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The reference for this question is Coates and Schmidt, Iwasawa theory for the symmetric square.

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 $\rho_r' : G\to \textrm{Sym}^2(H^1_\ell(E)^{I_r})$.

The primitive symmetric square of $E$ is the $L$-series defined by the Euler factors

$\mathcal D_r (X) = \textrm{det}(1-\rho_\ell(\textrm{Frob}_r^{-1})X)$

and the imprimitive symmetric square of $E$ is the $L$-series defined by the Euler factors $D_r (X) = \textrm{det}(1-\rho_\ell'(\textrm{Frob}_r^{-1})X).$

Since $\rho'_r$ is a submodule of $\rho_r$, we have that $D_r(X) | \mathcal{D}_r(X)$

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 $D_r(X) = \mathcal{D}_r(X)$ .

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.

My question is:

How do I find the imprimitive Euler factors $D_r(X)$ at primes where $E$ has bad reduction?

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Hard to parse the notation. To recap: $D_r(X)$ and $\rho_r′$ correspond to taking $I_r$ invariants on $H_1^l(E)$ and then taking the ${\rm Sym}^2$, while the more standard $\rho_r$ first takes ${\rm Sym}$ and then takes $I_r$ invariants. So I think the answer is: when $E$ is additive reduction, then $H^1_l(E)^{I_r}$ is trivial, and so is the symmetric square of it. When $E$ is multiplicative reduction, then $H_1^l(E)^{I_r}$ is a 1-dim subspace, and in fact is the same as with $\rho_r$ if I am not mistaken (it has degree 1, and divides the degree 1 Euler factor with $\rho_r'$, so must be same) –  Junkie Jun 10 '11 at 7:48
    
ah yes, thank you. –  Max Flander Jun 10 '11 at 13:01
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1 Answer

up vote 2 down vote accepted

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 $D_r$.

  1. 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 $D_r(X) = \mathcal{D}_r(X) = (1 - \alpha_r^2 X)(1 - \beta_r^2 X)(1 - r X)$.
  2. 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 $D_r(X) = (1 - X)$.
  3. If $r$ is a bad additive prime, then $H^1_\ell(E)^{I_r} = 0$ and hence $D_r(X)$ is identically 1.

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 $\mathrm{GL}_2(\mathbb{Z}_\ell)$ which has zero invariants but preserves some bilinear form.

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thank you that's exactly what I needed! –  Max Flander Jun 10 '11 at 9:43
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