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Serre's conjecture says that given any odd, irreducible, continuous representation $\rho:G_{\mathbb{Q}}\rightarrow GL_2(\overline{\mathbb{F}_p})$ there is some eigenform $f$ of weight $k(\rho)$, level $N(\rho)$, and nebentype $\epsilon(\rho)$, such that $\rho$ is isomorphic to the mod $p$ representation $\bar \rho_f$ associated to $f$. Up until now, I have intuitively thought of the Serre weight $k(\rho)$ as being the minimal weight among all weights of eigenforms whose representations are isomorphic to $\rho$. I am no longer sure whether this is correct: I am currently reading Edixhoven's survey of Serre's conjecture, and it seems like one might be able to cook up an example where the Serre weight $k(\bar\rho_f)$ of some weight $k$ eigenform $f$ is actually greater than the weight of $f$, which seems strange to me... Can this happen? I.e., are there examples where the Serre weight of a representation coming from a weight $k$ eigenform is actually greater that $k$?

To give this a bit more context, fix $p\nmid N$ and let $f \in S_{k}(\Gamma_0(N))$ be an eigenform with mod $p$ representation $\bar\rho_f$. Fix an inertia subgroup $I_p\subseteq G_{\mathbb{Q}}$ at $p$ and write $I_{p,w}\subseteq I_p$ for its wild subgroup. If $\bar\rho_f\mid_{I_{p,w}}$ is nontrivial, we can uniquely write $$ \bar\rho_f\mid_{I_{p}}=\begin{pmatrix}\chi^\beta &* \\ 0& \chi^{\alpha} \end{pmatrix} $$ where $\chi$ is the mod $p$ cyclotomic character and $\alpha$, $ \beta$ are integers such that $0\leq \alpha\leq p-2$, $1\leq\beta\leq p-1$. As stated in the Edixhoven article, setting $a=\min(\alpha,\beta)$ and $b=\max(\alpha,\beta)$, we define $k(\bar\rho_f)$ to be $1+pa+b+p-1$ if $\chi^{\beta-\alpha}=\chi$ and $\bar \rho_f\mid_{G_{\mathbb{Q}_p}} \otimes \chi^{-\alpha}$ is not finite at $p$, otherwise we define it to be $1+pa+b$.

Now consider the case where $p=3$, and suppose that we have some weight 6 eigenform $f$ on $\Gamma_0(N)$ such that $$ \bar\rho_f\mid_{I_{3}}=\begin{pmatrix}1 &* \\ 0& \chi \end{pmatrix} $$ and $\rho_f\mid_{G_{\mathbb{Q}_3}} \otimes \chi^{-1}$ is not finite at $3$. Then the formula above would say that $f$ has Serre weight $k(\bar \rho_f)=8$, which is larger than the weight of $f$. I am of course assuming that we can find such a form satisfying these conditions, but even so, I see no obvious reason why this shouldn't occur...

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  • $\begingroup$ The following article might be relevant (arxiv.org/abs/2004.07587) $\endgroup$ Commented Aug 21, 2020 at 13:01
  • $\begingroup$ The weight in Serre's conjecture is minimal. The above example fails because in this example $\alpha = 0$ and $\beta = 1$, giving a minimal weight of $4$. $\endgroup$
    – assaferan
    Commented Oct 25, 2020 at 3:56
  • $\begingroup$ Ah, sorry, there was a typo (now fixed): $\alpha$ and $\beta$ should be switched in the definition. (See Edixhoven's Defintion 1.7(b) in the FLT book.) Thus, in the example, we have $\beta=p-1$ and $\alpha=1$, in which case $a=1$ and $b=p-1$. So assuming $\rho_{G_{\mathbb{Q}_3}}\otimes \chi^{-1}$ is not finite, we get Serre weight $1+p+(p-1)+p-1=3p-1$. $\endgroup$
    – Arbutus
    Commented Oct 25, 2020 at 13:11

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I believe a modular form as you describe indeed cannot exist. I think it's easier to think about these issues if they're translated into the representation-theoretic language of Serre weights. Associated to $\overline{\rho}$ is a set of Serre weights, i.e., of irreducible mod $3$ representations of $\mathrm{GL}(2,\mathbf{F}_3)$. If $\overline{\rho}|_{D_3}$ is a tres ramifiee extension of $\chi$ by $\chi^2 = 1$, then this set is the singleton $\{ \mathrm{det} \otimes \mathrm{Sym}^2\}$, where $\mathrm{Sym}^a$ denotes the $a$th symmetric power of the standard representation. (Since we're in two dimensions you can think about this in terms of crystalline lifts: your $\overline{\rho}|_{D_3}$ will lift to a crystalline extension of cyclotomic by cyclotomic^4 (but not of cyclotomic by cyclotomic^2, because of the tres ramifiee condition). So it has a crystalline lift with Hodge-Tate weights $(s,t) = (4,1)$ [but not $(2,1)$] and the numerology is that the Serre weight associated to this lift is $\mathrm{det}^t \otimes \mathrm{Sym}^{s-t-1}$.)

Now the Breuil-Mézard conjecture for $\mathrm{GL}(2,\mathbf{Q}_3)$, which is known (and due to Shen-Ning Tung in this case), has the consequence that if $\overline{\rho}|_{D_3}$ has a crystalline lift of weights $(0,k-1)$ (in particular if $\overline{\rho}$ comes from a modular form of weight $k$ and level prime to $p$) then $\{ \mathrm{det} \otimes \mathrm{Sym}^2\}$ must be a Jordan-Hölder factor of $\mathrm{Sym}^{k-2}$.

If I haven't miscalculated, the Jordan-Hölder factors of $\mathrm{Sym}^4$ are $\{\mathrm{Sym}^2, \mathrm{det}, \mathrm{det}^2\}$, whereas the the Jordan-Hölder factors of $\mathrm{Sym}^6$ are $\{\mathrm{Sym}^2, \mathrm{det} \otimes \mathrm{Sym}^2, \mathrm{det}^2\}$, so indeed $k=8$ is the minimal weight where such $\overline{\rho}$ can be found.

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