System of eigenvalues of level $N$ weight $k$ mod $l$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T21:11:11Z http://mathoverflow.net/feeds/question/21603 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21603/system-of-eigenvalues-of-level-n-weight-k-mod-l System of eigenvalues of level $N$ weight $k$ mod $l$ Morawski 2010-04-16T18:44:14Z 2010-11-19T02:28:57Z <p>Let $M_k(N)$ be the space of modular forms of level $N$ and weight $k$ whose $q$-coefficients at infinity are rational and $l$-integral. Put $\tilde M_k(N) = M_k(N) \otimes \mathbb{\bar F}_l$ and let $\mathbb T_k$ be the subring of $\mathrm{End}_{\mathbb{\bar F}_l}(\tilde M_k(N))$ generated by Hecke operators $T_p$ for all prime $p$, prime to $N$. What is the number of different systems of eigenvalues for all $T_p$ with $(p,N)=1$? My feeling is that this is a number of maximal ideals in $\mathbb T_k$, but can not make it precise.</p> http://mathoverflow.net/questions/21603/system-of-eigenvalues-of-level-n-weight-k-mod-l/46572#46572 Answer by Joël for System of eigenvalues of level $N$ weight $k$ mod $l$ Joël 2010-11-19T02:28:57Z 2010-11-19T02:28:57Z <p>Your intuition is correct. Actually this has nothing to do with modular forms: there is a simple general statement which contains your question:</p> <p>Let $k$ be an algebraically closed field, $M$ a finite dimensional $k$-vector space, $(T_i)$ a family of commuting linear operators on $M$, and $\tau$ the sub-algebra of $End_R(M)$ they generate. Then there is a natural bijection between the maximal ideals of $\tau$ and the system of eigenvalues for all the $T_i$ in $M$.</p> <p>The bijection is easily constructed in one direction: suppose that $T_i \mapsto \lambda_i$ is a system of eigenvalues appearing in $M$ (that is to say, there is a $0 \neq v \in M$ such that $T_i v = \lambda_i v$ for all $i$). Then obviously, $v$ is an eigenvector for all $T \in \tau$ and the map $\tau \rightarrow k$ sending $T \in \tau$ on the unique $\lambda(T) \in k$ such that $T v = \lambda(T) v$ is a morphism of $k$-algebras. Its kernel is thus a maximal ideal of $T$. This defines a map "system of eigenvalues" $\rightarrow$ "maximal ideals of $\tau$".</p> <p>Conversely, if $m$ is a maximal ideal of $\tau$, then $\tau/m=k$ (since $\tau$ is finitely generated and $k$ is algebraically closed) so the map $\lambda : \tau \rightarrow \tau/m=k$ is a morphism of $k$-algebras. You want to prove that there is a $0 \neq v \in M$ such that $T v = \lambda(T)v$ for all $T \in \tau$. It suffices to do it for a family of generators $T_1,\dots,T_k$ of $\tau$, which we can choose finite. Now since $\lambda (T_1-\lambda(T_1))=0$, $T_1-\lambda(T_1)$ is not invertible in $\tau$, hence is not invertible in $End_k(M)$ (otherwise its inverse would be by Cayley-Hamilton a polynomial in $T_1$, hence in $\tau$) hence $V_1 = \ker (T_1 - \lambda(T_1))$ is not $0$. Now $T_2, \dots, T_k$ stabilize $V_1$ since they commute with $T_1$, and you can make the same reasoning for $T_2$ acting on $V_1$, and so on. At the end you get a non-zero subspace of $V$ on which each $T_i$ acts by the scalar $\lambda(T_i)$, QED.</p> <p>That's a very pedestrian proof. If you are willing to apply some basic commutative algebra (e.g. Artinian rings are semi-local) you can do it much faster.</p>