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](https://link.springer.com/chapter/10.1007/978-1-4612-1974-3_7), 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...



Is the weight in Serre's conjecture "minimal"?
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.


