ has CM). Yet it is still unknown whether any form of weight $\ge 4$ has even a single ordinary prime.

(ii) The analytic approach: What do the distributions of coefficients of weight $2$ and $4$ forms look like? Sato-Tate says that the normalized coefficients satisfy a precise distribution (now a theorem!). Yet the "normalized" coefficients of $f$ and $g$ are by construction exactly the same, so Sato-Tate says nothing. In particular, it is hard to see any analytic estimates of functions involving the $a_p$ being able to distinguish two classes of numbers with the same underlying distribution. A related argument: The Hasse bound in weight four is satisfied by $p b_p$ if and only if the weight two Hasse bound is satisfied by $b_p$.

Summary. Conjectures arise organically from heuristics and computation. I was "known" that Elliptic curves should be associated to weight two forms long before one could actually formally prove that they weren't associated to twists of weight $4$ forms. To prove the latter fact, one has to use $p$-adic Hodge theory.

(* I am not sure about a lot here, so I'm putting this community wiki. Also, there is no mention of weight 1 or half integers. Change whatever needs changing, or in the extreme cases, peacefully leave a comment to delete...)

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This is a great answer; I didn't know that "automorphic" generalised "modular". Thanks Dror! –  Barinder Banwait Aug 15 '11 at 10:39
I'm not sure that the following question makes sense, but "How" can an A.V. of dim d (> 1, say) give rise to a 2-dimensional representation? I mean, generically the action of Galois on the l-adic Tate module would give us a 2d dimensional rep. Is the issue simply that it may reduce? I sort-of see how one goes from 2d-dimensional to 2 dimensional for the A.Vs A_f, where f is a newform, but that is a special situation... –  Barinder Banwait Aug 19 '11 at 22:10
Barinder, the galois representation on the Tate module can be highly reducible when you extend scalars from $\mathbf{Q}_{\ell}$ to an extension field. Take a look at Ribet's article on GL2-type abelian varieties to see this in action. –  David Hansen Aug 20 '11 at 3:24
@Lavender: Regarding the related problem, strong multiplicity one for $GL_2$ proves that you can't have two such modular forms, since they would have for almost all primes (the spherical ones) the same Satake parameters. This is a representation theoretic solution without $p$-adic Hodge theory. –  Dror Speiser Sep 25 '12 at 0:17

There has been a lot written already about this question. but here is a simple answer. The Hodge--Tate weights of the Tate module of an elliptic curve are 0 and 1. The Hodge--Tate weights of the Galois representation associated to a weight $k$ modular form are 0 and $k-1$. So if the Tate module is associated to a modular form, it's associated to a weight 2 modular form.

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Thanks Kevin for this succinct answer. I wonder if the Hodge-Tate weights agreeing is equivalent to the L-functions agreeing... –  Barinder Banwait Aug 15 '11 at 17:24
Ah, and of course this should address the question using the different normalization that Matt Young brings up. You should be able to "see" the gap in the Hodge–Tate weights from the traces of Frobenius, right? –  Rob Harron Aug 15 '11 at 17:30
@Kevin: I don't know either, but I guess I'm asking: if you look at $\{a_p/p^{(k-1)/2}\}$ and $\{a^\prime_p/p^{(k^\prime-1)/2}\}$ coming from two normalized newforms of weights $k$ and $k^\prime$, respectively, can these two collections be equal? And my thought was that no, because all the normalization does is make the motives weight 0. So, now you have two motives of weight zero, but with different Hodge structures, can their traces of Frobenii really all be equal (in the $\ell$-adic realization for some $\ell$)? –  Rob Harron Aug 15 '11 at 18:28
No they can't be equal because the Hodge--Tate weights can't match up. Here's another proof -- the same proof in disguise -- look at the $L$-functions. The functional equations they satisfy will be different, because the Gamma factors coming from the infinite place can see the Hodge--Tate weights. The twists give them the same motivic weight, but the Gamma factors know a little more than this. –  Kevin Buzzard Aug 15 '11 at 20:55
@Dror: To get some more intuition for Hodge-Tate-Sen weights, it's well worth reading Sen's classic Inventiones 62 paper. –  David Hansen Aug 16 '11 at 22:39

The reason why the weight needs to be $k=2$ becomes clear when you consider this (equivalent) version of modularity: $E/\mathbb{Q}$ is modular if there there exists a normalized newform $f$ for $\Gamma_0(N)$ such that $L(f,s) = L(E,s)$, i.e., the $L$-function attached to $f$ coincides with the Hasse-Weil $L$-function attached to $E/\mathbb{Q}$.

