How do you calculate the Euler factors of the imprimitive symmetric square at primes with bad reduction? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T22:13:39Z http://mathoverflow.net/feeds/question/67409 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67409/how-do-you-calculate-the-euler-factors-of-the-imprimitive-symmetric-square-at-pri How do you calculate the Euler factors of the imprimitive symmetric square at primes with bad reduction? Max Flander 2011-06-10T07:04:40Z 2011-06-10T08:47:09Z <p>The reference for this question is Coates and Schmidt, Iwasawa theory for the symmetric square.</p> <p>Let $G = \textrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}))$ and let $D_r \supseteq I_r$ be a decomposition group and inertia group at $r$. Let $E/\mathbb{Q}$ be an elliptic curve, and for a prime $\ell \neq r$, consider the representations $\rho_r : G\to \textrm{Sym}^2(H^1_\ell(E))^{I_r}$ and $\rho_r' : G\to \textrm{Sym}^2(H^1_\ell(E)^{I_r})$. </p> <p>The primitive symmetric square of $E$ is the $L$-series defined by the Euler factors</p> <p>$\mathcal D_r (X) = \textrm{det}(1-\rho_\ell(\textrm{Frob}_r^{-1})X)$ </p> <p>and the imprimitive symmetric square of $E$ is the $L$-series defined by the Euler factors $D_r (X) = \textrm{det}(1-\rho_\ell'(\textrm{Frob}_r^{-1})X).$</p> <p>Since $\rho'_r$ is a submodule of $\rho_r$, we have that $D_r(X) | \mathcal{D}_r(X)$ </p> <p>for all $r$. Furthermore, if $r$ is a prime where $E$ has good reduction, since $H^1_\ell(E)$ is unramified at $r$ we have $D_r(X) = \mathcal{D}_r(X)$ .</p> <p>If $E$ has bad multiplicitive reduction at $r$, then by calculations in Coates and Schmidt, $\mathcal{D}_r(X) = 1-X$, and if $E$ has bad additive reduction at $r$, then $\mathcal{D}_r(X)$ is either equal to $(1-\alpha_r^2X)(1-\beta_r^2X)(1-rX)$, $1+rX$, $1-rX$ or $1$ depending on the image of the inertia group $I_r$, and I can calculate this using Mark Watkins's 'sympow' computer program.</p> <p>My question is:</p> <blockquote> <p>How do I find the imprimitive Euler factors $D_r(X)$ at primes where $E$ has bad reduction?</p> </blockquote> http://mathoverflow.net/questions/67409/how-do-you-calculate-the-euler-factors-of-the-imprimitive-symmetric-square-at-pri/67415#67415 Answer by David Loeffler for How do you calculate the Euler factors of the imprimitive symmetric square at primes with bad reduction? David Loeffler 2011-06-10T08:47:09Z 2011-06-10T08:47:09Z <p>The factor $D_r$ is easy to compute (much easier than $\mathcal{D}_r$). Basically, you just need to find the eigenvalues $\lambda_i$ of Frobenius on $H^1_\ell(E)^{I_r}$ (i.e. the reciprocal roots of the local L-factor of E itself), and then the eigenvalues of Frobenius on the symmetric square of that are the pairwise products $\lambda_i \lambda_j$. This gives you the reciprocal roots of $D_r$.</p> <ol> <li>If $r$ is a good prime, the eigenvalues of Frob on $H^1_\ell(E) = H^1_\ell(E)^{I_r}$ are $\alpha_r$ and $\beta_r$, so on the symm square you get $\alpha_r^2$, $\beta_r^2$ and $\alpha_r \beta_r = r$. So have <code>$D_r(X) = \mathcal{D}_r(X) = (1 - \alpha_r^2 X)(1 - \beta_r^2 X)(1 - r X)$</code>.</li> <li>If $r$ is a bad multiplicative prime, then $H^1_\ell(E)^{I_r}$ is 1-dimensional and the only eigenvalue of Frob is $\pm 1$ (depending whether the reduction is split or non-split). Either way the only eigenvalue on the symmetric square is $+1$, and $D_r(X) = (1 - X)$. </li> <li>If $r$ is a bad additive prime, then <code>$H^1_\ell(E)^{I_r} = 0$</code> and hence $D_r(X)$ is identically 1.</li> </ol> <p>The calculation of $\mathcal{D}_r$ is much harder, because you need to consider the case where the image of inertia is some finite subgroup of <code>$\mathrm{GL}_2(\mathbb{Z}_\ell)$</code> which has zero invariants but preserves some bilinear form. </p>