Timeline for How do you calculate the Euler factors of the imprimitive symmetric square at primes with bad reduction?

4 events

when toggle format	what		by	license	comment
Jun 10, 2011 at 9:42	vote	accept	Max Flander
Jun 10, 2011 at 8:47	answer	added	David Loeffler		timeline score: 2
Jun 10, 2011 at 7:48	comment	added	Junkie		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)
Jun 10, 2011 at 7:04	history	asked	Max Flander	CC BY-SA 3.0