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Let $\bar{\rho}$ be a residual ordinary and locally split galois representation associated to a weight $k$ level $1$ form (more specifically a mod $p$ companion form). In the sense of deformation theory, let $p$ be an unobstructed prime. Then $ dim H^{1}(G_{S}, Ad(\bar{\rho})) = 3$. Further $H^{1}(G_{S}, Ad(\bar{\rho})) = H^{1}(G_{S}, \mathbb{F}_{p}) \oplus H^{1}(G_{S}, Ad^{0}(\bar{\rho}))$.

What is the dimension of image of the map defined by

$$ H^{1}(G_{S}, Ad(\bar{\rho})) = H^{1}(G_{S}, \mathbb{F}_{p}) \oplus H^{1}(G_{S}, Ad^{0}(\bar{\rho})) \rightarrow H^{1}(I_{p}, Ad^{0}(\bar{\rho})) \rightarrow H^{1}(I_{p}, \mathbb{F}_{p}(\omega^{k-1}))$$

$$Ad^{0}(\bar{\rho}) \cong \mathbb{F}_{p} \oplus \mathbb{F}_{p}(\omega^{k-1}) \oplus \mathbb{F}_{p}(\omega^{1-k})$$

as an $I_p$ module.

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  • $\begingroup$ I fixed the formatting a bit, but I'm not quite sure I understand the question -- could you clarify? $\endgroup$ – David Loeffler Jun 21 '11 at 16:10
  • $\begingroup$ Hi David, thanks for formatting. By $Ad^{0}(\bar{\rho})$ i mean adjoint action on trace zero matrices. The first map is defined via restriction to $H^{1}(I_{p}, Ad^{0}(\bar{\rho}))$ and then projection onto $H^{1}(I_{p}, \mathbb{F}_{p}(\omega^{k-1}))$. $\endgroup$ – Guhan Venkat Harikumar Jun 22 '11 at 7:33
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Note that $\mathrm{dim}(H^{1}(G_S,\mathrm{Ad}^{0}))$ = 2. So, if your map is injective, then the image will be of dimension 2. Otherwise, dimension 1 or 0. Finding the injectiveness of the map is a seperate problem (not known yet). See for more details: Recent paper by E.Ghate and V.Vatsal (Locally indecomposable Galois representations).

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