# Slope of classical modular forms

A Celebrated theorem of Coleman says that any overconvergent eigenform of weight $k$ and slope $< k-1$ is classical (here $k \geq 2$ and the level is $\Gamma_1(N) \cap \Gamma_0(p)$). This result is almost optimal as the slope of a classical eigenform of weight $k$ is $\leq k-1$. This seems to be well known and probably easy, but I cannot find a proof. So the question is:

why is the slope of a classical modular eigenform of weight $k$ less or equal to $k-1$?

Thank you very much!

Robert

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This just follows from the characteristic polynomial. –  Arijit Apr 18 '13 at 13:50

There are several ways to see it, some low-level and some more conceptual. I assume that you are talking of an eigenform $g$ for $U_p$ of level $\Gamma_1(N) \cap \Gamma_0(p)$, with $p$ prime to $N$ (otherwise the result may be false: the slope may be infinite).
For a low-level proof, there are two cases to consider. The most important (because it is generic in a $p$-adic family sense) is when $g$ is old at $p$. That is, when $g$ is a linear combination of the form $f(z)$, $f(pz)$, for $f$ a form of level $\Gamma_1(N)$, eigenform for $T_p$, say with eigenvalue $a_p$. Then one can compute the matrix of $U_p$ in the basis $f(z)$, $f(pz)$. This is not hard since $T_p h$ is $U_p h$ plus an explicit scalar times $h(pz)$, by the very definitions of the Hecke operators $T_p$ and $U_p$, and one find that that the characteristic polynomial of $U_p$ is $X^2 - a_pX + p^{k-1} \epsilon(p)$, where $\epsilon$ is the nebentypus of $f$. (you can remove this term if $f$ is of level $\Gamma_0(N)$, say). Then the eigenvalue of $U_p$ on $G$ has to be a root $\alpha$ and $\beta$ of this polynomial, but since $a_p$ is an algebraic integer, so are the roots $\alpha$ and $\beta$; in particular $v_p(\alpha) \geq 0$ and $v_p(\beta) \geq 0$. Since $\alpha \beta = p^{k-1} \epsilon(p)$, and $\epsilon(p)$ is a root of unity, $v_p(\alpha)+v_p(\beta)=k-1$, and this implies $v_p(\alpha), v_p(\beta) \leq k-1$. (with possible equality only when one of those two slope is $0$, which is equivalent to $v_p(a_p)=0$.) For the case of a new form, a computation shows that the slope is always $(k-2)/2$.
Now for a more conceptual proof, in term of the geometric interpretation of $U_p$, I recommend reading carefully Kevin's Buzzard second paper (that is the one written by him alone, not him and Taylor) on the Artin conjecture.
Sorry a trivial typo, you meant to write the characteristic polynomial as $X^2 - a_{p}X + p^{k-1}\epsilon(p)$ –  Arijit Apr 18 '13 at 14:11