Now, the $L$-function of a normalized newform of weight $k$ for $\Gamma_0(N)$ has an Euler product of the form:

$$L(f,s) = \prod_{p|N} \frac{1}{1-\lambda_p p^{-s}} \prod_{p\nmid N} \frac{1}{1-\lambda_p p^{-s} + p^{k-1-2s}}.$$

The $L$-function of $E$ is defined as an Euler product:

$$L(E,s) = \prod_{p\geq 2}\frac{1}{L_p(p^{-s})},$$

where $L_p(T) = 1-a_pT+pT^2$ if $E$ has good reduction at $p$, $L_p(T)= 1-T$ if $E$ has split multiplicative reduction at $p$, $L_p(T) = 1+T$ if $E$ has non-split multiplicative reduction at $p$ and $L_p(T) = 1$ if $E$ has additive reduction at $p$.

In particular, $L_p(T)=1-a_pT+pT^2$ for almost all primes (for all primes $p\nmid N(E)$). If there is any hope that these two Euler products agree, then we must have $k=2$.

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Nice; your argument shows that k should be 1, which is OK, since your formula for L(f,s) is for when f has weight 2k. –  Barinder Banwait Aug 14 '11 at 22:07
I don't think this answer is convincing because both of the L-functions defined above could have been normalized to have central point $s=1/2$ (so $\sigma =1$ is the edge of absolute convergence). This is true for any modular L-function, and one can still wonder if $L(f,s) = L(E, s)$ with the "analytic" normalization. –  Matt Young Aug 15 '11 at 2:31
@Matt: But using your normalization, changes the definition of modular! You can view Álvaro's argument as simply taking the giving definition, throwing the data into some generating functions and factoring the latter into Euler products. One can indeed still wonder when your $L(s,f)$ equals your $L(s,E)$, but this is not the question posed here. Sure Álvaro has omitted all the work that goes into showing that his $L$-function definition is the same as the original question, but it seems to me to be an answer that would be okay with the OP. –  Rob Harron Aug 15 '11 at 4:23
@Álvaro: Matt is referring to the normalization where you basically replace the $\lambda_n$ with $\lambda_n/n^{(k-1)/2}$ and look at the corresponding Dirichlet series. Then, do a similar thing with $a_p$ (i.e. divide by $\sqrt{p}$). This simply shifts each of those $L$-functions by $(k-1)/2$ and $1/2$, respectively. Of course, equating $L(s,f)$ with $L(s,E)$ under these new normalizations is like equating $\lambda_p/p^{k/2}$ with $a_p$, so not the same definition of modular as the question mentions. –  Rob Harron Aug 15 '11 at 4:32
Perhaps I should add that Matt's exact same criticism can be given to Dror's answer. Indeed, you could say that instead of looking at $\rho_{f,\ell}$, you look at $\rho_{f,\ell}\otimes\chi_\ell^{(k-1)/2}$ (where $\chi_\ell$ is the $\ell$-adic cyclotomic character) and twist the Tate module of the elliptic curve, as well, and ask for those two to have matching characteristic polynomials of Frobenius. Again, this would be a different definition of modular, one which I have never seen discussed. A direct answer to the original question without a translation to other data is given in my answer. –  Rob Harron Aug 15 '11 at 13:58

In terms of the data of which you speak, you can use asymptotics of the Fourier coefficients on the modular form side and the Hasse bound on the elliptic curve side. Specifically, if $f$ is a normalized newform of weight $k$ (an integer $\geq2$, I'm not enough of an expert on this to know if more general $k$ are allowed), then $$\sum_{n\leq X}|a_n|^2/n^{k-1}=c_fX+O(X^{3/5})$$ (see section 14.9 of Iwaniec–Kowalski), but, on the other hand, the Hasse bound for an elliptic curve states that $$|a_p|<2\sqrt{p}$$ (see this mathoverflow answer to find out what this implies about the $a_n$ of the putative cusp form attached to the elliptic curve). Given this bound, $k$ must be at most 2.

Update: From reading the mathscinet review of an old paper of Selberg's, it looks like the asymptotic on the Fourier coefficients (I guess originally due to Rankin) works for any weight which is a positive real number (but the language in the review is rather old-fashioned and I don't have access to the paper right now). I'll take a look probably in a couple of days.

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@Rob: Your response does answer the question as it was asked. However, at a deeper level Matt Young's criticism applies. Namely the true question is not that what can be said about the weight of $f$ if, for some artificial normalization, we assume/demand that $L(s,E)=L(s,f)$. The true question is that what can be said about the weight of $f$ if, for a standard/natural/uniform normalization, we assume/demand that $L(s,E)=L(s,f)$. The standard/natural/uniform normalization of an $L$-function is the one in which the $n$-th Dirichlet coefficient is of size $n^{o(1)}$, at least on average. –  GH from MO Aug 16 '11 at 19:53
@GH: I don't agree with you that the normalization of the functional equation sending $s$ to $1-s$ is somehow natural (and certainly not standard). I think it's standard in analytic number theory since it's nice to always have your critical strip be between 0 and 1 and have a fixed central point at 1/2. But more generally, it is not the normalization that arises naturally, e.g. if you simply take the Mellin transform of $f$ you don't get your normalization. More generally, given an "object" $M$ (say a motive or an automorphic representation), one expects a functional equation... –  Rob Harron Aug 16 '11 at 22:40
(cont'd) relating $L(s,M)$ with $L(1-s,M^\vee$) (where $M^\vee$ is the dual). This is a uniform, and I'd argue natural, normalization, which, when specialized to weight $k$ cusp forms of trivial nebentype gives the normalization I use since the dual of $f$ is isomorphic to a twist of the original $f$. Furthermore, your normalization isn't very convenient when discussing special values of $L$-functions since half of the time these values will actually be at half-integers as opposed to integers. –  Rob Harron Aug 16 '11 at 22:40
@Rob: De gustibus non est disputandum, but still: in my view the set of automorphic $L$-functions is the widest and most natural family of $L$-functions occurring in arithmetic. Every member you can name is built out of them, at least conjecturally. Most automorphic $L$-functions are transcendental, they don't seem to come from algebraic objects. I am not talking about analytic number theory here, but arithmetic as a whole. When you study this wide family, that is when you take a truly global point of view, you must stick with some fixed normalization like $s$ is related to $-s$ or $1-s$. –  GH from MO Aug 17 '11 at 3:15
Holomorphic cusp forms parametrize very special $\mathrm{GL_2}$ cuspidal representations, but for $\mathrm{GL_2}$ cuspidal representations in general the concept of "weight" does not even make sense. I agree that studying special values (or other algebraic features) of special cuspidal representations does motivate particular normalizations, but these appear unnatural from a truly global perspective. –  GH from MO Aug 17 '11 at 3:24

I started to write this as a comment to the original question, but it became too long. It is not disjoint from the previous expert responses, but it emphasizes a particular viewpoint.

One can (and one should) shift normalize any automorphic (in particular any modular) $L$-function so that the functional equation relates $s$ to $1-s$. Then the gamma factors identify the archimedean component of the underlying automorphic form much like the Euler factors identify the non-archimedean components. In particular, if the $L$-function of a holomorphic cusp form is so normalized (i.e. $s$ is related to $1-s$), then the gamma factors determine the weight (and vice versa). I say gamma factors, because the usual single gamma factor can be factored into two gamma factors by the doubling formula for the gamma function (for a modular $L$-function the "true gamma factors" form a pair).

So you can ask your question as follows. If we shift normalize the $L$-function of an elliptic curve so that the functional equation relates $s$ to $1-s$, then why are the gamma factors always the same? In order to ask this question you need to assume already that the $L$-function obeys a rather specific functional equation with gamma factors, and then the question inquires what the gamma factors can be.

It is possible that the fairest answer to this question (i.e. your question) is as follows:

It was an experimental fact that the gamma factors are always the same, hence the precise form of the modularity conjecture was formulated, which then turned out to be right, namely it was proved by great efforts of great mathematicians.

Once again, normalization of all automorphic $L$-functions is key in this discussion, it should not be underestimated. It is not the usual normalization favored by algebraic people.

Added: In a leisurely style one could say the following. The first miracle is that the $L$-function of an elliptic curve is entire and satisfies some functional equation. The second miracle is that the $L$-function is automorphic, as suggested by the functional equation. More specifically, it looks like a $\mathrm{GL}_2$ automorphic $L$-function. Not only it is $\mathrm{GL}_2$, but it comes from a very specific modular form, namely a holomorphic form, let's call this the third miracle. Then, as a final miracle, this holomorphic form is always of weight 2 whose level can also be specified in terms of the elliptic curve.

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Thanks for your answer emphasising functional equations. But why does "gamma factors being the same" imply weight 2? And could you say more about this "experimental" fact? Who did these experiments, and when? –  Barinder Banwait Aug 16 '11 at 13:18
There are elliptic curves for which the mentioned analytic properties of the $L$-function are easier to establish. A specific example is the elliptic curve $E_r:y^2=x^3-r^2 x$, where $r$ is a square-free positive integer, see Section 8.2 in Iwaniec's Topics in Classical Automorphic Forms. In these examples the functional equation and the gamma factors involved in it are known, hence the modularity conjecture is that there is always a functional equation of a very similar shape (including with the same gamma factor, after normalization). Examples like this one can be regarded as "experiments". –  GH from MO Aug 16 '11 at 15:29
More precisely, it is known that modularity is equivalent to the right functional equation of all twists, this is a not-so-hard theorem of Weil. The point is that your question regarding the weight is a particular question about the gamma factors. Kevin Buzzard emphasized Hodge-Tate weights, which is saying pretty much the same in the language of Galois representations. The mystery or miracle is that these are always the same, they don't depend on the elliptic curve. I consider this part of the mystery or miracle of the whole modularity phenomenon. –  GH from MO Aug 16 '11 at 15:33
Note also that a more specific conjecture is sometimes easier to establish. I believe this was the case with the modularity conjecture as well (and also with the more recent proof of Serre's conjecture). –  GH from MO Aug 16 '11 at 15:34